Six Sigma and Beyond: Problem Solving and Basic Mathematics, Volume II

SIX SIGMA AND BEYOND Problem Solving and Basic Mathematics

SIX SIGMA AND BEYOND A series by D.H. Stamatis Volume I

Foundations of Excellent Performance

Volume II

Problem Solving and Basic Mathematics

Volume III

Statistics and Probability

Volume IV

Statistical Process Control

Volume V

Design of Experiments

Volume VI

Design for Six Sigma

Volume VII

The Implementing Process

D. H. Stamatis

SIX SIGMA AND BEYOND Problem Solving and Basic Mathematics

ST. LUCIE PRES S A CRC Press Company Boca Raton London New York Washington, D.C.

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Library of Congress Cataloging-in-Publication Data Stamatis, D. H., 1947Six sigma and beyond : problem solving and basic mathematics, volume II p. cm. — (Six Sigma and beyond series) Includes bibliographical references. ISBN 1-57444-310-0 (alk. paper) 1. Quality control—Statistical methods. 2. Production management—Statistical methods. 3. Industrial management. I. Title. II. Series. TS156 .S73 2001 658.5′62—dc21

2001041635 CIP

This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage or retrieval system, without prior permission in writing from the publisher. The consent of CRC Press LLC does not extend to copying for general distribution, for promotion, for creating new works, or for resale. Specific permission must be obtained in writing from CRC Press LLC for such copying. Direct all inquiries to CRC Press LLC, 2000 N.W. Corporate Blvd., Boca Raton, Florida 33431. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation, without intent to infringe.

Visit the CRC Press Web site at www.crcpress.com © 2002 by CRC Press LLC St. Lucie Press is an imprint of CRC Press LLC No claim to original U.S. Government works International Standard Book Number 1-57444-310-0 Library of Congress Card Number 2001041635 Printed in the United States of America 1 2 3 4 5 6 7 8 9 0 Printed on acid-free paper

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to Timothy, the entrepreneur

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Preface Problem solving is an attempt to understand the nature of human interaction and the differences in cognitive functioning of individuals which are observed again and again. This volume is specifically designed to address the issue of cognitive functioning by reviewing selected literature on the topic, to provide a generic approach to problem solving, and to review the basic mathematics that most people would use in the process. In all walks of life, at some time, all of us will likely use the process of problem solving. While we all talk about it and we all use it, chances are we all mean different things by it. In this volume, I have attempted to organize the topic and provide a structured approach based on the scientific method. In conjunction with six sigma methodology, I believe that it is fundamental to understand the process, not only because as a black belt one would use “an” approach to solve a specific concern, project, problem, etc., but because problem solving will help define the problem as well. This volume is intended to be a manual rather than a text for understanding the investigation process of problem solving. Toward that end, it is written to nontechnical as well as technical individuals. I have assumed that some readers may not have post high-school education or speak fluent English. The exception is the Introduction. In the Introduction, I dwell on the theoretical aspects of problem solving. If it is too cumbersome, the reader may skip this section without losing any understanding of the process. The literature review may enhance understanding, but it is not essential for use of the approach as explained in the following chapters. One may challenge this assumption and claim that anyone pursuing six sigma should have a basic understanding of problem-solving techniques and processes. From my experience, I can vouch that Fortune 500 companies have employees who are not familiar with the process or techniques. (In fact, very recently, I was with a client when I noticed that a training class of basic math was offered specifically for engineers. It was this experience that became the impetus for this volume, especially the inclusion of the basic mathematics in Part II.) The intent is not to embarrass or humiliate anyone; rather it is to help facilitate the process of solving problems with a basic commonality of understanding. If the reader is already proficient, obviously the math portion of this volume may be skipped without any loss of continuity. So, let us examine the issue of “problem solving.” The ability to solve problems is a central prerequisite for human survival, but the mechanics of the process itself often remain a puzzle for most of us — a puzzle because problem solving is a very complex cognitive process, not necessarily a behavior that can be observed. Perhaps the reason why several people find different, similar, or even the same solution to the same problem.

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This is a very important issue and it is the reason why this volume is in the form of a manual rather than a text. My hope is that readers will recognize the stage of the process they are involved with and as a consequence will pick the right spot for their analysis. In each step of the process, a variety of questions and flow charts are provided to facilitate understanding. By no means do I believe that all options have been exhausted; rather I believe that an initial generic guideline to provide a solution for a given problem has been created. D. H. Stamatis

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About the Author D. H. Stamatis, Ph.D., ASQC-Fellow, CQE, CMfgE, is currently president of Contemporary Consultants, in Southgate, Michigan. He received his B.S. and B.A. degrees in marketing from Wayne State University, his Master’s degree from Central Michigan University, and his Ph.D. degree in instructional technology and business/statistics from Wayne State University. Dr. Stamatis is a certified quality engineer for the American Society of Quality Control, a certified manufacturing engineer for the Society of Manufacturing Engineers, and a graduate of BSIIs ISO 9000 lead assessor training program. He is a specialist in management consulting, organizational development, and quality science and has taught these subjects at Central Michigan University, the University of Michigan, and Florida Institute of Technology. With more than 30 years of experience in management, quality training, and consulting, Dr. Stamatis has served and consulted for numerous industries in the private and public sectors. His consulting extends across the United States, Southeast Asia, Japan, China, India, and Europe. Dr. Stamatis has written more than 60 articles, presented many speeches at national and international conferences on quality. He is a contributing author in several books and the sole author of 12 books. In addition, he has performed more than 100 automotive-related audits and 25 preassessment ISO 9000 audits, and has helped several companies attain certification. He is an active member of the Detroit Engineering Society, the American Society for Training and Development, the American Marketing Association, and the American Research Association, and a fellow of the American Society for Quality Control.

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Acknowledgments In a typical book, an author has several, if not many, individuals who helped in the process of completing it. In this mammoth work, so many individuals have helped that I am concerned that I might forget someone. Sometimes, writing a book is a collective undertaking by many people. However, to write a book that conveys hundreds of thoughts, principles, and ways of doing things would truly be a Herculean task for one individual. Since I am definitely not Hercules or Superman, I depended on many people over the years to guide me and help me formulate my thoughts and opinions about many things, including this work. To thank everyone by name who has contributed to this work would be impossible, although I am indebted to all of them for their contributions. However, there are some organizations and individuals who stand out. Without them this series would not have been possible. There were some individuals who actually pushed me to write this series of books and have reviewed and commented on several of the drafts. There were also individuals who helped solidify some of the items covered in this work through lengthy discussions. Individuals who fall in these categories are M. Heaffy, H. Bajaria, J. Spencer, V. Lowe, L. Lemberson, R. Roy, R. Munro, E. Rice, and G. Tomlison. Their encouragement and thought-provoking discussions helped me tremendously to formalize not only the content, but also the flow as well as the depth of the material. I would like to thank the Six Sigma Academy for granting permission for use of some of its material in comparing the classical approach to the new approach of defects as well as the chart of significant differences between three sigma, four sigma, five sigma, and six sigma; the American Marketing Association for granting permission to summarize the data articles from Marketing News; the American Society for Training and Development for granting permission for use of the table, the Most Likely Influences of Program Development Practices, from Training and Development; Tracom Co. for granting permission for use of the material on the four social styles model; and the American Society for Quality (ASQ) for granting permission to summarize some key issues about teams from “Making Perfect Harmony with Teams” published in ONQ magazine and some definitions and characteristics of quality from The Certified Quality Manager Handbook (1999). Additionally, I would like to thank Mr. C. H. Wong for his persistence over the last four years to write this book. His faith in me and encouragement will never be forgotten; Dr. J. Farr for his thoughtful suggestions throughout the writing process and his insight on teams; Dr. W. Landrum for teaching me what teams are all about and why we must pursue the concept in the future. His futuristic insight has been an inspiration. His practice of teams has been a model for me to follow; my colleagues Dr. R. Rosa, Mr. H. Jamal, Dr. A. Crocker, and Dr. D. Demis, as well as

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Mr. J. Stewart and Mr. R. Start for their countless hours of discussions in formulating the content of these volumes in its final format; and J. Malicki, C. Robinson, and S. Stamatis for computer work on the early drafts and final figures in the text. I want to thank Ford Motor Company and especially, Mr. B. Kiger, Mr. R. H. Rosier, Mr. A. Calunas, and Ms. L. McElhaney for their efforts to obtain permission for using the questions of the Global 8D. Without their personal intercession, the book would be missing an important contribution to problem solving. As always, I would like to thank my personal inspiration, bouncing board, navigator, and editor, Carla, for her continued enthusiastic attitude during my most trying times. Especially for this work did she demonstrate her extraordinary patience, encouragement, and understanding by putting up with me. Special thanks also go to the editors of the series for their suggestions and for improvements to the text and its presentation. Finally, my greatest appreciation is reserved for the seminar participants and students of Central Michigan University. Through their input, concerns, and discussions, I was able to formulate these volumes to become a reality. Without their active participation and comments, these volumes would never have been completed.

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List of Figures FIGURE FIGURE FIGURE FIGURE FIGURE FIGURE FIGURE FIGURE FIGURE FIGURE FIGURE FIGURE FIGURE FIGURE FIGURE FIGURE FIGURE

1.1 Understanding “a” situation. 1.2 Team approach. 3.1 A pictorial view of a super generic five-step model. 3.2 The 14-step process improvement cycle model. 5.1 Something changed. 5.2 An overview of GPS0. 5.3 An overview of GPS1. 5.4 An overview of GPS2. 5.5 An overview of GPS3. 5.6 An overview of GPS4. 5.7 An overview of GPS5. 5.8 An overview of GPS6. 5.9 The events of GPS6. 5.10 An overview of GPS7. 5.11 An overview of GPS8. D.1 A typical problem solving worksheet. F.1 Path for root cause determination.

List of Tables TABLE 1.1 Generalized stages of problem solving. TABLE 3.1 Problem solving vs. process improvement selection chart. TABLE 4.1 Most likely influences of program development practices on sample core capabilities. TABLE 4.2 Typical quality tools. TABLE 5.1 Concern analysis report guidelines. TABLE 16.1 Squares, square roots, cubes, and cube roots. TABLE F.1 Comparison of root cause approaches.

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Table of Contents PART I Problem Solving Introduction ..............................................................................................................3 Chapter 1 Theoretical Aspects of Problem Solving...................................................................5 Definition of Terms ..........................................................................................9 Terms and Issues Related to Data Gathering ................................................14 Importance of the Sample..............................................................................25 References ......................................................................................................26 Selected Bibliography ....................................................................................28 Chapter 2 Overview of Key Elements to Problem Solving.....................................................33 The Road to Continual Improvement ............................................................33 Chronic vs. Sporadic Problems .....................................................................34 Three Typical Responses to Problems and the Antecedent ..........................35 Nine Common Roadblocks to Effective Problem Solving ...........................37 Six Key Ingredients Required to Correct Problems......................................39 Chapter 3 Problem-Solving and Process Improvement Cycles...............................................41 Why a Team Approach?.................................................................................41 Local Teams and Cross-Functional Teams....................................................42 General Guidelines for Effective Team Problem Solving and Process Improvement ................................................................................................43 What Makes a Team Work ............................................................................43 Examples of Problem-Solving Models..........................................................43 The Process Improvement Cycle ...................................................................46 Problem Solving vs. Process Improvement...................................................52 Chapter 4 The Quality Tools ....................................................................................................55 Tools for Problem Solving.............................................................................55 Quality Tools Inventory .................................................................................55 The Seven Basic Tools...................................................................................55 Using the Seven Basic Tools in Problem Solving ........................................57 Seven Quality Control Management Tools ...................................................59 Selected Bibliography ....................................................................................59

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Chapter 5 The Global Problem Solving Process .....................................................................61 General Overview ..........................................................................................61 Do’s and Do Not’s .........................................................................................65 Do’s ....................................................................................................66 Do Not’s .............................................................................................67 Concern Analysis Report ...............................................................................67 Root Cause Issues ..........................................................................................69 Verification .....................................................................................................71 Examples of “Snapshot” Verification (Reliability at 85% Confidence) ......................................................................................72 SPC Chart Subgroup and Sample Size .............................................73 GPS Application Criteria ...............................................................................74 Common Tasks...............................................................................................75 Change and Never-Been-There Situations ....................................................75 GPS Steps.......................................................................................................75 GPS0: Preparation..............................................................................77 GPS1: Establish the Team/Process Flow...........................................78 GPS2: Describe the Problem .............................................................82 GPS3: Develop the Interim Containment Action (ICA) ...................85 GPS4: Define and Verify Root Cause and Escape Point ..................88 GPS5: Choose and Verify Permanent Corrective Actions (PCAs) for Root Cause and Escape Point ...................................................93 GPS6: Implement and Validate Permanent Corrective Actions (PCAs)..............................................................................................95 GPS7: Prevent Recurrence.................................................................98 GPS8: Recognize Team and Individual Contributions....................102 References ....................................................................................................104 Chapter 6 Six Sigma Approach to Problem Solving .............................................................107 Overview ......................................................................................................107 First Week’s Project: Structure the Project — Goals, Objectives, and Scope..........................................................................................................109 Second Week’s Project: Structure the Project — Product-Based Estimating ..................................................................................................110 Third Week’s Project: Control the Project ..................................................112 Design for Six Sigma (DFSS) .....................................................................113

PART II Basic Mathematics Chapter 7 The Value of Whole Numbers...............................................................................119

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Chapter 8 Addition and Subtraction of Whole Numbers ......................................................123 Addition........................................................................................................123 Addition Exercises .......................................................................................124 Subtraction ...................................................................................................125 Application of Addition and Subtraction ....................................................127 Exercises.......................................................................................................128 Additional Exercises ....................................................................................129 Chapter 9 Multiplication and Division of Whole Numbers ..................................................131 Multiplication ...............................................................................................131 Multiplication Exercises...............................................................................133 Additional Multiplication Exercises ............................................................135 Division ........................................................................................................136 Exercises.......................................................................................................142 Additional Exercises ....................................................................................143 Chapter 10 Parts and Types of Fractions .................................................................................145 Exercises.......................................................................................................146 Additional Exercises ....................................................................................148 Chapter 11 Simple Form and Common Denominators of Fractions.......................................149 Simplest Form ..............................................................................................149 Simplest Form Exercises..............................................................................152 Additional Exercises ....................................................................................153 Common Denominators ...............................................................................154 Common Denominator Exercises ................................................................155 Additional Exercises ....................................................................................156 Changing Fractions ......................................................................................157 Changing Fractions Exercises......................................................................158 Chapter 12 Adding and Subtracting Fractions.........................................................................161 Addition of Fractions ...................................................................................161 Addition of Mixed Numbers........................................................................162 Addition of Fractions with Unequal Denominators....................................163 Sum of a Group of Mixed and Whole Numbers.........................................164 Applied Math Problems Using Addition of Fractions ................................164 Exercises.......................................................................................................165 Additional Exercises ....................................................................................166 Subtraction of Fractions...............................................................................168 Subtraction of Mixed Numbers ...................................................................169

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Subtraction of Fractions of All Kinds .........................................................172 Addition and Subtraction of Fractions ........................................................173 Exercises.......................................................................................................175 Additional Exercises ....................................................................................176 Chapter 13 Multiplication and Division of Fractions ..............................................................179 Multiplication of Proper and Improper Fractions .......................................179 Multiplication of Mixed Numbers...............................................................181 Division of Fractions....................................................................................182 Problems Calling for Any One of the Four Operations..............................185 Exercises.......................................................................................................186 Additional Exercises ....................................................................................187 Chapter 14 Ordering, Rounding, and Changing Decimals......................................................191 Decimals Using Fractions............................................................................191 Rounding Off ...............................................................................................194 Rounding Off Exercises ...............................................................................195 Additional Exercises ....................................................................................196 Changing Decimals ......................................................................................197 Exercises.......................................................................................................198 Chapter 15 The Four Operations in Decimals .........................................................................201 Adding and Subtracting ...............................................................................201 Multiplying Two Decimal Numbers ............................................................204 Division with Decimals................................................................................206 Division with Decimals Exercises ...............................................................211 Applied Problems Using Decimal Operations ............................................212 Exercises.......................................................................................................213 Change a Fraction into a Decimal...............................................................214 Exercises.......................................................................................................215 Chapter 16 Squares, Square Roots, Cubes, Cube Roots, and Proportions .............................217 Square and Cube Numbers ..........................................................................218 Exercises.......................................................................................................219 Squares and Cube Roots ..............................................................................220 Calculating the Square Root ........................................................................225 Exercises.......................................................................................................228 Additional Exercises ....................................................................................229 Application of Square Root .........................................................................231 Exercises.......................................................................................................234

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Review Test ..................................................................................................235 Solving Proportion Problems.......................................................................240 Exercises.......................................................................................................241 Additional Exercises ....................................................................................242 Chapter 17 Scientific Notation and Powers of Ten..................................................................245 Numbers Greater than One ..........................................................................245 Factoring and Writing Numbers ......................................................245 Factoring Exercises ......................................................................................247 Placing a Decimal Point ..............................................................................247 Numbers Less than One...............................................................................249 Changing Exponential Numbers to Standard Decimal Form .........249 Writing Numbers Less Than One in Standard Form ......................250 Chapter 18 Decimals ................................................................................................................253 Numbers Greater Than One.........................................................................253 Numbers Less Than One .............................................................................257 Addition of Decimal Numbers ....................................................................259 Same Exponents...............................................................................259 Different Exponents .........................................................................262 Exercises.......................................................................................................264 Subtraction of Decimal Numbers ................................................................265 Same Exponents...............................................................................265 Exercises.......................................................................................................265 Different Exponents .........................................................................266 Multiplication of Decimal Numbers............................................................268 Division of Decimal Numbers .....................................................................270 Chapter 19 The Metric System ................................................................................................273 Exercises.......................................................................................................274 Conversion of Measures...............................................................................275 Exercises.......................................................................................................275 Exercises.......................................................................................................276 More Conversions of the Metric System.....................................................277 Exercises.......................................................................................................283 Exercises.......................................................................................................286 Cumulative Exercises...................................................................................287 Chapter 20 International System of Units................................................................................289 Meter ............................................................................................................290

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Exercises.......................................................................................................291 Cubic Meter..................................................................................................292 Exercises.......................................................................................................293 Kilogram.......................................................................................................294 Exercises.......................................................................................................295 Kelvin ...........................................................................................................296 Exercises.......................................................................................................298 Review Exercises on Mass, Grams, and Temperature ................................300 Technical Units ............................................................................................302 Exercises.......................................................................................................304 Chapter 21 Conversion of English Units to Metric Units .......................................................305 Review Test ..................................................................................................305 Inches to Centimeters...................................................................................307 Exercises.......................................................................................................309 Yards to Meters ............................................................................................310 Exercises.......................................................................................................311 Quarts to Liters and Pints to Liters .............................................................312 Exercises.......................................................................................................313 Metric Units of Length to U.S. Units..........................................................316 Metric Units of Area to U.S. Units .............................................................317 Exercises.......................................................................................................319 Metric Units of Volume to U.S. Units.........................................................320 Exercises.......................................................................................................322 Cumulative Exercises for Length, Area, and Volume .................................323

PART III Appendices Appendix A A Typical Cover Sheet for the GPS Process ........................................................327 Appendix B GPS0. Preparation for Emergency Response Actions (ERAs): Assessment Questions .............................................................................................................329 Appendix C GPS1. Establish the Team/Process Flow: Assessment Questions ........................331 Appendix D GPS2. Describe the Problem: Assessment Questions ..........................................333 Appendix E GPS3. Develop Interim Containment Action (ICA): Assessment Questions.......341

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Appendix F GPS4. Define and Verify Root Cause and Escape Point: Assessment Questions .............................................................................................................343 Appendix G GPS5. Choose and Verify Permanent Corrective Actions (PCAs) for Root Cause and Escape Point: Assessment Questions................................................351 Appendix H GPS6. Implement and Validate Permanent Corrective Actions (PCAs): Assessment Questions .........................................................................................353 Appendix I GPS7. Prevent Recurrence: Assessment Questions ..............................................355 Appendix J GPS8. Recognize Team and Individual Contributions: Assessment Questions .............................................................................................................357 Selected Bibliography ..........................................................................................359 Index ......................................................................................................................361

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Part I Problem Solving

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Introduction

The six sigma methodology requires that a project be selected and an appropriate resolution be achieved in such a way that both the customer and organization will benefit — not a bad expectation. However, this is easier said than done. To select a project presupposes a need for improvement. The implication for that improvement is that either the customer and/or the organization is not satisfied with the product and/or service or status quo of whatever is happening “right now” in the organization. To select the problem is only half of the task. The other half is to resolve it. In Volume II, the focus is on the process of “how” to go about solving “a” problem, recognizing that more often than not, the solution is going to be through a team. Therefore, whoever is involved must understand the dynamics of team formation and performance. (This may be a good time to review Part II of Volume I.) From a problem-solving perspective, a team is an entity made up of individuals who have some ownership of the problem, complement each other on the skills necessary to resolve the problem, and have knowledge of the process (both the problem solving process and the process where the problem exists). For the team to have positive results, three basic strategies must be understood and planned for: • Solving the problem. Solving the problem comprises eight coping strategies (seek organizational support; focus on the specificity of the problem with excellent operational definition; physical interaction with all members; seek cooperative diversion; invest in excellent to good intrarelationships within the team; seek to belong in a winner environment; be willing to work hard and achieve results based on set criteria on a a priori basis; and focus on success) and represents a style of coping characterized by working at a problem while remaining optimistic, fit, relaxed, and organizationally connected. • Nonproductive coping. Nonproductive coping comprises eight strategies (worry; seek to belong; wishful thinking; to not cope, ignore the problem in the sense of postponing the due date for any reason; reduce tension; keep to self; blame self). These strategies reflect a combination of nonproductive avoidance behaviors which are empirically associated with an inability to cope. When this happens, the project falls apart.

3

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Six Sigma and Beyond: Problem Solving and Basic Mathematics, Volume II

• Reference to others. Reference to others contains four strategies (seek organizational support; seek professional help; seek spiritual support; seek social action) and can be characterized by turning to others for support whether they be peers, professionals, or deities. To facilitate these strategies, Chapter 1 provides a somewhat detailed rationale and a theoretical explanation of the problem-solving process. Also introduced are some generic terminology and a path for problem solving. In Chapter 2, the focus is on problem solving from a quality perspective. Introduced is the definition of a problem. Then progressively some of the issues involved with quality problem solving are addressed. Chapter 3, addresses some of the key concepts in team dynamics as they relate to problem solving. A discussion on the problem solving cycle follows and the point is made that most of us keep regurgitating problems without really resolving them. Also emphasized is the process improvement cycle. Some of the issues involved are reviewed and a flow chart to show the continual improvement cycle is introduced. Chapter 4 introduces some basic tools that are used in problem solving. Only an overview is given in this volume, recognizing that some of these tools will be fully developed in Volumes III and IV and some additional advanced tools will be introduced in Volumes V and VI. In Chapter 5, a version of the methodology used in the automotive industry, specifically at Ford Motor Company, and known as the 8D methodology is introduced. It is called the Global Problem Solving (GPS) process, since its application may be used in any industry. It is a very simple methodology, but demands a tremendous amount of time to identify and recommend a solution. However, it is very effective, when used properly. Chapter 6 introduces some concepts and approaches of problem solving specific to the six sigma project methodology. In Part II (Chapters 7 to 21), for the convenience of the reader who may not be familiar with some of the basic math required in problem-solving environments — especially at the root-cause level, or the “floor” level — some basic math will be explained. The intent of this volume is to ensure that the reader becomes familiar with the process of problem solving, not necessarily with the actual use of specific tools. Explanation and use will be described in Volumes III, IV, V, and VI. Therefore, in essence, this volume attempts to identify the problem-solving process and, in fact, to crystallize the notions that problem solving (1) must be based on fact; (2) must be creative; and (3) must be based on the experience of people.

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Theoretical Aspects of Problem Solving

This chapter introduces the problem solving process from a theoretical perspective. It addresses some fundamental issues that must be understood by anyone who is involved in solving a problem. It also establishes the rationale for a structure. Problem solving can be broadly defined as meeting challenges. Indeed, this is the core of what the six sigma methodology proposes to do, i.e., to reduce variability and increase profitability for the organization employing the initiative called “six sigma.” Looking further into the definition of a “problem solver,” the Oxford English Dictionary notes that a problem solver “is challenged to accomplish a specified result, often under prescribed conditions.” It is our task, then, to find a consistent approach that can be repeated again and again when a concern — an opportunity — arises. To appreciate this consistency, we must recognize that the way a situation is approached depends on a flow of information similar to Figure 1.1. In Figure 1.1, one can see that the response or decision is a function of the input (Sensory Input). But that is not all! That input has to be interpreted (Recognized) and that is the problem. Interpretation is a function of perceptual, cultural, and intellectual as well as emotional attributes that an individual brings with him/her to the table of problem solving. It is this interpretation of the inputs that will guide the team into a fruitful evaluation of the facts and data and ultimately the appropriate decision. (For the reader who is interested in more detail about the scientific background of the model of understanding, the following readings are suggested: Wundt, 1973; Koffka, 1935; Kohler, 1925; Wertheimer, 1929; Chomsky, 1965, 1967; Newell and Simon, 1972; and many others.) How do we go through this basic model? Generally, four steps are followed: Step 1. Preparation: When the problem solver becomes involved with the problem and searches for and accumulates relevant information Step 2. Incubation: When there is no conscious effort to deal with the problem, but although the subject is not aware of it, work on the problem continues Step 3. Illumination: The result of a successful incubation Step 4. Verification: When some elaboration of reality testing of the solution takes place 5

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Six Sigma and Beyond: Problem Solving and Basic Mathematics, Volume II

FIGURE 1.1 Understanding “a” situation.

These four stages have been empirically and experimentally validated through the work of many investigators including Wallas (1926), Anderson (1975), Aiken (1973), and Skemp (1971). Table 1.1 presents some typical examples of supported research generalizing the stages of problem solving. The term “problem solving process” has traditionally been applied to the characteristics of problem solving performance. However, this is troublesome because it is a matter of choice as to how one pursues this process. For example, the behaviorists usually focus on situational variables, the Gestaltists focus on built-in mechanisms in the subject, and information processing specialists study the characteristic requirements of the task itself. To be sure, all three dimensions are important, but to have the utmost result, all dimensions must occur simultaneously. How does one prepare and ultimately go through the process of problem solving? The literature is abundant with references from Plato and Aristotle, who focused on the notion of the human mind as the most important element of the process; to Medieval theology and Descartes, who pushed for the notion of the nature of man; to Hobbes in the 17th century; to Locke, Berkeley, and Hume in the 18th century; and to James and John Mill in the first part of the 19th century, who pushed the notion of reason as the foundation for problem solving. In 1927 it was Kohler who as one of the founders of Gestalt psychology sought to explain the notion of “insight.” He used the term to describe the capacity shown by members of the human race to restructure a given problem situation to their advantage or to solve a task through recognition of the relationships and interactions between its component parts. In this volume the Gestalt approach to problem solving has been selected as the most appropriate and user friendly of most applications. (By no means does this suggest that the other approaches are not significant.) The choice of the Gestalt

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Theoretical Aspects of Problem Solving

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TABLE 1.1 Generalized Stages of Problem Solving Researcher Helmholtz (1984)

Dewey (1910)

Wallas (1926)

Rossman (1931)

Young (1940)

Polya (1945)

Hutchinson (1949)

Mawardi (1960)

Osborn (1963)

Approach 1. 2. 3. 1. 2. 3. 4. 5. 1. 2. 3. 4. 1. 2. 3. 4. 5. 6. 1. 2. 3. 4. 5. 1. 2. 3. 4. 1. 2. 3. 4. 1. 2. 3. 4. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Investigation of performance in all directions Not consciously thinking about performance Appearance of “happy idea” Felt difficulty Location and definition Possible solutions Reasoning Acceptance or rejection Preparation Incubation Illumination Verification Observation of difficulty Analysis of the need Survey information Proposed solutions Birth of the new idea Experimentation to test promising solution; perfection by repeating some or all previous steps Assembly of material Assimilation of material Incubation Birth of the idea Development to usefulness Production Incubation Illumination Accommodation Preparation Frustration Insight Verification Abstract thoughts (A) Instrumental thoughts (I) Metaphonic ideas (M) Orientation (O) Think of all aspects Select subproblem Gather data Select relevant data Think of possible help Select attacks Think of possible tests Select soundest test Imagine all possibilities Decide final answer

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Six Sigma and Beyond: Problem Solving and Basic Mathematics, Volume II

TABLE 1.1 (Continued) Generalized Stages of Problem Solving Researcher Skemp (1971) Newell and Simon (1972)

Johnson (1972)

Anderson (1975) Sternberg (1980a)

Approach 1. 2. 1. 2. 3. 4. 5. 1. 2. 3. 4. 1. 2. 1. 2. 3. 4. 5. 6.

Assimilation Accommodation Input translation Internal representation Method selection Implement and monitor Reformulate Seek information Represent and transform Organize and reorganize Judgmental processes Preparation and production Incubation and eureka or “aha” experience Encoding Inference Mapping Application Justification (verification) Response (communication)

approach is based fundamentally on the notion of such greats as Kohler (1927), Maier (1970), Durkin (1937), and others who contend that the lack of success in problem solving is (1) biologically determined from an inability to integrate previous experience and (2) a result of functional fixedness, a tendency to think of objects as serving only in a limited “set,” which may have been established by previous experiences, but may be inappropriate to a particular problem situation. It is amazing that Durkin (1937) actually demonstrated that a solution to a complex puzzle was facilitated by prior experience with simpler ones. It would be unfair to not also mention the efforts of the behaviorists in this discussion. Indeed, C. L. Morgan (1894), Thorndike (1898, 1917), Woodworth and Schlosberg (1954), and Alexander Bain (1855, 1870) suggested that given a genuine problem, there must be some exploratory activity, more or less in amount and higher or lower in intellectual level. This, of course, is another way to express the old familiar statement about problem solving as the “trial and error” method. The works of Gofer (1961), Dulany (1968), Kendler and Kendler (1961, 1962), Skinner (1966), Staats (1968), and others support the notion that problem solving is dependent upon previously acquired stimulus. Although an ideal process of sequential operations might theoretically be assumed on the basis of the structure of the problem itself, at least for some welldefined tasks such as mathematical problems, this “ideal” process may bear little resemblance to the structural elements which operate when individuals solve problems. Processes which are presumed to operate on the basis of task requirements

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alone must be regarded as incomplete because they fail to take person- and environment-related process determinants into consideration. Different individuals might address the same task differently. They may differ in the amount of consideration and reflection on different solution possibilities and their utilization of external or internal cues or perhaps in the capacity or use of memory storage and retrieval processes. Therefore, different methodologies have been developed to accommodate each of these situations. The reader may want to explore some of these through the writings of Lazerte (1933), process tracing; Maier (1931b), introspection; Ghiselin (1952), retrospection; Binet (1903), Bloom and Broder (1950), Luer (1973), Newell and Simon (1972), and Simon (1976), protocol analysis (thinking aloud). Earlier a generalized format of problem solving was discussed; however, we must also recognize that research continues to identify the best and most efficient way of solving problems as well as the root causes. One such approach has been the innovation of Osborn (1963), who suggested that there is a difference between “idea creation” and “idea evaluation.” This, of course, is what is now called the “brainstorming” technique. According to Osborn, there are ten steps. Steps 1, 3, 5, 7, and 9 are the creative thought process and Steps 2, 4, 6, 8, and 10 are evaluative: Step Step Step Step Step Step Step Step Step Step

1. Think of all phases of the problem. 2. Select the subproblem to be attacked. 3. Think of what data might help. 4. Select the most likely sources of data. 5. Dream of all possible ideas as keys to the problem. 6. Select the ideas most likely to lead to a solution. 7. Think of all possible ways to test. 8. Select the soundest ways to test 9. Imagine all possible contingencies. 10. Decide on the final answer.

For the process of problem solving to be effective, perhaps the most important ingredient in that process is the operational definition. In problem solving terminology, an operational definition is one that specifies the meaning of the concept by denoting the measuring operations and suggests a criterion of whether or not a socalled empirical concept is a scientific concept (repeatable and reproducible by others) and whether or not it has been operationally defined. It must be understood that the definition in operational terms is not a theory, nor is it scientific in itself, but it provides the essential basis for the measurements which make possible initially the identification of the phenomena for subsequent scientific investigation. Excellent operations defining a particular phenomenon serve to differentiate it from other phenomena.

DEFINITION OF TERMS Just like anything else, problem solving has some important definitions. Everyone involved should be familiar with them. Therefore, this section provides some key definitions, general terms, terms related to the gathering of data, and terms related to the interpretation of the data.

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PROBLEM SOLVING The lack of a generally acceptable definition of this term in the literature and the broad and apparently indefinite scope of its use led Ernst and Newell (1969) to note: “Behind this vagueness … lies the absence of a science of problem solving that would support the definition of a technical term” (p. 1). Traditionally, the term problem solving has been used to describe the behaviors applied by a motivated subject, attempting to achieve a goal, usually in an unfamiliar context, after initial lack of success (Johnson, 1972). Definitions of the term tend to maintain that a problem exists when an individual is confronted by a “difficulty” (Dewey, 1933), a “gap” (Bartlett, 1958; Kohler, 1927), a “conflict” (Duncker, 1945), “disequilibrium” (Piaget, 1968), or a “deviation” from a familiar situation (Raaheim, 1974). While this type of description would seem to be equally suitable to define such terms as searching, understanding, or learning, “much contemporary research continues to reflect the basic Gestalt view that problem solving, by virtue of its emphasis on response discovery is something apart from learning” (Erickson and Jones, 1978, p. 62). Historically and to the present day, the term problem solving has been used with considerably greater frequency in reference to outcomes or products, particularly the success/failure aspect of the activity, rather than the process per se. Problem solving as a process became the focus of research with the weakening of interest of research workers in the perceptual and experiential aspects of thought, as had been pursued by associationists and Gestalt psychologists. As was noted earlier, the Gestalt approach to research into problem solving might be described as subject oriented. The more recent, important contributions of the information processing and artificial intelligence studies, though concerned with the investigation of the problem solving process itself, focused primarily on the demand characteristics and structure of certain problem solving tasks (e.g., Newell and Simon, 1972; Scandura, 1973, 1977; and Wickelgren, 1974). The term problem solving can refer to all overt and covert activities that take place to reach a solution or otherwise accomplish a goal or purpose in a problem solving situation. For the purposes of the six sigma methodology, the term problem solving is used to broadly describe the results of the interaction of components from the following five domains of variables: Step 1. The problem or task, T Step 2. The problem solver or subject, S Step 3. The situational circumstances or the environment in which the problem is presented or presents itself, E Step 4. The behaviors or processes which take place between the point of initial contact with the problem by S and the solution produced by S, X Step 5. The solution or product of the problem solving activity, P Regardless of the type of problem or the manner in which task (T), subject (S), environment (E), and process (X) variables interact, the product (P), whatever form it might take, is always a function of the interaction of variables from the remaining domains. This relationship can be expressed by the mathematical function:

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P(T) = f(T + S + E + X) where every variable may be represented by a number of domain components. It is imperative to not think of the product P as the dependent variable. Instead, the aim is to investigate structural components of the problem solving process X, which in turn must be expected to be a function of variables from the remaining domains; hence X = f(T + S + P + E) Although it is accepted that during problem solving the above suggested sets of variables may interact in many intricate ways, it seems legitimate and necessary to define each of them separately.

PROBLEM

OR

TASK

These terms are used interchangeably. The term problem has traditionally served as a label for a variety of phenomena, ranging from mathematical tasks to problems in real life. A common characteristic of all problems seems to be that they involve an aim which the problem solver wishes to accomplish, the means for which (i.e., the required knowledge, skills, techniques, or behaviors) are not at his or her disposal. To be confronted with a problem means to be faced with a difficulty or an obstacle that cannot be solved or dealt with in an already known or habitual manner; thus a reasonably general yet meaningful description of a problem or task might be that it may arise from any stimulus situation in which an appropriate response is not readily available. This definition is similar to many explanations of the term found in the research literature. A problem is described as: • Whatever — no matter how slight or commonplace in character — perplexes and challenges the mind so that it makes belief at all uncertain (Dewey, 1933, p. 13) • A question for which there is at the moment no answer (Skinner, 1966, p. 225) • When a person is motivated toward a goal and his first attempt to reach it is unrewarding (Johnson, 1972, p. 133) • A stimulus situation for which an organism does not have a ready response (Davis, 1973, p. 12) • When a system has or has been given a description of something but does not yet have anything that satisfies the description (Reitman, 1965, p. 126) The definition of a problem as a phenomenon that may arise in any stimulus situation in which an appropriate response is not readily available is broad enough to cover physical, emotional, intellectual, and social problems. It can refer to problems of varying complexity, to defined and ill-defined tasks, and to structured and unstructured problems. It serves to prevent the problem solver from avoiding or ignoring the problem, but does not require the individual to recognize the task

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“objectively.” The “objective” problem is the task as perceived by the individual. This position acknowledges that what is a problem for one person may not be for another. The latter individual, in this case, has an appropriate response readily available in a stimulus situation which might present a problem for the former person. In summary, it is suggested that the above presented definition of the term “problem” or “task” subsumes all types of problems without, however, obscuring the differences that may exist in different problem solving situations. A further point of definition may require clarification. It would follow from the above definition that a problem that does not elicit any reaction from the individual or a task that has been solved would cease to be a problem. The English language unfortunately does not provide an alternative term to cover these types of situations. Frequently occurring tasks, such as the simple arithmetic “problem” 1 + 1 = 2, which elicit well-rehearsed, often automatic responses, do not fall into the problem category defined above. Other languages provide alternative terms for these “problems” (e.g., in Greek one would use “askese,” in German one would use “die Aufgabe,” and in French one would use “la tache”) which permit a clearer distinction between a “problem” which requires at least some effort in terms of productive thinking on the part of the subject and what may be described as a simple exercise or routine stimulus-response association. The term “problem” as used in this volume does not include simple stimuli presented for automatic response.

PROBLEM SITUATION This concept was first used by Wertheimer (1923), who regarded it as consisting of two ingredients. These are the aim or solution, i.e., “that which is demanded,” and the stimulus or materials, knowledge, skills, etc., i.e., “that which is given.” The process of problem solving commences when what is given is brought into association with that which is demanded. This will include, for example, considerations of how the givens might lead to a solution. For example, the “functional value” (Duncker, 1945; Kohler, 1917) of the givens might be assessed as done by Kohler in the case of the sticks utilized by a chimpanzee to reach bananas.

SUBJECT This term refers to the individual who is attempting to solve a problem. The problem solver cannot be considered to be a neutral agent. All variables — many of them unknown, some known, but not yet measurable, others constant — that determine the subject’s behavior make up this domain. Obvious examples of these variables are motivation; memory; intelligence; general background and experience, including experience with problems of a certain type; and social and personality variables.

ENVIRONMENT The total problem solving environment includes physical, psychological, and sociological variables. The physical environment provides many perceptual cues and memory associations that might be used by the problem solver to define, analyze, and understand the problem and to enlarge the set of available approaches to the task.

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On the other hand, physical, psychological, and sociological variables and factors resulting from their interaction may place certain constraints on the task, the solution, or the problem solving activity and directly or indirectly influence the process as a whole. The problem environment contains, of course, the experimenter with all of his or her characteristics. Task directions, definitions, etc. form part of the problem or task.

PROCESS Problem solving is often referred to as a process. Yet the evidence for the occurrence of this process is based on an examination of the end product, performance. The outcome of problem solving, i.e., the results of the process (or processes), not the processes themselves, are described. The problem solving process, which begins with presentation of the problem, is terminated when the subject arrives at a correct solution; when the subject arrives at an incorrect solution believed to be correct; or when the experimenter ends the session.

PRODUCT

OR

SOLUTION

These terms are used interchangeably for the result of the problem solving activity. The term product would appear to be more suitable for research purposes because of the connotation of the term solution. One would, generally, expect a solution to be “correct.” In the six sigma methodology, both terms are used for the result or outcome of the problem solving attempted by subjects. The terms apply to both “correct” or “incorrect” outcomes.

PROBLEM-SOLVING BEHAVIOR STRATEGY

OR

OPERATION

AND

PROBLEM-SOLVING

Again, these terms are used interchangeably to denote any response or any part of a response pattern that is observed during the subject’s problem solving activity, i.e., can be identified on the basis of the subject’s problem solving protocol. It is acknowledged that many behaviors are unobservable and therefore will not be contained in the protocol. The difference in processing rate per time unit between thought and speech and the inability of speech to reflect parallel processes result in the fact that “thinking aloud” protocol contains a reduced version of the problem solving process, no matter how perfect the experimental conditions. Strictly speaking, use of the term “strategy” should perhaps be restricted to dynamic processes involved in performance which are made up of several operations or behaviors. On the other hand, the repeated use of a specific operation over time may well be conceived as a strategy rather than an operation. Dynamic processes that perform specific operations are called “strategies.” The idea is that elementary mental operations may be assembled into sequences and combinations that represent the strategy developed for a particular task. It is often difficult to determine whether the elementary mental operations isolated are strategies or whether they are structures.

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The chess master, for example, who has developed a strategy for analysis of the chess board over many years of practice, may be unable to modify it; thus it becomes more structural in character. Indeed, high levels of skill seem to be characterized by the development of a structural basis for what in most of us is a painfully assembled strategy. Perhaps, this is precisely the reason why “problem solvers” are generally sought after rather than planners. It is the cumulative experience that the “problem solver” brings into the situation to resolve the issue at hand. More often than not, a good problem solver operates from a pattern of processes rather than sequential ones. In fact, it is the pattern association of past experiences that drives new solutions.

TERMS AND ISSUES RELATED TO DATA GATHERING Definitions of the following terms are provided in an endeavor to increase the clarity of the description of the method and importance of data gathering.

STIMULUS PASSAGE This term refers to a single sheet of paper containing a diagram, a typed sentence, a paragraph, or paragraphs presenting the subject with the task. A typical stimulus passage may be a flow chart of the process or some measurements of a process. The rationale of stimulus passages is to provide the means by which the subject may be acquainted with each task.

VERBALIZATION

OR

“THINKING ALOUD”

The terms are used interchangeably to refer to the subject’s verbal expressions and descriptions of his or her ongoing problem solving activity. Typically this is a version of brainstorming.

PROTOCOL A verbatim written transcript of the cassette recording made of all verbalizations produced by subjects during the problem solving session is produced. The protocol contains a complete record of the description of problem solving provided by the subject’s “thinking aloud.” It also contains evidence of any other verbal or verbally reported activities that occurred. A protocol, then, provides a description, keeping time sequence intact, of problem solving performance as it occurred. On the other hand, not every description of performance of a task constitutes a protocol. A description of a task consisting of goals and outcomes is not sufficient. The problem solving protocols provide information, not only concerning the answers the subjects provided, but also, and more importantly, they provide an indication of the sequence in which the problem solving behaviors occurred. Subjects may ask questions, refer to the stimulus passage, compute, etc. in a particular order. This approach to gathering data may be very helpful in customer surveys and market research studies as well as when someone is interested in identifying very specific tasks and interpretations of a particular job, project, etc.

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Process components are identified on the basis of the amount of use made by each subject of certain problem solving behaviors or strategies and by investigating transition sequences of these behaviors at various temporal stages of the problem solving activity. Behavior or strategy use is measured by the activity of the process as a whole in addition to the individual participation of team members. This participation is of paramount importance, and it depends on the leadership of the team as well as on the appropriate cross-functionality and multidiscipline of the team members. It is this behavior and strategy that ultimately will (1) select the likely potential cause; (2) prevent recurrence; and (3) define the roles and team orientation. Select the Likely Potential Causes Once the problem has been described and the potential causes have been identified, the team should be evaluated. Are the right members on the team to investigate the potential causes? Are technical advisors required to assist in any special studies? Are new team members needed? Is the authority to pursue the analysis of the potential causes well defined? All of these questions must be answered to ensure that the team will be successful in investigating the potential causes and determining the root cause. A typical tool that is used is the cause-and-effect diagram, in conjunction with brainstorming, to identify the potential causes to be investigated. What is the probability that a potential cause could be responsible for the problem? Identify all potential causes that could have been present and might have caused the problem. Once all potential causes have been agreed on, choose several potential causes to investigate. If only one potential cause is investigated, a lot of time may be lost if that potential cause is not the cause of the problem. To expedite the investigation of potential causes, investigate several causes at the same time. Parallel actions on several potential causes will expedite the process. If the problem is a manufacturing process, begin to establish a stable process. Once the process is stable, definition of the potential cause will be clarified. On the other hand, if engineering design causes are identified, screening experiments may help identify the key variables which are affected by subsequent processes, and robust design actions may be appropriate. Four or five potential causes to investigate should be identified. Identifying several potential causes forces the team to address multiple causes rather than search for a single cause. An implicit part of the problem analysis is investigating potential causes in parallel rather than in series. Prevent Recurrence A second concern in the behavior and strategy phase is to make sure that the problem once solved will not recur. The analysis begins with understanding what in the process allowed the problem to occur. A cause-and-effect diagram can be used to outline the reasons the problem occurred. By asking “because?,” the cause-and-effect diagram could be constructed.

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Another effective tool is a process flow diagram. The process flow of the manufacturing or engineering process or even a service can be effective in identifying where in the process the problem could have been prevented. Most of the time, to prevent recurrence of the problem, a change to the management system will be required. Managers must understand why their system allowed a problem to develop. The same system will allow future problems to occur. Management systems, practices, and procedures need to be fully understood to be effective. Most of them are carry-overs from previous experiences and organizational structures. Some are outdated and need to be revised. Understanding the elements of a management system can be achieved by maintaining an up-to-date flow diagram of the system and process. Also, instructions should be easy to follow by those who are part of the system. Management systems, practices, and procedures should provide management support for “never-ending improvement” in all areas and activities. The system should encourage individuals to participate freely in the problem solving process. It should help them understand more about their job and how each individual’s effort affects the outcome of the final product on the customer satisfaction. The system should encourage everyone to learn something new, and it should recognize individual and team effort when these new skills are applied. Changes in the management system may require documenting new standard procedures, streamlining to remove obsolete procedures, and revising previous standards. Any changes in the management system need to be communicated clearly to all customers. To prevent recurrence, additional training is often required. Training may be needed in statistical tools, new engineering or manufacturing technologies or disciplines, better process, or project management. Some basic problem solving tools are 1. Process flow diagram. A process flow diagram (sometimes called mapping) is a graphic presentation of the flow and sources of variation in a process. In manufacturing, a process flow is a graphic presentation of the flow and sources of variation of machines, materials, methods, and operators from the start to the end of the manufacturing and/or assembly process. To graphically display the total process, some standard symbols are helpful in commonizing the process flow diagrams. While the symbols used will vary from one application to another, it should be noted that any process can be diagramed. 2. Control charts. The various types of control charts are graphic tools used to separate controlled and uncontrolled variation in a process. All such charts have the same two basic functions: • To establish whether or not a process is operating in a state of statistical control by identifying the presence of special causes of variation. This permits corrective actions to be taken. • To maintain the state of statistical control once it is achieved. This allows for periodic recalculation of the control limits which may, in turn, lead to reduced variability within the process being charted.

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There are two broad classifications of chart types. They are based on the type of data for which they are appropriate. Charts for quantitative data (measurements of length, weight, time, etc.) include: • • • •

Xbar and R charts Xbar and S charts Xbar and median charts Individual X and R charts (charts for individuals, run charts)

Attribute-type data have only two possible values (pass/fail, go/no go, present/absent, etc.). The principal types of charts for such data include: • • • •

P charts NP charts C charts U charts

3. Histograms. A histogram is a pictorial representation of the distribution of measured or counted items. It is a quick way to establish the average or middle value of a distribution, the variation or dispersion of the distribution, whether any outliers exist, and whether the distribution is bell shaped (normal). A histogram is a data collection device used to quantify and establish the distribution of the problem, to quantify the size of a problem, to highlight differences between the actual results and some standard, etc., and for before and after comparisons. 4. Stem and leaf plots. This tool is a special case of the histogram and serves the same purpose as the histogram, but has one significant advantage — it preserves the actual data values. In many cases this is a very desirable feature. While the typical histogram depicts the distribution of the data on which it is based, the data values cannot be reconstructed from the picture. They must be provided separately or they are lost. The stem and leaf plot, however, serves the same purpose, gives the same picture, and, simultaneously, preserves the original data values. 5. Check sheets. A check sheet is a form on which data are recorded as the data are collected. Check sheets are most often used to facilitate the data collection process and to simplify data analysis. Check sheets sort the data as the data are collected, providing construction of a histogram at the same time. However, this tool will not show changes over time. A check sheet is used to collect data to verify the effectiveness of interim actions, to analyze potential causes, and to verify the effectiveness of permanent solutions. As a data collection tool, similar to histograms and Pareto charts, a check sheet is often a logical starting point in problem solving. 6. Plots/graphs. Plots and graphs of various kinds are pictorials formed from what would otherwise be tables of numbers (data values). They are used primarily to allow the quick identification of trends or patterns in data and to suggest appropriate techniques for further analysis. Therefore, they are

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7.

8.

9.

10.

a very convenient way to summarize and visualize data. Plotting raw data is quick and simple and, by itself, provides no additional information not already in the data. The advantage of a plot or graph is that it highlights trends, etc. which are usually much more difficult to notice in an often intimidating table of numbers. With such a tool, the “reader” can quickly “see” a picture of what the data are “saying.” Plots and graphs also have a tendency to point out and quantify differences within the data and, in many cases, to identify the areas of concern to the “reader.” Frequently, stratifying or separating the data by categories such as shift, time, sequence, or supplier will be of help in understanding the differences which would otherwise be lost by “lumping’’ all the data together. Since there are no universal rules for plotting, the user may wish to plot the same data in several ways to see what “story” emerges from the different plots. Pareto diagrams. A Pareto diagram is a special type of vertical bar graph used to indicate which cause needs to be solved first in eliminating the problem. In problem solving, the Pareto diagram separates the “vital few” from the “trivial many.” Quite often this diagram will also identify the cumulative contribution of the characteristics identified as a percentage. Scatter diagrams. A scatter diagram is a plot between variables to illustrate a possible relationship between the variables. The plot is constructed so that the horizontal axis (x-axis) represents the measured values of one variable and the vertical axis (y-axis) represents the measured values of the second variable. The strength of the relationship between variables 1 and 2 is indicated by the direction and density of the cluster of plotted points. The scatter diagram can be used to show a correlation between a suspected cause and the effect being investigated. There must be, however, a rational (engineering, etc.) basis to suspect a cause-and-effect relationship since correlation alone does not imply cause and effect. Gage reproducibility and repeatability. A gage study is performed to investigate the amount of variation added to the total process by the measurement system. Most gage studies will quantify the amount of variation added to the process by the operator (reproducibility) and by the precision (repeatability). If the combined variability of the operator and the gage precision is “using up” too much of the blueprint tolerance, then the gage will need to be reworked. Gage studies are essential to problem solving for manufacturing processes. In many cases, the gage may be a large contributor to the production problem. There are several methods available for doing gage studies and each has its own appropriate area of application. Capability indices. Capability studies are performed on machines to establish the ability of the machine to produce parts to blueprint specifications. The most common indices cited for machine capabilities are Cp and Cpk. The first capability index, Cp, is the relationship between the engineering blueprint tolerance and the machine variability or process spread. The process capability index Cp is calculated by:

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Cp = tolerance/process spread = tolerance/6σ where σ is the standard deviation of the process. The second capability index, Cpk, is a measure of how well the process distribution is centered between the specification limits. The process location index, Cpk, is calculated by: Cpk = absolute value of [sl – Xbar]/3σ where sl is the closest specification limit to Xbar, the process average. Both of these indices are used to define the process capability. Capability studies can be used to define a problem or to verify permanent corrective actions in the problem solving process. Either way, the process must be stable before its capability can be studied. 11. Design of experiments (DOE). DOE is a discipline which is intended to produce experiments (tests) which are both effective and efficient. It is a replacement for the “one-change-at-a-time” method of experimentation. Designed experiments are effective because they are planned from the start to answer the specific questions for which answers are needed. These experiments are efficient because they get maximum use from minimum data. The result is that better experiments are performed rather than simply doing more of them and experiments produce the most information with the least cost in dollars and time. The DOE process usually requires the cooperation of the experimenter, test operator, and perhaps a statistician or consultant (thus a fringe benefit is the teamwork that results). The experimenter supplies the list of specific questions to be answered by the research and together they design an experiment which will ensure the collection of data sufficient to get these answers with a minimum expenditure of time and resources. The benefits of a designed experiment vis-à-vis the “one-at-a-time” approach include those already mentioned in addition to the ability to: • Establish cause-and-effect relationships between several different variables (simultaneously at different levels) and the outcome (effect) being studied. • Eliminate (block out a screen for) confusing or confounding the effects of two or more variables. • Evaluate the interactions of two or more variables. (In fact, a designed experiment is the only way to do this.) • Identify the variation due to planned changes while separating it from the remaining, ever-present, variation. The various kinds of designed experiments and the particular strengths and weaknesses of each are too numerous to list here. The scope of their applications, however, ranges from the social sciences to engineering and

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production applications to problem solving in general. A screening experiment is a type of designed experiment. It is most often used to evaluate many factors in an experiment. In general, the screening experiment will only look at the main effects and not investigate the possible “interaction” between variables. When first investigating a problem, a screening experiment is beneficial in identifying those potential causes that influence the effect the most. This type of experiment is also called a fractional factorial design. These experiments have been popularized by Taguchi. 12. Is/Is Not. This approach helps in the definition of the problem. It is used as part of the brainstorming session to focus on the true “problem” under consideration. 13. FMEAs. Failure mode and effect analysis (FMEA) is a semiquantitative analytical technique involving a well-disciplined, systematic approach structured around usual problem-solving techniques. However, the most important feature of FMEA is that it is done before an actual field problem occurs. Therefore, it is a preventive rather than a detective approach. FMEA provides a means to prioritize the identified potential problems for resolution. It also provides documentation of “due-care” in a legal sense. Total FMEA consists of concept, design, manufacturing, assembly, and other FMEAs at system, subsystem, or component level. The actual FMEA implementation level depends upon available resources, associated complexity, severity, etc. Usually, FMEAs are performed by cross-functional teams consisting of representatives from various activities. Proper implementation of the FMEA process will result in: • Improved product reliability due to prevention of problem occurrence • Improved validity of analytical methods through documentation and scrutiny • Retention of knowledge and experience • Sound knowledge and a database While there are several FMEA techniques, e.g., tabular FMEA, fault tree analysis (FTA), etc., most organizations use the tabular FMEA format. It is considered to be one of the best procedures. The important features of the procedure are • • • •

Identify part function and potential failure modes. Identify the severity and the effect of the failure. Identify the causes and their occurrences of these failure. Identify the controls that are in place to find and to make sure that the failure will not go to the customer. Calculate the risk priority number (RPN) which is the product of the following three ranking indices: RPN = Severity × Occurrence × Detection

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Each index ranges from 1 to 10 and specific values are assigned by the FMEA preparing team. The order of analysis may be in two forms: The first is • • • •

The highest severity ranking takes priority. The product of severity and occurrence takes the second priority. Detection (in a sense the RPN) takes the third priority. A higher RPN value indicates the need for corrective action.

The second is • Identify the highest RPN and proceed with appropriate corrective actions that will lower the RPN index. In both cases, reevaluation is essential and the goal is to drive all numerical values as low as possible. Reevaluation is based on the corrective action implementation. If the values of the analysis are still very high, identify further improvement actions and repeat the process as required. In the automotive industry, FMEA has undergone some major changes, such as making FMEA a critical part of the “robustness” attitude. In some cases, because of the changing technology and its complication, items such as “repairability” to reflect availability, maintainability, and “serviceability” are considered in the evaluation. When that occurs, the FMEA process itself remains unchanged except for the revised RPN which is calculated as: RPN = Occurrence × Severity × Detection × Repairability 14. Weibull analysis. This is a technique used to analyze and graphically present field failure and life test data. With it, one can also estimate (with or without confidence intervals) the useful life of an item. The process begins with collection of “run to failure” data which, with some relatively simple (and perhaps computer assisted) calculations, produces the information required for plotting on a special kind of graph paper called Weibull probability paper. The following information is then available from the resulting plot: • The percentage of the population which can be expected to fail below a specified life • The life value (Bq) below which q% are expected to fail (The user can select an appropriate value for q. For example, the B10 life is that value at which one can expect 10% of the population to fail.) • The slope of the failure distribution line which provides an indication of the type of failure (A slope of less than 1.0 indicates a decreasing failure rate — sometimes called the infant mortality period. A slope of approximately 1.0 indicates random failures and a slope greater than 1.0 is indicative of “wear out” failures.)

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An entire family of Weibull density functions can be individually distinguished based on this last item (the slope). This permits wide application of the technique. The method is also quite useful for many series systems and for comparing different designs, materials, or processes. Weibull probability paper is used for analysis. This graph paper is a special plotting paper: both scales are logarithmic and the horizontal scale is usually a function of time (hours, miles, etc.) to failure. The vertical scale is a cumulative percentage of failures. In the upper left corner, there is a preprinted compass-like scale with which to determine the slope of the Weibull distribution function for the particular item (part, etc.) under study. A Weibull distribution plot always identifies failures. However, if the failures are subtracted from 1, then one can actually identify the reliability, as well. An awareness of each of the above tools is a necessary minimum. Some persons will need considerably greater depth in one or more tools. These tools will be addressed in the next chapter and in more detail in Volume IV. If concerns develop regarding changes to the system, these issues will be addressed. A new team may need to be assigned with the authority to address the management system. Define the Roles and Team Orientation To maximize the effect of both behaviors and strategy in a problem solving environment, the team orientation and the roles of participants must be defined. Use the team approach: Establish a small group of people with process/product knowledge, allocated time, authority, and skill in the required technical disciplines to solve the problem and implement corrective actions. The group must have a designated champion. Describe the problem: Specify the internal/ external customer problem by identifying in quantifiable terms the who, what, when, where, why, how, and how many (5W2H) for the problem. • Who. Identify individuals associated with the problem. Characterize customers who are complaining. Which operators are having difficulty? • What. Describe the problem adequately. Does the severity of the problem vary? Are operational definitions clear (e.g., defects)? Is the measurement system repeatable and accurate? • Where. If a defect occurs on a part, where is the defect located? (Use location check sheet.) What is the geographic distribution of customer complaints? • When. Identify the time the problem started and its prevalence in earlier time periods. Do all production shifts experience the same frequencies of the problem? What time of the year does the problem occur? • Why. Any known explanation contributing to problem should be stated.

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• How. In what mode of operation did the problem occur? What procedures were used? • How many. What is the extent of the problem? Is the process in statistical control (e.g., P chart)? The above 5W2H questions are to help characterize the problem for further analysis. Some problems arise from customer complaints. An internal customer’s complaint could involve one department complaining that they cannot effectively use the output of another department. An external customer complaint could involve a customer complaining to a dealer that the car transmission “shifts funny.” Too frequently, the wrong problem is solved and the customer complaint is not addressed. It is imperative that the customer complaint be clearly understood. (That is the importance of operational definition.) The only method to ensure this is to have direct customer contact. For internal customers, it is advisable to have representatives from the complaining organization as part of the problem solving team. In many cases, this approach is the only way a problem can be truly solved. External customer complaints typically require direct interviews to understand why the customer is not satisfied. It is not unusual for a customer complaint to be misrepresented by a company reporting system that classifies problems in prearranged standard categories. Part of the 5W2H problem definition is to state the customer complaint clearly. Implement and verify interim (containment) actions: Define and implement containment actions to isolate the effect of problem from any internal /external customer until corrective action is implemented. Verify the effectiveness of the containment action. Define and verify root causes: Identify all potential causes which could explain why the problem occurred. Isolate and verify the root cause by testing each potential cause against the problem description and test data. Identify alternative corrective actions to eliminate root cause. Verify corrective actions: Through preproduction test programs, quantitatively confirm that the selected corrective actions will resolve the problem for the customer and will not cause undesirable side effects. Define contingency actions, if necessary, based on risk assessment. Implement permanent corrective actions: Define and implement the best permanent corrective actions. Choose ongoing controls to ensure that the root cause is eliminated. Once in production, monitor long-term effects and implement contingency actions, if necessary. Prevent recurrence: Modify the management systems, operating systems, practices, and procedures to prevent recurrence of this and all similar problems. Congratulate your team: Recognize the collective efforts of the team. To accomplish all of the tasks just mentioned, a team approach must be followed. The team approach is illustrated in Figure 1.2. Whereas the team process is important,

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FIGURE 1.2 Team approach.

it is just as important to be very conscious of the inhibitors that might creep into the problem solving process. Lack of effective problem solving exists in nonmanufacturing, manufacturing, and engineering environments. The following are a few selected characteristics to ensure that the reader is aware as to what might happen if one is not cognizant of the mechanics of the process. • Problem described incorrectly: A clear, thorough description of the problem is necessary. A problem must be adequately described and be narrow enough in scope for the team to handle effectively. • Problem-solving effort expedited: Steps were skipped in the problem solving process to obtain a quick solution. • Poor team participation: Not all team members participated effectively, so the team failed to consider all the causes of the problem. • No logical process: The team lacked a disciplined system to prioritize, analyze, and review problems. • Lack of technical skills: Team members were not trained in statistics and problem solving methods. • Management’s impatience: Management’s lack of knowledge of the problem solving process often results in all levels of management demanding

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to know exactly when a problem will be solved. This pressure causes the team to make an inadequate analysis. • Potential cause misidentified as a root cause: Sometimes a potential cause is quickly identified as a root cause, and the problem investigation is concluded. However, the problem often recurs because the root cause was not eliminated. • Permanent corrective actions not implemented: A root cause may be identified, but no action was taken to implement the permanent corrective actions. Permanent actions often require management approval of the costs and implementation of the action.

IMPORTANCE OF THE SAMPLE To have a solution for a problem, there must be data to evaluate. The beginning of this process is to select appropriate and applicable data for the occasion. In Volumes III and IV, this issue will be addressed in much more detail. However, this section will give a cursory view of what is a “sample” and how to go about asking appropriate questions of “stratification.” The size of a sample/subgroup is important/critical to any statistical procedure. Sample size is important because it determines how much information (or how little information) is being drawn from the process or population. The larger the sample size, the more information is drawn from the process. As more information is drawn from the process, fewer errors are made in the estimates of the process parameters. Taken to an extreme, the sample size may be increased so that it approaches the population size of the process or 100%. Although this would provide all of the information, it is extremely expensive and impractical. Besides the cost, a 100% inspection (a sampling plan that requires a 100% of the information) does not contribute to the process improvement effort — 100% of the products must be produced before the sampling plan is completed. This means that a tremendous investment must be made before a decision/interpretation may be made about the value of the process performance. Material will be consumed, energy will be used, and human resources will be wasted. Sometimes it is necessary to reconfigure the sample and data in a way so that the information comes “alive” and communicates what is really happening in the process. That reconfiguration is called stratification analysis. Through stratification analysis the extent of the problem for all relevant factors can be determined. A possible questioning scenario that might be used in formulating a stratification strategy is • Is the problem the same for all shifts? • Do all machines, spindles, and fixtures have the same problems? • Do customers in various age groups or parts of the country have similar problems? The important stratification factors will vary with each problem, but most problems will have several factors. Check sheets can be used to collect the data. Essentially,

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this analysis seeks to develop a Pareto diagram for the important factors. The hope is that the extent of the problem will not be the same across all factors. Then differences can lead to identifying a root cause. When 5W2H and stratification analyses are performed, it is important to consider a number of indicators. For example, a customer problem identified by warranty claims may also be reflected by various in-plant indicators. Also, customer surveys may be able to define the problem more clearly. In some cases, the analysis of a problem can be expedited by correlating different problem indicators to identify the problem clearly. It has been said that there are no new problems — only different manifestations of old problems. In problem definition, it is often useful to quantify the problem in similar situations. The criteria to match similar situations will vary with the type of problem. Identifying effective matches and evaluating the presence of the problem provides useful information to generate potential causes and possible problem solutions. If the similarity analysis identifies a comparable situation where the problem does not exist, the analysis can focus on the differences in where the problem is occurring and where it is not occurring. Once the three types of analysis have been completed, it is sometimes possible to divide the problem into separate problems. It is easier to address these smaller problems because fewer root causes are involved. In the ideal case, a single root cause would be responsible for each problem. If the problem is separated, different teams may be required to address each problem. All three elements of the problem definition are not used for every problem. However, collectively, the different analyses provide a comprehensive description. Complete specifications of the problem involve: • Measurability • Criteria for success • A given condition for existence

REFERENCES Aiken, L. R., Ability and creativity in mathematics. Review of Educational Research, 1973, 43, 405–432. Anderson, B. F., Cognitive Psychology: The Study of Knowing, Learning and Thinking. New York: Academic Press, 1975. Bain, A., Mental Science, A Compendium of Psychology and the History of Philosophy. New York: Appleton, 1870. Bain, A., The Senses and the Intellect. London: Parker, 1855. Bartlett, F., Thinking: An Experimental and Social Study. London: Allen and Unwin, 1958. Binet, A., L’Etude Experiimentale de l’Intelligence. Paris: Schleicher, 1903. Bloom, B. S. and Broder, L., Problem Solving Processes of College Students. Supplement to Education Monographs, 73. Chicago: University of Chicago Press, 1950. Byrne, R., Planning meals: problem-solving on a real data-base. Cognition, 1977, 5, 287–332. Chomsky, N., Aspects of the Theory of Syntax. Cambridge, MA: MIT Press, 1965. Chomsky, N., The general properties of language, in C. H. Millikan and F. L. Darley, Eds., Brain Mechanisms Underlying Speech and Language. New York: Grune & Stratton, 1967.

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Cofer, C. N., Ed., Verbal Behavior and Learning. New York: McGraw-Hill, 1961. Cronback, L. J., The two disciplines of scientific psychology. American Psychologist, 1957, 12, 671–684. Cronback, L. J., Beyond the two disciplines of scientific psychology. American Psychologist, 1975, 30, 116–128. Davis, G. A., Psychology of Problem Solving. New York: Basic Books, 1973. Dewey, J., How We Think. New York: Holt, Rinehart, & Winston, 1910. Dewey, J., How We Think: A Restatement of the Relation of Reflective Thinking to the Education Process. Boston: Heath, 1933. Dulany, D. E., Jr., Awareness, rules and propositional control: a confrontation with S-R behavior theory, in T. R. Dixon and D. L. Horton, Eds., Verbal Behavior and General Behavior Theory. Englewood Cliffs, NJ: Prentice-Hall, 1968. Duncker, K., On problem solving. Psychological Monographs, 1945, 58 (270). Durkin, H. E., Trial-and-error, gradual analysis and sudden reorganization, an experimental study of problem solving. Archives of Psychology, 1937, 30 (210). Erickson, J. R. and Jones, M. R., Thinking. Annual Review of Psychology, 1978, 29, 61–90. Ernst, G. W. and Newell, A., GPS: A Case Study in Generality and Problem Solving. New York: Academic Press, 1969. Ghiselin, B., Ed., The Creative Process. Berkeley: University of California Press, 1952. van Helmholtz, H., Vortrage and Reden, 5th ed. Braunschweig: Vieweg, 1894. Hutchinson, E. D., How to Think Creatively. New York: Abington Cokesbury, 1949. Johnson, D. M., Systematic Introduction to the Psychology of Thinking. New York: Harper & Row, 1972. Kendler, H. H. and Kendler, T. S., Problems in problem solving research, in Current Trends in Psychological Theory: A Bicentennial Program. Pittsburgh: University of Pittsburgh Press, 1961. Kendler, H. H. and Kendler, T. S., Vertical and horizontal processes in problem solving. Psychological Review, 1962, 69, 1–16. Koffka, K., Principles of Gestalt Psychology. New York: Harcourt, Brace & World, 1935. Kohler, W., Intelligenzprufungen an Menschenaffen. Berlin: Springer, 1917. Kohler, W., The Mentality of Apes (translated by E. Winter). New York: Harcourt, Brace & World, 1925. Kohler, W., The Mentality of Apes. London: Routledge, 1927. Lazerte, M. E., The Development of Problem Solving Ability in Arithmetic: A Summary of Investigations. Toronto: Clark Irwin, 1933. Luer, G., The development of strategies of solution in problem solving, in A. Elithorn and D. Jones, Eds., Artificial and Human Thinking. Amsterdam: Elsevier Scientific, 1973. Maier, N. R. F., Reasoning in humans. II. The solution of a problem and its appearance in consciousness. Journal of Comparative Psychology, 1931b, 12, 181–194. Mair, N. R. F., Problem Solving and Creativity in Individuals and Groups. Belmont, CA: Brooks/Cole, Wadsworth, 1970. Mawardi, B. H., Thought sequences in creative problem solving. American Psychologist, 1960, 15, 429. Morgan, C. L., An Introduction to Comparative Psychology. London: Scott, 1894. Neisser, U., Cognitive Psychology. New York: Appleton-Century-Crofts, 1967. Newell, A. and Simon, H. A., Human Problem Solving. Englewood Cliffs, NJ: Prentice-Hall, 1972. Osbom, A. F., Applied Imagination, 3rd ed., New York: Scribner, 1963. Pellegrino, J. W. and Glaser, R., Cognitive correlates and components in the analysis of individual difference. Intelligence, 1979, 3, 187–214.

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Piaget, J., Six Psychological Studies. London: University of London Press, 1968. Polya, G., How to Solve It. Princeton, NJ: Princeton University Press, 1945. Raaheim, K., Problem Solving and Intelligence. Oslo/Bergen/Tromso: Universitetsforlaget, 1974. Reitman, W. R., Cognition and Thought. New York: John Wiley & Sons, 1965. Rossman, J., The Psychology of the Inventor. Washington: Inventor’s Publishing, 1931. Scandura, J. M., On higher order rules in problem solving. Journal of Educational Psychology, 1973, 10, 159–160. Scandura, J. M., Problem Solving: A Structural/Process Approach with Instructional Implications. New York: Academic Press, 1977. Simon, H. A., Identifying basic abilities underlying intelligent performance on complex tasks, in L. B. Resnick, Ed., The Nature of Intelligence. Hillsdale, NJ: Lawrence Erlbaum Associates, 1976. Skemp, R. R., The Psychology of Learning Mathematics. Baltimore: Penguin Books, 1971. Skinner, B. F., An operant analysis of problem solving, in B. Kleinmuntz, Ed., Problem Solving: Research, Method and Theory. New York: John Wiley & Sons, 1966. Staats, A. W., Learning, Language and Cognition. London: Holt, Rinehart & Winston, 1968. Sternberg, R. J., Sketch of a componential subtheory of human intelligence. The Behavioral and Brain Sciences, 1980a, 3, 573–614. Thorndike, E. L., Animal intelligence: an experimental study of the associative processes in animals. Psychological Monographs, Review Supplements, 1898, 6 (8). Thorndike, L. L., Reading as reasoning: a study of mistakes in paragraph reading. Journal of Educational Psychology, 1917, 8, 323–332. Wallas, G., The Art of Thought. London: Jonathan Cape, 1926. Wertheimer, M., Untersuchungen zur Lehre von der Gestalt. Psychologische Forschung, 1923, 4, 301–351. Whitely, S. E., Information-processing on intelligence test items: some response components. Applied Psychological Measurement, 1977, 1, 465–476. Wickelgren, W. A., How to Solve Problems. Elements of a Theory of Problems and Problem Solving. San Francisco: W. H. Freeman, 1974. Woodworth, R. S. and Schlosberg, H., Experimental Psychology (Rev. ed.). London: Methuen, 1954. Wundt, W., Grundzuge der Physiologischen Psychologie. Leipzig: Breitkopf and Hartel, 1873. Young, J. W., Technique for Producing Ideas. Chicago: Advanced Publications, 1940.

SELECTED BIBLIOGRAPHY Allen, M. J. and Yen, M., Introduction to Measurement Theory. Belmont, CA: Wadsworth, 1979. Allport, D. A., The state of cognitive psychology: a critical notice of W. G. Chase, Ed., “Visual information processing.” Quarterly Journal of Experimental Psychology, 1975, 27, 141–152. Altman, J., Aspects of the criterion problem in small groups research. 11. The analysis of group tasks. Acta Psychologica, 1966, 25, 199–221. Anastasi, A., Individual differences. In D. L. Sills, Ed., Encyclopaedia of the Social Sciences, Vol. 7. New York: The Macmillan Co. & The Free Press, 1968. Asch, S. E., Effects of group pressure upon modification and distortion of judgments. In E. E. Maccoby, T. M. Newcomb, and E. L. Hartley, Readings in Social Psychology, 3rd ed. New York: Holt, Rinehart & Winston, 1958.

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Atkinson, R. C. and Shiffrin, R. M., Human memory: a proposed system and its control processes. In K. W. Spence and J. T. Spence, The Psychology of Learning and Motivation: Advances in Research and Theory, Volume 2. New York: Academic Press, 1968. Bainbridge, L., Analysis of verbal protocols from a process control task. In E. Edwards and F. P. Lees, Eds., The Human Operator in Process Control. London: Taylor & Francis, 1974. Bainbridge, L., Beishon, J., Hemming, J. H., and Splaine, M. A., A study of real-time human decision-making using a plant simulator. In E. Edwards and F. P. Lees, Eds., The Human Operator in Process Control. London: Taylor & Francis, 1974. Bakan, D., A reconsideration of the problem of introspection. Psychological Bulletin, 1954, 51, 105–118. Bannedi, R. B., Theory of Problem Solving. New York: Elsevier, 1969. Bartlett, F., Thinking: An Experimental and Social Study. London: Allen & Unwin, 1958. Battig, W. F., Some factors affecting performance in a word formation problem. Journal of Experimental Psychology, 1957, 55, 96–103. Benjafield, J., Evidence that “thinking aloud” constitutes an externalization of inner speech. Psychonomic Science, 1969, 15, 83–84. Benjafield, J., Evidence for a two-process theory of problem solving. Psychonomic Science, 1971, 23, 397–399. Berlyne, D. E., Structure and Direction in Thinking. New York: John Wiley & Sons, 1965. Berner, E. S., Hamilton, L. A., and Best, W. R., A new approach to evaluating problem solving in medical students. Journal of Medical Education, 1974, 49, 666–672. Bhaskar, R. and Simon, H. A., Problem solving in semantically rich domains: an example from engineering thermodynamics. Cognitive Science, 1977, 1, 193–215. Bloom, B. S., Taxonomy of Educational Objectives. New York: Longman Green, 1956. Bourne, L. E., Jr., Ekstrand, B. R., and Dominowski, R. L., The Psychology of Thinking. Englewood Cliffs, NJ: Prentice-Hall, 1971. Bower, G. and Trabasso, T., Concept identification, in R. C. Atkinson, Ed., Studies in Mathematical Psychology. Stanford: Stanford University Press, 1964. Braithwaite, R. B., Scientific Explanation: A Study of the Function of Theory, Probability and Law in Science. Cambridge: Cambridge University Press, 1953. Bree, D. S., The distribution of problem solving times: an examination of the stages model. British Journal of Mathematical and Statistical Psychology. 1975a, 28, 177–200. Bree, D. S., Understanding of structured problem solutions. Instructional Science, 1975b, 3, 327–350. Broadbent, D. E., Perception and Communication. New York: Pergamon Press, 1958. Brunswik, E., Perception and Representative Design of Psychological Experimentation. Berkeley, CA: University of California Press, 1956. Burack, B., The nature and efficacy of methods of attack on reasoning problems. Psychological Monographs, 1950, 63 (313). Burke, R. J. and Maier, N. R. F., Attempts to predict success on an insight problem. Psychological Reports, 1965, 17, 303–310. Byers, J. L. and Davidson, R. E., The role of hypothesizing in the facilitation of concept attainment. Journal of Verbal Learning and Verbal Behavior, 1967, 6, 595–600. Castellan, N. J., Pisoni, D. B., and Potts, G. R., Cognitive Theory, Volume 2. Hillsdale, NJ: Lawrence Erlbaum Associates. 1977. Chi, M T. H., Glaser, R., and Rees, E., Expertise in problem solving. In R. J. Sternberg, Ed., Advances in the Psychology of Human Intelligence. Hillsdale, NJ: Lawrence Erlbaum Associates, 1981.

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De Groot, A., Perception and memory versus thought. Some old ideas and recent findings, in B. Kleinmuntz, Ed., Problem Solving: Research and Theory. New York: John Wiley & Sons, 1966. De Kleer, J., Multiple representation of knowledge in a mechanics problem solver. Proceedings of the 5th Joint International Conference on Artificial Intelligence. Cambridge, MA: MIT Press, 1977. Dewey, J., Logic: The Theory of Inquiry. New York: Holt, 1938. Dienes, Z. P. and Jeeves, M. A., Thinking in Structures. London: Hutchinson, 1965. Dodwell, P. C., Visual Pattern Recognition. New York: Holt, Rinehart & Winston, 1970. Dominowski, R. L., How do people discover concepts? In R. L. Solso, Ed., Theories of Cognitive Psychology: The Loyola Symposium. Hillsdale, NJ: Lawrence Erlbaum Associates, 1974. Dreistadt, R., The use of analogies and incubation in obtaining insights in creative problem solving. Journal of Psychology, 1969, 71, 159–175. Duncan, C. P., Recent research on human problem solving. Psychological Bulletin, 1959, 56, 397–429. Elstein, A. S., Shulman, L. S., and Sprafka, S. A., Medical Problem Solving: An Analysis of Clinical Reasoning. Cambridge, MA: Harvard University Press, 1978. Fleishman, E. A., Toward a taxonomy of human performance. American Psychologist, 1975, 30, 1127–1149. Fleishman, E. A., Systems for describing human tasks. American Psychologist, 1982, 37, 821–834. Fleishman, E. A. and Bartlett, C. J., Human abilities. Annual Review of Psychology, 1969, 20, 349–380. Forehand, G. A., Constructs and strategies for problem solving research. In B. Kleinmuntz, Problem Solving: Research, Methods and Theory. New York: John Wiley & Sons, 1966. Fowler, W. and Fowler, F. G., The Concise Oxford Dictionary of Current English, 6th edition. Edited by J. B. Sykes. Oxford: Clarendon Press, 1978. Gagne, R. M., Human problem solving: internal and external events. In B. Kleinmuntz, Ed., Problem Solving: Research, Method, and Theory. New York: John Wiley & Sons, 1966. Green, B. F., Current trends in problem solving. In B. Kleinmuntz, Ed., Problem Solving: Research, Method and Theory. New York: John Wiley & Sons, 1966. Greeno, J. G., The structure of memory and the process of solving problems. In R. Solso, Ed., Contemporary Issues in Cognitive Psychology: The Loyola Symposium. Washington: Winston, 1973. Greeno, J. G., Indefinite goals in well-structured problems. Psychological Review, 1976, 83, 479–491. Greeno, J. G., Process of understanding in problem solving. In N. J. Castellan, D. B. Pisom, and G. R. Potts, Eds., Cognitive Theory, Volume 2. Hillsdale, NJ: Lawrence Erlbaurn Associates, 1977. Greeno, J. G., A study of problem solving. In R. Glaser, Ed., Advances in Instructional Psychology, Volume 1. Hillsdale, NJ: Lawrence Erlbaum Associates, 1978a. Hafher, J., Influence of verbalization on problem solving. Psychological Reports, 1957, 3, 360. Hutchinson, E. D., How to Think Creatively. New York: Abingdon Cokesbury, 1949. Johnson, E. S., An information-processing model of one kind of problem solving. Psychological Monographs: General & Applied, 1964, 78 (581). Kaiser, H. F., Image analysis. In C. W. Harris, Ed., Problems in Measuring Change, Madison: University of Wisconsin Press, 1963. Kleinmuntz, B., Ed., Problem Solving: Research, Method and Theory. New York: John Wiley & Sons, 1966.

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Kleinmuntz, B., Ed., Formal Representation of Human Judgment. New York: John Wiley & Sons, 1968. Larkin, J. H. and Reif, F., Understanding and teaching problem solving in physics. European Journal of Science Education, 1979, 1, 191–203. Lorge, I. and Solomon, H., Two models of group behavior in the solution of Eureka-type problems. Psychometrika, 1955, 20, 139–148. Luchins, A. S., Mechanization in problem solving: the effect of Einstellung. Psychological Monographs, 1942, 54 (248). Maier, N. R. F., Reasoning in humans. I. On direction. Journal of Comparative Psychology, 1930, 10, 115–143. Maier, N. R. F., The behavior mechanisms concerned with problem solving. Psychological Review, 1940, 47, 43–58. Mandler, J. M. and Mandler, G., Thinking: From Association to Gestalt. New York: John Wiley & Sons, 1964. Marks, M. R., Problem solving as a function of the situation. Journal of Experimental Psychology, 1951, 47, 74–80. Maslow, A. H., Motivation and Personality. New York: Harper & Row, 1954. Maxwell, A. E., Analyzing Qualitative Data. London: Chapman & Hall, 1975. Mayer, R. E., Comprehension as affected by structure of problem representation. Memory and Cognition, 1976, 4, 249–255. Mayer, R. E. and Greeno, J. G., Effects of meaningfulness and organization on problem solving and compatibility judgments. Memory and Cognition, 1975, 3, 356–362. Miller, G. A., The magical number seven, plus or minus two. Psychological Review, 1956, 63, 81–97. Piaget, J., Six Psychological Studies. London: University of London Press, 1968. Polya, G., How to Solve It, 2nd ed. New York: Doubleday, 1957. Reed, S. K., Facilitation of problem solving. In N. J. Castellan, D. B. Pisom, and G. R. Potts, Eds., Cognitive Theory, Volume 2. Hillsdale, NJ: Lawrence Erlbaum Associates, 1977. Roth, B., The Effects of Overt Verbalization on Problem Solving. Unpublished Doctoral Dissertation. New York University, 1966. Saugstad, P. and Raaheim, K., Problem solving, past experience and availability of functions. British Journal of Psychology, 1960, 51, 97–104. Simon, H. A., The functional equivalence of problem solving skills. Cognitive Psychology, 1975, 7, 268–288. Staats, A. W. and Staats, C. K., Complex Human Behavior. New York: Holt, Rinehart & Winston, 1963. Vernon, M. D., The Psychology of Perception, 2nd Edition. Harmondsworth, Middlesex: Penguin, 1971.

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Overview of Key Elements to Problem Solving

In this chapter, the essential elements of problem solving will be addressed from strictly a quality perspective. The definition of a problem will be introduced. Then progressively some of the issues involved with problem solving in quality will be addressed.

THE ROAD TO CONTINUAL IMPROVEMENT Everyone likes to improve in some way, shape, or form. To improve, among other definitions, means to: • • • • • • • • • • • • • • • • • • •

Eliminate the waste of waiting Eliminate wasted effort (motion) Minimize inventory Eliminate overproduction Eliminate correction (repair) Minimize material movement Eliminate wasted processing Minimize uniqueness Eliminate overburden Utilize the best-known method Support the Just In Time (pull) system Simplify the process Optimize the system Improve flexibility Provide consistent direction Focus on the process Reduce variation Reduce lead time Improve uptime 33

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• • • • • • • • • • •

Maximize throughput Improve understanding Support the operator Support management by sight Reduce cost Improve quality Eliminate bottlenecks or constraints Support small-lot production Improve responsiveness Reduce set-up time Involve all impacted parties

The way we improve, however, is a function of many tasks and approaches. It is unfortunate that there is no silver bullet that will guarantee improvement on its own. We have to be cognizant of that and be able to glean from several approaches and methodologies the best we can under specific circumstances for the specific problem at hand. In six sigma methodology, just like in any other methodology, problem solving is an important activity for attaining continual process improvement. There are two approaches to this improvement. Both are evolutionary rather than revolutionary: Kaizen: Kaizen is a gradual, relentless, incremental process improvement over time. Breakthrough: Breakthrough is a sudden, significant improvement in process performance achieved in a very short time interval. Continual process improvement is best achieved by a combination of Kaizen and breakthrough and, ideally, an organization should employ and encourage both. To improve, something must be changed. However, often we want to change things that present themselves as problems — or at least we see them as problems. What is a problem? In the Introduction, several definitions of a “problem” were given. However, in this chapter focus on the definition will be in the following manner: • A problem is a symptom, not a cause or a solution. • A problem is a gap between what is happening and what we want to happen. • A problem is something that does not allow a goal to be achieved. • A problem could be a mistake, a defect, an error, a missed deadline, a lost opportunity, or an inefficient operation. • A problem may be thought of as an opportunity.

CHRONIC VS. SPORADIC PROBLEMS There are two kinds of problems, each presenting different opportunities:

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• A sporadic problem is a sudden, elevated departure from a historic, status-quo level. • A chronic problem is the difference between the historic level and the optimum or zero level; a chronic problem is a chronic “disease” which can economically be eliminated. Which problem is easier to solve — sporadic or chronic? Chronic problems are much easier to solve because of their repetitive nature. Sporadic problems, on the other hand, are very difficult to solve because they are not consistent and do not follow a pattern. In both cases, one has to understand the process and the conditions that existed when the problem was identified. For both sporadic and chronic problems, data must be generated to support the claim of their existence, and appropriate planning to resolve these problems must be taken. Most problem solving effort usually deals with chronic problems. Chronic problems repeat, cause variation, and cause the process to be inconsistent. This volume will focus on presenting the reader with a process to solve chronic problems.

THREE TYPICAL RESPONSES TO PROBLEMS AND THE ANTECEDENT Problems are generally uncomfortable situations, demanding some form of change from the status quo. Because of that demand, we often try to circumvent problems by: 1. Viewing them as burdens and learning to live with them 2. Ignoring them in the hope that they will disappear 3. Accepting them as challenges or opportunities to improve In the six sigma methodology, and in the spirit of continual improvement, obviously the first and second options are not options at all. The focus should always be in a positive direction, approaching the problem as an “opportunity.” To do this, the environment must be structured in a way that is conducive to problem solving analysis. A typical approach is to follow something like the following process. Form • • •

a team with Cross-functional members Multidisciplinary members Members who have ownership of the problem

Define • • • • •

the problem with A process flow diagram to define the process A Pareto analysis to select priority problems Control charts to indicate special causes Check sheets to define 5W2H An action plan to coordinate problem definition actions

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Implement interim (containment) actions with • Check sheets to evaluate effectiveness of actions • Control charts and histograms with intensive sampling for process monitoring • An action plan to coordinate interim fixes Define and verify root causes Identify potential causes with • Brainstorming to develop the potential causes • Cause-and-effect diagrams to identify and organize potential causes • FMEA to identify potential causes from observed failure mode Analyze potential causes with • A check sheet to collect data • Comparison plots, histograms, and stratified graphs to evaluate stratification factors or different process or product parameters • Scatter plots to evaluate relationships between characteristics • Gage studies to evaluate the measurement system • An action plan to manage analysis steps Validate root causes with • Comparison plots, histograms, and stratified graphs to validate cause (e.g., with/without comparison) • Stratified graphs to validate presence of root cause factors • An action plan to manage validation actions Identify alternate solutions with • Brainstorming to solicit ideas • An alternative solution cause-and-effect diagram to address potential solutions Verify the effectiveness of the solution with • Control charts and histograms to evaluate process stability and capability • Check sheets to collect product or process evaluation information • FMEA Implement ongoing controls with • Control charts and check sheets to monitor process performance • Comparison plots to periodically ensure that stratification factors are not influencing process output • Dimensional control plan Prevent recurrence with • A process flow diagram to define the management system that did not prevent the problem

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• An action plan to coordinate needed changes • FMEA

NINE COMMON ROADBLOCKS TO EFFECTIVE PROBLEM SOLVING We all strive to remain in an equilibrium. Whether in our personal lives, work environment, or some other situation, equilibrium is built into our human genetics. Why, then, do we fail to solve problems effectively so that we do not experience them again and again? Many theories for the answer exist; however, the following are perhaps the most important in the work environment: 1. Lack of time. We seem to be optimistic about the time that is needed to resolve the problem or we are assigned too many projects at the same time. As a consequence, we fail. 2. Lack of ownership. For some strange reason, time and again, the wrong people are assigned to the problem. As a consequence, they are trying to learn the process rather than fixing the problem. By the time they are ready to contribute to the solution, they are moved to a new position and the cycle begins all over again. 3. Lack of recognition. Appropriate individuals who do perform are not recognized. They are quite often the silent heroes who are overshadowed by individuals who have been identified as “super stars” or ones in the “fast track.” By not recognizing the appropriate individuals, the wrong message is sent to the rest of the employees that “good work” does not really count — only connections and pedigree count. 4. Error as a way of life. This is a “cop out” option that has tremendous ramifications in the work environment. This method of thinking perpetuates the status quo and is one of the biggest inhibitors to progress. If errors are accepted as a way of life, our competitors will surpass us and our customers will abandon us. 5. Ignorance of the importance of a problem. If the problem is not recognized in time, the organization has a hidden problem. The problem is greater if it is directly correlated with customer expectations and the organization does not recognize it. In the six sigma methodology, great effort is given to recognizing that Y (customer’s wants, needs, and expectations) is indeed correlated with f(x) (the items that “should” appease the customer). The importance of the problem is of paramount significance and no one can afford to look the other way or ignore it. 6. Belief that no one can do anything about some problems. It is true that some problems do not have an easy answer. In fact, some problems may not have a solution in the short run. However, that does not mean we are off the hook and are not responsible. If indeed there is no solution in sight, perhaps the problem is too big or has been assigned to the wrong

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team. In any case, try to find a way to circumvent these obstacles. After all, if we do not, some one else will! 7. Poor balance by upper management concerning cost, schedule, and quality. In the early 1980s, Deming identified with laser accuracy the problem with management, i.e., “quarterly earnings.” As long as the emphasis is on quarterly earnings, there will be no true improvement. Cost, schedule, and quality will always be shuffled. It is sad when management cannot see the trap that has been created. Sure, in the short run we will look great, but in the long run, we will bleed to death. An example will prove the point. A Fortune 100 company for the year 2000 had over $5 billion in warranty costs, yet their target production quotes were met. This same company claims to have the customer as the driving force in deciding quality. Need we say more? 8. People who try to protect themselves. It is human nature to protect our turf, but not at the expense of the organization. When it comes to problem solving, quite often most people mean well, but are afraid to tell the truth. Again, it was Deming who blew the whistle on this situation in his 14 “obligations of management” (specifically number eight: eliminate fear). Yet in this new millennium we continue to intimidate, harass, and blame individuals for system failures. As long as fear exists, employees will continue to cover themselves. By doing this, the real problems, the real concerns, and the real issues that should be reviewed and fixed are not uncovered. The following responses may indicate that someone is trying to cover himself: • • • • • • •

We have never done it that way. We are not ready for it yet. We are doing all right without it. We tried it once and it did not work. It costs too much. It is not our responsibility. It will not work anyway.

9. Head-hunting by management. It is interesting to see how management reacts to problem solving. Management believes that problem solving is indeed the “cure all” of all the difficulties in the organization. So, when a problem arises, management “picks” an individual with the most experience in solving problems and assigns him/her the new task. Management fails to recognize time and time again that the best return on money is to plan before a problem occurs. However, a good planner is the one who already knows the process. What a contradiction. By the time one learns the process, he/she is ready to move on to a new problem. Needless to say, problem solvers are highly regarded in our culture, whereas planners are considered a fifth wheel — a necessary overhead.

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SIX KEY INGREDIENTS REQUIRED TO CORRECT PROBLEMS How then can the inhibitors just given be avoided? To resolve problems, there must be commitment to these six ingredients: 1. Awareness of the importance of eliminating errors (waste) and the cost of errors to the business 2. Desire to eliminate errors 3. Training in proven, effective methods to solve and prevent problems 4. Failure analysis to identify and correct the real root causes of problems 5. Follow-up in tracking problems and action commitments 6. Recognition to give liberal credit to all who participate In addition, unless management plays an important role in the implementation process, nothing will happen. So, what must management do or provide? Minimum requirements are • A motivating environment conducive to effective, companywide problem solving • A real opportunity • A systematic approach (road map) • Knowledge of problem solving tools

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Problem-Solving and Process Improvement Cycles

This chapter opens with a summary of some key concepts in team dynamics as they relate to problem solving. The problem-solving cycle will be discussed because most of us recycle problems without really resolving them. The process improvement cycle and some of the issues involved will be emphasized, and a flow chart to show the continual improvement cycle will be introduced. As already mentioned in the literature review and the previous chapter, problem solving is a process that is repeated primarily because a problem was not fixed properly and/or completely the first time. To do that, a system must be in place that is accurate, repeatable, and user friendly. That system turns out to be a structured approach. Why should a structured problem solving approach be used? A structured, systematic, problem solving approach: • Illustrates the relative importance of problems • Shows the real root causes of problems • Helps problem solvers get “UNSTUCK,” know where they are, keep on TRACK and persevere to results • Helps keep problems solved • Establishes accountability and a motivating environment for problem solving • Produces consistently better solutions than unstructured approaches However, for this structured approach to work, a team familiar with both the problem solving process as well as the process improvement method must be in place.

WHY A TEAM APPROACH? • A goal of employee involvement (EI) is to give employees increased influence over and responsibility for their own work. In recent years, a 41

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•

•

• • • • • • •

major route to achieving this goal was creation of problem solving teams: groups of employees who meet on a regular basis to identify and solve work-related problems. Managers now realize that there is a vast, untapped potential in their employees’ minds. To remain competitive, it is essential to make effective use of all the talents that the labor force has to offer, both physical and mental. Successful companies worldwide credit much of their success to the widespread use of employee teams to generate improvements at all levels in all departments. Employees need to be provided with challenging activities to keep their active minds fully utilized. Most employees want to participate in decision making and problem solving activities that affect them. No one individual has all of the process and product knowledge plus the special skills and experience required for optimal problem solving. The most effective, proven problem solving and process improvement methods and tools best lend themselves to a team approach. A team will invariably generate more problems, more causes, and more solutions than an individual can. Everyone affected by a problem should be a part of the solution. This assures a “buy-in” to the corrective action or solution put into place. Teams have a distinct advantage over solo efforts: the mutual support that arises among team members contributes to TEAM as an acronym (TEAM = Together Everyone Accomplishes More through synergy and consensus). Quality improvement is hard work and takes a long time. It is easy for one person’s commitment and enthusiasm to weaken during a long project. The synergy that comes from people working together toward a common objective is usually enough to sustain enthusiasm and support. Ray Kroc (founder of McDonald restaurants) said, “All of us are better than any one of us.”

LOCAL TEAMS AND CROSS-FUNCTIONAL TEAMS • Local or department improvement teams (DITs) are comprised of all members of a department. These employees typically work in close proximity, experience common problems, and form a “natural work group.” Their purpose is to provide a focus and a means for all employees to contribute to an ongoing activity aimed at improving the quality and productivity levels of the department. • Cross-functional or process improvement teams (PITs) are created to continuously improve quality, reduce waste, and improve productivity of a process that crosses many departmental lines. The PIT is made up of experienced, skilled problem solvers from all departments involved in and affected by the process.

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GENERAL GUIDELINES FOR EFFECTIVE TEAM PROBLEM SOLVING AND PROCESS IMPROVEMENT Requirements for a team to function successfully are • A team charter which specifies the team’s purpose and enumerates the team’s duties and responsibilities • Selection of the proper team makeup • Selection of an effective team leader and team coordinator • Knowledge of the organization’s mission, goals, and objectives • Learning to work together as a team (“team building”) • Adequate training in problem solving and process improvement methods and tools • Guidelines and ground rules for holding effective meetings and making decisions • A suitable meeting place with support services, such as typing, copying, information resources, etc. • Use of meeting agendas and minutes (recaps) • Liberal credit and recognition for team successes

WHAT MAKES A TEAM WORK For a team to work, the following attributes must be present: • • • • • • • • • •

A climate of trust A commitment to common goals Honest and open communication Common agreement on high expectations for the team Assumed responsibility for the work that must be done A general feeling that one can influence what happens Common access to information A win/win approach to conflict management Support for decisions that are made A focus on process as well as results

EXAMPLES OF PROBLEM-SOLVING MODELS In the introduction, a somewhat extensive literature review on the topic of problem solving was given. This section will focus on specific applications of problem solving processes that major companies have undertaken with measurable success. Successful, continuously improving companies around the world use structured problem solving systems, specifically: • XEROX (6 steps) • GM (4 steps)

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• DaimlerChrysler (7 steps) • Department of Defense (DoD) (6 steps) • Ford (8 steps) Regardless of what system is used, the foundation of a system lies in the basic scientific method or a variation of it. For our purposes, a very basic and effective way, approaching the problem solving process through a five-step problem solving (opportunity) cycle (PSC), will be used. The five distinct steps are • • • • •

Select the problem. Find the root cause. Correct the problem. Verify the correction. Prevent recurrence … then recycle.

This process is easy to understand and remember. Now look at the implementation process. • Each step should be fully completed before beginning the next step. A step should not be considered to be complete until the final sub-step is carried out. (Detail provided in Steps 1 through 5.) • There may be several iteration or feedback loops that must be properly followed to complete the cycle. Step 1. Select the problem (or opportunity) that represents a waste or bottleneck in the area. • Prepare a problem list. • Collect data to determine the magnitude of each problem. • Prioritize problems to determine which ones should be addressed first. Almost always a “vital few” problems account for the greatest proportion of losses. • Select the target problem. This is generally the problem whose solution will have the greatest beneficial impact. • Prepare a specific problem statement (e.g., reduce shipping errors from 3.1% to under 0.5% by the end of the third quarter). The problem statement represents the specific problem assignment or project commitment. Step 2. Find the root cause of the problem. This analysis step is usually the crucial work of the project. Problems must be solved at the root cause level if they are to remain completely and permanently corrected. • Identify all possible causes of the problem. • If necessary, collect data to ascertain the failure mechanism. • Select the most likely cause.

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Step 3. Correct the problem, following a plan that will eliminate the problem or at least reduce it to a level compatible with the team’s goal. • Determine a temporary fix, if appropriate, to protect customers until a permanent fix can be implemented. • Develop alternative solutions. • Select the best possible solution by narrowing down the alternatives using a priority-setting approach and making a final consensus decision. • Develop an implementation or action plan which includes a time schedule for implementation. Everyone who is involved in carrying out the solution should be consulted and should approve the plan. The plan should answer the basic questions: what? who? where? when? and how? • Establish the method for measuring the success of the proposed solution. Remember, the criterion for success was established as part of the problem statement in Step 1. • Gain management approval, if necessary, for implementation. • Implement the action plan. Step 4. Verify correction, using the criterion for success established in Step 1 and the measuring method established in Step 3. • Measure the impact of the solution. • Make a decision: Has the problem been solved to a level compatible with the team’s goal? Have any unforeseen new problems been created by the solution? • If the correction is unsatisfactory, the team must go back to an earlier step, selecting an alternative solution or possibly even selecting an alternative root cause and then proceeding. • If the correction is satisfactory, any temporary protective fix should be removed. Step 5. Prevent recurrence of the problem. This is an often overlooked aspect of problem solving. The benefits of preventing recurrence cannot be overstated. • Develop innovative methods for mistake-proofing the system (i.e., Japanese “Poka-Yoke”). • Alter systems and procedures to reflect the new, optimal practices. • Train all involved personnel in the new methods. • Publicize the results. Apply the knowledge gained to the rest of the product line and to other company activities with similar conditions. • Continue to monitor the problem. • Recognize the team’s success. • Prepare a final project report that describes the problem, the methods used to correct it, and the quality-productivity gains achieved. Also, a final team report should be made to management.

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1. SELECT THE PROBLEM ⇒ Prepare problem list ⇒ Collect data ⇒ Prioritize problems ⇒ Select target problem ⇒ Prepare problem statement 2. ROOT-CAUSE ANALYSIS ⇒ Identify possible root causes ⇒ Collect data ⇒ Select most likely cause 3. CORRECT THE PROBLEM ⇒ Make temporary fix ⇒ Develop alternative solutions ⇒ Select best solution ⇒ Develop implementation plan ⇒ Establish measuring method ⇒ Gain management approval ⇒ Implement solution

RECYCLE

4. VERIFY CORRECTION ⇒ Measure impact of solution ⇒ Decision: solution effective? ⇒ If necessary, loop back ⇒ Remove temporary fix 5. PREVENT RECURRENCE ⇒ Mistake proof system ⇒ Alter systems and procedures ⇒ Train involved personnel ⇒ Publicize results ⇒ Continue to monitor problem ⇒ Recognize team’s success ⇒ Prepare final project report

FIGURE 3.1 A pictorial view of the super generic five-step model.

• Recycle. Once the five steps have been completed, the team should return to Step 1, select another significant problem for correction, and continue improvement efforts. A pictorial view of the super generic five-step model is illustrated in Figure 3.1.

THE PROCESS IMPROVEMENT CYCLE When an organization undertakes six sigma methodology as an organizational initiative, one reason is to improve and/or change the status quo, showing significant gains for the organization and the customer. To do that, focus must not only be on specific tools, but also on the philosophy and main elements that will result in “real”

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improvement. Some basic tools and their application have been described. Now it is time to review the elements that will help in this initiative to improve the organization’s effectiveness and customer satisfaction. The elements are really selfexplanatory and do not need further analysis. However, they are critical for the success of the six sigma initiative. The basic elements are • • • • • • • • •

Process behavior or voice of the process or process control (vs. parts control) Quality at the source Factual and analytical decision making Decision making at the lowest appropriate level (presupposes empowerment) Focused, organizational teamwork Clear performance standards Suppliers developed as extensions of the production system Focus on prevention rather than detection Maximum value for all customers (through minimizing variability, minimizing waste, and controlling target)

Once these elements have been internalized, improvement may be pursued. However, this improvement must be structured and systematically approached. Why? Because in addition to what has already been said, this structured, systematic, process improvement approach: • Provides a more tightly focused process than the problem solving cycle (PSC), concentrating on improving the quality of a single process output (product or service) • Lends itself particularly well to customer issues, since it builds and strengthens ties between customer and supplier • Provides a catalyst for never-ending improvement through continuous refocusing on customer needs and expectations • Affords a prime opportunity to utilize statistical process control (SPC) methods for process analysis, control, and improvement • Focuses improvement efforts on product quality and waste reduction • Provides an enhanced motivational environment to improvement teams for their improvement efforts To optimize this process in conjunction with the problem solving cycle, a 14-step process improvement cycle (PIC, Figure 3.2), is recommended: Step 1. Identify the output. • Prepare a statement which answers the question “What is to be done?” and consists of two parts: a tangible, specific noun followed by a verb describing what to do to produce the output. • The statement should be neither too broad nor too specific, if it is to be useful (e.g., Shaft Machined, Doughnut Fried, Muffler Installed, Bicycle

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FIGURE 3.2 The 14-step process improvement cycle model.

Assembled, Forecast Developed, Order Taken, Machine Repaired, Order Shipped, Report Typed, etc.). Step 2. Identify the customer. • The (primary) customer is the next person in the work process — the person who will receive your output as input and act on it. • The customer may be either external or internal. • Secondary customers and end-users may be identified if their requirements are of significance.

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Step 3. Identify the customer requirements. • What does the customer want, need, or expect from the output? • Customer requirements may be general or very specific. • Customer requirements often fall into categories such as: cost, accuracy, timeliness, quantity, completeness, dimension, performance, appearance, etc. • Supplier/customer interaction is essential. Customer and supplier specify, negotiate, and evaluate information before reaching an agreement, e.g., Report needed ASAP; Expense cannot exceed the budget; Heat shield must withstand operating temperatures; Colors must match; Quick response time; etc. Step 4. Translate requirements into product specifications. • Put the customer’s needs into language that reflects the supplier’s active participation. • Specifications should be measurable, realistic, and achievable and should be reviewed with the customer to be certain both parties understand and agree about what the output should be, e.g., needed by 4:00 p.m. Friday; O.D. 1.625 ± .005 in.; Weight 5200 lb maximum; Type double-spaced, etc. Step 5. Identify the work process. • What are the process elements? A fishbone diagram used as a process analysis map (PAM) may help identify the equipment, people, input, methods, and environmental factors in the work process. • List step-by-step what must be done to achieve the output. Flow charts are especially helpful. • Identify or develop the needed documentation to describe or provide instructions for the work process. Step 6. Select the process measurements. • Select the critical measurements for process output. These measurements should be derived from customer requirements and product specifications. They should permit an objective evaluation of the quality of output. These measurements should provide for early identification of potential as well as actual problems, with stress on prevention rather than on detection of errors or defects, e.g., cylinder O.D. in inches; cake weight in grams; transit time in days; response time in seconds; number cycles before failure; circuit resistance in ohms; etc. • Also select measurements for key process inputs, e.g., raw materials, incoming parts, energy, supplies, information, etc., in-process factors (intermediate process outputs), e.g., in-process conditions that impact on the process output, such as temperature, pressure, time, etc.; process

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performance criteria, such as scrap, rework, downtime, yield, delays, efficiency, etc. These measurements become the basis for process analysis, control, and improvement. Step 7. Determine the process capability. • Can the existing process produce output which consistently meets the contractural product specifications with minimal waste? • The process is operated according to standard practices and output measurement data are collected and analyzed to provide a statistical evaluation of process control. • Is there consistency and process capability (ability to meet specifications)? • This step is an important activity of statistical process control (SPC). Step 8. Decide: Is the process capable? • If the evaluation indicates that specifications are not being met consistently and there is excessive waste (scrap, rework, delays, errors, etc.), the work process must be revamped or improved and problem solving is required. Proceed to Step 9. • If the process capability assessment of Step 7 determines that the existing process can produce output which consistently meets specification requirements and does so with minimal waste, proceed to Step 10. Step 9. Problem solve to improve the work process. • Use the 5-step problem solving cycle (PSC) to revamp or improve the existing process. • If the process capability problem indicated in Step 7 is a lack of process control (instability or inconsistency), indicated by sporadic deviations of the process average from the target, status quo, or historic level, the solution is one of identifying the destabilizing (special/assignable) causes and eliminating them. After “special-cause” problem solving is completed, loop back to Steps 7 and 8 (reassess process control). If the problem is solved, proceed to Step 10; if not, continue to repeat the PSC at Step 9, as needed, until process control is achieved. • If the process capability problem indicated in Step 7 is an inherent inability to meet specification requirements, the process system must be fundamentally changed. This requires identifying those key process factors (common or random causes) which have a major impact on the quality of process output and modifying them and their effects. After “common-cause” problem solving is completed, loop back to Step 5 to redefine the revised work process and proceed. If at Steps 7 and 8 the process capability problem is solved, proceed to Step 10; if not, continue to repeat the PSC at Step 9, as needed, until process capability is achieved.

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• If repeated attempts still result in an incapable process, contact the customer who needs to know what to expect. A renegotiation of requirements may be possible. If not, the customer may want to find another supplier. Step 10. Produce the output, continuing to follow standard practices (or the newly established practices which should be standardized for the new, revamped process). Step 11. Evaluate the results. • Evaluation of results must be based on product specifications that the customer and supplier agreed to as part of Step 4. Those specifications are a “template” against which results are compared. • There is a major difference between this step and Step 7 (Determine process capability). This step evaluates how well you actually did (not how well you are capable of doing). The emphasis is on results rather than on process. • There are two potential types of problems at this point: (1) the product output does not meet specification requirements and (2) the product output meets specification requirements, but the customer is dissatisfied with the product. Step 12. Decide: Is there a problem? • If evaluation of Step 11 indicates there is a problem, either type 1 or 2, problem solving must be carried out. Proceed to Step 13. • If no problem exists, continue production and proceed to Step 14. Step 13. Problem solve to improve results. • Use the 5-step problem solving cycle (PSC) to “troubleshoot” the problem. • If the problem with process output is type 1, i.e., the process output does not meet specification requirements, the cause of the problem is one of nonconformance to standard practices (since the process was shown to be capable in Step 7). The solution to this type of problem lies in identifying the nonconforming practice(s) and restoring operations according to established standard practices at Step 10. • If the problem with process output is type 2, i.e., the product output meets specification requirements, but the customer is dissatisfied with the product, the cause of the problem is usually one of (1) improper understanding of customer requirements; (2) improper translation of requirements into specifications; or (3) improper process monitoring (measurement). Any one of these causes requires looping back to Steps 3, 4, or 6 for corrective action. Step 14. Recycle. • At this point, there should be no evident problems in process capability or results. However, there may be opportunities for further quality

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improvement of the process output, e.g., in terms of reduced variability. This could result in not merely meeting, but in exceeding customer expectations. • Customer needs and expectations are rarely static. They are likely to change, especially in terms of becoming more restrictive. These opportunities for continuous process improvement require recycling to the PIC. • Under any circumstances, continue to monitor the work process and the results in order to maintain the required level of quality.

PROBLEM SOLVING VS. PROCESS IMPROVEMENT At this juncture, be sure there is understanding, that there are major differences between problem solving (PS) and process improvement (PI). A cursory comparison follows: • The 5-step PSC and the 14-step PIC are proven, systematic approaches for continual improvement. Both are extremely well-suited for team-oriented activity. • Which systematic methodology should be used? (See Table 3.1.) In general, use PIC to improve the quality of a particular, currently existing output or to produce new output. • Use PSC when there is a gap between what is currently happening and what should have happened. • PSC is an integral and vital part of PIC (at Steps 9 and 13). • It is not always obvious which of the two methods should be used to tackle a particular work issue. Skill and experience are required to make a better selection. The PSC/PIC selection chart (Table 3.1) compares the two methods and provides information to make the decision. • Fortunately, no matter which method is selected, it will quickly become apparent whether or not the method is helping accomplish the objective.

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TABLE 3.1 Problem Solving vs. Process Improvement Selection Chart Problem Solving Cycle (PSC) A general method for making a change in: • Systems • Results • Conditions • Work process • Management process This method fosters: • Definition of problems • Analysis of data • Understanding of causes • Creative ideas • More alternatives • Teamwork • Commitment Use this method when: • There is a gap between what is happening and what should be happening • Moving from vague dissatisfaction to a solvable, clearly defined problem • Unsure of how to approach an issue Switch to PIC when: • The problem identified is a lack of quality or an inability to assess quality • The recommended solution involves producing a specific output

Process Improvement Cycle (PIC) A tightly focused method for ensuring conformance of: • A specific product • A specific service

This • • • •

method fosters: Elimination of unneeded work Prevention of problems Shared responsibility Strong customer/supplier communication lines • Evaluation of work processes • Confidence in results Use this method when: • Improving the quality of a particular, currently existing output • There is no agreed-upon customer requirements for an output • Producing new output, the need for which has recently been determined Switch to PSC when: • Evaluation of process capability shows that the current work process cannot produce conforming output • Evaluation of results shows that the work process did not produce quality output

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4

The Quality Tools

This chapter introduces some basic tools that are used in problem solving. An overview is given, recognizing that some of these tools will be fully developed in Volumes III and IV and some additional, advanced tools will be introduced in Volumes V and VI.

TOOLS FOR PROBLEM SOLVING Teams need to learn and apply a number of quality tools, i.e., analytical techniques, to aid them in working through the five steps of the problem solving cycle. These tools are primarily used for collecting, analyzing, and understanding data or information. There are other tools that also help in the problem solving process. Not all of them are science based, but they do provide a very important input in the process. Table 4.1 summarizes some of these. The core capabilities shown in Table 4.1 are only samples. With other capabilities, the importance may vary for each discipline. In Table 4.1 one may conclude that business practices have the most influence on the program and as such appropriate tools should be used to evaluate the results for the project. The table is to be used in analysis of the tools early in the process to demonstrate the need of improvement in isolated cases. With this information, one can determine which capabilities need to be acquired or developed to become more effective in each discipline.

QUALITY TOOLS INVENTORY There are numerous quality tools that may be applied to problem solving. Some of these are very complex, advanced mathematical techniques, while others are quite simple to learn and apply. Table 4.2 displays 38 of the most commonly used tools. For each tool listed, each step of the problem solving cycle (PSC) in which that tool can be utilized is indicated. For example, the check sheet may be used in Step 1, Problem selection; Step 2, Root cause analysis; and Step 4 , Verification of correction.

THE SEVEN BASIC TOOLS • Tools indicated by * in Table 4.2 are widely considered to be one of the seven basic tools: 1. Flow chart 2. Check sheet 55

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TABLE 4.1 Most Likely Influences of Program Development Practices on Sample Core Capabilities The Nine Disciplines

Direction

Differentiation

Discovery

Diagnosis

Design

Development

Delivery

Practices Core Capabilities Analysis Phase Strategic visioning Mission development Purpose definition Product alignment and integrity Competency identification Pricing strategy Quality implementation Packaging concept Brand identity Market strategy

Art

Science

x x x

x x x x

x

x x

x x

x

x

Task or job analysis Audience analysis Competitive analysis Market analysis Customer feedback

x x x x

Conception Phase Investment planning Portfolio analysis Costing modeling Life-cycle analysis Strategic marketing

x x x

x x x x x

Issue or argument definition Content synthesis Adult-learning theory Instructional systems design Predesign and design

x x x

Print and video writing Editing Tailoring or customizing Project management Cultural adaptation

x x x

Implementation Phase Graphics and layout Electronic publishing Software or video production Alpha and beta testing Inventory and product management

Business

x

x

x x x x

x

x x x

x x x

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TABLE 4.1 (Continued) Most Likely Influences of Program Development Practices on Sample Core Capabilities The Nine Disciplines Documentation

Deduction

3. 4. 5. 6. 7.

Practices Core Capabilities Record-keeping File management Total quality management Feedback systems Organizational memory Program evaluation Cost/benefit analysis Research and analysis Continuous improvement Internal return on investment

Art

Science

x x x x x x

Business x x x x

x

x x

Pareto chart Cause-and-effect diagram (fishbone) Histogram Scatter diagram Control chart

• These tools can be easily learned and used by everyone in the organization; are very effective in achieving basic problem solving success; and are essential to any properly designed improvement strategy. • Basic instructions describing the tools and their use may be found in Volumes III and IV (as well as any SPC book and/or basic statistics book or in The Memory Jogger).

USING THE SEVEN BASIC TOOLS IN PROBLEM SOLVING Baking cookies is a rigid, step-by-step process. Similarly, the problem solving cycle (PSC) specifies a disciplined step-by-step approach for achieving the desired result. Each step of a cookie recipe specifies exactly which ingredients, utensils, and equipment are to be used and in precisely which order. However, the analogy between baking cookies and solving problems breaks down. In each step of PSC, the team must decide which “ingredients” (data and information) and which “utensils” (quality tools) are the most appropriate and effective to use. A team of carpenters, building a house, following a blueprint, and selecting and using the right tools at the right times, is a much better analogy to a team following PSC. For a team to become proficient at tool selection and use, considerable practice and experience are required. Patience and perseverance become important virtues — they will pay off!

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TABLE 4.2 Typical Quality Tools Problem Solving Cycle Step

Quality Tools Inventory ANOVA Audit Brainstorming Cause-and-effect diagram Check sheet Control Charta Customer complaint log and tracking Customer feedback survey Design of experiments Evolutionary operation (EVOP) Failure mode and effects analysis (FMEA) Failure analysis Flow chart Force field analysis Goal setting Group vote Histogram Key characteristic selection Measurement system evaluation (MSE) Mind map Mistake proofing Nominal group technique Pareto chart Pilot run Process capability study Process log/events log Process potential study Process standardization and/or Qualification cost of quality analysis Quality function deployment (QFD) Regression analysis Response surface methodology (RSM) Scatter diagram Statistical process control (SPC) Stratification Taguchi methods Team meeting guidelines Test of hypothesis

Root Problem Cause Prevent Basic Selection Analysis Correction Verification Recurrence Seven (1) (2) (3) (4) (5) Data

* * *

X X X X X X X X

X

X X X X X X

X X X X X

X X X

X

*

X X X X X

* X

A A A A VA V A V

X X

X X X X

X X

X X *

X X

X X

X X X

X X X

X X X X X X

X X

X

X X

X X X

* X

X

X X X X X X

X

X

X

X

X

X

Note: V, variable data; A, attribute data; *, one of the seven basic tools. a

X X X

Control charts (variable, attribute, run, multivariable, box plots, pre-control, etc.).

V A A A A VA A A VA V A

A — A A — VA A VA — V — V V V VA VA V — VA

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SEVEN QUALITY CONTROL MANAGEMENT TOOLS 1. 2. 3. 4. 5. 6. 7.

Affinity diagram Relationship diagram Tree diagram Matrix chart Matrix data analysis chart Arrow diagram Process decision program chart

Use of the basic and management tools in the problem solving process is to maximize the solution of the problem. This is done by following a prescribed methodology in the particular tool of choice so that a plan can be developed to deal with the fundamental questions and concerns of the five-step approach. When the right tool is chosen, there will be appropriate answers to these questions: 1. 2. 3. 4. 5.

What is the problem? What can be done about it? Can a “star” be put on the best plan? Can the plan be carried out? Did the plan work?

As important as these tools are, the underlying assumption is that management supports the effort for success. The following items are most important in the problem solving process and unless they are at least considered, failure is on the horizon. They are called factors for success: • • • • • • • • • • •

Management commitment Employee involvement (ownership) Cooperative worker/management relationship (not adversarial) Something in it for the people Time, energy, and determination Appropriate communication (two-way) Applicable training Ability and freedom to pinpoint “real” problems and identify “real” causes Appropriate team for “the problem” Appropriate (reachable, realistic) goals and measurable performance Appropriate and applicable frequent recognition, as well as feedback

SELECTED BIBLIOGRAPHY Brassard, M. and Ritter, D., The Memory Jogger II. Salem, NH: GOAL/QPC, 1994. Ishikawa, K., Guide to Quality Control. White Plains, NY: Asian Productivity Organization (UNIPUB), 1982.

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5

The Global Problem Solving Process

This chapter introduces a version of methodology used in the automotive industry, specifically at Ford Motor Company, known as 8D methodology. We have chosen to call it the global problem solving (GPS) process, since its application may be used in any industry. GPS is a very simple methodology, but it demands an appropriate amount of time to identify and recommend a solution. However, GPS is very effective when used properly.

GENERAL OVERVIEW When there is a major problem or the response to a particular problem may require significant time to reach a full solution, then serious consideration of a structured approach to the problem should be considered. Consider a structured approach because: • • • • •

It It It It It

provides a systematic approach. provides a common methodology. is a fact-based, scientific approach. uses the team approach. builds up a database to understand and solve future, similar problems.

Perhaps one of the most innovative approaches to problem solving is the global problem solving (GPS) process. It is modeled after the approach used by Ford Motor Company and, indeed, GPS has become the standard for analysis worldwide, at least in manufacturing. It is a powerful process that has nine stages. This section summarizes the process; however, the reader is encouraged to review Appendices B through J for general guidelines about questions that are necessary for the optimum solution. (The questions are obviously only thought starters and do not present an exhaustive approach.) GPS begins with questioning the process itself to determine whether or not there actually is a legitimate problem. Then it considers possible emergency action. The problem at hand, more often than not, is called a “defect” (today called a

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nonconformance item). A “defect” is a problem, a concern, an opportunity — an opportunity to improve. From the point of a “defect,” serious consideration of the process evaluates each step in a very formal way, so that a resolution can be reached. In the following paragraphs, the formal approach is summarized as a whole and then each step is highlighted with specific considerations. Problem description: In this step, the team (1) describes the problem and its symptoms, (2) presents the rationale for selecting a particular concern, and (3) sets the baseline against which evaluations of actions will be measured. It also summarizes customer-related and in-plant indicators of the magnitude of the problem and refers to Pareto charts of TGW (things gone wrong), warranty, and/or in-plant indicators to provide a perspective on the importance of the area relative to other areas which might have been chosen for improvement action. The idea is to demonstrate with data-based evidence that the project is of sufficient importance to merit devoting valuable engineering resources to it. For purposes of control, customer-oriented measures should be related to inplant indicators. These include SPC charts on selected characteristics, daily evaluation samples (sometimes called daily audits), and first run capability. Efforts should be made to isolate unique contributing factors, such as charting data by shift to isolate shift-to-shift differences. Attention should be given to precise agreed-upon operational definitions, particularly for “judgment” items, and care should be taken to ensure that the measurement system is adequate. In addition, data taken at different points in the production process may be best suited to answering different questions. For example, data taken on paint scratches would measure different things depending on whether data is taken before or after the repair booth. (If taken before the repair booth, data can measure the performance of the paint operation; if taken after the repair booth, data measures the paint and repair system.) When using summary statistics, one should generally indicate the size and time frame and precise source of the sample, e.g., “first run capability for shift #2 is 83% based on 500 pieces during week of April 8” or “#4 overall Pareto rank based on second quarter data.” Despite small sample sizes, certain other measures of the concern can also be used to indicate that there is an area that should be addressed, e.g., audit results, internal and external complains, etc. These data, however, should generally not be used for conclusive verification because their small sample size prevents them from being an adequate tool to estimate change when the event of interest has a small rate of occurrence. Keep in mind that this step summarized the evidence which establishes the baseline against which future progress due to actions taken will be measured. Specify potential root causes: The purpose of this step is to summarize the efforts of the group to understand the problem described in the previous step by asking the question why?, why?, why? until it cannot be asked again. To aid in this process, highly recommended documents are flowcharts (present — what is done now — and ideal — how the process would work if it were perfect) and cause and effect diagrams (also called fishbone or Ishakawa diagrams). In addition, Pareto charts to analyze the sources of the problem and additional run and SPC charts to determine

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the magnitude of contributing factors are most helpful. For example, if the primary component of poor paint quality was found to be dirt in the paint (from a Pareto analysis of the problem description stage), this step would indicate the group’s efforts to determine exactly where the dirt was appearing on the item and where in the process dirt might enter the production system so that it would end up on at a particular part. In other words, why was dirt in the paint? At this point, problem resolution sub-teams might be formed. A page of the GPS report should indicate the core and resource person membership, their goal, the initial level of the quality indicators, and the required resolution documents. After the potential root cause has been identified, a short summary of the actions is addressed. The action quite often is in the form of: • Interim action to stop the concern from getting out of the source • Permanent action to stop the concern from recurring • Specific service action to correct any concerns that may have gone to the customer, if required (Note: Special attention should be given to “service needs.”) Be as specific as possible in this step. Do not write “hole moved,” but instead write “drain hole moved 1/4 in. to right.” Be sure to indicate deviation and specific (whenever possible) drawing and service bulletin numbers. Include anticipated and actual completion dates. A corrective action chronology (CAC) is a useful method to relate symptoms, root causes, actions, and verification. Verification: This is the step in which GPS write-ups have been found to be most often lacking and the area where management is placing increasing emphasis. Evidence of effective action is required for both interim and the permanent action, but not service action (discussed on an exception basis only). Persons who are deeply involved in the solution of a problem are sometimes too close to be objective about what constitutes good evidence that a problem has been adequately addressed. There are three elements of the adequacy of evidence: • Common sense and consideration of natural variability of the process • Cause and effect (where and when to take samples) • Statistical theory (how many samples and what inferences should be made from incomplete data) Additionally, in this step the team must make clear if it is attempting to demonstrate: • “Snapshot” verification and/or experimentation • Ongoing control at a new level recognizing that both are frequently required for interim and permanent actions. Of the three elements of data-based evidence, common sense and the consideration of natural variation are the most important. For example, a baseball general manager who had decided to send a player to the minors for poor batting would not

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change his mind and offer the player a $2 million a year contract just because the player had three hits for four at bats in his most recent game. This could happen by chance even if the player had not improved at all. (Likewise the GPS team should put itself in the place of a suspicious external observer when determining “if” evidence passes the test of common sense.) Next to common sense, considerations of cause and effect are most important in determining what data to take to verify that an action is effective. In other words, when a desirable result is obtained from a system, data gathering must be planned in a way to be reasonably sure that the result was caused by the change made rather than a change in some other input in the system. A hypothetical example is useful in illustrating the importance of cause and effect: consider a decision to improve door closing efforts by a change in weatherstripping cross section. Sample parts at the “now” design level result in an improvement. The new design is adopted, but customer complaints do not decrease. The reason is that the sample parts were made with a batch of raw material compound with a rubber formulation at the extreme end of the specification. It was the difference in chemistry that made the difference in closing efforts rather than the change in design. It is very difficult to be reasonably sure that improvement seen in the trial was due to the design change rather than the formulation change (or many other conceivable differences between the sample parts and the regular production parts). One approach to prevent other factors from influencing results is to “block” the effect of these other change in the test plan (perhaps using the design of experiments approach). Additionally, more confirmation generally should be obtained. This is accomplished by monitoring the process over time and by taking steps to keep the system in statistical control at the new level. When the process is monitored over a period of time, the confidence level of members of the group that the design/process change has had the desired effect is increased. This is why ongoing verification is part of the GPS process for both interim and permanent actions. The questions of determining which actions to select for both interim or permanent solutions should be answered with the help of data whenever appropriate. Ideally this will be done with design of experiments (DOE), a rigorous methodology (see Volume V) based on statistical theory which helps the engineer understand the impact of changing a number of components simultaneously in a specific organized manner. With DOE, the engineer can often obtain the information sought by running only a fraction of the tests which would need to be run to obtain the same information if one-factor-at-a-time testing were used. In addition, using Taguchi’s concept of parameter design (robustness), the engineer can not only look at design and processing parameters which can be set, but he can also gain insight into how “robust” design alternatives are. A “robust” design is one which will consistently function close to target intent despite a wide range of uncontrollable factors — called noises (e.g., temperature, customer use, etc.). An example would be to design an installation procedure which is “robust” against worker fatigue, i.e., a new part which is (almost) as easy to install the first time as after doing it 400 times a day. There are a variety of data analysis methods to help achieve this result, including simple graphical techniques, separate analysis of averages and variance signal to noise ratios, and

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computerized nonlinear optimization techniques. Each approach has advantages and disadvantages. When DOE is not used, “snapshot” testing of a proposed action should generally be employed before putting a proposed design/process change into production. A common use of “snapshot” testing is a go/no go type of test (e.g., does a revised weatherstrip prevent water leaks better than the present design). The procedure, which requires a large amount of data, uses a chart which indicates the demonstrated estimated reliability at 85% confidence for a given sample size (total number of revised weatherstrips tested) and the number of parts which have leaks. This estimate is compared to the reliability estimate for the present design. Another form of “snapshot” testing is to look at measurement data for the new design (e.g., door velocity) and use statistical tables to compute confidence intervals for the difference in velocity between the two designs. This assists the engineer to decide if the difference in performance is likely to be due to the difference in designs or to random noise (chance). Note: The use of statistical techniques for “snapshot” verification assumes that common sense has been used and that proper consideration of the cause and effect influence has been made. In short, statistics can enhance, but never replace, engineering judgment (more about this in Volume III). Statistical theory is also important in determining subgroup size and frequency for SPC charts, which can be very useful in both analyzing and verifying the problem in a process (more about this in Volume IV). Prevention: List the changes that will prevent recurrence of this and similar concerns in the future, e.g., use FMEAs, changes in supplier certification, and specific actions to ensure engineering/manufacturing/supplier communications. These actions frequently require management action, sometimes upper management action, to produce the changes in the system required to prevent recurrence of the concern on any part of any product in any plant. Recognize your team: Mention when and from whom the team received timely, honest feedback on the impact of their solution. Frequently, it is also useful to evaluate how the group problem solving process worked and how the process might be improved. Additionally, some teams celebrate success with a group event such as a luncheon or attending a baseball game.

DO’S AND DO NOT’S In a classic sense, GPS is a process. However, each of the individual steps that make up this process is the “real” work. So, success as an output variable of the GPS process depends on the effort put into the individual steps. To be sure, GPS is a process that facilitates problem solution, but it is much more than that. It is also a document that others (managers and engineers) may use for different needs. Therefore, the following “Do” and “Do Not” lists are reminders:

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DO’S • Write sentences that are easily understood by nontechnical individuals. Presume no prior knowledge. Spell out acronyms, e.g., TGW (things gone wrong). • Ensure that it is clear which product, part, concern, root cause, or action has been selected. Use item numbers (A, B, C), buzzwords, or separate write-ups when multiple root causes are involved. • Provide enough information for an engineer to understand the thought process and solutions. Provide the engineer with information and thought starters for similar problems that might be encountered in the future. • Leave tracks so later (2 years from now) another engineer would be able to obtain more information: reference specific part numbers, deviations, test reports, service bulletins, and plant dates. • Distinguish fact from opinion. • If the concern has a prior history, indicate and explain. • Include timing risk estimate classification (TREC) if applicable. • Treat the GPS report as a living document. • Realize that only data collected over the long run can ensure that the solution is “robust.” • Indicate the outlook for improvement. • Be sure a structure is established which will provide ongoing control after the team leaves. • Ensure that the team leader has the necessary power to make things happen. • Have team members from all affected areas, i.e., those who will have to implement the solution, and have available diverse resource people who can generate and evaluate potential solutions or contribute unique skills. • List clear responsibilities of team members. • Ensure the measurement system is adequate before it is used on the system. • Have useful, agreed-upon operational definitions, e.g., first run. • Include cost estimates whenever available. • Include a cause-and-effect diagram. • Relate in-plant indicators to customer-oriented indicators. • Distinguish “snapshot” from ongoing verification. • Prioritize. • “Clean your own house first.” • Understand customer concern. Be customer driven. • Drive analysis and data collection upstream. • Make charts on the floor and use them as a control document (not only for display purposes)! • Use graphic data displays. • Ask why?, why?, why? (not who?, who?, who?) until why cannot be asked again. • If there is a roadblock, ask why and seek an alternative. • Involve hourly workers.

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• Use data that are already (secondary data) available, i.e., repairmens’ records. • Include backup documents such as fishbone diagrams, Pareto charts, and control charts. • Include action plans. • Use design of experiments (DOE). • Use charts rather than numbers alone. Capture the time dimension. • Use variable (vs. attribute) data whenever possible. • Understand the system by using flowcharts. • Use common sense! • Label each chart: part, plant, time, quantity, and metric. • Remember that a dubious observer is to be convinced. • Inform others of the results. • Confirm “snapshot” verification with statistical process control. • Make chronologies and summaries. • Call a statistical resource person to help with a sampling plan before collecting data.

DO NOT’S • • • • • • • • • • • •

“BS,” equivocate, or muddle. Let the report become an end in itself. Leave blanks. (Explain what has been done or what will be done.) Confuse root cause with symptom. Make charts for the sake of charts. Forget that the team is limited only to statistical tools frequently used. Fail to use many other useful techniques, e.g., exploratory data analysis and regression. Use warranty/TGW data for in-plant control. Use “operator retrained” prevention. Use P charts when U or C charts are called for. Be afraid to recommend management system changes in the prevention section. Use statistics as “a drunk uses a lamppost, for support rather than illumination.”

CONCERN ANALYSIS REPORT Perhaps one of the most useful tools in problem solving is use of a concern report. The report serves as a guideline about the path that the team will follow and the tools that may be used in the problem solving process. It may be modified to project the needs of the team members. There is no standard form; however, Table 5.1 is a thought starter.

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TABLE 5.1 Concern Analysis Report Guidelines Problem Solving Step Preparation

Team

Problem description

Root cause

Issue • Background information • Specific information about the issue • Selection of core members • Selection of resource persons • Definition of roles and responsibilities • Notification of everyone affected • Relate in-plant to customerrelated data • Define priority • Look for trends • Look for deviations from standard • Investigate who, what, where, when • Establish baseline

• • • • •

Study symptom and root cause Identify priority Identify top four or five Dive into root cause Review design history of this and similar parts for approaches which may be useful or useless

Tools • • • • •

Interviews Check sheets Communication skills Interviews Communication skills

• • • • • • • • • • •

Brainstorming Flow charts Good grammar Operational definitions Engineering judgment Data availability Loss function Market research Gage capability Analysis of variance Key characteristic dimensional calibration Pareto chart Survey results/test results Run charts SPC charts TGW, warranty Brainstorming Check sheets Flow charts Cause and effect diagrams Data collection plan Design of experiments Pareto chart Run chart SPC charts Scatter plots Correlation Regression Exploratory data analysis Repairmen’s records Design history EI team meetings Plant personnel surveys Professional literature Parameter design

• • • • • • • • • • • • • • • • • • • • • • • •

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TABLE 5.1 (Continued) Concern Analysis Report Guidelines Problem Solving Step

Issue

Actions — Interim

• Work plan • Deviation numbers • Plan for eliminating special causes • Corrective action chronology • Status report of pending modification or change

Actions — Permanent

• Work plan • Deviation numbers • Plan for eliminating special causes • Corrective action chronology • Change the documentation to reflect new or modified “way”

Verification

• “Snapshot”/screening experiments • Ongoing evidence of process control at new “breakthrough” level • Common sense, cause and effect and statistical theory • Fundamental system changes • Inform activities that might benefit from similar system change • Honest, accurate, and timely feedback

Prevention

Recognize the team

Tools • Engineering judgment based on root cause and parameter design analysis • Reliability analysis • Fractional factorial analysis • SPC • Internal and external statistical consultants • Cost benefit analysis • Engineering economics • Engineering judgment based on root cause and parameter design analysis • Reliability analysis • Fractional factorial analysis • SPC • Internal and external statistical consultants • Pilot study • SPC • Capability study

• SPC

• Interpersonal skills • Communication skills

ROOT CAUSE ISSUES Root cause is something that everyone talks about, but few understand. Even fewer do something about it. Because the root cause is ultimately the goal of any improvement initiative, take a closer look at the root cause by examining or becoming cognizant of some of the difficulties associated with this concept. • Specifying root cause is a very difficult process. • Potential root causes are identified in the early stages of the study. Brainstorming and cause-and-effect diagrams are useful at these stages.

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• To determine that a cause is a root cause, ask why?, why?, why? When the question why can no longer be asked, a root cause has been reached. • There can be more than one root cause. • There are frequently chains of causality. Root cause analysis drives one up the chain. • Sometimes several levels of causality should be addressed. • Root causes generally involve system changes — frequently changes in management systems that are beyond the immediate control of the team. If these system changes are not made, this concern or a similar concern will reoccur (or occur in another product or in another plant). • Generally, if actions do not address the root causes, future work is guaranteed. An example may prove the point. Often, the last link in a chain of causes is looked at as the root cause. For example, generally there is no objection to someone saying, “Joe won the pool tournament because he sank a difficult shot.” In going up the stream of causes, it could then be said: • Joe won the pool tournament because he sank a difficult shot. • He sank the shot because it came off the second bank at the proper angle and sufficient velocity. • It came off the second bank because it came off the first bank at the proper angle and sufficient velocity. • It came off the first bank at the proper angle because the target ball was struck by the cue ball at the proper angle and sufficient velocity. • The target ball was hit at the proper angle because the cue ball was struck by the cue at the proper angle and sufficient velocity. After going up the chain of causality, it would be tempting to say that the last statement is the root cause of Joe winning the tournament. If this were so, all that would be necessary to win the tournament would be to be able to strike the cue ball at a certain angle at a certain velocity to make this bank shot, i.e., anyone who could make the bank shot would win the tournament. Clearly this is not the case. Joe had to make a number of other shots to be in a position to have the bank shot described as the winning shot. Investigating further, it might be said that the root causes of Joe’s victory are • • • • •

Joe’s hand/eye coordination Joe’s practice Joe’s coaches and/or teachers Joe’s physical and mental state at the time of the game These same elements for each opponent Joe faced in the tournament

Looking at these potential root causes yields some basic prescriptions for anyone who wants to win a pool tournament:

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Seek out good teachers and coaches. Have a positive mental attitude. Come to the game rested and sober. Practice, practice, practice.

The example from a pool tournament resulted in prescriptions for success which are no surprise. Few people would think the way to win a tournament is to be able to sink a single shot. This is because the recipe for success — the root cause — in sports has been arrived at after centuries of trial and error and is now common knowledge. The root causes for a new challenge are frequently not as obvious and people do not spend time going up the chain of causality. For example, to say a part is not installed to specifications because the installer did not install it to specifications is like saying Joe won the tournament because he sank a shot. This, of course, leads to “mistake proofing” which will be addressed in Volume VI.

VERIFICATION As already discussed, there are three elements to verification: 1. Common sense 2. Cause and effect 3. Statistical theory Ultimately, the only verification that counts is whether or not the customer likes the product. However, the time it takes to get this information makes this data unsuitable for evaluations. In its place, statistical process control (SPC) of characteristics closely related to customer-oriented measures should be used for verification of both interim and permanent actions. Common sense is the basis of all verification. For example, if a football coach had a record of 0-7, no one would offer him a 5-year contract if he won his next game — at best he might not be fired. Essential to verifying the effectiveness of an action is having a baseline against which future progress may be measured. This emphasizes the importance of having a data gathering plan in place for key items. The question then becomes how to select these actions. Ideally this is done with design of experiments (DOE), a rigorous methodology based on statistical theory which helps an engineer understand the impact (i.e., the cause and effect relationship) of a number of design/manufacturing decisions by simultaneously changing them in a specific, organized manner. With DOE, the engineer can frequently obtain the information needed by running only a small fraction of the tests needed to obtain the same information using one-factor-at-a-time testing. Using Taguchi’s concept of parameter design allows the engineer to look not only at design and processing parameters, which can be set, but to also determine how “robust” his design might be. A “robust” design performs consistently in the face of widely varying “noise” variables which an engineer cannot control or chooses to not control. (Recall the

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design described earlier in this chapter which was “robust” against worker fatigue, i.e., the new part was as easy to install 400 times a day as it was to install once.) When a DOE approach cannot be used to determine if an action is likely to have the desired effect, the alternative is to use “snapshot” verification with care.

EXAMPLES OF “SNAPSHOT” VERIFICATION (RELIABILITY CONFIDENCE)

AT

85%

Interpretation of test results: For tests of a given sample size, use a classic sample table (Oliver et al., 1972; Schafer 1967a; Schafer 1967b; similar tables may be found in any statistic book and or reliability book). Because they are easily available, such tables are not included. (Volume III will cover sampling more thoroughly.) However, the interpretation is always the same, given a certain confidence level. For purposes, of example, an 85% confidence is assumed which also happens to be the most common. Use the tables by: • Reading the number of samples tested in the “n” column on the left-hand side. • Reading the number of nonconforming (sometimes called failures) across the column heading. • Finding the resulting demonstrated reliability at 85% confidence where the row and column intersect. Example 1: A 50-part trial has 3 nonconforming pieces (pieces that do not pass a go/no-go test). This demonstrates 87.9% reliability at 85% confidence. Example 2: All parts in a 100-part trial pass a go/no-go test. This demonstrates 7.9% reliability at 85% confidence. Determination of test size: • Determine the reliability at 85% confidence to be demonstrated. • Find this number (or next largest value) in the zero nonconforming column. • Read minimum sample size from “n” column. Example 1: To demonstrate 99.5% reliability at 85% confidence in a go/no-go test, 400 parts must be run and all must pass. If 1 part is nonconforming, demonstrated reliability drops to 99.1%. Example 2: To demonstrate 99.9% reliability at 85% confidence, 1500 parts must be run, all of which pass the go/no-go test. Note 1. Sample sizes can be greatly reduced if variable data (e.g., pounds) are used rather than go/no-go data. Different statistical techniques are needed. Note 2. Applying the results of “snapshot” testing, like all statistical methodologies, to future production assumes the tests are run against

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conditions which the production parts will actually encounter. If this is not the case, actual reliability may be very different. A connection made by an engineer who has time to finesse the operation may perform better than that one made by an operator who must perform the operation in 15 seconds. Note 3. One-at-a-time testing is an inefficient way to test design alternatives, especially for multiple changes. DOE is recommended. Note 4. “Snapshot” verification gives only an indication of what is likely to occur. Generally, it should be followed up by ongoing statistical techniques.

SPC CHART SUBGROUP

AND

SAMPLE SIZE

A special verification approach is done with SPC. Obviously, as already mentioned, statistical theory is important in determining the number of items to be looked at in a subgroup of an SPC chart. As a general rule, a p chart for attribute (go/no go) data should have subgroups such that np (bar) is greater than or equal to 5 if the chart is to be interpretable, i.e., the rules of determining out-of-control situations can be used directly (for some applications, np 9 is recommended). An alternative minimum sample size (n) which does not allow for control chart interpretation, but which ensures that at least one nonconforming unit in every subgroup with at least 90% confidence is n = –1/log[1 – p(bar)] This formula is a special case of n = log (1 – confidence)/log[1 – p(bar)] In these formulas, p is the average proportion nonconforming, n is the desired sample, log is the regular log of base 10, and “confidence” is the desired confidence level. For the other attribute charts, C and U, the “Rule of Five” also applies, i.e., there should be at least 5 nonconformities per subgroup for interpretability (again, for some applications c ≥ 9 is recommended). A few calculations quickly indicate that use of attribute charts requires a great deal of data (more on this in Volumes III and IV). For X-bar and R charts, the subgroup size and frequency determination are based on the relationship between the variation between the pieces which are produced consecutively, the period to period variability, and the capability of the process. In all determinations of subgroup size and frequency, the economics of the time to take data must also be considered, as well as sources of stratification, e.g., two sets of molds in an injection molding process. If only a “snapshot” is available at the time of the report, the plan to collect control chart information should be indicated. It is important to state the sample size and subgroup frequency when presenting verification. If in doubt about either the sample or the frequency or both, consult with a statistician or internal/external consultant.

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If verification is dependent on physical inspection, then cost considerations play a major role (Deming, 1982) and the following rule applies: if ci = cost of inspection on a per part basis, and cd = cost of finding and fixing the defect at end, then the ratio of ci/cd will determine the effectiveness of inspection. If the ratio is greater than the fraction defective (p), then inspect 100%; if the ratio is smaller, no inspection is necessary; and if it is “1,” it is the break-even point.

GPS APPLICATION CRITERIA Therefore, to be successful in the GPS process, the following must be considered and reviewed, as appropriate: • Control plan (This is actually a road map of the process when something has gone amiss.) • Attention to detail (Be sure the “current” process flow chart is understood.) • Accurate, timely, and meaningful data (the foundation of sound problem resolution) • Analysis and technical expertise • Appropriate and applicable training • Maintenance • Follow appropriate and applicable standards, procedures, policies, etc. • Tenacity and commitment to results • Accountability and feedback • Worker involvement • Appropriate and applicable incentives and rewards • Bad work unknowingly “passed” • Focus on customer satisfaction It is important to recognize that the GPS process is a summary process. Even though it may take several days and considerable work to complete the finding, analysis, and recommendations, the ultimate result should be written on one page. Working with this in mind will add even more to the discipline and to the process improvement as a whole. A typical format for this type of cover sheet may be found in Appendix A. For GPS to be a viable option to solve problems, the following criteria must be met: 1. There is a definition of the symptom(s). (The symptom has been quantified.) 2. GPS customer(s) who experienced the symptom(s) and, when appropriate, the affected parties have been identified. 3. Measurements taken to quantify the symptom(s) demonstrate that a performance gap exists and/or priority (severity, urgency, growth) of the symptom warrants initiation of the process. 4. The cause is unknown. 5. Management is committed to dedicate the necessary resources to fix the problem at the root cause level and to prevent recurrence.

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6. Management is committed to results at the root level. 7. Symptom complexity exceeds the ability of one person to resolve.

COMMON TASKS As one proceeds through the GPS process, it becomes obvious that there are common tasks in each of the individual steps. Tasks which require continual consideration throughout the GPS process are 1. 2. 3. 4. 5. 6.

Document the changes. Review team composition. Review measurable(s). Determine if a service action is required. Review assessment questions. Update the GPS report.

In fact, these tasks are so common, yet powerful, that each step should begin with a review of each task to be sure that a previous task was accomplished or if something must be done before the current task is undertaken.

CHANGE AND NEVER-BEEN-THERE SITUATIONS Things do go wrong. The question is why? More often than not, the reasons are • • • •

Something was omitted. (There was not have enough time.) Something was minimized. (It has always been done this way.) Something is insufficient. (It’s what was asked for, isn’t it?) Something was bypassed. (There was a check off, but the necessary related content was not implemented.)

In short, when any one of these items occurs in an organization, there are “problems.” The problems cause “change” and that type of change falls into three categories: (1) something changed gradually; (2) something changed abruptly; and (3) something changed in a way that is “new” — we have “never been there” (Figure 5.1). The level of investigation and problem solving will be different, depending on the change being experienced. However, the process for both will be the same. The process and each of its steps will be examined in detail in the next section.

GPS STEPS There are common tasks in each GPS step that require continual consideration throughout the GPS process. The tasks are so important that they should be reviewed at the beginning of each GPS step. Therefore, the common tasks are part of each step:

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FIGURE 5.1 Something changed. - - - Actual performance, ____ Expected performance or desired performance.

Common tasks: • • • • • •

Document the changes. Review team composition. Review measurables. Determine if a service action is required. Review assessment questions. Update the GPS report.

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GPS0: PREPARATION (FIGURE 5.2)

FIGURE 5.2 An overview of GPS0.

Background: Purpose. In response to a symptom, evaluate the need for the GPS process. If necessary, provide an emergency response action (ERA) to protect the customer and initiate the GPS process. Review the common tasks. Tasks that need continual consideration throughout the preparation process: 1. 2. 3. 4. 5. 6.

Document the changes. Review team composition. Review measurable(s). Determine if service action is required. Review assessment questions. Update the GPS report.

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Overview. Application criteria identify conditions that justify use of the GPS process. This justification is necessary to allocate resources because the demand for resources in the form of people, time, money, experience, etc. is usually greater than what is currently available. Many problem solving techniques and processes are available. Matching the right problem with the right process is one way to avoid wasting limited resources. Using resources effectively is also important. GPS requires the use of many resources, but it offers many benefits. The focus of GPS0 is to be sure that there is a preliminary analysis of whether or not a full-scale GPS or an ERA is necessary.

GPS1: ESTABLISH

THE

TEAM/PROCESS FLOW (FIGURE 5.3)

Input GPS Process Initiated

Identify Champion Identify Team Leader Determine Skills and Knowledge team will need Select Team Members Establish Team Goals and Membership Roles Establish Operating Procedures and Working Relationships of the Team

Output Team Established

FIGURE 5.3 An overview of GPS1.

Background: Purpose. Establish a small group of people with process and/or product knowledge, allocated time, authority, and skill in the required technical disciplines to solve the problem and implement corrective actions. The group must have a designated champion and team leader. This group begins the team building process.

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Review the common tasks. Tasks that need continual consideration throughout formation of the team and the process flow process: 1. 2. 3. 4. 5. 6.

Document the changes. Review team composition. Review measurable(s). Determine if a service action is required. Review assessment questions. Update the GPS report.

Overview. Problem solving implies that changes will be made to resolve an undesired condition. Comprehensive problem solving approaches acknowledge the need to make provisions for the entire problem solving effort. Such provisions may include blending the organization’s needs with the problem solvers’ skills, behaviors, and personality styles. GPS1 addresses some of the requirements for successful problem solving efforts. Certain problems can be tackled by one person, but generally speaking resolving a concern at the systemic level requires a cross-functional group — a team. A problem solving team may be created with specific tasks or objectives in mind. However, when persons with different experiences, skills, knowledge, and priorities are assembled, human relations’ issues tend to arise and interfere with completion of assigned tasks. The primary function of GPS1 is to prevent human relations’ issues from developing. During GPS1, roles, goals, and responsibilities are clarified. In group problem solving situations, the team is well advised to establish guidelines for group and individual behaviors. At a minimum, the owner of the concern (called the champion), the one with decision-making authority, must be strongly committed to the problem solving effort. Fixing the problem at the root cause level and subsequent prevention efforts are impossible without this commitment. The champion’s critical role is to empower the problem solver(s) to complete the prescribed problem solving steps. The champion also helps remove barriers and procures resources needed to resolve the concern. Champion. The champion has the authority to make (and implement) decisions regarding containment actions, permanent corrective actions, and prevention of recurrence. The Champion also supports the problem solving efforts of the team throughout the process. One key duty is to set priorities: • • • • • •

Task the team with the GPS method. Help remove organizational barriers to the GPS method. Help procure resources the team requires to complete the process. Review assessing questions with the team. Exercise authority to implement team recommendations. Support the organization as a point of contact regarding the GPS process.

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• Forward the GPS report to concerned departments. Team members. Team members are persons who are selected to participate on the GPS team due to their expertise and knowledge. They are the subject matter experts who do the work of each objective. Team members might change during the GPS process. They are generally responsible for: • • • • • •

Doing investigative work Developing plans Using judgment, skills, experience, and knowledge Finding answers Developing recommendations at each GPS step Executing implementation actions

Team leader. The team leader is the individual responsible for guiding team members through the GPS process. This person has leadership and interpersonal skills: • • • • •

Lead the teams to complete each GPS objective Develop agendas for meetings and team activities Conduct meetings Ask (generally does not answer) GPS process questions Manage the team in accordance with established group effectiveness guidelines • Support the champion as the primary point of contact

Recorder. The recorder is a team member who generates, holds, and publishes team reports (such as meeting minutes, agendas, action plans, etc.): • Control documents for the team • Tend to be an administrative support person for the team • Take responsibility for creating and distributing meeting minutes and reports in a timely manner Facilitator. The facilitator’s primary task is to assist the team with interpersonal relationship issues such as resolving conflict, arriving at consensus (when appropriate), confronting counterproductive group behaviors, etc. The facilitator’s role is optional. The facilitator: • Works with the team to resolve conflicts • Shows team members how to work together while maintaining individual self-esteem • Coaches individual team members, the leader, and the champion on how to be more effective in group situations • Helps the team develop group effectiveness guidelines • Provides feedback to the team on group performance behaviors

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• Helps the team apply techniques such as decision-making and creative processes, etc., but is neutral in decision-making situations Process guidelines. The purpose of GPS1 is to form and develop the team with the required experience for the GPS process. During GPS1, goals, roles, and responsibilities are developed and defined. Specific questions may be found in Appendix C. • The type of problem usually defines required experience and skills and, therefore, who should be on the team. • All members must contribute, use their experience, and help gather facts. • Team members may change as the team’s required experience changes. An ideal team size is about seven, generally, but not necessarily, ranging from four to ten members. • Team positions include leader, recorder, members, and facilitator. • The roles of persons on a GPS team are interdependent. No one person can do it by himself/herself. • The champion owns the problem, supports the team’s efforts, uses authority and influence to help the team with organizational barriers, and helps get resources needed by the team to complete each GPS step. • The entire GPS team works with the champion’s help, involvement, and agreement. • The champion should conduct the first meeting to describe what is needed, why the problem needs to be resolved, and to outline what is expected of the team. • The leader is expected to lead the team, not have all the answers. Critical competencies for the leader are people-management and leadership skills. • In GPS1, there are more questions than definitive answers about the problem. This is normal. Better data regarding the problem will be created as the team progresses through the GPS process. In GPS1, the team members, the champion, and others are selected using the best currently available information.

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GPS2: DESCRIBE

THE

Input: Team established Go to GPS0. Application criteria step for each symptom

PROBLEM (FIGURE 5.4)

Review available data

No Yes Can symptom be subdivided?

Yes

Does problem describe a “something changed” situation?

No

Does problem describe a “never been there” situation?

No

Yes

State symptom as an object and a defect.

Yes

No

Use GPS process.

Is the “real” cause known? (repeated whys)

Use alternative methods such as DOE, process improvement approaches, innovation.

Yes

No Document problem statement.

Suspend GPS

Initiate development of problem description (Is/Is Not).

Identify process flow.

Identify additional data required.

Collect and analyze additional data.

Supplement GPS process with alternative methods.

Yes

Use alternative methods.

No Review problem description with customer and affected parties.

OUTPUT Problem description.

FIGURE 5.4 An overview of GPS2.

Background: Purpose. Describe the internal/external customer problem by identifying “what is wrong with what?” and detail the problem in quantifiable terms. Review the common tasks. Tasks that need continual consideration throughout the problem identification process: 1. Document the changes. 2. Review team composition. 3. Review measurable(s).

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4. Determine if a service action is required. 5. Review assessment questions. 6. Update the GPS report. Overview to define the “problem.” The word problem may refer to a cause, concern, defect, or consequences of the defect’s occurrence. The word problem is too general a term. GPS2 makes a clear distinction between the cause and the effect. The function of GPS2 is to factually describe the problem. In GPS terminology, the problem is a deviation from expectation, a special cause, a distribution within control limits that are too wide, or any unwanted effect where cause is not known. Gathering factual data. The GPS process is a data-driven process. The focus of GPS2 is to obtain an accurate and unbiased description of the object and defect. The information collected during GPS2 is critical to successful problem resolution. Historically, problem solving proponents have advocated that defects be described using either rationally based or experientially based techniques. Each approach has advantages and limitations. The GPS process advocates a synergistic approach to problem description. The result is a factually based, team-generated, problem description. Time well spent. The time spent in GPS2 is always recovered (many-fold) by the time saved in subsequent GPS steps. Unfortunately, this feature of GPS2 is not obvious until later in the process. Teams that quickly pass through or skip GPS2 (and so do not adequately describe the problem) have solved the wrong problem. The initial GPS2 step develops a problem statement which is then expanded by a factual listing of the problem’s profile. Critical features of the problem are made visible using an accurate process flow diagram which depicts the entire process (including any existing informal fixes). This body of information is further enhanced by developing a complete cause-and-effect (fishbone) diagram. Supplemental, effect-specific techniques (often experienced-based) are useful. GPS2 includes various types of reports and description approaches. Completing all these fact-gathering techniques may seem excessive. In reality, the synergism created by combining these techniques is of enormous benefit for GPS3 through GPS7. Remaining objectives tend to be realized more quickly, more effectively, and with far less frustration. A significant byproduct of this synergism is the positive effect on behaviors of the team members. Team members appreciate each member’s experience, resist the tendency to jump to conclusions, and realize the benefits of teamwork. The need for a leader remains, but the team is prepared to proceed from a complete and factual (not opinionated) foundation. Process guidelines. The purpose of GPS2 is to define the problem for which the root cause will eventually be determined. • GPS2 is not intended to determine the root cause. The information will be used later (at GPS4) to determine the root cause.

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• GPS2 is the observations collection step. • GPS2 establishes the data that is the foundation for all other GPS steps. • GPS2 demands the team develop: – A problem statement – A problem description using Is and Is Not (what, where, when, how big?) – A process flow diagram – A cause-and-effect diagram (also known as a fishbone diagram) • The problem may also need to be defined using additional approaches (e.g., SPC data, customer satisfaction, and field reports). • To complete GPS2, the team may need to obtain missing information. • Recommended attachments to the GPS Report at GPS2 include Is/Is Not and the cause-and-effect diagram. • The team and champion use their judgment to determine how much information is sufficient. • The process flow diagram should include the initial actions taken to compensate for the problem when it first occurred. – Note all temporary or informal fixes on the process flow. Do not remove them — just note them. – Start where the defect can be first seen; then work back from there to the beginning of the process. • Use one GPS per problem. Do not use a single GPS on many different problems. Each problem probably has a different root cause.

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GPS3: DEVELOP THE INTERIM CONTAINMENT ACTION (ICA) (FIGURE 5.5) Input Problem Description

Is ICA required?

No

Go to GPS4.

Yes Evaluation ERA.

Identify and choose the “best” ICA and verify.

ICA Verified?

No

Yes Develop action plan, implement and validate ICA.

No

ICA validated?

Yes Continue monitoring ICA effectiveness.

OUTPUT ICAs established.

FIGURE 5.5 An overview of GPS3.

Background: Purpose. Define, verify, and implement the ICA to isolate effects of the problem from any internal/external customer until permanent corrective actions (PCAs) are implemented. Validate the effectiveness of the containment actions. Review the common tasks. Tasks that need continual consideration throughout development of the ICA process: 1. Document the changes. 2. Review team composition. 3. Review measurable(s).

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4. Determine if a service action is required. 5. Review assessment questions. 6. Update the GPS report. Overview. GPS3 is an optional step in the GPS process. Its purpose is to protect the customer until the problem is resolved at GPS6. The customer is whoever receives your output as their input. The customer could be internal to the company as well as external. If the customer cannot accept the defect(s), then an ICA is required. How an ICA works. An ICA is intended to protect the customer from the problem because a root cause is not yet known. An ICA is a temporary measure. It is an added non-value operation. Commitment to completing the GPS remains in force. An ICA is effective when, from the customer’s point of view, the defect is no longer evident. ICAs are often compared to “quick fixes” or “hidden factories,” but containment actions are actually: • Verified to work before implementation • Validation that the ICA is working after implementation • Validated independently, in addition to the customer’s confirmation of effectiveness • Documented to exist (i.e., made visible) by amendments to the organization’s process flow diagrams, procedures, process instructions, etc. • Replaced by a PCA at GPS6 • Implemented without creating other trouble downstream • Formal temporary fixes (Care is exercised to ensure that all supporting functional areas/departments are involved in planning and implementation.) • Implemented only with the clear approval of the champion and customer • Managed through comprehensive planning and follow up Another important feature of GPS3 is that prudence may require containment action (an ERA) before GPSl and GPS2 are completed. However, after GPS2 is completed, the emergency response action (ERA) should be reviewed. The GPS2 data may suggest a more effective ICA (GPS3). The champion’s authority over ICAs. An ICA should not be implemented by a team without the champion’s knowledge or involvement. A subtle but significant feature of the GPS process is that the authority to implement an ICA is retained by the champion. It is not delegated to the team. Likewise, the champion is responsible for cross-functional communications. Some team problem solving guidelines imply the team itself would assume this responsibility. The GPS process promotes action within the organization’s chain of command. It does not promote creation of an artificial command structure which bypasses the normal authority structure.

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Process guidelines. The purpose of GPS3 is to protect the customer while determining the cause of the problem. Steps and questions during GPS3 include: • • • • • • • • • • • • • • • • • • • •

• • •

Is a GPS3 step necessary? If implemented, can the GPS0 ERA be improved? Which action(s) is (are) best? Choose the best containment action. Will the containment action work (i.e., eliminate 100% of the problem from reaching the customer)? (Verification) Follow the management cycle. Plan (in detail) the containment action(s) and verify. Implement the containment action. Record the results. Evaluate the ICA. Did it work? (Validation) Document the actions and communicate how to do them to all who need to know. Continue to monitor the effectiveness of the ICA throughout the time it is in place. The champion must be involved before the ICA is implemented and during implementation. ICA is the only optional step in the entire GPS process. ICA is temporary, not permanent. It is costly, not cost effective. It is containment, not mistake proof. ICA works against the problem, not the root cause. Costs increase while the ICA is used. ICA may be done while other members of the team work on GPS4 (determining root cause). ICA may require additional members to be added to the team to complete all of GPS3. ICA looks like a quick fix, but its actions are documented, later removed, and checked for effectiveness. The team must ensure that the actions will not create other problems for departments and customers downstream of GPS3 action(s). More than one GPS3 action may be required to fully protect the customer (internal or external). After GPS3, the team’s membership might change. This is true at all steps. ICA may occur before GPSl or GPS2. If done before GPS2, then GPS3 ICA should be reviewed after the GPS2 data are compiled. Better GPS2 data may prescribe the need for an improved GPS3 ICA.

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GPS4: DEFINE AND VERIFY ROOT CAUSE AND ESCAPE POINT (FIGURE 5.6)

FIGURE 5.6 An overview of GPS4.

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FIGURE 5.6 Continued.

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FIGURE 5.6 Continued.

Background: Purpose. Isolate and verify the root cause by testing each possible cause against the problem description and test data. Also isolate and verify the place in the process where the effect of the root cause should have been detected and contained (escape point). Review the common tasks. Tasks that need continual consideration throughout the root cause verification and escape point determination process: 1. 2. 3. 4. 5. 6.

Document the changes. Review team composition. Review measurable(s). Determine if a service action is required. Review assessment questions. Update the GPS report.

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Overview. If a situation was once problem-free, but now is not, then the situation is considered a change-induced condition. Identifying this change is called problem solving. Various problem solving techniques exist. They are based on three basic strategies: 1. Fact-based, deductive approaches 2. Experientially based approaches 3. Creative approaches A working knowledge of various problem solving techniques is desirable. The single verified reason that accounts for the problem is the root cause. It is the cause that explains all the facts about the problem. The root cause is verified by its ability to make the problem come and go. Determining root cause. Information compiled at GPS2 is used to identify a set of possible causes. At GPS4, the root cause is discovered. Problem solving techniques are used to reduce the time and confusion to systematically deduce the cause. All subsequent GPS objectives depend on the accurate diagnosis of the root cause. Therefore, verification of the root cause is critical to the success of the GPS process. Few things are more damaging to the problem solving process than assigning blame. Blaming leads to defensiveness and facts are obscured or kept hidden. Misinformation is often generated as a defensive measure. Process guidelines. The purpose of GPS4 is to determine the root cause and escape point of the problem. The root cause and escape point are only identified and verified during GPS4s, but will be fixed at GPS6. • • • • •

• •

•

•

GPS4 is a problem-solving step. GPS4 uses information collected during GPS2. Root cause is the single verified reason that accounts for the problem. Escape point is the earliest location in the process, closest to the root cause, where the problem should have been detected but was not. Verification helps to make certain that PCAs are directed at the root cause and escape point. Time, money, effort, and resources are not wasted on false causes. Band-Aids™ can mask information needed to find the root cause. Watch for Band-Aid™ fixes. Usually, the root cause is one change that caused the problem. If previous problems were never fixed at the root cause level, then the problem may be the result of more than one change. Use deductive reasoning (first) to identify the possible causes. If this method does not work (due to missing information), then use causeand-effect diagrams and team members’ experience to pursue other possible causes. Root cause cannot be identified without facts. The Is/Is Not data must be correct and the process flow diagram must be correct and up to date.

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• The root cause should explain all known facts about the problem. Unexplained facts often indicate the presence of another root cause that creates a similar problem. Ideally, the problem should be made to come and go to prove the root cause has been identified. • When the root cause is uncovered, other troubles (which have gone unnoticed) are sometimes made visible. These other troubles create the need for an improved ICA in GPS3 and/or a return to GPS2. Success factors. In order to ensure success, the following must be present in the organization: • • • •

Management Management Management Management actions.

must create the right climate. must recognized the need to prioritize problems. must actively support the problem solving process. must be committed to a system to carry out corrective

Management creates the right climate by: • • • • •

Having patience for results Providing practice time Demonstrating interest — even when failure is experienced Stating expectations clearly Allowing for risk

Management sets priorities by: • • • •

Continuously improving the system Having realistic expectations in every respect Providing the appropriate resources Communicating the significance of the problem with the team and everyone concerned with the specific problem

Management demonstrates interest by: • Giving authority and responsibility to the appropriate individuals to carry out the assigned task • Always reviewing progress of project and asking coaching questions • Making sure that the process defined is adhered to • Being an implementation advocate • Providing a system to retain the “gained” knowledge • Providing the appropriate resources

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GPS5: CHOOSE AND VERIFY PERMANENT CORRECTIVE ACTIONS (PCAS) FOR ROOT CAUSE AND ESCAPE POINT (FIGURE 5.7)

FIGURE 5.7 An overview of GPS5.

Background: Purpose. Select the best PCA to remove the root cause. Also select the best PCA to eliminate escape. Verify that both decisions will be successful when implemented without causing undesirable effects. Review the common tasks. Tasks that need continual consideration throughout the PCA choice and verification process for root cause and escape point: 1. 2. 3. 4. 5. 6.

Document the changes. Review team composition. Review measurable(s). Determine if a service action is required. Review assessment questions. Update the GPS report.

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Overview. The GPS process is applied to fix a problem at the root cause level and prevent its recurrence. In change-induced conditions, the root cause(s) is introduced by a change. In an overwhelming number of cases, the change was introduced by someone’s decision. At the time of the decision, the problem was not anticipated. The GPS5 objective is designed to avoid a similar sequence of events. The historical function of GPS5 is to make the best decision on how to remove the root cause variable that created the defect. Because a decision can potentially lead to other effects, the decision needs to be thoroughly examined before its implementation at GPS6. Customers and managers must be involved. The decision is first examined for both benefits and risks. Then the decision is tested (verified) to ensure that it will be successful when implemented at GPS6. The GPS5 objective ends with a commitment to full-scale implementation of this decision at GPS6. The decision-making process depends on the experience of the decision makers and the criteria that is applied. This is especially true during the risk analysis phase of the decision-making process. Experience is also critical to the verification segment of GPS5. Experience guides how verification should be conducted and how much data are appropriate. Champion’s commitment. The champion must be committed to the decision. The decision is typically based on a synthesis of the champion’s criteria and the detailed experience of the team members. Process guidelines. The purpose of GPS5 is to select the best PCA(s) to eliminate the root cause and escape point. • More than one PCA may be required to resolve 100% of the problem. • The GPS5 decision is not implemented until GPS6. • Consider having risk analysis conducted by those who must implement the corrective action. • Implied actions of GPS5: – Check the composition of the team. Does the team have the right experience to make the decision(s)? – Decide what is the best permanent corrective action. – Evaluate each choice from both its benefits and liabilities. – Test, if necessary, the decision to determine that the choice (action) will, in fact, work. – If this PCA cannot be implemented quickly, will the GPS3 ICA last long enough? – Get approval (through the champion) to proceed with the choice. Implement the decision at GPS6. • Sometimes the best PCAs cannot be implemented due to cost, limited resources, etc. Sometimes the root cause cannot be eliminated. In these cases, the best decision is to continue using the GPS3 ICA. This measure is a compromise, but sometimes it is the only resolution. This should be considered an exception for GPS5, not the norm. • Even if the above action is appropriate, GPS6, GPS7, and GPS8 should still be completed.

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• Sometimes the best PCA will require time before implementation. In this case, the GPS3 ICA will continue until it is replaced by the GPS6 PCA. The GPS process permits a redesign of the ICA. This redesigned ICA might be more effective and less costly than the initial GPS3 ICA. • During GPS5 verification, the presence/evidence of another root cause which was not discovered during GPS4 may be uncovered. This is indicated when the verified PCA does not remove 100% of the unwanted effect.

GPS6: IMPLEMENT AND VALIDATE PERMANENT CORRECTIVE ACTIONS (PCAS) (FIGURE 5.8)

FIGURE 5.8 An overview of GPS6.

Background: Purpose. Plan and implement selected PCAs. Remove the ICA. Monitor the long-term results.

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Review the common tasks. Tasks that need continual consideration throughout the PCA implementation and validation process: 1. 2. 3. 4. 5. 6.

Document the changes. Review team composition. Review measurable(s). Determine if a service action is required. Review assessment questions. Update the GPS report.

Overview. In GPS5, the PCA was chosen and verified. In GPS6, the PCA is implemented and the problem is corrected. In GPS6, the team is responsible for detailed planning, implementation, and evaluation. Cross-functional participation. One of the major responsibilities of the champion’s authority is the requirement that the champion must authorize making changes to resolve the problem. Generally, even relatively minor changes will require cross-functional participation. For example, changing the length of a purchased bolt impacts the documentation of purchasing, quality assurance, material control, inventory control, engineering, and production departments. In this example, the champion might be the production department manager, but the production department still requires communication and close coordination. Cross-functional involvement at GPS6 also means cross-functional planning to implement the GPS6 correction. During GPS6, team membership may expand temporarily to facilitate complete planning. This planning may or may not require a group meeting, but some form of contact with other departments is essential for successful implementation. At a minimum, the existing team must identify who (in which department) needs to be involved in planning implementation of GPS6. Removing the ICA. Embedded in GPS6 is removal of the GPS3 ICA. The ICA is no longer needed because the PCA removes the root cause variable. Continuing the ICA wastes important resources. Furthermore, the ICA has masked the problem, so validation of the GPS6 PCA would be questionable. GPS6 planning must account for when to terminate the GPS3 ICA and how to validate results of the GPS6 PCA. Technically, this validation is continuous and ongoing, but validation is likely to be more frequent at first, tapering off over time. In simple terms, the events of GPS6 are expressed by the model illustrated in Figure 5.9. The team members’ and champion’s active involvement occur at each event, although the activities of each tend to be different. Reflecting the PCA. The critical task of updating all documents, procedures, process sheets, etc. is too frequently overlooked. The GPS6 PCA fixes a problem by making changes. If the PCA change is not documented in the normal systems, practices, and procedures, then additional corrections will likely be necessary. A problem recurs when someone uses an obsolete instruction. Related problems also occur, such as ordering obsolete parts

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FIGURE 5.9 The events of GPS6.

or being unable to order needed parts, because they were not recorded on the bill of materials in the material control system. Process guidelines. The purpose of GPS6 is to implement the decision made at GPS5. • PCA implementation is done via management systems. They are not bypassed. • The implied action of GPS6 includes: – Check the composition of the team. Is the needed experience on the team? – The decision needs to be planned, implemented, monitored, and evaluated for effectiveness. – Any changes implemented during GPS6 must be documented (formally) in process flow sheets, procedures, practices, instructions, standards, spec sheets, etc. in the departments which maintain such records or documents. • The three minimum items for any action plan are what?, who?, when due? • GPS6 requires detail-oriented, hands-on people to be involved in planning GPS6 actions. • The GPS process does not supersede an organization’s procedures to implement process changes or customer approval processes. • Some changes may require approval by the customer(s) before the changes can be made. • No changes should be made without the champion’s direct approval. • Some persons may require training before implementation of the GPS6 change because their procedures may be affected. • During GPS6, the GPS3 action should be stopped because the ICA is no longer necessary. Continued GPS3 actions may mask a portion of the defect that is still present. • GPS6 is not completed until the PCA is validated (the customer no longer experiences the problem). Prudence dictates that validation

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should occur at some point before the customer can experience the problem. • Long-term monitoring is appropriate after the GPS6 PCA is implemented. • Failure to document the correction created by the GPS6 PCA will eventually reintroduce the root cause. Failure to document the PCA typically results in overstocking obsolete parts and stock-outs of needed parts. • Consider adopting the following statement as a normal GPS6: Flawless implementation of the PCA, where things work right the first time.

GPS7: PREVENT RECURRENCE (FIGURE 5.10)

FIGURE 5.10 An overview of GPS7.

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Background: Purpose. Modify the necessary systems including policies, practices, and procedures to prevent recurrence of this problem and similar ones. Make recommendations for systemic improvements, as necessary. Review the common tasks. Tasks that need continual consideration throughout determination of the GPS process to prevent recurrence: 1. 2. 3. 4. 5. 6.

Document the changes. Review team composition. Review measurable(s). Determine if a service action is required. Review assessment questions. Update the GPS report.

Overview. GPS7 is frequently cited as perhaps the most important step of the GPS process. During GPS7, the root cause of the GPS4 root cause is fixed. Behind every change that creates the need to do a GPS is at least one system, practice, or procedure that introduced the change. If the system, practice, or procedure remains the same, the same or a similar problem causing change will be reintroduced. This can and should be prevented. The following are some of the reasons why GPS7 is not always completed. Being aware of these reasons helps to identify steps an organization needs to take to combat them. Factors that inhibit completion of GPS7: • GPS7 is not even attempted: After the problem is corrected, other priorities receive more attention. • GPS7 develops into a blame session: Rather than focus objectively on what happened, attention drifts to who did/did not do something. Defensiveness sets in and objective review stops. • The GPS process stopped at GPS3 containment: The root cause was never found. Prevention of the root cause is impossible because it remains unknown. • The GPS process was never done: Someone penciled in a form that became an GPS Report. • Politics: Fixing systems is a low priority because it requires risking (forcing) a change. • Fear: People perceive/know that spotlighting defective/inadequate system practices and procedures will result in reprisal. • Not enough structure/authority: The champion is too low in the organization’s power structure to implement necessary system revisions. Furthermore, this champion is unable to solicit involvement of those who do have the authority.

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• Those who own the system(s) do not have to live with the consequences: The organization is unwilling to create consequences at the proper points of systems management. • Systems maintenance and management is a low priority: Activity is more important than prevention. Process control is not understood. Basic questions. Some of these issues obviously exceed the scope of one champion’s authority. However, they can be corrected. Some take considerable effort and time. The first step of any GPS7 objective is to accurately identify (at least) the cause of the root cause — the system, process, etc. which introduced the root cause. This much the champion can prompt with the team’s understanding of the first six objectives. GPS7 often requires bringing together all former team members to answer two questions: 1. What went wrong to introduce the root cause in the first place? 2. What needs to be fixed, modified, or reinforced to prevent it from happening again? GPS7 also implies opportunities, including the chance to suggest what a good system should look like and to identify places where a defect might be created. The champion’s options in GPS7. Three paths may be followed after the systemic issues are identified: 1. The champion uses his/her authority to correct the system, practice, or procedure. 2. The champion takes the recommendation for systemic changes to those who have the authority to change the system, etc. 3. The need for systemic changes is acknowledged, but the organization chooses to actively monitor the day-to-day operation of the system. This is done to better understand the whole system, rather than overreact to a single incident. Other GPSs might also be reviewed for recommendations and observations of the system’s shortcomings. A word of caution, however, is that the GPS process requires input at GPS7. The champion retains 51% of the vote. GPS7 does not dictate that the champion implement every recommendation. However, the champion works against his/her own best interest if ideas are discounted without serious consideration. Process guidelines. The purpose of GPS7 is prevent recurrence of this root cause and escape point. • Approximately 95% of all root causes are introduced because of a procedure, practice, instruction, or management system malfunction. • These systems may have been ignored, they may have never worked initially, or they may be too cumbersome or obsolete.

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• If the system is not fixed, it will create a similar kind of problem and/or another root cause. • Typically, the team is best able to identify the details of the systemic problem. • The team is chartered with both the responsibility and the authority to provide feedback to management (through the champion) as to which systems contributed to the root cause and escape point. • The team members draw from all the knowledge and experience gained from the first six steps. The champion may have to take this information to other areas within management to implement systemic recommendations because he/she may not have full authority to make changes as recommended. • Not all team-identified prevent actions and systemic recommendations will be the best and they should not be implemented. However, in fairness to the team, reasons why these will not be implemented should be communicated. • If the team has skill, experience, and knowledge, the champion can ask members to design and implement their preventative actions. If the team does not have skill, etc., then the champion will request that other areas of the organization design and implement the change(s). • GPS7 works best when the people who use the systems have a hand in the changes. Think in terms of simple, easy, and quick system redesign. Usually, adding more words or instructions only adds to confusion rather than removing it. • GPS7 is best achieved by a face-to-face review between the champion and the team. • Team members should recognize that the champion may have to reflect on their recommendations before being able to suggest or approve changes. • The team should follow the format of observation first and recommendation second in the presentation to the champion. • The team and champion should avoid creating a blame session. • Sometimes the systems that contributed to the root cause are so large that no one individual in an organization or company owns all parts of that system, e.g., cost accounting systems, order entry systems, etc. may require a separate task force designated by the company to redesign them. • The champion and team require cooperation and objectivity from each other and from the rest of the organization to make GPS7 a success. • The champion’s reputation and actions often precede him/her in the GPS7 face-to-face phase. Most teams will only put as much effort into the GPS7 steps as they believe the champion will support. • It is not uncommon for team members to feel threatened at GPS7. Therefore, the champion must communicate, before GPS7, his/her interest in receiving legitimate feedback on systems, practices, and

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procedures. The champion must be careful to not create unrealistic expectations. • The team needs to realize that champions often do not know every detail and consequence of all the systems, practices, and procedures they are expected to manage. This is the team’s opportunity to objectively (not emotionally) point out flaws in day-to-day operations in procedures. • The leader may have to point out that most GPS7 actions make life easier in the workplace, not more difficult. • Use of the “repeated why?” method is an effective technique during GPS7. However, it may also be very frustrating for those trying to answer it. Use it carefully.

GPS8: RECOGNIZE TEAM AND INDIVIDUAL CONTRIBUTIONS (FIGURE 5.11)

FIGURE 5.11 An overview of GPS8.

Background: Purpose. Complete the team experience, sincerely recognize both team and individual contributions, and celebrate.

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Review the common tasks. Tasks that need continual consideration throughout the recognition process: 1. 2. 3. 4. 5. 6.

Document the changes. Review team composition. Review measurable(s). Determine if a service action is required. Review assessment questions. Update the GPS report.

Overview. GPS8 is an important step. At the very least, it is a clear message for everyone to return to their full-time job. GPS8 says to everyone that the task team is now disbanded and no further study of this concern is justified. More importantly, GPS8 is a way for the organization to provide feedback to those who did the work (GPS1 through GPS7). • How the message is conveyed to the team is almost as important as the message itself. It should be fit, focused, and timely. Recognition at GPS includes noting the team’s contribution and efforts at all earlier steps. If recognition is omitted, team members interpret that their contributions were not valued. If the GPS process was not implemented using people, GPS8 could be left out. • Volumes have been written on recognition — the need for it, how to do it, and how to do it well. This brief section is not an attempt to repeat or condense them. A more complete consideration of the subject is well advised, but a few highlights are helpful. Recognition. • • • •

• •

• •

Encourages a repeat of the behavior Works best when it is perceived to be sincere Is expected when extra effort has been expended Does not have to be expensive in dollars, but does require investment in time: time to find out exactly what was done, time to express the recognition (personal time) etc. Is situational Works best when the method of recognition is unique, a bit of a surprise, and considered valuable (Value does not mean money. Often value means personalized, customized.) Recharges a person’s battery (It feels good. People tend to do things for which they are recognized.) Several people can benefit from recognition: team members, the team’s leader, former team members who were released from the team, and the recorder and facilitator (Each provided important services and made contributions. Recognition is a way to encourage future participation. The more effectively recognition is used, the better the organization’s results.)

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Process guidelines. The purpose of GPS8 is to recognize the contributions made to the company or department by the GPS team. • There is a difference between recognition and reward. Either or both can be used at GPS8. • The responsibility for recognition is the champion’s. However, legitimate recognition opportunities exist between team members, the leader and the team, the team and the champion, etc. • Messages about organizational values are communicated to the rest of the organization by the GPS8 action. Failure to conduct GPS8 also sends a message throughout the organization. • Care must be taken that the GPS8 action is commensurate with the contribution of that GPS team. • Care must be taken that the GPS8 action is perceived by the team members to be a legitimate and sincere acknowledgment of their efforts. • It is recommended that variety and creativity be applied in identifying what the champion’s GPS8 actions are. Additional guidelines for success include: – Fit – Focus – Timeliness • The action should fit the contribution of the team. Verbal recognition should focus (accurately describe) the team’s actions and contributions. The actions should also be completed in a timely manner. • Make certain all members, both present and past, are included during GPS8. • GPS8 is an opportunity to encourage certain behaviors and actions. If the team’s contributions at Steps GPS5, GPS6, and GPS7 are not acknowledged, then the team would rightly presume that these actions do not matter. • If GPS8 actions are consistently the same and predictable, the GPS process will be perceived as another lip-service program. Generally speaking, people look for some legitimate, personal involvement of the champion for the GPS step to be interpreted as sincere recognition.

REFERENCES Deming, W. E., Quality, Productivity and Competitive Position. 1982. Boston, MA: MIT. Ford Motor Co., Team Oriented Problem Solving. (September 1987). Dearborn, MI: Ford Motor Company/Power Train. Oliver, L. R. and Springer, M. D., A General Set of Bayesian Attribute Acceptance Plans. American Institute of Industrial Engineers, Technical Papers, 1972. Orsini, J., Simple Rule to Reduce Total Cost of Inspection and Correction of Product in State of Chaos, Doctorate dissertation, Graduate School of Business Administration, New York University, 1982. Ann Arbor, MI: University Microfilms.

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Schafer, R. E., Bayesian Operating Characteristic Curves Reliability and Quality Sampling Plans. Industrial Quality Control, 14:118–122, 1967a. Schafer, R. E., Bayes Single Sampling Plans for Attribute Based on the Posterior Risk. Naval Research Logistical Quarterly, 14:81–88, 1967b.

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Volume I emphasized the importance of project selection and evaluation. This chapter will give an overview of what is required in the project based on 4 weeks of training and considerations from a problem solving perspective. An overview for the DFSS (design for six sigma) problem solving process will also be given.

OVERVIEW The life blood of the six sigma initiative in any organization is the “project.” How one handles this project is the difference between success and failure. So “handling” is obviously the approach that one takes to resolve the problem identified as a “project.” A schematic overview of the project skills is illustrated in Figure 6.1. Specifically, under planning the requirements of the project the focus is on the “what,” “who,” “how,” “when,” and “how much.” “What” is what the customer wants, translated into the project work breakdown structure. The more precise the definition of “what” is, the higher the probability of success will be. “Who” is what the organization is willing to do in allocating resources to complete the project. Another way to understand this is to think of it as a task/responsibility matrix. “How” is the technical requirements that will meet the “what” of the customer. “How” is very critical to the project because if there is no relationship between the project and the customer’s expectations, there will be a mismatch in expectations. “When” defines the schedule of the project using the tools of project management (typical examples may be the networking chart, a Gantt chart, etc.). Finally, “how much” defines the best cost/benefit for every one concerned through estimates of the outcome of each alternative. For uncertain events, concern is with the level of risk involved and whether or not that risk — as identified — is properly executed and documented. Areas of concern are Risk planning, which defines • Requirements • Responsibilities

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Facilitating the project

Production and inventory planning – Resource plans – Production – Scheduling

Uncertain events

Process Improvement

– Forecasting – Mitigating

– Problem solving – Systems learning

FIGURE 6.1 Overview of project skills.

• Operational definitions • Resources • Procedures Risk assessment, which defines • Risk events • Analysis • Update assessments • Document findings Risk handling, which defines • Mitigation tasks for completion • Metrics • Report Risk monitoring, which defines • Metrics • Track status • Report For the process improvement stage, concern is with systems engineering in the sense that the requirements cascade throughout the system. Issues in the “project” phase of the six sigma methodology are • • • • •

Requirement analysis (RA) Functional analysis/allocation (FA) System analysis and control (SA) Verification (V) Synthesis (S)

There is a continuing loop between RA and FA to optimize and understand the requirements. Also, there is a continuing loop between FA and SA, as well as FA and S to make sure there is balance and control, as well as optimum design, respectively. Finally, there is a continuing loop between S and RA to verify the requirements and the design.

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Now that an overview of the process has been given, let us get closer to project expectations. The project’s analysis will follow the 3 weeks of traditional training in any six sigma program. For the first week of training, the participants are expected to come with a preliminary project selected. The following is based on the first week’s input.

FIRST WEEK’S PROJECT: STRUCTURE THE PROJECT — GOALS, OBJECTIVES, AND SCOPE WHAT YOU WILL DO Using established business goals and objectives, determine the • Project objectives • Project route map • Project scope (Document logical, organizational, and deliverable scope.)

WHY YOU WILL DO IT The success of a project is measured by its ability to meet its goals. Clearly defined project objectives should support the organization’s business goals and objectives. That is where the operational definition comes into play. The more exact and precise the definition is, the better the project will be. At this stage, as project definition becomes more defined, identify a route map to establish the deliverables and tasks required to meet the objectives of the project. Defining the scope of a project is somewhat like keeping a tarp down in the wind. The more stakes that are driven into the ground to pin the tarp down, the better. If the scope is clearly defined, success in managing the project and achieving project goals and objectives is more likely.

HOW YOU WILL DO IT Translate established business goals and objectives into project objectives and scope by completing the following: • Review introduction and background materials regarding the project. Define the project objectives that will support some of the business objectives. • Review excerpts from the methodology and make sure that they are feasible. • Logical scope: Identify candidate subprocesses affected by the project. • Organizational scope: Identify the organizations affected as well as the personnel.

WHAT YOU WILL BRING BACK Be prepared to present:

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• Candidate project objectives and the business objectives they support • Logical and organizational scope • “First-cut” deliverables chosen from candidate route map phases Based on project identification, how does problem solving interact? In this phase, focus is on the define portion of the DMAIC model. In other words, identify the customer, define their needs, specify deliverables, identify the CTQs, map the process, and link the CTQs. The tools used in this phase are primarily brainstorming, check sheets, cause-and-effect diagram, histogram, quality function deployment, surveys, focus groups, interviews, process flow chart, cause-and-effect matrix, and correlation.

SECOND WEEK’S PROJECT: STRUCTURE THE PROJECT — PRODUCT-BASED ESTIMATING After the first week’s input, the initial project charter (IPC) was updated to include project goals and objectives and project scope. Also, as a result of investigation, the Y = f(x) has been identified and it is time to proceed with the next assignment which is the measure and analysis phase of the DMAIC model. Now complete the IPC by completing the effort estimate and develop a stage-level project work plan and schedule. In preparation for the estimate, meetings were held with appropriate personnel and issues and concerns were reviewed/discussed. As a result of these sessions, some estimating element assumptions were collected and documented on a project planning worksheet.

WHAT YOU WILL DO • Produce a product-based estimate using estimating templates. • Generate a stage-level work plan in the form of a Gantt chart. • Define the roles necessary to staff the project.

WHY YOU WILL DO

IT

• When structuring a project, take the opportunity to question whether or not the project should be undertaken (make a go/no-go decision). A product level estimate and stage level project work plan help to make the go/no-go decision. • Identify the roles required to force thinking about the skill sets required to complete the project and begin the process of obtaining the commitment of key project staff.

WHAT YOU WILL BRING BACK Be prepared to present a stage-level Gantt chart. The goal is to assess the feasibility of this request and determine the approximate number of full time equivalent (FTE) resources required for the project. Include the following:

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• Total number of hours of effort for each stage • Total number of FTEs for each stage • Recommended alternative ways to structure the project

HOW YOU WILL DO IT 1. Complete the product-based estimating template. Total the effort for each output deliverable. a. Review the project risk factors and determine the potential impact on producing the deliverables. Make adjustments to the product estimates as necessary and determine the total effort for each stage. Be prepared to discuss any assumptions made. Hints to make it go faster: • Make the assumption that the risks areas identified apply to all deliverables in the stage to which that risk area may apply. • Assume that the roles chosen will be applied to all deliverables for each stage. b. Estimate the overall duration (elapsed time) for each stage by adding up the deliverables for each stage and applying the appropriate number of FTEs based on the roles chosen. Adjust the duration of each stage by varying the number of FTE resources assigned. Document the approximate number of FTEs. c. Sketch a rough stage-level Gantt chart for the project showing effort hours, total FTEs, and the numbers of each role for each stage and be prepared to present the results to the champion of the project or anyone else with ownership. 2. Develop an organization chart comprised of the project roles to be included in the project. This does not have to include the name of the person who will be in that role. At this stage, focus is on establishing capability and prioritizing the improvement. Of specific importance are • • • • • • • • • • • •

Select a key product Create a product tree Define a performance variables Create a process map Measure performance variables Establish performance capability analysis Select performance variable Benchmark performance metric Discover best-in-class performance Conduct gap analysis Identify success factors Define performance goals

The tools and techniques used primarily in this phase are benchmarking, process flow chart, gauge repeatability and reproducibility (R & R), gap analysis, product

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tree diagram, FMEA, run charts, check sheets, sampling and stratification, Xbar and R chart, Xbar and S chart, ANOVA, Cp and Cpk, box plots, DPMO based on attribute charts (P, NP, C, and U), Pareto chart, histogram, and scatter plot.

THIRD WEEK’S PROJECT: CONTROL THE PROJECT At this stage of the project, some of the preliminary requirements such as capability issues, goals, priorities, and best-in-class categories have been completed. Now enter into the improvement and control phase of the DMAIC model. Focus in this phase is on identifying causes and implementing controls. Specifically, they are • • • • • • • • • • • •

Select a performance variable Diagnose variable performance Propose causal variables Confirm causal variables Establish operating limits Verify performance improvement Select a causal variable Define control system Validate control system Implement control system Audit control system Monitor performance metrics

WHAT YOU WILL DO Experience will be gained with project management techniques: monitoring activity (determine project status), managing issues (resolve project issues), and managing scope.

WHY YOU WILL DO IT An ongoing project results in many interrelated issues and problems which require close monitoring. It is necessary for the “black belt” (project manager) to have the ability to assess the status of the project and determine the cause of variances from project plans. This person must then choose appropriate alternatives for corrective action where necessary.

HOW YOU WILL DO IT • • • •

Review time schedules. Review the appropriateness of tools and techniques used. Review scope, objective, and results of the project. Review the things gone wrong (TGW) and determine problems with the project (if, indeed, there were any) and identify actions in the form of (1) issues, (2) change requests, and (3) plan adjustments. • Review the things gone right (TGR) and monitor the process.

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WHAT YOU WILL BRING BACK Prepare a brief critical comment (one paragraph) on the current status of the project, including: • Which week appeared to be a turning point? Was it a turning point for better or for worse? • Based only on the data available, what appears to be the underlying problem? How was it solved? • What additional information would be sought to determine if there are other underlying problems? • Establish at least three issues associated with the project and its current status. Also include • A record of the action plans for resolution of each issue. • A brief description of change requests, if any, that are necessary to get the project back on track. The tools and techniques used primarily in this phase are FMEA, project management, brainstorming, payoff matrix, Kano model, DOE, pilot study, validation of baseline capability, process mapping for the “should be” process, cost/benefit analysis, validation study, long-term measurement system analysis (MSA), mistake proofing, control plan reaction plan, and appropriate control charts.

DESIGN FOR SIX SIGMA (DFSS) The process for designing six sigma (see Volume VI) is based on: • • • •

Definition Characterization Optimization Verification

The design process is quite different than the traditional six sigma (SS) approach because the traditional DMAIC model focuses on elimination of the waste, whereas the design for six sigma (DFSS) is focused on prevention of the defect. In the definition stage, the attempt to capture the voice of the customer is made in such a way that selection of Y is critical to the customer and then progressively decomposes the Y to Y1 to y to y1 and so on as well as decomposing the f(X) in a specific definition of X to X1 to x to x1 and so on, until a complete understanding of customer requirements is accomplished and the appropriate operational definitions have been defined. In the characterization stage, two items are emphasized: measure and analysis. In the measure phase, these items are of concern:

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• • • • • •

Select a key product. Create a product tree. Define performance variables. Create process map. Measure performance variables. Establish performance capability.

In the analysis phase, these items are of concern: • • • • • •

Select performance variables. Benchmark performance metric. Discover best-in-class performance. Conduct gap analysis. Identify success factors. Define performance goal.

In the optimization stage, improving and controlling the design are the predominant factors. This is done through a reliability analysis which incorporates the change concept, strengthening the design, performing parameter design experimentation, and ultimately tolerance design experimentation. Perhaps the most crucial point to experimentation in this stage is projection of the most probable point in a given design. This is usually done with response surface methodology (RSM). The optimization stage also calls for a cost/benefit analysis. This is done to be sure that the proposed change is better than the status quo. An approach that may be used for such an analysis is • Layout the proposed process. • Identify where cost was removed from the process and where cost was added if appropriate. • Sum the benefits as well as all the new costs for all alternatives identified. • Report costs/benefits by unit produced or time period (whichever is most appropriate and applicable). It is of the utmost importance for this cost/benefit analysis to be on the same measurable scale as the one initially done when the problem was identified. A good example is identification of DPMO. The way the opportunities are measured may generate a new number, but that number may be different than the original number if the opportunities were not counted in the same way. In the verification stage, verify the design not only to meet customer’s needs, wants, and expectations, but also to verify the integrity of the hardware to meet the design intent. To do this, design appropriate and applicable testing to be sure the design under consideration is “robust” and also to assess the design based on predetermined scoring guidelines. In fact, it is these two characteristics that differ substantially from the traditional six sigma approach. (Remember: six sigma is an approach to problem solving using robustness thinking and statistical problem solving techniques and evaluating process variation and measuring process capability.)

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Typical advanced tools for DFSS are • Monte Carlo simulation • Design of experiments — traditional and Taguchi (L12, L18, L36, and L54 designs) • Axiomatic designs • Analytical reliability and robustness • TRIZ • Trade-off analysis • Pugh analysis In the repertoire of the “black belt,” there are additional tools which are considered to be quite advanced. In this chapter, some of the most common tools were listed; however, their explanation and application will be in Volumes III, V, and VI. Chi square Cross-tabulation tables Regression Performance tolerancing Taguchi’s robustness t-test

Analysis of variance Poison distribution Response surface experiment Full and partial factorials Exponential distribution F test

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Part II Basic Mathematics

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7

The Value of Whole Numbers

Objective: This chapter will demonstrate how to arrange whole numbers according to size. Definitions: 1. Place value: the value of a position in a whole number 2. Whole number: a number of whole objects being counted Rule 1: If two whole numbers have different place values filled, the one having the most places filled is larger. Note: The number in the place value farthest left must not be zero. Example: Which is larger, 143 or 98? Answer: 143 because it has 3 place values filled; and 98 has 2 place values filled. What if the two numbers have the same number of place values filled? Rule 2: If two whole numbers have the same number of place values filled, the number having the largest number in the left-most place value is larger. Example: Which is larger, 701 or 598? Answer: 701 because 7 is larger than 5. Example: Which is larger, 86,943 or 91,043? Answer: 91,043 because 9 is larger than 8. What if the two numbers have the same number of place values filled and the left-most numbers in each are the same? Rule 3: If two numbers have the same number of place values filled and the leftmost numbers are the same, then look at the second number from the left. If the 119

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second numbers are the same, look at the third numbers, and so on, until there is a difference. Example: Which is larger, 913 or 945? Answer: 945 because 4 is larger than 1. Example: Which is larger, 32,143 or 32,598? Answer: 32,598 because 5 is larger than 1. Now, use the rules to arrange a group of numbers by size. Example: Arrange the following numbers by size, with the the largest to the left. Select the correct answer.

(a) (b) (c) (d)

1,345

4,315

3,145

5,314

4,315 3,145 5,314 5,314

1,435 1,345 4,315 4,315

3,145 5,314 3,145 1,345

5,314 4,315 1,345 3,145

Answer: c [(c) and (d) are close until the third position; then 3,145 is larger than 1,345].

EXERCISES Directions: Select the arrangement of numbers that is in order by size, with the largest numbers to the left. If necessary, review the rules and exercises. 1.

(a) (b) (c) (d)

1,763 6,713 7,316 3,176

3,176 3,176 6,713 6,713

6,713 7,316 3,176 7,316

7,316 1,763 1,763 1,763

2.

(a) (b) (c) (d)

3,548 8,345 5,348 4,835

4,835 5,348 3,548 8,345

8,345 4,835 8,345 3,548

5,348 3,548 4,835 5,348

3.

(a) (b) (c) (d)

7,216 7,216 7,216 7,216

6,721 1,762 6,217 2,167

1,762 2,617 2,167 6,127

2,671 6,127 1,672 1,276

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4.

(a) (b) (c) (d)

9,547 9,547 9.547 9,547

4,795 7,549 5,974 7,549

5,974 4,795 4,795 5,974

7,549 5,974 7,549 4,795

5.

(a) (b) (c) (d)

8,241 8,241 8,241 8,241

4,281 4,281 4,281 4,281

1,842 2,148 2,841 1,284

2,841 2,482 1,284 2,481

6.

(a) (b) (c) (d)

9,653 9,536 9,536 9,653

6,953 6,953 6,953 3,953

5,963 6,963 5,693 5,693

3,965 5,965 5,695 3,596

7.

(a) (b) (c) (d)

8,754 8,754 8,754 8,754

7,584 7,854 8,854 7,458

5,874 5,874 5,784 5,478

4,875 4,875 4,875 6,785

8.

(a) (b) (c) (d)

9,321 9.321 9,321 9,321

2,921 3,921 3,921 3,921

3,913 2,391 2,931 1,139

2,932 2,392 1,932 2,293

9.

(a) (b) (c) (d)

85,430 85,430 85,430 85,430

85,403 85,340 85,403 85,340

85,304 85,403 85,340 85,043

85,434 85,043 85,304 85,403

10.

(a) (b) (c) (d)

97,621 97,612 97,621 97,612

97,612 97,621 97,612 97,621

96,721 96,721 96,712 96,217

96,712 96,712 96,721 96,127

2. b 7. b

3. c 8. c

4. d 9. c

5. c 10. a

Answers to the Exercises: 1. c 6. a

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8

Addition and Subtraction of Whole Numbers

Objective: This chapter will demonstrate how to add and subtract whole numbers. Definitions: 1. Applied problems: problems written in words, about real life situations 2. Borrow: a method used when subtracting numbers, where 10 is moved from one place value to the next lower place value 3. Carry: a method used in addition when groups of 10’s are moved from one place value to the next higher place value 4. One-place number: a number having only a number in 1’s place 5. Two-place number: a number having numbers in the 1’s place and the 10’s place 6. Sum: the answer to an addition problem 7. Difference: the answer to a subtraction problem

ADDITION When working with quantities, there are times when quantities are made larger. To keep a record of these changes, it is necessary to know how to add numbers. Rule 1: To add numbers, the units — the tens, the hundreds, and so on, in that order — must be summed: Units

Tens

Hundreds

Thousands

3 +5 -------= 8 (This is the sum.)

12 8 -----20

123 54 --------177

1,245 123 ------------1,368

123

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ADDITION EXERCISES Directions: Select the correct total (sum). If necessary, restudy the instructions and the examples. 1.

2. 3.

4. 5.

6. 7. 8. 9.

10.

7 5 +6 ------(a)16

(b) 18

(c) 12

(d) 22

8+1+3 (a) 23

(b) 12

(c) 14

(d) 10

9 5 +7 ------(a) 13

(b) 18

(c) 27

(d) 21

8+4+7+3 (a) 22

(b) 36

(c) 14

(d) 25

6 5 8 +3 ------(a) 22

(b) 15

(c) 36

(d) 27

3+2+8+7 (a) 30

(b) 24

(c) 17

(d) 20

8+7+3+6+9+1 (a) 39

(b) 34

(c) 27

(d) 43

4+6+0+1+8+3 (a) 41

(b) 18

(c) 22

(d) 36

82 15 + 63 ---------(a) 143

(b) 118

(c) 240

(d) 160

39 + 71 + 34 (a) 235

(b) 206

(c) 218

(d) 144

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11.

12. 13.

14.

15.

125

803 415 + 774 -------------(a) 1936

(b) 2015

(c) 1853

(d) 1992

993 + 163 + 475 (a) 1631

(b) 1805

(c) 2314

(d) 1436

5,413 1,854 8,470 + 6,234 -----------------(a) 43,186

(b) 19,825

(c) 21,971

(d) 63,150

9,345 8,465 1,903 + 2,463 -----------------(a) 18,410

(b) 27,443

(c) 19,318

(d) 22,176

54,643 18,459 70,213 + 94,654 --------------------(a) 237,969

(b) 384,615

(c) 810,341

(d) 454,310

Answers to the addition exercises: 1. b 6. d 11. d

2. b 7. b 12. a

3. d 8. c 13. c

4. a 9. d 14. d

5. a 10. d 15. a

SUBTRACTION This section will demonstrate how to subtract any number from any other number that is the same size or larger. Definition: The number 89 really means 8 tens and 9 ones, while the number 24 means 2 tens and 4 ones. If 24 were subtracted from 89, the process would look like this:

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89 – 24 ----------

8 tens – 2 tens -----------------6 tens

9 ones 4 ones --------------------------5 ones = 65

Rule: A bottom number is subtracted from the one above it which is in the same place value. In other words, the bottom number is smaller than the top number. 75 42 Example: –--------75 – 42 Answer: ---------33 75 39 Now try another problem: –--------Caution: When subtracting, the bottom number must be taken from the number above it. The order is important. In the example above, 9 cannot be subtracted from 5, so borrow from the tens place. Look at the example below. 75 – 39 ----------

7 tens 3 tens -------------6

7 tens 3 tens ---------------tens 6 tens 3 tens -------------3 tens

5 ones 9 ones --------------10

5 ones 9 ones ------------------ones 15 ones 9 ones ------------------ = 36 6 ones

To review, revisit the process: 75 – 39 ----------

532 185 Example: –------------

6

7

10 ""

5

3 9 ------------------

1

6 5 –3 9 --------------36

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2 1

5 3 2 – 1 8 5 One 10 or 10 is borrowed from the 10’s place and added -----------------to the 1’s place. 1

52 2 –1 8 5 ------------------ Now subtract 5 from 12 in the 1’s place. 7 4 1 1

522

– 185 ------------------- One 100 or ten 10s are borrowed from 100’s place and 7 added to the 10’s place. 1 1

4 2 2 –185 ------------------ Now subtract 8 from 12 in the 10’s place 47 1 1

4 2 2 –185 ------------------ Subtract 1 from 4 in the 100’s place. 347 Answer: The final answer is 347. 302 98 Example: –--------2 1 3 02 – 98 -------------

There can be no borrowing from the 10’s place. So borrow one 100 from the 100’s place to get 10 in the 10’s place.

2 1 91

30 2 – 98 ----------------- Now borrow 10 from 10’s place and add to 2 in the 1’s place.

2 1 91

30 2 – 98 ----------------- Now subtract in each column. 204 Answer: The final answer is 204.

APPLICATION OF ADDITION AND SUBTRACTION This section will begin the demonstration of how to find answers to applied mathematics problems using addition or subtraction.

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Directions: The applied problems in this lesson will be done either by adding, subtracting, or both. In most word problems there are terms that indicate whether to add or subtract. Words that mean to add:

sum and total plus increased by combined

Words that mean to subtract: difference decreased by minus take away less There are some suggested steps to solve a word problem: Step Step Step Step Step

1. 2. 3. 4. 5.

Read the problem carefully at least two times. Be sure to understand the question. Be sure to understand all the information given. Draw a picture of the problem if possible. Look for key words that indicate whether to add or subtract.

Example: If one machine made 832 parts and another machine made 750, how many parts were made in all? Answer: Question: How many total parts were made? Given: Parts were made by two different machines. Key Word: “How many … in all” means to add: 832 + 750 = 1,582

EXERCISES Directions: Find the answer to each of the applied problems. If necessary, restudy the instructions and examples. 1. If a man works 4 hours on Monday, 4 hours on Tuesday, 8 hours on Wednesday, 8 hours on Thursday, and 2 hours on Friday, how many hours a week does he work? (a) 20 (b) 22 (c) 26 (d) 18 2. If a man begins a trip with $325 and returns with $94, how much did he spend? (a) $420 (b) $231 (c) $218 (d) $196

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3. If plant A, which produces 125 cars a day, produced 35 cars more than plant B, how many cars does plant B produce? (a) 140 (b) 120 (c) 90 (d) 160 4. Matt’s gas tank can hold 25 gallons. It took 19 gallons to fill it. How many gallons were left in the tank before the refill? (a) 6 (b) 8 (c) 7 (d) 5 5. The number of parts shipped to four cities were 2718, 1845, 5240, and 973. What was the total of parts shipped? (a) 11,208 (b) 10,776 (c) 9843 (d) 11,316 6. A factory produced 485 fewer parts this month than it produced last month. This month the production was 1738 parts. How many parts were produced last month? (a) 2223 (b) 1253 (c) 2433 (d) 1903 7. If a tire costs $54, how much would four tires cost? (a) $258 (b) $216 (c) $89 (d) $325 8. If an inventory showed 12,500 sheets of blue index, what would the inventory show after filling an order of 1700 sheets? (a) 9800 (b) 9250 (c) 10,800 (d) 10,070 9. How many miles did a family travel during a two-week vacation to Florida? The distance to the Florida motel was 1475 miles and while in Florida they drove 573 miles. (a) 2918 (b) 4304 (c) 2813 (d) 3523 10. The mileage reading before a trip was 17,843 and after the trip the reading was 19,307. How many miles were traveled? (a) 1425 (b) 1382 (c) 1464 (d) 1275 Answers to the exercises: 1. c 6. a

2. b 7. b

3. c 8. c

4. a 9. d

5. b 10. c

ADDITIONAL EXERCISES Directions: Select the correct answer to each of the applied problems. 1. A paper counter read 2125 before starting a count. After counting paper for a period of time, the counter read 5089. How many sheets were counted during that period of time? (a) 7815 (b) 2635 (c) 2964 (d) 7214

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2. A truck delivered a load containing 53 cases of towels, 25 cases of toilet tissue, and 80 cases of table cloths. What was the total of pieces delivered? (a) 135 (b) 158 (c) 215 (d) 169 3. Resistors hooked up in a series were 200,000 ohm, 15,000 ohm, 65,000 ohm, and 150,000 ohms. What was the total resistance from the resistors? (a) 385,000 (b) 573,000 (c) 280,000 (d) 430,000 4. A new chair costs $135. The salesman told a customer that if he bought three chairs, the price would be reduced $85. How much would the customer pay for the three chairs? (a) $470 (b) $405 (c) $185 (d) $320 5. How many square yards of carpet are needed in the following rooms? • Two bedrooms, each with 35 square yards • A kitchen with 75 square yards • A living room with 185 square yards • A den with 90 square yards (a) 318 (b) 420 (c) 185 (d) 27 Answers to the additional exercises: 1. c

2. b

3. d

4. d

5. b

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9

Multiplication and Division of Whole Numbers

To add the same number many times would result in a long, difficult addition problem. Multiplication is a shorter process to find the answer to this type of addition problem. Similarly, subtraction of the same number many times would also be a long process. Division is a shorter process used to solve a long subtraction problem. Objective: This chapter demonstrates the operation of multiplication and division of whole numbers. Definitions: 1. Carrying: the method of putting part of a number into the next higher place value 2. Digit: the single symbol used to represent a number which is less than 10 (The ten digits used in this system are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.) 3. Division: a method used to tell how many times a number can be subtracted from a second number 4. Dividend: the number (total) being divided 5. Divisor: the number being divided into the dividend or the number being subtracted from the total 6. Multiples of ten: any number that 10 will divide evenly (no remainder) 7. Multiplication: a method used to add the same number many times 8. Product: the answer to a multiplication problem 9. Quotient: the answer to a division problem

MULTIPLICATION The correct way to multiply two numbers of any size will be demonstrated, beginning with a two-digit number times a two-digit number, e.g., 28 × 15, and then larger

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numbers. When numbers such as 20, 300, and 4000 are multiplied by a 1-digit number, notice the number of 0’s in the answer. 20 ×6 --------120

300 ×6 -----------1800

4000 ×6 ---------------24,000

The same number of 0’s are on the right side of the answer as there are in the number being multiplied. This information will be used in the method of multiplying, shown next. New word: Numbers such as 20, 200, 3000, and 40,000 are called multiples of ten (10). Rule: When a given number is multiplied by a multiple of 10, the steps are 1. Multiply the given number by the non-zero part of the multiple of 10. (The part left of the string of zeros.) 2. Add the same number of zeros to the product as are in the multiple of ten. Example: 4000 × 8 4 × 8 = 32 Now add the (3) 0’s. Answer: 32,000 Example: 50,000 × 3 5 × 3 = 15 Now add the (4) 0’s. Answer: 150,000 Start the method with examples. Example: 48 × 32 ----------1

48 ×2 -------96 Answer:

This really means

48 × 2 + 48 × 30 -----------------------

Now add

96 + 1440 ---------------1536

2

48 × 30 -----------1440

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Example: 146 × 297 --------------

This really means

3 4

146 × 7 plus 146 × 90 plus + 146 × 200 ---------------------------------

4 5

16 ×7 -----------1022

1

1 46 × 200 ---------------29,200

1 46 × 90 ---------------13,140 Now add

1022 13,140 + 29,200 --------------------43,362

Answer: Example: 4532 × 306 --------------

3 1 1

4 5 32 ×6 -----------------27,192

This really means

4532 × 6 plus 4532 × 0 plus 4532 × 300 --------------------------------

Skip 4532 × 0 (it is = to 0 and adds nothing to our answer)

Now add

3

4532 × 306 -----------------------1,359,600 27,192 + 1,359,600 ---------------------------1,386,792

Answer:

MULTIPLICATION EXERCISES Directions: Find the product of each of the problems. If necessary, restudy the examples and instructions. 1.

35 × 17 ----------(a) 385

(b) 715

(c) 485

(d) 595

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2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

83 × 41 ----------(a) 3403

(b) 2893

(c) 4033

(d) 3243

64 × 90 ----------(a) 7130

(b) 5760

(c) 8360

(d) 6160

29 × 55 ----------(a) 1845

(b) 3015

(c) 1595

(d) 2715

81 × 76 ----------(a) 5966

(b) 3816

(c) 6156

(d) 6846

913 × 45 ----------(a) 41,085

(b) 39,145

(c) 40,105

(d) 42,845

305 × 74 ----------(a) 24,030

(b) 28,460

(c) 22,570

(d) 21,140

920 × 19 ----------(a) 16,190

(b) 17,480

(c) 18,340

(d) 14,840

678 × 39 ----------(a) 26,432

(b) 25,842

(c) 26,182

(d) 26,442

137 × 85 ----------(a) 11,335

(b) 11,645

(c) 11,185

(d) 11,415

413 × 813 -------------(a) 335,769

(b) 335,719

(c) 336,149

(d) 335,109

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12.

13.

14.

15.

135

108 × 793 -------------(a) 85,644

(b) 85,144

(c) 87,844

(d) 86,044

538 × 913 -------------(a) 491,134

(b) 491,194

(c) 490,844

(d) 490,044

6709 × 832 -------------(a) 5,581,388

(b) 5,580,848

(c) 5,580,188

(d) 5,581,888

2849 × 754 -------------(a) 2,147,396

(b) 2,148,146

(c) 2,148,186

(d) 2,134,136

Answers to the multiplication exercises: 1. d 6. a 11. a

2. a 7. c 12. a

3. b 8. b 13. b

4. c 9. d 14. d

5. c 10. b 15. b

ADDITIONAL MULTIPLICATION EXERCISES Directions: Find the product of each of the problems. 1.

2.

3.

45 × 47 ----------(a) 1985

(b) 2115

(c) 2045

(d) 1915

81 × 36 ----------(a) 2886

(b) 2846

(c) 2916

(d) 2938

20 × 39 ----------(a) 850

(b) 780

(c) 960

(d) 640

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4.

5.

6.

7.

8.

9.

10.

130 × 24 ----------(a) 1840

(b) 3030

(c) 3120

(d) 2920

339 × 76 ----------(a) 62,414

(b) 25,764

(c) 63,844

(d) 64,444

731 × 46 ----------(a) 31,626

(b) 33,626

(c) 43,626

(d) 38,626

355 × 26 ----------(a) 9130

(b) 10,330

(c) 9030

(d) 9230

3842 × 93 -----------(a) 355,306

(b) 347,306

(c) 357,306

(d) 353, 306

1841 × 345 -------------(a) 635,145

(b) 636,145

(c) 645,145

(d) 640,145

4973 × 844 -------------(a) 4,147,232

(b) 4,197,212

(c) 4,287,212

(d) 4,187,212

Answers to the additional multiplication exercises: 1. b 6. b

2. c 7. d

3. b 8. c

4. c 9. a

5. b 10. b

DIVISION Multiplication was described earlier as an easy way to add the same number many times. Division is an easy way to tell how many times a given number can be subtracted from a total. An explanation of some words and symbols now follows. To divide 5 into 45 means to find how many 5’s can be subtracted from 45.

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Symbol: Symbol:

to divide 5 into 45 to divide 45 by 5

137

5 45 45 ÷ 5 quotient divisor dividend

Dividend: the number being divided into Divisor: the number being divided into the dividend Quotient: the answer to the division problem Very simple division such as 2 6

5 25

7 63

can be done just by knowing the multiplication tables. 3 2 6

because

2×3=6

5 5 25

because

5 × 5 = 25

9 7 63

because

7 × 9 = 63

Sometimes the divisor will not go into the dividend “evenly.” Therefore, use these steps: 1. Divide 2. Multiply 3. Subtract Example 1: Divide: Put the greatest number of 2’s that can be subtracted from 15 in the quotient over the 5. 7 2 15 Multiply: Every time a number is put in the quotient, multiply it by the divisor and write the answer under the part of the dividend that was divided into.

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7 2 15 14 Subtract: The answer to the subtraction is called the remainder. The answer is 7 2 15 14 1

7 R 1 (R is the remainder.)

Now take the next step. Example: 5 243 In this example, as in most, two steps are added to the method.

1. 2. 3. 4. 5.

Divide Multiply Subtract Bring down Repeat

Divide: Divide as many 5’s into the first number as possible; 5 will not go into 2, so divide 5 into 24 four (4) times. 4 5 243 Multiply: Multiply the number put in the quotient (4) by the divisor (5) and place the answer under the number divided into (24). 4 5 243 20 4 Subtract: 4 5 243 –20 4 Bring down: Bring down the next number in the dividend (3) and place it next to the remainder (4) of the subtraction.

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4 5 243 20 43 Repeat the process: Do this by dividing 5 into 43. Divide: 48 5 243 20 43 Multiply: 48 5 243 20 43 40 Subtract: 48 5 243 20 43 –40 3 There is nothing more to bring down; therefore, the answer is 48 R 3: 48 R 3 5 243 This means that 243 equals 5 × 48 plus 3. Check to see: 5 × 48 + 240 + 3 = 243 Example: 7 8139 Divide: 1 7 8139

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Multiply: 1 7 8139 7 Subtract: 1 7 8139 7 1 Bring down: 1 7 8139 7 11 Divide and multiply: 11 7 8139 7 11 7 Subtract: 11 7 8139 7 11 7 4 Bring down: 11 7 8139 7 11 7 43 Divide and multiply:

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116 7 8139 7 11 7 43 Subtract and bring down: 116 7 8139 7 11 7 43 42 19 Divide and multiply: 1162 7 8139 7 11 7 43 42 19 14 Subtract (and give answer): 1162 7 8139 7 11 7 43 42 19 14 5 The answer is 1162 R 5. There are two common errors made in these problems. A short example will demonstrate both types of errors. Example: 6 45

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Divide:

Quotient correct

7 6 45

Quotient too large

8 6 45

6 Quotient too small 6 45 Multiply and subtract:

Correct:

7 6 45 42 3

Too large:

8 6 45 48 Cannot subtract (clue to an error)

Too small:

6 6 45 36 9 9 is bigger than 6 (clue to an error)

EXERCISES Directions: Select the correct answer to each of the problems. If necessary, restudy the instructions and examples. 1. The weight of a freight shipment is 434 pounds. Each of the 7 boxes in the shipment weighs the same amount. How much does each box weigh? (a) 52 lb (b) 62 lb (c) 58 lb (d) 65 lb 2. There are 18 machines stored in a warehouse. Each machine weighs 43 pounds. What is the total weight of all the machines? (a) 774 lb (b) 854 lb (c) 1046 lb (d) 1932 lb 3. The number of parts produced were 218, 793, 452, and 516. How many parts were produced in all? (a) 1676 (b) 1979 (c) 1840 (d) 2138 4. A man drives a route each day which is 37 miles long. How many miles does he travel in a 5-day week on his route? (a) 185 miles (b) 94 miles (c) 135 miles (d) 192 miles

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5. If 1056 parts are produced on one 8-hour shift, how many parts are produced in 1 hour? (a) 138 parts (b) 114 parts (c) 125 parts (d) 132 parts 6. If a man can save 40 cents a day by changing his route to work, how much can he save in 1 month (22 days of work)? (a) 679 cents (b) 435 cents (c) 880 cents (d) 516 cents 7. If cities A, B, and C are on a line, and the distance from A to C is 875 miles and the distance from A to B is 251 miles, how far is it from B to C? A (a) 636 miles

B

(b) 635 miles

C (c) 518 miles

(d) 624 miles

8. How many yards are in 1 mile if there are 5280 feet in 1 mile and 3 feet in 1 yard? (a) 15,840 yards (b) 1760 yards (c) 1184 yards (d) 21,046 yards 9. A large yard needs 2,145 feet of chain-link fence. If the fence comes in 6-foot sections, how many sections are needed for the job (whole sections must be ordered)? (a) 276 sections (b) 358 sections (c) 154 sections (d) 484 sections 10. 678 street-light fixtures were replaced in a city in 1 year. At a cost of $4.00 each, what was the total cost of replacing the street-light fixtures? (a) $2712.00 (b) $169.00 (c) $2138.00 (d) $545.00 Answers to the exercises: 1. b 6. c

2. a 7. d

3. b 8. b

4. a 9. b

5. d 10. a

ADDITIONAL EXERCISES Directions: Select the correct answer to each of the problems. 1. When Pete bought his car, it had 23,015 miles. When he sold the car 4 years later, it had 73,924 miles. How many miles were put on the car when Pete owned it? (a) 50,909 miles (b) 32,185 miles (c) 41,563 miles (d) 68,413 miles 2. What was the total golf score for nine holes if the scores on each hole were 5, 6, 3, 6, 5, 6, 3, and 4? (a) 43 (b) 38 (c) 47 (d) 39

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3. What was the total cost of four tires if each one costs $54.00 and the total tax for all 4 was $9.00? (a) $273.00 (b) $239.00 (c) $184.00 (d) $225.00 4. The Marine Reserve collected 3380 toys to give to 845 children. How many toys would each child get? (a) 6 (b) 3 (c) 1 (d) 4 5. If a man bought a $900 piece of furniture by putting $100 down and paying the rest off in 20 equal payments, how much would each payment be? (a) $35.00 (b) $28.00 (c) $49.00 (d) $40.00 Answers to the additional exercises: 1. a

2. b

3. d

4. d

5. d

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10

Parts and Types of Fractions

Objective: Often it is necessary to work with numbers that are not whole numbers. For instance, a doorway was measured and it was found to be 33 3/4 inches wide. This chapter introduces fractional numbers. Their meaning and different forms of fractions will be demonstrated. When using fractions in problems, fractions must be in one form to multiply, but in a different form when presented as a final answer. This chapter will demonstrate how to recognize different fraction forms and also how to convert a fraction from one form to another. Definitions: 1. 2. 3. 4.

Denominator: the bottom number in a fraction Equal: having the same value or meaning the same amount Fraction: a number between two whole numbers in its value Improper fraction: a fraction whose top number is equal to or larger than the bottom number 5. Mixed number: a whole number with a proper fraction written beside it 6. Proper fraction: a fraction whose top number is smaller than its bottom number 7. Remainder: the number left over after a given number is divided by another number as many times as possible Parts: In applied problems, there are times when a fraction is used to describe a part of a whole quantity. This chapter will demonstrate how to take information about the part of a total and the total and then write a fraction to show the comparison between the two. Examples of the kinds of questions needing an answer will be presented in the next section.

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Example 1: In a whole day of 24 hours, a person slept 7 hours. What fraction of the day did the person sleep? Rule: If amounts are not in the same measurements, rewrite them so they are given in the same measuring units. (The smallest of the two units given.) Step 1. The denominator of the fraction is the number of units in one whole quantity. For Example 1, the denominator is 24. Step 2. The numerator is the number of units in the part of units asked for. In Example 1, the question asked for hours slept which was 7, so the numerator is 7. Caution: Be sure to understand what is asked for. Sometimes the given part is not the numerator. Answer: 7/24. Example 2: A test had 25 questions. If a person missed 4 questions, what fraction of the test did the person get right? Step 1. The denominator is 25. Step 2. The numerator is 21 because if the person missed 4 out of 25, then the person must have gotten 21 right. Answer: 21/25 Example 3: If 12 ounces were lost from a quart bottle, what fraction of the quart is still in the bottle? Rule: Amounts must be in the same units. 12 ounces lost 32 ounces total in 1 quart Step 1. The denominator is 32. Step 2. The numerator is 20 because if 12 ounces are lost 20 ounces remain. Answer: 20/32.

EXERCISES Directions: Select the fraction that gives the comparison of the part asked for to the total. If necessary, restudy the instructions and examples.

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1. If a board was 8 feet long and a carpenter cut off 5 feet, what fraction of the board did the carpenter cut off? (a) 5/8 (b) 3/8 (c) 3/5 (d) 5/3 2. A man worked a total of 24 hours in a work week of 40 hours. What fraction of the whole week did he work? (a) 24/40 (b) 16/24 (c) 24/16 (d) 16/40 3. A football player ran 80 yards for a touchdown. What fraction of the length of the football field did he run? (a) 20/80 (b) 20/100 (c) 80/100 (d) 80/20 4. A man had 100 dollars and spent 47 dollars. What fraction of his money did he spend? (a) 47/53 (b) 47/100 (c) 53/47 (d) 53/100 5. On a 10-question test, a girl missed 3. What fraction of the questions did she get right? (a) 7/10 (b) 7/3 (c) 3/7 (d) 3/10 6. What fraction of a whole day of 24 hours is from 8:00 a.m. to 1:00 p.m.? (a) 6/24 (b) 5/24 (c) 19/24 (d) 4/24 7. A battery was expected to last 3 years, but it lost its charge after 25 months. What fraction of its expected life did the battery last? (a) 25/36 (b) 11/36 (c) 11/25 (d) 25/11 8. A girl earned 200 dollars but 62 dollars was withheld for taxes. What fraction of her paycheck was take-home pay? (a) 138/200 (b) 62/138 (c) 138/62 (d) 168/200 9. A man was given 4 red flags, 3 blue flags, and 3 white flags. What fraction of the flags are blue? (a) 3/10 (b) 6/10 (c) 7/10 (d) 3/7 10. A car sticker price was $68,000. The dealer cost for the car was $42,000. What fraction of the sticker price is the dealer’s mark-up? (a) $26,000/$42,00 (b) $26,000/$68,000 (c) $26,000/$62,000 (d) $42,000/$26,000 Answers to the exercises: 1. a 6. b

2. a 7. a

3. c 8. a

4. b 9. a

5. a 10. b

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ADDITIONAL EXERCISES Directions: Select the fraction that gives a comparison of the part asked for to the total. 1. Out of a total inventory of 98 cars on a dealer’s lot, 38 are Mustangs. Mustangs are what fraction of the total inventory? (a) 60/98 (b) 98/60 (c) 38/98 (d) 38/60 2. One day a paint store had 38 customers and 24 were professional painters. What fraction of the total number of customers were nonprofessional painters? (a) 24/14 (b) 24/38 (c) 14/38 (d) 14/24 3. If a man’s monthly paycheck is 780 dollars and 180 dollars goes for the house payment, what fraction of his paycheck can the man spend for things other than the house payment? (a) 180/780 (b) 180/600 (c) 780/180 (d) 600/780 4. If in one order there were 5 defective parts and 80 good parts, what fraction of the total order is defective? (a) 5/75 (b) 5/85 (c) 80/85 (d) 5/80 5. If 14 quarts of oil are removed from a full 5-gallon can of oil, what fraction of the total oil is removed? (a) 14/20 (b) 5/14 (c) 6/14 (d) 6/20 Answers to the additional exercises: 1. c

2. c

3. d

4. b

5. a

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11

Simple Form and Common Denominators of Fractions

Objectives: This chapter demonstrates two processes related to the number one. Neither multiplying nor dividing a number by one changes the number. This fact allows changing the appearance of a fraction without changing its value. To present an answer to any problem involving a fraction, the numbers in the fraction are reduced as much as possible to make the final answer easier to read. This chapter will also demonstrate how to properly reduce the numbers in a fraction. To add fractions, the denominators must be equal. How to enlarge the numbers in the fraction so that the given fractions may be rewritten with the same denominators will also be demonstrated. Definitions: 1. Common denominator of a set of fractions: the number that can be divided evenly by the denominator of every fraction of the set 2. Equal fractions: fractions having the same value as a number 3. Reduced form of a proper fraction: a proper fraction whose numerator and denominator cannot be evenly divided by any (the same) whole number other than one 4. Simple form of a fraction: a fraction in which the numbers in the fraction (numerator and denominator) are whole numbers and read the same for everyone

SIMPLEST FORM Imagine the confusion for a lumber salesman if a customer asked for a 4/8-inch sheet of plywood instead of calling it 1/2 inch. Whenever there is a problem with a fraction, it is best to have that fraction written in the most simple way. This requires making the numerator and denominator as small as possible. This section will

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demonstrate how to write a proper fraction in its reduced form. Therefore, the task is to take a fraction such as 12/51 and reduce its numbers as much as possible. Rules: To reduce a fraction to its lowest numbers, divide the numerator and denominator evenly by the same number as many times as possible. To use this rule, a number must be found that will divide evenly into both 12 and 51. Rule 1. The number “2” will divide evenly into any number when its far right digit can be divided evenly by 2. Example: 12, 14, 18, and 84 can all be evenly divided by 2 because the far right digits (2, 4, 8, 4) can be evenly divided by 2. Non-example: 19, 37, 41, and 43 cannot be evenly divided by 2 because the far right digits (9, 7, 1, 3) cannot be evenly divided by 2. Rule 2. The number “5” will divide evenly into any number whose last digit is a 0 or a 5. Example: 10, 85, 30, and 45 can all be divided evenly by 5 because the last digits are either 0 or 5. Non-example: 12, 43, 81, and 64 cannot be evenly divided by 5 because the last digits are not 0 or 5. Rule 3. The number “3” will divide into any number when its repeated sum of digits is a number that can be divided by 3. Special note: “Repeated sum of digits” means to add the digits of the given number; to next add the digits of that sum; and so on until only one digit results. Example 1:

Answer:

39 Add digits 3 + 9 = 12 Add digits 1 + 2 = 3 Since 3 can be divided by 3, 39 can be divided by 3.

Example 2: 276 Add digits 2 + 7 + 6 = 15 Add digits 1 + 5 = 6 Answer: Since 6 can be divided by 3, 276 can be divided by 3. Example:

Answer:

85 Add digits 8 + 5 = 13 Add digits 1 + 3 = 4 Since 4 cannot be divided by 3, neither can 85.

Now, go back to the main task of reducing fractions:

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Example 3:

151

4 -----18 4 can be divided by 2, 4 18 can be divided by 2, 3, 6, 9, 18 Notice that both can be divided by 2 4 2 2 ------ ÷ --- = --18 2 9

Answer:

Example 4:

The fraction is “reduced” because no number divides into 2 and 9. 12 -----40 12 can be divided by 2, 3, 4, 6, 12 40 can be divided by 2, 4, 5, 8, 10, 20, 40 Notice that both numbers can be divided by 2 and 4. Choose the larger one. 12 4 3 ------ ÷ --- = -----40 4 10

Answer:

Example 5:

The fraction is “reduced” because no number divides into 3 and 10. 180 --------460

Note: In this problem, because the numbers are so large, several reductions will be done. 180 can be divided by 2, 5, 10 (maybe more) 460 can be divided by 2, 5, 10 180 10 18 --------- ÷ ------ = -----460 10 46 18 can be divided by 2, 3, 6, 9, 18 46 can be divided by 2, 23, 46 18 2 9 ------ ÷ --- = -----46 2 23 Answer:

The fraction is “reduced” because no number divides into 9 and 23.

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SIMPLEST FORM EXERCISES Directions: Select the correct reduced form of each of the proper fractions. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

4/12 (a) 4/12

(b) 2/6

(c) 1/3

(d) 3/6

10/40 (a) 5/8

(b) 1/4

(c) 5/20

(d) 3/12

15/30 (a) 5/6

(b) 3/10

(c) 5/10

(d) 1/2

9/33 (a) 9/33

(b) 3/11

(c) 1/8

(d) 2/22

18/40 (a) 9/20

(b) 9/15

(c) 3/10

(d) 18/40

20/33 (a) 20/33

(b) 5/6

(c) 10/11

(d) 4/8

24/100 (a) 12/50

(b) 6/25

(c) 8/33

(d) 24/100

15/54 (a) 15/54

(b) 3/5

(c) 5/9

(d) 5/18

85/225 (a) 5/9

(b) 17/45

(c) 15/45

(d) 17/50

63/70 (a) 8/10

(b) 7/10

(c) 9/10

(d) 63/70

30/450 (a) 1/15

(b) 8/12

(c) 3/45

(d) 5/90

44/96 (a) 8/15

(b) 2/9

(c) 11/24

(d) 4/18

93/117 (a) 18/23

(b) 93/117

(c) 31/39

(d) 13/37

120/840 (a) 6/42

(b) 1/7

(c) 8/19

(d) 12/84

75/300 (a) 1/4

(b) 4/14

(c) 7/42

(d) 3/12

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Answers to simplest form exercises: 1. c 6. a 11. a

2. b 7. b 12. c

3. d 8. d 13. c

4. b 9. b 14. b

5. a 10. c 15. a

ADDITIONAL EXERCISES Directions: Select the correct reduced form of the proper fractions. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

5/15 (a) 1/5

(b) 5/15

(c) 1/4

(d) 1/3

4/14 (a) 3/8

(b) 2/7

(c) 4/8

(d) 3/7

12/30 (a) 2/5

(b) 3/5

(c) 2/9

(d) 6/15

24/40 (a) 3/5

(b) 7/9

(c) 12/20

(d) 6/10

32/80 (a) 8/40

(b) 6/11

(c) 4/16

(d) 2/5

18/63 (a) 2/7

(b) 8/14

(c) 3/19

(d) 18/63

42/120 (a) 21/40

(b) 7/20

(c) 14/40

(d) 8/25

55/200 (a) 55/200

(b) 15/42

(c) 5/14

(d) 11/40

130/145 (a) 23/24

(b) 13/39

(c)130/145

(d) 26/29

147/300 (a) 38/120

(b) 147/300

(c) 53/95

(d) 49/100

Answers to the additional exercises: 1. d 6. a

2. b 7. b

3. a 8. d

4. a 9. d

5. d 10. d

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COMMON DENOMINATORS Before fractions can be added or subtracted, the denominators of those fractions must be equal. If two fractions with different denominators are to be added, the form of the two fractions must be changed so that the denominators are equal. This section demonstrates the method for changing the numbers in a fraction to obtain an equal fraction, but with a different denominator. Directions: Find the answer to a question: 3 ? --- = -----4 24 Here the fraction 3/4 is rewritten as a fraction with 24 in the denominator. Remember from the section on simplest form that the numbers in a fraction could be reduced by dividing both the numerator and denominator by the same number. It is also safe to increase the numbers in a fraction by multiplying the numerator and denominator by the same number. To write 3/4 as ?/24, multiply the numerator and denominator of 3/4 by 6 because 4 × 6 = 24. 3×6 18 --= -----4×6 24 Example 1: Change 4/9 into ?/18. Multiply numerator and denominator of 4/9 by 2 because 9 × 2 = 18. 4×2 8 --= -----9×2 18 Example 2: Change 2/3 into ?/147. Multiply numerator and denominator of 2/3 by 49 because 3 × 49 = 147. 2 × 49 98 --= --------3 × 49 147 There is also another method that can be used to find the answer. Choose either one. Example 3: Change 3/5 into ?/35. Step 1. Divide the denominator of the old fraction into the denominator of the new fraction. 7 5 35

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Step 2. Multiply the answer of Step 1 by the numerator of the old fraction. 7 × 3 = 21 Step 3. The numerator of the new fraction is the answer to Step 2 or 3/5 = 21/35. Example 4: Change 5/8 into ?/48. 6 Step 1. 8 48 Step 2. 6 × 5 = 30 Step 3. 5/8 = 30/48

COMMON DENOMINATOR EXERCISES Directions: Select the correct number for (x) which is the numerator of the “new fraction.” If necessary, restudy the directions and examples. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

1/2 = x/6 (a) 2

(b) 3

(c) 4

(d) 5

2/5 = x/20 (a) 8

(b) 5

(c) 4

(d) 2

3/8 = x/48 (a) 18

(b) 22

(c) 16

(d) 24

5/7 = x/21 (a) 12

(b) 15

(c) 8

(d) 19

7/8 = x/64 (a) 39

(b) 42

(c) 56

(d) 18

6/10 = x/90 (a) 38

(b) 48

(c) 54

(d) 72

8/9 = x/63 (a) 64

(b) 81

(c) 40

(d) 56

5/12 = x/48 (a) 18

(b) 8

(c) 32

(d) 20

4/6 = x/72 (a) 34

(b) 26

(c) 48

(d) 55

3/4 = x/120 (a) 90

(b) 72

(c) 100

(d) 84

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11. 12. 13. 14. 15.

5/7 = x/49 (a) 9

(b) 24

(c) 18

(d) 35

1/5 = x/45 (a) 9

(b) 11

(c) 7

(d) 4

2/6 = x/24 (a) 8

(b) 6

(c) 12

(d) 15

3/12 = x/144 (a) 24

(b) 36

(c) 33

(d) 39

5/18 = x/54 (a) 18

(b) 10

(c) 20

(d) 15

Answers to the common denominator exercises: 1. b 6. c 11. d

2. a 7. d 12. a

3. a 8. d 13. a

4. b 9. c 14. b

5. c 10. a 15. d

ADDITIONAL EXERCISES Directions: Select the number which is the numerator of the “new fraction.” 1. 2. 3. 4. 5. 6. 7. 8.

1/3 = x/9 (a) 4

(b) 3

(c) 2

(d) 5

4/5 = x/40 (a) 32

(b) 18

(c) 24

(d) 28

3/4 = x/12 (a) 10

(b) 7

(c) 8

(d) 9

5/6 = x/54 (a) 45

(b) 50

(c) 38

(d) 48

2/9 = x/18 (a) 3

(b) 4

(c) 2

(d) 5

3/7 = x/28 (a) 12

(b) 14

(c) 16

(d) 18

3/8 = x/56 (a) 33

(b) 24

(c) 18

(d) 21

5/11 = x/55 (a) 31

(b) 25

(c) 14

(d) 28

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9. 10.

157

4/13 = x/52 (a) 16

(b) 18

(c) 12

(d) 10

9/20 = x/140 (a) 18

(b) 49

(c) 56

(d) 63

4. a 9. a

5. b 10. d

Answers to the additional exercises: 1. b 6. a

2. a 7. d

3. d 8. b

CHANGING FRACTIONS When adding or subtracting fractions, the denominators of all fractions must be equal. If the fractions being added or subtracted do not have equal denominators, they must be rewritten into fractions with equal denominators. Earlier in this section, changing a fraction into a fraction of equal value but with a larger denominator was demonstrated. In this section, rewriting fractions that have different denominators into fractions that have the same denominator will be demonstrated, e.g., rewrite 2/3 and 1/2 so that each have the same denominator. Definition: 1. Common denominator: When fractions are rewritten to have the same denominator, it is said that they have a common denominator. Directions: Step 1. Decide what the common denominator will be. A common denominator can be any number into which all of the given denominators divide evenly. In the case of 2/3 and 1/2, any of the following can be used as a common denominator: 6, 12, 18, 24, 30, 36, 42, etc. Always use the smallest common denominator if it is easy to find. In the example, use 6. Special note: A common denominator (not always the smallest) can be found by multiplying the given denominators together. Step 2. Change each fraction so that it has the common denominator as its denominator. 2/3 = ?/6 2/3 = 4/6

1/2 = ?/6 1/2 = 3/6

Therefore, 2/3 and 1/2 are now 4/6 and 3/6.

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Example 1: Rewrite 3/5 and 4/7 so that each has the same denominator. Step 1. Choose a common denominator: 35 is the smallest number both 5 and 7 will divide into evenly. Step 2. Change each fraction: 3/5 = ?/35 3/5 = 21/35 Answer:

and and

4/7 = ?/35 4/7 = 20/35

3/5 and 4/7 are now 21/35 and 20/35

Example 2: Rewrite 5/12 and 1/6 so that each has the same denominator. Step 1. The common denominator is 12 because both 6 and 12 divide into 12. Step 2. Only one fraction needs to be changed (1/6) to get a common denominator. 1/6 = ?/12 1/6 = 2/12 Answer:

5/12 and 1/6 are now 5/12 and 2/12

Example 3: Rewrite 2/3, 3/4, and 1/8 as fractions with the same denominator. Step 1. The smallest number that 3, 4, and 8 will all divide into is 24, so 24 will be the common denominator. Step 2. Rewrite each fraction: 2/3 = ?/24 2/3 = 16/24 Answer:

3/4 = ?/24 3/4 = 18/24

1/8 = ?/24 1/8 = 3/24

2/3, 3/4, and 1/8 become 16/24, 18/24, and 3/24

CHANGING FRACTIONS EXERCISES Directions: Select the set of fractions that is equal to the given set of fractions, but has a common denominator. 1. 2.

1/3; 1/2 (a) 2/6; 3/6

(b) 3/12; 4/12

(c) 4/6; 3/6

(d) 4/12; 3/12

2/5; 1/6 (a) 7/11; 3/11

(b) 4/30; 8/30

(c) 12/30; 5/30

(d) 4/20; 3/20

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3. 4. 5.

6. 7. 8. 9. 10. 11.

12.

13.

14.

15.

7/10; 3/5 (a) 4/5; 3/5

159

(b) 7/10; 6/10

(c) 13/20; 12/20

(d) 21/30; 19/30

3/5; 4/7 (a) 21/35; 20/35 2/7; 5/8

(b) 10/12; 9/12

(c) 15/32; 11/32

(d) 10/35; 9/35

(a) 16/15; 35/15

(b) 2/15; 5/15

(c) 2/56; 5/56

(d) 16/56; 35/56

4/7; 2/14 (a) 15/28; 4/28

(b) 8/14; 2/14

(c) 16/28; 5/28

(d) 4/7; 5/7

6/11; 3/4 (a) 24/44; 33/44

(b) 17/15; 7/15

(c) 24/11; 33/11

(d) 4/44; 11/44

3/5; 4/9 (a) 8/14; 13/14

(b) 27/45; 20/45

(c) 21/36; 16/36

(d) 18/30; 12/30

3/10; 3/7 (a) 13/17; 10/17

(b) 21/35; 15/35

(c) 21/70; 30/70

(d) 12/40; 15/40

3/13; 2/5 (a) 9/35; 14/35

(b) 15/18; 26/18

(c) 16/18; 7/18

(d) 15/65; 26/65

1/3; 2/5; 1/10 (a) 10/30; 12/30; 3/30

(b) 5/15; 6/15; 2/15

(c) 7/20; 8/20; 2/20

(d) 13/40; 16/40; 4/40

2/3; 5/6; 1/12 (a) 21/24; 20/24; 2/24

(b) 12/18;15/18; 2/18

(c) 8/12;10/12; 1/12

(d) 10/30; 25/30; 2/30

3/8; 1/4; 3/16 (a) 6/16; 4/16; 3/16

(b) 12/32; 9/32; 6/32

(c) 9/24; 6/24; 5/24

(d) 5/16; 4/16; 2/16

2/9; 13/27; 1/3 (a) 8/36; 3/36; 7/36

(b) 6/27;13/27; 9/27

(c) 2/9; 4/9; 3/9

(d) 4/18; 3/18; 6/18

5/8; 2/3; 1/4 (a) 10/16; 9/16; 4/16

(b)18/24; 16/24; 5/24

(c) 10/16;7/16; 3/16

(d) 15/24; 16/24; 6/24

Answers to the changing fractions exercises: 1. a 6. b 11. a

2. c 7. a 12. c

3. b 8. b 13. a

4. a 9. c 14. b

5. d 10. d 15. d

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Objective: Addition and subtraction are two of the four operations used with any type of number. Because fractions are used in so many types of problems, such as measurement problems, it is important to know how to perform these operations. This chapter will demonstrate how to add and subtract fractions in several forms. It will also strengthen skills in reducing fractions and changing them from one form to another. Definitions: 1. Sum: the answer to an addition problem

ADDITION OF FRACTIONS To add fractions, the denominators of the fractions must be equal. This section demonstrates how to add proper fractions which already have common denominators. Rules: Step a. Add the numerators to get the numerator of the answer. Step b. The denominator of the answer is the same as the denominator of the given fractions. Step c. Write the answer in simple form. Example 1: 2/7 + 3/7 = ? Step a. The numerator is 5. Step b. The denominator is 7. Step c. The answer is 5/7. Example 2: 1/8 + 3/8 + 5/8 = ? Step a. The numerator is 9. 161

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Step b. The denominator is 8. Step c. The answer is 9/8. The simple form is 1 1/8. Example 3: 5/10 + 8/10 + 3/10 = ? Step a. The numerator is 16. Step b. The denominator is 10. Step c. The answer is 16/10. The simple form is 1 3/5.

ADDITION OF MIXED NUMBERS This section demonstrates how to add mixed numbers when the proper fraction parts have common denominators. Example:

3 2/7 + 4 3/7 + 1 5/7 = ?

Rules: Step a. Add all the proper fraction parts of the mixed number and write the answer in simple form. Step b. Add all the whole number parts of the mixed numbers. Step c. Add the whole number parts of Step a to the results of Step b. Step d. The final answer is a mixed number which is found by using the whole number from Step c and the proper fraction part of the answer in Step a. Example 1: 3 2/7 + 4 3/7 + 1 5/7 = ? Step a. Add all the proper fractions: 2/7 + 3/7 + 5/7 = 10/7. The simple form is 1 3/7. Step b. Add all the whole number parts of the mixed numbers: 3 + 4 + 1 = 8 Step c. Add the answer of Step b to the whole number part of the answer to Step a: 8 + 1 = 9 Step d. Find the answer: 9 3/7 answer to Step c

proper fraction part of Step a

Example 2: 4 3/8 + 5/8 + 2 4/8 = ? Step a. 3/8 + 5/8 + 4/8 = 12/8. The simple form is 1 1/2. Step b. 4 + 2 = 6 Step c. 6 + 1 = 7 Step d. The answer is 7 1/2

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Example 3: 14 + 3 2/13 + 8 10/13 + 9/13 = ? Step a. 2/13 + 10/13 + 9/13 = 21/13 = 1 8/13 Step b. 14+ 3 + 8 = 25 Step c. 25 + 1 = 26 Step d. The answer is 26 8/13

ADDITION OF FRACTIONS WITH UNEQUAL DENOMINATORS In order to add fractions, the denominators of those fractions must be equal. In the first two sections, how to add fractions whose denominators were already equal was demonstrated. The next two sections will demonstrate how to add fractions whose denominators are not equal. Directions: Step 1. Rewrite as many of the given fractions as necessary so that each given fraction has the same denominator. (If you have forgotten how to do this; restudy Mod 6, lesson 1). Step 2. Add the fractions and write the answers in simple form. (If you have forgotten how to do this; restudy Mod 6, lesson 1). Example 1: 2/3 + 3/4 = ? Step 1. 12 is the common denominator. 2/3 is equal to 8/12. 3/4 is equal to 9/12. Step 2. 8/12 + 9/12 = 17/12. The simple form is 1 5/12. Example 2: 7/10 + 3/8 = ? Step 1. 40 is the common denominator. 7/10 is equal to 28/40. 3/8 is equal to 15/40. Step 2. 28/40 + 15/40 = 43/40. The simple form is 1 3/40. Example 3: 5/8 + 3/4 + 7/16 = ? Step 1. 16 is the common denominator. 5/8 is equal to 10/16. 3/4 is equal to 12/16. 7/16 is equal to 7/16. Step 2. 10/16 + 12/16 + 7/16 = 29/16. The simple form is 1 13/16.

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SUM OF A GROUP OF MIXED AND WHOLE NUMBERS To add mixed numbers, the denominators of the proper fraction parts must be equal. This section will demonstrate how to find the sum of a group of mixed and whole numbers when the denominators of the proper fraction part of the mixed number are different. Directions: Step 1. Rewrite the proper fraction part of as many mixed numbers as necessary so that each given mixed number has the same denominator in the proper fraction part. (Review previous sections if necessary.) Step 2. Add the whole and mixed numbers, and write answers in simple form. (Review previous sections if necessary.) Example 1: 5 1/2 + 7 2/3 = ? Step 1. 6 is the common denominator for 1/2 and 2/3. 5 1/2 is equal to 5 3/6. 7 2/3 is equal to 7 4/6. Step 2. 5 3/6 + 7 4/6 = 12 7/6. The simple form is 13 1/6. Example 2: 1 3/8 + 4 1/4 = ? Step 1. 8 is the common denominator for 3/8 and 1/4. 1 3/8 is equal to 1 3/8. 4 1/4 equal to 4 2/8. Step 2. 1 3/8 + 4 2/8 = 5 5/8. The simple form is 5 5/8. Example 3: 5 2/3 + 4 3/8 + 9 5/12 = ? Step 1. 24 is the common denominator for 2/3, 3/8, and 5/12. 5 2/3 is equal to 5 16/24. 4 3/8 is equal to 4 9/24. 9 5/12 is equal to 9 10/24. Step 2. 5 16/24 + 4 9/24 + 9 10/24 = 18 35/24. The simple form is 19 11/24.

APPLIED MATH PROBLEMS USING ADDITION OF FRACTIONS This section contains applied math problems using addition of fractions. The problems are written in words. One of the difficulties encountered is changing the words in applied problems into numbers.

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Directions: Words that help decide if a problem can be solved by adding are sum, total, and plus. Suggested steps to solve math problems given in words or pictures are Step 1. Read the problem carefully at least two times. Step 2. Be sure to understand what is to be found. Step 3. Be sure to understand all the information given. Step 4. Draw a picture of the problem if possible. Step 5. Look for key words that tell what mathematics operation is needed to find the answer. Example:

A chef mixed 1/3 cup of sugar and 1/4 cup of flour. In cups, how much mixture did the chef have? Question: Given: Key words: Operation:

Answer:

How many cups of mixture of sugar and flour? 1/3 cup sugar and 1/4 cup flour “Mixture” meaning the combination of 2 things Add 1/3 + 1/4 is 4/12 + 3/12 = 7/12 7/12 cup mixture.

EXERCISES Directions: Select the correct answer, written in simple form, to each of the addition problems. 1. A certain pipe is 3 3/8 inches long. How long would 2 of these pipes be if lying end to end? (a) 6 3/4 in. (b) 5 1/3 in. (c) 3 3/4 in. (d) 6 3/8 in. 2. George mixed 3/10 of a gallon of oil with 4 1/2 gallons of gasoline. How many gallons of liquid mixture does he have? (a) 4 4/5 gal (b) 4 1/5 gal (c) 4 3/4 gal (d) 4 3/8 gal 3. One piece of metal weighs 5/6 of a pound and a second piece of metal weighs 3/8 of a pound. What is the total weight of the two pieces of metal? (a) 9/20 lb (b) 1 5/24 lb (c) 1 3/8 lb (d) 4/7 lb 4. What would the total thickness of wall covering be if the wallboard was 1/2 inch thick and the paneling was 3/8 inch thick? (a) 1/4 in. (b) 2/5 in. (c) 1/2 in. (d) 7/8 in. 5. What would be the total distance around a triangle with sides of: 4 1/2, 6 3/4, and 8 3/8 inches. (a) 18 13/24 in. (b) 18 7/8 in. (c) 19 5/8 in. (d) 19 1/8 in.

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6. One piece of metal weighs 5 3/10 ounces and a second piece of metal weighs 13 9/10 ounces. What is the combined weight of the two pieces of metal? (a) 19 1/5 oz (b) 18 2/5 oz (c) 18 3/5 oz (d) 19 3/8 oz 7. What is the total horsepower delivered by three motors of 3/4, 1/2, and 1/3 horsepower if all are working together (in parallel)? (a) 5/9 hp (b) 1 3/5 hp (c) 2 1/12 hp (d) 1 7/12 hp 8. What length of a bolt is covered by a washer of 1/16 inch width and a nut of 1/4 inch width? (a) 1/10 in. (b) 3/8 in. (c) 5/16 in. (d) 7/32 in. 9. What is the total length of the boxes shown in the illustration?

(a) 1 7/8 in.

(b) 1 11/24 in.

(c) 2 3/8 in.

(d) 1 11/32 in.

10. A brush 27/32 inch thick is covered with a 3/64 inch coating of copper on both sides. What is the total thickness of the brush and the coatings? (a) 15/16 in. (b) 17/64 in. (c) 39/64 in. (d) 57/64 in. Answers to the exercises: 1. a 6. a

2. a 7. d

3. b 8. c

4. d 9. c

5. c 10. a

ADDITIONAL EXERCISES Directions: Select the correct answer, written in simple form, to each of the word problems. 1. Two packages weigh 3 2/5 ounces and 15 1/2 ounces. What is their combined weight? (a) 18 7/10 oz (b) 18 3/7 oz (c) 18 3/10 oz (d) 18 9/10 oz 2. What is the distance around a rectangular lot 50 1/2 feet wide and 9 3/4 feet long? (There are two widths and two lengths.) (a) 59 3/8 ft (b) 120 1/2 ft (c) 60 1/2 ft (d) 118 3/4 ft

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3. The height of a door is 6 3/8 feet, and from the top of the door to the ceiling, it is 2 1/2 feet. How far from the floor is the ceiling? (a) 8 1/2 ft (b) 8 7/8 ft (c) 8 3/4 ft (d) 8 7/16 ft 4. How many total gallons can be carried in three cans that hold 5 gallons, 4 2/3 gallons, and 2 1/2 gallons? (a) 11 7/6 gal (b) 11 7/12 gal (c) 12 1/6 gal (d) 12 3/8 gal 5. How many feet of electrical cord are needed to make a lamp if 1 1/8 feet of cord are inside the lamp and 6 1/3 feet of cord are outside the lamp? (a) 7 11/24 ft (b) 7 5/24 ft (c) 7 2/11 ft (d) 8 1/3 ft 6. How many total hours were worked if on the first day a man worked 6 2/3 hours and on the second day he worked 4 3/10 hours? (a) 10 29/60 hr (b) 11 1/30 hr (c) 11 1/2 hr (d) 10 29/30 hr 7. On the first day, a stock rose 1 3/8 points, on the second day it rose 1/4 points, and on the third day it rose 4 1/2 points. What was the total rise in points in the three days? (a) 6 1/8 points (b) 6 1/4 points (c) 5 5/14 points (d) 5 3/8 points 8. What would be the total weight of two parts, one weighing 4 5/6 pounds and the other weighing 14 7/8 pounds? (a) 18 6/7 lb (b) 19 3/8 lb (c) 12/14 lb (d) 19 17/24 9. A carpenter has three boards that measure 7 7/32 inches, 4 5/16 inches, and 5 3/8 inches. What is the total length of all three boards? (a) 18 3/32 in. (b) 17 9/32 in. (c) 16 29/32 in. (d) 16 7/32 in. 10. Find the distance between the arrows shown in the figure:

2 3/16” 2 1/8”

(a) 4 5/16 in.

(b) 4 1/2 in.

(c) 4 1/6 in.

(d) 4 3/8 in.

Answers to the additional exercises: 1. d 6. d

2. b 7. a

3. b 8. d

4. c 9. c

5. a 10. a

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SUBTRACTION OF FRACTIONS Subtraction is one of the four operations used with any type of number. Subtraction of fractions is a very common operation in any group of problems with measurements. Objective: The rules and requirements for subtracting fractions are very much the same as for addition of fractions. Common denominators are needed. This section demonstrates how to subtract one proper fraction from another when both have the same denominator. Directions: An example of the type of problem in this lesson is 10/15 – 3/15 (This problem could also be written as 10/15 – 3/15.) --------------Notice two things: 1. The denominators are the same. 2. The top fraction’s numerator is larger than the numerator of the bottom fraction. Steps: For both ways to write the subtraction problem: Step 1. Subtract the numerator of the bottom (or the right) fraction from the numerator of the top (or the left) fraction to get the numerator of the answer. Step 2. The denominator in the answer is the same as the common denominator of the two given proper fractions Step 3. Write the answer in simple form. Example 1: Done both ways.

10/15 – 3/15

or

10/15 – 3/15 ---------------

Step 1. The numerator is 7. 10 –3 ------7

or

Step 2. The denominator is 15. Step 3. The answer is 7/15. Example 2: 13/20 – 7/20 = ?

10 – 3 = 7

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Step 1. The numerator is 6. 13 – 7 = 6 Step 2. The denominator is 20. 6/20 Step 3. The simple form is 3/10.

Example 3:

20/35 – 10/35 ------------------

Step 1. The numerator is 10. 20 – 10 ---------Step 2. The denominator is 35. Step 3. The answer is 10/35. The simple form is 2/7.

SUBTRACTION OF MIXED NUMBERS In subtracting mixed numbers, as in adding mixed numbers, the fractions must have common denominators. Be careful in subtraction, unlike addition, to subtract the bottom or right fraction from the top or left fraction. In subtraction, order counts. This presents a special problem when subtracting fractions, solved by using a fancy type of “borrowing.” Directions: This is a complicated concept. The discussion will be divided into two parts.

PART 1 An example is

10 4/7 – 3 1/7

or

10 4/7 – 3 1/7 -----------------

Steps: Step 1. Subtract the right (or bottom) proper fraction from the left proper fraction (or top).

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Step 2. Subtract the right (or bottom) whole number from the left whole number (or top). Step 3. Write the answer in simple form. Example 1: 10 4/7 – 3 1/7 = ? Step 1. 4/7 – 1/7 = 3/7 Step 2. 10 – 3 = 7 Step 3. The answer is 7 3/7 in simple form.

Example 2:

12 9/10 – 4 3/10 --------------------

Step 1. 9/10 – 3/10 = 6/10 Step 2. 12 – 4 = 8 Step 3. The answer is 8 6/10. The simple form is 8 3/5.

PART 2 In Part 1, it was always possible to subtract the bottom proper fraction from the top proper fraction. In Part 2, it is not possible to subtract until a special type of borrowing takes place, e.g., it is not possible to subtract the bottom proper fraction from the top: 10 1/7 – 8 4/7 ----------------Special caution: Because the order of subtraction is important, never subtract the top numerator from the bottom number. The above example will be used to demonstrate how to borrow from the whole number 10 to make the proper fraction on top big enough to subtract 4/7 from it. 10 1/7 means 10 + 1/7 10 1/7 same as 9 + 1 + 1/7 10 1/7 same as 9 + 1 1/7

10 1/7 same as 9 + 8/7 10 1/7 same as 9 8/7

Meaning of mixed number. Take a 1 from 10. Write mixed numbers. Change to improper fraction. Write as mixed number.

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So the problem is

10 1/7 – 8 4/7 -----------------

becomes

9 8/7 – 8 4/7 -----------------

which can be solved: 9 8/7 – 8 4/7 ----------------1 4/7 Steps: Step 1. Take a 1 away from the whole number. Step 2. The 1 joins the proper fraction to form a mixed number. Step 3. The mixed number is changed to an improper fraction. Step 4. The original whole number is rejoined to the improper fraction to make a mixed number. Step 5. Subtract as in Part 1.

Example 1: Step Step Step Step Step

1. 2. 3. 4. 5.

5 3/8 – 3 7/8 ----------------5 3/8 same 5 3/8 same 5 3/8 same 5 3/8 same Subtract.

as as as as

4 4 4 4

+ 1 + 3/8 + 1 3/8 + 11/8 11/8

4 11/8 – 3 7/8 ----------------1 4/8 (The simple form is 1 1/2.)

Example 2:

12 4/10 – 8 7/10 --------------------

Step 1. 12 4/10 same as 11 + 1 + 4/10 Step 2. 12 4/10 same as 11 + 1 4/10 Step 3. 12 4/10 same as 11 + 14/10 Step 4. 12 4/10 same as 11 14/10 Step 5. Subtract

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11 14/10 – 8 7/10 --------------------3 7/10 (The simple form is 3 7/10.)

Example 3:

14 – 3 3/8 -----------------

Step 1. 14 same as 13 + 1 Step 2. 14 same as 13 + 8/8 Step 3. 14 same as 13 8/8 Step 4. Subtract 13 8/8 – 3 3/8 ---------------10 5/8 (The simple form is 10 5/8.)

SUBTRACTION OF FRACTIONS OF ALL KINDS This section will demonstrate how to subtract fractions of all types with different denominators. This section is a combination of items already mastered. Directions: Step 1. If necessary, rewrite fractions to have common denominators. Step 2. Subtract and give the simple form of answer; the steps are brief because some rules and steps learned earlier have been combined. Several examples follow.

Example 1.

9/10 – 3/4 ---------------

Step 1. The common denominator is 20. 9/10 same as 18/20. 3/4 same as 15/20. 18/20 – 15/20 Step 2. -----------------3/20 (The simple form is 3/20.)

Example 2.

5 6/9 – 1/3 ------------

Step 1. The common denominator is 9.

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5 6/9 stays 5 6/9. 1/3 same as 3/9. 5 6/9 – 3/9 Step 2. -----------5 3/9 (The simple form is 5 1/3.)

Example 3.

9 3/7 – 1 2/3 -----------------

Step 1. The common denominator is 21. 9 3/7 same as 9 9/21. 1 2/3 same as 1 14/21.

Step 2.

9 9/21 – 1 14/21 ----------------------8 30/21 – 1 14/21 ----------------------7 16/21 (The simple form is 7 16/21.)

Example 4. 12 – 4 3/8 = ? Step 1. The common denominator is 8. 12 – 4 3/8 Step 2. -----------------

11 8/8 Borrowed 1 from 12 to make 8/8 – 4 3/8 ----------------7 5/8 (The simple form is 7 5/8.)

ADDITION AND SUBTRACTION OF FRACTIONS This section demonstrates applied math problems with addition and subtraction of fractions. Addition and subtraction are mixed in this section to gain practice, not only in subtracting fractions, but also in knowing when to subtract. Directions: The biggest problem in solving applied problems is to be able to read either the given words or labeled picture and to set up the math problem correctly. A word or two in a word problem usually indicates whether to add or subtract. Words that mean to add: sum total and plus more increased by

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Words that mean to subtract: minus less take away difference remove reduced by Some suggested steps to solve a math problem given in words or labeled pictures are: Step 1. Read the problem carefully at least two times. Step 2. Be sure to understand what is to be found. Step 3. Be sure to understand all the information given. Step 4. Draw a picture of the problem if possible. Step 5. Look for key words that tell what mathematics operation is needed to find the answer. Example: If a board 35 3/4 inches long is cut from an 8-foot board, how many inches of board are left? Step 1. Read. Step 2. Find the number of inches left on a board. Step 3. The whole board is 8 feet; 35 3/4 inches is to be cut off. Step 4. Drawing with key words. Cut off

Amount left

35 3/4 in.

? 8 ft

Solution: Subtract: 8 feet – 35 3/4 inches (First rewrite 8 feet as inches, 8 × 12 = 96.) 96 – 35 3/4 -------------------95 4/4 – 35 3/4 -------------------60 1/4 Answer: The simple form is 60 1/4 inches.

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EXERCISES Directions: Select the correct answer to each of the problems. If necessary, restudy the instructions and examples. 1. At 8:00 a.m. Wednesday, the water level was 48 3/4 inches. By 10:00 a.m., it had dropped to 32 1/2 inches. How far did the water level drop in inches? (a) 16 2/4 (b) 16 1/2 (c) 16 1/4 (d) 15 3/4 2. A new motor brush was 1 5/8 inches long. Find the length of the brush after it had been worn down by 9/16 inches. (a) 1 1/2 in. (b) 1 4/8 in. (c) 1 1/16 in. (d) 1 1/8 in. 3. How long does a piece of drill rod (a rod from which a drill is made) need to be to make 2 drill bits if each drill bit is 1 7/8 inches long? (a) 3 1/4 in. (b) 2 14/16 in. (c) 3 3/4 in. (d) 2 7/8 in. 4. A full gas tank will hold 20 3/10 gallons. If 13 9/10 gallons was needed to refill the tank after a trip, how much gas was burned on the trip? (a) 7 1/10 gal (b) 6 2/5 gal (c) 6 1/10 gal (d) 7 3/10 gal 5. A woman worked 2 1/4 hours in the morning and 3 4/5 hours in the afternoon. How long, in total, did she work that day? (a) 5 5/9 hr (b) 6 1/20 hr (c) 5 3/4 (d) 6 3/8 hr 6. A door was 31 3/8 inches wide. A carpenter planed off 11/16 inches. How wide is the door now? (a) 30 11/16 in. (b) 31 8/16 in. (c) 30 5/8 in. (d) 31 1/2 in. 7. If it is 4 5/8 miles from the beach to a house and if the trip were made three times, how many miles were traveled? (a) 13 7/8 mi (b) 12 5/8 mi (c) 13 1/8 mi (d) 12 15/24 mi 8. Sue had 7/12 of a yard of cloth before using 1/6 of a yard. How much cloth (in yards) does she have left? (a) 2/12 yards (b) 1 yard (c) 5/12 yards (d) 1/2 yards 9. A contractor will add a 42 1/3-foot section of stockade fence to his present fence. If the present fence is 72 3/4 feet long, how long will it be with the addition? (a) 114 13/24 ft (b) 115 1/12 ft (c) 114 8/12 ft (d) 114 2/3 ft 10. A section of insulated telephone cable 45 9/10 feet long was cut from a 1,000-foot reel. How much cable is left on the reel? (a) 954 1/10 ft (b) 954 9/10 ft (c) 955 9/10 ft (d) 145 9/10 ft

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Answers to the exercises: 1. c 6. a

2. c 7. a

3. c 8. c

4. b 9. b

5. b 10. a

ADDITIONAL EXERCISES Directions: Select the correct answer, written in simple form, to each of the applied problems. 1. The figure shows an end view of a piece of steel cut from a strip of given width and bent to make a U-bracket. How long does a piece of steel need to be to make this bracket? Add 3/16 inch for each bend to the measures already given. • How long does a piece of steel need to be to make the bracket? • Add 3/16 inch for each bend to the measures already given. (a) 21 5/8 in.

(b) 21 3/4 in.

(c) 20 13/48 in.

(d) 21 10/16 in.

2. A piece of plywood 80 3/16 inches long must be trimmed by 9 3/4 inches to fit between two shelf supports. What will be the finished length? (a) 70 7/16 in. (b) 71 3/4 in. (c) 71 7/16 in. (d) 70 in. 3. If 17 7/20 gallons were drained from a full 50-gallon tank of fuel oil, how much oil would still be in the tank? (a) 33 7/20 gal (b) 32 3/8 gal (c) 33 gal (d) 32 13/20 gal 4. Find the length of pipe needed to fill a 27 1/4-inch space if the pipe must also have a 2 3/8-inch overlap on each end. (a) 22 1/2 in. (b) 31 8/24 in. (c) 32 3/4 in. (d) 31 1/3 in. 5. A copper contact was insulated with the following: one 5/8-inch piece of mica, one piece of 1/4 inch fiber, and one piece of 1/16-inch press board. What was the total thickness of the insulation? (a) 1/4 in. (b) 3/8 in. (c) 15/16 in. (d) 7/28 in. 6. A piece of metal 7/16 inches thick was ground down to a thickness of 3/32 inch. In inches, how much metal was ground off of the 7/16-inch piece? (a) 1/2 in. (b) 4/16 in. (c) 11/32 in. (d) 1/4 in.

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7. What was the loss in value of a stock that opened at 37 points and closed at 35 5/8 points? (a) 2 1/8 points (b) 2 3/8 points (c) 1 5/8 points (d) 1 3/8 points 8. A shelf has to be built to allow room for three boxes of the following lengths: 27 1/2 inches, 15 3/8 inches, and 17 1/16 inches. How wide must the inside dimension of the shelf be? (a) 60 1/8 in. (b) 59 5/16 in. (c) 59 5/26 in. (d) 59 15/16 in. 9. An electronics kit includes a wire 17 1/2 feet long. If the first project in the kit calls for 2 7/8 feet, how much is left for the other project? (a) 14 7/8 ft (b) 14 5/8 ft (c) 15 1/8 ft (d) 15 ft 10. What is the inside diameter (distance across a circle) of the conduit in the illustration if the outside diameter is 2 3/8 inches and the walls are 3/16 inches thick?

(a) 2 1/4 in.

(b) 2 3/8 in.

(c) 1 7/8 in.

(d) 2 in.

Answers to the additional exercises: 1. b 6. c

2. a 7. d

3. d 8. d

4. a 9. b

5. c 10. d

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13

Multiplication and Division of Fractions

Objective: Multiplication and division in any number system are operations used to do addition and subtraction more easily. To perform multiplication and division of fractions, the fractions do not have to be rewritten to have common denominators. However, some form changes have to be made. This chapter demonstrates how to multiply and divide fractions and how to discover when these operations are called for in applied problems. Definitions: 1. Of: “of” between a fraction and another number indicates to multiply 2. Over: indicates to divide

MULTIPLICATION OF PROPER AND IMPROPER FRACTIONS When adding and subtracting fractions, the fractions must have common denominators. This is not true when multiplying fractions. However, fractions must be in proper or improper form before they can be multiplied. This section will demonstrate how to multiply proper and improper fractions. As always, the answer to any fraction problem must be written in simple form. Directions: Step 1. The numerator of the answer is found by multiplying the numerator of the given fractions. Step 2. The denominator of the answer is found by multiplying the denominator of the given fraction. Step 3. The answer is written in simple form. Some examples are

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Example 1: 3/7 × 4/5 = ? Step 1. 3 × 4 = 12 Step 2. 7 × 5 = 35 Step 3. The answer is 12/35; the simple form is 12/35. Example 2: 10/12 × 4/5 = ? Step 1. 10 × 4 = 40 Step 2. 12 × 5 = 60 Step 3. The answer is 40/60; the simple form is 2/3. Short-cut to reducing the answer: There is a short-cut that can help obtain the simple form of the answer without doing a large reducing step after multiplying numerators and denominators. Before the numerators and denominators are multiplied, reduce the numerator of any of the fractions and any of the denominators as many times as possible. Repeat Example 2 using the short-cut: 10/12 × 4/5 = ? Short-cut:

102/12 × 4/51 Reduce 10/5 to 2/1 by 5 102/123 × 14/51 Reduce 4/12 to 1/3 by 4 Now 10/12 × 4/5 has become 2/3 × 1/1

Answer: 2/3. Note: If the short-cut is used, the answer after multiplying numerators and denominators will be in the simple form. No more reducing will be necessary. Example 3: 8/10 × 9/50 = ? 8/10 × 9/5025 82/105 × 9/25 2/5 × 9/25

Short-cut:

4

Answer:

18/125 (The answer is in simple form.)

Reduce by 2 Reduce by 2

Example 4: 15/8 × 10/3 = ? Short-cut:

155/8 × 10/31 5/84 × 105/31 5/4 × 5/1

Answer:

25/4 (The simple form is 6 1/4.)

Reduce by 3 Reduce by 2

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MULTIPLICATION OF MIXED NUMBERS This section will demonstrate how to multiply fractions when the fractions are given in the form of mixed numbers and how to multiply fractions given in any form by whole numbers. Directions: Four examples will be used in this lesson: Example Example Example Example

1: 2: 3: 4:

2 5 4 6

1/4 × 3/8 3/5 × 8 4/7 × 9/10 5/8 × 10

Note: When multiplying with mixed numbers, first change the mixed numbers into improper fractions and multiply. Example 1: 2 1/4 × 3/8 = ? Step 1. Rewrite and change 2 1/4 to an improper fraction. Step 2. 9/4 × 3/8 = 27/32 Step 3. The simple form is 27/32. Example 2: 5 3/5 × 8 4/7 = ? Step 1. Rewrite, changing mixed numbers to improper fractions. 28/5 × 60/7 Step 2. Reduce, using the short-cut. 28/51 × 6012/7 Reduce by 5 284/51 × 6012/71 Reduce by 7 Step 3. 4/1 × 12/1 = 48/1 (The simple form is 48.) Note: To multiply fractions by whole numbers, rewrite the whole number with the given whole number as the numerator and 1 as the denominator; then multiply. Example 3: 4 × 9/10 = ? Step 1. Rewrite 4 as a fraction. 4/1 × 9/10 Step 2. Reduce, using the short-cut. Reduce by 2 22/1 × 9/105

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Step 3. 2/1 × 9/5 = 18/5 (The simple form is 3 3/5.) Example 4. 6 5/8 × 10 = ? Step 1. Rewrite 10 as a fraction and 6 5/8 as an improper fraction. 53/8 × 10/1 Step 2. Reduce, using the short-cut. 53/84 × 105/1

Reduce by 2

Step 3. 53/4 × 5/1 = 265/4 (The simple form is 66 1/4.) Example 5. 1 3/8 × 6 × 9/10 = ? Step 1. Rewrite and change 1 3/8 to improper fraction and change 6 to a fraction. 11/8 × 6/1 × 9/10 Step 2. Reduce, using the short-cut. 11/84 × 63/1

Reduce by 2

Step 3. 11/4 × 3/1 × 9/10 = 297/40 (The simple form is 7 17/40.)

DIVISION OF FRACTIONS Dividing fractions, like multiplying, does not require the fractions to have common denominators. In this section, division of fractions is by making several changes in the way the problem is written and then by multiplying. Directions: Break the method into Part 1 and Part 2. Part 1 will be division of proper and improper fractions, e.g., • 9/10 ÷ 3/4 • 3/8 ÷ 1/4 • 8/20 ÷ 2/3 Part 2 will be division of mixed numbers, whole numbers, and other fractions, e.g., • 4 1/2 ÷ 3/4 • 10 3/8 ÷ 3

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• 15 ÷ 2 1/3 • 9 3/5 ÷ 2 1/2 Before beginning, there must be a clear meaning for the symbol “÷” — “÷” indicates divided by. Example 1: 10 ÷ 5 means 10 divided by 5 or 5 into 10 or 5 10 Example 2: 3/4 ÷ 1/2 means 3/4 divided by 1/2 Note: The number left of ÷ is the dividend (or the left number) and the number right of ÷ is the divisor (or the right number) . Therefore, in Example 2, 3/4 is the dividend and 1/2 is the divisor. Part 1: To divide proper or improper fractions: Step 1. Rewrite the division problem and turn the right fraction up-side-down; change the ÷ symbol into ×. Step 2. Multiply the fractions. Step 3. Write the answer in simple form. Some more examples will now be provided. Example 1: 9/10 ÷ 3/4 = ? Step 1. 9/10 × 4/3 Step 2. 93/105 × 42/31 3/5 × 2/1 = 6/5

Reduce by 3 and 2

Step 3. The simple form is 1 1/5. Example 2: 3/8 ÷ 1/4 = ? Step 1. 3/8 × 4/1 Step 2. 3/82 × 41/1 3/2 × 1/1 = 3/2 Step 3. The simple form is 1 1/2. Example 3: 8/20 ÷ 2/3 = ?

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Step 1. 8/20 × 3/2 Step 2. 84/20 × 3/21 41/205 × 3/1 1/5 × 3/1 = 3/5 Step 3. The simple form is 3/5. Part 2: To divide with mixed numbers or whole numbers: Step 1. Change all mixed numbers to improper fractions. Change the whole number to an improper fraction with: (a) the whole number as the numerator (b) “1” as the denominator. Steps 2 to 4. Now there is a division problem like those in Part 1: follow the steps in Part 1. Example 1: 4 1/2 ÷ 3/4 = ? Step 1. 9/2 ÷ 3/4 Step 2. 9/2 × 4/3 Step 3. 39/21 × 42/31 3/1 × 2/1 Step 4. The simple form is 6 (3/1 × 2/1 = 6/1). Example 2: 10 3/8 ÷ 3 = ? Step Step Step Step

1. 2. 3. 4.

83/8 ÷ 3/1 83/8 × 1/3 83/8 × 1/3 = 83/24 The simple form is 3 11/24.

Example 3: 15 ÷ 2 1/3 = ? Step Step Step Step

1. 2. 3. 4.

15/1 ÷ 7/3 15/1 × 3/7 15/1 × 3/7 = 45/7 The simple form is 6 3/7.

Example 4: 9 3/5 ÷ 2 1/2 Step Step Step Step

1. 2. 3. 4.

48/5 ÷ 5/2 48/5 × 2/5 48/5 × 2/5 = 96/25 The simple form is 3 21/25.

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PROBLEMS CALLING FOR ANY ONE OF THE FOUR OPERATIONS This section provides instruction and practice in solving applied problems. So far, the “how” to multiply and divide fractions have been discussed and demonstrated. The last chapter described “how” to add and subtract fractions. This section will provide applied problems calling for any one of the four mathematics operations. The exercises include only problems for multiplying and dividing fractions. The extra exercises include problems for adding, subtracting, multiplying, or dividing fractions. Directions: Problems calling for multiplication of fractions are very common. Division of fractions is a less common practice. There are several terms found in word problems that indicate multiplying: product, times, and or. It is common for “of” to appear between a fraction and another number. Words that indicate division might be quotient, divided by, into, or over. The steps to follow to solve any word problem in mathematics are: Step 1. Read the problem carefully at least two times. Step 2. Be sure to understand what is to be found. Step 3. Be sure to understand all the information given. Step 4. Draw a picture of the problem if possible. Step 5. Look for key words that tell what mathematics operation is needed to find the answer. Example 1: If a person makes $240 a week take-home pay, and 3/4 of it goes toward living expenses, how much is left for savings and pocket money? Step 1. Read twice. Step 2. How much money can be used for savings and pocket money? Step 3. $240 is take-home pay and 3/4 of $240 is for bills. Step 4. Bills

Bills

Savings in Pocket

Bills

Step 5. “of” appears between 3/4 and $240. Solution:

3/4 × 240/1 3/14 × 26400/1 3/1 × 60/1 = 180/1 $180 is used for bills. $60 is used for savings.

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EXERCISES Directions: Select the correct answer to each of the applied problems. Each problem in this set of exercises is solved either by multiplying or dividing with fractions. If necessary, restudy the instructions and examples. 1. A computer teletype can print out 13 2/3 lines in 1 minute. At this rate, how many lines can be printed out in 4 1/2 minutes? (a) 61 1/2 lines (b) 64 3/4 lines (c) 57 2/3 lines (d) 52 1/3 lines 2. A strip of molding 32 1/4 inches long is to be cut into 6 pieces of equal lengths. What is the length of each piece? (a) 10 3/4 in. (b) 3 1/2 in. (c) 1 3/8 in. (d) 5 3/8 in. 3. A drill press operator needs 1 1/3 minutes to drill through 1 inch of metal. How long would it take the operator, in minutes, to drill through 7/8 inch of metal? (a) 13/20 min (b) 1 11/21 min (c) 1 1/6 min (d) 2 1/6 min 4. How many whole pieces of cable 9 1/4 feet long can be cut from a piece of cable 39 3/4 feet long? (a) 2 (b) 3 (c) 4 (d) 5 5. A carpenter uses an auger bit to drill a hole 1 5/16 inches deep. If the bit advances 3/32 inches on each turn, how many turns are needed to drill the hole? (a) 12 (b) 10 (c) 14 (d) 8 6. A standard barrel contains 31 1/2 gallons. How many gallons are in a barrel that is 3/5 full? (a) 14 2/3 gal (b) 18 9/10 gal (c) 24 gal (d) 21 3/4 gal 7. The height of 1 step in a flight of stairs was 9 3/4 inches high. If the flight had 13 steps, how high, in inches, was the total flight? (a) 137 5/8 in. (b) 94 1/2 in. (c) 126 3/4 in. (d) 148 3/8 in. 8. Weight bars are used to hold sheets of paper together in a tablet before the ends are glued. How many bars 7/16 inches wide can be placed side by side in a space 5 1/4 inches wide? (a) 4 bars (b) 8 bars (c) 12 bars (d) 14 bars 9. Find the total height and width of the storage space in this set of drawers. The height of each storage space is 9 3/8 inches and the width of each box is 6 3/4 inches.

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(a) height 65 5/8 in. (b) height 36 3/8 in. width 20 1/4 in. width 18 3/4 in. (d) height 46 1/2 in. width 18 3/4 in.

(c) height 36 3/8 in. width 20 1/4 in.

10. Parts A, B, C, and D are of equal length. How long is A?

(a) 4 3/4 in.

(b) 2 1/4 in.

(c) 16/87 in.

(d) 5 7/16 in.

Answers to the exercises: 1. a 6. b

2. d 7. c

3. c 8. c

4. c 9. a

5. c 10. d

ADDITIONAL EXERCISES Directions: Select the correct answer to each of the applied problems. Each problem in this set can be solved by adding, subtracting, multiplying, or dividing fractions. 1. How many boards 3/4 inch thick are in a pile 20 1/4 inches high? (a) 12 boards (b) 38 boards (c) 15 boards (d) 27 boards 2. A bushing is to be force-fitted into a gear. The hole in the gear is 2 9/16 inches in diameter and the bushing is 1/32 inch larger. What is the diameter of the bushing? (a) 2 5/16 in. (b) 2 5/8 in. (c) 2 19/32 in. (d) 2 9/16 in.

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3. A hole was drilled 5/16 inch deep into a piece of metal 1 1/4 inches thick. What is the distance from the bottom of the drilled hole to the other side of the metal? (a) 1 1/16 in. (b) 1 9/16 in. (c) 9/16 in. (d) 15/16 in. 4. If an auger advances 3/32 of an inch on every turn, how far does it advance on 10 turns? (a) 13/32 in. (b) 30/32 in. (c) 3/8 in. (d) 15/16 5. An order of 27,000 bolts was damaged in shipping to a customer. If 3/250 of the shipment was ruined, how many bolts were wasted because of damage? (a) 543 bolts (b) 412 bolts (c) 118 bolts (d) 324 bolts 6. In making a bolt 2 3/8 inches long, 1/8 inch extra is needed because of waste. How many inches of bars are needed to make 25 bolts? (a) 59 3/4 in. (b) 62 1/2 in. (c) 54 in. (d) 50 3/8 in. 7. A wiring job needed 14 pieces of 1/2 inch conduit, each 6 1/2 feet long, and 10 pieces of conduit, each 4 3/8 feet long. Find the total number of feet of conduit needed. (a) 134 3/4 ft (b) 184 1/2 ft (c) 163 9/16 ft (d) 144 ft 8. A certain alloy is 2/3 aluminum. How many pounds of aluminum are in 64 pounds of alloy? (a) 42 2/3 lb (b) 15 3/8 lb (c) 61 3/4 lb (d) 53 2/3 lb 9. In a certain job, cable lengths of 4 1/2 feet, 15 2/3 feet, and 10 5/8 feet were needed. What is the length of cable needed? (a) 30 19/24 ft (b) 31 2/3 ft (c) 29 8/13 ft (d) 29 3/4 ft 10. Find the length of distance D from the figure.

(a) 1 3/8 in.

(b) 1 15/16 in.

(c) 1 3/8 in.

(d) 1 7/16 in.

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Answers to the additional exercises: 1. d 6. b

2. c 7. a

3. d 8. a

4. d 9. a

5. d 10. d

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14

Ordering, Rounding, and Changing Decimals

Chapter 14 will describe how to use decimal numbers to solve mathematical problems. Objective: Understanding decimals and how to use them is important for anyone who works with numbers: (1) in the United States, there has been a shift to the metric system of measures and problems in metrics can be solved with decimal numbers, and (2) calculators owned by many companies and persons use decimals only. Decimals may be used when there are numbers other than whole numbers. Fractions are also used for numbers other than whole numbers. Fractions and decimals are two ways to write these “in-between” types of numbers. Definitions: 1. Decimal (or decimal number): a number written as a whole number with a decimal point next to one of the digits 2. Decimal point: a dot placed in the number to locate the starting point for the place values 3. Key digit: one digit right of the “last digit” in any decimal number; a term used when rounding off numbers 4. Last digit: the digit in the place value to which a given decimal number is being rounded 5. Place value: the number value given to a location in a decimal number 6. Rounding off: a method of giving a simple number close to the next given number

DECIMALS USING FRACTIONS This section will explain decimals by using fractions. Every decimal can be written as a fraction. This fact helps in understanding and reading decimals. Directions: The U.S. money system is a decimal system and it will be used to explain decimal numbers.

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Facts about the U.S. Money System 1. There are 100 cents in 1 dollar. 2. 1 cent is 1/100 of a dollar. 3. There are 10 dimes in 1 dollar. 4. 1 dime is 1/10 of a dollar. Therefore: • A cash register receipt of $ .04 means 4 cents, and 4 cents is 4/100 of a dollar: .04 is the same as 4/100. • A cash register receipt of $ .38 means 38 cents which is 38/100 of a dollar: .38 is the same as 38/100. • A cash register receipt of .60 means two things: (a)6 dimes or 6/10 of a dollar (b)60 cents or 60/100 of a dollar. 6 is the same as 6/10 and 60/100. Place values of the numbers to the right of the decimal point in decimal numbers are illustrated in the chart:

1/100

1/1,000

1/10,000

1/100,000

1/1,000,000

ten thousandths

hundred thousandths

millionths

ones

tens

1/10

thousandths

.

hundreths

1

tenths

10

Methods to write a decimal as a fraction: Method 1. Write a decimal number less than one as a fraction. Step 1. The numerator of the fraction is the set of numbers to the right of the decimal point: (a) Beginning with the left-most non-zero digit (b) Ending with the right-most non-zero digit Step 2. The denominator of the fraction is the place value number of the right-most non-zero digit. Step 3. Reduce the fraction. Example:

.102

Step 1. The numerator is 102.

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Step 2. The denominator is 1,000 because the “2” is in the 1/1000’s place. 102 51 Step 3. Reduce: ------------ = --------1000 500 Example:

.085

Step 1. The numerator is 85. Step 2. The denominator is 1000. 85 17 Step 3. ------------- = --------1,000 200 Example:

.0004

Step 1. The numerator is 4. Step 2. The denominator is 10,000. 4 1 Step 3. Reduce: ---------------- = ------------10,000 2,500 Method 2. Write a decimal number that is larger than one as a fraction. Step 1. The whole number part of a fraction (a mixed number) is the entire whole number to the left of the decimal point in the decimal number. Step 2. The proper fraction part of the mixed number is found as described in Method 1. Example:

Change 4.3 to a mixed number.

Step 1. The whole number part is 4. Step 2. .3 means 3/10: 4.3 is the same as 4 3/10. Example:

Change 12.25 to a mixed number.

Step 1. The whole number part is 12. Step 2. .25 is 25/100 which equals 1/4: 12.25 is the same as 12 1/4. Special note: Any whole number, such as 758, automatically has a decimal point just to the right of the digit farthest to the right. So, 758 could be written 758.0 (with a decimal). Example:

17 is the same as 17.0.

Example:

124 is the same as 124.0.

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ROUNDING OFF There are times when an exact answer is not required, but rather an “about” number or an approximate number is needed. For example, if the almanac says a city has a population of 120,000, this is a number that is only close to the real population which might actually be 121,168. In this section, “rounding off” an exact number to an approximate number, close to the real number, will be demonstrated. Directions: In the method presented for rounding off numbers, two special terms are used. Work through an example. Example 1: Round 1284 off to the 100’s place. The last digit is 2 (2 is in the 100’s place) The key digit is 8. Steps to round off any decimal number: Step 1. Recopy all digits to the left of the last digit. 1_ _ _ Step 2. Replace all digits to the right of the last digit with zeros. 1_00 Step 3. Put the number into the last digit’s place by: (a) Putting in the last digit itself if the key digit is 4 or less. (b) Putting in one number larger than the last digit if the key digit is 5 or larger. 1300 (A “3” must be put in because the key digit “8” was bigger than 5.) Example 2: Round 67,549 off to the 10’s place. Step 1. Recopy all digits to the left of the 10’s place. 67,5_ _ Step 2. Digits to the right of the 10’s place. 67,5_0 Step 3. Digit in the 10’s place. (This digit must be a “5” because the key digit (next to the one to the right) is bigger than 5, so the “last digit” is increased by 1. The answer is 67,550.)

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Example 3: Round 843,107 off to the 10,000’s place. Step 1. Digits to the left of the 10,000’s place. 8_ _,_ _ _ Step 2. Digits to the right of the 10,000’s place. 8_0,000 Step 3. Digit in the 10,000’s place. (Must be a “4” because the next digit to its right is less than 5. The answer is 840,000.)

ROUNDING OFF EXERCISES Directions: Select the number equal to the given number, rounded off to the given place value. If necessary, restudy the directions and examples. 1. 87.5 rounded to 10’s place (a) 90 (b) 80

(c) 88

(d) 87

2. 94.13 rounded to 1/10’s place (a) 95 (b) 94.1

(c) 94

(d) 94.2

3. 123.43 rounded to 10’s place (a) 120 (b) 100

(c) 130

(d) 123.4

4. 42.75 rounded to one’s place (a) 43 (b) 42.8

(c) 42

(d) 42.7

5. 9.9581 rounded to 1/100’s place (a) 9.958 (b) 9.96

(c) 10

(d) 9.95

6. 8,745,134 rounded to 10,000’s place (a) 8,740,000 (b) 8,700,000 (c) 8,750,000

(d) 8,800,00

7. 3.1932 rounded to tenth’s place (a) 3.19 (b) 3.1

(c) 3.2

(d) 3.21

8. 543.4 rounded to hundred’s place (a) 550 (b) 540

(c) 600

(d) 500

9. 84.1754 rounded to thousandth’s place. (a) 84.176 (b) 84.18 (c) 84.1754

(d) 84.175

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10. 571.04 rounded to ten’s place. (a) 571.0 (b) 571

(c) 570

(d) 580

Answers to the rounding off exercises: 1. a 6. c

2. b 7. c

3. a 8. d

4. a 9. d

5. b 10. c

ADDITIONAL EXERCISES Directions: Select the number equal to the given number, rounded off to the given place value. 1. 38.4 rounded to 1’s place (a) 38 (b) 39

(c) 40

(d) 38.4

2. 1.79 rounded to 1/10’s place (a) 1.7 (b) 1.79

(c) 2

(d) 1.8

3. 4.348 rounded to hundredth’s place (a) 4.35 (b) 4.3 (c) 4.4

(d) 4.34

4. 625.039 rounded to hundred’s place (a) 600 (b) 700 (c) 625.04

(d) 630

5. 37.13 rounded to one’s place (a) 40 (b) 37

(c) 38

(d) 37.1

6. 37,184 rounded to 1000’s place (a) 38,000 (b) 37,200

(c) 38,200

(d) 37,000

7. 417.893 rounded to 1/100’s place (a) 417.893 (b) 417.89

(c) 417.9

(d) 400

8. 857.13 rounded to 100’s place (a) 900 (b) 860

(c) 857.13

(d) 800

9. 914,318 rounded to ten thousandth’s place (a) 914,000 (b) 900,000 (c) 920,000 10. 84.743 rounded to one’s place (a) 85 (b) 84.7

(c) 84

(d) 910,000

(d) 84.8

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Answers to the additional exercises: 1. a 6. d

2. d 7. b

3. a 8. a

4. a 9. d

5. b 10. a

CHANGING DECIMALS When working with metric measurements, decimal numbers are used. It is necessary to determine which of two decimal numbers is larger, e.g., is .89 meters larger or smaller than .099 meters? This section will demonstrate how to write a group of decimal numbers in order of size. Directions: The rules for ordering decimals will be given three sections. The rules are simple and easy to understand. Example:

Which is larger: 0843 or .91?

Rule 1. The decimal number having a non-zero digit in the place value farthest to the left is larger. Answer:

.91 is larger than .0843 because the “9” in .91 is in a place value farther to the left than the “8” in .0843

Example: Answer:

Which is larger: 1.004 or .9834? 1.004 because “1” is further left than “9”

Example:

Which is larger: .0783 or .0535?

Rule 2. If two decimal numbers have their left-most non-zero digits in the same place value positions, then the decimal number with the largest left-most non-zero digit is larger. Answer:

.0783 is larger than .0535 because the “7” in .0783 is larger than the “5” in .0535

Example: Answer:

Which is larger: 12.349 or 40.999? 40.999 because the “4” in 40.999 is larger than the “1” in 12.349

Example:

Which is larger: 1.37 or 1.934 ?

Rule 3. If two decimal numbers have the same left-most non-zero digit in the same place value position, then look one place to the right in each number until a higher digit is reached.

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Answer:

1.934 is larger than 1.37 because “9” in the tenth’s place of 1.934 is larger than “3” in the tenth’s place of 1.37

Example: Answer:

Which is larger: 37.438 or 37.449? 37.449 because the “4” in the 1/100’s place of 37.449 is larger than the “3” in the 1/100’s place of 37.438

EXERCISES Directions: Select the proper order by size, with the largest number on the left. If necessary, restudy the instructions and examples. 1.

(a) (b) (c) (d)

.0432 .432 4.32 43.2

2.

(a) (b) (c) (d)

.196 1.96 196 19.6

3.

(a) (b) (c) (d)

4.

(a) (b) (c) (d)

5.

587 58.7 5.87 .587

43.2 4.32 .0432 4.32 196 19.6 19.6 196

4.32 43.2 .432 .432 19.6 .196 1.96 .196

432 432 432 .0432 1.96 196 .196 1.96

5.87 5.87 587 58.7

58.7 .587 .587 .0587

.0587 .0587 58.7 587

417 417 417 417

41.7 4.17 .417 .0417

4.17 41.7 41.7 41.7

.417 .417 4.17 .417

(a) (b) (c) (d)

69.3 693 69.3 693

6.93 6.93 693 69.3

693 69.3 6.93 6.93

.693 .693 .693 .693

6.

(a) (b) (c) (d)

54.1 514 5.41 514

.541 54.1 541 51.4

51.4 51.4 51.4 54.1

5.41 5.41 54.1 5.14

7.

(a) (b) (c) (d)

972 972 97.2 927

97.2 97.2 9.27 92.7

92.7 9.27 9.72 97.2

9.72 9.72 .927 9.27

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8.

(a) (b) (c) (d)

9.38 27.3 32.17 8.54

1.46 17.41 4.85 7.32

5.38 .98 .094 9.18

.097 .435 .715 .44

9.

(a) (b) (c) (d)

7.314 9.38 7.314 8.24

3.82 4.093 7.013 9.36

4.09 3.604 6.38 8.39

.38 8.714 5.21 4.24

3. b 8. b

4. a 9. c

5. d

Answers to the exercises: 1. d 6. b

2. c 7. a

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The Four Operations in Decimals

Objective: This chapter demonstrates the four operations of addition, subtraction, multiplication, and division with decimal numbers. These operations should be done with pencil and paper for times when a calculator is not available. Therefore, do not use a calculator for this chapter. In addition to demonstrating the decimal operations, exercises with applied problems using decimals will be provided. Definitions: 1. Centimeter (cm): A metric unit of length, about 3/8 inch 2. Column: A vertical (up and down) row 3. Diameter: The distance from one side of a circle to the other through the center 4. Factors: Numbers that are being multiplied 5. Millimeter (mm): A metric unit of length, about 1/16 inch 6. Place holders: Zeros used to fill in place values in decimal numbers 7. Powers of ten: Numbers such as 10; 100; 1,000; 10,000; etc. 8. Product: The answer to a multiplication problem 9. Vertical: Up and down

ADDING AND SUBTRACTING All four mathematic operations can be used with decimal numbers. In fact, calculators use only decimal numbers. Directions to add decimals: Step 1. Write all numbers in a vertical column with all the decimal points aligned under each other. Step 2. Insert zeros on the right end of any numbers to make all right ends even. Step 3. Insert a decimal point under all other decimal points where the answer will be. Step 4. Add the numbers as if adding whole numbers. 201

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Caution: Be sure all same-place values are aligned over each other before adding!! Example 1: 2.74 + 18.3 = ? 2.74 + 18.3 Step 1. ------------------

Write numbers in a column.

2.74 18.30 Step 2. + ------------------

Insert zeros to make right ends even.

2.74 18.30 Step 3. + -----------------.

Insert decimal point in answer.

11

2.74 18.30 Step 4. + -----------------21.04 Answer:

Add.

21.04

Example 2: 48 + 7.84 + .196 = ? 48. 7.84 Step 1. + .196 --------------------48.000 7.840 Step 2. + .196 --------------------48.00 7.840 .196 Step 3. + --------------------. 1 1

Write numbers in a column.

Insert zeros.

Insert decimal point in answer.

1

4 8 .0 00 7.8 00 + .196 Step 4. ----------------------56.036 Answer:

56.036

Add.

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Directions to subtract decimals: Step 1. Write the two numbers in a vertical column with the two decimal points aligned under each other. Step 2. Insert zeros on the right ends of any numbers to make the right ends even. Step 3. Insert a decimal point under the other two decimal points where the answer will be. Step 4. Subtract the numbers as if subtracting whole numbers. Example 1: 10.89 – 4.71 = ? 10.89 – 4.71 Step 1. --------------

Write numbers in a column.

10.89 – 4.71 Step 2. ---------------

Insert zeros (none needed).

10.89 – 4.71 Step 3. --------------.

Insert decimal point in answer.

10.89 – 4.71 Step 4. --------------6.18

Subtract.

Answer:

6.18

Example 2: 17.8 – 4.37 = ? 17.8 4.37 Step 1. –--------------

Write numbers in a column.

17.80 – 4.37 Step 2. --------------

Insert zeros.

17.80 – 4.37 Step 3. --------------.

Insert decimal points in answer.

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17. 8 0 – 4. 37 Step 4. -----------------13. 43 Answer:

Subtract.

13.43

MULTIPLYING TWO DECIMAL NUMBERS This section will demonstrate multiplying two decimal numbers together. When decimal numbers were added and subtracted it was important to keep all decimal points in a vertical line. This is not the case when multiplying decimal numbers. There is a new rule for locating the decimal point in the product of two decimal numbers.

Example 1:

7.45 × 3.8 ---------------

Step 1. Multiply the factors as if they were both whole numbers. In the example, multiply 745 by 38. 745 × 38 -----------------5960 2235 -----------------28310 Step 2. To put the decimal point in the product: (a) Add the number of digits to the right of the decimal point in each factor. (b) Starting to the right of the far right digit in the product from Step 1, count over a number of times equal to the number found in (a) and put a decimal point. In Example 1, factor 7.45 has 2 digits to the right of the decimal point: factor 3.8 has 1 digit to the right of the decimal point: 2+1=3 Therefore, count over 3 times and put a decimal point: 28.310 3 21

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Answer:

28.31

Example 2:

14.7 × .008 ---------------

147 ×8 Step 1. -----------1176

205

Multiply.

14.7 .008 Step 2. ------------.1176

1 +3 -------4

Count digits and move decimal point.

4 321

Answer:

.1176

Example 3:

.375 × .54 ---------------

375 × 54 Step 1. --------------1500 1875 --------------20250 .375 × .54 Step 2. ---------------.20250

Multiply.

3 +2 ------5

5 43 2 1

Answer:

.2025

Example 4:

.018 × .004 ---------------

18 ×4 Step 1. -------72

Multiply.

Count digits and move decimal point.

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Step 2.

.018 × .004 ------------------.000072

3 +3 ------6

Count digits and move decimal point.

6 54 32 1

Answer:

.000072

Note: Four zeros needed to be added to fill in the place values. The zeros are called place holders.

DIVISION WITH DECIMALS This section demonstrates division with decimal numbers. Division with decimals is much like long division of whole numbers; however, there are two additional rules for the decimal points. Directions: First, review the meaning of three terms for the numbers in any division problem: 2 6 12

or

12 ÷ 6 = 2

Definitions: 1. Dividend: The number being divided (12 in the example) 2. Divisor: The number being divided into the dividend (6 in the example) 3. Quotient: The answer to the division problem (2 in the example) This set of instructions will be divided into three parts: Part 1: Divisors are whole numbers and the quotient has no remainder. Part 2: Divisors are decimal numbers and the quotient has no remainder. Part 3: The quotient has a remainder.

PART 1 The examples in this section are Example 1:

6 1.38

Example 2:

37 46.25

Example 3: 309.54 ÷ 134

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Example 1: Step 1. Put the decimal point in the quotient above the decimal point in the dividend. . 6 1.38 Step 2. Divide as if dividing whole numbers. .23 6 1.38 12 18 18 Answer:

.23

Example 2:

37 46.25

. Step 1. 37 46.25 1.25 Step 2. 37 46.25 37 92 74 85 85 Answer:

Put decimal point in quotient.

Divide.

1.25

Example 3: 309.54 ÷ 134 = ? Put decimal point in quotient. . Step 1. 134 309.54 2.31 Step 2. 134 309.54 Divide. 268 415 402 134 134 Answer:

2.31

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PART 2 The examples in this section are Example 1:

.05 .215

Example 2:

1.6 3.84

Example 3: 8.9388 ÷ 2.34 Example 1: Step 1. Move the decimal point in the divisor over next to the division symbol or to the right end of the divisor. .05 .215

Step 2. Move the decimal point in the dividend the same number places right as the decimal point in the divisor was moved. .05 .215

2 places

Now the example is 5. 21.5 Step 3. Divide as in Part 1. 4.3 5. 21.5 20 15 15 Answer:

4.3

Example 2:

1.6 3.84

Steps 1-2. 1.6 3.84

16 38.4

Move the decimals.

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. Step 3. 16 38.4

209

Divide. 2.4 16 38.4 32 64 64

Answer:

2.4

Example 3: 8.9388 ÷ 2.34 = ? Steps 1-2. 2.34 8.9388 2

2

. 234 893.88 702 1918 1872 468 468

Step 3.

Answer:

Move the decimals.

3.82

PART 3 The examples are Example 1:

9 1.3

Rounded off to 1/100’s

Example 2:

4.3 7.9 Rounded off to 1/10’s

Example 3: .816 ÷ 4.23 Rounded off to 1/100’s The examples are of division when the divisor process never ends and if it were stopped, there would be a remainder. Example 1: Step 1. Move decimal points in the divisor and dividend if necessary. 9 1.3

(No moves necessary.)

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Step 2. Place the decimal point in the quotient above the decimal point in the dividend. . 9 1.3 Step 3. Divide until there is a digit in one place beyond the place value where the answer must be rounded. Note: Zero may be added to the right end of the dividend to continue the division beyond the given dividend. .144 9 1.300 9 40 36 40 36

(Two zeros were added.)

Step 4. Round the quotient to the stated place value: round .144 to 1/100s. Answer:

.14

Example 2:

4.3 7.9 Round to 1/10s.

Step 1. 4.3 7.9

Move decimals.

1 43 79. . Step 2. 43 79. 1.83 Step 3. 43 79.00 43 360 344 160 129

Put decimal in quotient.

Divide.

Step 4. Round 1.83 to 1/10s. Answer:

1.8

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Example 3: .816 ÷ 4.23 Round to 1/100s 4.23 .816 Step 1. 4.23 .816

Move decimals. 423 81.6 2 2

. Step 2. 423 81.6

Step 3.

.192 423 81.600 42 3 39 30 38 07 1 230 846

Put decimal in quotient.

Divide.

Step 4. Round .192 to 1/100s. Answer:

.19

DIVISION WITH DECIMALS EXERCISES Directions: Select the correct answer to each of the following division problems. If necessary, restudy the instructions and the examples. 1.

7 10.5 (Round to 1/10s place.) (a) .15 (b) 13.

(c) 1.3

(d) 1.5

2. 8 100.8 (Round to 1/10s place.) (a) 11.9 (b) 12

(c) 13

(d) 12.6

3. 13 4.55 (Round to 1/100s place.) (a) .4 (b) .35 (c) .36

(d) .38

4. 72.45 ÷ 23 (Round to 1/100s place.) (a) 3.09 (b) 3.15 (c) 2.84

(d) .314

5. 2.4 8.544 (Round to 1/100s place.) (a) 18.73 (b) 31.41 (c) 2.78

(d) 3.56

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6. .84 4.3764 (Round to 1/100s place.) (a) 5.19 (b) 4.86 (c) 5.43

(d) 5.21

7. 9 1.43 (Round to 1/100s place.) (a) .15 (b) .18

(d) .16

8.

1.4 7.34 (a) 5.24

(c) .17

(Round to 1/100s place.) (b) 5.3 (c) 5.25

9. .75 81.04 (Round to 1/100s place.) (a) 108.05 (b) 106.34 (c) 107.09 10. 1.49 4.34 (Round to 1/1000s place.) (a) 2.90 (b) 2.92 (c) 2.913

(d) 4.81

(d) 108.06

(d) 2.938

Answers to the division with decimals exercises: 1. d 6. d

2. d 7. d

3. b 8. a

4. b 9. a

5. d 10. c

APPLIED PROBLEMS USING DECIMAL OPERATIONS So far adding, subtracting, multiplying, and dividing decimals have been demonstrated. This section will use the operations to solve applied problems. Directions: Some suggested steps to solve any written applied problem. Step 1. Read the problem carefully at least two times Step 2. Be sure to understand what is to be found. Step 3. Be sure to understand all the given information. Step 4. Draw a picture of the problem if possible. Step 5. Look for key words that indicate what mathematics operation is needed to find the answer. Several terms found in written applied problems indicate a certain operation: Add

Subtract

sum and total more than plus increase combine

less difference minus less than reduce

Multiply

product of times by

Divide

quotient over into

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EXERCISES Directions: Select the correct answer to each of the following applied problems. If necessary, restudy the instructions and examples. 1. Find the total amount of current used by an iron using 5.3 amps and a radio using .7 amps. (a) 5.8 amps (b) 7.4 amps (c) 6.0 amps (d) 3.71 amps 2. What is the difference in the widths of two wires if one wire has a width of .373 inches and the other wire has a width of .173 inches? (a) .31 in. (b) .2 in. (c) .546 in. (d) .417 in. 3. If a certain light bulb used .84 amps, how many amps would 12 of these bulbs use? (a) 10.08 amps (b) 10.008 amps (c) 10.8 amps (d) 18 amps 4. If a farmer places 15 fence posts evenly spaced over a distance of 188 feet, how many feet are between each post? (Round to 1/10s.) (a) 12.5 ft (b) 13 ft (c) 11.67 ft (d) 13.2 ft 5. If the distance between the outside circle and the inside circle is .0023 inch, what is the diameter (distance across) of the inside circle?

3.75”

(a) 3.7514 in.

(b) 3.518 in.

(c) 3.7454 in.

(d) 3.7461 in.

6. A man who works 38.25 hours a week and is paid $6.72 per hour earns how much money before taxes are taken out? (a) $263.76 (b) $245.57 (c) $257.04 (d) $284.75 7. How wide is distance d as shown in the diagram?

(a) 5.481 in.

(b) 5.184 in.

(c) 4.184 in.

(d) 7.251 in.

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8. How many boards .375 inch thick are stacked in a pile that is 22.875 inches high? (Round to 1’s place.) (a) 63 (b) 75 (c) 58 (d) 61 9. A board is 81.3 centimeters long. If 23.7 centimers are cut off of the board, how long is the remaining piece? (a) 57.6 cm (b) 48.2 cm (c) 17.9 cm (d) 63.2 cm 10. A certain metal drill bit advances .013 millimeters with each turn. How many turns are necessary to drill through a piece of metal 4.34 millimeters thick? (Round to 1/10s place.) (a) 438.4 (b) 333.8 (c) 812.1 (d) 356 Answers to the exercises: 1. c 6. c

2. b 7. b

3. a 8. d

4. a 9. a

5. c 10. b

CHANGE A FRACTION INTO A DECIMAL A special application of the division of decimals is changing a fraction into a decimal. This section will demonstrate how to change any fraction into a decimal number rounded to any stated place value. Directions: 1. To change a proper fraction into a decimal, divide the denominator into the numerator and round it off to the stated place value. 2. To change a mixed number to a decimal, write the whole number that is left of the decimal point, and write the decimal for the proper fraction part to the right of the decimal. Example 1: Change 3/4 to a decimal rounded to 1/100s place. Divide.

4 3 .75 4 3.00 28 20 20 Answer:

.75

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Example 2: Change 7/11 to a decimal rounded to 1/100s place. 11 7

Divide. .636 11 7.000 66 40 33 70 66 4

Answer:

.64

Example 3: Change 4 3/8 to a decimal rounded to 1/100s. 8 3

Rounded to .38.

375 8 3.000 24 60 56 40 40 Answer:

Divide.

4.38

Below is a table of the most common equal fraction-decimal values. They should be memorized. Fraction

Decimal

1/4 1/3 1/2 2/3 3/4

.25 .333 (approx.) .5 .667 (approx.) .75

EXERCISES Directions: Select the correct decimal, rounded to the stated place value, equal to each of the fractions. If necessary, restudy the instructions and examples. 1. 2/5 (Round to 1/10s place.) (a) .3 (b) .2

(c) .4

(d) .5

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2. 4/7 (Round to 1/100s place.) (a) .56 (b) .47

(c) .46

(d) .57

3. 3/8 (Round to 1/1,000s place.) (a) .38 (b) .375

(c) .37

(d) .385

4. 5/12 (Round to 1/10s place.) (a) .6 (b) .2

(c) .4

(d) .5

5. 7/11 (Round to 1/100s place.) (a) .64 (b) .63

(c) .72

(d) .71

6. 1 3/4 (Round to 1/100s place.) (a) 1.75 (b) 1.43

(c) 1.67

(d) 1.34

7. 3 4/9 (Round to 1/100s place.) (a) 3.33 (b) 3.67

(c) 3.49

(d) 3.44

8. 10 2/15 (Round to 1/100s place.) (a) 10.14 (b) 10.12

(c) 10.11

(d) 10.13

9. 14 5/7 (Round to 1/100s place.) (a) 14.64 (b) 14.71

(c) 14.67

(d) 14.57

10. 43 2/3 (Round to 1/100s place.) (a) 43.33 (b) 43.24

(c) 43.75

(d) 43.67

Answers to the exercises: 1. c 6. a

2. d 7. d

3. b 8. d

4. c 9. b

5. a 10. d

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Squares, Square Roots, Cubes, Cube Roots, and Proportions

Objective: The focus of this section is to find the square and square root of numbers. Because these are advanced skills, this section was placed near the end of the basic math material. Finding the square or cube of a number is a common application in many area and volume formulas as well as in statistics. Squares and square roots are also used in the “Rule of Pythagoras” which is a simple method to find the length of the hypotenuse of a right triangle when the other two sides are known. Another common application is the calculation of variance and standard deviation. In many problems, a change in one quantity causes a change in a related quantity. Examples are problems such as gear/ratio, pulleys, scale models, gas mileage, etc. This chapter will demonstrate how to determine if a problem can be solved with a proportion and if so, how to do it. Definitions: 1. Cube: the result after multiplying three of the given numbers together 2. Cube root: the number of which a given number is the cube 3. Rule of Pythagoras: a method used to find the length of a hypotenuse when the lengths of the other two sides are given 4. Square: the product found by multiplying two of the given numbers together 5. Square root: the number which when squared will produce a given number 6. Proportion: a math statement that two ratios are equal 7. Proportional figures: figures of the same shape and same size 8. Proportional quantities: quantities that act the same or exactly the opposite by the same factor 9. Ratio: a comparison of two numbers 10. Similar figures (or scale modes): figures of the same shape and same size; same as proportional figures

217

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SQUARE AND CUBE NUMBERS This section demonstrates how to square a number and how to cube a number. These two skills are often used in statistics as well as in finding areas and volumes. Square of a number: The square of a number is the product of that number times itself one time. Example 1: What is the square of 5? 5 × 5 = 25 Answer:

25

Example 2: What is the square of 4 2/3? 4 2/3 × 4 2/3 14/3 × 14/3 = 196/9 Answer:

21 7/9

Squaring a number: Squaring a number is to obtain the square of the number. Cube of a number: The cube of a number is the product of the square of the number and the number itself or the product of multiplying three of the numbers together. Cubing a number: Cubing a number is to obtain the cube of the number. Example 3: Find the cube of 8. 8 × 8 × 8 = 512 Answer:

512

Example 4: Find the cube of 4.7. 4.7 × 4.7 × 4.7 = 103.823 Answer:

103.823

Example 5: Find the cube of 3 1/4. 13/4 × 13/4 × 13/4 = 2197/64 Answer:

34 21/64 or 34.328

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EXERCISES Directions: Select the correct value of the square or cube of each of the following numbers. Round any decimal answers off to the 1/100s place. If necessary, restudy the instructions and examples. 1. Find the square of 9. (a) 81 (b) 27

(c) 18

(d) 729

2. Find the cube of 3. (a) 18 (b) 6

(c) 9

(d) 27

3. What is the result of cubing 11? (a) 22 (b) 121

(c) 33

(d) 1331

4. What is the square of .8? (a) .64 (b) .51

(c) 1.6

(d) 6.4

5. Find the square of 1/2. (a) 1 1/2 (b) 1/4

(c) 3/4

(d) 1/8

6. Find the cube of 1/2. (a) 1/16 (b) 1/4

(c) 3/4

(d) 1/8

7. What is the square of 4.3? (a) 341.88 (b) 8.6

(c) 18.49

(d) 12.9

8. What is the cube of 1.8? (a) 3.6 (b) 5.83

(c) 3.24

(d) 5.4

9. What is the result of squaring 6 3/4? (a) 20 1/4 (b) 45 9/16 (c) 83 9/16

(d) 13 1/2

10. What is the result of cubing 2 1/3? (a) 3 1/27 (b) 24 17/29 (c) 18 3/8

(d) 12 19/27

11. What is the square of 8? (a) 64 (b) 512

(c) 24

(d) 16

12. What is the cube of 10? (a) 100 (b) 30

(c) 20

(d) 1000

13. Find the square of 4.5. (a) 91.13 (b) 9

(c) 20.25

(d) 13.5

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14. Find the cube of 9 1/3. (a) 813 1/27 (b) 27

(c) 729 1/27

15. What is the result of squaring 19.4? (a) 376.36 (b) 38.8 (c) 7301.38

(d) 87 1/9

(d) 58.2

Answers to the exercises: 1. a 6. d 11. a

2. d 7. c 12. d

3. d 8. b 13. c

4. a 9. b 14. a

5. b 10. d 15. a

SQUARES AND CUBE ROOTS The opposite of finding the square of a number is to find the square root. The same is true for cubing a number and finding the cube root. Finding the square root or cube root of a number is a skill used mostly in algebra problems and statistics. Square root of a number: The square root is a number which when squared will produce a given number. Example 1: What is the square root of 9? Answer:

3 because the square of 3 is 9.

Example 2: What is the square root of 36? Answer:

6 because the square of 6 is 36.

New symbol:

symbol for square root 16

square root of 16

75

square root of 75

Example 3: What number is Answer:

9 because the square of 9 is 81.

Example 4: What number is Answer:

81 equal to?

40 equal to?

This problem is different because no whole number squared equals 40 (6 squared is 36; 7 squared equals 49).

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Therefore, the square root of 40 is between 6 and 7. The method below describes how to find the square root of any whole number from 1 to 100 using Table 16.1. (Example 4 will be the model.) Step 1. Locate the given number in one of the columns labeled “No.” (For the example, find 40 under “No.”) Step 2. Read the number two columns to the right of the given number under “Square Root.” This number is the square root of the given number. (For the example, 6.325 is the square root of 40.) Note: All the square roots in Table 16.1 are rounded off to the 1/100s place. Example 5: Find the value of Answer:

8.367

Example 6: Find the value of Answer:

70

91

9.539

Another feature of Table 16.1 is the column marked “Sq.” This column gives the square of the numbers to the left of it in the “No.” column. 1. Look under “Sq.” to 441. This is the square of 21. 2. Look under “Sq.” to 5476. This is the square of 74. Note: If 441 is the square of 21, then 21 must be the square root of 441. Therefore, every number under “No.” is the square root of the number next to it under “Sq.” To find the square root of whole numbers more than 100: Step 1. Find the given number that is more than 100 under the column labeled “Sq.” in Table 16.1. (a) If the given number is found, go to Step 2. (b) If the given number is not found, stop. This table cannot be used. Step 2. Look at the number just to the left of the given number in the column labeled “No.” — the square root of the given number. Example 7: Find the value of

841

Step 1. Find 841 under “Sq.” Step 2. Look left under “No.” and see 29. Answer:

29

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TABLE 16.1 Squares, Square Roots, Cubes, and Cube Roots (1–100) No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

Sq. 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 625 676 729 784 841 900 961 1,024 1,039 1,156 1,225 1,296 1,369 1,444 1,521 1,600 1,681 1,764

Square Root 1.000 1.414 1.732 2.000 2.236 2.449 2.646 2.828 3.000 3.162 3.317 3.464 3.606 3.742 3.873 4.000 4.123 4.243 4.359 4.472 4.583 4.690 4.796 4.899 5.000 5.099 5.196 5.291 5.385 5.477 5.568 5.657 5.745 5.831 5,916 6,000 6.083 6,184 6,245 6,325 6.403 6.481

Cube 1 8 27 64 125 216 343 512 729 1,000 1,331 1,728 2,197 2,744 3,375 4,096 4,913 5,832 6,859 8,000 9,261 10,648 12,167 13,824 15,625 17,576 19,683 21,952 24,389 27,000 29,791 32,765 35,937 39,304 42,875 46,656 50,653 54,872 59,319 64,000 68,921 74,088

Cube Root 1.000 1.260 1.442 1.587 1.710 1.817 1.913 2.000 2.080 2.154 2.224 2.289 2.351 2.410 2.466 2.520 2.571 2.621 2.668 2.714 2.759 2.802 2.844 2.884 2.924 2.952 3.000 3.037 3.072 3.107 3.141 3.175 3.203 3.240 3.271 3.302 3.332 3.362 3.391 3.420 3.443 3.476

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TABLE 16.1 (Continued) Squares, Square Roots, Cubes, and Cube Roots (1–100) No. 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84

Sq. 1,849 1,936 2,025 2,116 2,029 2,304 2,401 2,500 2,601 2,704 2,880 2,916 3,025 3,136 3,249 3,364 3,481 3,600 3,721 3,844 3,969 4,096 4,225 4,356 4,489 4,624 4,761 4,900 5,041 5,184 5,329 5,476 5,625 5,776 5,929 6,084 6,241 6,400 6,561 6,724 6,839 7,056

Square Root 6.557 6.633 6.703 6.782 6.856 6.928 7.00 7.071 7.141 7.211 7.280 7.348 7.416 7.483 7.550 7.616 7.631 7.746 7.810 7.874 7.937 8.000 8.062 8.124 8.185 8.246 8.307 8.367 8.426 8.485 8.544 8.602 8.660 8.718 8.775 8.832 8.888 8.944 9.000 9.055 9.110 9.165

Cube 79,507 85,184 91,125 97,336 103,823 110,592 117,649 125,000 132,651 140,603 148,877 157,464 166,375 175,616 185,193 195,112 205,379 216,000 226,981 238,328 250,047 262,144 274,625 287,496 300,763 314,432 328,509 343,000 357,911 373,248 389,017 405,224 421,875 438,976 456,533 474,552 493,039 512,000 531,441 551,368 571,787 592,704

Cube Root 3.503 3.530 3.551 3.583 3.609 3.631 3.659 3.684 3.708 3.732 3.756 7.780 3.803 3.826 3.848 3.871 3.893 3.915 3.935 3.958 3.979 4.000 4.021 4.041 4.052 4.082 4.102 4.121 4.141 4.160 4.179 4.198 4.217 4.236 4.254 4.273 4.291 4.309 4.327 4.344 4.362 4.380

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TABLE 16.1 (Continued) Squares, Square Roots, Cubes, and Cube Roots (1–100) No.

Sq.

85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

7,225 7,396 7,569 7,744 7,921 8,100 8,281 8,464 8,649 8,836 9,025 9,216 9,409 9,604 9,801 10,000

Square Root 9.220 9.274 9.327 9.381 9.434 9.487 9.539 9.592 9.611 9.695 9.747 9.798 9.849 9.899 9.950 10.00

Cube 614,125 636,056 658,503 681,472 704,969 729,000 753,571 778,638 804,357 830,584 857,375 884,736 912,673 941,192 970,299 1,000,000

Cube Root 4.397 4.414 4.431 4.448 4.465 4.481 4.493 4.514 4.531 4.547 4.563 4.579 4.595 4.610 4.626 4.632

Note: No., number; Sq., square.

Example 8: Find the value of Answer:

92

Example 9: Find value of Answer:

8464

3000

Cannot be found using Table 16.1.

Cube root: The cube root of a given number is that number whose cube is the given number. Symbol:

3

symbol for cube root

To find the cube root of a given whole number from 1 to 100 using Table 16.1: Step 1. Locate the given number in one of the columns marked “No.” Step 2. See the number four columns to the right of the given number under “Cube Root.” Example 10: Find the cube root of 45.

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Step 1. Locate 45 under “No.” Step 2. Read 3.557 under Cube Root. Answer:

3.557

Example 11: Find the value of

3

66

Step 1. Locate 66 under “No.” Step 2. See 4.041 under Cube Root. Answer:

4.041

CALCULATING THE SQUARE ROOT There are several methods for finding the square root of a number. One is to use Table 6.1, another is to use a calculator with a square root function, and the third is to actually calculate the number. This calculation method will be presented in this section. This method is difficult to remember. The steps in this method will include an example. Example 1: Find the square root of 579.1. Round answer to the 1/10s place. To find the square root of any decimal number, with the result rounded to a given place value: Step 1. Write the

symbol over the given number. 579

Step 2. Mark off pairs of digits in both directions from the decimal point. 5, 79.10, 00, Step 3. Put a decimal point in the answer above the decimal point in the given number. . 5, 79.10, 00, Step 4. Over the group of numbers farthest left (a pair or single), put the largest number whose square is not larger than the number in the left group. . 2 5, 79.10, 00, 2 is the largest number whose square (4) is not larger than 5.

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Step 5. Square the first number on top (2) and put the result (4) under the number in the far left group and subtract. . 2 5, 79.10, 00, 4 1 Step 6. Add the barrier (a line) to the left of the answer (1) to the subtraction.

barrier

. 2 5, 79.10, 00, 4 1

Step 7. Double the entire number on top (2), write it (4) to the left of the barrier, leaving one space (place value) to the left of the barrier empty (). Step 8. Bring down the next pair of numbers (79), putting them to the right of the answer to the subtraction (1) of Step 5. . 2 5, 79.10, 00, 4 4  1 79 Step 9. Divide the number to the left of the barrier (4 |1) into the number to the right of the barrier (179). (a) The result (4) goes over the pair of numbers last brought down (79). (b) The number in the blank spot () left of the barrier (4) must be the same as the new answer on top (4). (c) Multiply the answer (4) times the whole number to the left of the barrier (44); put down the product (like long division). 24 . 5, 79.10, 00, 4 4 4  1 79 1 76 3 All other steps. Repeat steps 6 to 9 over and over until the answer has numbers in the place values asked for. Answer:

24.06

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Example 2: Find the square root of 8.16. Round answer to the 1/100s place. Steps 1–3. . 8. 16, 00, 00, Steps 4–6. 2. 8. 16, 00, 00, 4 4 Steps 7–8. 2. 8. 16, 00, 00, 4 4  4 16 Step 9. 2. 8 8. 16, 00, 00, 4 4  4 16 3 84 32 Other steps: 2. 8 5 6 8. 16, 00, 00, 4 4 8  4 16 3 84 560 3200 2825 5706 37500 34236 3264 Answer:

2.86

Example 3: Find Steps 1–5.

4178 . Round answer to 1/10s place.

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6 . 41, 78.00, 00 36 5 Steps 6–9. 6 4. 41, 78.00, 00 36 12  5 78 4 96 82 All other steps: 6 4.63 41, 78.00, 00 36 12  5 78 4 96 82 00 12 8  77 16 4 84 00 12 92  3 87 69 96 31 Answer:

64.6

EXERCISES Directions: Select the correct value of the square root, rounded to the given place value, for each number. Do not use a calculator. If necessary, restudy the instructions and examples. 1.

2.

3.

4.

47 to 1/10s place. (a) 5.9 (b) 6.71

(c) 6.9

(d) 5.8

538.5 to 1/10s place. (a) 24.1 (b) 269.1

(c) 26.9

(d) 23.2

17360 to 1s place. (a) 132 (b) 130

(c) 131

(d) 129

9.38 to 1/100s place. (a) 3.06 (b) 4.02

(c) 4.18

(d) 3.19

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5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

229

.43 to 1/1000s place. (a) .695 (b) .656

(c) .614

(d) .718

85 to 1/10s place. (a) 9.8 (b) 9.4

(c) 8.7

(d) 9.2

4.93 to 1/1000s place. (a) 2.220 (b) 3.104

(c) 3.106

(d) 2.008

8143 to 1s place. (a) 85 (b) 90

(c) 89

(d) 87

191.25 to 1/1000s place. (a) 14.213 (b) 14.108

(c) 13.829

(d) 13.146

.982 to 1/1000s place. (a) .985 (b) .914

(c) .991

(d) .875

90 to 1/10s place. (a) 9.5 (b) 8.1

(c) 9.7

(d) 9.8

3.17 to 1/1000s place. (a) 1.715 (b) 1.675

(c) 1.780

(d) 1.438

417 to 1/10s place. (a) 19.9 (b) 20.4

(c) 20.18

(d) 19.4

93417 to 1s place. (a) 302 (b) 309

(c) 315

(d) 306

.845 to 1/100s place. (a) .90 (b) .92

(c) .91

(d) .93

Answers to the exercises: 1. c 6. d 11. a

2. d 7. a 12. c

3. a 8. b 13. b

4. a 9. c 14. d

5. b 10. c 15. b

ADDITIONAL EXERCISES Directions: Select the correct value for the square root or the cube root for each of the following whole numbers from Table 16.1. In the answers, the choice “not in” means the square or cube root is not in the table. If necessary, restudy the instructions and examples.

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1. Find the square root of 4. (a) 16 (b) 1

(c) 2

(d) 64

2. What is the square root of 49? (a) not in (b) 2401

(c) 3.659

(d) 7

3. Find the cube root of 8. (a) 2 (b) 64

(c) 2.828

(d) 4

4. What is the value of 64 ? (a) 4 (b) 8

(c) not in

(d) 32

5. Find the square root of 18. (a) 2.621 (b) 9

(c) 4.243

(d) not in

6. Find the cube root of 22. (a) 4.690 (b) 6.104

(c) 2.802

(d) 7.333

7. Find the square root of 84. (a) 21 (b) 9.165

(c) 42

(d) 4.380

8. What is the value of 242 ? (a) 16.917 (b) 121

(c) 18.213

(d) not in

9. What is the value of 3 83 ? (a) 4.362 (b) 8.741

(c) 9.110

(d) 27.231

10. Find the square root of 784. (a) not in (b) 28

(c) 19.4

(d) 392

11. Find the square root of 16. (a) 4 (b) 2.520

(c) 8

(d) 2

12. What is the value of 36 ? (a) not in (b) 6

(c) 18

(d) 3.302

13. What is the cube root of 27? (a) 5.196 (b) 9

(c) 3

(d) not in

14. Find the square root of 14. (a) 7 (b) 5.106

(c) 2.410

(d) 3.742

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15. What is the square root of 79? (a) 9.27 (b) 4.291

(c) 8.888

(d) 38.51

16. What is the value of 3 46 ? (a) 6.782 (b) 5.804

(c) 3.583

(d) 15.351

17. Find the square root of 6241. (a) 810.4 (b) 79

(c) 3120

(d) not in

18. What is the value of 95 ? (a) 4.563 (b) 4841

(c) not in

(d) 9.747

19. What is the cube root of 25? (a) 8.333 (b) 5

(c) 625

(d) 2.924

20. Find the square root of 8000. (a) 90.18 (b) not in

(c) 4000

(d) 40

Answers to the additional exercises: 1. 6. 11. 16.

c c a c

2. 7. 12. 17.

d b b b

3. 8. 13. 18.

a d c d

4. 9. 14. 19.

b a d d

5. 10. 15. 20.

c b c b

APPLICATION OF SQUARE ROOT This section illustrates square roots used in an applied situation involving any right triangle. To illustrate the application, the names of the sides of a right triangle must be known. A right triangle is a triangle which has one right angle and special names for the sides.

Sides of a right triangle.

Hypotenuse: The side opposite the right angle. Legs: The two sides of the triangle that form the right angle.

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Example 1: Find the length of the hypotenuse in the right triangle:

To find the length of the hypotenuse of a right triangle when the lengths of both legs are known: Step 1. Find the square of each leg and add the squares. Step 2. Find the square root of the answer from Step 1. Solve Example 1. Step 1. The square of 2 is 4. The square of 5 is 25. Add the squares = 29. Step 2. Find the square root of 29. (Round to 1/10s place.) Answer:

5.4

Note: To find the needed square roots use either (a) Table 16.1 or (b) manual calculation (use the method in Calculating the Square Root). Example 2: Find the length of the hypotenuse in the triangle. (Round the answer to 1/10s place.)

Step 1. The square of 17 is 289. The square of 6 is 36. Add the squares = 325. Step 2. Find the square root of 325. Answer:

18.0

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Example 3: Find the length of the hypotenuse in the triangle. (Round answer to nearest 1 inch.)

Step 1. 1 ft 9 in. is 21 in. 2 ft 2 in. is 26 in. 21 squared is 441. 26 squared is 676. Add the squares = 1117. Step 2. Find the square root of 1117. Answer:

33 inches (2 feet, 9 inches)

Example 4: Find the length of the hypotenuse in the triangle below. Provide answer to nearest 1/10 inch written as a decimal.

Step 1. Change both leg lengths to decimals. (Round to 1/100s place.) 3 3/8 = 3.38. 2 1/4 = 2.25. The square of 3.38 is 11.42. The square of 2.25 is 5.06. The sum of the squares is 16.48. Step 2. Find the square root of 16.48. (Round to 1/10s place.) Answer:

4.1 inch

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EXERCISES Directions: Select the correct value for the length of the hypotenuse in each of the following right triangles. If necessary, restudy the instructions and examples. Round all decimal answers off to the 1/10s place. Use Table 16.1. Solve all exercises before looking at the answers.

1. (a) 5.3 ft

(b) 7.2 ft

(c) 10.0 ft

(d) 8.4 ft

(a) 12.1 in.

(b) 14.6 in.

(c) 13.5 in.

(d) 11.2 in.

(a) 7.1 in.

(b) 6.4 in.

(c) 8.3 in.

(d) 5.9 in.

(a) 9.3 ft

(b) 10.5 ft

(c) 11.8 ft

(d) 12.4 ft

2.

3.

4.

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? Round to 1/16”

4 1/2”

5. (a) 5 7/8 in.

(b) 4 7/8 in.

(c) 6 3/4 in.

(d) 6 1/4 in.

Answers to the exercises: 1. b

2. d

3. a

4. b

5. a

REVIEW TEST Directions: The test is to determine if the skills in this chapter have been mastered. I. Select the correct value for the square or cube in each exercise. Round any decimal answers off to the 1/100s place. 1. Find the square of 14. (a) 196 (b) 874 (c) 169 (d) 2,744 2.

3.

4.

5.

6.

7.

8.

What is the cube of 12? (a) 24 (b) 36

(c) 144

(d) 1,728

What is the result of squaring 23.4? (a) 848.45 (b) 547.56 (c) 1,474.9

(d) 70.2

Find the cube of 1/2. (a) 1 (b) 1/8

(c) 3/2

(d) 1/4

Find the square of 4 7/8. (a) 18 19/64 (b) 21 15/64

(c) 16 21/64

(d) 23 49/64

Find the square of 18. (a) 5832 (b) 54

(c) 36

(d) 324

Find the cube of 6. (a) 36 (b) 216

(c) 12

(d) 18

What is the square of 9.4? (a) 88.36 (b) 18.8

(c) 28.2

(d) 318.10

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9.

What is the result of cubing 5.3? (a) 148.88 (b) 45.9 (c) 84.65

10. Find the square of 5 3/8. (a) 28 57/64 (b) 25 9/64

(c) 32 19/64

(d) 28.09

(d) 18 15/64

II. Select the correct answer to each exercise. Use Table 16.1. 11. Find the square root of 81. (a) 9 (b) 4.327 (c) 40.5 (d) 6,561 12. What is the value of 42 ? (a) 21 (b) 3.476

(c) 6.481

(d) 1,674

13. Find the cube root of 17. (a) 8.103 (b) 2.571

(c) 5.67

(d) 4.123

14. Find the square root of 75. (a) 8.66 (b) not in

(c) 8.165

(d) 4.217

15. Find the value of 3 62 . (a) 30.667 (b) 3.958

(c) 7.874

(d) not in

16. Find the square root of 25. (a) 9.81 (b) 5

(c) 6.104

(d) 2.924

17. What is the value of 3 64 ? (a) 2 (b) 8

(c) not in

(d) 4

18. What is the square root of 55? (a) 3.803 (b) not in

(c) 4.175

(d) 7.416

19. Find the cube root of 78. (a) 26 (b) 8.832

(c) not in

(d) 4.273

20. What is the value of 3844 ? (a) 62 (b) 45.108

(c) not in

(d) 1922

III. Select the correct value of the square root rounded to the given place value of each exercise. Do not use a calculator. 21. 318 to 1/10s place. (a) 17.8 (b) 16.3 (c) 17.3 (d) 16.9 22.

7500 to 1s place. (a) 87 (b) 86

(c) 85

(d) 81

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23.

24.

25.

26.

27.

28.

29.

30.

237

8.34 to 1/ 1000s place. (a) 2.079 (b) 2.888

(c) 2.715

(d) 2.984

52.4 to 1/100s place. (a) 6.94 (b) 6.58

(c) 7.18

(d) 7.24

814163 to the 1s place. (a) 894 (b) 902

(c) 917

(d) 951

275 to 1/10s place. (a) 16.3 (b) 16.4

(c) 16.6

(d) 16.5

7 to 1/1000s place. (a) 3.019 (b) 2.646

(c) 2.184

(d) 3.451

.184 to 1/100s place. (a) .31 (b) .41

(c) .43

(d) .39

27, 148 to 1s place. (a) 172 (b) 165

(c) 164

(d) 171

7, 085 to 1/10s place. (a) 84.0 (b) 84.2

(c) 84.1

(d) 84.3

IV. Select the correct value of the square root, rounded to the given place value, of each exercise. Do not use a calculator.

31. (a) 10.1 in.

(b) 9.3 in.

(c) 7.4 in.

(d) 8.6 in.

(a) 11.0 in.

(b) 7.8 in.

(c) 10.4 in.

(d) 8.9 in.

(a) 6.9 in.

(b) 7.1 in.

(c) 6.8 in.

(d) 6.6 in.

32.

33.

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34. (a) 12.7 ft

(b) 13.8 ft

(c) 10.7 ft

(d) 11.4 ft

(a) 8 1/8 in.

(b)7 1/4 in.

(c) 7 3/8 in.

(d) 8 3/16 in.

(a) 13.1 in.

(b) 10.8 in.

(c) 12.4 in.

(d) 11.6 in.

(a) 10.3 in.

(b) 7.4 in.

(c) 8.4 in.

(d) 9.9 in.

(a) 17.1 in.

(b) 16.3 in.

(c) 15.6 in.

(d) 12.1 in.

35.

36.

37.

38.

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239

39. (a) 14.4 in.

(b) 10.1 in.

(c) 10.8 in.

(d) 11.6

(a) 12.83 in.

(b) 8.11 in.

(c) 13.75 in.

(d) 13.38 in.

40.

Answers to the review test: Question

Answer

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

a d b b d d b a a a a c b a b b d d d a a

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Question

Answer

22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

a b d b c b c b b d d d a a b d d b b

SOLVING PROPORTION PROBLEMS Objective: This section describes a proportion and how to find a missing number in a proportion. This section requires solving a simple equation such as .38x = 4.54. To solve for “x” in this case, divide both sides of the equation by .38. The result is x = 4.54/.38 = 11.95. Examples: 1/2 = 3/6, 4/12 = 1/3, 3/8 = 9/24 In each example, the two ratios are actually the same number if both are reduced. The number in the left top times the number in the right bottom equals the number in the right top times the number in the left bottom, e.g., 1/2 = 3/6 1×6=3×2

4/12 = 1/3 4 × 3 = 1 × 12

3/8 = 9/24 3 × 24 = 9 × 8

The process is called cross multiplication. From this, the answer for a missing number (x) in a proportion in the examples below may be found: 1. 3/x = 7/5

2. 5/8 = x/10

3. 4/9.3 = 7.5/x

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Squares, Square Roots, Cubes, Cube Roots, and Proportions

To find the value of the missing number in a proportion: Step 1. Cross multiply and write an equation. a. Multiply left top by right bottom. b. Multiply right top by left bottom. c. Write the result of (a) equal to the result of (b). Step 2. Solve the equation for the value of the missing number. Example 1: 3/x = 7/5 Step 1. 3 × 5 = 7 × x 15 = (7) (x) or just 7x 15 = 7x Step 2. 15 ÷ 7 = x Answer:

2 1/7 = x

Example 2: 5/8 = x/10 Step 1. 5 × 10 = x × 8 50 = 8x Step 2. 50 ÷ 8 = x Answer: 6 1/4 = x Example 3: 1.4/5.3 = 7.5/x Step 1. 1.4 × x = 7.5 × 5.3 1.4x = 39.75 Step 2. x = 39.75 ÷ 1.4 Answer:

x = 28.4 (Rounded off to 1/10s place.)

EXERCISES Directions: Select the correctly written ratio for each of the comparisons. 1. Output to time in minutes if a machine can produce 42 times per hour. (a) 42/1 (b) 7/10 (c) 42/60 (d) 60/42

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2. Area of triangle A to area of triangle B.

(a) 4/1

(b) 1/4

(c) 1/2

(d) 2/1

3. Cost in cents to weight in ounces if oranges cost $1.50 for 6 pounds. (a) 5/2 (b) 1/25 (c) 25/16 (d) 2/5 4. Output to labor costs in dollars if in 1 day 4 men working 8 hours at $7.00 per hour make 18 units. (a) 4/7 (b) 9/14 (c) 9/112 (d) 9/28 5. Cost of owning a car in dollars to time of ownership in years if a car costs $6,000 and all other costs in a 4-year period were $800. (a) 3/425 (b) 1700/1 (c) 425/3 (d) 1/200 Answers to the exercises: 1. b

2. d

3. c

4. c

5. b

ADDITIONAL EXERCISES Directions: Select the correct value for the missing number in each of the proportions. Round any decimal answer to the 1/10s place. If necessary, restudy the instructions and examples. 1. 3/x = 5/8 (a) 10 3/8

(b) 1 7/8

(c) 13 1/3

(d) 4 4/5

2. 7/10 = n/9 (a) 7 7/9

(b) 8 3/4

(c) 6 3/10

(d) 12 6/7

3. 85/40 = 35/x (a) 5 2/7

(b) 9 7/18

(c) 25 3/40

(d) 16 8/17

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4. p/20 = 1/8 (a) 10 7/8

(b) 2/5

(c) 2 1/2

(d) 60

5. 3 1/2 /5 = 7/x (a) 5 1/4

(b) 10

(c) 2 3/4

(d) 15 1/2

T 1 3/8 6. ------------ = -----------3 1/4 9 2/3 (a) 10 3/8

(b) 4 7/78

(c) 75 1/4

(d) 7/178

3 1/2 10 7/8 7. ------------ = --------------5 3/4 S (a) 8 9/10

(b) 7 3/4

(c) 52 9/14

(d) 17 97/112

8. .8/M = 2.4/12 (a) 12.8

(b) 36

(c) .2

(d) 4

(c) 1.3

(d) 24.8

(c) 23.2

(d) 19.0

9. U/3.5 = 4.6/12.3 (a) 16.2 (b) 9.4 10. 24.1/8 = 8.3/V (a) 2.8

(b) 25.0

Answers to the additional exercises: 1. d 6. b

2. c 7. d

3. d 8. d

4. c 9. c

5. b 10. a

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17

Scientific Notation and Powers of Ten

When working with numbers there will be times when it is necessary to work with very large and very small numbers. A method will now be demonstrated to make this task easier. The method is called scientific notation or powers of ten. Definitions: 1. Scientific notation: a method to keep track of the decimal point in large or small numbers or when changing from one metric unit to another 2. Powers of ten: another name for scientific notation 3. Factor: any number that is multiplied by another number to obtain a final answer which is called a product 4. Factoring: the process of multiplying numbers together 5. Exponent: a number written above and to the right of another number, telling how many times the number is used as a factor 6. Exponential numbers: any decimal number written using an exponent

NUMBERS GREATER THAN ONE FACTORING

AND

WRITING NUMBERS

Objective: This section demonstrates the correct method of factoring and writing numbers using exponents. Practice: If 50 can be written as 5 × 10, how can the following numbers be written? 1. 30 (a) 3 × 1 Answer: b

(b) 3 × 10

(c) 3 × 100

2. 500 (a) 5 × 10 Answer: b

(b) 5 × 100

(c) 50 × 100

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3. 2000 (a) 2 × 100 Answer: c

(b) 21 × 10

(c) 2 × 1000

4. 100 (a) 10 × 10 Answer: a

(b) 10 × 1

(c) 10 × 100

Because 100 can be written as 10 × 10, 100 can be factored into 2 equal parts. Also 1000 can be written 10 × 10 × 10. Therefore, 1000 can be factored into 3 equal parts. These equal parts can be indicated by using exponents. An exponent is a number written above and to the right of another number and it tells how many times the number is used as a factor. Examples: 1 10 100 1000

= = = =

100 101 102 103

or 10 × 10 or 10 × 10 × 10

If 1000 can be written as 103 and 2000 can be written as 2 × 1000, then 2000 can also be written as 2 × 103 (1000 = 103). Note: There are 3 zeros in 1000. The exponent of 10 is also 3. Practice: How can the following numbers be written? 1. 300 (a) 3 × 101 Answer: b

(b) 3 × 102

(c) 3 × 103

2. 500 (a) 5 × 102 Answer: a

(b) 50 × 102

(c) 5 × 101

3. 3000 (a) 30 × 101 Answer: c

(b) 30 × 103

(c) 3 × 103

4. 80000 (a) 800 × 103 Answer: c

(b) 80 × 102

(c) 8 × 104

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FACTORING EXERCISES Directions: Select the correct answers. If necessary, restudy the instructions and examples. 1. 70 (a) 7 × 1

(b) 7 × 101

(c) 7 × 100

2. 40 (a) 4 × 100

(b) 4 × 1

(c) 4 × 101

3. 600 (a) 6 × 101

(b) 6 × 100

(c) 6 × 1

4. 900 (a) 9 × 100

(b) 9 × 101

(c) 9 × 100

5. 2000 (a) 2 × 1000

(b) 2 × 100

(c) 2 × 101

6. 700 (a) 7 × 101

(b) 7 × 102

(c) 7 × 103

7. 4000 (a) 4 × 102

(b) 4 × 103

(c) 4 × 104

8. 90 (a) 9 × 101

(b) 9 × 102

(c) 9 × 103

9. 80000 (a) 8 × 102

(b) 8 × 103

(c) 8 × 104

10. 5000 (a) 5 × 102

(b) 5 × 103

(c) 5 × 104

Answers to the factoring exercises: 1. b 6. b

2. c 7. b

3. b 8. a

4. c 9. c

5. a 10. b

PLACING A DECIMAL POINT Objective: This section will demonstrate the correct method of placing a decimal point in standard numbers and standard exponents.

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Remember: If a decimal point is not shown in a number, always assume that there is a decimal point at the far right end of the number. Examples: 15 = 15.

650 = 650.

Practice: Using the examples as reference, how should the following numbers be written? 1. 15 (a) 1.5 Answer: c

(b) .15

(c) 15.

2. 65 (a) .65 Answer: b

(b) 65.

(c) 6.5

3. 725 (a) 72.5 Answer: b

(b) 725.

(c) .725

4. 1050 (a) 1050 Answer: a

(b) 10.50

(c) 105.0

The number 15 can also be written as 1.5 × 101 (1.5 × 101 = 15). The number 65 can also be written as 6.5 × 101 (6.5 × 101 = 65). Also 150 can be written as 1.5 × 100 and 650 as 6.5 × 100 (1.5 × 100 = 150 and 6.5 × 100 = 650). Therefore, if 100 can be written as 102 (10 × 10), 150 can be written as 1.5 × 102 and 650 can be written as 6.5 × 102. Note: The decimal point has been moved to the left the same number of places shown by the exponent. 150. = 1.50 × 10

2

650. = 6.50 × 10

2

1.5 × 102 and 6.5 × 102 are examples of numbers written in standard exponential form. Standard exponential form always has one number to the left of the decimal point.

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Examples: 1.5 × 102

6.5 × 102

Practice: Using the examples as a reference, how should the following numbers be written in standard exponential form? 1. 450 (a) 45 × 101 Answer: b

(b) 4.5 × 102

(c) .45 × 103

2. 275 (a) 2.75 × 103 Answer: c

(b) 27.5 × 101

(c) 2.75 × 102

3. 8050 (a) 8.05 × 103 Answer: a

(b) 80.50 × 102

(c) 805 × 101

4. 525 (a) 5.25 × 104 Answer: c

(b) 5.25 × 103

(c) 5.25 × 102

Remember: Standard exponential form always has one number to the left of the decimal point.

NUMBERS LESS THAN ONE CHANGING EXPONENTIAL NUMBERS

TO

STANDARD DECIMAL FORM

Objective: This section demonstrates the correct method of changing exponential numbers back to standard decimal form. Remember: There will be times when exponential numbers need to be rewritten into standard form. To change 6.5 × 102 back to standard form, first look at the exponent. In this case, it is 2. This means that the decimal point was shifted to the left 2 places. If the decimal point is moved back to its original place, it would have to be moved to the right 2 places. 2

6.5 × 10 = 6 50.

Note: The decimal point has been moved to the right the same number of places as shown by the exponent.

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Examples: 2

2

1.5 × 10 = 1 50.

7.25 × 10 = 7 25.

Practice: Using the examples as a reference, change the following exponential numbers to standard decimal form. 1. 7.5 × 102 (a) 75 Answer: b

(b) 750

(c) 7500

2. 6.25 × 103 (a) 6250 Answer: a

(b) 625

(c) 62.5

3. 5 × 103 (a) 50 Answer: c

(b) 500

(c) 5000

4. 9.1 × 104 (a) 910 Answer: c

(b) 9100

(c) 91000

WRITING NUMBERS LESS THAN ONE

IN

STANDARD FORM

Objective: This section will demonstrate the correct method of writing numbers less than one in standard exponential form. Remember: Standard exponential form always has one number to the left of the decimal point. Examples: 1.5 × 10–2

6.5 × 10–2

Remember: Use negative exponents (10–2) to represent numbers less than one. 1 = 100 .1 = 10–1 or 1/10

.01 = 10–2 or 1/100 .001 = 10–3 or 1/1000

To change a number less than one to standard exponential form, move the decimal point to the right enough places to get the number to standard exponential form. The number of places that the decimal point has to be moved will always be equal to the exponent.

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251

Examples: 01.5 = 1.5 × 10

–2

06.5 = 6.5 × 10

–2

Practice: Using the examples as a reference, change the following numbers to standard exponential form. 1. .0035 (a) 3.5 × 10–3 Answer: a

(b) 35 × 10–3

(c) 350

2. .0275 (a) 2.75 × 10–2 Answer: a

(b) 2750

(c) 27.50 × 10–3

3. .06 (a) 60 × 10–3 Answer: c

(b) 6.0 × 10–3

(c) 6 × 10–2

4. .0072 (a) 7.2 × 10–2 Answer: b

(b) 7.2 × 10–3

(c) 7.2 × 10–4

Practice: Using the examples as a reference, change the following numbers to standard decimal form. 1. 7.5 × 10–2 = (a) .75 Answer: b

(b) .075

(c) .0075

2. 6.25 × 10–3 = (a) .00625 Answer: a

(b) .625

(c) .000625

3. 5 × 10–3 = (a) .005 Answer: a

(b) .05

(c) .0005

4. 9.1 × 10–4 = (a) .0091 Answer: c

(b) .0910

(c) .00091

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Whenever working with scientific notation, or powers of ten, it is sometimes necessary to change numbers from their standard form to perform many basic mathematics operations. These changes are accomplished by shifting the decimal point. Definitions: 1. Addition: the sum of two numbers 2. Subtraction: a process of taking a smaller amount from a larger amount to find the difference between the two amounts (numbers) 3. Multiplication: the process to find the result of adding a certain number to itself a certain amount of times 4. Product: a number that is the result of multiplying (In 4 × 8 = 32, 32 is the product.) 5. Digit: any number from 0 through 9 6. Division: the process of determining how many times one number is contained in another 7. Divisor: the number by which another number is divided (In 32 ÷ 8 = 4, 8 is the divisor.) 8. Dividend: the number into which another is divided (In 32 ÷ 8 = 4, 32 is the dividend.) 9. Quotient: the answer to a division problem (In 32 ÷ 8 = 4, 4 is the quotient.)

NUMBERS GREATER THAN ONE Objective: This section will demonstrate the correct method to compare exponential numbers greater than one and to shift the decimal point without changing the value of the number. It is often necessary to compare numbers written in exponential form and to determine which number is larger and which number is smaller. Keep in mind that it is usually the exponent that determines whether a number is larger or smaller: 105 is larger than 104 and 1.5 × 103 is larger than 2.5 × 102

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When the exponents are the same, look at the decimal number to determine which number is larger: 26.5 × 102 is larger than 10.5 × 102 Practice: Using the examples as a reference, which numbers are largest? 1. (a) 6.27 × 102 Answer: c

(b) 8.59 × 102

(c) 4.1 × 103

2. (a) 9.826 × 104 Answer: b

(b) 5.29 × 106

(c) 1.1 × 105

3. (a) 1.5 × 105 Answer: a

(b) 8.724 × 103

(c) 4.79 × 104

4. (a) 3.462 × 104 Answer: c

(b) 9.2 × 104

(c) 2 × 105

So far, all of the numbers in this section have been written in standard exponential form; however, there are times when numbers are used that are not written in standard form. In the number 225.5 × 103, notice that it is written in exponential form, but it is not written in standard form. Practice: Standard exponential form always has one number to the left of the decimal point. To change this number to standard form, shift the decimal point two places to the left: 2.25 5 × 10

3

Is 2.255 × 103 the same as 225.5 × 103? It is not. By shifting the decimal point, the value of the number has been changed. To keep the number value the same, the exponent must be changed by the same amount. The decimal point was moved two places to the left: 2.25 5 × 10

3

If the decimal point is shifted to the left, the exponent must increase by the same amount: 103 must now become 105 or 2.255 × 105 = 225.5 × 103. If 49.5 × 103 were to be changed to standard form, the decimal point would have to be shifted 1 place to the left. What would the correct answer be? (a) 4.95 × 104

(b) 4.95 × 103

(c) 4.95 × 102

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The correct answer is a. Remember: When the decimal point is shifted, the exponent must change by the same amount. Example:

Change .6452 × 103 to standard form. (a) 6.452 × 104

(b) 6.452 × 102

(c) 6.452 × 105

Answer: b Practice: Using the examples as a reference, change the following exponential numbers into standard form. 1. 395.4 × 102 = (a) 3.954 × 103 Answer: b

(b) 3.954 × 104

(c) 3.954 × 101

2. .463 × 106 = (a) 4.63 × 104 Answer: b

(b) 4.63 × 105

(c) 4.63 × 107

3. 2725.1 × 104 = (a) 2.7251 × 107 (b) 2.7251 × 101 (c) 2.7251 × 108 Answer: a 4. .0325 × 106 = (a) 3.25 × 105 Answer: b

(b) 3.25 × 104

(c) 3.25 × 108

Sometimes it will be necessary to change numbers from standard form to an alternate form of scientific notation. To change 5.555 × 106 to an exponential number needing an (× 104) exponent, it cannot be written as 5.555 × 104 because 5.555 × 104 does not equal 5.555 × 106. To change the exponent, the decimal point must also be changed. When changing the exponent, the decimal number has to be changed by the same amount. To change (× 106) into (× 104), the exponent has been decreased by 2. This means our decimal must be increased by 2 places. 6

5.555 × 10 = 5 55.5 × 10

4

Practice: Using the examples as a reference, change the numbers written in standard form into an alternate form.

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1. 2.52 × 105 = (a) 2.52 × 104 Answer: b

(b) 252 × 103

(c) 25.2 × 103

2. 7.923 × 106 = (a) 7923 × 103 Answer: a

(b) 792.3 × 103

(c) 79.23 × 104

3. 6.362 × 107 = (a) 636.2 × 106 Answer: c

(b) 636.2 × 103

(c) 6362 × 104

4. 86 × 104 = (a) 860 × 103 Answer: a

(b) 86 × 103

(c) 8600 × 104

To change an exponent from a smaller number to a larger number, the decimal must also be changed by the same amount. To change 5.555 × 104 to an exponential number needing an (× 106) exponent, it cannot be written as 5.555 × 106 because 5.555 × 106 does not equal 5.555 × 104. To change × 104 into × 106, the exponent has been increased by two. This means the decimal must be decreased by two. 4

5.555 × 10 = .05 55 × 10

6

Practice: Using the examples as a reference, change the numbers written in standard form into an alternate form. 1. 2.52 × 105 = (a) 25.2 × 107 Answer: b

(b) .252 × 106

(c) 252 × 108

2. 7.923 × 106 = (a) .7923 × 107 Answer: a

(b) 7923. × 107

(c) 792.3 × 105

3. 6.326 × 107 = (a) 63.62 × 107 Answer: b

(b) .6362 × 108

(c) .6362 × 106

4. 8.6 × 105 = (a) 86 × 106 Answer: c

(b) .86 × 104

(c) .86 × 106

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NUMBERS LESS THAN ONE Objective: This section will demonstrate the correct method of comparing exponential numbers less than one and shifting the decimal point without changing the value of the number. It may also be necessary to compare numbers less than one which are written in exponential form and to determine which number is larger and which number is smaller. Remember: It is usually the exponent that determines whether the number is larger or smaller. Also remember: When working with numbers less than one, use negative exponents. (If negative numbers are troublesome review Chapter 17, Numbers Less Than One.) Compare the following two numbers and determine which one is larger: (a) –5 or (b) –9 The answer is a. Hint: In terms of temperature, (a) would be 5 degrees below zero. Because 5 below zero is warmer than 9 below zero, 5 below zero is a higher temperature, or –5 is a larger number than –9. –9

–8

–7

–6

–5

–4

–3

–2

–1

0

1

2

Smaller

(a) –1 or (b) –4 The answer is b. Practice: Use the same idea when working with exponents.

Which is smaller? (a) 10–1 (b) 10–4 Answer: b

4

5

6

7 8

Larger

Using the illustration, which of the numbers is smaller?

Which is larger? (b) 10–9 (a) 10–5 Answer: a

3

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Which is larger? (a) 2.65 × 10–6 (b) 3.75 × 10–3 Answer: b Which is smaller? (a) 3.27 × 10–5 (b) 6.73 × 10–2 Answer: a Practice: Using the examples as a reference, determine which of the numbers is the largest. 1. (a) 6.24 × 10–5 Answer: c

(b) 8.75 × 10–6

(c) 4.39 × 10–4

2. (a) 2.375 × 10–2 (b) 8.673 × 10–6 Answer: a

(c) 1.45 × 10–3

3. (a) 5.75 × 10–8 Answer: c

(c) 5.27 × 10–2

(b) 1.27 × 10–4

4. (a) 2.365 × 10–1 (b) 4.72 × 10–5 Answer: a

(c) 6.314 × 10–8

All of the numbers above are written in standard exponential form. However, there will be times when numbers not written in standard form are used. In the number 437.5 × 10–4, notice that it is written in exponential form, but it is not written in standard form. To change this number into standard form, shift the decimal point two places to the left: 4.37 5 × 10

–4

Is 4.375 × 10–4 the same as 437.5 × 104? It is not. By shifting the decimal point, the value of the number has been changed. To keep the number value the same, the negative exponent must also be changed by the same amount. Move the decimal point two places to the left: 4.37 5 × 10

–4

If the decimal point is shifted to the left, the exponent must increase by the same amount: 10–4 must now become 10–2. 4.375 × 10–2 = 437.5 × 10–4

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Remember: –2 is larger than –4. If 55.5 × 10–3 were changed to standard form, the decimal point would be shifted 1 place to the left. What would the correct answer be? (a) 5.55 × 10–4

(b) 5.55 × 10–3

(c) 5.55 × 10–2

The answer is c. Remember: When the decimal point is shifted, the exponent must change by the same amount. Change .7651 × 10–3 to standard form. What would the answer be? (a) 7.651 × 10–2

(b) 7.651 × 10–4

(c) 7.651 × 10–3

The answer is b. Practice: Using the examples as a reference, change the exponential numbers into standard form. 1. 927.5 × 10–3 = (a) 9.275 × 10–1 (b) 9.275 × 10–5 Answer: a

(c) 9.275 × 10–2

2. .723 × 10–6 = (a) 7.23 × 10–5 Answer: c

(c) 7.23 × 10–7

(b) 7.23 × 10–8

3. 8756.2 × 10–7 = (a) 8.7562 × 10–3 (b) 8.7562 × 10–4 (c) 8.7562 × 10–10 Answer: b 4. .8 × 10–5 = (a) 8 × 10–4 Answer: b

(b) 8 × 10–6

(c) 8 × 10–5

ADDITION OF DECIMAL NUMBERS SAME EXPONENTS Objective: This section will demonstrate the correct method to add numbers in scientific notation when the numbers have the same exponents. There are two important rules to remember when adding numbers written in scientific notation: 1. Decimal points must line up with each other. 2. Exponents must be the same.

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First work with Rule 1. It has already stated that if a decimal point is not shown in a number, always assume that there is one at the far right end of the number. Examples: 15 = 15.

650 = 650.

When adding 2 + 2 = 4, it is actually 2. + 2. = 4. or

2. + 2. --------4.

The decimal points are lined up. If adding 23.2 + 1.35 = 24.55 or 23.2 + 1.35 --------------24.55 The decimal points are lined up. Not 23.2 + 1.35 --------------367 The decimal points are not lined up. Practice: Using the examples as a reference, choose the correct answer to the addition problems. 1. 26.5 + 18.6 = (a) 45.1 Answer: a

(b) 4.51

(c) 27.96

2. 39.7 + 2.81 = (a) 67.8 Answer: c

(b) 6.78

(c) 42.51

3. 101.32 + 1.01 = (a) 102.33 (b) 202.32 Answer: a

(c) 111.42

4. .0235 + 47.3 = (a) 282.3 Answer: c

(c) 47.3235

(b) 708

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If 2.54 × 103 were to be written as a standard number, it would be 2540. If 6.32 × 103 were to be written as a standard number, it would be 6320. Therefore, 2540 + 6320 would equal 8860 or 2540 + 6320 ----------------8860 Also, adding 2.54 × 103 + 6.32 × 103 would equal 8.86 × 103 or 2.54 × 10

3 3

6.32 × 10 ------------------------3 8.86 × 10

3

8.86 × 10 = 8860

When adding exponential numbers, be sure decimal points are lined up and exponents are the same (× 103 = × 103). When adding exponents, the answer will have the same exponent. Examples: 2.54 × 10

3

13.7 × 10

3

+ 1.3 × 10 ---------------------------–4 15.0 × 10

+ 6.32 × 10 -----------------------------38.86 × 10

–4 –4

In both examples, the decimals line up. Practice: Using the examples as a reference, choose the correct answer to the addition problems. 1. 15.6 × 103 + 12.2 × 103 = (a) 27.8 × 106 (b) 27.8 × 103 Answer: b

(c) 278 × 103

2. .135 × 104 + 13.2 × 104 = (a) 13.335 × 104 (b) 267 × 104 Answer: a

(c) 26.7 × 108

3. .025 × 10–2 + 1.13 × 10–2 = (a) 113.25 × 10–2 (b) 11.3025 × 10–4 (c) 1.155 × 10–2 Answer: c 4. 1.271 × 10–6 + .312 × 10–6 = (a) 1.583 × 10–6 (b) 1.583 × 10–12 (c) 1583 × 10–6 Answer: a

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DIFFERENT EXPONENTS Objective: This section will demonstrate the correct method to add numbers written in scientific notation when the numbers do not have the same exponents. Two important rules to remember when adding numbers written in scientific notation are 1. Decimal points must line up with each other. 2. Exponents must be the same. This section uses Rule 2. How to shift decimal points and how to change exponents so exponential numbers do not change in value has already been demonstrated. Remember: Any time an exponent is changed, the decimal point must be shifted by the same amount. Examples: 1. To change 37.45 × 104 to an exponent of 103, shift the decimal point one place to the right: 37.4 5 × 10

3

2. To change 863.95 × 104 to an exponent of 106, shift the decimal point two places to the left: 8.63 95 × 10

6

3. To change 9.035 × 10–3 to a negative exponent of 10–4, shift the decimal point one place to the right: 9 0 35 × 10

4

4. To change 3.76 × 10–6 to a negative exponent of 10–4, shift the decimal point two places to the left: .03 76 × 10

–4

Practice: Using the examples as a reference, choose the correct answer for the problems.

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1. 8.47 × 107 = (a) 84.7 × 108 Answer: b

(b) .847 × 108

(c) .0847 × 105

2. 96.31 × 104 = (a) 9.631 × 103 Answer: b

(b) 963.1 × 103

(c) 9631 × 103

3. .0175 × 10–4 = (a) .000175 × 10–2 (b) 1.75 × 10–2 Answer: a

(c) .0017 × 10–5

4. 1.267 × 10–6 = (a) 12.67 × 10–5 (b) 126.7 × 10–4 Answer: c

(c) 12.67 × 10–7

Keeping in mind the two rules, exponential numbers with different exponents may be added. To add 3.57 × 104 + 6.36 × 105, first change one of the exponents so that both of them will be the same. Change × 104 to × 105. 4

3.57 × 10 = .3 57 × 10

5

(104 was changed to 105 and the decimal was shifted one place to the left.) Now .357 × 105 + 6.36 × 105 = 6.717 × 105 or .357 × 10

5

5

+ 6.36 × 10 --------------------------------5 6.717 × 10 Note: Decimal points are lined up and exponents are the same. Practice: Using the examples as a reference, solve the addition problems. (Always change the first exponent when necessary.) 1. 4.63 × 104 + 83.17 × 106 = (a) 83.2163 × 106 (b) 8.32163 × 104 (c) 87.80 × 106 Answer: a 2. 25.4 × 102 + 38.65 × 104 = (a) 41.19 × 104 (b) 38.904 × 104 (c) 411.9 × 104 Answer: b

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3. .014 × 10–3 + 3.51 × 10–2 = (a) 3.65 × 10–2 (b) 3.524 × 10–2 Answer: c

(c) 3.5114 × 10–2

4. .273 × 10–4 + 4.73 × 10–6 = (a) 32.03 × 10–6 (b) 7.46 × 10–6 Answer: a

(c) 4.7573 × 10–6

EXERCISES Directions: In each exercise, select the correct answer. If necessary, restudy the instructions and examples. (Always change the first exponent when necessary.) 1. 2.59 × 104 + 6.73 × 105 = (a) 6.989 × 105 (b) 32.63 × 105

(c) 9.32 × 105

2. 26.95 × 107 + 8.24 × 105 = (a) 277.74 × 105 (b) 2703.24 × 105 (c) 10.935 × 105 3. .043 × 102 + 6.37 × 104 = (a) 10.67 × 104 (b) 6.37043 × 104 (c) 49.37 × 104 4. 28.3 × 10–4 + .814 × 10–6 = (a) 1.097 × 10–6 (b) 3.614 × 10–6

(c) 2830.814 × 10–6

5. 5.37 × 10–1 + .37 × 10–4 = (a) 53.737 × 10–4 (b) 5370.37 × 10–4 (c) .0907 × 10–4 6. 27.23 × 105 + 18.43 × 103 = (a) 2115.3 × 103 (b) 2741.43 × 103 (c) 18.45723 × 103 7. 476.1 × 102 + 37.43 × 105 = (a) 37.9061 × 105 (b) 476.43 × 105 (c) 3790.61 × 105 8. 8.3 × 104 + 4.7 × 106 = (a) 37.70 × 104 (b) 13.00 × 1010

(c) 4.783 × 106

9. .123 × 10–4 + .432 × 10–1 = (a) .432123 × 10–1 (b) 123.432 × 10–1 (c) 456 × 10–1 10. 31.41 × 10–3 + 2.17 × 10–6 = (a) 2.74141 × 10–6 (b) 31412.17 × 10–6 (c) 314.1217 × 10–6 Answers to the exercises: 1. a 6. b

2. b 7. a

3. b 8. c

4. c 9. a

5. b 10. b

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SUBTRACTION OF DECIMAL NUMBERS SAME EXPONENTS Skills to remember are how to shift the decimal point to make all numbers have identical exponents and how to subtract decimal numbers. First, do a simple subtraction of a decimal number. (Review the process in Chapters 7, 11, and 14.)

EXERCISES Directions: In each exercise, select the correct answer. If necessary, restudy the instructions and examples. 1. 47.6 – 21.35 = (a) 26.25

(b) 262.5

(c) 45.465

2. 87.25 – 6.37 = (a) 8.088

(b) 80.88

(c) 23.55

3. 925.01 – 31.3 = (a) 61.201 (b) 612.01

(c) 893.71

4. .136 – .0275 = (a) 13.5725

(c) 1.085

(b) .1085

5. 2601.01 – 6.937 = (a) 19.064 (b) 190.64

(c) 2594.073

6. 814.01 × 104 – 21.375 × 104 = (a) 792.635 × 104 (b) 600.26 × 104 (c) 600.26 × 108 7. 143.075 × 109 – 19.62 × 109 = (a) 141.113 × 109 (b) 123.455 × 209 (c) 12.3455 × 109 8. 600.35 × 106 – 1.0754 × 106 = (a) 49.281 × 106 (b) 589.596 × 106 (c) 599.2746 × 106 9. .147 × 10–2 – .0641 × 10–2 = (a) .0829 × 10–2 (b) .829 × 10–2

(c) .14059 × 10–2

10. 16.301 × 10–8 – .01395 × 10–8 = (a) 16.1615 × 10–8 (b) 149.06 × 10–16 (c) 16.28705 × 10–8 Answers to the exercises: 1. a 6. a

2. b 7. b

3. c 8. c

4. b 9. a

5. c 10. c

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DIFFERENT EXPONENTS Objective: This section will demonstrate the correct method to subtract numbers written in scientific notation when the numbers do not have the same exponents. Two important rules to remember when subtracting numbers written in scientific notation are: 1. Decimal points must line up with each other. 2. Exponents must be the same. This section will use Rule 2. How to shift decimal points and how to change exponents so that exponential numbers did not change in value was demonstrated earlier. Remember: Any time an exponent is changed, the decimal point must be shifted by the same number. Examples: 1. To change 63.27 × 105 to an exponent of 104, shift the decimal point one place to the right: 63 2.7 × 10

4

2. To change 435.71 × 106 to an exponent of 108, shift the decimal point two places to the left: 4.35 71 × 10

8

3. To change 7.057 × 10–5 to a negative exponent of 10–6, shift the decimal point one place to the right: 7 0 57 × 10

–6

4. To change 9.34 × 10–4 to a negative exponent of 10–2, shift the decimal point two places to the left: .09 34 × 10

–2

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Practice: Using the examples as a reference, choose the correct answer for the problems. 1. 9.45 × 105 = (a) .945 × 106 Answer: a

(b) 94.5 × 106

(c) .945 × 107

2. 26.45 × 107 = (a) 2645 × 106 Answer: c

(b) 2.645 × 106

(c) 264.5 × 106

3. .0247 × 10–6 = (a) 2.57 × 10–4 Answer: b

(b) .000247 × 10–4 (c) .00247 × 10–7

4. 2.643 × 10–3 = (a) 26.43 × 10–4 (b) 26.67 × 10–2 Answer: a

(c) 126.7 × 10–1

Keeping in mind the two rules, exponential numbers with different exponents may be subtracted. To subtract 6.23 × 103 from 8.27 × 104, first change one of the exponents to make both of them the same. Change × 103 to 104 3

6.23 × 10 = .6 23 × 10

4

(103 was changed to 104 and the decimal was shifted one place to the left.) Now 8.27 × 104 – .623 × 104 = 7.647 × 104 or 8.270 × 10

4 4

– .623 × 10 -----------------------------4 7.647 × 10 Note: Decimal points line up and exponents are the same. Practice: Using the examples as a reference, solve the subtraction problems. (Always change the first exponent when necessary.) 1. 6.24 × 103 – 8.71 × 102 = (a) 53.69 × 102 (b) 615.29 × 102 (c) 2.47 × 102 Answer: a

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2. 33.4 × 103 – .3101 × 104 = (a) 33.0899 × 104 (b) 3.0299 × 104 (c) 30.299 × 104 Answer: b 3. .215 × 10–3 – 1.37 × 10–4 = (a) .078 × 10–4 (b) 20.13 × 10–4 Answer: c

(c) .78 × 10–4

4. 8.76 × 10–6 – .0123 × 10–4 = (a) .0753 × 10–4 (b) .753 × 10–4 Answer: a

(c) 8.7477 × 10–4

MULTIPLICATION OF DECIMAL NUMBERS Objective: This section will demonstrate the correct method to add the exponent part of numbers written in scientific notation when solving problems in multiplication. To facilitate understanding, the multiplication process will be broken down into two separate parts. One part is adding the exponents together and the other part is multiplication of the decimal portion. This section will present the exponent part. All that has to be done with the exponent part is to add the exponents together. Example: To multiply 2 × 104 by 2 × 104, simply separate the exponent part from the decimal part and add the exponents together: 104 × 104 = 108 It is that simple. Practice: Using this example as a reference, solve the problems. 1. 103 × 102 = (a) 106 Answer: b

(b) 105

(c) 102

2. 107 × 104 = (a) 103 Answer: c

(b) 1028

(c) 1011

3. Multiply 108 by 103 = (a) 1024 (b) 105 Answer: c

(c) 1011

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4. 102 times 105 = (a) 105 Answer: c

269

(b) 1010

(c) 107

The same process is used with negative exponents. Example: To multiply 2 × 10–3 by 2 × 10–4, simply separate the exponent part from the decimal part and add the exponents together. 10–3 × 10–4 = 10–7 Practice: Using this example as a reference, solve the problems. 1. Multiply 10–5 by 10–2 = (a) 10–3 (b) 10–7 Answer: b

(c) 10–10

2. 10–7 × 10–4 = (a) 10–28 Answer: c

(b) 10–3

(c) 10–11

3. 10–1 × 10–5 = (a) 10–6 Answer: a

(b) 105

(c) 10–5

4. 10–2 × 10–5 = (a) 10–7 Answer: a

(b) 10–3

(c) 10–10

To work with both positive and negative exponents, still add them together, but the answer must have the sign of the larger exponent. Example: To multiply 2 × 10–5 by 2 × 103, again simply separate the exponent part from the decimal part and add the exponents together. 10–5 × 103 = 10–2 –5 is the larger numeral of the two, so the answer is also (–) negative. Exercise: Using this example as a reference, solve the problems. 1. 10–7 × 103 = (a) 104 Answer: c

(b) 10–10

(c) 10–4

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2. Multiply 105 by 10–3 = (a) 10–2 (b) 102 Answer: b

(c) 10–15

3. 103 times 10–6 = (a) 1018 (b) 103 Answer: c

(c) 10–3

4. 10–2 × 108 = (a) 106 Answer: a

(c) 1010

(b) 10–16

DIVISION OF DECIMAL NUMBERS Objective: This lesson is to demonstrate the correct method to subtract the exponent part of numbers written in scientific notation when solving division problems. The division process will be much easier if it is in two separate parts. One part is subtracting the exponents and the other part is division of the decimal portion. This section will only present the exponent part. Simply subtract the exponents from one another. Always subtract the exponent of the number being used to “divide by” (divisor) from the number being “divided into” (dividend). Example: Looking only at the number with an exponent to divide 4 × 104 by 2 × 102, simply separate the exponent part from the decimal part and subtract the divisor exponent from the dividend exponent: 104 ÷ 102 = 102 The rule for subtracting exponents is easy to remember. Just change the sign of the number being subtracted and add. The answer always has the sign of the larger digit. 104 ÷ 102 = 102

(4 – 2 = 2)

4 is the larger digit and 4 is positive, therefore, the answer is positive. Exercise: Based on this example, solve the problems. 1. 105 ÷ 103 = (a) 108 Answer: b

(b) 102

(c) 10–2

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271 –7

10 2. ---------310 (a) 10–10 Answer: a

(b) 10–4

(c) 1010

3. 104 divided by 10–2 = (a) 102 (b) 10–6 Answer: c

(c) 106

4. 10–3 ÷ 10–5 = (a) 10–8 Answer: b

(c) 10–2

(b) 102

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19

The Metric System

Now that there has been practice using the powers of ten or scientific notation, a system of measurement based entirely on the number 10 will be described. This system is known as the metric system. The metric system is a decimal system. The metric system originated in France about 1800. Only in the last few years has there been much interest in changing to the metric system in the United States. The United States is the only major country in the world that does not already use the metric system, making measurements in foreign trade more complicated. Definitions: 1. 2. 3. 4. 5. 6. 7. 8.

Prefix: letters added to the beginning of a word which change its meaning Symbol: a letter that represents a word or a prefix Length: the measure of how long something is Mass: the measure of how heavy something is (“Mass” is the same as “weight.”) Weight: the measure of how heavy something is (“Weight” is the same as “mass.”) Volume: the amount of space inside something (“Volume” is the same as “Capacity.”) Capacity: the amount of space inside something (“Capacity” means the same as “ Volume.”) Area: the size of a surface

The metric system uses 10 as its base, and the units of measure are determined by adding prefixes to the terms for the standard measures. The following table provides the most commonly used prefixes and their symbols, meanings, and values. (Less common prefixes will be introduced later.) Prefix

Symbol

Value

kilodecicentimilli-

k d c m

1000 or 103 .1 or 10–1 .01 or 10–2 .001 or 10–3

273

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When writing symbols for metric measures, always use lower-case (small) letters (unless the prefix is a person’s name; then it is capitalized.) Consider the following metric measures. Note that as prefixes are added, the value of the standard measure changes to the value given in the table above. Length Measure (standard unit: 1 meter or 1 m) 1 kilometer (km) = 1000 m 1 decimeter (dm) = .1 m 1 centimeter (cm) = .01 m 1 millimeter (mm) = .001 m Weight (mass) Measure (standard unit: 1 gram or 1 g) 1 kilogram (kg) = 1000 g 1 decigram (dg) = .1 g 1 centigram (cg) = .01 g 1 milligram (mg) = .001 g Volume Measure (standard unit: 1 liter or 1 L) 1 kiloliter (kL) = 1000 L 1 deciliter (dL) = .1 L 1 centiliter (cL) = .01 L 1 milliliter (mL) = .001 L

EXERCISES Directions: In each exercise select the correct answer. If necessary, restudy the instructions and examples. 1. A decimeter is equal to (a) 10 meters (b) .10 meters

(c) .01 meters

2. A kilogram is equal to (a) .1000 grams (b) 10,000 grams (c) 1000 grams 3. Which is the largest measure? (a) milliliter (b) liter

(c) kiloliter

4. A centigram is equal to (a) 10–2 grams (b) 10–1 grams

(c) 10 grams

5. The prefix “deci-” means (a) one (b) ten

(c) tenth

Answers to the exercises: 1. b

2. c

3. c

4. a

5. c

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CONVERSION OF MEASURES Objective: This section will demonstrate conversion from higher or lower measures of value by dividing by powers of ten or scientific notation. Repeating the table from the previous section, convert from higher to lower values or lower to higher values, using the table. Prefix

Symbol

Value

kilodecicentimilli-

k d c m

1000 or 103 .1 or 10–1 .01 or 10–2 .001 or 10–3

To determine how many millimeters are in 1 meter, simply look at the table and see that 1 millimeter is equal to 10–3 or 1/1000 of a meter. To determine how many millimeters are in 1 meter, simply divide 10–3 into 1 (1 = 100). Remember: When dividing exponents, change the sign of the exponent in the divisor and add the exponents. 0

10 0 – ( –3 ) 0+3 3 --------- = 10 = 10 = 10 –3 10

(1000 mm = 1 m)

To determine how many milliliters there are in a kiloliter, again simply look at the table and see that 1 mL = 10–3 L and 1 kL = 103 L: 3

10 3+3 6 --------- = 10 = 10 –3 10 There are 1,000,000 mL in 1 kL.

EXERCISES Directions: In each exercise, select the correct answer. If necessary, restudy the instructions and examples. 1. How many centiliters are in 25 liters? (a) 2500 cL (b) 250 cL (c) 25000 cL 2. How many kilograms are in 800 g? (a) 800 × 103 kg (b) 8 × 10–1 kg (c) 8 × 10 kg 3. How many grams are in 3.6 × 103 milligrams? (a) 3.6 g (b) 360 g (c) .360 g

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4. Convert 400 km to millimeters. (a) 400 × 103 km (b) 400 × 106

(c) 400 × 10–3

5. How many millimeters are in 90 centimeters? (a) 90 × 102 (b) 9 × 10–1 (c) 9 × 102 Answers to the exercises: 1. a

2. b

3. a

4. b

5. c

The same method is used to determine how many kilograms are in 10,000 grams. Looking at the chart (if necessary), a kilogram is 103 grams and 1 gram is 100 gram. Therefore, 104 (remember 10,000 = 104) ÷ 103 = 4

10 4 ––3 1 --------3 = 10 = 10 10 There are 10 kg in 10,000 g.

EXERCISES Directions: Using the examples as a reference, solve the problems. Use the table if necessary. 1. How many centimeters are in a meter? (a) 10 (b) .1 (c) 100 Answer: c 0

10 --------- = 100 –2 10 2. How many mL are in 40 cL? (a) 400 (b) 4 × 10–1 Answer: a

(c) 40

–2

10 - = 40 × 10 or 400 40 × --------3 10 3. How many kilograms are in 300 milligrams? (a) 300 (b) 300 × 10–6 (c) 300 × 106 Answer: b

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277 –3

10 –6 - = 300 × 10 300 × --------3 10

(3 × 10

4. How many g are in 3 kg? (a) 300 (b) 30,000 Answer: c

–4

= standard form)

(c) 3000

3

10 3 3 × --------0 = 3 × 10 or 3000 10

MORE CONVERSIONS OF THE METRIC SYSTEM Objective: This section will build on the basic material presented in the previous section, adding more detail and specifications, such as linear and weight measure. Review Test: I. Select the correct answer. 1. A metric volume measure: (a) liter (b) pint 2.

3.

4.

5.

A U.S. measure of length: (a) ounce (b) meter

(c) yard

A metric measure of area: (a) kilometer (b) square centimeter

(c) milliliter

A U.S. measure of weight: (a) gram (b) pound

(c) kilogram

A metric linear measure: (a) centiliter (b) centimeter

(c) kilogram

II. Select the correct answer. 6. Convert a megameter into kilometers. (a) 100 km (b) 1000 km 7.

8.

(c) millimeter

How many micrograms are in 45 centigrams? (a) 450 µg (b) 45 µg

(c) 10 km (c) 450,000 µg

80 decameters is equal to how many centimeters? (b) 8 × 103 (c) 8 × 10–4 (a) 8 × 104

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9.

How many grams are in 35 metric tons? (a) 3500 g (b) 35,000,000 g

(c) 35000 g

10. Convert 15 decimeters into decameters? (a) 15 dam (b) 1.5 dam

(c) .15 dam

11. How many grams equal 75 kilograms? (a) 7500 g (b) 750 g

(c) 75000 g

12. Express in grams the sum of 100 hectograms, 450 metric tons, and 86 decagrams. (a) 4510960 g (b) 451086 g (c) 45,010,860 g 13. Convert into centimeters the difference between 3 meters and 25 decimeters. (a) 5 cm (b) 50 cm (c) 500 cm 14. In hectometers, what is the sum of 80 meters, 1 kilometer, and 400 centimeters. (a) 1084 hm (b) 108.4 hm (c) 10.84 hm 15. What is the difference between 2 tons and 2000 kg? (a) 0 (b) 20 kg (c) 1 ton III. Select the correct answer. 16. How many hectoliters are in 50 decaliters? (a) 5 hL (b) .5 hL

(c) .05 hL

17. Convert 20 square decameters into square centimers. (b) 12 × 106 cm2 (c) 2 × 107 cm2 (a) 12 × 103 cm2 18. Convert a deciliter into milliliters. (a) 10 mL (b) 100 mL

(c) .100 mL

19. 25 × 108 m2 is equal to how many square kilometers? (a) 25 km2 (b) 250 km2 (c) 2500 km2 20. Change 45,000 centiliters into liters. (a) 4.5 L (b) 450 L

(c) .45 L

21. How many square millimeters are in 5 square kilometers? (a) 5 × 106 (b) 5 × 10–12 (c) 5 × 1012 22. Express in liters the difference between 200 decaliters and 15 hectoliters. (a) 500 L (b) 185 L (c) 50 L

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23. Convert into square decameters the difference between 4 square kilometers and 325 square hectometers. (a) 7.5 × 104 dam2 (b) 7.5 × 105 dam2 (c) 7.5 × 103 dam2 24. Express in milliliters the sum of 86 milliliters, 110 centiliters, and 45 deciliters. (a) 1636 mL (b) 4696 mL (c) 5686 mL 25. Express in square meters the sum of 50 square decameters, 8.5 × 105 square centimers, and 35 square meters. (a) 5120 m2 (b) 512 m2 (c) 5885 m2 Answers to the review test: 1. a 6. b

2. c 7. c

3. b 8. a

4. b 9. b

5. b 10. c

11. c 16. a

12. a 17. c

13. b 18. b

14. c 19. c

15. a 20. b

21. c

22. a

23. c

24. c

25. a

The metric system appears to be more difficult to use than the U.S. system. However, by “thinking metric,” it actually is easy to use. There are 9 used units in the metric system compared to 11 in the U.S. system. Both systems are compared: U.S.

Metric

Linear Measure

12 in. = 1 ft 36 in. = 1 yd 3 ft = 1 yd 5280 ft = 1 mi 1760 yd = 1 mi

10 mm = 1 cm 1000 mm = 1 m 100 cm = 1 m 1000 m = 1 km

Volume

16 oz = 1 pt 32 oz = 1 qt 2 pt = 1 qt 4 qt = 1 gal 8 pt = 1 gal

1000 mL = 1 L

Weight (mass)

16 oz = 1 lb 2000 lb = 1 ton

1000 g = 1 kg 1000 kg = 1 ton

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U.S.

Metric

144 in.2 = 1 ft2 9 ft2 = 1 yd2 1296 in.2 = 1 yd2 3097,600 yd2 = 1 mi2

100 mm2 = 1 cm2 10,000 cm2 = 1 m2 1,000,000 mm2 = 1 m2 1,000,000 m2 = 1 km2

Area

In the metric system all measurements are multiples of ten. This important characteristic makes understanding and calculations much easier. Example: A man needs lumber in the following sizes: 2 ft 10 1/2 in., 7 ft 7 13/16 in., 5 ft 6 in., and 3 ft 9 5/8 in. To find the total length of lumber needed, first change the fractions of inches to sixteenths, and then add the measurements: 2 ft 10 8/16 in. 7 ft 7 13/16 in. 5 ft 6 in. 3 ft 9 10/16 in. ---------------------------------------------17 ft 32 31/16 in. = 19 ft 9 15/16 in. Using metric measurements, 5 pieces of lumber in the following sizes would be needed: 876, 2332, 1676, and 1159 mm. To find the total, add: 876 2332 1676 1159 -----------6043 mm or 6.043 m This example illustrates how easy it is to use the metric system. Using the metric system is also easy for weight, volume, and area measurements. The next sections contain additional problems using metric measurements for length, weight (mass), volume, and area. Measurements are often given for large quantities. In the metric system, large quantities are multiples or fractions of the 10 base. Several tables will provide the multiples for the practice and exercise problems. Common Length (Linear) Measures Unit

Abbreviation

Megameter Myriameter

Mm mym

Meter Equivalent

1,000,000 or 106 10,000 or 104

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Common Length (Linear) Measures (Continued) Unit

Abbreviation

Kilometer Hectometer Decameter Meter Decimeter Centimeter Millimeter Micrometer

Meter Equivalent

km hm dam m dm cm mm µm

1000 or 103 100 or 102 10 or 101 1 or 100 1/10 or 10–1 1/100 or 10–2 1/1000 or 10–3 1/1,000,000 or 10–6

Practice: Using the method in the last section, solve the problems. 1. How many km are in 10 dam? (a) 10 × 10–4 (b) 1 × 10–1 Answer: b

(c) 1 × 10–2

10 × 10 2 –3 –1 -----------------3- = 10 × 10 = 1 × 10 km 1 × 10 2. Convert 1 mym into mm. (a) 10–6 (b) 108 Answer: c

(c) 107

4

10 7 --------- = 10 –3 10 3. 20 mm is equal to how many dm? (a) .2 (b) .02 Answer: a

(c) 2

–3

20 × 10 –2 --------------------- = 20 × 10 = .2 –1 1 × 10 Common Weight (Mass) Measures Unit

Abbreviation

Metric ton Kilograms Hectogram Decagram

Mt or t kg hg dag

Gram Equivalent

1,000,000 or 106 1000 or 103 100 or 102 10 or 101

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Common Weight (Mass) Measures Unit

Abbreviation

Gram Decigram Centigram Milligram Microgram

g dg cg mg µg

Gram Equivalent

1 or 100 1/10 or 10–1 1/100 or 10–2 1/1000 or 10–3 1/1,000,000 or 10–6

Practice: Using the method in the previous section, solve the problems. 1. How many grams are in 4 metric tons? (a) 4,000,000 (b) 400 (c) 4000 Answer: a 6

4 × 10 1 ---------------- = 4 × 10 or 40 5 10 2. How many decigrams are in 400 milligrams? (a) .4 (b) 40 (c) 4 Answer: c –3

400 × 10 –2 ------------------------ = 400 × 10 or 4 –1 10 3. Convert 100 kilograms into decigrams. (a) 100 × 104 (b) 10 × 104 (c) 10 × 106 Answer: a 3

100 × 10 4 ---------------------- = 100 × 10 –1 10 To work problems involving metric measures of the same kind, it is necessary to convert all of them into a common measurement. Example: To add the following length measures, 10 mm, 45 dm, and 150 cm, first convert them to a common length. For this example use centimeters. To do this, use the same method as in earlier sections. 1. 10 mm equals how many cm? –3

10 × 10 –1 --------------------- = 10 or 1 –2 10

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2. 45 dm equals how many cm? –1

45 × 10 --------------------- = 45 × 10 or 450 cm –2 10 3. 150 cm equals 150 cm. 1 + 450 + 150 = 601 cm or 1 450 + 150 -------------601 cm Practice: Using the examples as a reference, solve the problems. 1. Express in grams the sum of 30 dg, 25 hg, and 50 kg. (a) 7503 g (b) 30,030 g (c) 52,503 g Answer: c 2. Express in decimeters the sum of 200 cm, .85 hm, and 800 mm. (a) 860 dm (b) 878 dm (c) 1058 dm Answer: b 3. Express in meters the difference between 800 hm and 4000 dam. (a) 400 m (b) 4000 m (c) 40,000 m Answer: c

EXERCISES Directions: Select the correct answer. Feel free to restudy the instructions and examples. 1. How many dam are in 600 mm? (a) .06 (b) 6

(c) 60

2. Convert 8 hg into mg. (a) 8 × 10–1 (b) 8 × 105

(c) 800

3. 1 dm equals how many m? (a) 10 (b) 100

(c) .1

4. How many meters are in 5 km? (a) 50 (b) 5000

(c) 500

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5. Convert 100 cg into kg. (a) .001 (b) .01

(c) .1

6. 10 kg equals how many dg? (a) 10,000 (b) 100,000

(c) 1000

7. Express in grams the sum of 2 kg, 4 dag, and 100 cg. (a) 2041 g (b) 241 g (c) 2410 g 8. Express in decimeters the sum of 1,000 mm, 100 cm, and 2 m. (a) 400 dm (b) 4 dm (c) 40 dm 9. What is the difference between 50 km and 300 hm? (a) 2000 m (b) 20,000 m (c) 200,000 m 10. Express in grams the difference between 100 hg and 900 dag. (a) 100 g (b) 800 g (c) 1000 g Answers to the exercises: 1. a 6. b

2. b 7. a

3. c 8. c

4. b 9. b

5. a 10. c

Other common measures in the metric system will now be introduced. Square feet, square yards, and square miles are area measures in the U.S. system. Below is a table of common metric area measures. Common Area Measures Unit

Square Square Square Square Square Square

Abbreviation

Square Meter Equivalent

km2 hm2 dam2 m2 cm2 mm2

1,000,000 or 106 10,000 or 104 100 or 102 1 or 100 1/10,000 or 10–4 1/1,000,000 or 10–6

kilometer hectometer dekameter meter centimeter millimeter

Remember: A meter is the standard measure of length in the metric system. Practice: Using the method in the previous section, solve the problems. 1. Convert 10 hm2 into m2. (a) 1000 (b) 10,000 Answer: c

(c) 100,000

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10 × 10 ------------------- = 100,000 0 10 2. How many mm2 are in .45 dam2? (a) 4.5 × 107 (b) .45 × 10–5 Answer: a

(c) 45 × 107

2

.45 × 10 8 7 --------------------= .45 × 10 or 4.5 × 10 –6 10 3. Change 250 dam2 into m2. (a) 25 × 102 m2 (b) 2500 m2 Answer: c

(c) 25,000 m2

2

250 × 10 ---------------------- = 25,000 0 10 To work problems with metric measures of the same kind, it is necessary to convert all of them to a common measurement. Example: To add the following volume measurements, 4 dkL, 80 dL, and 150 cL, first convert to a common volume. For this example, change them to centiliters. 1. 4 daL equals how many cL? 1

4 × 10 3 ---------------- = 4 × 10 or 4000 cL –2 10 2. 80 dL equals how many cL? –1

80 × 10 --------------------- = 800 cL –2 10 3. 150 cL equals 150 cL. Therefore, 4000 + 800 + 150 = 4950 cL or 4000 800 + 150 -------------4950 cL

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Practice: Using the above examples as a reference, solve the problems. 1. Express in liters the sum of 2 kL, 400 cL, and 3000 dL. (a) 2304 L (b) 234 L (c) 504 L Answer: a 2. Express in dam2 the difference between 86 km2 and 2300 hm2. (a) 6300 dam2 (b) 63,000 dam2 (c) 6.3 × 105 dam2 Answer: c 3. 210 daL + 4000 dL + 9000 mL = _________ cL. (a) 250,900 cL (b) 259,000 cL (c) 214,900 cL Answer: a

EXERCISES Directions: In each of these exercises, select the correct answer. If necessary, restudy the instructions and examples. 1. Convert 750 L into hL. (a) 75 hL (b) 7.5 hL

(c) .75 hL

2. How many deciliters are in 84 decaliters? (a) 8.4 × 103 (b) 84 × 103 (c) 8.4 × 102 3. 2.5 × 106 mL equals how many kL? (a) 25 kL (b) 250 kL (c) 2.5 kL 4. Express 19 km2 in hm2. (a) 1900 hm2 (b) 19,000 hm2

(c) 190 hm2

5. How many mm2 are in 1 m2? (a) 1000 mm2 (b) 10,000 mm2

(c) 1,000,000 mm2

6. Convert 10 hm2 into cm2. (a) 10 × 108 cm2 (b) 10 × 104 cm2 (c) 1,000,000 cm2 7. Express in daL the difference between 14,500 daL and 4750 daL. (a) 9750 daL (b) 97,500 daL (c) 120,250 daL 8. Express in hm2 the sum of 25 m2, 4500 cm2, and 250 dam2. (a) 5025 × 1018 hm2 (b) 2.502545 hm2 (c) 5,025 × 1012 hm2 9. What is the difference between 20 hL and 450 dL? (a) 255 L (b) 1955 L (c) 25.5 dL

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10. In square meters express 2 km2, 40 dam2, and 100,000 cm2. (a) 200,410 m2 (b) 204,010 m2 (c) 2,004,010 m2 Answers to the exercises: 1. b 6. a

2. a 7. a

3. c 8. b

4. a 9. b

5. c 10. c

CUMULATIVE EXERCISES Directions: Solve the following problems. I. Select the correct answer. 1. A U.S. measurement of volume: (a) quart (b) inch 2.

3.

4.

5.

A metric measurement of length: (a) kiloliter (b) foot

(c) millimeter

A U.S. measurement of area: (a) gallon (b) square inch

(c) cubic inch

A metric measurement of mass: (a) liter (b) area

(c) gram

A metric measurement of area: (a) square (b) square hectometer

II. Select the correct answer. 6. How many m are in 30 hm? (a) 300 m (b) 3000 m 7.

8.

9.

(c) milliliter

(c) 30,000 m

Convert 25 g into kg. (a) 2.5 kg (b) .25 kg

(c) .025 kg.

Convert a cm into µm. (a) 1000 µm (b) 10,000 µm

(c) 100 µm

8 t is equal to how many dag? (b) 8 × 106 (a) 8 × 105

(c) 8 × 104

10. Change 2500 centimeters into meters. (a) 25 m (b) 2500 m

(c) 25,000 m

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III. Select the correct answer. 11. How many cL are there in 5 daL? (a) 50 cL (b) 500 cL

(c) 5,000 cL

12. Convert 4 square kilometers into centares. (a) 4 × 105 (b) 4 × 106

(c) 4 × 10–6

13. 75 hL is equal to how many dL? (a) 7.5 × 104 dL (b) 7.5 × 103 dL

(c) 7.5 × 10–3 dL

14. How many square centimeters are in a square meter? (a) 1000 cm2 (b) 100 cm2 (c) 10,000 cm2 15. Convert 225 dL into daL. (a) 22.5 daL (b) 2.25 daL

(c) .225 daL

IV. Select the correct answer. 16. Express in meters the difference between 400 dam and 25 hm. (a) 15 m (b) 150 m (c) 1500 m 17. Express in hg the sum of 15 dag, 4000 cg, and 65 hg. (a) 66.90 hg (b) 6690.0 hg (c) 669.0 hg 18. Express in mL the sum of 25 mL, 95 mL, and 25 dL. (a) 3475 mL (b) 2620 mL (c) 1225 mL 19. Convert into dg the difference between 20 kg and 180 hg. (a) 2000 dg (b) 20,000 dg (c) 200 dg 20. What is the total area in hectares of 2 km2, 40 ha, and 65 ha? (a) 2654 ha (b) 269 ha (c) 2069 ha Answers to the cumulative exercises: 1. 6. 11. 16.

a b c c

2. 7. 12. 17.

c c b a

3. 8. 13. 18.

b b a b

4. 9. 14. 19.

c a c b

5. 10. 15. 20.

b a b b

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International System of Units

This chapter introduces the International System of measurement units. The system is usually abbreviated as S.I. for the original French Systeme International d’Unités or International System of Units. S.I. units are part of a superior system of measurement and calculation. The S.I. system was developed for two reasons. 1. To revise and extend the original metric system 2. To eliminate units no longer needed The S.I. system consists of: 1. Basic units 2. Two supplementary units 3. Sixteen derived units with special names This chapter will introduce the basic units: 1. 2. 3. 4. 5. 6. 7.

Length (meter) Mass (gram) Time (second) Electric current (ampere) Temperature (kelvin) Light intensity (candela) Molecular substance (mole)

Definitions: 1. 2. 3. 4. 5.

Square: a flat figure with four equal sides and four right angles Rectangle: a flat figure with four right angles and four sides Width: the measure of an object from side to side Height: the measure of an object from top to bottom Adjacent: things that touch

289

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6. Weight: the measure of how heavy an object is (“Weight” means almost the same as “mass.”) 7. Mass: the measure of how heavy an object is (“Mass” means almost the same as “weight.”) 8. Temperature: the measure of how hot an object is 9. Degree: a unit used to measure temperature (The symbol for degree is °.) 10. Energy: the power of certain forces of nature to do work 11. Acceleration: the rate at which speed is increased 12. Force: power or energy that can do or make something 13. Pressure: the force exerted against a surface 14. Circuit: the complete path of an electrical current 15. Current: the flow of electricity in a wire or other conductor 16. Resistance: the opposition of one force or thing to another 17. Cycle: a complete set of events, returning to the original state 18. Frequency: the number of complete cycles of current 19. Density: the amount of mass of an object for each unit of area or volume

METER The S.I. unit for length and area is the meter (m). By using prefixes, shorter or longer lengths can be expressed. Originally, the meter was one ten-millionth of the distance of the line passing from the equator to the North Pole through Paris. Today, the meter is defined as the length of 1,650,763.73 wavelengths in a vacuum of the orange-red line in the spectrum of krypton 86. A standard bar of this length is safely stored at the National Bureau of Standards in Washington, D.C. A meter is 39.37 in. or about 3 1/3 in. longer than a yard. The prefixes from the previous chapter make it easy to express very long and very short lengths. The S.I. unit for area is the square meter (m2). Simply, a square meter is a square whose sides measure 1 m each. If a surface is a square and has sides measuring 2 m each, it is said to have an area of 4 m2. The area of a square surface is found by squaring (or multiplying by itself) the measure of one of the sides. Therefore, if each side of a square measures 6 m, the area is 36 m2, because 6 × 6 = 36. By using prefixes, areas of very small squares may be found. For example, if a square has sides measuring 1 mm each, the square has an area of 1 square millimeter (mm2); also if each side of a square measures 5 cm, the area of the square is 25 cm2 or (5 cm × 5 cm = 25 cm2). Practice: Using the examples as references, solve the problems: 1. What is the area of a square with sides measuring 7 mm each? (a) 14 mm2 (b) 49 mm (c) 49 mm2 Answer: c (7 mm × 7 mm = 49 mm2) 2. What is the area of a square with sides measuring 3 cm each? (a) 9 cm (b) 9 cm2 (c) 6 cm2 Answer: b (3 cm × 3 cm = 9 cm2)

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The area of a rectangular surface may be found using a similar method. If a rectangle has two sides measuring 3 m and two sides measuring 4 m, it has an area of 12 m2. To find the area of a rectangle, multiply the measure of one of the sides by the measure of an adjacent side (sides touch each other). If one side of a rectangle measures 2 cm, and an adjacent side measures 5 cm, the area of the rectangle is 10 cm2: 2 cm × 5 cm = 10 cm2. When finding the area of a rectangle, be sure the measures of adjacent sides are expressed in the same unit. Consider a rectangle that has one side measuring 50 cm and an adjacent side measuring 3 m. Change one of the measures so that both are expressed in either meters or centimeters: 50 cm could be changed to .5 meters; then multiply by 3 m. The area of a rectangle is 1.5 m2. This problem could also be solved by changing 3 m to 300 cm and then multiplying by 50 cm. The area of the rectangle is 1500 cm2. Both answers are correct. If one side of a rectangle measures 20 dm and an adjacent side measures 3 m, what is the area of the rectangle expressed in square meters? The answer is 6 m2 20 dm = 2 m: 2 m × 3 m = 6 m2. Practice: Using the examples, answer the questions: 1. What is the area of a rectangle with two sides measuring 3 m and two sides measuring 12 m? (b) 15 m2 (c) 36 m (a) 36 m2 Answer: a (12 m × 3 m = 36 m2) 2. What is the area of a rectangle with one side measuring 5 cm and an adjacent side measuring 3 cm? (b) 15 cm2 (c) 15 cm (a) 8 cm2 Answer: b (5 cm × 3 cm = 15 cm2) 3. What is the area in m2 of a rectangle with two sides measuring 3 km and two sides measuring 3 m? (b) 9000 m2 (c) 6 m2 (a) 9 m2 Answer: b (3 km = 3000 m; 3000 m × 3 m = 9000 m2) 4. What is the area in mm of a rectangle with one side measuring 3 cm and an adjacent side measuring 5 mm2? (b) 150 mm2 (c) 1.5 mm2 (a) 15 mm2 Answer: b (3 cm = 30 mm; 30 mm × 5 mm = 150 mm2)

EXERCISES Directions: Select the correct answer. If necessary, restudy the instructions and examples. 1. What is the area of a square with sides measuring 4 cm each? (b) 4 cm2 (c) 16 cm2 (a) 8 cm2

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2. What is the area of a rectangle with two sides measuring 5 m and two sides measuring 7 m? (a) 35 m2 (b) 12 m2 (c) 24 m2 3. What is the area in cm2 of a rectangle with one side measuring 2 m and an adjacent side measuring 3 cm? (a) 600 cm2 (b) 6 cm2 (c) 60 cm2 4. The square meter is a measure of (a) volume (b) length

(c) area

5. What is the area in m2 of a rectangle with two sides measuring 5 m and two sides measuring 3 km? (a) 15 m2 (b) 15000 m2 (c) 1500 m2 Answers to the exercises: 1. c

2. a

3. a

4. c

5. b

CUBIC METER In the S.I. system, the cubic meter (m3) is the standard measure of volume. A cubic meter is a square box with each side measuring 1 meter. To find the volume of a box, multiply together the length of the box, the width of the box, and the height of the box. Look at a few examples:

Example 1: a square box. The length is 3 m, the width is 3 m, and the width is 3 m. Volume is length × width × height: 3 m × 3 m × 3 m. The volume of the box is 27 m3. Example 2: a rectangular box. The length is 30 mm, the width is 2 cm, and the height is 1 cm. Before multiplying length × width × height, all three measurements must be expressed in the same unit. Change 30 mm to 3 cm and multiply length × width × height to find the volume. The volume of the box is 3 cm × 2 cm × 1 cm or 6 cm3. Example 3: a square box. The length is 1 dm, the width is 1 dm, and the height is 1 dm. The volume of the box is 1 dm × 1 dm × 1 dm or 1 dm3 or “one cubic decimeter.” Cubic decimeter is a unique measure of volume. Another name for a cubic decimeter is liter. A liter (1000 cm3 = 1 liter) is a little larger than a quart. A liter

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is used as the standard measure of liquid volume in the metric system and can be expressed as very small or very large units of volume by using prefixes: 1 cubic centimeter equals 1 milliliter or 1 cm3 = 1 mL. Practice: 1. What is the volume of a square box with sides measuring 5 m each? (a) 10 m3 (b) 25 m3 (c) 125 m3 Answer: c (5 m × 5 m × 5 m = 125 m3) 2. What is the volume in cm3 of a rectangular box with a length of 2 cm, a width of 2 mm, and a height of 2 m? (a) 8 cm3 (b) 80 cm3 (c) 800 cm3 Answer: b (2 mm = .2 cm; 2 m = 200 cm; 2 cm × .2 cm × 200 cm = 80 cm3) 3. What is the volume in m3 of a rectangular box with a length of 3 m, a width of 2 m, and a height of 400 cm? (b) 240 m3 (c) 24 m3 (a) 2400 m3 Answer: c (400 cm = 4 m; 3 m × 2 m × 4 m = 24 m3)

EXERCISES Directions: Select the correct answer. If necessary, restudy the instructions and examples. 1. What is the volume of a square box with sides measuring 3 cm each? (a) 9 cm3 (b) 27 cm3 (c) 6 cm3 2. What is the volume of a rectangular box with a length of 3 cm, a height of 4 cm, and a width of 5 cm? (a) 12 cm3 (b) 60 cm3 (c) 60 cm2 3. What is the S.I. standard unit of volume? (a) square meter (b) cubic decimeter (c) cubic meter 4. What is the volume in m3 of a rectangular box with a length of 2 m, a width of 20 cm, and a height of 200 mm? (a) 8 m3 (b) .8 m3 (c) .08 m3 5. What is the volume in cm3 of a rectangle with a length of 3 cm, a width of 2 mm, and a height of .5 m? (a) 30 cm3 (b) 5.5 cm3 (c) 3 cm3 Answers to the exercises: 1. b

2. b

3. c

4. c

5. a

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KILOGRAM “Gram” is derived from the Latin word “gramma,” meaning “a small weight” (1 gram is about .04 ounces). All units of weight in the metric system are based on the gram. The gram unit of mass is the kilogram: a kilogram is equal to 1000 grams and 1 kilogram is about 2.2 pounds. The gram is such a small unit of weight that it is often not practical to use it. Therefore, the kilogram is the standard unit. The gram is often used to weigh small things. For instance, a teaspoon contains about 5 grams. The kilogram is used to measure bigger things. For example, 150 pounds is about 68 kilograms. To weigh extremely large objects, such as an elephant, use the metric ton. One metric ton equals 1000 kg. An elephant weighs about 5 metric tons or 5000 kilograms. Many prefixes can be added to the word “gram” to indicate larger weights: 10 100 1000 1,000,000

grams grams grams grams

(g) (g) (g) (g)

= = = =

1 1 1 1

dekagram (dag) = 101 hectogram (hg) = 102 kilogram (kg) = 103 metric ton (t or MT) = 106

Very small weights may also be expressed by adding prefixes to the word “gram:” 1/10 1/100 1/1000 1/1,000,000

gram gram gram gram

(g) (g) (g) (g)

= = = =

1 decigram (dg) = 10–1 1 centigram (cg) = 10–2 1 milligram (mg) = 10–3 a microgram (µg) = 10–6

Still another way of conversion is from pounds to ounces and vice versa: 1 pound = 16 ounces 1 kilogram = 1000 grams Example 1: There are approximately 2.2 pounds in a kilogram. To know how many kilograms are in 17 pounds, simply multiply 2.2 by 17. There are 37.4 kilograms in 17 pounds. Example 2: To know how many pounds are in 52 grams, first determine how many kilograms are in 52 grams. Because 1,000 grams is equal to 1 kilogram, 52 grams would equal .052 kg. The number of pounds can now be determined using the method described above: .052 kg × 2.2 lb/kg = .1144 lb. There are .1144 pounds in 52 grams. Example 3: How many ounces are in 12 kilograms? There are 26.4 pounds in 12 kilograms (12 × 2.2 = 26.4). To change from pounds to ounces, multiply the number of pounds by 16: 26.4 × 16 = 422.4 oz. There are 422.4 ounces in 12 kilograms.

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Example 4: How many ounces are in 4 grams? 4 grams = .004 kilograms .004 kg × 2.2 = .0088 pounds .0088 × 16 = .1408 ounces There are .1408 ounces in 4 grams. Practice: Refer to the examples, solve the following problems: 1. How many pounds are in 3 kilograms? (a) 5.2 (b) 6.6 (c) 66 Answer: b (2.2 × 3 = 6.6) 2. How many ounces are in 4 kilograms? (a) 8.8 (b) 64 (c) 140.8 Answer: c (2.2 × 4 = 8.8 lb, 8.8 × 16 = 140.8 oz) 3. How many pounds are in 300 grams? (a) 660 (b) .3 (c) .66 Answer: c (300 g = .3 kg, .3 × 2.2 = .66) 4. How many ounces are in 14 grams? (a) .0308 (b) .4928 (c) 492.8 Answer: b (14 g = .014 kg, .014 × 2.2 = .0308, .0308 × 16 = .4928)

EXERCISES Directions: Select the correct answer. If necessary, restudy the instructions and examples. 1. A gram weighs (a) more than a pound

(b) less than a pound

(c) the same as a pound

2. A pair of men’s shoes has a mass of about (a) 1 g (b) 1 kg (c) 50 kg 3. A car has a mass of about (a) 50 kg (b) 100 kg

(c) 1 MT

4. 500 grams equals (a) 50 dag (b) 2 kg

(c) .5 hg

5. 1 milligram equals (a) 10–1 g (b) 10–2 g

(c) 10–3 g

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6. How many pounds are in 17 kilograms? (a) 37.4 lb (b) 19.2 lb (c) 3.74 lb 7. How many ounces are in 6 kilograms? (a) 13.2 oz (b) 212 oz (c) 211.2 oz 8. How many ounces are in 11 grams? (a) .0242 oz (b) .3872 oz (c) 387.2 oz 9. How many pounds are in 400 grams? (a) .88 lb (b) 880 lb (c) 8.8 lb 10. How many ounces are in 3.3 grams? (a) .00726 oz (b) .11616 oz (c) 116.16 oz Answers to the exercises: 1. b 6. a

2. b 7. c

3. c 8. b

4. a 9. a

5. c 10. b

KELVIN In the S.I. system, Kelvin is the standard unit for temperature, named after the English physicist, Lord William Thomson Kelvin. The scale is based on the average kinetic energy per molecule of a perfect gas. Zero is equal to –273.16° Centigrade or –459.69° Fahrenheit. Zero temperature on the Kelvin scale represents absolute zero, meaning molecular motion has ceased. There is no upper limit on temperature because none is known. To convert a Celsius temperature to Kelvin, simply add 273.16 to the Celsius temperature. The symbol used for the Kelvin temperature scale is “K.” The four temperature scales are 1. The Celsius scale (from which the centigrade scale was developed) is based on the boiling point of water of 0° and the freezing point of 100°. This scale is no longer in use as originally designed. Instead the letter “°C” is used to designate the centigrade scale. 2. The Centigrade scale is based on 100. The freezing point of water is 0° and the boiling point 100°. This scale is the opposite of the Celsius scale. The distance between the freezing and boiling points in Fahrenheit and Reaumur is 180° and 80°, respectively. 3. The Fahrenheit scale is a temperature scale in which zero is defined by a mixture of equal weight of snow and common salt. The freezing point is 32° and the boiling point is 212° and the symbol for is “°F.” Both freezing and boiling are measured under standard atmospheric pressure. 4. The Reaumur scale defines 0° as the temperature of melting ice and 80° as the temperature of boiling water. This scale is not used.

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In 1948, the International Committee on Weights and Measures and the National Bureau of Standards decided to discontinue the Celsius scale and to use the centigrade scale (now known as the Celsius or the centigrade scale). To convert Celsius to Fahrenheit, use: C° = (F° – 32)5/9. To convert Fahrenheit to Celsius, use: F° = (9/5 C°) + 32. Example 1: What is 65 degrees Fahrenheit on the centigrade scale? C = (65 – 32)5/9 = 18.338°C Example 2: What is 105 degrees Celsius on the Fahrenheit scale? F = 9/5(105) + 32 = 221°F Example 3: What is 86 degrees Celsius on the Kelvin scale? K = 86 + 273.16 = 359.16 K Example 4: What is 150 K on the °C scale? To convert from Kelvin to Celsius, use the formula from the previous page in reverse. Use C = K – 273.16, C = 150 – 273.16 = –123.16°C. Another method to change from a temperature expressed in degrees Celsius to one expressed in degrees Fahrenheit is to multiply the number of degrees Celsius by 1.8. Then add 32 to the answer. This is then the number of degrees Fahrenheit. Example 5: How many degrees Fahrenheit are in 60°C? Multiply 60 by 1.8: 60 × 1.8 = 108. Then add 32 to the answer: 108 + 32 = 140. If the temperature is 60°C, it can be expressed as 140°F. Practice: Refer to examples, solve the problems: 1. How many degrees Fahrenheit are in 75°C? (a) 135°F (b) 167°F (c) 192.6°F Answer: b (75 × 1.8 = 135, 135 + 32 = 167°F) 2. How many degrees Fahrenheit are in –5°C? (a) 23°F (b) –9°F (c) 27°F Answer: a (–5 × 1.8 = –9, –9 + 32 = 23°F) To change from a temperature expressed in Kelvin to one expressed in degrees Fahrenheit, first change from Kelvin to degrees Celsius. Do this by subtracting 273.16 from the Kelvin number. Then change this new answer from Celsius to Fahrenheit using the method explained at the beginning of this section.

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Example 6: To change from 328.16 K to degrees Fahrenheit, first change to degrees Celsius: 328.16 – 273.16 = 55°C. Then change 55°C to degrees Fahrenheit: 55 × 1.8 = 99 99 + 32 = 131°F Answer: 131°F in 55°C. Example 7: Change 358.66 K to degrees Fahrenheit: 358.66 – 273.16 = 85.50°C 85.5 × 1.8 = 153.9° 153.9 + 32 = 185.9°F Answer: 358.66 K = 185.9°F. Practice: Refer to the examples, solve the following problems. 1. How many °F are in 284.66 K? (a) 20.7°F (b) 52.7°F (c) 78.3°F Answer: b (284.66 – 273.16 = 11.5, 11.5 × 1.8 = 20.7, 20.7 + 32 = 52.7°F) 2. How many °F are in 300 K? (a) 8.0312°F (b) 48.312°F (c) 80.312°F Answer: c (300 – 273.16 = 26.84, 26.84 × 1.8 = 48.312, 48.312 + 32 = 80.312°F)

EXERCISES Directions: Select the correct answer. If necessary, restudy the introduction and the examples. 1. The coldest something can be is (a) 0°C (b) 0°F

(c) 0 K

2. In the Celsius system, water boils at (a) 100°C (b) 212°C (c) 273.16°C 3. In the Kelvin system, how many units are between the boiling point of water and the freezing point of water? (a) 273.16 (b) 100 (c) 373.16 4. –459.69°F equals (a) 0°K (b) 0°C

(c) 0 K

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5. Which two systems are most similar? (a) Celsius and (b) Celsius and (c) Fahrenheit and Fahrenheit Kelvin Celsius 6. The standard S.I. measure of temperature is (a) Fahrenheit (b) Celsius (c) Kelvin 7. Absolute zero is (a) 0°C (b) 0 K

(c) 0°F

8. Zero degrees Celsius equals (a) 212°F (b) 0°F

(c) 273.16 K

9. The boiling point of water is (a) 373.16 K (b) 212 K

(c) 100 K

10. The freezing point of water is (a) 0 K (b) 273.16 K

(c) 32 K

11. Convert 25°C to °F. (a) 77°F (b) –414.688°F

(c) 45°F

12. Convert 276 K to °C. (a) 37.112°C (b) 28.4°C

(c) 2.84°C

13. How many °F are in 52.16 K? (a) –365.8°F (b) 429.8°F

(c) 365.8°F

14. How many °F are in 712.66 K? (a) 439.5°F (b) 823.1°F

(c) 791.1°F

15. Convert 521.56 K to °F. (a) 248.4°F (b) 447.12°F

(c) 479.12°F

16. Change 280 K to °C. (a) 44.312°C (b) 6.84°C

(c) 68.4°C

17. Change 272 K to °F. (a) 29.912°F (b) –2.088°F

(c) 34.008°F

18. Change 65°C to °F. (a) –342.688°F (b) 149°F

(c) 117°F

19. How many °F are in 290 K? (a) 554°F (b) 30.312°F

(c) 62.312°F

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20. How many °F are in 305 K? (a) 31.84°F (b) 57.312°F

(c) 89.312°F

Answers to the exercises: 1. 6. 11. 16.

c c a b

2. 7. 12. 17.

a b c a

3. 8. 13. 18.

b c a b

4. 9. 14. 19.

c a b c

5. 10. 15. 20.

b b c c

REVIEW EXERCISES ON MASS, GRAMS, AND TEMPERATURE Directions: Try to answer each problem correctly from memory alone. I. Select the correct answer. 1. A gram weighs (a) more than an ounce (b) less than an ounce (c) the same as an ounce 2.

A kilogram weighs (a) more than a pound

(b) less than a pound

(c) the same as a pound

3.

A whale would most likely be weighed in what unit? (a) grams (b) kilograms (c) metric tons

4.

A key would most likely be weighed in what unit? (a) grams (b) kilograms (c) metric tons

5.

A newborn baby would have a mass of about (a) 3 grams (b) 3 kilograms

(c) 3 milligrams

6.

Which of the following units would most likely be used to weigh a chair? (a) grams (b) kilograms (c) decigrams

7.

Which of the following is not a unit of mass? (a) quintal (b) metric ton

(c) liter

1,000,000 grams is equal to (a) one microgram (b) one metric ton

(c) one quintal

8. 9.

17 centigrams is equal to (a) .17 grams (b) 1.7 milligrams

(c) .000017 metric tons

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10. 10 grams is equal to how many kilograms? (a) 1 kilogram (b) .01 kilogram II. Select the correct answer. 11. What is the standard S.I. unit of temperature? (a) Fahrenheit (b) Celsius

(c) .001 kilogram

(c) Kelvin

12. The freezing point of water is (a) 0° Fahrenheit (b) 0° Celsius

(c) 0 Kelvin

13. The boiling point of water is (a) 100° Fahrenheit (b) 212° Celsius

(c) 373.16 K

14. Absolute zero is (a) 0° Fahrenheit

(c) 0 K

(b) 0° Celsius

15. Zero degrees Celsius is equal to (a) 0° Kelvin (b) 273.16 Kelvin 16. The coldest temperature is (a) absolute zero (b) Kelvin

17. Zero Kelvin equals (a) –459.69°F

(c) 32° Kelvin

(c) the freezing point

(b) 0°C

(c) –100°C

18. In the Celsius system, how many degrees are between the boiling point and the freezing point of water? (a) 180°C (b) 273.16°C (c) 100°C 19. Which is the coldest? (a) 0°C

(b) 0°K

(c) 0°F

20. 100 K is the same as (a) 373.16°C

(b) 212°C

(c) –173.16°C

Answers to the review exercises: 1. 6. 11. 16.

b b c a

2. 7. 12. 17.

a c b a

3. 8. 13. 18.

c b c c

4. 9. 14. 19.

a a c b

5. 10. 15. 20.

b b b c

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TECHNICAL UNITS This section introduces some of the derived units in the S.I. system. Many derived units are technical and hard to comprehend. However, it is important to recognize and understand them. There are many derived S.I. units. Simply, a derived unit is one that has at least two characteristics. It is formed by multiplying or dividing a base unit by one or more other units. Square meter is a derived unit (the meter) multiplied by another unit (again, the meter). This section describes 16 derived units which have special names. When studying derived units, remember the standard S.I. units: 1. 2. 3. 4. 5. 6. 7.

Length (meter, m) Mass (kilogram, kg) Time (second, s) Electrical current (ampere, A) Temperature (Kelvin, K) Light intensity (candela, c) Molecular substance (mole, mol)

The 16 derived units are 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

Energy — work, heat, quantity (joule, J) Force (newton, N) Pressure (pascal, Pa) Power (watt, W) Electrical charge — quantity of electricity (coulomb, C) Electrical potential difference — voltage, electromotive force (volt, V, E) Electrical resistance (ohm, Ω) Electrical conductance (siemens, S) Electrical capacitance (farad, F) Electrical inductance (henry, H) Frequency (hertz, Hz) Magnetic flux (weber, Wb) Magnetic flux density (tesla, T) Luminous flux (lumen, lm) Illumination (lux, lx) Customary temperature (degree celsius, °C)

Each of the derived units will now be briefly described. 1. Joule: measures energy. The joule is used to measure the amount of work done. It is a very small amount of energy. For everyday use, kilowatthours are more practical. 2. Newton: the amount of force which will give a mass of 1 kilogram an acceleration of 1 meter per second squared. If an apple is tossed into the air and then caught, the amount of force that hits the hand is about 1 newton.

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3. Pascal: the same as newton per square meter. A pascal describes the force in a certain area, such as tire pressure. It is an extremely small amount of pressure. 4. Watt: measures power or the rate at which work is being done. A watt is equal to the flow of 1 ampere at a pressure of 1 volt. A watthour is the amount of electrical energy used to keep a 1 watt unit working for 1 hour. About 746 watts equal 1 horsepower. 5. Coulomb: measures the quantity of electricity flowing past a section of a circuit in 1 second when the current is 1 ampere. 6. Volts: a “push” that moves a current of 1 ampere through a resistance of one ohms. 7. Ohm: measures resistance to the passage of an electrical current. The ohm measures how well something “conducts.” A good conductor of electricity has little resistance to the flow of electricity. A poor conductor has significant resistance. 8. Siemens: measures conductance. Conductance is the opposite of resistance which is measured by the ohm. The siemens was formerly called “mho” (ohm spelled backward). 9. Farad: measures electrical capacity. Because farad is a very large unit, often in work done, the microfarad is used (one millionth of a farad). 10. Henry: measures electrical inductance. Inductance is when a current produces an electrical effect on a force in a circuit. 11. Hertz: measures frequency. A hertz is the frequency of one cycle per second. 12. Weber: measures magnetic flux. Magnetic flux are the lines of force that surround a magnet. 13. Tesla: measures flux density (magnetic field strength). A tesla is the density of 1 weber per square meter. 14. Lumen: measures luminous flux. To better understand lumen, consider the following example. If a candle is placed in the center of a globe, a lumen is the amount of light that falls on one square foot of the globe’s inside surface.

15. Lux: measures illumination of how bright or dull an object is. Footcandle is an older unit of illumination. 16. Degree Celsius: See the section on Kelvin.

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EXERCISES Directions: Select the correct answer. If necessary, restudy the overview and definitions. 1. The abbreviation for ohm is (a) 0 (b) °

(c) Ω

2. A measure of frequency is (a) hertz (b) henry

(c) farad

3. A watt is (a) more power than (b) less power than a horse power a horsepower

(c) the same amount as a horsepower

4. The opposite of resistance is (a) conductance (b) ohm

(c) volt

5. Which is not a measure of illumination? (a) lux (b) lumen

(c) footcandle

Answers to the exercises: 1. c

2. a

3. b

4. a

5. b

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The previous chapter introduced the S.I. system of measurements and conversion from a smaller unit to a larger unit. This chapter presents conversions from inches to centimeters, from yards to meters, and from quarts to liters, including squared and cubic units. Definitions: 1. 2. 3. 4. 5.

Decimal point: the period in a number indicating the value of the number Yard: a measurement of length (36 inches; 3 feet) Pint: a measurement of volume equal to half of a quart Quart: a measurement of volume (2 pints; 1/4 gallon) Inch: a measurement of length

REVIEW TEST I. Select the correct answer: 1. How many centimeters are in 7 inches? (a) 17.5 cm (b) 177.8 cm 2.

3.

4.

5.

(c) 17.78 cm

How many millimeters are in 2 inches? (a) 5.08 mm (b) 50.08 mm

(c) 50.8 mm

How many meters are in 75 inches? (a) 190.50 m (b) 19.050 m

(c) 1.9050 m

How many cm2 are in 2 in.2? (a) 13 cm2 (b) 12.90 cm2

(c) 129.0 cm2

How many mm2 are in 16 in.2? (a) 10,320 mm2 (b) 1032 mm2

(c) 103.20 mm2

305

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6.

7.

8.

9.

Convert 17 cm2 into mm2. (a) 109.65 mm2 (b) 1700 mm2

(c) 170 mm2

How many m2 are there in 235 in.2? (a) 1,515.75 m2 (b) 15.1575 m2

(c) .151575 m2

How many cm3 are in 12 in.3? (a) 1968 cm3 (b) 196.8 cm3

(c) 19.68 cm3

How many m3 are in 22 in.3? (a) 360.8 m3 (b) .3608 m3

(c) .0003608 m3

10. A box has a volume of 252 in.3. How many dm3 does the box contain? (a) 4.1328 dm3 (b) 413.28 dm3 (c) 4132.8 dm3 II. Select the correct answer: 11. How many meters are there in 3.5 yards? (a) 3.2004 m (b) 32.004 m

(c) 3.24 m

12. How many kilometers are in 859 yards? (a) 785.4696 km (b) .7854696 km

(c) 7.854696 km

13. How many kilometers are in 21 yards? (a) .0192024 km (b) 192.024 km

(c) 19.2024 km

14. How many square meters are in 13 square yards? (b) 10.868 m2 (c) 18.68 m2 (a) 13.836 m2 15. Convert 11 square yards to square meters. (b) 11.836 m2 (a) 9.196 m2

(c) .9196 m2

16. Convert 122 square yards to square kilometers. (a) 101.992 km2 (b) 101,922 km2

(c) .000101992 km2

17. Convert 17 square yards to square decimeters. (a) 1421 dm2 (b) 142.12 dm2

(c) 14.212 dm2

18. How many cubic meters are in 7 cubic yards? (a) 5.3515 m3 (b) 53.515 m3

(c) 7.7645 m3

19. How many cm3 are in .06 yd3? (a) .04587 cm3 (b) 45,873 cm3

(c) 4.587 cm3

20. If a box has a height of 2 yards, a width of 3 yards, and a length of 4 yards, how many cubic meters does it contain? (b) 1834.8 m3 (c) 24 m3 (a) 18.3480 m3

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III. Select the correct answer: 21. How many liters are in 8 quarts of milk? (a) 3.8463 L (b) 7.5704 L

(c) .75704 L

22. How many liters are in 5 quarts? (a) 5.9643 L (b) 47.315 L

(c) 4.7315 L

23. Convert 58 pints to liters. (a) 274.442 L (b) 27.4442 L

(c) 27.5442 L

24. Convert 17 quarts to liters. (a) 8.04372 L (b) 16.0871 L

(c) 16.871 L

25. Convert 15 pints to liters. (a) 7.0974 L (b) 70.974 L

(c) .70974 L

Answers to the review test: I.

1. c 6. b

2. c 7. c

3. c 8. b

4. b 9. c

5. a 10. a

II.

11. a 16. c

12. b 17 a

13. a 18. a

14. b 19. b

15. a 20. a

III.

21. b

22. c

23. b

24. b

25. a

INCHES TO CENTIMETERS Objective: This section will demonstrate how to convert a measure expressed in inches to a measure expressed in centimeters. To change from U.S. inches to centimeters, multiply the number of inches by 2.54. A 12-in. ruler has a length of 30.48 cm because 12 × 2.54 = 30.48. Inches may also be changed to millimeters. To determine how many millimeters are in a 12-in. ruler, remember that the ruler contained 30.48 cm and convert 30.48 to mm: a 12-in. ruler contains 304.8 mm. Practice: Refer to the examples and solve the problems. 1. How many centimeters are in 7 inches? (a) 17.5 cm (b) 177.8 cm (c) 17.78 cm Answer: c (7 × 2.54 = 17.78 cm) 2. How many decimeters are in 14 inches? (a) 35.56 dm (b) 3.556 dm (c) 3556 dm Answer: b (14 × 2.54 = 35.56, 35.56 cm = 3.556 dm)

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3. How many meters are in 18 feet (216 inches)? (a) 548.64 m (b) .54864 m (c) 5.4864 m Answer: c (216 × 2.54 = 548.64, 548.64 cm = 5.4864 m) If a unit of area is expressed in in.2, how can the unit be changed to cm2? Do not multiply by 2.54; instead multiply by 6.45 (2.54 × 2.54 = 6.45). A rectangle that is 7 in. long and 4 in. wide has an area of 28 in.2. To determine out many cm2, multiply 28 by 6.45. This rectangle has an area of 180.60 cm2. To determine how many dm2 are in a rectangle that measures 38 in.2, you would convert your answer from cm2 to dm2 (180.60 cm2 = 1.8060 dm2). Looking closely at the conversion just made, notice the decimal point has been moved two places instead of just one. When using squared measurements, always move the decimal point two more places than when using measurements of length. Examples: 25 cm2 = .25 dm2 25 cm2 = .0025 m2 25 cm2 = .00025 dam2 Practice: Refer to the examples and solve the following problems. 1. How many cm2 are there in 2 in.2? (b) 12.90 cm2 (c) 129.0 cm2 (a) 13 cm2 2 Answer: b (6.45 × 2 = 12.90 cm ) 2. How many mm2 are in 6 cm2? (a) 60 mm2 (b) 600 mm2 Answer: b (6 × 100 = 600)

(c) .06 mm2

3. How many mm2 are in 6 in.2? (b) 387.0 mm2 (c) 3870 mm2 (a) 38.70 mm2 Answer: c (6.45 × 6 = 38.70, 38.70 cm2 = 3870 mm2) To change a unit of volume expressed in.3, multiply the number of in.3 by 16.4. If a box has a volume of 8 in.3, the volume in cm3 is 131.2 cm3: 8 × 16.4 = 131.2 cm3 (.06102 × cm3 = in.3) To determine the mm3 in this same box, convert 131.2 cm3 to mm3: 131.2 cm3 = 131,200 mm3. When using cubic measurements, move the decimal point three more places than when using measurements of length. 25 cm3 = .025 dm3 25 cm3 = .000025 m3 Practice: Refer to the examples and solve the problems.

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1. If a box has a volume of 72 in.3, how many cm3 does it contain? (a) 1180.8 cm3 (b) 118.08 cm3 (c) 1188 cm3 Answer: a (72 × 16.4 = 1180.8 cm3) 2. How many mm3 are in 8 cm3? (a) 80 mm3 (b) 700 mm3 Answer: c (8 × 1000 = 8000)

(c) 8000 mm3

3. How many dm3 are in 12 in.3? (a) 196.8 dm3 (b) 1.968 dm3 (c) .1968 dm3 Answer: c (16.4 × 12 = 196.8, 196.8 cm3 = .1968 dm3)

EXERCISES Directions: Select the correct answer. If necessary, restudy the instructions and examples. 1. How many centimeters are in 17 inches? (a) 43.18 cm (b) 431.8 cm (c) 42.5 cm 2. How many millimeters are in 2 inches? (a) 5.08 mm (b) 50.08 mm (c) 50.8 mm 3. How many meters are in 48 inches? (a) 121.92 m (b) 1.2192 m (c) .12192 m 4. How many cm2 are in 14 in.2? (a) 90.30 cm2 (b) 903 cm2

(c) 9050 cm2

5. How many mm2 are in 16 in.2? (a) 103.20 mm2 (b) 1032 mm2

(c) 10,320 mm2

6. How many dm2 are in 4 in.2? (a) 25.80 dm2 (b) .2580 dm2

(c) 10.16 dm2

7. How many m3 are in 120 cm3? (a) 1968 m3 (b) 1.2 m3

(c) .00012 m3

8. How many cm3 are in 13 in.3? (a) 213.2 cm3 (b) 2.132 cm3

(c) 2132 cm3

9. How many m3 are in 22 in.3? (a) 360.8 m3 (b) .3608 m3

(c) .0003608 m3

10. How many dm3 are in a rectangular box with a length of 12 in., a width of 2 in., and a height of 3 in.? (a) 72 dm3 (b) 1180.8 dm3 (c) 1.1808 dm3

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Answers to the exercises: 1. a 6. b

2. c 7. c

3. b 8. a

4. a 9. c

5. c 10. c

YARDS TO METERS Objective: This section will introduce the method used to change measurements expressed in yards to measurements expressed in meters. The method used to change a measurement from yards to meters is similar to that used to change from inches to centimeters. The only difference is that multiplication is by different numbers. When changing a measurement of length expressed in yards to a measurement expressed in meters, multiply the number of yards by .9144. If an object is 3 yards long, it is 2.7432 meters long because 3 × .9144 = 2.7432. Yards can also be changed to other metric units, such as decimeters. If an object is 3 yards long, it is 27.432 decimeters long because 2.7432 meters = 27.432 decimeters. Practice: Refer to the examples and solve the problems. 1. How many meters are in 3.5 yards? (a) 3.2004 m (b) 32.004 m (c) 3.24 m Answer: a (3.5 × .9144 = 3.2004 m) 2. Convert 6 yards to decimeters. (a) 6.9114 dm (b) 5.4864 dm (c) 54.864 dm Answer: c (6 × .9114 = 5.4864, 5.4864 m = 54.864 dm) 3. How many kilometers are in 800 yards? (a) 731.52 km (b) .73152 km (c) .073152 km Answer: b (800 × .9144 = 731.52, 731.52 m = .73152 km) To change a unit of area expressed in square yards to one expressed in square meters, multiply the number of square yards by .863. For example, 17 square yards = 14.212 square meters because 17 × .836 = 14.212 m2. To determine square decimeters, convert 14.212 m2 to 1421.2 dm2. Remember: Move the decimal point two extra places. Practice: Refer to the examples and answer the problems. 1. How many m2 in 13 square yards? (a) 13.836 m2 (b) 10.868 m2 (c) 18.38 m2 2 Answer: b (13 × .836 = 10.868 m )

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2. How many km2 in 725 square yards? (a) .0006061 km2 (b) 60.61 km2 (c) 725.836 km2 2 Answer: a (725 × .836 = 606.1 m = .0006061 km2) 3. Convert 2 square yards to cm2. (a) 1.672 cm2 (b) 167.2 cm2 (c) 16,720 cm2 Answer: c (2 × .836 = 1.672, 1672 m2 = 16,720 cm2) To change a unit of volume from cubic yards to cubic meters, multiply the number of cubic yards by .7645: 24 cubic yards equals 18.348 cubic meters because .7645 × 24 = 18.348 m2. To change to a different unit, such as cubic centimeters, move the decimal point the necessary number of places: 24 yd3 equals 18,348,000 cm3. Remember: Move the decimal point three extra places. Practice: Refer to the examples and answer the problems. 1. How many m3 are in 12 cubic yards? (a) 9.174 m3 (b) 91.74 m3 (c) 12.7645 m3 3 Answer: a (.7645 × 12 = 9.174 m ) 2. Convert 117 cubic yards to km3. (a) 89.4465 km3 (b) .894465 km3 (c) .0000000894465 km3 Answer: c (117 × .7645 = 89.4465, 89.4465 m3 = .0000000894465 km3) 3. How many dm3 are in 3 cubic yards? (a) 2.2935 dm3 (b) 2,293.5 dm3 (c) 22.935 dm3 Answer: b (.7645 × 3 = 2.2935, 2.2935 m3 = 2,293.5 dm3)

EXERCISES Directions: Select the correct answer. If necessary restudy the instructions and examples. 1. How many meters are in 7 yards? (a) 7.9144 m (b) 6.4008 m

(c) 64.008 m

2. How many kilometers are in 859 yards? (a) 785.4696 km (b) .7854696 km (c) 7.854696 km 3. How many decimeters are in 4 yards? (a) 36.576 dm (b) 4.9144 dm (c) 3.6576 dm 4. Convert 11 yd2 into m2. (a) 9.196 m2 (b) 11.836 m2

(c) .9196 m2

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5. Convert 17 yd2 into dm2. (a) 14.212 dm2 (b) 142.12 dm2

(c) 1,421.2 dm2

6. How many km2 are in 5,721 square yards? (a) 4782.756 km2 (b) 4.782756 km2 (c) .004782756 km2 7. How many m3 are in 8 cubic yards? (a) 6.116 m3 (b) 8.7645 m3 (c) 61.16 m3 8. How many dm3 are in .7 cubic yards? (a) .53515 dm3 (b) 5.3515 dm3 (c) 535.15 dm3 9. How many m3 are in .06 cubic yards? (a) .04587 m3 (b) 45,870 m3 (c) 4.587 m3 10. If a box is 9 yards long, 3 yards wide, and 2 yards high, how many dm3 does it contain? (a) 54 dm3 (b) 41.283 dm3 (c) 41,283 dm3 Answers to the exercises: 1. b 6. c

2. b 7. a

3. a 8. c

4. a 9. a

5. c 10. c

QUARTS TO LITERS AND PINTS TO LITERS Objective: This section will introduce the method to change U.S. liquid quarts to liters and U.S. liquid pints to liters. To change a measurement of liquid expressed in quarts to a measurement expressed in liters, multiply the number of quarts by .9463: 3 quarts of water is 2.8389 liters of water because .9463 × 3 = 2.8389. Practice: Refer to the examples and solve the problems. 1. How many liters are in 2 quarts of milk? (a) 2.9463 L (b) 1.8926 L (c) 18.926 L Answer: b (2 × .9463 = 1.8926 L) 2. How many liters are in 8 quarts of water? (a) 8.9463 L (b) 7.5704 L (c) .75704 L Answer: b (8 × .9463 = 7.5704 L) 3. Convert 4 pints to liters. (a) 1.89264 L (b) 4.47316 L (c) 18.9264 L Answer: a (4 × .47316 = 1.89264 L)

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4. Convert 15 pints to liters. (a) .70974 L (b) 70.974 L (c) 7.0974 L Answer: c (15 × .47316 = 7.0974 L)

EXERCISES Directions: Solve the problems. I. Select the correct answer: 1. Convert 17 inches into centimeters. (a) 43.18 cm (b) 20.32 cm 2.

3.

4.

5.

(c) 4.318 cm

How many millimeters are in 12 inches? (a) 30.48 mm (b) 304.8 mm

(c) 3048 mm

How many meters are in 78 inches? (a) 1.9812 m (b) 198.12 m

(c) 89.54 m

How many cm2 are in 45 in.2? (a) 290.25 cm2 (b) 2902.5 cm2

(c) 29.25 cm2

How many m2 are in 300 in.2? (a) .1935 m2 (b) 193.5 m2

(c) 1935 m2

6.

Convert 17 square inches to square decimeters. (a) 109.65 dm2 (b) 10.965 dm2 (c) 1.0965 dm2

7.

Convert 12 in.3 to cm3. (a) 196.8 cm3 (b) 1968 cm3

(c) 19.68 cm3

How many dm3 are in 19 in3? (a) 311.6 dm3 (b) 31.16 dm3

(c) .3116 dm3

Convert 712 in.3 to m3. (a) 116.768 m3 (b) 11,676.8 m3

(c) .0116768 m3

8.

9.

10. If a box has a height of 5 inches, a width of 2 inches, and a length of 4 inches, what is the volume in cubic decimeters? (a) .0656 dm3 (b) .656 dm3 (c) 6.56 dm3 II. Select the correct answer: 11. How many meters are in 3 yards? (a) 2.7432 m (b) 27.432 m 12. Convert 21 yd to km. (a) .0192024 km (b) 19.2024 km

(c) 3.9144 m

(c) 1.92024 km

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13. Convert 30 yd to dm. (a) 2.7432 dm (b) 273.42 dm

(c) 27.432 dm

14. How many m2 are in 5 yd2? (a) 41.8 m2 (b) 5.836 m2

(c) 4.18 m2

15. Convert 12 yd2 to dm2. (a) 10.032 dm2 (b) 100.32 dm2

(c) 1,003.2 dm2

16. Convert 700 yd2 to km2. (a) .0005852 km2 (b) .5852 km2

(c) 585.2 km2

17. Convert 5 yd3 to m3. (a) 3.8225 m3

(c) 38.225 m3

(b) 5.7645 m3

18. Convert .06 yd3 to dm3. (a) .04587 dm3 (b) 45,870 dm3

(c) 45.87 dm3

19. Convert 1,776 yd3 to km3. (a) 1.357742 km3 (b) .000001357752 km3 (c) 1357.752 km3 20. A box has a height of 5 yards, a width of 4 yards, and a length of 6 yards. What is the volume of this box in cubic meters? (a) 120 m3 (b) 91.74 m3 (c) 917.4 m3 III. Select the correct answer: 21. How many liters are in 5 quarts? (a) 4.718 L (b) 2.3658 L

(c) 4.7315 L

22. How many liters are in 7 pints? (a) 6.6241 L (b) 3.31212 L

(c) 33.1212 L

23. Convert 12 pints to liters. (a) 5.67792 L (b) 11.3556 L

(c) 56.7792 L

24. Convert 13 quarts to liters. (a) 61.5108 L (b) 12.319 L

(c) 12.3019 L

25. Convert 2 pints to liters. (a) 9.4632 L (b) .94632 L

(c) 1.8926 L

26. How many feet are in 15 decimeters? (a) 4.9212 ft (b) 16.404 ft

(c) 1.6404 ft

27. Convert 1.3 meters into inches. (a) 1.42168 in (b) 4.26504 in.

(c) 51.1810 in.

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28. Convert 55 hm into ft. (a) 6014.8 ft (b) 60.148 ft

(c) 18045.5 ft

29. How many yards are in 17 meters? (a) 1.85912 yd (b) 18.5912 yd

(c) 15.5448 yd

30. Convert 27 cm to in. (a) 10.62900 in.

(c) 29.5272 in.

(b) .295272 in.

31. How many yd2 are in 200 dam2? (a) 23,920 yd2 (b) 71,760 yd2

(c) 717.6 yd2

32. How many in.2 are in 2 m2? (a) 2.392 in.2 (b) 86.112 in.2

(c) 3100.032 in.2

33. How many yd2 are in 3 hm2? (a) 35,880 yd2 (b) 358.8 yd2

(c) .0003588 yd2

34. Convert 12 square meters into square yards. (a) 14.352 yd2 (b) 143.52 yd2

(c) 1.4352 yd2

35. Convert 12 dam2 into ft2. (a) 12,916.8 ft2 (b) 129.168 ft2

(c) 1.29268 ft2

36. Change 5 cubic meters to cubic inches. (a) 176.5 in.3 (b) 2118 in.3

(c) 305,121 in.3

37. Convert 17 cubic decimeters to cubic feet. (a) 60.01 ft3 (b) 6001 ft3

(c) .6004 ft3

38. Convert 1 m3 to in.3. (a) 35.3 in.3

(c) 61024.32 in.3

(b) 423.6 in.3

39. How many in.3 are in 25 dm3? (a) 88.25 in.3 (b) 1059 in.3

(c) 1525.5 in.3

40. Convert 12 cubic decameters to cubic feet. (a) 423,780 ft3 (b) 423.6 ft3

(c) 42.36 ft3

41. Convert 12 liters to pints. (a) 12.6804 pt (b) 24 pt

(c) 25.3608 pt

42. Convert 20 liters to quarts. (a) 2.134 qt (b) 21.134 qt

(c) 211.34 qt

43. How many gallons are in 28 liters? (a) 29.5876 gal (b) 7 gal

(c) 7.3976 gal

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44. How many gallons are in 8 liters? (a) 8.4536 gal (b) 2.1136 gal

(c) 2 gal

45. How many pints are in 5 liters? (a) 5.2835 pt (b) 10 pt

(c) 10.567 pt

Answers to the exercises: I.

1. a 6. c

2. b 7. a

3. a 8. c

4. a 9. c

5. a 10. b

II.

11. a 16. a

12. a 17. a

13. b 18. c

14. c 19. b

15. c 20. b

21. 26. 31. 36. 41.

22. 27. 32. 37. 42.

23. 28. 33. 38. 43.

24. 29. 34. 39. 44.

25. 30. 35. 40. 45.

III.

c a a c c

b c c c b

a c a c c

c b a c b

b a a a c

METRIC UNITS OF LENGTH TO U.S. UNITS Objective: This section introduces the method of changing from a metric unit of length to a U.S. unit of length. It will be helpful to remember: 36 in. = 1 yd 3 ft = 1 yd 1760 yd = 1 mi Also, keep in mind the metric units of length. To change a measurement from meters to U.S. yards, multiply the number of meters by 1.0936, e.g., 7 meters = 7.6552 yards because 7 × 1.0936 = 7.6552. Example: To determine how many feet are in 7 meters, multiply the number of yards by 3 because there are 3 feet in a yard. Therefore, 7 meters = 22.9656 feet (7.6552 × 3 = 22.9656). Example: To determine many inches are in 7 meters, multiply the number of yards by 36 because there are 36 inches in a yard. There are 275.5872 inches in 7 meters (7.6552 × 36 = 275.5872). Example: To change 215 km to miles, first, change 215 km to 215,000 meters: 215,000 × 1.0936 = 235,124 yards. Divide 235,124 by 1760 to determine the number of miles: 1760 yards equals 1 mile (235.24 ÷ 1760 = 133.59318 miles).

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Example: To change 17 cm to inches, first, change 17 cm to .17 m: 17 × 1.0936 = .185912 yards; .185912 × 36 = 6.692832 in. Practice: Refer to the examples and solve the problems. 1. How many yards are in 12 meters? (a) 13.0936 yd (b) 13.1232 yd (c) 12.1232 yd Answer: b (12 × 1.0936 = 13.1232 yd) 2. How many feet are in 15 decimeters? (a) 4.9212 ft (b) 16.404 ft (c) 1.6404 ft Answer: a (15 dm = 1.5 m, 1.5 × 1.0936 = 1.6404 yards, 1.6404 × 3 = 4.9212 ft) 3. How many miles are in 14 kilometers? (a) 15,310.4 mi (b) 8.6990909 mi (c) 26,946,304 mi Answer: b (14 km = 14,000 m, 14,000 × 1.0936 = 15,310.4 m, 15,310.4 ÷ 1760 = 8.6990909 mi)

METRIC UNITS OF AREA TO U.S. UNITS Objective: This section demonstrates how to change from a metric unit of area to a U.S. unit of area. The following facts will be helpful. 1296 square inches = 1 square yard 9 square feet = 1 square yard 3,097,600 square yards = 1 square mile These numbers are different from the ones given at the beginning of this section. One square yard is the area of a square with all sides measuring one yard: 1 yard = 36 inches. One square yard is the area of a square with all sides measuring 36 in. To find the area of a square, multiply the measure of one side by itself. The area of the square is 36 in. × 36 in. = 1296 in.2. The area of this same square is equal to one square yard. Therefore, 1 square yard must equal 1296 square inches. By a similar method, other equivalents may be obtained. Become familiar with the following: Unit

Square Square Square Square Square Square Square

kilometer hectometer decameter meter decimeter centimeter millimeters

Abbreviation

Square Meters

km2 hm2 dam2 m2 dm2 cm2 mm2

1,000,000 square meters 10,000 square meters 100 square meters 1 square meter 1/100 square meter 1/10,000 square meters 1/1,000,000 square meters

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Most of the information in the matrix should be familiar; however, some of the terms are new. To change square meters (m2) to U.S. yards, simply multiply the number of square meters by 1.196. If something, has an area of 5 m2 and the number of square yards is needed, multiply 5 × 1.196: 5 × 1.196 = 5.98. There are 5.98 square yards in 5 square meters. Example: To change 7 m2 to square feet, first, convert square meters to square yards: 7 × 1.196 = 8.372. There are 8.372 square yards in 7 square meters. Now determine square feet. Do this by multiplying the number of square yards by 9: 8.372 × 9 = 75.348. There are 75.348 square feet in 7 m2. Now change from 12 square decimeters to square yards. First, change square decimeters to square meters:. 1. 12 dm2 = .12 m2. Now, it is simply a process of changing .12 m2 to square yards. 2. 12 × 1.196 = .14352. There are .14352 square yards in 12 square decimeters. Example: How can 5 km2 be changed to mi2? First, express 5 km2 in m2: 5 km2 = 5.000.000 m2. Then multiply the number of m2 by 1.196: 5,000,000 × 1.196 = 5,980,000. The answer is now in square yards. To determine the number of square miles, divide this answer by 3,097,600: 3,980,000 ÷ 3,097,600 = 1.9305263. There are 1.9305268 mi2 in 5 km2. Practice: Refer to the examples and tables and solve the problems. 1. Change 17 hm2 to yd2. (a) .00203320 yd2 (b) 20.3320 yd2 (c) 203,320 yd2 Answer: c (17 hm2 = 170,000 m2; 170,000 × 1.196 = 203,320 yd2) 2. Change 7 square centimeters to square inches. (a) .0008372 in.2 (b) 1.0850112 in.2 (c) 10,850.112 in.2 Answer: b (7 cm2 = .0007 m2, .0007 × 1.196 = 0008372 yd2, .0008372 × 1296 = 1.0850112 in.2) 3. How many ft2 are in 200 dam2? (a) 23,920 ft2 (b) 215,280 ft2 (c) 717.6 ft2 Answer: b (200 dam2 = 20,000 m2, 20,000 × 1.196 = 23,920 yd2, 23,920 × 9 = 215,280 ft2) Additional helpful information is 1728 cubic inches = 1 cubic foot 46656 cubic inches = 1 cubic yard 27 cubic feet = 1 cubic yard

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and Unit

Cubic Cubic Cubic Cubic Cubic Cubic Cubic

Abbreviation

Cubic Meters

km3 hm3 dam3 m3 dm3 cm3 mm3

1,000,000,000 100,000 1000 1 1/1000 1/1,000,000 1/1,000,000,000

kilometer hectometer decameter meter decimeter centimeter millimeter

EXERCISES Directions: Select the correct answer. If necessary, restudy the instructions and examples. 1. Convert 16 meters to yards. (a) 14.6304 yd (b) 17.4976 yd

(c) 16.4976 yd

2. How many feet are in 22 meters? (a) 24.0592 ft (b) 721.776 ft

(c) 72.1776 ft

3. Convert 1.3 meters to inches. (a) 1.42168 in. (b) 4.26504 in.

(c) 51.18048 in.

4. Convert 25 dm into ft. (a) 2.734 ft (b) 82.02 ft

(c) 8.202 ft

5. Convert 7 km into yd. (a) 7655.2 yd (b) 765.52 yd

(c) 76.552 yd

6. How many ft are in 700 cm? (a) 7.6552 ft (b) 22.9656 ft

(c) 229.656 ft

7. Convert 55 hm into ft. (a) 6014.8 ft (b) 60.148 ft

(c) 18,044.4 ft

8. How many miles are in 3000 meters? (a) 3280.8 mi (b) 1.8640909 mi (c) 1.5586363 mi 9. How many miles are there in 75 kilometers? (a) 46.602272 mi (b) 82.019998 mi (c) 466.02272 mi 10. Convert 23 mm to in. (a) .9055008 in. (b) .7571232 in. (c) .0160272 in.

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11. Convert 17 m2 to yd2. (a) 14.212 yd2 (b) 20.332 yd2

(c) 23.32 yd2

12. Convert 2.3 dam2 to yd2. (a) .027508 yd2 (b) 2.7508 yd2

(c) 275.08 yd2

13. How many in.2 are in 2 m2? (a) 2.392 in.2 (b) 86.112 in.2

(c) 3,100.032 in.2

14. Convert 17 dm2 to ft2. (a) 6099.6 ft2 (b) 18,298.8 ft2

(c) 1.82988 ft2

15. Convert 300 mm2 to in.2. (a) .4650048 in.2 (b) 4.650048 in.2 (c) 465.0043 in.2 16. How many mi2 are in 25 km2? (a) 16,988.636 mi2 (b) 9.6526342 mi2 (c) .00962526 mi2 17. How many yd2 are in 3 hm2? (a) 35,880 yd2 (b) 358.8 yd2

(c) .0003588 yd2

18. How many in.2 are in 35 cm2? (a) .150696 in.2 (b) 5.425056 in.2 (c) .004186 in.2 19. Convert 11 m2 to ft2. (a) 13.156 ft2 (b) 39.468 ft2

(c) 118.404 ft2

20. Convert 500 km2 to mi2. (a) 19.305268 mi2 (b) 193.05268 mi2 (c) 598 mi2 Answers to the exercises: 1. 6. 11. 16.

b b b b

2. 7. 12. 17.

c c c a

3. 8. 13. 18.

c b a b

4. 9. 14. 19.

c a c c

5. 10. 15. 20.

a a a b

METRIC UNITS OF VOLUME TO U.S. UNITS Objective: This section demonstrates changing from a metric unit of volume to a U.S. unit of volume. Remember: the facts from the last two sections.

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To change a measurement of 7 cubic meters to cubic feet, multiply the number of cubic meters by 35.3157: 7 cubic meters = 247.2 cubic feet because 7 × 35.316 = 247.2 cubic feet. To change 16 cubic meters to cubic yards, first find the number of feet: 16 × 35.315 = 565.04. There are 565.04 feet in 16 cubic meters. Now convert 565.04 feet to cubic yards by dividing the number of cubic feet by 27: 565.04 ÷ 27 = 20.9274. There are 20.9274 cubic yards in 16 cubic meters. Example: To change 4 cubic meters to cubic inches, use a similar method. First, determine the number of cubic feet: 4 × 35.315 = 141.26 cubic feet. Then change 141.26 cubic feet to cubic inches by multiplying the number of cubic feet by 1728: 141.26 × 1728 = 244,097.28 cubic inches. Example: To change 2 cubic decameters to cubic yards, first change cubic decameters into cubic meters: 2 dam3 = 2000 m3. Next change 2000 m3 into cubic feet: 2000 × 35.315 = 70,630 ft3. Finally, convert 70,630 ft3 to cubic yards: 70,630 ÷ 27 = 2615.926 yd3. Practice: Refer to the examples and solve the problems. 1. Change 5 cubic meters to cubic inches. (a) 176.5 in.3 (b) 2,118 in.3 (c) 305121.6 in.3 3 Answer: c (5 m × 35.315 = 176.575 ft3, 176.575 × 1728 = 305121.6 in.3) 2. Change 12 cubic decimeters to cubic feet. (a) .42378 ft3 (b) 423.6 ft3 (c) 4236 ft3 3 3 Answer: a (12 dm = .012 m , .012 × 35.315 = .42378 ft3) 3. Change 573 cm3 to cubic inches. (a) 34.966935 in.3 (b) .0202269 in.3 (c) 349.520832 in.3 Answer: a (573 cm3 = .000573 m3, 000573 × 35.315 = .0202354 ft3, .0202354 × 1728 = 34.966935 in.3) Sometimes it is necessary to change from a metric unit of volume to U.S. unit of liquid volume. To make the conversion, the following information is needed. 1 quart = 2 pints 4 quarts = 1 gallon To change from metric liters to U.S. liquid quarts, multiply the number of liters by 1.0567. To change 17 liters to quarts, multiply 17 by 1.0567. There are 17.9639 quarts in 17 liters. Example: To change 23 liters to pints, first determine how many quarts are in 23 liters: 23 × 1.0567 = 24.3041 quarts. Then multiply the number of quarts by 2 to get the number of pints: 24.3041 × 2 = 48.6082. There are 48.6082 pints in 23 liters.

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To change 36 liters to gallons, first determine the number of quarts: 36 × 1.0567 = 38.0412. To convert quarts to gallons, divide by 4: 38.0412 ÷ 4 = 9.5103. There are 9.5103 gallons in 36 liters. Practice: Refer to the examples and solve the problems. 1. How many quarts are in 7 liters? (a) 7.3969 qt (b) 8.372 qt (c) 73.969 qt Answer: a (7 × 1.0567 = 7.3969 qt) 2. How many gallons are in 16 liters? (a) 4 gal (b) 16.9072 gal (c) 4.2268 gal Answer: c (16 × 1.0567 = 16.9072, 16.9072 ÷ 4 = 4.2268 gal) 3. Convert 12 liters to pints. (a) 12.6804 pt (b) 24 pt (c) 25.3608 pt Answer: c (12 × 1.0567 = 12.6804, 12.6804 × 2 = 25.3608 pt)

EXERCISES Directions: Select the correct answer. If necessary, restudy the instructions and examples. 1. Convert 1 cubic decameter to cubic feet. (a) 35.3 ft3 (b) 35,315 ft3 (c) 3.53 ft3 2. Convert 1 cubic meter to cubic inches. (a) 35.3 in.3 (b) 423.6 in.3 (c) 61,024.32 in.3 3. Change 31 cubic decimeter to cubic yards. (a) .40529629 yd3 (b) 109.43 yd3 (c) 405.29629 yd3 4. How many cubic yards are in 3 cubic meters? (c) 11.8333 yd3 (a) 3.92222 yd3 (b) 3.9444 yd3 5. How many cubic feet are in 17 cubic decimeters? (a) .6001 ft3 (b) 6001 ft3 (c) 600.1 ft3 6. Convert 20 liters to quarts. (a) 2.134 qt (b) 21.1340 qt

(c) 211.34 qt

7. Convert 19 liters to pints. (a) 20.0773 pt (b) 40.1546 pt

(c) 38 pt

8. Convert 24 liters to gallons. (a) 6 gal (b) 6.3402 gal

(c) 25.3608 gal

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9. How many gallons are in 28 liters? (a) 29.5876 gal (b) 7 gal (c) 7.3969 gal 10. How many pints are in 3 liters? (a) 6 pt (b) 3.1701 pt

(c) 6.3402 pt

Answers to the exercises: 1. b 6. b

2. c 7. b

3. a 8. b

4. a 9. c

5. a 10. c

CUMULATIVE EXERCISES FOR LENGTH, AREA, AND VOLUME Directions: Select the correct answer. If necessary, restudy the instructions and examples. I. Select the correct answer: 1. How many feet are in 4 decimeters? (a) 1.31232 ft (b) .43744 ft 2.

3.

4.

5.

(c) 13.1232 ft

How many in. are in 11 cm? (a) 433.0656 in. (b) .120296 in.

(c) 4.3307 in.

Convert 12 dam to yd. (a) 13.1232 yd (b) 131.232 yd

(c) 1.31232 yd

How many yards are in 1 mile? (a) 1760 yd (b) 1296 yd

(c) 1728 yd

Convert 17 dm to in. (a) 66.929 in.

(c) 669.2832 in.

(b) 1.85912 in.

II. Select the correct answer: 6. How many square yards are in 1 square mile? (a) 1760 yd2 (b) 3,097,600 yd2

(c) 1296 yd2

How many yd2 are in 1 dam2? (a) 1.196 yd2 (b) 11.96 yd2

(c) 119.6 yd2

How many ft2 are in 1 m2? (a) 1.196 ft2 (b) 3.588 ft2

(c) 10.764 ft2

How many in.2 are in 1 dm2? (a) 4.3056 in.2 (b) 15.50016 in.2

(c) 430.56 in.2

7.

8.

9.

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10. Convert 1.2 hm2 to yd2. (a) 14,352 yd2 (b) 1.4352 yd2

(c) 143.52 yd2

III. Select the correct answer: 11. Convert 1 cm3 to m3. (a) 1/1,000,000 m3 (b) 1,000,000 m3

(c) 1000 m3

12. Change 5 dm3 to in.3. (a) 17.65 in.3 (b) 2,541.6 in.3

(c) 305.122 in.3

13. Change 17 dm3 to ft3. (a) 6.001 ft3 (b) .600355 ft3

(c) 600355 ft3

14. Convert 12 dam3 to yd3. (a) 4236 yd3 (b) 470.666 yd3

(c) 15696 yd3

15. How many in.3 are in 1.5 m3? (a) 52.95 in.3 (b) 91,536.48 in.3

(c) 635.4 in.3

IV. Select the correct answer, 16. How many pints are in 3 liters? (a) 6.3402 pt (b) 3.1701 pt

(c) 6 pt

17. How many gallons are in 32 liters? (a) 8 gal (b) 33.8144 gal

(c) 8.4536 gal

18. How many quarts are in 7 liters? (a) 7.2969 qt (b) 7.3969 qt

(c) 7.4169 qt

19. How many gallons are in 56 liters? (a) 14.7938 gal (b) 14 gal

(c) 4.2268 gal

20. Convert 17 liters to pints. (a) 34 pt (b) 17.9639 pt

(c) 35.9278 pt

Answers to the cumulative exercises: I.

1. a

2c

3. b

4. a

5. a

II.

6. b

7. c

8. c

9. b

10. a

III.

11. a

12. c

13. b

14. c

15. b

IV.

16. a

17. c

18. b

19. a

20. c

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Part III Appendices

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Appendix A: A Typical Cover Sheet for the GPS Process A cover sheet summarizes the individual steps of the GPS process. There are many variations and no particular cover sheet is better than any other. The goal of the cover sheet is to present information in a structured way. Obviously, detailed backup information of the work should be kept for future reference. Identification number:

Department:

Target date:

Completion date:

GPS0. Describe the situation and the emergency response action: GPS1. Team: GPS2. Describe the problem: GPS3. Develop the interim containment action: GPS4. Define and verify root cause and escape point: GPS5. Choose and verify permanent corrective action (root cause/escape point): GPS6. Implement and validate permanent corrective actions: GPS7. Prevent recurrence: GPS8. Recognize team:

327

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Appendix B: GPSO. Preparation for Emergency Response Actions (ERA): Assessment Questions Typical questions are • • • •

Are ERAs necessary? Is a service action required as part of the emergency response? How was the ERA verified? How was the ERA validated?

GPS application criteria: How well does the proposed GPS meet the application criteria? • Is there a definition of the symptom(s)? Has the symptom been quantified? • Have the GPS customer(s) who experienced the symptom(s) and the affected parties (when appropriate) been identified? • Have measurements which were taken to quantify the symptom(s) demonstrated that a performance gap exists and/or has the priority (severity, urgency, growth) of the symptom warranted initiation of the process? • Is the cause unknown? • Is management committed to dedicating the necessary resources to fix the problem at the root cause level and to prevent recurrence? • Does the symptom complexity exceed the ability of one person to resolve the problem? Other: Will the new GPS duplicate an existing GPS? Common Tasks: • Have all changes been documented (e.g., FMEA, control plan, process flow)?

329

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• Is the team composed of the appropriate members to proceed to the next step? • Have the measurable(s) been reviewed? • Has it been determined if a service action is required? The essence of GPS0 is to be sure an investigation is needed and, if so, to determine the extent of the investigation.

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Appendix C: GPS1. Establish the Team/Process Flow: Assessment Questions Warm up: • When and where will the team meet? • What has been done to make the meeting room user-friendly? • What has been done to encourage team members to build relationships with each other? • What has been done to encourage team members to focus on the team’s activity? • Has the purpose of the meeting been stated? • Has the team received the agenda for the meeting? Membership: Are people affected by the problem represented? • How is the GPS customer’s viewpoint represented? • Does each person have a reason to be on the team? • Is the team large enough to include all necessary input, but small enough to act effectively? • Does team membership reflect the problem’s current status? • Do team members agree on membership? Product/process knowledge: What special skills or experience does the team require to function effectively? Operating procedures and working relationships: Have team goals and membership roles been clarified? • Does the team have sufficient decision-making authority to accomplish its goals? • How will information from the team be communicated internally and externally? • Do all members agree with and understand the team’s goals?

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Roles: • • • •

Has the designated champion of the team been identified? Has the team leader been identified? Are team members’ roles and responsibilities clear? Is a facilitator needed to coach the process and manage team consensus?

Common tasks: Have all changes been documented (e.g., FMEA, control plan, process flow)? • Is the team composed of the appropriate members to proceed to the next step? • Have the measurable(s) been reviewed? • Has the team determined if a service action is required? The essence of GPS1 is to define the team. Specifically, the team must be crossfunctional and multidisciplinary and the appropriate roles must be assigned for optimum results.

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Appendix D: GPS2. Describe the Problem: Assessment Questions Symptom: Can the symptom be subdivided? Problem statement: • • • •

Has a specific problem statement been defined (object and defect)? Have “repeated whys” been used? What iswrong with what? Is it known for certain why the problem is occurring?

Problem description: • • • • • • • • • • •

Has “Is/Is Not analysis” been performed (what, where, when, how big)? When has this problem appeared before? Where in the process does this problem first appear? What, if any, pattern(s) is (are) there for this problem? Are similar components and/or parts showing the same problem? Has the current process flow been identified? Does this process flow represent a change? Have all required data been collected and analyzed? How does the ERA affect the data? Is there enough information to evaluate for potential root cause? Is there physical evidence of the problem? Has a cause-and-effect diagram been completed?

Type of problem: Does this problem describe a “something-changed” or a “neverbeen-there” situation? Review of problem description: • Has the problem description been reviewed for completeness with the GPS customer and affected parties? (Has concurrence been obtained from the GPS customer and the affected parties?)

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• Should this problem be reviewed with executive management? • Should financial reserves be set aside? • Should any moral, social, or legal obligations related to this problem be considered? Common tasks: • Have all changes been documented (e.g., FMEA, control plan, process flow)? • Does the team have the appropriate members to proceed to the next step? • Have the measurable(s) been reviewed? • Has the team determined if a service action is required? The essence of GPS2 is to force the team to come up with the “real” problem. The real problem must be unambiguous, valid, objective, measurable, repeatable, and verifiable. To focus on this definition, the following discussion may be of help. Problem description — To develop the correct problem statement: Ask: “What is wrong with what?” Ask next: “Is it known why this is happening?” If the answer is “no,” the answer to “What is wrong with what?” is the problem statement. If the answer is “yes,” repeat the question, “What’s wrong with what?” until reaching the statement where the cause is unknown. To develop the problem description, apply the Is/Is Not method using the following questions: Is

Is Not

What • Name the object having trouble. • Name the trouble/problem the object is having.

What • Name objects that are similar in shape, composition, form, or function that could have the same trouble, but do not. • Name other kinds of trouble or problems this object could be experiencing, but is not. Where • Name other places where this object can be found having no trouble. • Name places where the problem could have first appeared, but did not. • Name similar places where this problem has never occurred.

Where • Name the place where this object having trouble can be found. • Name the place where this problem first appeared. • Name all other places where this problem occurs (or has occurred).

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Appendix D: GPS2.

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Is

Is Not

Where (continued) • Where on the object does the problem occur (inside, outside, top, bottom, etc.)? • Where in the process flow did the problem first develop? • Describe all other places where the object and the trouble can be found.

Where (continued) • Where on the object could the problem occur, but does not? • Where in the process could the problem have first developed, but did not? • Use similar units of measure to describe where the object and the trouble are absent. When • Using the same unit of time, when could the trouble and/or problem have first occurred, but did not? • Describe the above information, matching day, month, year, time of day, etc. to when the trouble did not occur. • Use corresponding units of measure to describe when the trouble did not occur. • Describe when in the process the trouble/problem could have first occurred, but did not. • Describe when in the life cycle the problem could have first occurred, but did not. • Describe other places in the process or life cycle when it could have been observed, but was not.

When • With respect to time, when did the trouble/problem first occur? • Describe the above information with regard to day, month, year, time of day, etc. • Describe any patterns with regard to time. • Describe when in the process the trouble/problem first occurred. • Describe when in the life cycle the problem first occurred. • Describe other places in the process and life cycle where the trouble/problem was observed. Note: Consider all units of time, e.g., hours, days, minutes, seconds, shifts, periods, quarters, years, etc. Also consider sequences in operations as they would occur in a process flow diagram. How Big • Describe the size of the problem/trouble/effect. • Describe the number of objects that have or have had the trouble/problem/effect. • Describe the magnitude of the trouble in terms of percentages, rates, patterns, trends, yield, dimensions, etc. • Describe the number of defects per object. • Describe the physical dimensions of the defect or problem.

How Big • Using a similar unit of measure, describe the limits of the problem/trouble/effect. • Describe the number of objects that could have had the trouble/problem/effect, but did not. • Describe what the magnitude could have been, but was not. • Describe what the defects could have been, but were not. • Describe the physical dimensions the defect could demonstrate, but has not.

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Ask: “Are there any other facts that do not appear on the Is/Is Not tool?” Ask next: “Who can describe the Is/Is Not facts?” Developing the problem description — process overview: Step 1. Develop the problem statement. • Develop a simple, concise statement to describe who or what object illustrates the specific defect that has been chosen for determination of the root cause. • The statement is intended to be a starting point for further description; the statement must be very concise and clearly perceived. • Use the “repeated whys” to indicate the lowest rung in what might be a series of events, a chain reaction, etc. Step 2. Develop the problem description. • Describe in quantifiable, factual terms the extent of the defect. List all facts with regard to who, what, where, when, and how big (magnitude). • Describe in quantifiable, contrasting dimensions where the symptom is not manifested or visible. Step 3. Identify information which needs to be confirmed or collected. Obtain information from knowledgable people by asking the questions in Step 2. Developing the problem description — process detail: Step 1: Develop the problem statement. Ask: “What is wrong with what?” Then identify the effect/defect on the object. Ask next: “Why is that happening on that object?” Verify that the answer which has been provided is correct and not just an assumed reason. Repeat the first question over and over until the question cannot be answered with certainty. If the cause is unknown and needs to be found, the root cause then the last object and defect with unknown cause become the operational statement. Proceed to Step 2 to define the full extent (or profile) of the defect in factual terms. Example of “Repeated Whys” Technique: 1. 2. 3. 4.

I had an automobile accident. Why? I couldn’t see clearly. Why? My diet is poor (and it has affected my vision). Why? My teeth hurt (so I eat the wrong food). Why? I don’t know why my teeth hurt.

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Problem statement: My teeth hurt. Step 2: Develop the problem description. Not every question will apply to every problem. What Is • • • •

What is the specific name(s) of the object that has the defect? What kind of defect is it? Is there more than one kind of defect? (If so, separate.) Is the reason for this defect known? (Are you certain?)

What Is Not • What objects are similar to “Is” in shape, composition, size, function, etc., but do not have the same defect? • What other objects could show the same kind of defect, but do not? Where Is • Point out exactly where the defect can be found. What is the location? Is it necessary to go to more than one place? What is the name of this place or places? • Where in the procedure or process is the defect first seen? • Where in the life cycle of this object is this defect first seen? • Where on the object is the defect observed? Is this the only place? (Think inside/outside, end to end.) How is the location defined? When Is • Was this object ever free of defects? Has this problem existed since the start (day one deviation/common cause conditions)? • When did this defect first occur? (Use year, month, week, day, hour, minute, second, if possible.) • Does the defect occur all the time or just part of the time? Look for and define any patterns with respect to time. • When else did it happen (year, month, day, etc.)? • Does it still happen? Is it continuous? When did it stop? • Is it stopping and starting? Define frequency (e.g., only on third shift, only begins after seven hours). How Big Is • What is the magnitude of this defect? How can this defect be measured (e.g., length, width, weight, decibels, units of force, etc.)? • How many defects per object?

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1. Statement: What is wrong with what? 2. Problem description (Use a form similar to the following.)

Problem description

Is

Is Not

Get Information

What Where When How Big

FIGURE D.1 A typical problem solving worksheet.

• Compare results (parts, scrap reports, percents, averages, anything else that tells about quality with quantity). • Are there any patterns or trends with regard to the number of objects affected or the unit of measure of this defect? Notes: • Who has more information about this defect? • Use a question about the process. Avoid using questions that can be answered with “yes” or “no.” To facilitate the questions in Step 2, it is recommended that the team develop a problem solving worksheet. A typical worksheet is shown in Figure D.1. The contents of this worksheet should cover the IS and IS NOT conditions as well as the information needed. Comparative analysis: Step 3A: Uncover differences. Differences are • Facts • Unique to the IS • Facts that have not already been stated in the IS column Ask: To uncover differences ask “What is different, unique, peculiar, special or true only of the Is when compared to the companion Is Not?” The guidelines to develop differences are 1. List all facts without prejudice to cause or potential cause.

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Appendix D: GPS2.

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2. Consider categories of people, methods, materials, machines, measurement, and mother nature (environment). Step 3B: Identify changes. Changes are • Facts • Related to differences Ask: to uncover changes ask “What has changed in, on, around, or about the difference and when did the change occur?” The guidelines to develop changes are 1. List all changes that occurred which are related to the difference regardless of date or the potential for cause. 2. Consider the categories of people, methods, materials, machines, measurement and mother nature (environment). Step 4: Develop possible causes (change-how theories). Theories are • Statements of ways that the changes may have created the trouble • A limited form of brainstorming • A simple listing of possible causes, not probable causes Ask: to develop theories ask “How could this change have caused the effect on the object?” or, “in what ways might this change have caused the effect on the object?” The guidelines for developing change-how theories are 1. 2. 3. 4. 5.

List each theory individually. Do not reject or qualify the theory based on its practicality or probability. List single change/single variability theories first. Develop more complex variability theories second. Continue to prompt the problem-solving group with the above question until all possible theories are developed. 6. Defer critical thinking until the next step. Step 5: Make a trial run of the theories. Trial runs are • Critical evaluations of theories against the sets of Is/Is Not data • A test of the plausibility, not remote possibility, of each theory • A test of the likelihood of the theory • A process of elimination Ask: to test the theory ask “Does the change-how theory explain both the IS and the IS NOT?” Guidelines for a trial run: 1. Test the theory against each individual set of Is and Is Not. 2. If the theory accounts for both Is and Is Not phenomena, record with a plus (+) symbol.

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3. If the theory cannot plausibly explain either the Is or the Is Not phenomenon, record the result with a minus (–) symbol and note the reason for the rejection. 4. Test the theory against all sets of Is and Is Nots. Do not stop testing a theory prematurely. 5. Test all the theories in and of themselves. Consider each “change-how” theory a separate theory. 6. Avoid testing interactive changes and highly complex theories until the end of the trial runs. Test simple theories first. 7. If the plausibility of the theory is uncertain ( + or –), use a question mark (?) to record the uncertainty. Note why the theory is uncertain. Step 6: Verify the root cause. • Describe the optimum method to passively, and then actively, verify the root cause. • Conduct verification in the appropriate setting. • Record the results.

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Appendix E: GPS3. Develop Interim Containment Action (ICA): Assessment Questions Before ICA implementation: • • • •

Are ICAs required? Is a service action required as part of the ICA? What can be learned from the ERA that will help select the “best” ICA? Based on consultation with the GPS customer and champion, have criteria been established for ICA selection? • Based on the criteria established, does the ICA provide the best balance of benefits and risks? • How does choice of this ICA satisfy the following conditions? 1. The ICA protects the customer 100 percent from the effect. 2. The ICA is verified. 3. The ICA is cost-effective and easy to implement.

Planning: • Have the appropriate departments been involved in planning this decision? • Have appropriate advanced product quality planning (APQP) tools (e.g., FMEA, control plans, instructions) been considered? • Have plans, including action steps, been identified (who needs to do what by when)? • Has a validation method been determined? • Does the customer have a concern with this ICA (is customer approval required)? • Has what could go wrong with the plan been identified and have preventive and contingency actions been considered? • Are implementation resources adequate? Post implementation: Does validation data indicate that the GPS customer is being protected? 341

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Common tasks: Can the ICA effectiveness be improved? • Have all changes been documented (e.g., FMEA, control plan, process flow)? • Does the team have the appropriate members to proceed to the next step? • Have the measurable(s) been reviewed? • Has the team determined if a service action is required?

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Appendix F: GPS4. Define and Verify Root Cause and Escape Point: Assessment Questions General: • Has factual information in the problem description been updated? • What sources of information have been used to develop the potential root cause list? Root cause: • Is there a root cause (a single verified reason that accounts for the problem)? • What factor(s) changed to create this problem? What data are available that indicate any problem in the manufacturing or design process? • How was this root cause verified? • Does this root cause explain all facts compiled at GPS2? • Has the root cause analysis gone far enough (is it necessary to know why this root cause happened)? Potential root cause: • Is there more than one potential root cause? • Does each item on the potential root-cause list account for all known data? Has each item been verified (used to make the defect come and go)? • How was assignment of percent contribution determined? • Combined, do the items on the potential root-cause list account for 100 percent of the problem (is the desired performance level achievable)? • If the level is achievable, has the team considered and reviewed with the champion the benefit of developing a separate problem description (and, by definition, a separate GPS) for each potential root cause?

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• If the level is not achievable, has the team considered and reviewed with the champion the benefit of alternate problem-solving approaches? 1. Approaches independent of an GPS? 2. Approaches as a compliment to an GPS? Escape point: • Does a control system exist to detect the problem? • Has the current control system been identified? Does this control system represent a change from the original design? • Is there verification that the control system is capable of detecting the problem? • Is the identified control point closest to the root cause/potential root cause? • Is there a need to improve the control system? Common tasks: • Have all changes been documented (e.g., FMEA, control plan, process flow)? • Does the team have the appropriate members to proceed to the next step? • Have the measurable(s) been reviewed? • Has the team determined if a service action is required? Root cause analysis: Root cause analysis is based on a questioning process focusing on the what, where, when, and how big and answering the following concerns about the cause — Is/Is Not, differences, changes, and test. Determine the root cause of the problem (Figure F.1): Determination of the root cause: In cases when “should” and “actual” were once the same, but now are different, the root cause will be a change of some type. The definition of root cause is the single verified reason that accounts for the problem. Approaches for determining root cause: Two deductive approaches are available to direct the problem solver to the root cause. Both approaches are based on a thorough defect profile built around the problem statement. Both deductive approaches are a series of questions that yield answers to which another question is applied. The result is a steady reduction in the number of possible causes to be investigated. For comparison, the steps of the two approaches are shown in Table F.1, having been anchored by the scope (problem statement and problem description). Method A should be used when all changes are known. Method B’s advantage is that it provides a hint about where to investigate for hidden changes. The additional step eliminates consideration of changes common to both the Is and Is Not comparisons listed in the problem description.

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Appendix F: GPS4.

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1. Develop the problem statement

2. Develop the problem description using Is/Is Not method

3A. List changes on timeline

3B. Comparative analysis List differences List changes in differences

4. Develop theories

5. Trial run theories

6. Verify root cause FIGURE F.1 Path for root cause determination.

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TABLE F.1 Comparison of Root Cause Approaches Approach A 1. 2. 3. 4. 5. 6.

Establish a problem statement. Develop a problem description. List ALL known changes on a timeline. Develop theories based upon the changes. Trial run the theories. Verify the root cause/potential root cause.

Approach B 1. Establish a problem statement. 2. Develop a problem description. 3a. List differences. 3b. List changes in differences (only). 4. Develop theories based upon the changes. 5. Trial run the theories. 6. Verify the root cause/potential root cause.

Note: The two approaches are identical except Method B has an addition inserted in the third step.

Using either Method A or B, the problem solver may have to collect missing information. The following list of questions may be used to gather information from people who do the work and are familiar with details of the problem. Frequently, these people have essential information that they do not recognize as significant. Problem-solving questions and guidelines: Step 1: Develop the problem statement. Ask: 1a. “What is wrong with what? “ 1b. “Is the cause of that defect on that object known (for certain)?” (see note) If the answer to 1b is “no,” use the answer to 1a as the problem statement. Note: Use “repeated whys” to reach the correct problem statement by repeating 1b until a “cause is unknown” statement has been reached. Step 2: Develop the problem description using the Is/Is Not method. Ask: 2a. “What … Is the effect?” “Where … Is the effect?” “When … Is the effect?” “How Big … Is the effect?” 2b. “What … Is Not the effect?” “Where … Is Not the effect?” “When … Is Not the effect?” “How Big … Is Not the effect?”

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Notes: Answers to 2b are contrasting information/data based on the specific items in 2a. All answers must be facts. Answers to 2b should be similar in category, shape, function, form, and composition to its counterpart in 2a. (If 2a states an increment in time, then 2b should also be the same increment in time when the problem or effect is not visible or present.) Step 3: List all known changes on a timeline (Method A). Ask: “What is unique, peculiar, different, distinctive, unusual about Is?” Notes: The information must be factual, true only about the Is, not already listed in the Is column of the problem scope section. Consider features such as people, methods, material, machines, and mother nature. Do not rule out any facts that are valid answers to the question. If it is a fact, write it down. Step 3: Comparative analysis (Method B). Step 3a: List differences. Ask: “What is unique, peculiar, different, distinctive, unusual about Is?” Notes: The information must be factual, true only about the Is; and not already listed in the IS column of the problem scope section. Consider features such as people, methods, material, machines, and mother nature. Do not rule out any facts that are valid answers to the question. If it is a fact, write it down. Step 3B: List changes in differences. Ask: “What has changed in, on, around, or about this difference?” Notes: When asking a question, name the difference. (If the difference is an electric motor, ask: “What has changed inside, on, around, or about the electric motor?”) Consider changes in people, methods, material, machines, and mother nature. The rest of the steps are identical for Method A and Method B. Step 4: Develop theories based on the changes (Method A or B). Ask: “In what ways might this change create the defect on the object?” or “How could this change create the defect on the object?” Write the answer in a “change-how” format. Notes: Problem solvers should use the actual names of the changes, defect, and object when asking the question. This is superior to using generic words for the defect, change, and object.

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In Step 4 (theory development), a recommended technique is to list all possibilities, no matter how strange they may sound initially. With groups, it better than debating the practicality; this debate will occur in Step 5. During Step 4, the problem solver is theorizing about the specific mechanism or variable that caused the defect. Beware of using generalities such as “out of spec,” “poor quality,” or “something is wrong” as the variable introduced by the change which in turn, created the defect on the object from Step 1, the problem statement. Step 5: Trial run the theories using the test matrix. Test each theory against each discrete set of Is/Is Nots listed in the Problem Description (Step 2) for every element of “what,” “where,” “when,” and “how big.” Ask: “Does this “change-how “ theory completely explain both the Is and Is Not?” Notes: The question asked is “if this theory is the cause of the effect, do the factual data in the problem description fully explain why the effect manifests itself in the Is dimension, but never manifests itself in the Is Not dimension?” There are three possible outcomes: 1. Explains fully: If the data fully explain why the effect manifests itself in the Is dimension, but never manifests itself in the Is Not dimension, then a plus (+) symbol should be entered in the test matrix for that element. This theory fully explains this element. 2. Cannot explain: If the data cannot explain why the effect manifests itself in the Is dimension and/or never manifests itself in the Is Not dimension, then a minus (–) symbol should be entered in the test matrix for that element. This theory definitely does not explain this element, and therefore cannot be the cause. 3. Insufficient data: If the theory can explain the effect, but there are insufficient data to fully explain why the effect manifests itself in the Is dimension and/or never manifests itself in the Is Not dimension, then a question mark (?) symbol should be entered in the test matrix for that element. A comment should be added at that point to indicate the further data collection or analysis that is required. This theory could explain this element, but more data or analysis are required. Repeat the test question against each pair of Is/Is Nots until a minus symbol is encountered for a theory; then it should be discounted and a further theory examined. Use of the minus (–) symbol is to exclude nonfeasible theories from further consideration. If only one theory passes through the trial run (with all pluses or a combination of pluses and questions marks), then proceed to Step 6 to verify the roof cause included in this theory. In practice, particularly where multiple potential root causes are considered, more than one theory may pass the trial run with a combination of pluses and question marks. In those cases, and where it is feasible and practical to do so, collect

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and analyze the missing data and reexamine the theory to resolve the question marks (?) to either pluses (+) or minuses (–) and proceed as above. If it is not feasible or practical to collect and analyze the missing data, proceed to Step 6, starting with the theory with the most pluses (+). Step 6: Verify root cause. Verification of possible causes can only occur where the defect is occurring. Root cause verification cannot be done on a problem-solving form. Verification can be done in two ways: passive or active. Passive verification is by observation. With passive verification, look for the presence of the root cause/potential root cause without changing anything. If the presence of the root cause/potential root cause variable cannot be proved, then chances are great that this possible cause is not the root cause. Active verification is a process in which the problem solver seeks to make the defect come and go using the variable thought to be the root cause/potential root cause. Both coming and going are important tests to confirm root cause/potential root cause. Helpful supplemental tools: The two deductive approaches to determine root cause are based on the availability of all relevant facts. Also assume the two deductive approaches contain “myths” or information that are not fact. Sometimes, despite the best efforts of asking the right questions to the appropriate people, the necessary facts cannot be found. Additional information may be obtained from: • • • • • • • • •

Statistical process control data Process flow diagrams Process flow cause-and-effect diagram Possible causes listed on a cause-and-effect diagram An expert’s experience Part analysis of the defect Distinction analysis Group brainstorming Force field analysis

There are benefits and liabilities for each when used with a purely deductive method.

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Appendix G: GPS5. Choose and Verify Permanent Corrective Actions (PCAs) for Root Cause and Escape Point: Assessment Questions Before the PCA decision: • What criteria have been established for choosing a PCA for the root cause and escape point? Does the champion agree with these criteria? • Is a service action required as part of the PCA? • What choices have been considered for the PCAs? • Have better choices been overlooked? • What features and benefits would the perfect choice offer? How can these benefits be preserved? • Do members of this team have the appropriate experience on this team to make this decision? • What risks are associated with this decision and how should they be managed? • Does the champion concur with the PCA selections? Verification: • What is the evidence (proof) that this will resolve the concern at the root-cause level? • Did verification approaches make allowances for variations in the frequency (or patterns) created by the cause? • Which variables were measured during the verification step? Do these indicators constitute sound verification? After the PCA decision: • What are the possibilities that this choice, once implemented, will create other troubles? 351

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• Can the customer live with this resolution? • Will containment continue to be effective until the choice can be implemented? • What resources will be required for PCA implementation? Are these resources available? • What departments will need to be involved in the planning and implementation of this decision? • Have actions been considered to improve the ICA prior to PCA implementation? Common tasks: • Have all changes been documented (e.g., FMEA, control plan, process flow)? • Does the team have the appropriate members to proceed to the next step? • Have the measurable(s) been reviewed? • Has the team determined if a service action is required?

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Appendix H: GPS6. Implement and Validate Permanent Corrective Actions (PCAs): Assessment Questions Planning: • What departments are needed to implement the PCAs? • Are representatives of those departments on the team to plan and implement their parts? • What customer and/or supplier involvement is needed? • Who will do the planning for the customer? For the supplier? • Has an action plan been defined (responsibilities assigned, timing established, required support determined)? • Are the necessary resources available to implement this plan? What is needed? • At what point(s) is this plan vulnerable to “things-gone-wrong?” What can be done to prevent them? • What are appropriate contingent actions? • What will trigger contingent actions? • How is completion of the plan being monitored? • When will the ICA be removed? • How will this plan be communicated to those who have a need to know? What training will be required? • What measurable(s) will be used to validate the outcome of the PCAs (short-term and long-term)? Validation: • • • •

Has the ICA been discontinued? Has the unwanted effect been totally eliminated? How can this be conclusively proved? How will long-term results continue to be monitored? What is the measurable? Is this the best way to prove the root cause has been eliminated?

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• How have the findings been confirmed with the customer? • Have all systems, practices, procedures, documents, etc. been updated? Do they accurately reflect what is to be done from here on? Common tasks: • Have all changes been documented (e.g., FMEA, control plan, process flow)? • Does the team have the appropriate members to proceed to the next step? • Have the measurable(s) been reviewed? • Has the team determined if a service action is required?

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Appendix I: GPS7. Prevent Recurrence: Assessment Questions Problem history: • How and where did this problem enter the process? • Why did the problem occur at this place/point and how did it escape? (Why was it not detected?) • Did confusion or lack of knowledge contribute to the creation of this root cause? Did it contribute to the escape? • What policies, methods, procedures, and/or systems allowed this problem to occur and escape? • Were Band-Aid™ fixes uncovered in the processes? Where? What do they compensate for? • Have the affected parties been identified? Prevent actions (this problem and similar problems): • What needs to be done differently to prevent recurrence of the root cause? Of the escape? • Is a service action required as part of the prevent actions? • What evidence exists to indicate the need for a process improvement approach (i.e., focused improvement, reengineering)? • Who is best able to design improvements in any of the systems, policies, methods, and/or procedures that resulted in this root cause and escape? • What is the best way to perform a trial run with these improvements? • What practices need standardization? • What plans have been written to coordinate prevent actions and standardize the practices: who, what, and when? • Does the champion concur with the identified prevent actions and plans? • How will these new practices be communicated to those affected by the change? • Have the practices been standardized? • What progress check points have been defined to assess system improvements?

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Systemic prevent recommendations: • What management policy, system, or procedure allowed this problem to occur or escape? • Are these practices beyond the scope of the current champion? • Who has responsibility for these practices? • Does the current champion agree with the systemic prevent recommendations of the team? Lessons learned: What data have been submitted to the organization’s lessons learned database? Common tasks: • Have all changes been documented (e.g., FMEA, control plan, process flow)? • Does the right team have the right members to proceed to the next step? • Have the measurable(s) been reviewed? • Has the team determined if a service action is required?

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Appendix J: GPS8. Recognize Team and Individual Contributions: Assessment Questions GPS report: • Has the GPS report been updated and published? • Is the GPS paperwork complete and have other members of the organization (who have a need to know), and the customer, been informed of the status of this GPS? • Are the GPS report and its attachments retained in the historical file system? Recognition planning: • Is there a complete list of all team members, current and past? • Were there significant contributions by individual team members? What were they? • Are there opportunities to provide recognition from leader to team, team member to team member, team to leader, and team to champion? • What are some different ways to communicate the recognition message? • Are there any non-team members whose contributions to the GPS process justify inclusion at recognition time? Recognition implementation: • Are all current and past team members being recognized? • Do the results achieved by the team warrant some (e.g., company newsletter)? • Does the recognition satisfy the fit, focus, timely criteria? • GPS8 is intended to be positive. What are the chances that it might backfire and become negative?

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Lessons learned: • What has been learned as individuals and as a team? About each other? About problem-solving? About teamwork? • How did the organization benefit by the completion of this GPS process? • Review each GPS objective. What was done well? • What sort of things should be repeated if conditions bring them together to escape on another GPS? • Are there changes to business practices that should be considered, based on information learned in this GPS? Warm down: • Has all unfinished business been finalized? • Has each person had an opportunity to express appreciation to other team members? • Has each person made a final statement? • Has there been an appropriate celebration? What was it?

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Selected Bibliography

Adams, M. and Spann, M. (May 2000). Continuous process improvement when it counts most. Quality Progress, pp. 74–80. Anderson, M. J. and Kraber, S. L. (July 1999). Eight keys to successful DOE. Quality Digest, pp. 39–44. Backman, G. (January 2000). Brainstorming deluxe. Training & Development, pp. 15–17. Chaudhry, A. M. (June 1999). To be a problem solver, be a classicist. Quality Progress, pp. 53–59. Cotnareanu, T. (December 1999). Old tools — new uses. Quality Progress, pp. 48–52. Draper, E. and Ames, M. (February 2000). Enhanced quality tools. Quality Progress, pp. 41–46. Natarajan, R. N., Matz, R. E., and Kurosaka, K. (February 1999). Applying QFD to internal service system design. Quality Progress, pp. 65–70. Wilkins, J. O. and Plsek, P. E. (May 2000). Putting Taguchi Methods to work to solve design flaws. Quality Progress, pp. 55–60.

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Index

A Addition decimals, 201–204 fractions, 161–174 whole numbers, 123–128 Affinity diagrams, 59 Application criteria, for global problem solving (GPS), 74–75, 78 Area metric–English conversions, 317–320 S.I. measurements, 290–292 Arrow diagrams, 59 Assessment (problem description), 333–340

B Band-Aid™ fixes, 91 Behaviorism, 8

C Capability indices, 18–19 Cause-and-effect, correlation versus, 18 Cause-and-effect (fishbone) diagrams, 15, 57 Cause selection, 15 Celsius temperature scale, 296–297 Centigrade temperature scale, 296–297 Centimeter–inch conversions, 307–310 Chain of causality, 69–71 Champions, 79, 86, 100 Change situations, 75 Charts control, 16–17, 57 flow, 55–56

Gannt, 111 matrix, 59 matrix data analysis, 59 Pareto, 18, 57 process decision program, 59 Check sheets, 17, 55–56 Chronic versus sporadic problems, 34–35 Common tasks, for global problem solving (GPS), 75, 82, 85, 90, 93, 99, 103 Concern analysis report, 67–69 Continual improvement concept, 33–34 Control charts, 16–17, 57 Coping nonproductive, 3 problem-solving, 3 Core capabilities, 56–57 Correlation versus cause/effect, 18 Cover sheet, for global problem solving (GPS) process, 327 Cross-functional participation, 96 Cube numbers, 218–220 Cube roots, 220–226 Cubic meter (volume), 292–293 Customer identification, 48 Customer requirements, 49

D DaimlerChrysler 7-step cycle, 44 Data gathering, 14–25, 83 potential-cause selection, 15 preventing recurrence, 15–22 protocol, 14–15 stimulus passage, 14 team approach, 22–25 verbalization or “thinking aloud,” 14

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Decimal point placement, 240–247 Decimals addition, 201–204, 259–264 applied problems, 212 division, 206–212, 270–271 multiplication, 204–206, 268–270 numbers greater than one, 253–256 numbers less than one, 257–259 ordering, rounding, and changing, 191–199 subtraction, 201–202, 265–268 Deming, W. Edwards, 38 Density function, Weibull, 22 Department improvement teams (DITs), 42 Department of Defense (DoD) 6-step cycle, 44 Design for six sigma (DFSS) approach, 107–115, see also Six sigma (DFSS) approach Design of experiments (DOE), 19–20, 71–72 Diagrams affinity, 59 arrow, 59 cause-and-effect (fishbone), 15, 57 flow, 16 Pareto, 18, 57 process flow, 16 relationship, 59 scatter, 18, 57 tree, 59 Distribution, Weibull, 22 Division decimals, 270–271 fractions, 179–185 whole numbers, 131–140 DMAIC model, 110–112, 113 DOE (design of experiments) process, 19–20, 71–72

E 8D methodology, 4, see also Global Problem Solving (GPS) Emergency response actions (ERAs), 329–330 Employee involvement, 22–25, 41–53, see also Team approach English–metric conversions, 305–324, see also Metric–English conversions Environment (problem-solving), 12–13 Escape point identification, 88–92 Experiments, design of (DOE), 19–20, 71–72

F Facilitator, 80–81 Factoring, 245–247

Fahrenheit temperature scale, 296, 297–298 Failure mode and effect analysis (FMEA), 20–21 Fear, 38 Fishbone (cause-and-effect) diagrams, 15, 57 5W2H approach, 22–23, 25, 107 Floor (root cause) level, 4, see also Root cause Flow charts, 55–56 Flow diagram, 16 FMEA (failure mode and effect analysis), 20–21 Ford Motor Company, 4, 44 global (8D) problem solving process, 61–104, see also Global problem solving (GPS) process Fractions addition and subtraction, 161–174 multiplication and division, 179–185 parts and types of, 145–148 simplest form and common denominators, 149–159 Full time equivalent (FTE) resources, 110–112 Functional analysis/allocation (FA), 108

G Gage studies, 18 Gannt chart, 111 General Motors (GM) 4-step cycle, 43 Gestalt approach, 6–8 Global problem solving (GPS) process, 61–104 application criteria, 74–75 assessment (problem description), 333–340 change and never-been-there situations, 75 common tasks, 75 concern analysis report, 67–69 cover sheet for, 327 do’s and do not’s, 65–67 emergency response actions (ERAs), 329–330 general overview, 61–65 implementation/validation assessment questions, 353–354 individual and team recognition assessment questions, 357–358 interim containment action (ICA) assessment questions, 341–342 permanent corrective actions (PCAs) assessment questions, 351–352 recurrence prevention assessment questions, 355–356 root cause and escape point assessment questions, 343–349 root cause issues, 69–71 steps, 75–104 1: establish team/process flow, 78–81

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Index 2: describe problem, 82–84 3: develop interim containment actions (ICAs), 85–87 4: define and verify root cause and escape point, 88–92 5: choose and verify permanent corrective actions (PCAs), 93–95 6: implement and validate permanent corrective actions (PCAs), 95–98 7: prevent recurrence, 98–102 8: recognize team and individual contributions, 102–104 team/process flow assessment, 331–332 verification, 71–74 Gram/kilogram, 294–296 Graphs, 17–18

H Head-hunting, 38 Histograms, 17, 57

I Illumination step, 5 Implementation, of permanent corrective actions (PCAs), 95–98 Implementation/validation assessment questions, 353–354 Importance, ignorance of, 37 Improvement, continual, 33–34 Inch–centimeter conversions, 307–310 Incubation step, 5 Indices, capability, 18–19 Initial project charter (IPC), 109–110 Insight, 6 Interim containment actions (ICAs), 85–87, 96 assessment questions, 341–342 International system of units (S.I. units), 289–304 cubic meter, 292–293 kelvin, 296–300 kilogram, 294–296 meter, 290–292 technical (derived) units, 302–304 Is/is not analysis, 20

K Kelvin (temperature), 296–300 Kilogram/gram (weight), 294–296

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L Length (linear) measurements English–metric conversions, 307–310, 316–317 metric, 280–281 S.I., 290–292 Liquid, metric measurements, 285–286 Liter–quart/pint conversions, 312–313

M Management in creating problem-solving climate, 92 demonstration of interest by, 92 fear and self-protection in, 38 head-hunting by, 38 quarterly earnings emphasis of, 38 in setting priorities, 92 Management systems, 16 Mass (weight) measurements metric, 281–283 S.I., 294–296 Mathematics, see also Measurements and individual subtopics decimals addition, 201–204, 259–264 applied problems, 212 division, 206–212, 270–271 multiplication, 204–206, 268–270 numbers greater than one, 253–256 numbers less than one, 257–259 ordering, rounding, and changing, 191–199 subtraction, 201–204, 265–268 fractions addition and subtraction, 161–174 multiplication and division, 179–185 parts and types of, 145–148 simplest form and common denominators, 149–159 proportion, 240–241 scientific notation and powers of 10, 245–251 decimal point placement, 240–247 factoring, 245–247 numbers greater than one, 245–247 numbers less than one, 240–251 square and cube numbers, 218–220 square and cube roots, 220–226 square root applications, 231–234 square root calculation, 225–229 whole numbers addition and subtraction, 123–128 multiplication and division, 131–140 value of, 119–121

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Matrix charts, 59 Matrix data analysis charts, 59 Measurement cubic meter (volume), 292–293 definitions and principles, 289–290 English–metric conversions, 305–324, see also Metric–English conversions area units, 317–320 cumulative exercises, 323–324 inches to centimeters, 307–310 length units, 316–317 quarts and pints to liters, 312–316 review test, 305–307 volume units, 320–323 yards to meters, 310–312 international system of (S.I.), 289–304 cubic meter, 292–293 kelvin, 296–300 kilogram, 294–296 meter, 290–292 technical (derived) units, 302–304 kelvin (temperature), 296–300 kilogram/gram (weight), 294–296 meter (length and area), 290–292 metric system, 273–288, see also Metric system process, 49–50 technical (derived) units, 302–304 Meter, see also Metric system cubic, 292–293 as S.I. unit, 290–292 Meter–yard conversions, 310–312 Metric–English conversions, 305–324 area units, 317–320 inches to centimeters, 307–310 length units, 316–317 quarts and pints to liters, 312–316 review test, 305–307 volume units, 320–323 yards to meters, 310–312 Metric system, 273–288 common linear measures, 280–281 common weight (mass) measures, 281–283 conversion of measures within, 275–280 Multiplication decimals, 204–206, 268–270 fractions, 179–185 whole numbers, 131–140

N Never-been-there situations, 75 Nonproductive coping, 3

O Ordering, rounding, and changing, 191–199 Output identification, 47–48 Ownership, lack of, 37

P Pareto charts, 57 Pareto diagram, 18 Permanent corrective actions (PCAs) assessment questions, 351–352 choosing and verifying, 93–95 implementation and validation, 95–98 Plots, 17–18 stem and leaf, 17 Powers of ten, 245–251 Preparation step, 5 Problem (task) definition, 11, 83 Problem description, 82–84, 333–340 Problems chronic versus sporadic, 34–35 six key ingredients for corection of, 39 three typical responses to, 35–37 Problem situation, 11, 12 Problem solving basic model, 5–9 as compared with process improvement, 52 data gathering for, 14–25, see also Data gathering definition of, 10–11 design for six sigma (DFSS) approach, 107–115, see also Six sigma (DFSS) approach generalized stages of, 7–8 global (Ford Motor Company, 8D) approach, 61–104, see also Global problem solving (GPS) process key elements in, 33–39 nine common roadblocks to effective, 37–38 quality tools, 55–59 sample for, 25–26 steps of, 5 strategies for, 3 terminology of, 9–25, see also Data gathering; Terminology theoretical aspects, 5–26, see also Theoretical aspects Problem-solving behavior (operation, strategy), defined, 13–14 Problem-solving process, 13 Problem-solving subject, 11

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Index Problem statement, 44 Process capability, 50 Process decision program charts, 59 Process flow diagram, 16 Process guidelines, 81, 83–84, 87, 91–92, 94–95, 100–102 Process improvement, as compared with problem solving, 52 Process-improvement cycle, 46–52, see also Team approach Process improvement teams (PITs), 42 Process measurements, 49–50 Process output, 49 Product (solution) defined, 13 Product specifications, 49 Proportion, 240–241 Protocol, 14–15 Purpose statements, 84, 90, 93, 95, 99

Q Quality tools, 55–59 application of, 57–58 inventory of, 55 management, 59 seven basic, 55–57 Quarterly earnings emphasis, 38 Quart/pint–liter conversions, 312–313

R Reamur temperature scale, 296 Recognition, 65, 102–104 assessment questions, 357–358 lack of, 37 Recorder, 80 Recurrence prevention, 15–22, 98–102 assessment questions, 355–356 Recycling of process, 51–52 Relationship diagrams, 59 Reliability at 85% confidence, 72–73 Repeatability and reproducibility (R&R), 111–112 Requirement analysis (RA), 108 Response surface methodology (RSM), 114 Risk assessment, 108 Risk handling, 108 Risk monitoring, 108 Risk planning, 107–108 Root cause and escape point, 4, 69–71 assessment questions, 343–349 identification, 88–92

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S Sample, importance of, 25–26 Sample size, 72–73 Scatter diagrams, 18, 57 Scientific notation and powers of ten, 245–251 decimal point placement, 247–240 factoring, 245–247 numbers greater than one, 245–247 numbers less than one, 240–251 Self-protective approaches, 38 Sensory Input, 5–6 Six sigma (DFSS) approach, 107–115 design process, 113–115 overview, 107–109 week 1: structuring: goals, objective, and scope, 109–110 week 2: structuring: product-based estimating, 110–112 week 3: controlling the project, 112–113 Snapshot verification, 72–73 Solution (product), 13 SPC verification, 73–74 Sporadic versus chronic problems, 34–35 Square and cube numbers, 218–220 Square and cube roots, 220–226 Square root applications, 231–234 Square root calculation, 225–229 Stem and leaf plots, 17 Stimulus passage, 14 Strategies nonproductive coping, 3 problem solving, 3 reference to others, 4 Subject (problem solver), 12 Subtraction decimals, 201–204 fractions, 161–174 whole numbers, 123–128 Success factors, 92 Synthesis, 108 System analysis and control (SA), 108 Système Internationale (S.I.) units, 289–304, see also International system of units (S.I.)

T Task (problem) defined, 11 Team(s) local (department-improvement) and crossfunctional (process improvement), 42 required attributes of, 43 TEAM acronym, 42

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Team approach, 22–25, 41–53 general guidelines for, 43 importance of, 41–42 problem-solving models for, 43–46 Team leader, 80 Team members, 80 Team/process flow, 78–81, 331–332 Technical (derived) S.I. units, 302–304 Temperature scales of measurement, 296 S.I. measurements, 296–300 Terminology, 9–14 behavior, operation, or strategy, 13–14 of data gathering, 14–25, see also Data gathering environment, 12–13 problem or task, 11–12 problem situation, 12 problem solving, 10–11 process, 13 product or solution, 13 subject, 12 Theoretical aspects, 5–26 Things gone wrong/right (TGW/TGR), 112 “Thinking aloud” (verbalization), 14 Time, lack of, 37 Tree diagrams, 59

Verification, 5, 71–74, 108 elements of, 71–72 snapshot, 72–73 SPC chart subgroup and sample size, 73–74 Volume metric–English conversions, 320–323 metric measurements, 285–286 S.I. measurements, 292–293

W Weibull analysis, 21–22 Weight (mass) measurements metric, 281–283 S.I., 294–296 5W2H approach, 22–23, 25, 107 Whole numbers addition and substraction, 123–128 multiplication and division, 131–140 value of, 119–121 Work process identification, 49 Work process improvement, 50–51

X Xerox (6-step) system, 43

V Validation, of permanent corrective actions (PCAs), 95–98 Verbalization (“thinking aloud”), 14

Y Yard–meter conversions, 310–312