CLEP College Mathematics

THE BEST TEST PREPARATION FOR THE

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Mel Friedman, M.S.

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Research & Education Association

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CONTENTS Chapter 1 Passing the CLEP College Mathematics Exam About This Book and TESTware® About the Exam How to Use This Book and TESTware® Format and Content of the Exam About Our Course Review Scoring Your Practice Tests Studying for the CLEP Test-Taking Tips The Day of the Exam

1

3

3

6

6

7

7

8

8

10

Chapter 2 Sets Sets Subsets Union and Intersection of Sets Laws of Set Operations Cartesian Product Drill Questions

11

13

15

17

18

20

22

Chapter 3 The Real Number System Properties of Real Numbers Components of Real Numbers Fractions Odd and Even Numbers Factors and Divisibility Numbers Absolute Value Integers Inequalities Drill Questions

27

vi

29

31

31

32

.32

33

34

36

41

Chapter 4 Alge Exponents.... Logarithms .. Equations..... Simultaneow Absolute Vall Inequalities .. Complex NUl Quadratic Eq Advanced AI. Drill Questio

Chapter 5 FUM Elementary I

Translations.

Drill Questio

Chapter 6 Geol Triangles . The Pythagol Quadrilateral Similar Po1YJ Circles .. Formulas for Drill Questi(J

Chapter 7 Prot The Fundam Permutation! Combinatior Probability ..

Probability " Measures of Measures of Data Ana1ys Analyzing P Drill Questic

Contents

Exam 1

............................3

............................3

............................6

............................6

............................7

............................7

............................8

............................8

.......................... 10

.........................11

.......................... 13

.......................... 15

.......................... 17

.......................... 18

, 20

.........................22

........................ 27

.........................29

.........................31

.........................31

.........................32

.........................32

.........................33

.........................34

.........................36

.........................41

Chapter 4 Algebra Topics Exponents Logarithms Equations Simultaneous Linear Equations Absolute Value Equations Inequalities Complex Numbers Quadratic Equations Advanced Algebraic Theorems Drill Questions

45

47

50

52

57

64

65

67

70

76

77

Chapter 5 Functions and Their Graphs Elementary Function : Translations, Reflections, and Symmetry of Functions Drill Questions

83

85

90

93

Chapter 6 Geometry Topics Triangles The Pythagorean Theorem Quadrilaterals Similar Polygons Circles Formulas for Area and Perimeter Drill Questions Chapter 7 Probability and Statistics The Fundamental Counting Principle Permutations Combinations Probability Probability Word Problems Measures of Central Tendency Measures ofVariability Data Analysis Analyzing Patterns in Scatterplots Drill Questions

'"

99

10 1

l 07

110

116

118

124

126

131

133

134

13 5

136

142

144

149

151

156

160

vii

CLEP College Mathematics

Chapter 8 Logic Sentences Statements Basic Principles, Laws, and Theorems Necessary and Sufficient Conditions Deductive Reasoning Truth Tables and Basic Logical Operations Logical Equivalence Sentences, Literals, and Fundamental Conjunctions Drill Questions

165 167 170

Practice Test 1 Answer Key Detailed Explanations ofAnswers

187

Week

203 205

1

Practice Test 2 Answer Key Detailed Explanations of Answers

217 233 235

Answer Sheets

249

Installing REA's TESTware®

262

172 173 l 74 175 178 182 183

The followi CLEP College :N it can be reducel riod into one. & day-to study. l' more time you 51 on the day of the

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viii

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CHAPTER 1

Passing the

CLEP College

Mathematics Exam

Chapter 1

Passing the CLEP College Mathematics Exam ABOUT THIS BOOK AND TESTware@ This book provides you with complete preparation for the CLEP College Mathematics exam. Inside you will find a targeted review of the subject mat­ ter, as well as tips and strategies for test taking. We also give you two practice tests, featuring content and formatting based on the official CLEP College Mathematics exam. Our practice tests contain every type of question that you can expect to encounter on the actual exam. Following each practice test you will find an answer key with detailed explanations designed to help you more completely understand the test material. All CLEP exams are computer-based. The practice exams in this book and software package are included in two formats: in printed format in this book, and in TESTware® format on the enclosed CD. We strongly recom­ mend that you begin your preparation with the TESTware® practice exams. The software provides the added benefits of instant scoring and enforced time conditions.

ABOUT "rHE EXAM

Who takes CLEP exams and what are they used for?

CLEP (College-Level Examination Program) examinations are typically taken by people who have acquired knowledge outside the classroom and wish to bypass certain college courses and earn college credit. The CLEP is designed to reward students for learning-no matter where or how that knowledge was acquired. The CLEP is the most widely accepted credit-by-examination program in the country, with more than 2,900 colleges and universities granting credit for satisfactory scores on CLEP exams. Although most CLEP examinees are adults returning to college, many graduating high school seniors, enrolled college students, military person­ nel, and international students also take the exams to earn college credit or to demonstrate their ability to perform at the college level. There are no pre­ requisites, such as age or educational status, for taking CLEP examinations. 3

CLEP College Mathematics

HOW TO USE THIS BOOK AND TESTware@

The approxin

What do I study first?

10% Sets 10% Logic 20% Real Nll 20% FunctiO! 25% Probabi 15% Additiol

To begin your studies, read over the introduction and the suggestions for test taking. Take Practice Exam 1 on CD-ROM to determine your strengths and weaknesses, and then study the course review material, focusing on your specific problem areas. The course review includes the information you need to know when taking the exam. Make sure to follow up your diagnostic work by taking the remaining practice exam on CD-ROM to become famil­ iar with the format and feel of the CLEP College Mathematics exam. To best utilize your study time, follow our Independent Study Sched­ ule, which you'll find in the front of this book. The schedule is based on a four-week program, but can be condensed to two weeks if necessary by collapsing each two-week period into one.

ABOUT OUR (

The review i all the important reinforce the facts standing of the d with the practice t Mathematics eXaJ

When should I start studying? It is never too early to start studying for the CLEP College Mathemat­ ics exam. The earlier you begin, the more time you will have to sharpen your skills. Do not procrastinate! Cramming is not an effective way to study, since it does not allow you the time needed to learn the test material. The sooner you learn the format of the exam, the more time you will have to familiarize yourself with it.

FORMAT AND CONTENT OF THE EXAM The CLEP College Mathematics exam covers the material one would find in a college-level class for nonrnathematics majors and majors in fields not requiring knowledge of advanced mathematics. The exam places little emphasis on arithmetic and calculators are not allowed. A nongraphing cal­ culator is provided to the test taker during the examination as part of the testing software. The exam consists of 60 questions to be answered in 90 minutes.

l

6

SCORING YOI

How do I scon

The CLEP( To score your pm your total raw Sc( version table on only an estimate time, and in no c test performance level of perfofDll

When will I re4

The test ad! you immediatel) lish Compositio YQu ask to have college or other Since your scor scripts from Edl

Chapter 1: Passing the CLEP College Mathematics Exam

The approximate breakdown of topics is as follows:

.the suggestions for nine your strengths lterial, focusing on he information you .up your diagnostic ,f to become famil­ natics exam. ldent Study Sched­ b.edule is based on :ks if necessary by

:olJege Mathemat­ ill have to sharpen =ctive way to study, : test material. The e you will have to

aaterial one would

nd majors in fields

exam places little l nongraphing cal­ tion as part of the

l

10% Sets

10% Logic

20% Real Number System

20% Functions and Their Graphs

25% Probability and Statistics

15% Additional Topics from Algebra and Geometry

ABOUT OUR COURSE REVIEW The review in this book provides you with a complete background of all the important mathematical formulas relevant to the exam. It will help reinforce the facts you have already learned while better shaping your under­ standing of the discipline as a whole. By using the review in conjunction with the practice tests, you should be well prepared to' take the CLEP College Mathematics exam.

SCORING YOUR PRACTICE TESTS How do I score my practice tests? The CLEP College Mathematics exam is scored on a scale of 20 to 80. To score your practice tests, count up the number of correct answers. This is your total raw score. Convert your raw score to a scaled score using the con­ version table on the following page. (Note: The conversion table provides only an estimate of your scaled score. Scaled scores can and do vary over time, and in no case should a sample test be taken as a precise predictor of test performance. Nonetheless, our scoring table allows you to judge your level of performance within a reasonable scoring range.)

When will I receive my score report? 90 minutes.

The test administrator will print out a full Candidate Score Report for you immediately upon your completion of the exam (except for CLEP Eng­ lish Composition with Essay). Your scores are reported only to you, unless you ask to have them sent elsewhere. If you want your scores reported to a college or other institution, you must say so when you take the examination. Since your scores are kept on file for 20 years, you can also request tran­ scripts from Educational Testing Service at a later date. 7

CLEP College Mathematics

PRACTICE-TEST RAW SCORE CONVERSIONTABLE*

29 27

47

A A A

25

45

46

1

44

I

66

I

44

I

64

I

I

62

42 40

I

60

I I

I

A A A

19 18 17 16 15

B

13

A

B B B

43

23 21

46

49

A A A

I

9

I

D

38

D

37 36 35

D

29

28 38 36

I

58

I

56

I

B

I

B

I

I

C 0 C C C 0 C 0 C

39

33 32 31

11

I

41

7 6 5

27 26 25

3 2 1

23 22 21

I

D

D

I

•

D

52

I

B

50

I

B C

20

Although Y tests such as tht to acquaint you your test-taking tomed to the CI tests as well.

F F

F F F F

'This table is provided for scoring REA practice tests only. The American Council on Education recommends that colleges use a single across-the-board credit-granting score of 50 for all CLEP computer-based exams. Nonetheless, on account of the different skills being measured and the unique content requirements of each test, the actual number of correct answers needed to reach 50 will vary. A 50 is calibrated to equate with performance that would warrant the grade C in the cor­ responding introductory college course.

8

As you COI1 review the explaJ do not review to time by reviewin until you are con

TEST-TAKIN'

0 F F F

B B

It is very in ing that works lx of hours every m going to sleep. 0 line, or even whi your study time' wisely. Work out

Whenyout much like the ac1 sit down at a qui Start off by settiJ be sure to reset tl a new section.

0

D

STUDYING F(

Read all 011

the correct respc Read through ~ jumping to cone

I

Use the p"

and eliminate ~ two answer cho correct, since th

Chapter 1: Passing the CLEP College Mathematics Exam

SIONTABLE*

~7

C C C

6

c

-5

C

19

18

14 3

C

12

C

·1

C

0

C

9

D

8

D

;7

D

6

D

5

D

lit

3 2 1 0 9 IS

D

When you take the practice tests, try to make your testing conditions as much like the actual test as possible. Tum your television and radio off, and sit down at a quiet table free from distraction. Make sure to time yourself. Start off by setting a timer for the time that is allotted for each section, and be sure to reset the timer for the appropriate amount of time when you start a new section. As you complete each practice test, score your test and thoroughly review the explanations to the questions you answered incorrectly; however, do not review too much at one time. Concentrate on one problem area at a time by reviewing the question and explanation, and by studying our review until you are confident that you completely understand the material.

D D

TEST-TAKING TIPS

F

Although you may not be familiar with computer-based standardized tests such as the CLEP College Mathematics exam, there are many ways to acquaint yourself with this type of examination and to help alleviate your test-taking anxieties. Listed below are ways to help you become accus­ tomed to the CLEP, some of which may be applied to other standardized tests as well.

F

7

F F

5 4

F

3

F

F

2

Just because you think you have found the correct response, do not automatically assume that it is the best answer. Read through each choice to be sure that you are not making a mistake by jumping to conclusions. Read all of the possible answers.

F

tJ

Council on Education ;core of 50 for all CLEP eing measured and the IterS needed to reach 50 the grade C in the cor-

1

It is very important for you to choose the time and place for study­ ing that works best for you. Some students may set aside a certain number of hours every morning, while others may choose to study at night before going to sleep. Other students may study during the day, while waiting on a line, or even while eating lunch. Only you can determine when and where your study time will be most effective. But be consistent and use your time wisely. Work out a study routine and stick to it!

D

6

1

STUDYING FOR THE CLEP

Go through each answer to a question and eliminate as many ofthe answer choices as possible. By eliminating just two answer choices, you give yourself a better chance of getting the item correct, since there will only be three choices left from which to make your Use the process of elimination.

9

CLEP College Mathematics

guess. Remember, your score is based only on the number of questions you answer correctly. Work quickly and steadily. You will have only 90 minutes to work on 60 questions, so work quickly and steadily to avoid focusing on anyone ques­ tion too long. Taking the practice tests in this book will help you learn to budget your time. Acquaint yourself with the computer screen. Familiarize yourself with the CLEP computer screen beforehand by logging on to the College Board website. Waiting until test day to see what it looks like in the pretest tutorial risks injecting needless anxiety into your testing experience. Also, familiarizing yourself with the directions and format of the exam will save you valuable time on the day of the actual test. Be sure that your answer registers before you go to the next Item. Look

at the screen to see that your mouse-click causes the pointer to darken the proper oval. This takes less effort than darkening an oval on paper, but don't lull yourself into taking less care!

THE DAY OF THE EXAM On the day of the test, you should wake up early (hopefully after a decent night's rest) and have a good breakfast. Make sure to dress comfort­ ably, so that you are not distracted by being too hot or too cold while taking the test. Also plan t<;> arrive at the test center early. This will allow you to collect your thoughts and relax before the test, and will also spare you the anxiety that comes with being late. As an added incentive to make sure you arrive early, keep in mind that no one will be allowed into the test session after the test has begun. Before you leave for the test center, make sure that you have your admission form and another form of identification, which must contain a recent photograph, your name, and signature (i.e., driver's license, student identification card, or current alien registration card). You will not be admit­ ted to the test center if you do not have proper identification. If you would like, you may wear a watch to the test center. However, you may not wear one that makes noise, because it may disturb the other test-takers. No dictionaries, textbooks, notebooks, briefcases, or packages will be permitted and drinking, smoking, and eating are prohibited. Good luck on the CLEP College Mathematics exam!

10

er of questions you Ilutes to work on 60 g on anyone ques­ 1 help you learn to mriliarize yourself ~ on to the College s like in the pretest g experience. Also, the exam will save the next Item. Look ~inter to darken the on paper, but don't

, (hopefully after a 'e to dress comfort­ o cold while taking s will allow you to also spare you the 'e to make sure you nto the test session

hat you have your ich must contain a :r's license, student 1 will not be admit­ tion.

st center. However,

y disturb the other cases, or packages prohibited.

if

~--­

CHAPTER 2

Sets

Chapter 2

Sets You will see the topics of set theory in most ofthe chapters ofthis book; in fact, you use set theory in many of your everyday activities. But since it is not labeled as "set theory" in most cases, you are unaware that set theory is the basis for most of your mathematical and logical thought. This chapter introduces the set theory vocabulary you should know as well as such topics as Venn diagrams for the union and intersection of sets (used in logic), laws of set operations (similar to those for operations on the real number system), and Cartesian products (used in graphs of linear functions). So let's set the stage for sets.

SETS A set is defined as a collection of items, Each individual item belOnging to a set is called an element or member of that set. sets are usually represented by capital letters, and ele­ ments by lowercase letters. If an item k belongs to a set A. .we write k E A (" k is an element of A"). If k is not in A. we write k fl A (" k is not an element of A").

The order of the elements in a set does not matter: {I, 2, 3} = {3, 2, I} = {I, 3, 2}, etc. A set can be described in two ways: 1. element by element. 2. a rule characterizing the elements. For example, given the set A of the whole numbers starting with 1 and ending with 9, we can describe it either asA = {I, 2, 3,4,5,6,7,8, 9} or as A = {whole numbers greater than 0 and less than lO}. In both methods, the description is enclosed in brackets. A kind of shorthand is often used for

13

CLEP College Mathematics

the second method of set description, so instead of writing out a complete sentence between the brackets, we can write instead A = {k I 0

SUBSETS

Given every I

< k < 10, k a whole number}

This is read as "the set of all elements k such that k is greater than 0 and less than 10, where k is a whole number." A set not containing any members is called the empty or null set. It is written either as cP or { }. A set is finite if the number of its elements can be counted.

Example:

{2, 3, 4, 5} is finite since it has four elements.

Example:

{3, 6, 9, 12, ..., 300} is finite since it has 100 elements.

Note: The empty set, denoted by cP, is finite since we can count the number of elements it has, namely zero. Any set that is not finite is called infinite.

Example:

{ I, 2, 3, 4, ... }

A is aprope writeA ~ B if A Two sets an if A = B, thenA

Example:

= {I, B = {I,

LetA

C = {I, 1. A equals C, subsets.

2. B ~A,B ~ ( B is a proper:

Two sets an

Example:

0= {3, 7, since each one b

Example:

{..., -7, -6, -5, -4}

Example:

F= {l,3,~ lent sets, since e Example: ~

{x I x is a real number between 4 and 5}

14

Note: Iftw4

Chapter 2: Sets

ing out a complete

SUBSETS A is said to be a subset of B if also a member of set B.

r}

s greater than 0 and

IPtY or null set. It is

A is aproper subset of B if B contains at least one element not inA. We writeA ~ B if A is a subset of B, andA C B if A is a proper subset of B.

:ounted.

Two sets are equal if they have exactly the same elements; in addition, if A = B, then A ~ B andB ~ A.

:nts.

e we can count the

Example:

LetA = {I, 2, 3,4, 5}

B = {I, 2}

C

= {l, 4, 2, 3, 5}

1. A equals C, and A and C are subsets of each other, but not proper subsets. ~

~

C, B C A, Bee (B is a subset of both A and C. In particular, B is a proper subset of A and C).

2. B

A, B

Two sets are equivalent if they have the same number of elements.

Example:

0= {3, 7, 9, 12} and E = {4, 7, 12, 19}. 0 and E are equivalent sets, since each one has four elements.

Example:

F= {I,3,5, 7, ...,99}andG= {2,4,6,8, ..., 100} Fand Gareequiva­ lent sets, since each one has 50 elements.

Note: If two sets are equal, they are automatically equivalent.

15

CLEP College Mathematics

A universal set U is a set from which other sets draw their members. If A is a subset of U, then the complement of A, denoted A /, is the set of all elements in the universal set that are not elements of A.

Example:

IfU= {l, 2, 3,4, 5, 6, ... } andA

=

{l, 2, 3}, then A'

= {4, 5, 6, ... }.

Figure I illustrates this concept through the use of a simple Venn diagram.

UNION AND I

The" of all t

Figure 3 is: given operation. Figure 1

-,

A Venn diagram is a visual way to show the relationships among or between sets that share something in common. Usually, the Venn diagram consists of two or more overlapping circles, with each circle representing a set of elements, or members. If two circles overlap, the members in the overlap belong to both sets; if three circles overlap, the members in the over­ lap belong to all three sets. Although Venn diagrams can be formed for any number of sets, you will probably encounter only two or three sets (circles) when working with Venn diagrams. As shown in Figure I, the circles are usually drawn inside a rectangle called the universal set, which is the set of all possible members in the universe being described. Venn diagrams are organizers. They are used to organize similarities (overlaps) and differences (non-overlaps of circles) visually, and they can pertain to any subject. For example, if the universe is all animals, Circle A may represent all animals that live in the water, and Circle B may represent all mammals. Then whales would be in the intersection of Circles A and B, but lobsters would be only in Circle A, humans would be only in Circle B, and scorpions would be in the part of the universe that was outside of Circles A and B. These relationships are shown in Figure 2. 16

The II

theS4

Chapter 2: Sets

raw their members. 1A', is the set of all

A' = {4, 5, 6, ... }.

U Animals

: of a simple Venn

tionships among or ~ the Venn diagram circle representing the members in the lembers in the over­ 1 be formed for any r three sets (circles) 'e I, the circles are , which is the set of

rganize similarities ually, and they can II animals, Circle A :le B may represent of Circles A and B, ~ only in Circle B, IS outside of Circles

Scorpions

Figure 2

UNION AND INTERSECTION OF SETS The union of two sets A and S, denoted AU' S, of all elements that are either in A or S or both.

Figure 3 is a Venn diagram for A U B. The shaded area represents the given operation.

U

AUB Rgure 3

The Int&rsectlon of two sets A and B, denoted the set of all elements that belong to both A and

17

CLEP College Mathematics

Figure 4 is a Venn diagram for A given operation.

n B. The shaded area represents the

nB=

=V

4a. A U A'

IfA = {I, 2, 3,4, 5} andB = {2, 3,4, 5, 6}, thenA U B = {I, 2, 3,4, 5, 6} andA n B = {2, 3,4, 5}. IfA

Complement La

4b. A

n A' =

cjJ

5a. cjJ' = V

cjJ,A and B are disjoint.

5b. V' = cjJ

Commutative L 6a. A U B

6b. An B

=B L =B (

u Ana Rgure 4

Associative La,

7a. (A U B) U C

7b. (A

LAWS OF SET OPERATIONS If V is the universal set, andA, B, C are any subsets of V, then the fol­ lowing hold for union, intersection, and complement:

n B) n C

Figures 5 ar ure 5, the intersec result with C. In the intersection ( hatched region) ~

Identity Laws

lao A U cjJ = A lb. A

n cjJ =

cjJ

2a. AU V= V

2b. An V= A Idempotent Laws

3a. A UA = A

3b. A nA = A

Distributive l..a

8a. A U (B 8b. A

18

n (B

n C. U

C.

Chapter 2: Sets

area represents the

Complement Laws

=U

4a. A UA'

B= {1,2,3,4,5,6}

s of U, then the fol-

n A' =

4b. A

cJ>

5a. cJ>' = U 5b. U' = cJ>

Commutative Laws

6a. AU B = B UA

nB = B nA

6b. A

Associative Laws 7a. (A U B) U C

7b. (A

= A U (B U C)

n B) n C = A n (B n C)

Figures 5 and 6 illustrate the associative law for intersections. In Fig­ ure 5, the intersection ofA and B is done first, and then the intersection of this result with C. In Figure 6, the intersection of Band C is done first, and then the intersection of this result with A. In both cases, the end result (double hatched region) is the same.

(A n B) n C Figure 5

An (B n C) Figure 6

Distributive Laws 8a. A U (B

8b. A

n C) =

n (B U C) =

(A U B) (A

n (A

U C)

n B) U (A n C)

19

CLEP College Mathematics

Example:

De Morgan's Laws 9a. (A U B)' 9b. (A

M={1,3,5]

= A' n B'

n B)' = A'

The Cartesian pro< We can easily see 1 elements of M X J

U B'

"The differeneeof two sets, A and B, written as A-B. is the set ofaHelernents that belong to A but do not belong to B.

Example:

J = {l0, 12, 14, 16}, K = {9, 10, 11, 12, 13}

J - K = {14, 16} . Note that K - J = {9, 11, 13}.

In general, J - K

*" K -

J.

Example:

W = {a, b, c The Cartesian pro

Example:

T

=

{a, b, c}, V = {a, b, c, d, e,f}.

T- V=

cP, whereas V -

(c, a), (c, g), (c,

T = {d, e,f}.

h:

The elements for 1

Note, in this example, that T is a proper subset of V. In general, when­

ever set A is a proper subset of set B, A - B = cP.

If set P is any set, then P any sets, P - Q = P n Q'.

cP = P and cP -

P

= cP. Also if P and Q are

CARTESIAN PRODUCT Given two sets M and N. the Cartesian product, clenoted M X N, is the set of all ordered pairs of elements in which the first component is a member of M and the second com­ ponent is a member of N.

Often, the elements of the Cartesian product can be found by making a table with the elements of the first set as row headings and the elements of the second set as column headings, and the elements of the table the pairs form from these elements. 20

In the first eJl MX Nhas3 X 2 elements and Y h if the first set has. product will have

Chapter 2: Sets

Example: M= {1,3,5},N= {2,8} The Cartesian product M X N = {(1, 2), (1, 8), (3, 2), (3, 8), (5, 2), (5, 8)} We can easily see that these are all of the elements of M X N and the only elements of M X N by looking at Table I: A - B, is klt belong

Table 1-M N

2

8

1,2

1,8

3

3,2

3,8

5

5,2 5,8

Example: W= {a,b,c}, Y= {a,g,h}

The Cartesian product W X Y = {(a, a), (a, g), (a, h), (b, a), (b, g), (b, h), (c, a), (c, g), (c, h)}

The elements for this Cartesian product are shown in Table 2.

v. In general, when­ Table 2-W Y

. Also if P and Q are

t, denoted s in which condcom­

e found by making a and the elements of )f the table the pairs

a

g

h

a

a

g

a,.h

b

b, a

b, g

b, h

c

c,a

c, g

c,

In the first example above, since Mhas 3 elements and Nhas 2 elements, M X N has 3 X 2 = 6 elements. In the second example above, since W has 3 elements and Y has 3 elements, W X Y has 3 X 3 = 9 elements. In general, ifthe first set has x elements and the second set has y elements, the Cartesian product will have xy elements.

21

(B) {all negative numbers} (C) {all negative even integers} (D) {all negative odd integers}

6. If the universa {l, 5, 7}, and (A) 15

(B) 13 (C) 7 (D) 2

22

Chapter 2: Sets

4. Consider the Venn diagram shown below. Which one of the following correctly describes the shading? dements. How many

which one of the fol­

(A)E

nF

(B) D

nEnF

(C) D

nE

(D)D

nF

; 7.

s 17.

;10.

= {all negative num­

5. Which one of the following is an example of disjoint sets? (A) {O, 1,2,3} and {3, 2, 1, O} (B) {O, 2, 4, 6} and {2, 4,6, 8} (C) {O, 3, 6, 9} and {9, 16,25, 36}

~

(D) {O, 4,8, 12} and {6, 10, 14, 18} ~ers}

6. If the universal set U = {x I x is a positive odd integer less than 30}, R = {I, 5, 7}, and S = {l, 3, 7, 11, 13}, how many elements are in (R n S)'? (A) 15 (B) 13 (C) 7 (D) 2

23

CLEP College Mathematics

7. If PC Q, which one ofthe following conclusions must be true?

10. Consider the Vern

(A) P is either equal to Q or P is a proper subset of Q. (B) P is a proper subset of Q. (C) Q is a proper subset of P. (D) P is either equal to Q or P is the empty set.

8. Given any two sets F and G, which one of the following is not necessar­ ily true? (A)FU G= G UF

(B) F

n G = G nF

(C) F- G = G - F (D) F

n F' = ¢

9. Which one of the following is a finite set? (A) {x I x is an irrational number less than O} (B) {x I x is a positive rational number less than 20} (C) {x I x is a negative fraction greater than - 8} (D) {x I x is a natural number less than 50}

24

If the given num1 how many elemeJ (A) 10

(B) 9 (C) 7 (D) 6

Chapter 2: Sets

must be true?

10. Consider the Venn diagram shown below.

fQ.

JWing is not necessar-

If the given numbers represent the number of elements in each region, how many elements are in T - S? (A) 10

(B) 9 (C) 7 (D) 6

)}

25

9l

N

(s)

'or

(0)

'~

(0)

(s) '9

(:) 'z

(y) 'L

(:) 'f

(:) '8

(y) 'P

(0) '6

'r

sOfJewaL/Jew af3allo~ d37~

Chapter 3

The Real Number System Real numbers provide the basis for most precalculus mathematics topics. Real numbers are all

(see Figure 1).

In fact, a nice way to visualize real numbers is that they can be put in a one-to-one correspondence with the set of all points on a line. Real numbers include positives, negatives, square roots, 1T (Pi), and just about any number you have ever encountered.

-6 -5 -4

-3 -2 -1

0 1 Figure 1

2

3

4

5

6

PROPER"rlES OF REAL NUMBERS Similar to the Laws of Set Operations presented in the last chapter, real numbers have several properties that you should know. You have used some ofthese properties ever since you could count. You intuitively knew that 3 + 2 gives the same result as 2 + 3, or that if you add 0 to a number it remains unchanged. These properties deal with addition and multiplication. They do not work for subtraction and division. For example, you also intuitively know that 3 - 2 is not the same as 2 - 3. Perhaps you are not familiar with the names of these properties. The following list provides the names for these properties. Learn them-you will encounter these property names in the next chapter and on mathematics tests. The examples use the numbers 2,3, and 4, but the rules apply to any real numbers. Commutative Property

The numbers commute, or move: Addition

2+3=3+2

Multiplication

2x3=3x2 29

CLEP College Mathematics

Associative Property

COMPONENTS OF

The numbers can be grouped, or associated, in any order:

+ (3 + 4) = (2 + 3) + 4

Addition

2

Multiplication

2 X (3 X 4) = (2 X 3) X 4

Later in this chapter, you will see that operations in parentheses are always done first, but the associative property says that you can move the parentheses and it won't make a difference.

The set of all real] N

W= {O,I,2,3, ... },tb

1= { ... , -3, -2, -l,l

Q = {~I a, bEl and 1 S

Distributive Property

= {x I x has a decimal

block}, the set of alII..,

The first number gets distributed to the ones in parentheses. 2 X (3

= {l, 2, 3, ... }, the s

+ 4) = (2

X 3)

+ (2

X 4)

The following properties have to do with the special numbers 0 and 1.

It is obvious that j does not hold betwee for elements of Q are repeating block.

Examples of ratio

This means that'

Identity Property

Adding 0 or multiplying by 1 doesn't change the original value. Addition Multiplication

+0=3 3X I =3

3

The inverse of addition is subtraction and the inverse of multiplication is division.

Multiplicative Inverse

All real numbers means that every real I

FRACTIONS

Inverse Property

Additive Inverse

Examples of irrat

3

+ (-3) = 0

3X

1

'3

=

1

All rational numl numerator (on the top:

..

numerator ~

Note that the multiplicative inverse doesn't work for 0 because division by 0 is not defined.

~

numerator 1 'i

30

Chapter 3: The Real Number System

COMPONENTS OF REAL NUMBERS my order:

The set of all real numbers (designated as R) has various components:

r4

N = {1, 2, 3, ... }, the set of all natural numbers

(4

W

in parentheses are hat you can move the

IllS

= {O, 1, 2, 3, ... }, the set of all whole numbers

1= {... , -3, -2, -1,0,1,2,3, ... },thesetofallintegers

Q = {~I a, bEl and b t=-

o} }, the set of all rational numbers

S = {x Ix has a decimal that is nonterminating and does not have a repeating block}, the set of all Irrational numbers. mentheses. ~)

cia! numbers

°and 1.

It is obvious that N C; W, W C; I, and I C; Q, but a similar relationship does not hold between Q and S. More specifically, the decimal names for elements of Q are either (l) terminating or (2) nonterminating with a repeating block.

. . 1 Examples of ratIOnal numbers mclude '2

1

= .5 and "3 = .333....

This means that Q and S have no common elements.

:original value.

Examples of irrational numbers include .101001000... , 7T, and.J2.

All real numbers are normally represented by Rand R = Q U S. This

means that every real number is either rational or irrational.

FRACTIONS

'f!1"se of multiplication

All rational numbers can be displayed as fractions, which consist of a numerator (on the top) and a denominator (on the bottom).

°

for because division

31

CLEP College Mathematics

Improper fractions are also called mixed numbers because they can be written as a whole number with a fractional part. Examples of improper

2 4

19

List the factors

~

.

fractions are -, -, and - . The first ofthese IS actually a whole number (2); 17 I 2 I 3 the others are equivalent to the mixed numbers I 3" and 1 , respectively.

17

ODD AND EVEN NUMBERS When dealing with odd and even numbers keep in mind the following:

The factors of21 56,84, and 112. Note

ABSOLUTE VAU

Adding:

+ even = even odd + odd = even even + odd = odd

even

The absolute va

A if.

Multiplying: even X even even X odd odd X odd

IAI = { -A if. = even

= even = odd

Examples:

151 = 5 FACTORS AND DIVISIBILITY NUMBERS

1-81 =

-

(-8)

Absolute value

3.

I-AI=IA IA \:> 0, eq IAI Ii = IAI \BI,j

4.

IABI

1. 2.

The factors of 20 are I, 2, 4, 5, 10, and 20. AhynUmber that~an

rernaind~,.is called a

Examples of multiples of 20 are 20, 40, 60, 80, etc.

32

5.

= IA 2 \A f =A

Chapter 3: The Real Number System

ers because they can List the factors and multiples of 28.

txamples of improper f

a whole number (2); 2

.

[117' respectIvely. The factors of28 are 1,2,4, 7, 14, and 28. Some multiples of28 are 28, 56,84, and 112. Note that the list of multiples is endless.

o mind the following:

unberwith

ABSOLUTE VALUE oIute va lines aro egardless of sign.

The absolute value of a real number A is defined as follows:

IA I =

{

A if A > 0 -A if A < 0

Examples:

151 = 5

I -81 = ­

(-8)

=8

Absolute values follow the given rules: 1. 2. 3.

I-A I = jA I I A I > 0, equality holding only if A

= 0

IBI' I ~I=~B*O B

4.

IAB I = / A I X IB I

5.

I A /2 = A2

te.

33

CLEP College Mathematics

Consecutive Integen

Calculate the value of each of the following expressions: 1.

112 - 51 + 6 -141

2.

I-51

X

The set of integer. {n,lI

1-121

141 + - ­

4

Prime Numbers

Before solving this problem, one must remember the order of opera­ tions: parenthesis, multiplication and division, addition and subtraction. 1.

II -

2.

12 (5 X 4) + - = 20 + 3 = 23

31 + 6 - 141 = 13 + 6 - 141 =

19 -

141 =

I-

The set of posifu and themselves:

51 = 5

Composite Numben

4

The set of intege

INTEGERS There are various subsets of I, the set of all integers:

Classify each of possible.

Negative Integers

Example: real, ~

The set of integers starting with - 1 and decreasing:

1.0

{-I, -2, -3, ... }.

2.9 Even Integers

The set of integers divisible by 2:

{... , -4, -2,0,2,4,6, ... }.

1. 0 is a real numbe1

2. 9 is a real numt number. Odd Integers

3.

,J6 is a real num

4.

"2 IS a real numbc:

5.

"3 is a real numb

The set of integers not divisible by 2:

{... , -3, -1, 1,3,5,7, ... }.

34

1. 2

Chapter 3: The Real Number System

Consecutive Integers ~IOns:

The set of integers that differ by 1: {n, n

+

1, n

+ 2, ... } (n

= an integer).

Prime Numbers r the order of opera­ 1 and

The set of positive integers greater than 1 that are divisible only by 1

and themselves:

subtraction.

{2, 3, 5, 7, 11, ... }.

. 141 =

I- 5 I =

5

Composite Numbers The set of integers, other than 0 and ± 1, that are not prime. :IS:

tg:

Classify each of the following numbers into as many different sets as possible. Example: real, integer ... 2

"3

1.0

5.

2.9

6.1.5

7.11

1. 0 is a real number, an integer, a whole number, and a rational number. 2. 9 is a real number, an odd number, a natural number, and a rational number. 3. ,J6 is a real number, and an irrational number. 1 4. 2 is a real number, and a rational number. 2 5. "3 is a real number, and a rational number.

35

CLEP College Mathematics

Example:

6. 1.5 is a real number, a decimal, and a rational number. 7. 11 is a prime number, an odd number, a real number, a natural number, and a rational number.

x>2

~ -4

INEQUALITIES

-

2 is not a solution

If x and y are real numbers, then one and only one of the following statements is true.

x > y, x = y, or x < y.

The endpoint is a to or (2) greater than ( point is indicated by a

This is the order property of real numbers. If a, b, and c are real numbers, the following statements are true: If a

< band b < c, then a < c.

If a

> b and b > c, then a > c.

5>x2:2

~ -1

This is the transitive property of Inequalities. If a, b, and c are real numbers and a > b, then a + c a - c> b ­ c. This is the addition property of Inequality.

Example:

> b + c and

An inequality is a state quantity or expression greater than or eq not equal to th

(*>

In this case, 2 ~ should be representee:

Example:

x < 2 or x> 5

• Example:

5 > 4. This expression means that the value of 5 is greater than the value of 4. The graph of an Inequality in one variable is represented by either a ray or a line segment on the real number line. The endpoint is not a solution if the variable is strictly less than or greater than a particular value. In those cases, the endpoint is indicated by an open circle.

36

In this case, nei shown at x = 2 and:

Chapter 3: The Real Number System

ber.

ber, a natural number,

Example:

x>2

-4

. one of the following

-3

-2

-1

0

234

2 is not a solution and should be represented as shown. The endpoint is a solution ifthe variable is either (1) less than or equal to or (2) greater than or equal to a particular value. In those cases, the end­ point is indicated by a closed circle.

tements are true:

Example: 5>x~2

-1

I

a + c > b + c and

Ity.

r iIIIue of one

• than «), III to (:S), or

~

234

0

5

678

In this case, 2 is a solution and 5 is not a solution, and the solution should be represented as shown.

Example:

x<2orx>5

. . ......------~$.....-----$I---- .......~

o 5 is greater than the

2

5

In this case, neither 2 nor 5 is a solution. Thus, an open circle must be shown at x = 2 and at x = 5.

:sented by either a ray

s strictly less than or ~int is indicated by

37

CLEP College Mathematics

Example:

Example:

x

-<

2 and x

..

>-

{xlx<4}.The~

5

.

~

In this case, there is no solution. It is impossible for a number to be both no greater than 2 and no less than 5.

Example:

Example:

x

>-

2 or x

-<

{x

5

•

12 < x < 6}.

.

•

In this case, the solution is all real numbers. Any number must belong to at least one of these inequalities. Some numbers, such as 3, belong to both inequalities. If you graph these inequalities separately, you will notice two rays going in opposite directions and which overlap between 2 and 5, inclusive.

Example:

I

{x x is any real

Intervals on the number line represent sets of points that satisfy the conditions of an inequality.

.

A closed intern

Example:

{xl-5-<x~:

Example:

{x Ix > - 3}, read as "the set of values x such that x would appear as:

•

38

+

-3

I a

> - 3." The graph

II

...

...

-5

A half~pen inle

Chapter 3: The Real Number System

Example: {x I x

< 4}. The graph would appear as:

. . . . .--------~---+I--------1.~

o

Ie for a number to be

4

Example: {x

12 < x < 6}. The graph would appear as:

. . ... --------1H---+--------·~

o

2

6

r number must belong such as 3, belong to rately, you will notice rIap between 2 and 5,

Example: {x I x is any real number}. The graph would appear as:

)Dints that satisfy the

It x

o

A closed interval includes two endpoints.

Example: {x

> - 3." The graph

I - 5 <: x :s; 2}. The graph would appear as:

..

+----HI--------1·~

-5

0

2

A half-open interval includes one endpoint.

39

CLEP College Mathematics

Example:

{x I x

DRILL QUESTION >-

3}. The graph would appear as:

+

I

III

o

•

3

(A) ~ 2 (B) 3c ­ 1

Example:

{x Ix

1. If c is any odd ill integer?

<:

6}. The graph would appear as:

..

I

o

+------­

(C) c2

+2

(D) 2c

+1

6

2. What is the value Example:

(A) 16

{xl-4<x<: -I}.

~

+-1

$ -4

-1

0

•

(B) 5 (C) -5 (D) -16

Example:

{x

3. Which one of til between 7 and 8~

1-2 <: x < I}. ~

+

-2

H1

0

•

(A) ~400 (B)

7.2

(C)~ (D) 25 3

40

Chapter 3: The Real Number System

DRILL QUESTIONS

1. If c is any odd integer, which one of the following must be an even integer? (A) ~ 2 (B) 3c - 1 (C) c2

+2

(D) 2c

+1

2. What is the value of I 7 - 13

1 -

1 -

2 - 9 I?

(A) 16

(B) 5 (C) -5 (D) -16

3. Which one of the following is an irrational number whose value lies between 7 and 8? (A) J400 (B)

7.2

(C)~

(D) 25

3

41

CLEP College Mathematics

4. Which one of the following numbers is a prime number between 90 and 100?

(A) 91

7. The number 0.4 is 4 (A) ­ 5

(B) 94

(B) ~ 7

(C) 95

(C)

(D) 97

(D)~

~

9 11

5. Which one of the following inequalities describes a graph on the number line that includes all real numbers?

(A) x

>-

1 and x

<

4

8. Not including the (A) 7

(B) x >- 1 or x < 4

(B) 6

(C) x

(C) 5

<

1 or x

>-

4

(D) x < 1 and x >- 4

6. If 15 is a factor of x, which one of the following is true for any x? (A) 30 must be a factor of x. (B) Each of 3 and 5 must be factors of x. (C) x must be a prime number greater than 15. (D) The only prime factors of x are 3 and 5.

(D) 4

9. n2 is a proper frcu: be true? (A) n is a proper (B) n is an imprc (C) n is a positiv (D) n is a negati,

10. Which one ofthl

(A) Ix

I> 0 (B) Ix 1= 0 (C) Ix I = -x (D) I x 1< 0

42

Chapter 3: The Real Number System

m.OOr between 90 and

7. The number 0.4 is equivalent to which fraction?

(A)

~

(B)

~

(C)

~

5

7

9

(D)~ 11

l graph

on the number

.true for any x?

8. Not including the number 16, how many factors of 16 are there?

(A) 7 (B) 6

(C) 5 (D) 4

9. n 2 is a proper fraction in reduced form. Which one ofthe following must be true?

(A) n is a proper fraction. (B) n is an improper fraction. (C) n is a positive number. (D) n is a negative number.

10. Which one ofthe following has no solution for x?

(A) Ix

I> 0 (B) Ix 1= 0 (C) Ix I = -x (D) Ix 1< 0 43

(0)

'01

(a)

'~

(y) '£

'8

'v

(y) '6 (0)

(0)

(a)

'9

(0)

(0) 'L (a)

'z 'I

Chapter 4

Algebra Topics This chapter discusses some interesting algebra topics that you should know, such as exponents, logarithms, complex numbers, and inequalities. In addition, it presents applications from algebra so you can see the usefulness of this subject.

EXPONENTS

In the expression 32 ,3 is the base and 2 is the exponent. This means that 3 appears 2 times when multiplied by itself(3 X 3), and the product is 9. An exponent can be either positive or negative. A negative exponent implies a fraction, such that if n is a negative integer a

So 2 ,

-4

I 24

-n

I

= - a=tD an'

I 16

=-=­

We see that a negative exponent yields a fraction that is the reciprocal of the original base and exponent, with the exponent now positive instead of negative. Essentially, any quantity raised to a negative exponent can be

47

CLEP College Mathematics

"flipped" to the other side of the fraction bar and its exponent changed to a positive exponent. For example, 3

I

1. Here the exponent

1

_y

4- = 43 = 64

x

An exponent that is 0 gives a result of 1, assuming that the base itself is not equal to O. aO

= 1, a

~

=/= 0

(­

-3 ali

1

2. In this case, the ex

An exponent can also be a fraction. If m and n are positive integers, m

=­

nr-;;

= '\lam

3. 4­

The numerator remains the exponent of a, but the denominator tells what root to take. For example,

1

m' -3

=

--1­

1

4. -16 -2 --

-

=_

_1_1 = 162

4~=~=~=8 3~=~=h1=9

General Laws Cl

If a fractional exponent was negative, the operation involves the recip­ rocal as well as the roots. For example,

27-~ = _1_ = _1_ = _1_ = 27~

~272

~729

! 9

tfJdI

(c/J

(/J

Simplify the following expressions:

a'l

1. -3- 2 2. (-3)-2

-3

3.­ 1 41

4. -16-2

48

(ab)

Chapter 4: Algebra Topics

q>Onent changed to a 1. Here the exponent applies only to 3. Since xIIg

y

=

~,

1 so -3- 2 = _(3)-2 = - (312 ) = --

9

that the base itself 2. In this case, the exponent applies to the negative base. Thus,

(-3)-2 = _1_ = 1 (-3)2 (-3)(-3) c positive integers,

1

9

-3 -3 -3 -3 -3 4 3 -= = = = -x-=-12 1 . 41 ~ ! 1 1

4 41 4

(!)

the denominator tells

I

4. -16-2

1 =-= 16!

1

1

v%

4

- - - =-­

General Laws of Exponents

00 involves the recip­

aPaq = aP + q, bases must be the same.

4243 = 42 + 3 = 45 = 1,024 1

9

(aP)q

= aPq (2 3)2 = 26 = 64

aP

- q = aP- q , bases must be the same, a i= 0 a

(ab)P

= aPb P

(3

X

2)2 = 32 X 22 = (9)(4) = 36

49

CLEP College Mathematics

aP (ba)p =bP,b*,O

Find the value of

4)2 _ 42 _ 16 ---(5 52 25 -

logs 25 = x is e<: logs 25 = 2.

LOGARITHMS

log4x = 2 is equ

Logarithm Prop If M, N,p, and b 10gb 1 = 0 The exponential functions with base b can be written as

10gb b = 1 10gb fr =

y

= f(x) = fr.

X

10gb (MN) = 10gb M ­ 10gb (MIN) = 10gb M

10gb MP = P 10gb M

Inverse is denoted byr l ( ); this doesn't mean a negative exponent.

IfloglO 3 = .47~

Since 12 = 4(3J Write the following equations in logarithmic fonn: 34 = 81 and Me = 5.

The expression y = fr is equivalent to the logarithmic expression IOgbY = x. Therefore, 34 = 81 is equivalent to the logarithmic expression, log381 = 4.

Me = 50

5 is equivalent to the logarithmic expression logM 5 = k.

Remember, lo~ Therefore, logu

Chapter 4: Algebra Topics

Find the value of x for logs 25 = x and log4X = 2.

logs 25 = x is equivalent to 5x = 25. Thus x = 2, since 52 = 25, and logs 25 = 2. log4 X = 2 is equivalent to 4 2 = x, or x = 16.

f

~

......

itten as

Logarithm Properties If M, N, p, and b are positive numbers and b = 1, then

=0 10gb b = 1 10gb b = X 10gb 1

X

10gb (MN) = 10gb M 10gb (MIN)

= 10gb M

+ 10gb N - 10gb N

10gb !vJP = P 10gb M

I

negative exponent.

Iflog IO 3 = .4771 and log IO 4 = .6021, find log IO 12.

Since 12 = 4(3), log IO 12 = log IO (4) (3).

Ill:

Remember, log b(MN) =

log~

Therefore, log IO 12 = log IO 4

+ log~.

+ log IO 3

= .6021

+ .4771

= 1.0792

Dgaritbmic expression ilg3ritbmic expression,

logv5

= k. 51

CLEP College Mathematics

EQUATIONS

Multiplication or DI

We can divide b

For example, in the equation 3x = 18, 6 is the solution since 3(6) = 18. Depending on the equation, there can be more than one solution. Equations with the same solutions are said to be equivalent equations. An equation without a solution is said to have a solution set that is the empty or null set, represented by 4>. Replacing an expression within an equation by an equivalent expres­ sion will result in a new equation with solutions equivalent to the original equation. For example, suppose we are given the equation

So 3x = 6 is eqt

Solve for x, justi

3x + Y + x + 2y = 15. By combining like terms, we get

3x + Y + x + 2y = 4x + 3y.

3x - 8 = 7x + J

Since these two expressions are equivalent, we can substitute the sim­ pler form into the equation to get

3x - 8 + 8 = 7J

3x + 0 = 7x + 1

4x + 3y = 15

3x = 7x + 16

Performing the same operation to both sides of an equation by the same expression will result in a new equation that is equivalent to the original equation.

3x - 7x = 7x

+

-4x = 7x - 7x -4x = 0 + 16

Addition or Subtraction

-4x = 16 y+6=10

We can add ( - 6) to both sides

y + 6 + (- 6)

x= -4 =

10 + (- 6)

y+O=1O-6=4 52

-4x 16 ---4 -4

Chapter 4: Algebra Topics

Multiplication or Division

~

..... 6 •.• . CIJIItain-' sep "a . ra . t!. ..

...

. equaliotfi .;

3x = 6 We can divide both sides by 3:

3x

r

Ilion since 3(6) = 18. le solution. Equations 1IIIIIons. An equation the empty or null set,

m equivalent expres­ iwlent to the original lion

an substitute the sim-

equation by the same ivalent to the original

I

6

--3 3 x=2 So 3x

= 6 is equivalent to x = 2.

Solve for x, justifying each step.

3x - 8 = 7x

3x - 8

= 7x + 8

3x - 8 + 8 = 7x 3x

+ 0 = 7x +

3x

=

7x

3x - 7x

+

+8+8

Additive identity property

= 7x + 16 - 7x

-4x = 0

+

= 16

-4x 16 ----4 -4

16

Add 8 to both sides Additive inverse property

16

16

-4x = 7x - 7x

-4x

+8

+

16

Add (-7x) to both sides Commutative property Additive inverse property Additive identity property Divide both sides by -4

x= -4

53

CLEP College Mathematics

Replacing x with -4 in the original equation: Cross-multiply as

3x - 8 = 7x + 8

3(-4) - 8

-12 - 8 -20

= 7(-4) + 8

= -28 + 8

=

to obtain:

-20

Linear Equations

This is equivalent

which can be solv

2lx + 6 = l5x + -b

To solve a linear equation means to transform it into the form x = - . a

2lx - l5x + 6 = 6x

+ 6 - 6 = 20

6x

= 14

6x

14

6

6

A. If the equation has unknowns on both sides of the equality, it is convenient to put similar terms on the same sides. Refer to the following example.

4x 4x

+ 3 = 2x + 9 +3-

(4x - 2x)

2x

= 2x + 9 -

+ 3 = (2x

2x

- 2x)

+9

Add - 2x to both sides Commutative property

---

7

x=­ 3

2x+3=0+9

Additive inverse property

2x+3-3=0+9-3

Add - 3 to both sides

2x = 6

Additive inverse property

6

--2 2 x=3 2x

x= cisal Divide both sides by 2

B. If the equation appears in fractional form, it is necessary to transform it using cross-multiplication, and then repeat the same procedure as in (A). For example,

54

Factor Theorem

factorCi -

~

Chapter 4: Algebra Topics

3x + 4

3

+2

7x

=

5

Cross-multiply as follows:

3X+4

3

X

7x+2

5

to obtain:

3(7x

+ 2) = 5(3x + 4).

This is equivalent to:

21x

+6=

15x

+ 20,

which can be solved as in (A).

21x

into the form x

-b

= -. a

+6=

21x - 15x 6x

15x

+ 20

+6=

+ 6 - 6 = 20

1pIlity, it is convenient ~gexample.

:JCh sides

property

15x - 15x - 6

+ 20

Add -15x to both sides Combine like terms and add - 6 to both sides

6x = 14

Combine like terms

14

6x - - -

Divide both sides by 6

6

6

7

x=­ 3

seproperty

Ih sides

"SC

Factor Theorem

property

des by 2

:eessary to transform it lIle procedure as in (A).

55

CLEP College Mathematics

SIMULTANEOUS I

Example: 2

Letf(x) = 2.x - 5x - 3. By inspection, we can determine that 2(3)2 ­ (5)(3) - 3 = (2)(9) - (5)(3) - 3 = O. In this example, C = 3, so (x - 3) is also a factor ofU - 5x - 3.

Two or more equ stants and a, b -=1= 0 all or simultaneous equat

Equations with I you have as many equ Example:

Letf(x) = x 3 + 3~ - 4. By inspection, we can determine that (-2)3 + 3(-2i - 4 = -8 + (3)(4) - 4 = O. In this example, C = -2, so (x + 2) is also a factor of x 3 + 3x2 - 4.

Method 1: Substl other. Substitute this'

Method 2: Addltli by numbers that will] equations numerica11) subtract the equatioru unknown; we solve it, the unknown that we :

Remainder Theorem

Example:

Given a polynomial p(x) = 2.x3 - ~ + x remainder is P(l) = -2(1)3 - (1)2 + 1 + 4 = 6. That is, 2.x 3 - x 2

There are severa variables. Three of th€

+ x + 4 = q(x) +

6

+

4, divided by x-I, the

Method 3: Grapi the drawn lines is a ! dinates correspond b addition!substraction

, where q(x) is a polynomial.

(x - 1)

A system~

Note that in this case a = 1.

anySl

+ x + 2.

Also, by using long division, we get q(x) = U

Inconsistent eq1 later in this chapter. Example:

Given a polynomial p(x) = x 4 der isp(-3) = 28. That is, x 4

+ x-50 = q(x) +

+ x-50 divided by x + 3, the remain­

(x

28

+ 3)

, where q(x) is a polynomial.

Note that in this case a = - 3.

Also, by using long division, we get q(x) = x 3

56

-

3~

+ 9x -

26

Chapter 4: Algebra Topics

SIMULTANEOUS LINEAR EQUATIONS lIetermine that 2(3)2 ­ Ie, c = 3, so (x - 3) is

Two or more equations of the form ax + by = c, where a, b, c are con­ stants and a, b 0 are called linear equations with two unknown variables, or simultaneous equations.

"*

Equations with more than one unknown variable are solvable only if you have as many equations as unknown variables. leteimine that (- 2)3 + ~ C = - 2, so (x + 2) is

There are several ways to solve systems of linear equations with two variables. Three of the basic methods are: Method I: Substitution-Find the value of one unknown in terms of the other. Substitute this value in the other equation and solve. Method 2: Addition or subtraction-If necessary, multiply the equations by numb~rs that will make the coefficients of one unknown in the resulting equations numerically equal. If the signs of equal coefficients are the same, subtract the equations, otherwise add. The result is one equation with one unknown; we solve it and substitute the value into the other equations to find the unknown that we first eliminated.

, divided by x - I, the

Method 3: Graph--Graph both equations. The point of intersection of the drawn lines is a simultaneous solution for the equations, and its coor­ dinates correspond to the answer that would be found by substitution or addition/substraction.

re q(x) is a polynomial.

+x+2.

d by x

Inconsistent equations represent parallel lines, which are discussed later in this chapter.

+ 3, the remain-

l(x) is a polynomial.

- 3.1 + 9x - 26 57

CLEP College Mathematics

Subtract x from b Solve the system of equations 1.

x+y=3

2.

3x ­ 2y = 14

Subtract 3x from'

Divide by -2:

Method 1 (Substitution): From equation (1), we gety = 3 - x. Substi­ tute this value into equation (2) to get

3x - 2(3 - x) = 14

+ 2x =

3x - 6

14

The graph of eac

5x = 20

ting only two points. F

x=4 Substitute x = 4 into either of the original equations to find y = - 1. The answer is x

= 4,y = -1.

+ 2y =

+ 3x 5x

= 1, theny = 2. Th4

Y = -7

3

+ -x

2 '

let x =

points on this second 1

Method 2 (Addition or subtraction): If we multiply equation (1) by 2 and add the result to equation (2), we get

2x

x

To find the point

6

2y = 14

+0=

20

x=4 Then, as in Method 1, substitute x = 4 into either of the original equa­ tions to findy = -1. The answer isx

= 4,y = -1.

Method 3 (Graphing): Find the point of intersection of the graphs of the equations. To graph these linear equations, solve for y in terms of x. The equations will be in the form y = mx + b, where m is the slope and b is the intercept on the y-axis. This is the slope-Intercept form of the equation.

x+y=3 58

AB is the graph (J

point of intersection j coordinates of P satis

Chapter 4: Algebra Topics

Subtract x from both sides:

y=3-x

3x - 2y = 14

Subtract 3x from both sides:

-2y get y

14 - 3x

Divide by -2:

= 3 - x. Substi­

3 y = -7 +-x 2 The graph of each of the linear functions can be determined by plot­ ting only two points. For example, for y

=-

1.

=3-

x, let x

= 0, theny = 3. Let

1, theny = 2. The two points on this first line are (0,3) and (1, 2). For 3 y = -7 + 2"x, let x = 0, theny = -7. Let x = 2, theny = -4. The two

x

ions to find y

=

points on this second line are (0, -7) and (2, -4).

tiply equation (1) by 2

I:r of the

=

To find the point of intersection P of the two lines, graph them. D

original equa­

:dion of the graphs of for y in terms of x. The s the slope and b is the II of the equation.

c AB is the graph ofequation (1), and CD is the graph ofequation (2). The point of intersection P of the two graphs is the only point on both lines. The coordinates of P satisfy both equations and represent the desired solution of

59

CLEP College Mathematics

the problem. From the graph, P seems to be point (4, - I). These coordinates satisfy both equations, and hence are the exact coordinates of the point of intersection of the two lines.

To show that (4, - 1) satisfies both equations, substitute this point into both equations.

x+y=3

3x - 2y = 14

4+(-1)=3

3(4) - 2(-1) = 14

4-1=3

12 + 2 = 14

3= 3

14 = 14

Solve the equatioI

We have two eqw

As with all siro solution. Since equatic method of substitutioll

Dependent Equations

Substitute -

Since there are an infinite number of points on a line, there are an infi­ nite number of simultaneous solutions.

(~

Distribute:

Example: 2x+y=8 4x

+ 2y =

Although the m No matter what real equation (l) willalw:

16

These equations are dependent. Since they represent the same line, all points that satisfy either of the equations are solutions of the system.

The reason for at the equation y = ­

+ 3y =

6.

In other words. x and y that satisfies

60

Chapter 4: Algebra Topics

- 1). These coordinates :dinates of the point of Solve the equations 2x

IIbstitute this point into

aline, there are an infi­

esent the same line, all IS ofthe system.

+ 3y = 6 and y

= -

(2;) + 2

simultaneously.

We have two equations and two unknowns. 2x

+ 3y = 6 and

y=-(~)

+2

As with all simultaneous equations, there are several methods of solution. Since equation (2) already gives us an expression for y, we use the method of substitution. Substitute -

(~)

+ 2 for y in equation (1): 2x +

3(- ~ + 2) 6 =

Distribute:

2x-2x+6=6 6=6 Although the result 6 = 6 is true, it indicates no single solution for x. No matter what real number x is, if y is determined by equation (1), then equation (1) will always be satisfied. The reason for this peculiarity may be seen if we take a closer look at the equation y = -

(~)

+ 2. It is equivalent to

3y = - 2x

+ 6, or 2x

+ 3y = 6. In other words, the two equations are equivalent. Any pair of values of x and y that satisfies one satisfies the other.

61

CLEP Col/ege Mathematics

It is hardly necessary to verify that in this case the graphs of the given equations are identical lines, and that there are an infinite number of simul­ taneous solutions to these equations.

Multiply equatiOl

Multiply equatioJ

Add equations (3

Parallel Lines

We obtain a pecu

Actually, what . simultaneous solution there is no simultaneo satisfying both.

The straight line5 if they never intersect of these equations.

Example:

Line It : 2x - 7y = 14, Line 12 : 2x - 7y = 56 In the slope-intercept form, the equation for II is y = equation for 12 is y

= ~x 7

Solve the equati9ns 2x

~x -

2 and the

8. Each line has a slope of~.

7

+ 3y = 6 and 4x + 6y = 7 simultaneously.

•

We have two equations and two unknowns, 2x

+ 3y = 6

and

4x + 6y = 7 Again, there are several methods to solve this problem. We have chosen to multiply each equation by a different number so that when the two equa­ tions are added, one of the variables drops out. Thus,

62

Perpendicular l

Chapter 4: Algebra Topics

Ile graphs of the given Ioite number of simul­

4x

Multiply equation (1) by 2:

+ 6y =

+ -4x ­

Multiply equation (2) by -1:

12

6y = -7

(3) (4)

0=5

Add equations (3) and (4):

We obtain a peculiar result!

~ ~--­

r° ..•.

Actually, what we have shown in this case is that there is no simultaneous solution to the given equations because 0 =1= -5. Therefore, there is no simultaneous solution to these two equations, and hence no point satisfying both.

they

The straight lines that are the graphs ofthese equations must be parallel if they never intersect, but not identical, which Can be seen from the graph of these equations. y

is y =

r~.

~x ­ 7

2 and the

7

: 7 simultaneously.

~-+--+---1I--+--+-t--.>IIk-~~+-""'X

2x + 3y

=6

Perpendicular Lines

""em. We have chosen at when the two equa­

63

CLEP College Mathematics

An example of two numbers that are negative reciprocals of each other 1 2 are 2 and -"2' (Remember: 2 = 1 .)

Example:

15-3xl=7is~

=7 -3x = 2

5 - 3x Example:

13 : 5x

+ 6y = 30, 14 : 6x - 5y = 90

x= -­

3

~x + 5 and the

The solution set i

= ~ x - 18. The slope of 13, which is - ~ is the negative

Remember, the a equation I 5x + 4 I =

In the slope-intercept form, the equation for 13 is y = equation for 14 is y

2

5

reciprocal of the slope of 14 , which is

6

~. Therefore, 13 is perpendicular to 14 . 5

To summarize:

• Parallel lines have slopes that are equal.

INEQUALI1"IES

The solution of a ues of x for which the

• Perpendicular lines have slopes that are negative reciprocals of each other.

ABSOLUTE VALUE EQUATIONS

The sentence 3 . real numbers, since it others, or y < is the

°

When the definition ofabsolute value is applied to an equation, the quan­ tity within the absolute value symbol may have two values. This value can be either positive or negative before the absolute value is taken. As a result, each absolute value equation actually contains two separate equations.

The sentence x expression on the lefl

When evaluating equations containing absolute values, proceed as follows: A sentenOlj vari$bles ~

64

Chapter 4: Algebra Topics

iprocals of each other

Example:

I5 -

3x I = 7 is valid if either

5 - 3x = 7

or

5 - 3x = -7

-3x = -12

-3x = 2 2

x= -3

. 5 y = - 6x + 5 and the .

5.

IS - - IS

6

h . t e negative

The solution set is therefore x = ( -

x=4

~,

4)

Remember, the absolute value of a number cannot be negative. So the equation I 5x + 4 I = - 3, would have no solution.

is perpendicular to [4.

INEQUAI.ITIES The solution of a given inequality in one variable x consists of all val­ ues of x for which the inequality is true.

iprocals of each other.

~8S

..

!

I

an equation, the quan­ mlues. This value can e is taken. As a result, IJ3l8te equations.

The sentence 3 - y > 3 + y is a conditional inequality for the set of real numbers, since it is true for any replacement less than 0 and false for all others, or y < 0 is the solution set.

The sentence x + 5 > x + 2 is an absolute inequality because the expression on the left is greater than the expression on the right.

e values, proceed as false when its

65

CLEP College Mathematics

The sentence x + 10 < x + 5 is inconsistent because the expression on the left side is always greater than the expression on the right side.

•

~

The sentence 5y < 2y + Y is inconsistent for the set of non-negative real numbers. For any y greater than 0, the sentence is always false. Two inequalities are said to have the same sense if their signs of inequality point in the same direction. The sense of an inequality remains the same ifboth sides are multiplied or divided by the same positive real number.

Solve the inequali1

Example:

For the inequality 4 > 3, if we multiply both sides by 5, we will obtain:

4X5>3x5

Add - 5 to both si Additive inverse p

Additive identity I Combine terms:

20> 15 Multiply both side The sense of the inequality does not change. Ifeach side ofan inequality is multiplied or divided by the same negative real number, however, the sense of an inequality becomes opposite.

same as dividing 1

The solution set iJ Example:

X= {xlx>2}

For the inequality 4 > 3, if we multiply both sides by - 5, we would obtain:

(that is, all x, sud

4 X (- 5) < 3 X (- 5)

COMPLEX NUMB

-20 < -15

As indicated abo' Ius mathematics topic: are not enough to expl were developed.

The sense of the inequality becomes opposite. If a > b and a, b, and n are positive real numbers, then an> bn and a- n < b- n

> y and q > p, then x + q > Y + p. Ifx > y > 0 and q > P > 0, thenxq > yp. If x

66

Chapter 4: Algebra Topics

lOSe the

expression on lie right side. .e set of non-negative ,always false.

_

if their signs of

Ib. sides are multiplied

Solve the inequality 2x

+5>

9.

Additive identity property:

+ 5 > 9. 2x + 5 + (- 5) > 9 + (- 5) 2x + 0 >: 9 + (- 5) 2x> 9 + (-5)

Combine terms:

2x >4

2x

l

sides by 5, we will

d by the same negative JDCS opposite.

ides by -5, we would

s,then

Add - 5 to both sides: Additive inverse property:

Multiply both sides by! (this is the 2 same as dividing both sides by 2):

I I -(2x) > 2 2 x>2

X 4

The solution set is

X= {x

Ix > 2}

(that is, all x, such that x is greater than 2).

COMPLEX NUMBERS As indicated above, real numbers provide the basis for most precalcu­ lus mathematics topics. However, on occasion, real numbers by themselves are not enough to explain what is happening. As a result, complex numbers were developed.

67

CLEP College Mathematics

(3

+

Returning momentarily to real numbers, the square of a real number cannot be negative. More specifically, the square of a positive real number is positive, the square of a negative real number is positive, and the square of 0 is O.

i is defined to be a number with a property that F = - 1. Obviously, i is not a real number. C is then used to represent the set of all complex numbers:

Division of Comil

Division of two ce

cial procedure that invo C

= {a + bi I a and b are real numbers}.

Addition, Subtraction, and Multiplication of Complex Numbers Here are the definitions of addition, subtraction, and multiplication of complex numbers. Suppose x thati2 = - 1):

+ yi and z + wi are complex numbers. Then (remembering

Simplify (3

(x

+ yi) + (z + wi) = (x + z) + (y + w)i

(x

+ yi)

- (z

+ wi) =

(x

+ yi)

X (z

+ wi) = (xz

+ i)(2 + i).

(x - z)

+ (y -

- lry)

gate of a

+ bi is denok

Also,

The usual proced1 jugate as shown beloVll same quantity leaves tI

w)i

+ (xw + yz)i

If a is a real nWII Hence, every real nun

All the propertie complex numbers, so

68

Chapter 4: Algebra Topics

(3

+ i)(2 + i)

are of a real number l positive real number 1Sitiye, and the square l ;2

= -

1. Obviously, be set of all complex

= 3(2

+ i) + i(2 + i)

= 6

+ 3i + 2i + P

= 6

+ (3 + 2)i + (-1)

= 5

+ 5i

Division of Complex Numbers Division of two complex numbers is usually accomplished with a spe­ cial procedure that involves the conjugate of a complex number. The conju­

crs}.

gate of a

+ bi is denoted by a + bi and defined by a 1- bi = a - bi.

Also,

(a

+ bi)(a - bi)

= a2

+ b2

, and multiplication of

L

Then (remembering

The usual procedure for division is to multiply and divide by the con­ jugate as shown below. Remember that multiplication and division by the same quantity leaves the original expression unchanged.

x

+ yi

x

--= z + wi z

.. w)i

- w)i (~+

+ yi z - wi x-­ + wi z - wi

(xz

yz)i

+ yw) + (-xw + yz)i

=--'----;:----;:---'-­ z2 +w2

=

xz+yw z2 + w2

+

-xw+yz. l z2 + w2

If a is a real number, then a can be expressed in the form a = a Hence, every real number is a complex number and R ~ C.

+ Oi.

All the properties of real numbers described in Chapter 2 carry over to complex numbers, so those properties will not be stated again.

69

CLEP College Mathematics

QUADRATIC EQUATIONS

We need two nUll would be + 3 and +4, (x + 4) = O.

x

There are several methods to solve quadratic equations, some of which are highlighted here. The first two are based on the fact that if the product of two factors is 0, either one or the other of the factors equals O. The equation can be solved by setting each factor equal to O. If you cannot see the factors right away, however, the quadratic formula, which is the last method presented, always works.

=

Therefore, x + 3 : -3,x = -4.

Substitute the vall For x = -3, (-3 x = - 3 is a solution. Likewise, for x =

= O. So x = -4 is a sc

Solution by Factoring We are looking for two binomials that, when multiplied together, give the quadratic trinomial

ar + bx + c = 0 This method works easily if a = 1 and you can find two numbers whose product equals.c and sum equals b. The signs of band c need to be considered:

• If c is positive, the factors are going to both have the same sign, which is b's sign. • If c is negative, the factors are going to have opposite signs, with the larger factor having b's sign.

Example:

Suppose the quae: the sign of b is negatn

We need two num be - 3 and -4, and the

Therefore, x - 3 are x = 3, x = 4.

Once you have the two factors, insert them in the general factor format (x

+ _)(x + _) = O.

To solve the quadratic equation, set each factor equal to 0 to yield the solution set for x.

Substitute the va For x = 3, solution.

(3i ­

Likewise, for x : X

Example:

Solve the quadratic equation 70

r

+ 7x + 12 = O.

= 4 is a solution.

Chapter 4: Algebra Topics

We need two numbers whose product is + 12 and sum is +7. They would be +3 and +4, and the quadratic equation would factor to (x + 3) (x

+ 4) = O.

Therefore, x + 3 x = -3,x = -4.

: equations, some of DO the fact that if the rthe factors equals O. 131 to O. If you cannot nola. which is the last

Substitute the values into the original quadratic equation: For x = -3, (-3)2 + 7(-3) + 12 x = - 3 is a solution.

= dtiplied together, give

= 0 or x + 4 = 0 would yield the solutions, which are

= 0, or9

Likewise, for x = -4, (-4)2 + 7(-4) + 12 o. So x = -4 is a solution.

+ (-21) + 12

= O.

So

= Q, or 16 + (-28) + 12

Example: Suppose the quadratic equation is similar to the previous example, but the sign of b is negative. Solve the quadratic equation x 2 - 7x + 12 = o.

:an find two numbers

of b and c need to be same sign, which is

We need two numbers whose product is + 12 and sum is -7. They would be - 3 and -4, and the quadratic would factor to (x - 3)(x - 4) = O.

posite signs, with the

Therefore, x - 3 = 0 or x - 4 = 0 would yield the solutions, which are x = 3,x = 4.

Ie

:general factor format Substitute the values into the original quadratic equation:

equal to 0 to yield the

For x solution. x

=

= 3, (3)2

Likewise, for x 4 is a solution.

- 7(3) + 12

= 4, (4)2

= 0, or 9

- 7(4) + 12

So x

= 3 is a

- 28 + 12

= O. So

- 21 + 12

= 0, or 16

= O.

71

CLEP College Mathematics

Example: As a final example, solve the quadratic equation Y?

+ 4x - 12 = 0.

For x For x

We need two numbers whose product is -12 and sum is +4. They would be +6 and -2 (note that the larger numeral gets the + sign, the sign of b). The quadratic equation would factor to (x + 6)(x - 2) = 0.

°

Therefore, x + 6 = are x = -6,x = 2.

or x - 2 =

°

Likewise, for x x = 2 is a solution.

+ 4(-6) - 12 = 0, or 36 + (-24) - 12 = 0. So

= 2,

(2i

Example: Solve the quadrat

would yield the solutions, which

Substitute the values into the original quadratic equation: For x = -6, (-6i x = -6 is a solution.

= 4, (4)2 ­ = -4 , (-4'.

This is the diffeR roots are 3x and 6. So 1 or 3x - 6 = 0, and the

+ 4(2) - 12 = 0, or 4 + 8 - 12 = 0. So For x

Sum of Two Squares

For x = -2,9(­

If the quadratic c~nsists of the difference of only two terms of the form aY? - c, and you can recognize them as perfect squares, the factors are sim­ ply the sum and difference of the square roots of the two terms. Note that a is a perfect square, but not necessarily 1, for this method.

Quadratic Forml quad! of quadratic. as shoWn b

Example: Solve the quadratic equation Y?

= 2, 9(2)2 .

-

16

= 0. (til-

This is the difference of two perfect squares, Y? and 16, whose square roots are x and 4. So the factors are (x + 4)(x - 4) = 0, and the solution is x = ±4.

72

Chapter 4: Algebra Topics

-l + 4x ­

12 = O.

For x For x

md sum is +4. They Is the + sign, the sign ~ - 2) = O.

= 4, (4)2 - 16 = 16 - 16 = O. = -4, (-4)2 ­ 16 = 16 - 16 = O.

Example:

Solve the quadratic equation

I the solutions, which

~on:

. (-24) ­ 12 = O. So

.. + 8 -

9~ - 36

= O.

This is the difference of two perfect squares, 9x4 and 36, whose square roots are 3x and 6. So the factors are (3x + 6)(3x - 6) = 0, then 3x + 6 = 0 or 3x - 6 = 0, and the solution is x = ±2.

12 = O. So For x

= 2, 9(2)2 -

For x

= -2, 9( -2)2 ­ 36 = 36 - 36 = O.

36

= 36 -

36

= O.

two terms of the form

~ the factors are sim­ two terms. Note that a

Quadratic Formula

IOd.

and 16, whose square : 0, and the solution is

where (if - 4ac~) iscaUed the 4:1Isc~rIm_nt of thequadra1tic equation.

73

CLEP College Mathematics

• Ifthe discriminant is less than zero (b 2 - 4ac < 0), the roots are complex numbers, since the discriminant appears under a radical and square roots of negatives are imaginary numbers. A real number added to an imagi­ nary number yields a complex number. • Ifthe discriminant is equal to zero (b 2 equal.

-

4ac

= 0), the roots are real and

• If the discriminant is greater than zero (b 2 - 4ac > 0), then the roots are real and unequal. The roots are rational if and only if a and b are rational and (b 2 - 4ac) is a perfect square; otherwise, the roots are irrational.

Since the discrimin unequal.

1. 4xl - l2x

2. 3x

2

3. 5xl

-

+9=0

9

=0

Here, a, b, and c an a

= 5, b = 2, and c

Therefore, b2

-

4ac

= 22 - 4C

Since the discrimin roots are irrational.

+ 3x + 5 =

4. xl

Compute the value ofthe discriminant and then determine the nature of the roots of each of the following four equations:

+ 2x -

3. 5xl

0

Here, a, b, and c aI1 a

= 1, b = 3, and c

Therefore,

7x - 6 = 0

b2

+ 2x - 9 = 0

-

4ac = 32

-

4

Since the discrimin

4. xl + 3x + 5 = 0

t~~ Solve the equatic 2

1. 4x -

l2x

+ 9 = 0,

Here, a, b, and c are integers:

= 4, b = -12, and c = 9.

a

In this equation, l formula.

Therefore, b2

-

4ac

= (-12)2 -

4(4)(9)

=

144 - 144

=0

Since the discriminant is 0, the roots are rational and equal. 2. 3xl - 7x - 6 = 0

Here, a, b, and c are integers: a

= 3, b = -7, and c = -6.

Therefore, b2

74

-

4ac

= (-7)2 -

4(3)(-6)

= 49 + 72 =

121

=

11 2.

J

Chapter 4: Algebra Topics

the roots are complex meal and square roots ~ added to an imagi-

Since the discriminant is a perfect square, the roots are rational and unequal. 3. 5x2

+ 2x -

9 = 0

Here, a, b, and c are integers: I

the roots are real and

• 0), then the roots are if a·and b are rational oots are irrational.

= 5, b = 2, and c =

a

Therefore, b2

IDd equal.

-

4ac

= 22 -

4(5)(-9)

=4+

= 184.

180

Since the discriminant is greater than zero, but not a perfect square, the roots are irrational and unequal. 4. x 2

letennine the nature of

-9

+ 3x + 5 = 0

Here, a, b, and c are integers:

= 1, b = 3, and c = 5

a

Therefore, b2

-

4ac

= 32 -

4(1)(5)

=9-

20

=

-11

Since the discriminant is negative, the roots are complex.

Solve the equation x 2

In this equation, a formula.

-

x

+ 1 = O.

= 1, b = -

x=

1, and c

-( -1) ±

= 1. Substitute into the quadratic

J( _1)2 -

4(1)(1)

2(1) 1 ±.JI=4

=---­ 2

±.J=3

1 =--­ 2

75

CLEP College Mathematics

DRILL QUESTION

1 ± ,,[3i

2

x=

1 + ,,[3i

2

1 - ,,[3i orx=--­ 2

ADVANCED ALGEBRAIC THEOREMS A. Every polynomial equationf(x) = 0 of degree greater than zero has at

least one root either real or complex. This is known as the fundamental theorem of algebra.

B. Every polynomial equation of degree n has exactly n roots.

C. If a polynomial equationf(x) = 0 with real coefficients has a root a + bi, then the conjugate of this complex number a - bi is also a root of f(x) = O. D. If a +../b is a root of polynomial equationf(x) = 0 with rational coef­ ficients, then a - ../b is also a root, where a and b are rational and ../b is irrational.

E. If a rational fraction in lowest terms ~ is a root of the equation a"x" + c an_1XZ- 1 + ... +alx+aO = 0, ao =1= 0, and the aj are integers, then b is a factor of ao and c is a factor of an'

F. Any rational roots of the equation XZ + qlXZ- 1 +q2XZ-2 + ... + qn-IX + qn = 0 must be integers and factors of qn. Note that qI, q2, ..., qn are integers.

76

1. Which one of the j of5x + 9y = 14? 5 (A) y= -x+ 14 9 5 (B) y=--x-l' 9 9 (C) y = - -x + 1, 5 9 (D) y = - -x - 1, 5

2. The solution set of

(A) X = {x Ix

-< ;

(B) X = {x I x

>:

(C) X = {xlx-<; (D)X=

{xIx>:

Chapter 4: Algebra Topics

DRILL QUESTIONS

1. Which one ofthe following is an equation of a line parallel to the graph of5x + 9y = 14? 5 (A) y = 9x + 14 r:ater than zero has at m as the fundamental

'II

9 (C) y= -S-x+ 14

roots.

cients has a root a + - bi is also a root of

: 0 with rational coef­ are rational and ,Jb is

9

(D) y= -S-x-14

2. The solution set of the inequality 5 - 7x > -9 is (A) X = {x I x -< 2}

f the equation

Ire

5 (B) y= --x-14 9

an>!' +

integers, then b is a

_~-2 ~ 1hat

+ ... + qn-l x

qI, q2, ... , qn are

(B) X = {x I x > 2}

(C) X = (D) X

{XIX -< ;

}

= {xIX ~;}

77

CLEP College Mathematics

3.

Whatisthevalueof(:-~)

(-3)-2?

6. What is the simplif

(A) lli (A)

-~

(B) -lli

3 1 (B) - ­ 9

(C)

~

(D)

~

(C) i (D) -i

9

7. If x

+ 2 is a factor

3

(A) 22

4. If log 10 P = 0.5 and lOglO (f = 1.2, what is the value of lOglO PQ ?

(B) 16

(A) 1.9

(C) -16

(B) 1.7

(D) -22

(C) 1.1 (D) 0.6

5. What is the solution set for x in the following inequality? 2

(A) 28 - 4x

3

(B) 'f8 - 5x

-6<-x+6<2 (A) -12 < x < 0 (B) -6

<x<0

(C) -18 <x < -12 (D) -18

78

< x < -6

8. A piece oflumber The length ofthe fi is four times the 14 the length of the tt

(C) 4x ­ 28 (D) 5x ­ 28

Chapter 4: Algebra Topics

6. What is the simplified expression for

-sF + (2i)(i2 -2)?

(A) Iii (B) -lli (C) i (D) -i

7. Ifx

+ 2 is a factor of3x3 -

x

+ k, what is the value of k?

(A) 22

lue oflog lO PQ ?

(B) 16 (C) -16 (D) -22

8. A piece oflumber that is 28 inches in length is divided into three pieces. The length of the first piece is x inches and the length ofthe second piece is four times the length of the first piece. Which expression represents the length of the third piece? (A) 28 - 4x (B) 28 - Sx

(C) 4x ­ 28 (D) Sx - 28

79

CLEP College Mathematics

9. What is the value of y in the following system of equations?

4x + 2y = 5

3x + 5y = 9

(A)

~

(B)

~

(C)

~

(D)

2

2

2

2

2

10. 2 5 = 32 is equivalent to which logarithmic expression? (A) log5 32 = 2

(B) log3z 5 = 2

(C) logz 32 = 5

(D) logz 5

80

= 32

~

Chapter 4: Algebra Topics

ptions?

.

1. (8)

6. (D)

2. (A)

7. (A)

3. (A)

8. (8)

4. (C)

9. (8)

5. (D)

10. (C)

?

IIOD.

81

1 1:1

----~---­

CHAPTER.S

Functions and

Their Graphs

Chapter 5

Functions a

ELEMENTARY FUI

Is

1

1 = x a functi~

Graph the equatil = x is not a funCtiOI

Chapter 5

Functions and Their Graphs ELEMENTARY FUNCTIONS

Is I = x a function?

I

Graph the equation. Note that x can have two values of y. Therefore, = x is not a function. y

---+---~x

85

CLEP College Mathematics

J?

Find the domain and range for y = 5 -

Let f(x) functions: 1.

First determine if there are any values that would make the function undefined (i.e., division by 0). There are none. Thus, the domain is the set of real numbers. The range can be found by substituting some corresponding values for x in the equation. x

I

2

1

0

y

I 1

4

5

2·I-g 3. fXg

4.

£g

1.

(I + g) (x) = f(x)

4

-f

2. (f- g) (x)

= f(x)­

3. (IX g) (x)

= f(x) g

= f(x) = 5x + 2.

+2 . = 5(1) + 2

f(x) = 5x

f(l)

1+ g

-1-2

The range is the set of real numbers less than or equal to 5.

Evaluatef(l) for y

= 22 ­

=5+2

4.

=7

(£) g

(x)

= I(x) = g(x)

Note: The domair

Operations on Functions

fraction is indetermina

Functions can be added, subtracted, multiplied, or divided to form new functions.

Composite Func

a. (I + g) (x) = I(x)

+ g(x)

= I(x)

- g(x)

b. (I - g) (x)

c. (IX g) (x) = I(x) g(x)

d. (£)(x) = I(x) g g(x) 86

The COIIIIIIII f(g(x). .. i

Chapter 5: Functions and Their Graphs

Let f(x) = ~ functions:

1 and g(x) = 5x

+

3. Determine the following

1. f+ g

del make the function be domain is the set of

2. f-g

Bsome corresponding

3. fXg 4.

equal to 5.

IF divided

£.g

1. (f + g)(x)

= f(x) + g(x) = ~ ­ = ~

2. (f - g) (x)

= f(x)

- g(x)

~

+3

+ 5x + 2

=~­ =

1 + 5x

1 ­ (5x

+ 3)

­ 1 ­ 5x - 3

= ~ ­ 5x - 4

3. (fX g) (x)

= f(x) g(x) = (~ - 1) (5x + 3) = 1o~ + 6r ­ 5x - 3

4. (£.)(x) g

2

= f(x) =

(2x

g(x)

(5x

-

1)

+ 3)

Note: The domain of (4) is for all real numbers except ­

~ because the 5

fraction is indeterminate if the denominator = O. to form new

Composite Function

87

CLEP College Mathematics

Givenf(x)

Two functions f ane x. To findgwhenf is gn solve for y = g(x).

= 3x and g(x) = 4x + 2.

Find (fo g) (x) and (g 0 f) (x).

Find the inverse of (fo g) (x) = f(g(x)) (g 0 f) (x)

Note that (fo g) (x)

= 3(4x + 2) = 12x + 6

= g(f(x)) = 4(3x) + 2 = 12x + 2 =1=

1. f(x)

= 3x + 2

2. f(x)

= x2 -

1. f(x)

= y = 3x + 2

3

(g of) (x).

Find (fo g) (2) iff(x) = x 2

3 and g(x) = 3x

-

+1

To findf- 1(x), intl

(fo g) (2)

= f(g(2))

g(x)

= 3x + 1

Solve for y:

Substitute the vcrlue of x = 2 in g (x). g(2)

= 3(2) + 1 = 7

f(x)

=~ -

2. f(x)

3

To findf- 1(x), in1

Substitute the value of g(2) inf(x). f(7)

= (7i

- 3 = 49 -

3

= 46 Solve for y:

Inverse lverse of a function, ("1. is obtained from f g the x and y (= 1{x») and then solving for y.

88

= y = x2 - 3

Chapter 5: Functions and Their Graphs

Two functions f and g are inverses of one another if g of = x and f o g = x. To find g whenfis given, interchange x and g in the equationy = f(x) and solve for y = g(x).

Find the inverse of the functions

llx

+6

llx + 2

.. 1

1. f(x) = 3x 2. f(x) =

+2

r- - 3

1. f(x) = y

= 3x + 2

To findf-l(x), interchange x and y. x = 3y

3y = x - 2

Solve for y:

x-2

y=-­

3

2. f(x)

= y = r-

To find f

imf~ fory.

-3

-1 (x),

interchange x and y.

x=l- 3 I =x+ 3

6

l

+2

Solve for y: y = Jx+3

89

CLEP Col/ege Mathematics

TRANSLATIONS, REFLECTIONS, AND SYMMETRY OF FUNCTIONS

The letter 0 repre tion is counterclockwis clockwise, each point (

f(x)

Given the function/(x) = x 2 - 5, what is the resulting function g(x) in which each point of/ex) is moved 2 units to the left and I unit up?

Suppose a funeti~ nates of this point if tI:

Since each point is moved 2 units to the left, for g(x) we replace each x in/ex) by x + 2. Also, because each point is moved I unit up, we add I to the expression for lex) to get g(x). Thus, g(x) = (x + 5 + 1 = (x + 4.

(4, 3) will becom

To verify that this result is correct, let's choose a point on/ex), such as (3, 4). Following the rules for obtaining g(x), the corresponding point on g(x) is (1, 5). Note that, by substitution, (1, 5) does lie on the graph of g(x) = (x + 2)2 - 4.

Suppose a functi nates of (4, 3) if the fi

2i -

2i -

(4,3) will becoD

90

Chapter 5: Functions and Their Graphs

IMETRY

The letter 0 represents the origin, which is located at (0, 0). Ifthe rota­ tion is counterclockwise, each point (x, y) becomes (-y, x). Ifthe rotation is clockwise, each point (x, y) becomes (Y, -x).

p'

Adting function g(x) in IIId 1 unit up?

l, for g(x) we replace s moved 1 unit up, we

r) = (x

+ 2i

Suppose a function contains the point (4,3). What are the new coordi­ nates of this point if the function is rotated 90° counterclockwise?

(4,3) will become (-3,4).

-5+1

• point on/ex), such as

orresponding point on ie on the graph of g(x)

Suppose a function contains the point (4,3). What are the new coordi­ nates of(4, 3) if the function is rotated 90° clockwise?

(4,3) will become (3, -4).

91

CLEP College Mathematics

DRILL QUESTIONS

A reflection about the x-axis changes point (x, y) into point (x, -y). A reflection about the y-axis changes point (x,y) into point (-x,y). A reflec­ tion about the line y = x will move each point P to a new point P' so that the line y = x is the perpendicular bisector of PP'. Each point (x, y) becomes (Y, x) after the reflection. P

1. Ifj(x) = 2x

+ 3 anl

(A) -61 (B) -32 (C) 25 (D) 47

5 2. If hex) = - - , wl

o

x-2

h (x)? (A) x - 2 5 P'

5 (B)­

x+2

A function contains the point (- 6,3). If this function is reflected about the line y = x, what will be the new coordinates for (-6, 3)?

(c) 2x 5­

5

(D) 2x + 5 i

X

(-6,3) will become (3, -6).

3. What is the domaiJ . (A) x >- 2

(B) x

<:

2

(C) x >- -2 (D) x

92

<:

-2

Chapter 5: Functions and Their Graphs

DRILL QUESTIONS

y) into point (x, -y). point (-x,y). A reflec­ r:w point P' so that the II point (x, y) becomes

1:r~~J1:~E,+

lC'tion is reflected about -6,3)?

= 2x + 3 and g(x) = x3 -

1. If/(x)

5, what is the value of/(g (- 3»?

(A) -61 (B) -32

(C) 25 (D) 47

5 2. If hex) = x _ 2' which of the following is

equiv~lent

to the inverse of

h (x)? (A) x ­ 2 5 (B) _5_

x+2 (C) 2x ­ 5 5

+5

(D) 2x

x

3. Whatisthedomainof/(x) = J6-3x? (A) x

>

2

(B) x <2 (C) x > -2 (D) x < -2

93

CLEP College Mathematics

4. If {(4, -1), (5, -3), (6, -9), (_, -I6)} represents a function, which one of the following cannot be filled in the blank space? (A) 5 (B) 3 (C) -4 (D) -9

5. Each point of the functionf(x) = ~ + 10 is moved 3 units to the left and 2 units down to create a new function g(x). What would be the y-coordinate of a point on the graph of g(x) whose x-coordinate is -1 on the graph off(x)? (A) 6

I -

I I

-

(B) 7

7. The function g(x) i zero? Which one 01 (A)4~ - 4

(B) ­

4 x

(C) ­

4 x2

(D) -v'x2 - 4

8. The point (5, -8) location? (A) (-5,8) (B) (-5, -8)

(C) 8

(C) (-8,5)

(D) 9

(D) (-8, -5) 6. Iff(x) is a linear function such thatf(l) = 5 andf(4) = -7, then what is the value off( -I)?

(A) 11 (B) 12 (C) 13 (D) 14

~

-

9. The point (-2, - 3 new location? (A) (-3,2) (B) (-3, -2) (C) (3, -2) (D) (3,2)

94

Chapter 5: Functions and Their Graphs

:nts a function, which space?

7. The function g(x) is known to have a range of all real numbers except zero? Which one of the following expressions could represent g(x)? (A) 4~ - 4 4 (B) x 4 (C) - 2 x

!Wed 3 units to the left ~ What would be the se x-coordinate is -1

(D)

-Jx2 -

4

8. The point (5, -8) is reflected across the line y. = x. What is the new location? (A)(-5,8) (B) (-5, -8)

(C) (-8,5) (D) (-8, -5)

(4) = -7, then what

9. The point (- 2, - 3) is rotated 90° clockwise about the origin. What is its new location? (A)( -3,2) (B) (-3,-2) (C) (3, -2) (D) (3,2)

95

CLEP College Mathematics

10. For which one ofthe following functions isf(l) = f( -1) = 2?

r- - 3x + 4 (B) f(x) = 4r- - 2

(A)f(x)

96

Ans"

=

(C)f(x)

= x2

(D)f(x)

=

-

2x - 1

3r- - x

1

~

Chapter 5: Functions and Their Graphs

=/(-1) = 2?

Answers to Drill Questions 1. (A)

6. (C)

2. (D)

7. (B)

3. (B)

8. (C)

4. (A)

9. (A)

5. (D)

10. (B)

97

Chapter 6

Geometry Topics Plane geometry refers to two-dimensional shapes (that is, shapes that

can be drawn on a sheet of paper), such as triangles, parallelograms, trap­ ezoids, and circles. Three-dimensional objects (that is, shapes with depth) are the subjects of solid geometry.

TRIANGLES A closed three-sided geometric figure is called a triangle. The points of the intersection of the sides of a triangle are called the vertices of the triangle.

Ii!l

B

"

I I"

!i:

:'1

i111

A

c

A side of a triangle is a line segment whose endpoints are the vertices of two angles of the triangle. The perimeter of a triangle is the sum of the measures of the sides of the triangle. An interior angle of a triangle is an angle formed by two sides and includes the third side within its collection ofpoints. The sum ofthe measures of the interior angles of a triangle is 180 0 • A scalene triangle has no equal sides.

101

CLEP College Mathematics

An acute trianl

An isosceles triangle has at least two equal sides. The third side is called the base of the triangle, and the base angles (the angles opposite the equal sides) are equal.

An obtuse tria"

A

c

B

An equilateral triangle has all three sides equal. AB = AC

= BC. An

equilateral triangle is also eqUiangular, with each angle equaling 60°. A

A right triangle

a right triangle is ca sides are called the I Theorem, the lengths formula

where c is the hypotE Pythagorean Theorel

102

Chapter 6: Geometry Topics

An acute triangle has three acute angles (less than 90°). A

s. The third side is angles opposite the

c

B

An obtuse triangle has one obtuse angle (greater than 90°). B

-

-

-

B=AC=BC.An

qualing 60°.

c

A

A right triangle has a right angle. The side opposite the right angle in a right triangle is called the hypotenuse of the right triangle. The other two sides are called the legs (or arms) of the right triangle. By the Pythagorean Theorem, the lengths of the three sides of a right triangle are related by the formula c2

= a2 + b2

where c is the hypotenuse and a and b are the other two sides (the legs). The Pythagorean Theorem is discussed in more detail in the next section.

103

CLEP College Mathematics

A

b

c

a

L

B

B

An altitude, or height, of a triangle is a line segment from a vertex of the triangle perpendicular to the opposite side. For an obtuse triangle, the altitude sometimes is drawn as a perpendicular line to an extension of the opposite side.

A line that bise a perpendicular bisec

A

8

I I I

an altitude

I I

I~ I I I I

~

G

C

0

L . -_ _

A

C

8

an altitude

B '-------1

An angle bisectl to the opposite side (

The area of a triangle is given by 1

A = -bh 2 where h is the altitude and b is the base to which the altitude is drawn. A line segment connecting a vertex of a triangle and the midpoint of the opposite side is called a median of the triangle.

B

The line segmel called a midline of th,

104

Chapter 6: Geometry Topics

A

AD is a median

.. B

e segment from a vertex of For an obtuse triangle, the line to an extension of the

A line that bisects and is perpendicular to a side of a triangle is called a perpendicular bisector of that side. A

A I

I

1 I

line I' is the perpendicular bisector of Be

I

I 11-I

an altitude

I

1

B

II

1

II

II

"c

-- - - ~

D

An angle bisector of a triangle is a line that bisects an angle and extends to the opposite side of the triangle. A

the altitude is drawn.

La =Lf3

19le and the midpoint ofthe

B

c

The line segment that joins the midpoints of two sides of a triangle is called a midline of the triangle.

105

------------lt~----------

_C_L_E_P_C_o_"_e9_e_M_a_th_e_m_a_tl_·c_s

We can solve the measures of thi

A AD = DC

BE

=

EC

An exterior angle of a triangle is an angle formed outside a triangle by one side of the triangle and the extension of an adjacent side.

Therefore, the sures 80 0 •

A

THE PYTHAGOI

The Pyth which, as The Pytha

hypotenU! squares

G

C

B

PROBLEM

The measure of the vertex angle of an isosceles triangle exceeds the measure of each base angle by 30 0 • Find the value of each angle of the triangle.

SOLUTION

In an isosceles triangle, the angles opposite the congruent sides (the base angles) are, themselves, congruent and of equal value. Therefore, 1. Let x = the measure of each base angle 2. Then x

106

+ 30

=

the measure of the vertex angle

The Pythagorea any two sides of a ri side.

outside a triangle by .t side.

triangle exceeds the If each angle of the

:ongruent sides (the .tue.

Chapter 6: Geometry Topics

We can solve for x algebraically by keeping in mind that the sum of all the measures of the angles of a triangle is 1800 • x

+ x + (x + 30) =

180

+ 30 =

180

3x

3x = 150 x = 50

Therefore, the base angles each measure 50 0 , and the vertex angle mea­ sures 80 0 •

THE PYTHAGOREAN THEOREM The Pythagorean Theorem pertains to a right triangle, which, as we saw, is a triangle that has one 90° angle. The Pythagorean Theorem tells you that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides, or

c2

=

a2 + b 2

a

b

The Pythagorean Theorem is useful because if you know the length of any two sides of a right triangle, you can figure out the length of the third side.

107

CLEP College Mathematics

PROBLEM

In a right triangle, one leg is 3 inches and the other leg is 4 inches. What is the length of the hypotenuse?

3

4

PROBLEM

What is the \-a] SOLUTION

c2

= 9 + 16

c2 = 25

c=5 SOLUTION PROBLEM

If one leg of a right triangle is 6 inches, and the hypotenuse is 10, what is the length of the other leg?

SOLUTION

First, write down the equation for the Pythagorean Theorem. Next, plug in the information you are given. The hypotenuse c is equal to 10 and one of the legs b is equal to 6. Solve for a.

108

To answer this The problem is ash

Chapter 6: Geometry Topics

leg is 4 inches. What

c2 = a 2

+ b2

a2

= c2

-

a2

= 102 - 6 2

b2

a 2 = 100 - 36 2

= 64

a

= 8 inches

a

PROBLEM What is the value of b in the right triangle shown below?

12

b

SOLUTION To answer this question, you need to use the Pythagorean Theorem. The problem is asking for the value of the missing leg.

)otenuse is 10, what

c2 = a

2

b2 = c2

+ b2

b2 = 13 2

-heorem. Next, plug Jal to 10 and one of

a2

-

-

122

b2 = 169 - 144 b2 = 25

b=5

II'"III' !I!!

iii

I,;

1

109

-------t----------­

_C_L_E_P_C_o_"_e9_e_M_a_th_e_m_a_t_ic_s

I

where b is the ba~ Note that the heigl

QUADRILATERALS A polygon is any closed figure with straight line segments

as sides. A quadrilateral is any polygon with four sides.

The points where the sides meet are called vertices (sin­ gular: vertex).

Example:

The area of th

Parallelog rams A parallelogram is a quadrilateral whose opposite sides are parallel. B

o

c

Two angles that have their vertices at the endpoints of the same side of a parallelogram are called consecutive angles. So LA is consecutive to LB; L.._B is consecutive to L C; L C is consecutive to LD; and LD is consecutive to LA.

Rectangles A rectangle is

I

The perpendicular segment connecting any point of a line containing one side of a parallelogram to the line containing the opposite side of the parallelogram is called the altitude of the parallelogram. B A , -.............,.-------~-

----

....- - - - - . , / - - - altitude

• The diagonals of< • If the diagonals rectangle.

o

c

A diagonal of a polygon is a line segment joining any two nonconsecu­ tive vertices. The area of a parallelogram is given by the formula A = bh,

110

i

• If a quadrilateral h

• The area of a reet length and w is the

Chapter 6: Geometry Topics

where b is the base and h is the height drawn perpendicular to that base. Note that the height is the same as the altitude of the parallelogram. segments )ur sides. tices (sin­

Example:

The area of the parallelogram below is: A = bh

A ;ides are parallel.

= (10)(3)

A = 30

Ls 7

,1, l' I

'I

,I

10

'!'

of the same side of consecutive to LB; I ~D is consecutive

,I

.

Rectangles

[

I

A rectangle is a parallelogram with right angles.

)f a line containing lpposite side of the

A

i<'

Ji

I,

ii

B i::>i

D

,I'

altitude

~I

11

II Ii

• The diagonals of a rectangle are equal, AC = BD. • If the diagonals of a parallelogram are equal, the parallelogram is a rectangle.

:1

• If a quadrilateral has four right angles, then it is a rectangle. • The area of a rectangle is given by the formula A = lw, where I is the length and w is the width.

Ytwo nonconsecu­ ~ formula A = bh,

111 .ll

CLEP College Mathematics

Example: The area of the parallelogram below is: A = lw A

= (4) (9)

A

= 36

ABCD is a rhe

(~) (AC)(BD) =

• The diagonals of

4

• If the diagonals a rhombus.

9

0

• If a quadrilateral

Rhombi

• A parallelogram bisects the angles

A rhombus (plural: rhombi) is a parallelogram that has two adjacent sides that are equal.

Squares A square is a rl

• All sides of a rhombus are equal.

• A square is an equ

• The diagonals of a rhombus are perpendicular bisectors of each other.

• A square has all th

• The area of a rhombus can be found by the formula A where d 1 and d 2 are the diagonals.

I

= 2(d 1 X d2),

• In a square, the me ing the length of ar

• The area of a squar the square. • Since all sides of a

112

Chapter 6: Geometry Topics

~1r------,jY

o

c

ABCD is a rhombus. AC

= 4 and BD = 7. The area of the rhombus is

(~) (AC)(BD) = (~) (4)(7) =

14.

• The diagonals of a rhombus bisect the angles of the rhombus. • If the diagonals of a parallelogram are perpendicular, the parallelogram is a rhombus. • If a quadrilateral has four equal sides, then it is a rhombus. • A parallelogram is a rhombus if either diagonal of the parallelogram bisects the angles of the vertices it joins.

at has two adjacent

Squares A square is a rhombus with a right angle. A

S

K

S

S

o

B i

S

c

• A square is an equilateral quadrilateral. • A square has all the properties of rhombi and rectangles.

Irs of each other. la A

1

= -(dl 2

• In a square, the measure of either diagonal can be calculated by multiply­ ing the length of any side by the square root of 2.

X d2),

• The area of a square is given by the formula A the square.

= i, where s is the side of

• Since all sides of a square are equal, it does not matter which side is used.

113 A

CLEP College Mathematics

------------------------t-------­ Example:

Trapezoids

The area of the square shown below is: A

=!.,.z

A

= 36

A trape;

sides. T The me of the nl

6

1 2 length of the diagonal squared. This comes from a combination of the facts

c

The area of a square can also be found by taking - the product of the

that the area of a rhombus is

(~) d] d2 and that d]

=

dz for a square.

The perpendic one base of the trap of the trapezoid.

A pair of angh of base angles.

Example:

The area of the square shown below is:

A=

~d2

A=

~(8)2

A

32

=

2

2

B

Chapter 6: Geometry Topics

Trapezoids A trapezoid is a quadrilateral with two and only two parallel sides. The parallel sides of a trapezoid are called the bases. The median of a trapezoid is the line joining the midpoints of the nonparallel sides.

B

A

I

\

c

1

-.., the product of the

median of trapezoid

o

The perpendicular segment connecting any point in the line containing one base of the trapezoid to the line containing the other base is the altitude of the trapezoid.

Ibination of the facts

: for a square.

A pair of angles including only one of the parallel sides is called a pair Ii

of base angles.

Pairs of base angles

B

b1

•

A

b2

b1

\

altitude ----.1 h

b2

115

CLEP College Mathematics

• The median of a trapezoid is parallel to the bases and equal to one-half their sum. • The area of a trapezoid equals one-half the altitude times the sum of the I bases, or "2h(b1 + b2). • An isosceles trapezoid is a trapezoid whose non-parallel sides are equal. A pair of angles including only one of the parallel sides is called a pair of base angles.

Pairs of base angles

Theorem 1

The perimeter: sure of any pair of (

Theorem 2

The ratio of th polygons is equal tc of the polygons.

Theorem 3

The areas of 1\, the measures of an\" • The base angles of an isosceles trapezoid are equal. • The diagonals of an isosceles trapezoid are equal. • The opposite angles of an isosceles trapezoid are supplementary.

SIMILAR POLYGONS Two polygons are similar if there is a one-to-one corre­ spondence between their vertices such that all pairs of corresponding angles are congruent and the ratios of the measures of all pairs of corresponding sides are equal.

Theorem 4

Two polygons each, and similarly I Note that when nan match: A to A " B to . A

Note that although similar polygons must have the same shape, they may have different sizes. B

CJ

PROBLEM

The lengths of 4 and 7. If the perimi the larger polygon.

116

Chapter 6: Geometry Topics

md equal to one-half

Theorem 1

times the sum of the

The perimeters of two similar polygons have the same ratio as the mea­ sure of any pair of corresponding line segments of the polygons.

-allel sides are equal. jes is called a pair of

Theorem 2

se

The ratio of the lengths of two corresponding diagonals of two similar polygons is equal to the ratio of the lengths of any two corresponding sides of the polygons.

Theorem 3

The areas of two similar polygons have the same ratio as the square of the measures of any pair of corresponding sides of the polygons.

plementary.

1e corre­ pairs of )s of the equal.

~

Theorem 4

Two polygons composed of the same number of triangles similar to each, and similarly placed, are similar. Thus, ABCD is similar to A 'B'CD'. Note that when naming similar polygons, the corresponding letters must match: A to A I, B to B ', C to C, and D to D'. A

A'

same shape, they

7

B

c

B'

0'

PROBLEM

The lengths of two corresponding sides of two similar polygons are 4 and 7. If the perimeter of the smaller polygon is 20, find the perimeter of the larger polygon.

117

CLEP College Mathematics

•

SOLUTION

The circumfe

We know, by theorem, that the perimeters of two similar polygons have the same ratio as the measures of any pair of corresponding sides. If we let sand p represent the side and perimeter of the smaller polygon and s' and p' represent the corresponding side and perimeter of the larger one, we can then write the proportion s .• s' = p . . p", or ~I = S

L

where r is the radiI approximately equ

The are.

I

P

By substituting the given values, we can solve for p'.

4 20 --­ 7

4p' !I,' k"

p'

A full circle is

A line that im

= 140

p' = 35 Therefore, the perimeter of the larger polygon is 35.

'~; .1:

."" '11 "

CIRCLES A circle is a set of points in the same plane equidistant from a fixed point, called its center. Circles are often named by their center point, such as circle 0 below.

A line segmen circle.

A radius of a circle is a line segment drawn from the center of the circle to any point on the circle.

a A chord tJ called a di is twice th

118

Chapter 6: Geometry Topics

The circumference of a circle is the length of its outer edge, given by milar polygons have ding sides.

C = 'TT'd = 2'TT'r

.the smaller polygon imeter of the larger

where r is the radius, d is the diameter, and 'TT'(pi) is a mathematical constant approximately equal to 3.14. The area of a circle is given by

A=

7Tr

2

D.

A full circle is 360°. The measure ofa semicircle (haIfa circle) is 180°. A line that intersects a circle in two points is called a secant.

!I'~l i ii .,'I"I')

luidistant 3re often

:1" !I'I

)W.

i

le center

i!l:

'Iii

A line segment joining two points on a circle is called a chord of the circle.

"lii

Ii

I

I:

, ,

A

A chord that passes through the center of the circle is called a diameter of the circle. The length of the diameter is twice the length of the radius, d = 2r.

119

CLEP College Mathematics

•

A--------4I..---...... B

o

The line passing through the centers of two (or more) circles is called the line of centers.

~,

I:

.'".

"

An angle whose vertex is on the circle and whose sides are chords of the circle is called an inscribed angle (LBAC in the diagrams).

~i'

B

c

B

B

~---

The measure intercepts that arc.

The length of to the circle's circu

circle. Therefore. ar the central angle.

c A sector is the Its area is given by A

A

An angle whose vertex is at the center of a circle and whose sides are radii is called a central angle. The portion of a circle cut off by a central angle is called an arc of the circle.

the radii. The distance fr

from that point to tt

with endpoints at thl P to the diagrammec

segment PO.

120

Chapter 6: Geometry Topics

A~B I

,

'" \

,

\

I I

,

I I

\

,

\ \

'" , .....

,re) circles is called

I

""

_--""

The measure of a minor arc is the measure of the central angle that intercepts that arc. The measure of a semicircle (half a circle) is 180°.

J--­

I

'Iii!

mAB =

sides are chords of

:ams).

0:

=

mLAOB

The length of an arc intercepted by a central angle has the same ratio to the circle's circumference as the measure of the arc is to 360°, the full n

circle. Therefore, arc length is given by 360 ~i

I

:ld \\"hose sides are ut off by a central

X 27rr,

where n = measure of

the central angle. A sector is the portion of a circle between two radii (sector AOB here). Its area is given by A = ~(7rr2), where n is the central angle formed by 360 the radii. The distance from an outside point P to a given circle is the distance from that point to the point where the circle intersects with a line segment with endpoints at the center of the circle and point P. The distance of point P to the diagrammed circle with center 0 is the line segment PB, part of line

segment PO.

121

...

-------tr---------------j

_C_L_E_P_C_o_II_e9.::....e_M_a_th_e_m_a_t_ic_S

Circles that h,

P

tric circles.

B

A line that has one and only one point of intersection with a circle is called a tangent to that circle, and their common point is called a point of tangency.

A circumscritM

polygon. The polyg,

In the diagram, Q and P are each points oftangency. A tangent is always perpendicular to the radius drawn to the point of tangency.

i Points of

Tangency

PROBLEM

A and B are po

length of side AB = Congruent circles

122

are circles whose radii are congruent.

Chapter 6: Geometry Topics

Circles that have the same center and unequal radii are called concen­ tric circles.

ion with a common

A circumscribed circle is a circle passing through all the vertices of a polygon. The polygon is said to be inscribed in the circle.

A. tangent is always ~\.

II

>

PROBLEM A and B are points on a circle Q such that 6AQB is equilateral. If the length of side AB = 12, find the length of arc AB.

em.

lit

I

.a

123

.II.

CLEP College Mathematics

SOLUTION

PROBLEM

To find the length of arc AB, we must find the measure of the central angle LAQB and the measure of radius QA. LAQB is an interior angle of the equilateral triangle LAQB. Therefore, m LAQB = 60°.

Points P and, is the area of secto

Similarly, in the equilateral LAQB, AQ =AB = QB = 12 = r. Given the radius, 1', and the central angle, n, the arc length is given by 60 1 X 2'ITT = X 217 X 12 = - X 217 X 12 = 417. 360 360 6

n

Therefore, the length of arc AB = 417.

FORMULAS FOR AREA AND PERIMETER Figures

SOLUTION

Areas

Area (A) of a:

t l'

"

'.,"'

~:::

A

= i·, where s = side

rectangle

A

= Iw; where 1 = length, w = width

Letting X = a

parallelogram

A

= bh; where b = base, h = height

2

triangle

A

=

1 -bh; where b 2

circle

A

=

m).; where

sector

A

= (~)

l'

trapezoid

= base, h =

17 = 3.14,

l'

17(18

. heIght

= radius

(171'2), where n = central angle, 360 = radius, 17 = 3.14

A=

(~) (h)(b i + h); where h

=

height, b i

and b2 = bases Figures

Perimeters

Perimeter (P) of a:

124

Area c Area

square

r, ~'

120° 360°

-----

square

P

= 4s; where s = side

rectangle

P

= 21 + 2w; where 1 = length, w = width

triangle

P

= a + b + c; where a, b, and c are the sides

Circumference (C) of a circle

C = 17d; where 17 = 3.14, d = diameter

)

= 32417, We

Chapter 6: Geometry Topics

PROBLEM asure of the central an interior angle of 0:.

Points P and R lie on circle Q, m LPQR is the area of sector PQR?

= 120°, and PQ = 18. What

1]=12=1: c length is given by

.

.R

SOLUTION 120° _ Area of sector PQR 360° Area of circle Q

---

\' = width

= height

II

=

radius

. =

1

height

=

~re

central angle,

h

gth,

=

1'V

Letting X

= area of sector PQR. and replacing area of circle Q with

1T(l8 2 ) = 3241T, we get

"

120° X ---360° 3241T

,I

Then X = (l200)(3241T) = 1081T 3600

height, hi

= width

md c are the sides =

.

"

diameter

125

CLEP College Mathematics

DRILL QUESTIONS

3. The center 01 circle. What i

1. What is the area of the following right triangle?

48

(A) 47T (B) 4v17T

(A) 1200 (C) 27T

(B) 672

(D) 2v1rr

(C) 336 (D) 112 I. ,

...

I.:

r.

2. Parallelogram RSTU is similar to parallelogram

wxyz. If LRST =

60°,

4. If the shoft sic twice as long. ' area as this ree

LXYz=

(A)3v1

(A) 60°

(B) 6

(B) 90°

(C) 18

(C) 120°

(D) 36

(D) Not enough information is given 5. Which of the fc (A) 2, 5, 6 (B) 3,4,5

(C) 4, 5, 6 (D) 4,5, 10

126

Chapter 6: Geometry Topics

3. The center of circle 0 is at the origin, as shown. Point (2, 2) is on the circle. What is the circumference of circle O?

(A)47T (B) 4v21T

(C) 27T (D) 2v2n

4. If the short side of a rectangle measures 3 inches, and its long side is twice as long, what is the length of a diagonal of a square with the same area as this rectangle?

Z. If LRST = 60°,

(A) 3v2

(B) 6 (C) 18 (D) 36

5. Which of the following cannot be the lengths of the sides of a triangle? (A) 2, 5, 6

(B) 3,4,5 (C) 4, 5, 6

(D) 4,5,10

127 ..I.

I

~

<

'"

I

---t~.

_C_LE_P_C_o_'_le:::..ge_M_at_h_em_at_ic_s

---------1

6. Find the length of the missing side in this right triangle.

Ii

8. This figure 51' scribed about .

6V3

(A) 6

(A) 16(27T - 1

(B)~

(B)

(C) 12

(C) 87T

(D) J306

(D) 8(7T - 2)

7. Given the intersecting lines and angle measurement in the figure, x =

! .

8-Jh -

I

9. An isosceles ri

(A) 30°, 60:. 9(

(B) 45°,45°. 9t ---_----::~..----------

[,

(C) 0°, 90 c , 90: --------~::..:.....-----£2

(D) 60°,60°. 6(

10. An angle of me (A) straight (A) 55°

(B) supplemeTIl (B) 60°

(C) obtuse (C) 65°

(D) reflex (D) 70°

128

Chapter 6: Geometry Topics

I.'

191e.

8. This figure shows sector ADE equal to a quarter of a circle, circum­ scribed about a square. What is the area of the shaded region? A

OB = 4v2 B

(A) l6(27T - 1)

(B) 8~-16 (C) 87T (D) 8(7T - 2)

in the figure, x

=

9. An isosceles right triangle has angles of (A) 30°, 60°, 90° (B) 45°,45°,90° (C) 0°,90°, 90°

!, II f 2

(D) 60°, 60°, 60°

10. An angle of measure 180° is termed (A) straight (B) supplementary (C) obtuse (D) reflex

I I·'i"

'Ii

129

1

CLEP College Mathematics

Answers to Drill Questions

1. (C)

6. (C)

2. (C)

7. (A)

3. (B)

8. (0)

4. (B)

9. (B)

5. (0)

10. (A)

p .,

.,

"

I

130

Chapter 7

Probability and Statistics The first part ofthis chapter discusses counting principles and how they relate to permutations and combinations, which are the building blocks of probability. The remainder ofthe chapter addresses statistics, introduces data measurements-such as mean, median, mode, and standard deviation-and data analysis of various charts and graphs.

THE FUNDAMENTAL COUNTING PRINCIPLE The fundamental counting principle deals with identify­ ing the number of outcomes of a given experiment and encompasses the counting rule: If one experiment can be performed in m ways, and a sec­ ond experiment can be performed in n ways, then there are m x n distinct ways both experiments can be performed in this specified order. The counting principle can be applied

to more than two experiments.

iI,

:!j

<.

!~ :

PROBLEM

A new line of children's clothing is color-coordinated. Niki's mother bought her five tops, three shorts, and two pairs of shoes for the summer. How many different outfits can Niki choose from?

SOLUTION

How many choices are there for a top?

5

How many choices are there for shorts?

3

How many choices are there for shoes?

2

Apply the counting principle: 5 X 3 X 2 or 30.

Thus, Niki has 30 different outfits.

133

CLEP College Mathematics

PERMUTATIONS

SOLUTION Order is of del 720. Thus, there are third.

A permutation is an arrangement of specific objects in which order is of particular importance. To determine the number of possible permutations, the following formula can be used:

n! nPr= - - ­ (n - r)!

where n is the number of objects in the given set, r is the number of objects being chosen, and! is the notation used for factorial. The factorial of a number is the product of that number and all the numbers less than it down to 1, or n! = n x (n - 1) x (n - 2) x ... x 3 x 2 x 1.

For example, 5! = 5 X 4 X 3 X 2 X 1 = 120. The formula for permutations is a consequence of the counting rule described above.

.,, :, ,

, I , 'I

Example:

Let's look at th

,, ,

SOLUTION How many chi! How many chil How many chl1 Apply the coun

COMBINATIONS

A combin which ordl the numt>­ formula:

6! 6X 5X4X 3X 2X 1 = . Note that 4! cancels out part of 6! 4! 4X 3X2X 1 and we are left with only 6 X 5 = 30. So you really don't have to do that much 6P 2 = -

multiplication. Permutations cancel down to n X (n - 1) X (n - 2) X ... (n - r

+ 1), so even a permutation with large numbers, such as 12P3 becomes

where n i~ the numbE used for fc

12 X 11 X 10 = 1320.

PROBLEM Ten children in Ms. Berea's fifth grade class compete in the school's science fair. First, second, and third place trophies will go to the top three projects. In how many ways can the trophies be awarded?

Note that this fc order is not imponar nations, ABC is the : divide the permutatic

As with permUl combinations. After

1

which the first r factol

and then further canc, 134

--

I

Chapter 7: Probability and Statistics

SOLUTION Order is of definite importance here, so we use loP3 = lOx 9 X 8 = 720. Thus, there are 720 ways these 10 children could finish first, second, or third.

: objects in termine the ing formula

Let's look at the same problem using the counting principle:

SOLUTION How many children have a chance at the first place trophy?

>et, r is the tation used duct of that ) 1, or n! =

10

How many children have a chance at the second place trophy? 9

How many children have a chance at the third place trophy?

8

Apply the counting principle: 10 X 9 X 8 = 720.

of the counting rule

COMBINATIONS A combination is an arrangement of specific objects in which order is not of particular importance. To determine the number of possible combinations, use the following formula:

meels out part of 6!

nCr

=

n! r!(n - r)!

haw to do that much 11 X (n - 2)

x ...

lch as J2P 3 becomes

Jete in the school's go to the top three

)

where n ;s the number of objects in the given set, r is

the number of objects being ordered, and! is the notation

used for factorial.

Note that this formula is similar to the one for permutations, but since order is not important, we have to factor out all the duplications. In combi­ nations, ABC is the same as ACB, BAC, BCA, CAB, and CBA. So we must divide the permutation by the duplications, which occur r! times. As with permutations, we don't have to do too much calculation for combinations. After the cancellations, combinations reduce to a fraction in which the first r factors of n! are in the numerator and r! is in the denominator, 8X7X6 and then further cancellations can take place. For example, 8C3 = , 3x2Xl

'.

II

!i

,

1--------­ ;I

_C_LE_P_C_o_"e_9_e_M_a_t_h_em_at_ic_s

'i

The mathematical formula associated with the probability of a simple m event is peE) = -, where m is the number of favorable outcomes relative to n event E, and n is the total number of possible outcomes.

ji

When you th either a spade or a

,II

II

I

I

Note: If e\"enl the formula abo\e

I

PROBLEM Given that Jeffrey gets two hits out of every three times at bat at every T-ball game, what is the probability that Jeffrey will get a hit his next time up to bat?

PROBLEM

On a field trip information in wid If there were 133 0 fries?

SOLUTION The number of favorable outcomes (i.e., the number of hits) is 2. The 2 number of possible outcomes is 3. Thus, the probability of a hit is 3'

SOLUTION

Before using t:

P( fries), and P( cola

Probability of A or B If the probability of one event is not affected by the probability of another, then the probability of either one ofthem occurring in an experiment is the sum of their individual probabilities; however, one of the instances of both occurring has to be subtracted out to prevent counting twice the prob­ ability of both events. The mathematical formula that allows us to find the probability of obtaining either event A or event B is:

}

peA or B) = peA) + PCB) - peA and B). For example, the probability of selecting a spade or an ace from a deck of cards

wOU~d be tlhe probability of a spade (~~ ~ ~) plus the probability

of an ace (52

=

13),

but each of these probabilities includes the ace of

spades, so we have to subtract one ofthe probabilities that the card is the ace 1 of spades 2). Therefore, the formula is:

(5

P(spade or ace)

= P(spade) + peace) - Peace of spades) 1

= -

4

138

1

1

13

52

+­ - -

16

=-.

52

Pte

Note that you d and since the denorr subtraction problem.

Chapter 7: Probability and Statistics

robability of a simple ~

outcomes relative to

Note: If events A and B are mutually exclusive, peA and B)

'S.

= 0, then

the formula above simplifies to peA or B) = peA) + PCB).

~

times at bat at every ~er a hit his next time

ber of hits) is 2. The ~. ofa hit is

When you think about it, 16 (and not 17) of the cards in a deck are either a spade or an ace or both.

2

3"'

PROBLEM

On a field trip, the teachers counted the orders for a snack and sent the information in with a few people. The orders were for 94 colas and 56 fries. If there were 133 orders, what was the probability of an order for a cola and fries?

SOLUTION

Before using the formula presented above, we must determine P(cola),

P(fries), and P( cola or fries).

, :

,I

I

'y the probability of

ing in an experiment Ie of the instances of lting twice the prob­ allows us to find the

I

94 P(cola) = ­ 133

56 P(fries) = 133

,

':I

I

r

133 P(cola or fries) = 133

,,

peA or B) = peA) + PCB) - peA and B)

BI.

133 94 56 .

= - + - -P (cola and fnes) 133 133 133

133 150 .

= - P (cola and fnes) 133 133

150 133

P(cola and fries) = 133 - 133

-

r an ace from a deck plus the probability includes the ace of

at the card is the ace

of spades)

17

P(cola and fries) = 133

Note that you do not have to actually get a decimal for each fraction, and since the denominators are the same, this reduces to an addition and subtraction problem.

139

CLEP College Mathematics

Probability of One of Two Exclusive Events

Conditional F

Given two events, A and B, if we want the probability that exactly one of these occurs, then the formula becomes: P(exactly one of A or B) = peA)

+ PCB)

- 2 X peA and B).

The factor of 2 in this equation eliminates all possibility of both events taking place, since you want exactly one of events A or B to occur, not both. In the last example, it eliminates the probability of an ace and a spade (the ace of spades) when considering peace) plus the probability of a spade and an ace (also the ace of spades) when considering P( spade).

PROBLEM An ordinary coin is tossed and a fair 6-sided number cube is rolled. What is the probability of getting tails on the coin or getting a 4 on the top face of the number cube, but not both of these events? ,,,

.

, '"

From th

SOLUTION

where P occurrec

We can deri\c is not necessary to as you are able to a

PROBLEM

If Kyle has ei! one red). What is ' white socks to wear

Let A = Event of getting tails on the coin.

...

, ,.,

SOLUTION

Let B = Event of getting a 4 on the number cube.

!I\I

What is the pro white socks:

I I peA) = - and PCB) = ­

2

Then P( exactly one of A, B occurring) =

6

~ + l- (2) (~) (l)

I 2

CHECK To verify this result, let H = heads and T = tails. If the coin is tossed first, the twelve possible outcomes for the coin and number cube are written as HI, H2, H3, H4, H5, H6, TI, T2, T3, T4, T5, T6. The favorable result would consist of six outcomes, namely Tl, T2, T3, T5, T6, H 4. Thus, the 6 1 corresponding probability is 12 or 2'

140

Given that Kyle socks, what is r another pair of, This is because socks left to ch( Thus, the proba white socks is: If the Occurren non-occurrence of e\ If events A and Bare ing to the counting n

Chapter 7: Probability and Statistics

nts

Conditional Probability

iliry that exactly one From the conditional probability formula:

PIA and B).

P(AIB)

ibiliry of both events B to occur, not both. aCe and a spade (the biliry of a spade and

del.

=

peA and B) PCB) ,

where P(AIB) means "the probability of A, given that B has

occurred."

We can derive the multiplication rule: peA and B) = peA IB) X PCB). It is not necessary to fully understand what conditional probability is as long as you are able to apply the counting rules discussed earlier.

TIber cube is rolled. erring a 4 on the top

PROBLEM

I

If Kyle has eight pairs of socks (four white, two black, one blue, and one red). What is the probability that he randomly chooses two pairs of white socks to wear on consecutive days.

SOLUTION What is the probability of selecting one pair of white socks?

,I

(~) (~)

I

2

Given that Kyle already chose a pair of white socks, what is the probability that he chooses another pair of white socks? This is because he now has 3 pairs of white socks left to choose out of 7 pairs of socks left.

If the coin is tossed ber cube are written The favorable result

:. T6, H4. Thus, the

Thus, the probability of drawing two pairs of white socks is:

4

8

3 7

433

"8

X

7' or 14

If the occurrence of event A in no way affects the occurrence or non-occurrence of event B, then events A and B are said to be independent. If events A and B are independent, then peA and B) = peA) X PCB), accord­ ing to the counting rule.

141

"

'I

:1 :1 "

CLEP College Mathematics

Using Tables 1

PROBLEM The probability that a male child is born is 0.5. What is the probabil­ ity that the next three unrelated children born at Memorial Hospital are all boys?

Tables presen down of the stude] president.

St

SOLUTION The birth of three unrelated male children in a row would be considered an independent event.

Male Female

P(3 males) = (0.5) (0.5) (0.5)

=0.125 Thus, the probability ofthree consecutive male unrelated births is 0.125, which is 12.5%.

PROBLEM If a student \\h probability that that

PROBLEM An ordinary coin is tossed and a fair 6-sided cube is rolled. What is the probability of getting tails on the coin and getting a 4 on the top face of the number cube?

SOLUTION Since A and B represent independent events, peA and B) = P (get­ ting tails on the coin and getting a 4 on the number cube) = peA) . PCB) = 1 1 1 -. - - ­ 2 6 12

SOLUTION

By locating the from sophomore rna

PROBLEM

What is the pr, senior?

SOLUTION Locate the seni(

PROBABILITY WORD PROBLEMS As with all prior real-world problems, the context of the problem can vary widely. A table or graph will provide the necessary information to cal­ culate a single outcome, multiple outcomes, conditional probability, or an expected value.

142

9% males + II Nonseniors: !Of Thus, the proba

Chapter 7: Probability and Statistics

Using Tables for Probability Tables present data in an easy-to-read format. This table is a break­ down of the student vote for the winner of the election for student body president.

.nat is the probabil­ )rial Hospital are all

Student Votes for Student Body President t,"

~\-ould be

considered

Male

9%

17%

10%

9%

Female

16%

13%

15%

11%

il PROBLEM

:lated births is 0.125,

If a student who voted for the winner is selected at random, what is the probability that that person is a sophomore and a male?

l

i

,

:'

SOLUTION

is rolled. What is the ,n the top face of the

By locating the correct cell in the table, we see that 17% of votes were from sophomore males. This gives a probability of 0.17.

PROBLEM What is the probability of a randomly selected student not being a senior?

-1 and B) = P (get­ ,,~) =

peA) . PCB) = SOLUTION Locate the senior cells. 9% males

of the problem can information to cal­ .al probability, or an

t

+

11 % females

= 20% seniors

Nonseniors: 100% - 20% = 80%

~.

Thus, the probability of not being a senior is 0.80.

143

.'

::-=-= CDllege Mathematics PROBLEM :h~

Knowing that the selected person is a junior, what is the probability that person is a female?

numbers are measu "center" of the data

Mean SOLUTION

This is a conditional probability problem. To solve such a problem, we need to identify how many are in the given category, in this case juniors. Locate the junior cells: 10% males + 15% females = 25% juniors. The numerical value associated with the given information, that the person is a junior, is the denominator for the fraction.

The mean is the by the total number

Example:

The mean of -L

Of the juniors, identify the percent who are female: 15%. This is the numerator. Thus, the probability that the person is female, given that the person is 15 ajunior, is 25 or 0.60. Note that you don't have to know how many students are juniors or female juniors. Let's say the whole student population is x. Then, according to the table, .25x are juniors and .15x are female juniors, so P(femaleljunior) .15x = --. The x's cancel out, and the answer is .60. .25x

PROBLEM

Find the mean] $5/hour, $8/hour. S1~

SOLUTION

The mean hour]' PROBLEM

$5 ­

Ifin this election there were 500 votes for the winner, how many females cast votes for the winner?

Median SOLUTION

Total number of females: 16%

+ 13% + 15% + 11% = 55%

55% of 500 = 275. Thus, 275 females cast votes for the winner.

MEASURES OF CENTRAL TENDENCY There are three ways to describe the tendency of a set of data, meaning what a "typical" value is: the mean, median, and mode. All three of these

144

The median is th to be put in order. SIT number ofvalues in th is an equal number of set has an even numb the median.

Chapter 7: Probability and Statistics

numbers are measures of central tendency. They describe the "middle" or "center" of the data.

is the probability that

Mean ~

The mean is the arithmetic average. It is the sum ofthe variables divided by the total number of variables.

such a problem, we this case juniors.

, = 25% juniors. The . that the person is a

Example:

The mean of 4, 3, and 8 is Ie: 15%. This is the

4+3+8 15 ---=-=5 3 3

en that the person is "

:1

PROBLEM

ients are juniors or

; x.

Then, according

I~I

"

I

Find the mean hourly wages for four company employees who make $5/hour, $8/hour, $12/hour, and $15/hour.

so P(femaleljunior)

SOLUTION The mean hourly wage is the average. $5

. how many females

+ $8 + $12 + $15 = 4

$40 = $10jhour

4

Median

lOo=55%

r the winner.

The median is the middle value in a set. The set of numbers first needs to be put in order, smallest to largest or vice versa. When there is an odd number of values in the set, the median is simply the middle value, and there is an equal number of values larger and smaller than the median. When the set has an even number of values, the average of the two middle values is the median.

~t

of data, meaning .-\11 three of these

145

:1

:, "

-; = College Mathematics &le:

SOLUTION

The median of (2,3,5,8,9) is 5.

The mean is tt dividing by the num

5+9

The median of(2, 3, 5, 9, 10, II) is -2- = 7. Note that the median doesn't have to be an element of the set. In fact, with an even number of values, it often is not.

500 + 600 + 800 ­

Note: The above rules apply, even if some numbers are repeated. For example, the median of (2, 3, 3, 4, 6, 8, 9) is 4.

Mode

I'

Ii

The median is 1 and they are already

The mode is the most frequently occurring value in the set of values.

:I

Example:

The mode of (4,5,8,3,8,2) would be 8, since it occurs twice whereas the other values occur only once. There can be two modes (in which case the set is called bimodal), or in fact as many modes as you have values.

Example:

The set 2, 2, 3, 3, 5, 5, 8, 8, 9, 9 has five modes, since each value is mentioned twice. If each number in a set of numbers appears only once, there is no mode. Example: (2, 3, 5, 8, 9) has no mode. Of course, in this case the mode isn't a very useful measure of central tendency, and the mean or median would describe the data better.

PROBLEM For this series of observations, find the mean, median, and mode. 500,600,800,800,900,900,900,900,900,1,000,1,100

The mode is th> which appears five ti

: I

Comparing the r of Distributions

It is sometimes 1 the mean, median. ar having to calculate th Step 1: If a bar graph

in the problem. The graph will approximately norma the other. • A graph skewed te The order of the th order). • A graph skewed to The order of the 1 alphabetical order) • A graph that is ap: like Figure 3, and t

146

Chapter 7: Probability and Statistics

SOLUTION

The mean is the value obtained by adding all the measurements and dividing by the number of measurements. m of the set. In fact,

500 + 600

+ 800 + 800 + 900 + 900 + 900 + 900 + 900 + 1,000 + 1,100 11

~rs

are repeated. For 9,300 = 845.45

= 11

1

the set of values.

The median is the value appearing in the middle. We have 11 values, and they are already in order, so here the sixth, 900, is the median. The mode is the value that appears most frequently. That is also 900, which appears five times.

:curs twice whereas

llled bimodal), or in

Comparing the Mean, Median, and Mode in a Variety of Distributions It is sometimes useful to make comparisons about the relative values of

the mean, median, and mode. This section presents steps to do that without having to calculate the exact values of these measures. If a bar graph is not provided, sketch one from the information given in the problem.

Step 1.:

since each value is appears only once, . Of course, in this ency. and the mean

The graph will either be skewed to the left, skewed to the right, or approximately normal. A skewed distribution has one of its tails longer than the other. • A graph skewed to the left will look like Figure 1; its left tail is longer. The order of the three measures is mean < median < mode (alphabetical order).

m. and mode. 1.100

• A graph skewed to the right will look like Figure 2; its right tail is longer. The order of the three measures is mode < median < mean (reverse alphabetical order). • A graph that is approximately normal (also called bell-shaped) will look like Figure 3, and the mean = median = mode.

147

CLEP College Mathematics

PROBLEM

Estimate the 1 how the mean com] - ­ Mode

Mode-­

Figure 1

Figure 2

Mean Median Mode Figure 3

Step 2: Write the word "mode" under the highest column of the bar graph,

because the mode is the most frequent. If the graph is skewed right or left, the positioning ofthe mode establishes the order for the remaining two terms according to the information provided above. If the graph is approximately symmetrical, the value of all three terms are approximately equivalent.

3

2

1

PROBLEM

On a trip to the Everglades, students tested the pH ofthe water at differ­ ent sites. Most ofthe pH tests were at 6. A few read 7, and one read 8. Select the statement that is true about the distribution of the pH test results.

o

1

2

3

A. The mode and the mean are the same.

R The mode is less than the mean.

e. The median is greater than the mean. I

D. The median is less than the mode.

SOLUTION

Sketch a graph. The graph is skewed to the right. The mode is furthest left. Thus, mode

< median < mean. Choices (A), (C), and (D) do not coincide with what has been estab­ lished in terms of relative order. (B) is the only choice that does follow from our conclusions. Thus, (B) is the correct response.

SOLUTION

Graph (A) is sy be about 0, since it ~l with a median of abc of 4-7 than it will Graph (C) is left-ske1 toward the lower Yall few extreme high Yal slightly larger due to

MEASURES OF ,

In addition to IT described with meaSl where the middle of manufacturer of light

148

Chapter 7: Probability and Statistics

~

PROBLEM Estimate the median of each distribution shown below and describe how the mean compares to it.

'.lean

'.~edian

'.lode igure 3

of the bar graph, right or left, ;:>maining two terms )h is approximately ely equivalent. ill

k~wed

C

j

the water at differ­ one read 8. Select t~st results.

st left. Thus, mode

32101234 (A)

o

012345678 (C)

-2

1

2

3

4

5

678

(~

o

2 (0)

4

6

SOLUTION Graph (A) is symmetric with a median at about 0. The mean will also be about 0, since it approximates a normal curve. Graph (B) is right-skewed with a median of about 3. The mean will be pulled more toward the values of 4 -7 than it will be toward 1- 2, so the mean will be greater than 3. Graph (C) is left-skewed with a median of about 5. The mean will be pulled toward the lower values, so it is less than 5. Graph (D) is symmetric with a few extreme high values. The median is about 0, and the mean will be just slightly larger due to the outliers around 6.

at has been estab­ [ does follow from

MEASURES OF VARIABILITY In addition to measures of central tendency, distributions need to be described with measures of variability, or spread. It is not enough to know where the middle of a distribution is, but also how spread out it is. A manufacturer of light bulbs would like small variability in the amount of

149

CLEP College Mathematics

hours the bulbs will likely burn. A track coach who needs to decide which athletes go on to the finals may want larger variability in heat times because it will be easier to decide who are truly the fastest runners.

The range of a data set is simply the difference between the maximum value and the minimum value.

The range is rarely a good choice to represent the data set, especially because it can be affected by outliers. Look, for example, at graph (D) ofthe last problem. Its range is - 3 to + 7, but that doesn't show that almost all of the data lie between - 2 and + 2. For data that are fairly symmetric, the standard deviation and variance are useful measures of variability.

Mar 2.1

Compute the \,

SOLUTION To compute the the mean. In this cia follows: S2

= _I_

n - 1

The variance tells us how much variability exists in a dis­ tribution. It is the "average" of the squared differences between the data values and the mean. The variance is calculated with the formula

1

--12 ­

+ ... ­ 1

= _[Ol

11

.

= 0.08'+

where n is the number of data points, Xi represents each data value, and x is the mean. The standard deviation is the square root of the variance. The formula for the standard deviation is therefore

The standard

j

,)0.084 = 0.290. Ihu

rainfall in Birminaha 1::

from that each month

The standard deviation is used for most applications in statistics. It can be thought of as the typical distance an observation lies from the mean.

DATA ANALYSIS

Data analysis oft.

such as bar graphs. lir more intuitive underst

PROBLEM The average monthly rainfall in inches in Birmingham, England, is shown in the table below.

Bar Graphs

Bar graphs are It

The following bar gr. 150

Chapter 7: Probability and Statistics ~ds to h~at

[1

decide which times because

crs. ~

between

Compute the variance and standard deviation of the monthly rainfall.

SOLUTION

, data set, especially ~. at graph (D) ofthe )\\ that almost all of ~iation and

To compute the variance and standard deviation, we must first compute 7 · data set, x = 26.5 Th e vanance . .IS compute d as = 2.22. the mean. In th IS follows: 12

variance s2

s in a dis­ fferences 3riance is

1 n-1L-­ 1

= - - '" (Xi

- x)2

-[(2.3 - 2.225)2 12 - 1

+ ... + (2.6

+ (1.9 -

2.225)2

+ (2.1

- 2.225)2

- 2.225)2]

1 11 0.084

= -[0.9225] =

mts each ~ation is

The standard deviation is the square root of the vanance, or JO.084 = 0.290. Thus, from this data set, we can say that the monthly mean

"e

rainfall in Birmingham is 2.225 inches and it varies by about 0.29 inches from that each month, sometimes more, sometimes less.

DATA ANALYSIS

s in statistics. It can from the mean.

Data analysis often involves putting numerical values into picture form, such as bar graphs, line graphs, and circle graphs. In this manner, we gain a more intuitive understanding of the given information.

19ham, England, is

Bar Graphs Bar graphs are used to compare amounts of the same measurements.

The following bar graph compares the number of bushels of wheat and

151

l

_C_L_E_P_C_o_"_eg_e_M~a_th_e_m_a_tl_'c_s

~_ _~

corn produced on a farm from 1975 to 1985. The horizontal axis for a bar graph consists of categories (e.g., years, ethnicity, marital status) rather than values, and the widths of the bars are uniform. The emphasis is on the height of the bars. Contrast this with histograms, discussed next.

PROBLEM According to the graph below, in which year was the least number of bushels of wheat produced?

J

~_------------.j Histograms

A histogram i primarily for contil wide spread. The h be of uniform size. the bar denotes the

The histogram on a ship during a r

300..,----------------------­

...

250

+---------r...,--~------.-:::r-----­

200

+E'I_--B_----f~_____E'I____,...",-B______f3______E'I____B-B_­

Example: 10 'c.O

c

"0 C

8

Q)

D. U) '+­

o

6

>,

100

+€::t--___€3r-.g....,.---j:3_-€::t--~,.....,__8'~3__:-I==I----€3,.....,-.g..f:-­

()

c

Q)

I-:

4

::J

cr

~ u..

2

rr1 19751976 1977 1978 1979 1980 1981 198219831984 1985

I

Wheat

0

Pa

Corn

Number of bushels (to the nearest 5 bushels) of wheat and corn produced by Farm RQS, 1975-1985

SOLUTION By inspecting the graph, we find that the shortest bar representing wheat production is the one for 1976. Thus, the least number of bushels of wheat produced in 1975-1985 occurred in 1976.

The intervals h; $500, two spent ben' graph the precise arr The distributior to $620. The data arc observations will be is about $180, but th $620. There are no e

Line Graphs Line graphs ar related subjects. Lin 152

Chapter 7: Probability and Statistics

izontal axis for a bar tal status) rather than lhasis is on the height ~" t.

the least number of

Histograms A histogram is an appropriate display for quantitative data. It is used primarily for continuous data, but may be used for discrete data that have a wide spread. The horizontal axis is broken into intervals that do not have to be of uniform size. Histograms are also good for large data sets. The area of the bar denotes the value, not the height, as in a bar graph. The histogram below shows the amount of money spent by passengers on a ship during a recent cruise to Alaska.

Example: F'"

~

10 'QJ)

c

'6 c

8

OJ

Q.

(f)

'0

6

>,

u

c

OJ

4

:::J

cr ~

I.L..

2

480 500 520 540 560 580 600 620 640 660 '984 1985

Spending

Passenger spending during cruise to Alaska

corn produced by

it bar representing mber of bushels of

The intervals have widths of $20. One person spent between $480 and $500, two spent between $500 and $520, and so on. We cannot tell from the graph the precise amount each individual spent. The distribution has a shape skewed to the left with a peak around $600 to $620. The data are centered at about $590-this is about where half of the observations will be to the left and half to the right. The range of the data is about $180, but the clear majority of passengers spent between $560 and $620. There are no extreme values present or gaps within the data.

Line Graphs Line graphs are very useful in representing data on two different but related subjects. Line graphs are often used to track the changes or shifts 153

---------t----------­

_C_LE_P_C_o_"e_9_e_M_a_t_h_em_at_ic_s

in certain factors. In the next problem, a line graph is used to track the changes in the amount of scholarship money awarded to graduating seniors at a particular high school over the span of several years.

how a family's bu percentages.

PROBLEM PROBLEM

According to the line graph below, by how much did the scholarship money increase between 1987 and 1988?

Using the bue month would plan

$250,000

Ho",

$200,000

$150,000

$100,000

$50,000

Gifts/Crla

81

82 83 84

85 86 87 88 89 90

Amount of scholarship money awarded to graduating seniors, West High, 1981-1990

3

ii

SOLUTION

To find the increase in scholarship money from 1987 to 1988, locate the amounts for 1987 and 1988. In 1987, the amount of scholarship money is halfway between $50,000 and $100,000, or $75,000. In 1988, the amount of scholarship money is $150,000. The increase is thus $150,000 - 75,000 = $75,000.

Pie Charts Circle graphs (or pie charts) are used to show the breakdown of a whole picture. When the circle graph is used to demonstrate this breakdown in terms of percents, the whole figure represents 100% and the parts of the circle graph represent percentages of the total. When added together, these percentages add up to 100%. The circle graph in the next problem shows

154

SOLUTION

To find the am centage allotted to h The family plans to

Stemplots

A stemplot. als, variate data as well. forms a plot much It a class of 32 student:

Chapter 7: Probability and Statistics

is used to track the to graduating seniors rs.

how a family's budget has been divided into different categories by using percentages.

PROBLEM 1

did the scholarship

Using the budget shown below, a family with an income of $3,000 a month would plan to spend what amount on housing? Auto - 15%

r­

19

90

Miscellaneous - 8%

ng seniors,

987 to 1988, locate . scholarship money n 1988, the amount S150,000 - 75,000

:akdown of a whole this breakdown in ld the parts of the ded together, these ~xt problem shows

Family budget

SOLUTION To find the amount spent on housing, locate on the pie chart the per­ centage allotted to housing, or 30%. Then calculate 30% of $3,000 = $900. The family plans to spend $900 on housing.

Stemplots A stemplot, also called stem-and-leaf plot, can be used to display uni­ variate data as well. It is good for small sets of data (about 50 or less) and forms a plot much like a histogram. This stemplot represents test scores to:­ a class of 32 students.

-:..: ~

CLEP College Mathematics

3

3

4

the response variable been observed.

5 6

379

7

22 57

8

126888999

900013345567 10 Key: 6

0 0 0 0 3 represents a score of 63 Test scores

The values on the left of the vertical bar are called the stems; those on the right are called leaves. Stems and leaves need not be tens and ones-they may be hundreds and tens, ones and tenths, and so on. A good stemplot always includes a key for the reader so that the values may be interpreted correctly.

PROBLEM

Describe the distribution of test scores for students in the class using the stemplot above.

SOLUTION

The distribution of the test scores is skewed toward lower values (to the left). It is centered at about 89 with a range of 67. There is an extreme low value at 33, which appears to be an outlier. Without it, the range is only 37, about half as much. The test scores have a mean of appropriately 85.4, a median of 89, and a mode of 100.

PROBLEM

Data collected j grams of fat per sery: which are the explar tionship between fat

SOLUTION

The explanator: calories. The number calories in the snack.

Scatterplots can examined univariate I tion's shape, center. : we focus on its shap~ unusual features. Bel history of the Nationa 1 of the 30 players. games, is noted.

'0

~

o (/; ~

'6 Q.

ANALYZING PATTERNS IN SCATTERPLOTS consist of two variables. Typically, we are looking for an association between these two variables. The variables may be categori­ calor quantitative; in this section, we focus on quantitative bivariate data. Scatterplots are used to visualize quantitative bivariate data.

40000· 35000 . 30000 . 25000 . 20000· 15000 • 10000 • 5000 ­ 4

Bivariate data

The two variables under study are referred to as the explanatory variable (x) and the response variable (y). The explanatory variable explains or predicts

156

The shape of a pI The direction of a scat the explanatory varial of the data. The streni scatterplot are.

d rhe stems; those on . rens and ones-they :m. A. good stemplot s may be interpreted

!rs in rhe class using

lTd lower values (to There is an extreme it. rhe range is only .ppropriately 85.4, a

we are looking for :5 may be categori­ ni\'e bivariate data. lara. ~xplanatory variable

e.iplains or predicts

Chapter 7: Probability and Statistics

the response variable. The response variable measures the outcomes that have been observed.

PROBLEM Data collected from the labels on snack foods included the number of grams of fat per serving and the total number ofcalories in the food. Identify which are the explanatory and response variables when looking for a rela­ tionship between fat grams and calories.

SOLUTION The explanatory variable is grams of fat and the response variable is calories. The number of grams of fat would be a predictor of the number of calories in the snack. Scatterplots can tell us if and how two variables are related. When we examined univariate data in the preceding sections, we described a distribu­ tion's shape, center, spread, and outliers/unusual features. In a scatterplot, we focus on its shape, direction, and strength, and we look for outliers and unusual features. Below is a scatterplot of the top 30 leading scorers in the history ofthe National Basketball Association (NBA). Each point represents 1 of the 30 players. Michael Jordan, who scored 32,292 points in 1,072 games, is noted. Michael Jordan

40000 1:l 35000 ~ 30000 o (A 25000 ~ 20000

,

§ 15000 ~

c.. 10000 5000

"'" .'

.:•.•. .. '" i i i

I

400

600

800

1000

1200

I

I

1400

1600

Games Played NBA Top 30 Scorers

The shape of a plot is usually classified as linear or nonlinear (curved). The direction of a scatterplot tells what happens to the response variables as the explanatory variable increases. This is the slope of the general pattern of the data. The strength describes how tight or spread out the points of a scatterplot are. 157

1

------=---------------------.;t CLEP College Mathematics

,I

,I

The three scatterplots below show comparisons of various directions of the data. The top two have a clear linear trend, whereas the third scatter­ plot shows a random type of distribution with no clear association among points.

The scatterpll the general pattern the outlier, which r

.

' • <0.

.. . .. . . . ...

:

'

linear positive

.. '"

...

PROBLEM

..

..

.

In the scattefl: beginning of this se variable. Describe t

linear negative

".. "

..

'

..

..

SOLUTION

. . no association

When analyzing a scatterplot, it is also a good idea to look for outliers, clusters, or gaps in the data. The scatterplot below has an obvious gap. There is an overall positive, linear association, but we should find out the reason for the gap. o o o

o

158

0

o

0

0 0

The explanato variable is the numl tive. There are no 0 games. The two ext careers, appear to t(

Chapter 7: Probability and Statistics

The scatterplot below has an obvious outlier. An outlier falls outside the general pattern of the data. There could be several possible reasons for the outlier, which merits investigation.

If \"arious directions ~as ~

the third scatter­ association among

o

o o

o

o

o

0

o 0

000

PROBLEM

In the scatterplot of the top 30 scorers in NBA history shown at the beginning of this section, identify the explanatory variable and the response variable. Describe the association between the two variables.

SOLUTION

The explanatory variable is the number of games played. The response variable is the number of points scored. The relationship is linear, and posi­ tive. There are no outliers, but there is a large gap between 1, I00 and 1,400 games. The two extreme points, probably a couple of players with very long careers, appear to follow the general pattern of the data.

o look for outliers, )byious gap. There ind out the reason

159

>.

CLEP College Mathematics

DRILL QUESTIONS

1. What is the value of the median in the following sample? {19, 15,21,24, 11 }

3. Mrs. Smith tea tered to her c1< between 80 an 79. Which one dents' exam sc (A) The media

(A) 18

(B) The mean

(B) 19

(C) The media

(C) 20

(D) The medial

(D) 21

2. The sample standard deviation, s, of a group of data is given by the

=

L?- 1 (Xi -

X)2

.

_

4. A jar contains ~ that a randomh

X

(A) _1 20

represents the mean, and n represents the number of data points. What is

(B) _1 10

the sample standard deviation for the following data (rounded off to the

(C) ~ 5

formula S

I -

n-1

,

where

Xi

represents each data pomt,

nearest hundredth)? {12, 15,20,21} (D)

(A) 3.67

~

4

(B) 3.81

(C) 4.24 (D) 8.25

5. From four textl physics, a resea textbook. How r (A) 12 (B) 18

(C) 24 (D) 36

160

Chapter 7: Probability and Statistics

.. - - - - - - - - - - - - - - - - - - - - - - - ­

~

3. Mrs. Smith teaches a class of 150 students. On a recent exam adminis­ tered to her class, 100 students scored 90 or better, 40 students scored between 80 and 89, and the remaining students scored between 70 and 79. Which one of the following statements is correct concerning the stu­ dents' exam scores?

pIc? {19, 15,21,24,

(A) The median equals the mean. (B) The mean is less than the mode.

(C) The median is less than the mean. (D) The median is greater than the mode.

4. A jar contains 20 balls, numbered 1 through 20. What is the probability that a randomly chosen ball has a number on it that is divisible by 4?

ata is given by the ~

each data point, i

(A) _1

20

iata points. What is

(B) _1

10 (rounded off to the

(C) ~ 5 1 (D) ­ 4 5. From four textbooks of college math and three textbooks of college physics, a researcher will choose two math textbooks and one physics textbook. How many different groupings of three textbooks are there? (A) 12 (B) 18

i

!~

I

~M

(D) 36

11

ii fI

I~

i~

161

1

j ".!'

'I

!

CLEP College Mathematics

6. Four out of five dentists surveyed recommended professional teeth cleaning twice a year. Of the dentists who made this recommendation, 10% said they are underpaid. If a dentist from this survey is randomly selected, what is the probability that this selected dentist has recom­ mended professional teeth cleaning twice a year and also said that he or she is underpaid?

(A)~ (B)

!

8. In a class of 30 majors, and 40: female students (A) 16

(B) 14

25

(C) 12

4

(D) 10

15

(C) ~ 5 (D)

1

2­

10

7. An "unusual" number cube has five faces, numbered 1 through 5. This number cube will be rolled twice. Assuming that each of the five num­ bers has the same likelihood of appearing, what is the probability that at least one of the following events will occur? Event 1: The number 3 will appear on the first roll.

9. The mean weigl When two peopl is 166 pounds. \ who leave the ro (A) 136

(B) 145 (C) 154 (D) 163

Event 2: An even number will appear on the second roll.

(A)

~

5

(B) ~ 5

10. At the MNP cor number contains sive. The left-me how many differ

(C) ~ 25

(A) 32,768

(D)~

(B) 8192

25

(C) 6720 (D) 4096

162

Chapter 7: Probability and Statistics

d professional teeth his recommendation, s suryey is randomly j dentist has recom­ d also said that he or

8. In a class of 30 students, 24 are female. There are a total of 12 history majors, and 4 of these history majors are male. How many non-history female students are there? (A) 16 (B) 14

(C) 12 (D) 10

9. The mean weight of a group of ten people in a room is 160 pounds. When two people leave, the mean weight of the remaining eight people is 166 pounds. What is the mean weight, in pounds, of the two people who leave the room? (A) 136

'd 1 through 5. This ch of the five num­ e probability that at

(B) 145 (C) 154 (D) 163

roll. 10. At the MNP company, each person gets an identification number. The number contains five digits, chosen from the digits 1 through 8, inclu­ sive. The left-most digit must be 2. If repetition of any digit is allowed, how many different identification numbers are possible? (A) 32,768 (B) 8192

(C) 6720 (D) 4096

i Ii

i

163

CLEP College Mathematics

Answers to Drill Questions

1.

(B)

6. (A)

2.

(C)

7. (C)

3. (B)

8. (A)

4. (D)

9. (A)

5. (B)

10. (D)

Chapter 8

Logic The topic of logic encompasses a wealth of subjects related to the principles of reasoning. Here, we will be concerned with logic on an ele­ mentary level, and you will recognize the principles introduced here because not only have you been using them in your everyday life, you also have seen them in one form or another in this book. The chapter is designed to famil­ iarize you with the terminology you might be expected to know as well as thought processes that you can use in everyday life. Sentential calculus is the "calculus of sentences," a field in which the truth or falseness of assertions is examined by using algebraic tools. We will approach logic from a "true" or "false" perspective here. This chapter con­ tains many examples of sentences to illustrate the terms that are defined.

SENTENCES A sentence is any expression that can be labeled either true or false.

Examples:

Expressions to which the terms "true" or "false" can be assigned include the following: 1. "It is raining where I am standing." 2. "My name is George."

3. "1 + 2

=

3"

767

CLEP College Mathematics

i

--------------------------f---------­ '!

Examples:

Logical Truth

Expressions to which the terms "true" or "false" cannot be assigned include the following:

A sentence is false; that is, the del

1. "I will probably be healthier if I exercise." 2. "It will rain on this day, one year from now." 3. "What I am saying at this instant is a lie." Sentences can be combined to form new sentences using the connec­ tives AND, OR, NOT, and IF-THEN.

Examples:

Example:

Either Mars is ,

Logical Falsity

A sentence is II true; that is, the sem

The sentences 1. "John is tired." 2. "Mary is cooking."

Example:

Mars is a plane!

can be combined to form 1. "John is tired AND Mary is cooking."

Logical Indetermin

2. "John is tired OR Mary is cooking."

A sentence is II neither logically true

3. "John is NOT tired." 4. "IF John is tired, THEN Mary is cooking."

Example:

Logical Properties of Sentences

Einstein was a r

Consistency

A sentence is consistent if and only if it is possible that it is true. A sentence is inconsistent if and only if it is not consistent; that is, if and only if it is impossible that it is true.

Example:

"At least one odd number is not odd" is an inconsistent sentence.

168

Logical Equivalent

Two sentences a one ofthe sentences t only if it is impossibl

Example:

"Chicago is in I equivalent to "Pittsbu

Chapter 8: Logic 1100

Logical Truth

cannot be assigned

A sentence is logically true if and only if it is impossible for it to be false; that is, the denial of the sentence is inconsistent.

Example:

Either Mars is a planet or Mars is not a planet. cos using the connec­ Logical Falsity

A sentence is logically false if and only if it is impossible for it to be true; that is, the sentence is inconsistent.

Example:

Mars is a planet and Mars is not a planet.

Logical Indeterminacy (Contingency)

A sentence is logically indeterminate (contingent) if and only if it is neither logically true nor logically false.

Example:

Einstein was a physicist and Pauling was a chemist.

'e that it is true. A that is, if and only

~nt

sentence.

Logical Equivalent of Sentences

Two sentences are logically equivalent if and only if it is impossible for one ofthe sentences to be true while the other sentence is false; that is, if and only ifit is impossible for the two sentences to have different truth values.

Example:

"Chicago is in Illinois and Pittsburgh is in Pennsylvania" is logic':":'::­ equivalent to "Pittsburgh is in Pennsylvania and Chicago is in Illinois'-­

1 iii!

CLEP College Mathematics

•

Inverse

STATEMENTS

The inverse of ( A statement is a sentence that is either true or false. but not both.

The following terms and their definitions should become familiar to you. Their logic is probably familiar, even though you haven't given it a label before.

Biconditional

The statement I called a biconditiona

Validity Conjunction

If a and b are statements, then a statement of the form "a and b" is called the conjunction of a and b, denoted by a 1\ b.

An argument i:,; sions must also be tr

Intuition Disjunction

Intuition

The disjunction of two statements a and b is shown by the compound statement "a or b," denoted by a V b.

The negation of a statement q is the statement "not q," denoted by ~q.

PROBLEM

1. If a person is steal

2. If a line is perpenc lar bisector of the

Implication

The compound statement "if a, then b," denoted by a - b, is called a conditional statement or an implication. "If a" is called the hypothesis or premise of the implication, "then b" is called the conclusion of the implica­ tion. Further, statement a is called the antecedent of the implication, and statement b is called the consequent of the implication. Converse

The converse of a - b is b - a. Contrapositive

170

~

Write the imel whether the inverse i

Negation

The contrapositive of a - b is

is the

~b

-

~a.

3. Dead men tell no .

SOLUTION

The inverse of both the hypothesis a

1. The hypothesis of is "he is breaking· not stealing." The. the law." The inverse is fa stealing. Clearly. a the law.

Chapter 8: Logic

Inverse

The inverse of a

!

-+

b is

~a -+ ~b.

or false, but

Biconditional

The statement of the form "p if and only if q," denoted by p called a biconditional statement.

lId become familiar to hayen't given it a label

+-+

q, is

Validity

An argument is valid if the truth of the premises means that the conclu­ sions must also be true.

:he form "a and b" is

Intuition Intuition

is the process of making generalizations on insight.

wn by the compound PROBLEM

Write the inverse for each of the following statements. Determine whether the inverse is true or false. "not q," denoted by

1. If a person is stealing, he is breaking the law. 2. If a line is perpendicular to a segment at its midpoint, it is the perpendicu­ lar bisector of the segment. 3. Dead men tell no tales.

Jy a -+ b, is called a ~d the hypothesis or !Sion of the implica­ :he implication, and

SOLUTION

The inverse of a given conditional statement is formed by negating both the hypothesis and conclusion of the conditional statement. 1. The hypothesis of this statement is "a person is stealing"; the conclusion is "he is breaking the law." The negation of the hypothesis is "a person is not stealing." The inverse is "if a person is not stealing, he is not breaking the law." The inverse is false, since there are more ways to break the law than by stealing. Clearly, a murderer may not be stealing but he is surely breaking the law. 171 J.

CLEP College Mathematics

2. In this statement, the hypothesis contains two conditions: a) the line is perpendicular to the segment; and b) the line intersects the seg­ ment at the midpoint. The negation of (statement a and statement b) is (not statement a or not statement b). Thus, the negation of the hypothesis is "The line is not perpendicular to the segment or it doesn't intersect the segment at the midpoint." The negation of the conclusion is "the line is not the perpendicular bisector of a segment." The inverse is "if a line is not perpendicular to the segment or does not intersect the segment at the midpoint, then the line is not the perpendicu­ lar bisector of the segment." In this case, the inverse is true. If either ofthe conditions holds (the line is not perpendicular; the line does not intersect at the midpoint), then the line cannot be a perpendicular bisector. 3. This statement is not written in if-then form, which makes its hypothesis and conclusion more difficult to see. The hypothesis is implied to be "the man is dead"; the conclusion is implied to be "the man tells no tales." The inverse is, therefore, "If a man is not dead, then he will tell tales." The inverse is false. Many witnesses to crimes are still alive but they have never told their stories to the police, probably out of fear or because they didn't want to get involved.

NECESSARY AI

Let P and Q rE ment in which P is sary condition for 1

Example:

Consider the 5t it rains" is a sufficiE to the movies" is a r

Note that for tt dition for which Jan, Likewise, "Jane wil from a rainy weath~ another likely conel is a necessary eondi In the biconditi sufficient condition

BASIC PRINCIPLES, LAWS, AND THEOREMS Example:

I. Any statement is either true or false. (The Law of the Excluded Middle) 2. A statement cannot be both true and false. (The Law of Contradiction) 3. The converse of a true statement is not necessarily true. 4. The converse of a definition is always true. 5. For a theorem to be true, it must be true for all cases.

Consider the 5t, gets paid" is both a 51 Rick's working is a s Thus, we ha"e from the preceding s

7. The inverse of a true statement is not necessarily true.

11. If a given statem the hypothesis 01 conclusion of thi

8. The contrapositive of a true statement is true and the contrapositive of a false statement is false.

12. If a given statem are sufficient bUI

9. If the converse of a true statement is true, then the inverse is true. Like­ wise, if the converse is false, the inverse is false.

13. If a given statem are neither suffic

6. A statement is false if one false instance of the statement exists.

10. Statements that are either both true or both false are said to be logically equivalent.

172

Chapter 8: Logic

mditions: a) the line e intersects the seg­ a and statement b) is rion of the hypothesis t doesn't intersect the :lusion is "the line is

NECESSARY AND SUFFICIENT CONDITIONS Let P and Q represent statements. "If P, then Q" is a conditional state­ ment in which P is a sufficient condition for Q, and similarly Q is a neces­ sary condition for P.

~ segment or does not s nor the perpendicu-

Example:

Consider the statement: "If it rains, then Jane will go to the movies." "If it rains" is a sufficient condition for Jane to go to the movies. "Jane will go to the movies" is a necessary condition for rain to have occurred.

Iltions holds (the line ~ midpoint), then the

Note that for the statement given, "Ifit rains" may not be the only con­ dition for which Jane goes to the movies; however, it is a sufficient condition. Likewise, "Jane will go to the movies" will certainly not be the only result from a rainy weather condition (for example, "the ground will get wet" is another likely conclusion). However, knowing that Jane went to the movies is a necessary condition for rain to have occurred.

makes its hypothesis is implied to be "the n tells no tales." The ill tell tales." ~ still alive but they It of fear or because

IS ~ Excluded Middle)

In the biconditional statement "P if and only if Q," P is a necessary and sufficient condition for Q, and vice versa.

I

Example: 11

II "1

\,

. of Contradiction)

,'I

,I

[Ii :,:

lie.

Consider the statement "Rick gets paid if and only if he works." "Rick gets paid" is both a sufficient and necessary condition for him to work. Also, Rick's working is a sufficient and necessary condition for him to get paid. Thus, we have the following basic principles to add to our list of ten from the preceding section:

rlent exists.

11. If a given statement and its converse are both true, then the conditions in the hypothesis of the statement are both necessary and sufficient for the conclusion of the statement.

:ontrapositive of a

12. If a given statement is true but its converse is false, then the conditions are sufficient but not necessary for the conclusion of the statement.

-erse is true. Like-

13. If a given statement and its converse are both false, then the conditions are neither sufficient nor necessary for the statement's conclusion.

aid to be logically

173

CLEP College Mathematics

DEDUCTIVE REASONING An arrangement of statements that would allow you to deduce the third one from the preceding two is called a syllogism. A syllogism has three parts: 1. The first part is a general statement concerning a whole group. This is called the major premise. 2. The second part is a specific statement which indicates that a certain indi­ vidual is a member of that group. This is called the minor premise. 3. The last part of a syllogism is a statement to the effect that the general statement which applies to the group also applies to the individual. This third statement of a syllogism is called a deduction.

The flaw in this a condition on peop] are interchanged, thl

In the followin: tences, and develop linking them with th as operations transfi: describe them in gre, will find that differc different references.

TRUTH TABLES Example:

The truth possible Ie

This is an example of a properly deduced argument.

The logica and false (

A. Major Premise: All birds have feathers.

B. Minor Premise: An eagle is a bird. C. Deduction: An eagle has feathers.

The technique of employing a syllogism to arrive at a conclusion is called deductive reasoning. If a major premise that is true is followed by an appropriate minor premise that is true, a conclusion can be deduced that must be true, and the reasoning is valid. However, if a major premise that is true is followed by an inappropriate minor premise that is also true, a conclusion cannot be deduced.

Negation

If X is a sentenci contradiction of X T where ~ is called the

Example:

This is an example of an improperly deduced argument. A. Major Premise: All people who vote are at least 18 years old.

B. Improper Minor Premise: Jane is at least 18. C. Illogical Deduction: Jane votes.

174

Example:

For X = "Jane is ~ X = "Jane is

Chapter 8: Logic

ou to deduce the third , syllogism has three whole group. This is

res that a certain indi­ minor premise.

ffect that the general ) the individual. This

The flaw in this example is that the major premise in statement A makes a condition on people who vote, not on a person's age. If statements Band C are interchanged, the resulting three-part deduction would be logical. In the following we will use capital letters X, Y, Z, ... to represent sen­ tences, and develop algebraic tools to represent new sentences formed by linking them with the above connectives. Our connectives may be regarded as operations transforming one or more sentences into a new sentence. To describe them in greater detail, we introduce symbols to represent them. You will find that different symbols representing the same idea may appear in different references.

TRUTH TABLES AND BASIC LOGICAL OPERATIONS The truth table for a sentence X is the exhaustive list of possible logical values of X.

It.

The logical value of a sentence X is true (or and false (or F) if X is false.

~

at a conclusion is

appropriate minor lUst be true, and the true is followed by nclusion cannot be

l

n if X is true,

Negation If X is a sentence, then ~X represents the negation, the opposite, or the contradiction of X. Thus, the logical values of ~ X are as shown in Table 1, where ~ is called the negation operation on sentences. Table 1.-Truth Table for Negation

T

F

F

T

lent.

ars old. Example:

For X ~X

= "Jane is eating an apple," we have

= "Jane is not eating an apple."

175

-------t~------~-~-

_C_L_E_P_C_o_"_eg_e_M_a_th_e_m_a_t_ic_s

The negation operation is called unary, transforming a sentence into a unique image sentence.

IFF We use the symbol IFF to represent the expression "if and only if."

AND For sentences X and Y, the conjunction "X AND Y," represented by X A Y, is the sentence that is true IFF both X and Yare true. The truth table for A (or AND) is shown in Table 2, where A is called the conjunction operator.

Ii

I

As with the cc tion, transforming rl XVY.

Table 2-Truth Table for AND

Example:

For X = "Jane have X V Y = "Jane

T

T

T

T

F

F

F

T

F

IF-THEN

F

F

F

For sentences .\ "IF X THEN Y." .r ­ true. The truth table

The conjunction A is a binary operation, transforming a pair of sen­ tences into a unique image sentence.

cation operator.

Example:

For X = "Jane is eating an apple" and Y = "All apples are sweet," we have X A Y = "Jane is eating an apple AND all apples are sweet."

AND/OR For sentences X and Y, the disjunction "X AND/OR Y," represented by X V Y, denotes the sentence that is true if either or both X and Yare true. The truth table for V is shown in Table 3, where V is called the disjunction operator.

176

Implication is a and Y into a unique iT

Chapter 8: Logic

mg a sentence into a

Table 3-Truth Table for AND/OR

"if and only if."

T

T

T

T

F

T

F

T

T

F

F

F

. y.,. represented by

are true. The truth lIed the conjunction

ling a pair of sen-

Ies are sweet," we ~ sweet."

f':' represented by X and Yare true.

ed the disjunction

As with the conjunction operator, the disjunction is a binary opera­ tion, transforming the pair of sentences X, Y into a unique image sentence XVY.

Example:

For X = "Jane is eating the apple" and Y = "Marvin is running," we have X V Y = "Jane is eating the apple AND/OR Marvin is running."

IF-THEN For sentences X and Y, the implication X ---+ Y represents the statement "IF XTHEN Y." X ---+ Y is false IFF X is true and Y is false; otherwise, it is true. The truth table for ---+ is shown in Table 4. -~ is referred to as the impli­ cation operator. Table 4-Truth Table for IF-THEN

I

T

I

T

I

T

I

T

F

F

F

T

T

F

F

T

Implication is a binary operation, transforming the pair of sentences X and Y into a unique image sentence X ---+ Y.

177

CLEP College Mathematics

LOGICAL EQUIVALENCE

Fundamental F

For sentences X and Y, the logical equivalence X - Y is true IFF X and Y have the same truth value; otherwise, it is false. The truth table for - is

The next three duced in another fo: number system.

shown by Table 5, where - represents logical equivalence, "IFF." Table S-Truth Table for Equivalence

THEOREM 2-Pro~

T

T

T

T

F

F

F

T

F

F

F

T

For any sentenci 1. Commutativity: X 2. Associativity: X .

THEOREM 3-Prop

For any sentence Example:

1. Commutativity: X

For X = "Jane eats apples" and Y = "apples are sweet," we have X - Y = "Jane eats apples IFF apples are sweet."

2. Associativity: X .

Equivalence is a binary operation, transforming pairs of sentences X and Y into a unique image sentence X - Y. The two sentences X, Y for which X - Yare said to be logically equivalent.

THEOREM 4-Distri

For any sentence 1. Xv (Y I\Z) - (X

Logical Equivalence versus "Meaning the Same"

2. X 1\ (Yv Z) -

(X

Logical equivalence (-) is not the same as an equivalence of mean­ ings. Thus, if Jane is eating an apple and Barbara is frightened of mice, then for X = "Jane is eating an apple" and Y = "Barbara is frightened of mice," X and Yare logically equivalent, since both are correct. However, they do not have the same meaning. Statements having the same meaning are, for example, the double negative ~ ~ X (not-not) and X itself.

2. ~(Xv Y) -

THEOREM i-Double Negation Equals Identity

Proof of Part 1 of Th

For any sentence X, ~~X-X

178

THEOREM 5-DeM(

For any

sentence~

1. ~(X 1\ Y) - (~XI (~.o':n

We can prove ~L over all possible com! assumed by the senten expression ~(X 1\ Y) iJ

Chapter 8: Logic

Fundamental Properties of Operations

Y is true IFF X and e truth table for - is ence, "IFF."

'-4

The next three theorems will look familiar because they were intro­ duced in another format in Chapter 2 for sets and Chapter 3 for the real number system.

THEOREM 2-Properties of Conjunction Operation

For any sentences X, Y, Z, the following properties hold: 1. Commutativity: X /\ Y -

Y /\ X

2. Associativity: X /\ (Y /\ Z) - (X /\ Y) /\ Z

THEOREM 3-Properties of Disjunction Operation

For any sentences X, Y, Z, the following properties hold: 1. Commutativity: Xv Y -

YV X

eet," we have X - Y

2. Associativity: Xv (YV Z) - (XV Y) V Z

airs of sentences X ~nces X, Y for which

THEOREM 4-Distributive Laws

~

For any sentences X, Y, Z, the following laws hold: 1. Xv (Y /\Z) -

Same"

Jiyalence of mean­

[ened of mice, then

ightened of mice,"

However, they do

e meaning are, for

f

(XV Y) /\ (XV Z)

2. X /\ (YV Z) - (X /\ Y) V (X /\ Z)

THEOREM 5-DeMorgan's Laws for Sentences

For any sentences X, Y, the following laws hold: 1.

~(X /\

Y) -

(~X)

V (~Y).

2.

~(XV

Y) -

(~X)

/\

(~Y).

Proof of Part 1 of Theorem 5:

We can prove ~(X /\ Y) - (~X) V (~ Y) by developing a truth table over all possible combinations of X and Y and observing that all values assumed by the sentences are the same. To this end, we first evaluate the expression ~(X /\ Y) in Table 6a.

179

CLEP College Mathematics

Table 6a-Truth Table for Negation of Conjunction

Table 7i

y T

T

T

F

T

T

T

F

F

T

T

F

F

T

F

T

F

T

F

F

F

T

F

F

Now we evaluate ( ~ X) V ( ~ Y) in Table 6b.

For any sentenci equivalent, i.e.,

Table 6b-Truth Table for Disjunction of Negation ........................

This is pro\"en il

Table

T T

The last columns of the truth tables coincide, proving our assertion.

F

THEOREM 6-Two Logical Identities

F

For any sentences X, Y, the sentences X and (X A Y) V (X A logically equivalent. That is, (X A Y) V (X A

~

Y)

+-+

X

~ Y)

are

For any sentence

This is pro\"en in

This is proven in Table 7a.

f T

F F

180

Chapter 8: Logic

Ijunction

Table 7a-Truth Table for (X /\ Y) V (X /\ ~ Y)

<--+

X

(X 1\ y),

F T T

T

~ation

T

T

F

T

F

T

T

F

T

F

T

T

F

T

F

F

F

F

F

F

T

F

F

F

For any sentences X; Y, the sentences X and Xv (Y /\ equivalent, i.e., Xv (Y /\

~ Y) +-+

~

Y) are logically

X

This is proven in Table 7b.

Table 7b-Truth Table for X V ( Y/\ ~ Y)

<--+

X

This is proven in Table 7c. Table 7c

T

T

T

F

T

T

F

F

F

F

F

T

T

T

T

F

F

T

T

T

181

I

CLEP College Mathematics

DRILL QUESTIO

THEOREM 7-Proof by Contradiction For any sentences X, Y, the following holds:

1. If P and Q repre~ to "Not P and nc

To prove this, we consider Table 8.

(A) Not P or not Table 8-Truth Table for Proof by Contradiction

(B) Not P or Q

(C) Not (P or Q.

T

T

T

F

F

T

T

F

F

T

F

F

F

T

T

F

T

T

F

F

T

T

T

T

(D) Not (P and ~

2. Let Rand 5 repr~

I. If R then 5 II. Not Rand 5

SENTENCES, LITERALS, AND FUNDAMENTAL CONJUNCTIONS We have seen that logically equivalent sentences may be expressed in different ways, the simplest examples being that a sentence is equal to its double negation,

III. If S then R

Which of the abc a necessary cond

(A) Only I (B) I and II

~~X-X,

(C) II and III and by DeMorgan's theorem, (D) Only III

The significance of sentential calculus and the algebra oflogic is that it provides us with a method ofproducing a "standard" form for representing a statement in terms ofthe literals. This is indeed unique and, although usually the simplest representation, it does serve as a standard form for comparison and evaluation of sentences.

3. What is the inwr indoors"? (A) Ifpeople St3\

(B) If it is not snc

(C) Ifpeople do r

(D) Ifit is not sno

182

Chapter 8: Logic

DRILL QUESTIONS

1. If P and Q represent statements, which one ofthe following is equivalent to "Not P and not Q"? (A) Not P or not Q iction

(B) Not P or Q (C) Not (P or Q) T

F

(D) Not (P and Q)

T

T

2. Let Rand S represent statements. Consider the following: I. If R then S

II. Not Rand S

J\L

nay be expressed in tence is equal to its

III. If S then R Which of the above statements is (are) equivalent to the statement "R is a necessary condition for S"? (A) Only I (B) I and II (C) II and III (D) Only III

)ra oflogic is that it n for representing a ld although usually mn for comparison

3. What is the inverse of the statement "If it is snowing, then people stay indoors"? (A) Ifpeople stay indoors, then it is snowing. (B) If it is not snowing, then people do not stay indoors. (C) Ifpeople do not stay indoors, then it is not snowing. (D) If it is not snowing, then people stay indoors.

183

CLEP College Mathematics

4. What is the negation for the statement "Image is important or personal­ ity matters"?

8. "All of P is in ment, which on

I

(A) Image is important and personality does not matter.

(A) Some of R

(B) Image is not important or personality does not matter.

(B) All of R is i

(C) Image is important or personality does not matter.

(C) Some of R

(D) Image is not important and personality does not matter.

(D) None of R i

5. Given any two statements P and Q, where Q is a false statement, which one of the following must be false?

!

9. Given three stat Which one of th

(A) Not P and Q

(A) (P A Q)-­

(B) Not (P and Q)

(B) R

(C) P or not Q

(C) (P

(D) P implies Q

(D) (R

6. What is the contrapositive of the statement "If Johnny is lying, then Marie will be angry"?

--*

(p

--* --*

Q

Q) Q)

10. For which one

}

0

(A) If you liw iT (A) If Johnny is not lying, then Marie will be angry. (B) If your pet

i~

(B) If Marie is not angry, then Johnny is not lying. (C) If a number (C) If Johnny is not lying, then Marie will not be angry.

(D) If a geometr (D) If Marie is angry, then Johnny is lying.

7. Which of the following is an example of a biconditional statement? (A) A triangle has an angle less than 90° if it is acute. (B) A rectangle is a square if and only if it has four equal sides. (C) A circle has both a radius and a diameter. (D) A cube has either six edges or it has more than six edges. 184

Chapter 8: Logic

mportant or personal-

8. "All of P is in Q and some of R is in P." Based on the previous state­ ment, which one of the following is a valid conclusion?

latter.

(A) Some of R is not in Q.

: matter.

(B) All of R is in Q.

nero

(C) Some of R is in Q.

~r

matter.

(D) None of R is in Q.

3.1se statement, which

9. Given three statements, P, Q, and R, suppose it is known that R is true. Which one of the following must be true? (A) (P /\ Q) (B) R

(C)

R

(P V Q)

(P~Q)/\R

(D) (R

Jhnny is lying, then

~

~

~

Q) V P

10. For which one of the following statements is the converse true?

(A) If you live in Toledo, then you live in Ohio. (B) If your pet is a cat, then it has a tail. (C) If a number is negative, then its square is positive.

19ry. (D) If a geometric figure has three sides, then it is a triangle.

onal statement? e. equal sides.

~ix

edges.

185

i

CLEP College Mathematics

Answers to Drill Questions 1. (C)

6. (8)

2. (D)

7. (8)

3. (8)

8. (C)

4. (D)

9. (A)

5. (A)

10. (D)

i i·

This test i: TESTil'Qrt It is highl~ on compu features ar instant. ac,

186