Unformatted text preview: MATH REVIEW
for
Practicing to Take the
GRE
General Test
® Copyright © 2003 by Educational Testing Service. All rights reserved.
EDUCATIONAL TESTING SERVICE, ETS, the ETS logos, GRADUATE RECORD EXAMINATIONS,
and GRE are registered trademarks of Educational Testing Service. MATH REVIEW
The Math Review is designed to familiarize you with the mathematical skills and
concepts likely to be tested on the Graduate Record Examinations General Test.
This material, which is divided into the four basic content areas of arithmetic,
algebra, geometry, and data analysis, includes many definitions and examples
with solutions, and there is a set of exercises (with answers) at the end of each
of these four sections. Note, however, this review is not intended to be comprehensive. It is assumed that certain basic concepts are common knowledge to all
examinees. Emphasis is, therefore, placed on the more important skills, concepts,
and definitions, and on those particular areas that are frequently confused or
misunderstood. If any of the topics seem especially unfamiliar, we encourage
you to consult appropriate mathematics texts for a more detailed treatment of
those topics. TABLE OF CONTENTS
1. ARITHMETIC
1.1 Integers ..................................................................................................... 6
1.2 Fractions ................................................................................................... 7
1.3 Decimals ................................................................................................... 8
1.4 Exponents and Square Roots .................................................................. 10
1.5 Ordering and the Real Number Line ...................................................... 11
1.6 Percent .................................................................................................... 12
1.7 Ratio ....................................................................................................... 13
1.8 Absolute Value ........................................................................................ 13
ARITHMETIC EXERCISES ........................................................................ 14
ANSWERS TO ARITHMETIC EXERCISES.............................................. 17
2. ALGEBRA
2.1 Translating Words into Algebraic Expressions ....................................... 19
2.2 Operations with Algebraic Expressions.................................................. 20
2.3 Rules of Exponents ................................................................................. 21
2.4 Solving Linear Equations ....................................................................... 21
2.5 Solving Quadratic Equations in One Variable ........................................ 23
2.6 Inequalities ............................................................................................. 24
2.7 Applications ............................................................................................ 25
2.8 Coordinate Geometry ............................................................................. 28
ALGEBRA EXERCISES ............................................................................. 31
ANSWERS TO ALGEBRA EXERCISES ................................................... 34
3. GEOMETRY
3.1 Lines and Angles .................................................................................... 36
3.2 Polygons ................................................................................................. 37
3.3 Triangles ................................................................................................. 38
3.4 Quadrilaterals ......................................................................................... 40
3.5 Circles ..................................................................................................... 42
3.6 ThreeDimensional Figures .................................................................... 45
GEOMETRY EXERCISES .......................................................................... 47
ANSWERS TO GEOMETRY EXERCISES ............................................... 50
4. DATA ANALYSIS
4.1 Measures of Central Location ................................................................ 51
4.2 Measures of Dispersion .......................................................................... 51
4.3 Frequency Distributions ......................................................................... 52
4.4 Counting ................................................................................................. 53
4.5 Probability .............................................................................................. 54
4.6 Data Representation and Interpretation .................................................. 55
DATA ANALYSIS EXERCISES .................................................................. 62
ANSWERS TO DATA ANALYSIS EXERCISES ....................................... 69 ARITHMETIC
1.1 Integers
The set of integers, I, is composed of all the counting numbers (i.e., 1, 2,
3, . . .), zero, and the negative of each counting number; that is, : ? I = . . . ,  3,  2 ,  1, 0, 1, 2, 3, . . . .
Therefore, some integers are positive, some are negative, and the integer 0 is
neither positive nor negative. Integers that are multiples of 2 are called even
integers, namely . . . ,  6 ,  4 ,  2, 0, 2 , 4 , 6 , . . . . All other integers are called
odd integers; therefore . . . ,  5,  3,  1, 1, 3, 5, . . . represents the set of all
odd integers. Integers in a sequence such as 57, 58, 59, 60, or − 14, − 13, − 12, − 11
are called consecutive integers.
The rules for performing basic arithmetic operations with integers should be
familiar to you. Some rules that are occasionally forgotten include:
(i) Multiplication by 0 always results in 0; e.g., (0)(15) = 0.
(ii) Division by 0 is not defined; e.g., 5 ÷ 0 has no meaning.
(iii) Multiplication (or division) of two integers with different signs yields
a negative result; e.g., ( 7) (8) = 56 and ( 12 ) ( 4) = 3.
(iv) Multiplication (or division) of two negative integers yields a positive
result; e.g., ( 5)( 12) = 60 and ( 24) ( 3) = 8 .
The division of one integer by another yields either a zero remainder, sometimes called “dividing evenly,” or a positiveinteger remainder. For example,
215 divided by 5 yields a zero remainder, but 153 divided by 7 yields a remainder of 6. : : 43
5 215
20
15
15
0 = Remainder ? ? 21
7 153
14
13
7
6 = Remainder When we say that an integer N is divisible by an integer x, we mean that N
divided by x yields a zero remainder.
The multiplication of two integers yields a third integer. The first two integers
are called factors, and the third integer is called the product. The product is said
to be a multiple of both factors, and it is also divisible by both factors (providing
the factors are nonzero). Therefore, since ( 2 ) ( 7) = 14, we can say that
2 and 7 are factors and 14 is the product,
14 is a multiple of both 2 and 7,
and 14 is divisible by both 2 and 7.
Whenever an integer N is divisible by an integer x, we say that x is a divisor
of N. For the set of positive integers, any integer N that has exactly two distinct
positive divisors, 1 and N, is said to be a prime number. The first ten prime
numbers are
2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.
The integer 14 is not a prime number because it has four divisors: 1, 2, 7, and 14.
The integer 1 is not a prime number because it has only one positive divisor. 6 1.2 Fractions
a
, where a and b are integers and b 0.
b
The a is called the numerator of the fraction, and b is called the denominator.
7
is a fraction that has 7 as its numerator and 5 as its denomiFor example,
5
a
nator. Since the fraction means a b, b cannot be zero. If the numerator
b
a
and denominator of the fraction are both multiplied by the same integer, then
b
a
the resulting fraction will be equivalent to . For example,
b
A fraction is a number of the form ( 7) ( 4)
7
28
=
=
.
5
(5)( 4)
20
This technique comes in handy when you wish to add or subtract fractions.
To add two fractions with the same denominator, you simply add the
numerators and keep the denominator the same. 8
5
8 + 5
3
+
=
=
11
11 11
11
If the denominators are not the same, you may apply the technique mentioned
above to make them the same before doing the addition. (2)( 4)
5
2
5
5
8
5 + 8 13
=
+=
+
=
+
=
12 3 12 (3)(4)
12 12
12
12
The same method applies for subtraction.
To multiply two fractions, multiply the two numerators and multiply the two
denominators (the denominators need not be the same). 10
170 31 = (10)(()1) = 21
( 7) 3 To divide one fraction by another, first invert the fraction you are dividing by,
and then proceed as in multiplication. 17)
85
5 = ((8)((35)) = 24
3 17 3
17
=
8
5
8
An expression such as 4 3
3
is called a mixed fraction; it means 4 + .
8
8 Therefore, 4 3
3
32 3
35
=4+ =
+=
.
8
8
8
8
8 7 1.3 Decimals
In our number system, all numbers can be expressed in decimal form using
base 10. A decimal point is used, and the place value for each digit corresponds
to a power of 10, depending on its position relative to the decimal point. For
example, the number 82.537 has 5 digits, where
“8” is the “tens” digit; the place value for “8” is 10.
“2” is the “units” digit; the place value for “2” is 1. 1
.
10
1
“3” is the “hundredths” digit; the place value for “3” is
.
100
1
“7” is the “thousandths” digit; the place value for “7” is
.
1000
“5” is the “tenths” digit; the place value for “5” is Therefore, 82.537 is a short way of writing
(8)(10) + ( 2)(1) + (5) 1
1
1
10 + (3) 100 + (7) 1000 , or 80 + 2 + 0.5 + 0.03 + 0.007.
This numeration system has implications for the basic operations. For addition and subtraction, you must always remember to line up the decimal points: 126.5
+ 68.231
194 .731 126.5
 68.231
58.269 To multiply decimals, it is not necessary to align the decimal points. To determine the correct position for the decimal point in the product, you simply add
the number of digits to the right of the decimal points in the decimals being multiplied. This sum is the number of decimal places required in the product.
15.381
0.14
61524
15381
2 .15334 (3 decimal places)
(2 decimal places) (5 decimal places) To divide a decimal by another, such as 62.744 ÷ 1.24, or
1. 24 62 .744 , first move the decimal point in the divisor to the right until the divisor becomes
an integer, then move the decimal point in the dividend the same number of
places;
.
124 6274.4 This procedure determines the correct position of the decimal point in the quotient (as shown). The division can then proceed as follows: 8 50.6
124 6274.4
620
744
744
0
Conversion from a given decimal to an equivalent fraction is straightforward.
Since each place value is a power of ten, every decimal can be converted easily
to an integer divided by a power of ten. For example,
841
10
917
9.17 =
100
612
0.612 =
1000
84.1 = The last example can be reduced to lowest terms by dividing the numerator
and denominator by 4, which is their greatest common factor. Thus, 0.612 = 612
612 4
153
=
=
( in lowest terms).
250
1000
1000 4 a
b
means a b , we can divide the numerator of a fraction by its denominator to
3
convert the fraction to a decimal. For example, to convert to a decimal, divide
8
3 by 8 as follows.
Any fraction can be converted to an equivalent decimal. Since the fraction 0.375
8 3.000
24
60
56
40
40
0 9 1.4 Exponents and Square Roots
Exponents provide a shortcut notation for repeated multiplication of a number
by itself. For example, “34 ” means (3)(3)(3)(3), which equals 81. So, we say that
34 = 81; the “4” is called an exponent (or power). The exponent tells you how
many factors are in the product. For example,
2 5 = (2 )(2)(2)(2)(2) = 32
10 6 = (10)(10)(10)(10)(10)(10) = 1,000,000
( 4) 3 = ( 4)( 4)( 4) =  6 4 1
2 4 = 1 1 1 1
2222 = 1
16 When the exponent is 2, we call the process squaring. Therefore, “52 ” can be
read “5 squared.”
Exponents can be negative or zero, with the following rules for any nonzero
number m.
m0 = 1
m 1 = 1
m m 2 = 1
m2 m 3 = 1
m3 m n = 1
for all integers n.
mn If m = 0, then these expressions are not defined.
A square root of a positive number N is a real number which, when squared,
equals N. For example, a square root of 16 is 4 because 42 = 16. Another square
root of 16 is –4 because (–4)2 = 16. In fact, all positive numbers have two
square roots that differ only in sign. The square root of 0 is 0 because 02 = 0.
Negative numbers do not have square roots because the square of a real number
cannot be negative. If N > 0, then the positive square root of N is represented by
N , read “radical N.” The negative square root of N, therefore, is represented
by  N .
Two important rules regarding operations with radicals are:
If a > 0 and b > 0, then
(i)
(ii) 10 1 a 61 b 6 =
a
=
b ab ; e.g., a
; e.g.,
b 1 561 206 = 192 =
4 48 = 100 = 10 (16 )(3) = 1 1661 36 = 4 3 1.5 Ordering and the Real Number Line
The set of all real numbers, which includes all integers and all numbers with
2
values between them, such as 1.25, , 2 , etc., has a natural ordering, which
3
can be represented by the real number line: Every real number corresponds to a point on the real number line (see examples
shown above). The real number line is infinitely long in both directions.
For any two numbers on the real number line, the number to the left is less
than the number to the right. For example,
 5< 3
2
2 1.75 <
5
< 7 .1
2 Since 2 < 5, it is also true that 5 is greater than 2, which is written “5 > 2.”
If a number N is between 1.5 and 2 on the real number line, you can express
that fact as 1.5 < N < 2. 11 1.6 Percent
The term percent means per hundred or divided by one hundred. Therefore,
43
= 0.43
100
300
300% =
=3
100
0.5
0.5% =
= 0.005
100
43% = To find out what 30% of 350 is, you multiply 350 by either 0.30 or 30
,
100 30% of 350 = (350) (0.30) = 105
or 30% of 350 = (350) 050
30
3
100 = (350) 10 = 1,10 = 105. To find out what percent of 80 is 5, you set up the following equation and
solve for x:
5
x
=
80
100
x= 500
= 6.25
80 So 5 is 6.25% of 80. The number 80 is called the base of the percent. Another
way to view this problem is to simply divide 5 by the base, 80, and then multiply
the result by 100 to get the percent.
If a quantity increases from 600 to 750, then the percent increase is found by
dividing the amount of increase, 150, by the base, 600, which is the first (or the
smaller) of the two given numbers, and then multiplying by 100: 150 (100)% = 25%.
600 If a quantity decreases from 500 to 400, then the percent decrease is found by
dividing the amount of decrease, 100, by the base, 500, which is the first (or the
larger) of the two given numbers, and then multiplying by 100: 100 (100 )% = 20 %.
500 Other ways to state these two results are “750 is 25 percent greater than 600”
and “400 is 20 percent less than 500.”
In general, for any positive numbers x and y, where x < y, y  x (100 ) percent greater than x
x
y  x
x is
y (100 ) percent less than y
y is Note that in each of these statements, the base of the percent is in
the denominator. 12 1.7 Ratio
The ratio of the number 9 to the number 21 can be expressed in several ways;
for example,
9 to 21
9:21
9
21
Since a ratio is in fact an implied division, it can be reduced to lowest terms.
Therefore, the ratio above could also be written:
3 to 7
3:7
3
7 1.8 Absolute Value
The absolute value of a number N, denoted by N , is defined to be N if N
is positive or zero and – N if N is negative. For example,
1
1
= , 0 = 0, and  2 .6 =  (  2.6 ) = 2.6 .
2
2 Note that the absolute value of a number cannot be negative. 13 ARITHMETIC EXERCISES
(Answers on pages 17 and 18)
1. Evaluate:
(a) 15 – (6 – 4)(–2) (e) (–5)(–3) – 15 (b) (2 – 17) ÷ 5 (f) (–2)4 (15 – 18)4 (c) (60 ÷ 12) – (–7 + 4) (g) (20 ÷ 5)2 (–2 + 6)3 (d) (3)4 – (–2)3 (h) (–85)(0) – (–17)(3) 2. Evaluate:
(a)
(b) 7
8  4
5
3
27
(d)
 8 32
2 11
1
+
2 3 12 (c) 3 + 1 52
47 3. Evaluate:
(a) 12.837 + 1.65 – 0.9816
(b) 100.26 ÷ 1.2 (c) (12.4)(3.67)
(d) (0.087)(0.00021) 4. State for each of the following whether the answer is an even integer or
an odd integer.
(a)
(b)
(c)
(d)
(e)
(f) The sum of two even integers
The sum of two odd integers
The sum of an even integer and an odd integer
The product of two even integers
The product of two odd integers
The product of an even integer and an odd integer 5. Which of the following integers are divisible by 8 ?
(a) 312 (b) 98 (c) 112 (d) 144 6. List all of the positive divisors of 372.
7. Which of the divisors found in #6 are prime numbers?
8. Which of the following integers are prime numbers?
19, 2, 49, 37, 51, 91, 1, 83, 29
9. Express 585 as a product of prime numbers. 14 10. Which of the following statements are true?
(g) 16 = 4 (b) 9<0 (h) (a) –5 < 3.1 21
3
=
28
4 (c) 7 ÷ 0 = 0 (i)   23 = 23 (d) 0 <  1 . 7 (j) 1
3 (e) 0 . 3 < 1
1
>
2 17 (k) (59)3 (59) 2 = (59) 6 (f) (–1)87 = –1 (l)  25 <  4 11. Perform the indicated operations.
(a) 5 3 +
(b)
(c)
(d) 27 1 6 61 306
1 300 6 1 12 6
1 5 61 26  90 12. Express the following percents in decimal form and in fraction form
(in lowest terms).
(a) 15% (b) 27.3% (c) 131% (d) 0.02% 13. Express each of the following as a percent.
(a) 0.8 (b) 0.197 (c) 5.2 (d) 3
8 (e) 2 1
2 (f) 3
50 14. Find:
(a) 40% of 15
(b) 150% of 48 (c) 0.6% of 800
(d) 8% of 5% 15. If a person’s salary increases from $200 per week to $234 per week, what is
the percent increase?
16. If an athlete’s weight decreases from 160 pounds to 152 pounds, what is
the percent decrease? 15 17. A particular stock is valued at $40 per share. If the value increases 20 percent
and then decreases 25 percent, what is the value of the stock per share after
the decrease?
18. Express the ratio of 16 to 6 three different ways in lowest terms.
19. If the ratio of men to women on a committee of 20 members is 3 to 2, how
many members of the committee are women? 16 ANSWERS TO ARITHMETIC EXERCISES
1. (a)
(b)
(c)
(d) 19
–3
8
89 (e)
(f)
(g)
(h) 0
1,296
1,024
51 2. (a) 1
4 (c) 9
1,600 (b)  5
14 (d)  4
9 3. (a) 13.5054
(b) 83.55 (c) 45.508
(d) 0.00001827 4. (a) even
(b) even
(c) odd (d) even
(e) odd
(f) even 5. (a), (c), and (d)
6. 1, 2, 3, 4, 6, 12, 31, 62, 93, 124, 186, 372
7. 2, 3, 31
8. 19, 2, 37, 83, 29
9. (3)(3)(5)(13)
10. (a), (b), (d), (e), (f ), (h), ( j), (l)
11. (a) 8 3
(b) 6 5 3
20
2 73
(b) 0.273,
1,000 12. (a) 0.15, (c) 5
(d) 2 10
(c) 1.31, 131
100 (d) 0.0002, 1
5,000 17 13. (a) 80%
(b) 19.7%
(c) 520% (d) 37.5%
(e) 250%
(f) 6% 14. (a) 6
(b) 72 (c) 4.8
(d) 0.004 15. 17%
16. 5%
17. $36
18. 8 to 3, 8:3,
19. 8 18 8
3 ALGEBRA
2.1 Translating Words into Algebraic Expressions
Basic algebra is essentially advanced arithmetic; therefore much of the
terminology and many of the rules are common to both areas. The major difference is that in algebra variables are introduced, which allows us to solve problems using equations and inequalities.
If the square of the number x is multiplied by 3, and then 10 is added to that
product, the result can be represented by 3 x 2 + 10. If John’s present salary S is
increased by 14 percent, then his new salary is 1.14S. If y gallons of syrup are to
be distributed among 5 people so that one particular person gets 1 gallon and the
rest of the syrup is divided equally among the remaining 4, then each of these
y 1
4 people will get
gallons of syrup. Combinations of letters (variables) and
4
y 1
numbers such as 3 x 2 + 10, 1.14S, and
are called algebraic expressions.
4
One way to work with algebraic expressions is to think of them as functions,
or “machines,” that take an input, say a value of a variable x, and produce a
2x
, the input x = 1
corresponding output. For example, in the expression
x6
2(1)
2
produces the corresponding output
=  . In function notation, the
16
5
2x
is called a function and is denoted by a letter, often the letter f
expression
x6
or g, as follows:
2x
f ( x) =
.
x6
We say that this equation defines the function f. For this example with input
2
2
2
x = 1 and output  , we write f (1) =  . The output  is called the
5
5
5
value of the function corresponding to the input x = 1. The value of the function
corresponding to x = 0 is 0, since
f (0 ) = 2( 0 )
0
=  = 0.
06
6 In fact, any real number x can be used as an input value for the function f,
except for x = 6 , as this substitution would result in a division by 0. Since
x = 6 is not a valid input for f, we say that f is not defined for x = 6 .
As another example, let h be the function defined by
h( z ) = z 2 + z + 3. Note that h( 0 ) = 3 , h(1) = 5 , h(10) = 103 + 10 106.2 , but h( 10 ) is not
defined since 10 is not a real number. 19 2.2 Operations with Algebraic Expressions
Every algebraic expression can be written as a single term or a series of terms
separated by plus or minus signs. The expression 3 x 2 + 10 has two terms; the
y 1
expression 1.14S is a single term; the expression
, which can be written
4
y1
 , has two terms. In the expression 2 x 2 + 7 x  5 , 2 is the coefficient of
44
the x 2 term, 7 is the coefficient of the x term, and 5 is the constant term.
The same rules that govern operations with numbers apply to operations
with algebraic expressions. One additional rule, which helps in simplifying
algebraic expressions, is that terms with the same variable part can be combined.
Examples are:
2 x + 5 x = (2 + 5) x = 7 x
x 2  3 x 2 + 6 x 2 = (1  3 + 6) x 2 = 4 x 2
3 xy + 2 x  xy  3 x = (3  1) xy + (2  3) x = 2 xy  x Any number or variable that is a factor of each term in an algebraic expression can be factored out. Examples are:
4 x + 12 = 4( x + 3) 0 5 15 y 2  9 y = 3 y 5 y  3 7 x + 14 x
7 x ( x + 2)
7x
=
=
2
2x + 4
2( x + 2 )
2 0if x 25 Another useful tool for factoring algebraic expressions is the fact that
a  b 2 = ( a + b )( a  b ). For example,
2 0 5 ( x + 3)( x  3)
x2  9
x+3
=
=
if x 3 .
4 x  12
4
4( x  3) To multiply two algebraic expressions, each term of the first expression is
multiplied by each term of the second, and the results are added. For example, ( x + 2)(3 x  7) = x (3 x ) + x (7) + 2(3 x ) + 2( 7)
= 3 x 2  7 x + 6 x  14
= 3 x 2  x  14
A statement that equates two algebraic expressions is called an equation.
Examples of equations are: 3 x + 5 = 2
x  3 y = 10
20 y 2 + 6 y  17 = 0 20 0linear equation in one variable5
0linear equation in two variables5
0quadratic equation in one variable5 2.3 Rules of Exponents
Some of the basic rules of exponents are:
1
(a) x  a = a ( x 0)
x
1
1
.
Example: 4 3 = 3 =
64
4
(b) 3 x 83 x 8 = x +
Example: 33 833 8 = 3 + = 3
3 x 83 y 8 = 0 xy5
Example: 3 2 833 8 = 6 = 216 .
a b ab 2 (c) a 4 (e) 3 = 729 . 3 xa
1
= x a  b = b  a ( x 0)
b
x
x
57
1
1
1
43
Examples: 4 = 5 7  4 = 5 3 = 125 and 8 = 8  3 = 5 =
.
1, 024
5
4
4
4 x
y 0 y 05
3
3
Example: =
4
a = xa
ya 2 2 4 (f) 6 a a 3 (d) 24 3x 8 ab 2 = 9
.
16 = x ab 38 Example: 2 5 2 = 2 10 = 1, 024 . (g) If x 0 , then x 0 = 1 .
Examples: 7 0 = 1; ( 3) 0 = 1; 0 0 is not defined. 2.4 Solving Linear Equations
(a) One variable.
To solve a linear equation in one variable means to find the value of the
variable that makes the equation true. Two equations that have the same solution
are said to be equivalent. For example, x + 1 = 2 and 2 x + 2 = 4 are
equivalent equations; both are true when x = 1 and are false otherwise.
Two basic rules are important for solving linear equations.
(i) When the same constant is added to (or subtracted from) both sides of
an equation, the equality is preserved, and the new equation is
equivalent to the original.
(ii) When both sides of an equation are multiplied (or divided) by the same
nonzero constant, the equality is preserved, and the new equation is
equivalent to the original. 21 For example,
3x  4
3x  4 + 4
3x
3x
3
x (b) =8
=8+4
4 added to both sides
= 12
12
both sides divided by 3
=
3
=4 0 0 5 5 Two variables. To solve linear equations in two variables, it is necessary to have two
equations that are not equivalent. To solve such a “system” of simultaneous
equations, e.g.,
4 x + 3 y = 13
x + 2y = 2 there are two basic methods. In the first method, you use either equation to
express one variable in terms of the other. In the system above, you could express
x in the second equation in terms of y (i.e., x = 2  2 y ), and then substitute
2  2 y for x in the first equation to find the solution for y: 0 5 4 2  2 y + 3 y = 13
8  8 y + 3 y = 13
8 y + 3 y = 5
5 y = 5
y = 1 08 subtracted from both sides5
0terms combined5
0both sides divided by 55 Then 1 can be substituted for y in the second equation to solve for x:
x + 2y
x + 2( 1)
x2
x =
=
=
= 2
2
2
4 (2 added to both sides) In the second method, the object is to make the coefficients of one variable
the same in both equations so that one variable can be eliminated by either
adding both equations together or subtracting one from the other. In the same
example, both sides of the second equation could be multiplied by 4, yielding
4 x + 2 y = 4(2) , or 4 x + 8 y = 8 . Now we have two equations with the same
x coefficient: 0 5 4 x + 3 y = 13
4 x + 8y = 8 If the second equation is subtracted from the first, the result is 5 y = 5 .
Thus, y = 1 , and substituting 1 for y in either one of the original equations
yields x = 4. 22 2.5 Solving Quadratic Equations in One Variable
A quadratic equation is any equation that can be expressed as
ax + bx + c = 0 , where a, b, and c are real numbers ( a 0 ) . Such an
equation can always be solved by the formula:
2 x= b b 2  4ac
.
2a For example, in the quadratic equation 2 x 2  x  6 = 0 , a = 2 ,
b = 1, and c = 6 . Therefore, the formula yields
x=
= ( 1)
1 ( 1) 2  4(2)( 6)
2( 2 ) 49 4
= 17
4 1+7
17
3
= 2 and x =
=  . Quadratic equations
4
4
2
can have at most two real solutions, as in the example above. However, some
quadratics have only one real solution (e.g., x 2 + 4 x + 4 = 0 ; solution: x = 2 ),
and some have no real solutions (e.g., x 2 + x + 5 = 0 ).
Some quadratics can be solved more quickly by factoring. In the original
example,
So, the solutions are x = 2 x 2  x  6 = (2 x + 3)( x  2 ) = 0 . Since (2 x + 3)( x  2 ) = 0 , either 2 x + 3 = 0 or x  2 = 0 must be true.
Therefore,
2x + 3 = 0
x2=0
2 x = 3
OR
x=2
3
x=2
Other examples of factorable quadratic equations are:
(a) x 2 + 8 x + 15 = 0
( x + 3)( x + 5) = 0
Therefore, x + 3 = 0; x = 3
or x + 5 = 0; x = 5 (b) 4x2  9 = 0
(2 x + 3)(2 x  3) = 0 23 Therefore, 2 x + 3 = 0; x = or 2 x  3 = 0; x = 3
2 3
2 2.6 Inequalities
Any mathematical statement that uses one of the following symbols is called
an inequality.
“not equal to”
“less than”
“less than or equal to”
“greater than”
“greater than or equal to” <
>
For example, the inequality 4 x  1 7 states that “ 4 x  1 is less than or equal
to 7.” To solve an inequality means to find the values of the variable that make
the inequality true. The approach used to solve an inequality is similar to that
used to solve an equation. That is, by using basic operations, you try to isolate the
variable on one side of the inequality. The basic rules for solving inequalities are
similar to the rules for solving equations, namely:
(i) When the same constant is added to (or subtracted from) both sides of
an inequality, the direction of inequality is preserved, and the new
inequality is equivalent to the original.
(ii) When both sides of the inequality are multiplied (or divided) by the
same constant, the direction of inequality is preserved if the constant
is positive, but reversed if the constant is negative. In either case the
new inequality is equivalent to the original.
For example, to solve the inequality 3x + 5 17 ,
3 x + 5 17
3 x 12
12
3 x
3
3 05 subtracted from both sides5
( both sides divided by  3, which
reverses the direction of the inequality) x 4 Therefore, the solutions to 3x + 5 17 are all real numbers greater than
or equal to  4 . Another example follows:
4x + 9
>5
11
4 x + 9 > 55 0both sides multiplied by 115
09 subtracted from both sides5
0both sides divided by 45 4 x > 46
46
x>
4
x > 11 24 1
2 2.7 Applications
Since algebraic techniques allow for the creation and solution of equations
and inequalities, algebra has many realworld applications. Below are a few
examples. Additional examples are included in the exercises at the end of this
section.
Example 1. Ellen has received the following scores on 3 exams: 82, 74, and 90.
What score will Ellen need to attain on the next exam so that the
average (arithmetic mean) for the 4 exams will be 85 ?
Solution: If x represents the score on the next exam, then the arithmetic
mean of 85 will be equal to 82 + 74 + 90 + x
.
4
So,
246 + x
= 85
4
246 + x = 340
x = 94 Therefore, Ellen would need to attain a score of 94 on the next exam.
Example 2. A mixture of 12 ounces of vinegar and oil is 40 percent vinegar
(by weight). How many ounces of oil must be added to the mixture
to produce a new mixture that is only 25 percent vinegar?
Solution: Let x represent the number of ounces of oil to be added. Therefore, the total number of ounces of vinegar in the new mixture
will be (0.40)(12), and the total number of ounces of new
mixture will be 12 + x . Since the new mixture must be
25 percent vinegar,
(0.40)(12)
= 0.25 .
12 + x Therefore,
(0.40)(12)
4.8
1.8
7.2 =
=
=
= (12 + x )(0.25)
3 + 0.25 x
0.25 x
x Thus, 7.2 ounces of oil must be added to reduce the percent of
vinegar in the mixture from 40 percent to 25 percent.
Example 3. In a driving competition, Jeff and Dennis drove the same course
at average speeds of 51 miles per hour and 54 miles per hour,
respectively. If it took Jeff 40 minutes to drive the course, how long
did it take Dennis? 25 Solution: Let x equal the time, in minutes, that it took Dennis to drive the
course. Since distance ( d ) equals rate (r) multiplied by time (t),
i.e.,
d = (r )(t ) , the distance traveled by Jeff can be represented by (51)
and the distance traveled by Dennis, (54)
distances are equal, 40
60 , x
60 . Since the 40
x
60 = (54) 60 (51) 34 = 0.9 x
37.8 x Thus, it took Dennis approximately 37.8 minutes to drive the
course. Note: since rates are given in miles per hour, it was
necessary to express time in hours (i.e., 40 minutes equals
40
2
, or , of an hour.)
60
3
Example 4. If it takes 3 hours for machine A to produce N identical computer
parts, and it takes machine B only 2 hours to do the same job,
how long would it take to do the job if both machines worked
simultaneously?
Solution: Since machine A takes 3 hours to do the job, machine A can do 1
1
of the job in 1 hour. Similarly, machine B can do of the job
3
2
in 1 hour. And if we let x represent the number of hours it would
take for the machines working simultaneously to do the job, then the
two machines would do 1
of the job in 1 hour. Therefore,
x 11
1
+=
32
x
23
1
+=
66
x
5
1
=
6
x
6
=x
5 6
hours, or
5
1 hour and 12 minutes, to produce the N computer parts. Thus, working together, the machines take only 26 Example 5. At a fruit stand, apples can be purchased for $0.15 each and pears
for $0.20 each. At these rates, a bag of apples and pears was purchased for $3.80. If the bag contained exactly 21 pieces of fruit,
how many were pears?
Solution: If a represents the number of apples purchased and p represents
the number of pears purchased, two equations can be written as
follows:
0.15a + 0.20 p = 3.80
a + p = 21 From the second equation, a = 21  p . Substituting 21  p into
the first equation for a gives 0 5 0.15 21  p + 0.20 p
(0.15)(21)  0.15 p + 0.20 p
315  0.15 p + 0.20 p
.
0.05 p
p =
=
=
=
= 3.80
3.80
3.80
0.65
13 pears 0 5 Example 6. It costs a manufacturer $30 each to produce a particular radio
model, and it is assumed that if 500 radios are produced, all will be
sold. What must be the selling price per radio to ensure that the
profit (revenue from sales minus total cost to produce) on the
500 radios is greater than $8,200 ? 0 5 Solution: If y represents the selling price per radio, then the profit must be
500 y  30 . Therefore, 0 5 > 8,200 500 y  30
500 y  15,000
500 y
y > 8,200
> 23,200
> 46.40 Thus, the selling price must be greater than $46.40 to make the
profit greater than $8,200. 27 2.8 Coordinate Geometry
Two real number lines (as described in Section 1.5) intersecting at right
angles at the zero point on each number line define a rectangular coordinate
system, often called the xycoordinate system or xyplane. The horizontal
number line is called the xaxis, and the vertical number line is called the yaxis.
The lines divide the plane into four regions called quadrants (I, II, III, and IV)
as shown below. Each point in the system can be identified by an ordered pair of real numbers,
(x, y), called coordinates. The xcoordinate expresses distance to the left (if
negative) or right (if positive) of the yaxis, and the ycoordinate expresses
distance below (if negative) or above (if positive) the xaxis. For example, since
point P, shown above, is 4 units to the right of the yaxis and 1.5 units above the
xaxis, it is identified by the ordered pair (4, 1.5). The origin O has coordinates
(0, 0). Unless otherwise noted, the units used on the xaxis and the yaxis are the
same. 28 To find the distance between two points, say P(4, 1.5) and Q( 2,  3), represented by the length of line segment PQ in the figure below, first construct a
right triangle (see dotted lines) and then note that the two shorter sides of the
triangle have lengths 6 and 4.5. Since the distance between P and Q is the length of the hypotenuse, we can
apply the Pythagorean Theorem, as follows: PQ = (6) 2 + ( 4.5) 2 = 56.25 = 7.5 (For a discussion of right triangles and the Pythagorean Theorem, see Section 3.3.)
A straight line in a coordinate system is a graph of a linear equation of the
form y = mx + b, where m is called the slope of the line and b is called the
yintercept. The slope of a line passing through points P x1 , y1 and Q x 2 , y 2
is defined as 0 slope = y1  y2
x1  x 2 0x 1 5 0 5 5 x2 . For example, in the coordinate system shown above, the slope of the line passing
through points P(4, 1.5) and Q( 2,  3) is
slope = 1.5  ( 3)
4.5
=
= 0.75 .
4  ( 2)
6 The yintercept is the ycoordinate of the point at which the graph intersects the
yaxis. The yintercept of line PQ in the example above appears to be about
1.5, since line PQ intersects the yaxis close to the point ( 0, 1.5). This can be 29 confirmed by using the equation of the line, y = 0.75 x + b, by substituting the
coordinates of point Q (or any point that is known to be on the line) into the
equation, and by solving for the yintercept, b, as follows:
y
3
b
b =
=
=
= 0.75 x + b
(0.75)( 2) + b
3 + ( 0.75)(2)
1.5 The xintercept of the line is the xcoordinate of the point at which the graph
intersects the xaxis. One can see from the graph that the xintercept of line PQ
is 2 since PQ passes through the point (2, 0). Also, one can see that the
coordinates (2, 0) satisfy the equation of line PQ, which is y = 0.75 x  1.5. 30 ALGEBRA EXERCISES
(Answers on pages 34 and 35)
1. Find an algebraic expression to represent each of the following.
(a) The square of y is subtracted from 5, and the result is multiplied by 37.
(b) Three times x is squared, and the result is divided by 7.
(c) The product of ( x + 4 ) and y is added to 18.
2. Simplify each of the following algebraic expressions by doing the indicated
operations, factoring, or combining terms with the same variable part.
(a) 3 x 2  6 + x + 11  x 2 + 5 x
(b) 3(5 x  1)  x + 4
x 2 + 9  25
( x 4)
(c)
x4
(d) (2 x + 5)(3x  1) 3 8 3. What is the value of the function defined by f ( x ) = 3 x 2  7 x + 23 when
x = 2 ? 05 4. If the function g is defined for all nonzero numbers y by g y = y
, what
y is the value of g(2 )  g( 200 ) ?
5. Use the rules of exponents to simplify the following.
(a)
(b) 3n 83n  8
3s 83t 8
5 7 3 7 r 12
(c)
r4
(d) 2ba (e)
(f) 3w 8 35 83d 8
3 x 83 y  8
3 x  83 y 8
3x 1
y
y
5 3 0 3 10 (g) 1 5 5 2 5 (h) 5 6. Solve each of the following equations for x.
(a)
(b)
(c)
(d)
(e)
(f) 5 x  7 = 28
12  5 x = x + 30
5( x + 2 ) = 1  3 x
( x + 6 )(2 x  1) = 0
x 2 + 5 x  14 = 0
3 x 2 + 10 x  8 = 0 31 7. Solve each of the following systems of equations for x and y.
(a) x + y = 24
x  y = 18
(b) 3 x  y = 20
x + 2 y = 30
(c) 15 x  18  2 y = 3 x + y
10 x + 7 y + 20 = 4 x + 2 8. Solve each of the following inequalities for x.
(a) 3x > 7 + x
(b) 25 x + 16 10  x
(c) 16 + x > 8 x  12
9. Solve for x and y.
x = 2y
5x < y + 7 10. For a given twodigit positive integer, the tens digit is 5 greater than
the units digit. The sum of the digits is 11. Find the integer.
11. If the ratio of 2x to 5y is 3 to 4, what is the ratio of x to y ?
12. Kathleen’s weekly salary was increased 8 percent to $237.60. What was
her weekly salary before the increase?
13. A theater sells children’s tickets for half the adult ticket price. If 5 adult
tickets and 8 children’s tickets cost a total of $27, what is the cost of
an adult ticket?
14. Pat invested a total of $3,000. Part of the money yields 10 percent interest
per year, and the rest yields 8 percent interest per year. If the total yearly
interest from this investment is $256, how much did Pat invest at 10 percent
and how much at 8 percent?
15. Two cars started from the same point and traveled on a straight course in
opposite directions for exactly 2 hours, at which time they were 208 miles
apart. If one car traveled, on average, 8 miles per hour faster than the other
car, what was the average speed for each car for the 2hour trip?
16. A group can charter a particular aircraft at a fixed total cost. If 36 people
charter the aircraft rather than 40 people, then the cost per person is greater by
$12. What is the cost per person if 40 people charter the aircraft? 32 17. If 3 times Jane’s age, in years, is equal to 8 times Beth’s age, in years, and
the difference between their ages is 15 years, how old are Jane and Beth?
18. In the coordinate system below, find the
(a)
(b)
(c)
(d) coordinates of point Q
perimeter of PQR
area of PQR
slope, yintercept, and equation of the line passing through
points P and R 19. In the xyplane, find the
(a) slope and yintercept of a graph with equation 2 y + x = 6
(b) equation of the straight line passing through the point (3, 2) with
yintercept 1
(c) yintercept of a straight line with slope 3 that passes through the
point ( 2, 1)
(d) xintercepts of the graphs in (a), (b), and (c) 33 ANSWERS TO ALGEBRA EXERCISES 3 8 1. (a) 37 5  y 2 , or 185  37 y 2
(b) (3 x )
9x 2
, or
7
7
2 05 (c) 18 + ( x + 4) y , or 18 + xy + 4 y
2. (a) 2 x 2 + 6 x + 5
(b) 14 x + 1 (c) x + 4
(d) 6 x 2 + 13 x  5 3. 49
4. 2
5. (a) n 2 (e) 1
w15 (b) ( st ) 7 (f) d 3 (c) r 8 (g) (d) 32a 5
b5 6. (a) 7 x 15
y6 (h) 9 x 2 y 3
(d)  6, 1
2 (b) 3
9
(c) 8 (e) 7, 2
2
(f)
, 4
3 7. (a) x = 21
y=3 (c) x = 1
2
y = 3 (b) x = 10
y = 10 7
4
3
(b) x 13 8. (a) x <  9. x < 34 14
7
,y<
9
9 (c) x < 4 10. 83
11. 15 to 8
12. $220
13. $3
14. $800 at 10%; $2,200 at 8%
15. 48 mph and 56 mph
16. $108
17. Beth is 9; Jane is 24.
18. (a) ( 2, 0 )
(b) 13 + (c) 21 19. (a) slope = (b) y = 30
6
,
, yintercept =
7
7
6
30
y=
x+
, or 7 y + 6 x = 30
7
7 (d) slope = 85 1
, yintercept = 3
2 x
+1
3 (c) 7
(d) 6, 3,  7
3 35 GEOMETRY
3.1 Lines and Angles
In geometry, a basic building block is the line, which is understood to be a
“straight” line. It is also understood that lines are infinite in length. In the figure
below, A and B are points on line l. That part of line l from A to B, including the endpoints A and B, is called a
line segment, which is finite in length. Sometimes the notation “AB” denotes
line segment AB and sometimes it denotes the length of line segment AB.
The exact meaning of the notation can be determined from the context.
Lines l1 and l 2 , shown below, intersect at point P. Whenever two lines
intersect at a single point, they form four angles. Opposite angles, called vertical angles, are the same size, i.e., have equal measure. Thus, µ APC and µ DPB have equal measure, and µ APD and µCPB
also have equal measure. The sum of the measures of the four angles is 360 .
If two lines, l1 and l 2 , intersect such that all four angles have equal measure
(see figure below), we say that the lines are perpendicular, or l 1 ^ l 2 , and each
of the four angles has a measure of 90 . An angle that measures 90 is called
a right angle, and an angle that measures 180 is called a straight angle. 36 If two distinct lines in the same plane do not intersect, the lines are said to be
parallel. The figure below shows two parallel lines, l1 and l 2 , which are intersected by a third line, l 3 , forming eight angles. Note that four of the angles have
equal measure (x° ) and the remaining four have equal measure ( y° ) where
x + y = 180. 3.2 Polygons
A polygon is a closed figure formed by the intersection of three or more line
segments, called sides, with all intersections at endpoints, called vertices. In this
discussion, the term “polygon” will mean “convex polygon,” that is, a polygon in
which the measure of each interior angle is less than 180 . The figures below are
examples of such polygons. The sum of the measures of the interior angles of an nsided polygon is
(n  2)(180 ). For example, the sum for a triangle (n = 3) is
(3  2)(180 ) = 180 , and the sum for a hexagon (n = 6) is
(6  2)(180 ) = 720 .
A polygon with all sides the same length and the measures of all interior
angles equal is called a regular polygon. For example, in a regular octagon
(8 sides of equal length), the sum of the measures of the interior angles is
(8  2 )(180 ) = 1,080 . Therefore, the measure of each angle is
1,080 8 = 135 . 37 The perimeter of a polygon is defined as the sum of the lengths of its sides.
The area of a polygon is the measure of the area of the region enclosed by the
polygon.
In the next two sections, we look at some basic properties of the simplest
polygons—triangles and quadrilaterals. 3.3 Triangles
Every triangle has three sides and three interior angles whose measures sum
to 180 . It is also important to note that the length of each side must be less
than the sum of the lengths of the other two sides. For example, the sides of
a triangle could not have lengths of 4, 7, and 12 because 12 is not less
than 4 + 7.
The following are special triangles.
(a) A triangle with all sides of equal length is called an equilateral triangle.
The measures of three interior angles of such a triangle are also equal
(each 60).
(b) A triangle with at least two sides of equal length is called an isosceles
triangle. If a triangle has two sides of equal length, then the measures of
the angles opposite the two sides are equal. The converse of the previous
statement is also true. For example, in ABC below, since both µABC
and µBCA have measure 50 , it must be true that BA = AC. Also,
since 50 + 50 + x = 180, the measure of µBAC must be 80 . (c) A triangle with an interior angle that has measure 90 is called a right
triangle. The two sides that form the 90 angle are called legs and the
side opposite the 90 angle is called the hypotenuse. 38 For right DEF above, DE and EF are legs and DF is the hypotenuse. The
Pythagorean Theorem states that for any right triangle, the square of the length
of the hypotenuse equals the sum of the squares of the lengths of the legs. Thus,
in right DEF,
( DF ) 2 = ( DE ) 2 + ( EF ) 2 . This relationship can be used to find the length of one side of a right triangle
if the lengths of the other two sides are known. For example, if one leg of a right
triangle has length 5 and the hypotenuse has length 8, then the length of the other
side can be calculated as follows: Since x 2 = 39 and x must be positive, x = 39, or approximately 6.2.
The Pythagorean Theorem can be used to determine the ratios of the
sides of two special right triangles:
An isosceles right triangle has angles measuring 45, 45, 90. The
Pythagorean Theorem applied to the triangle below shows that the lengths
of its sides are in the ratio 1 to 1 to 2 . A 30  60  90 right triangle is half of an equilateral triangle, as the
following figure shows. 39 So the length of the shortest side is half the longest side, and by the Pythagorean
Theorem, the ratio of all three side lengths is 1 to 3 to 2, since
x 2 + y 2 = (2 x ) 2
x 2 + y2 = 4x2
y2 = 4x 2  x 2
y 2 = 3x 2
y= 3x The area of a triangle is defined as half the length of a base (b) multiplied
by the corresponding height (h), that is,
Area = bh
.
2 Any side of a triangle may be considered a base, and then the corresponding
height is the perpendicular distance from the opposite vertex to the base (or an
extension of the base). The examples below summarize three possible locations
for measuring height with respect to a base. In all three triangles above, the area is (15)(6)
, or 45.
2 3.4 Quadrilaterals
Every quadrilateral has four sides and four interior angles whose measures
sum to 360 . The following are special quadrilaterals.
(a) A quadrilateral with all interior angles of equal measure (each 90) is
called a rectangle. Opposite sides are parallel and have equal length,
and the two diagonals have equal length. A rectangle with all sides of equal length is called a square. 40 (b) A quadrilateral with both pairs of opposite sides parallel is called a
parallelogram. In a parallelogram, opposite sides have equal length,
and opposite interior angles have equal measure. (c) A quadrilateral with one pair of opposite sides parallel is called
a trapezoid. For all rectangles and parallelograms the area is defined as the length of the
base (b) multiplied by the height (h), that is
Area = bh
Any side may be considered a base, and then the height is either the length of an
adjacent side (for a rectangle) or the length of a perpendicular line from the base
to the opposite side (for a parallelogram). Here are examples of each: 0 5 The area of a trapezoid may be calculated by finding half the sum of the
lengths of the two parallel sides b1 and b2 and then multiplying the result
by the height (h), that is,
Area = 0 5 1
b + b2 (h).
21 For example, for the trapezoid shown below with bases of length 10 and 18, and
a height of 7.5, 41 3.5 Circles
The set of all points in a plane that are a given distance r from a fixed
point O is called a circle. The point O is called the center of the circle, and the
distance r is called the radius of the circle. Also, any line segment connecting
point O to a point on the circle is called a radius. Any line segment that has its endpoints on a circle, such as PQ above, is
called a chord. Any chord that passes through the center of a circle is called a
diameter. The length of a diameter is called the diameter of a circle. Therefore,
the diameter of a circle is always equal to twice its radius.
The distance around a circle is called its circumference (comparable to the
perimeter of a polygon). In any circle, the ratio of the circumference c to the
diameter d is a fixed constant, denoted by the Greek letter p: c
=p
d
The value of p is approximately 3.14 and may also be approximated by the
22
c
= p, so the circumference
fraction
. If r is the radius of the circle, then
2r
7
is related to the radius by the equation c = 2 pr.
Therefore, if a circle has a radius equal to 5.2, then its circumference
is (2 )( p )(5.2) = (10.4)( p ), which is approximately equal to 32.7. 42 On a circle, the set of all points between and including two given points is
called an arc. It is customary to refer to an arc with three points to avoid ambiguity. In the figure below, arc ABC is the short arc from A to C, but arc ADC is
the long arc from A to C in the reverse direction. Arcs can be measured in degrees. The number of degrees of arc equals the
number of degrees in the central angle formed by the two radii intersecting the
arc’s endpoints. The number of degrees of arc in the entire circle (one complete
revolution) is 360. Thus, in the figure above, arc ABC is a 50 arc and arc ADC
is a 310 arc.
To find the length of an arc, it is important to know that the ratio of arc
length to circumference is equal to the ratio of arc measure (in degrees) to 360.
In the figure above, the circumference is 10 p. Therefore,
length of arc ABC
50
=
10 p
360
length of arc ABC = 50
360 (10p ) = 25p
18 The area of a circle with radius r is equal to pr 2 . For example, the area
of the circle above is p (5) 2 = 25p. In this circle, the pieshaped region bordered
by arc ABC and the two dashed radii is called a sector of the circle, with central
angle 50 . Just as in the case of arc length, the ratio of the area of the sector to
the area of the entire circle is equal to the ratio of the arc measure (in degrees)
to 360. So if S represents the area of the sector with central angle 50 , then
S
50
=
360
25p
50
125p
S=
(25p ) =
360
36 43 A tangent to a circle is a line that has exactly one point in common with the
circle. A radius with its endpoint at the point of tangency is perpendicular to the
tangent line. The converse is also true. If each vertex of a polygon lies on a circle, then the polygon is inscribed
in the circle, or equivalently, the circle is circumscribed about the polygon.
Triangle RST below is inscribed in the circle with center O. If each side of a polygon is tangent to a given circle, then the polygon is
circumscribed about the circle, or equivalently, the circle is inscribed in the
polygon. In the figure below, quadrilateral ABCD is circumscribed about the
circle with center O. Two or more circles with the same center are called concentric circles. 44 3.6 ThreeDimensional Figures
Basic threedimensional figures include rectangular solids, cubes, cylinders,
spheres, pyramids, and cones. In this section, we look at some properties of
rectangular solids and right circular cylinders.
(a) A rectangular solid has six rectangular surfaces called faces (see figure
below). Each line segment shown is called an edge (there are 12 edges),
and each point at which the edges meet is called a vertex (there are
8 vertices). The dimensions of a rectangular solid are length ( l),
width (w), and height (h). A rectangular solid with l = w = h is called a cube. The volume V of
a rectangular solid is the product of the three dimensions,
V = lwh.
The surface area A of a rectangular solid is the sum of the areas of the
six faces, or
A = 2( wl + lh + wh).
For example, if a rectangular solid has length 8.5, width 5, and height 10,
then its volume is
V = (8.5)(5)(10) = 425,
and its surface area is
A = 2[(5)(8.5) + (8.5)(10) + (5)(10)] = 355. 45 (b) A right circular cylinder is shown in the figure below. Its bases are
circles with equal radii and centers P and Q, respectively, and its height
PQ is perpendicular to both bases. The volume V of a right circular cylinder with a base radius r and
height h is the area of the base multiplied by the height, or
V = pr 2 h.
The surface area A of a right circular cylinder is the sum of the two
base areas and the area of the curved surface, or 38 A = 2 pr 2 + 2 prh. For example, if a right circular cylinder has a base radius of 3 and a
height of 6.5, then its volume is
V = p (3) 2 (6.5) = 58.5p , and its surface area is
A = (2)( p )(3) 2 + (2)( p )(3)(6.5) = 57p. 46 GEOMETRY EXERCISES
(Answers on page 50)
1. Lines l and m below are parallel. Find the values of x and y. 2. In the figure below, AC = BC. Find the values of x and y. 3. In the figure below, what relationship must hold among angle measures x,
y, and z ? 4. What is the sum of the measures of the interior angles of a decagon (10sided
polygon) ?
5. If the polygon in #4 is regular, what is the measure of each interior angle?
6. The lengths of two sides of an isosceles triangle are 15 and 22, respectively.
What are the possible values of the perimeter? 47 7. In rectangle ABDE below, AB = 5, BC = 7, and CD = 3. Find the
(a) area of ABDE
(b) area of triangle BCF
(c) length of AD
(d) perimeter of ABDE 8. In parallelogram ABCD below, find the
(a) area of ABCD
(b) perimeter of ABCD
(c) length of diagonal AC 9. The circle with center O below has radius 4. Find the
(a) circumference
(b) length of arc ABC
(c) area of the shaded region 48 10. The figure below shows two concentric circles each with center O. If the
larger circle has radius 12 and the smaller circle has radius 8, find the
(a) circumference of the larger circle
(b) area of the smaller circle
(c) area of the shaded region 11. For the rectangular solid below, find the
(a) surface area
(b) length of diagonal AB 49 ANSWERS TO GEOMETRY EXERCISES
1. x = 57, y = 138
2. x = 70, y = 125
3. z = x + y
4. 1,440
5. 144
6. 52 or 59
7. (a) 50
(b) 17.5
8. (a) 48 (c) 5 5
(d) 30
(c) 2 29 (b) 24 + 4 5
9. (a) 8p
(b) 16 p
9 8p
9 10. (a) 24 p
(b) 64 p
11. (a) 208
(b) 3 17 50 (c) (c) 80 p DATA ANALYSIS
4.1 Measures of Central Location
Two common measures of central location, often called “average,” for a
discrete set of numerical values or measurements are the arithmetic mean and
the median.
The average (arithmetic mean) of n values is defined as the sum of the
n values divided by n. For example, the arithmetic mean of the values 5, 8,
8, 14, 15, and 10 is 60 6 = 10 .
If we order the n values from least to greatest, the median is defined as the
middle value if n is odd and the sum of the two middle values divided by 2 if
n is even. In the example above, n = 6, which is even. Ordered from least to
greatest, the values are 5, 8, 8, 10, 14, and 15. Therefore, the median is 8 + 10
= 9.
2
Note that for the same set of values, the arithmetic mean and the median need
not be equal, although they could be. For example, the set of values 10, 20, 30,
40, and 50 has arithmetic mean = median = 30.
Another measure of central location is called the mode, which is defined
as the most frequently occurring value. For the six measurements above, the
mode is 8. 4.2 Measures of Dispersion
Measures of dispersion, or spread, for a discrete set of numerical values or
measurements take many forms in data analyses. The simplest measure of dispersion is called the range, which is defined as the greatest measurement minus the
least measurement. So, in the example in 4.1 above, the range for the six values
is 15 minus 5, or 10.
Since the range is affected by only the two most extreme values in the set
of measurements, other measures of dispersion have been developed that are
affected by every measurement. The most commonly used of these other
measures is called the standard deviation. The value of the standard deviation for
a set of n measurements can be calculated by (1) first calculating the arithmetic
mean, (2) finding the difference between that mean and each measurement,
(3) squaring each of the differences, (4) summing the squared values, (5) dividing
the sum by n, and finally (6) taking the nonnegative square root of the quotient.
The following demonstrates this calculation for the example used in 4.1. 51 x
5
8
8
10
14
15 x  10
−5
−2
−2
0
4
5 ( x  10) 2
25
4
4
0
16
25 standard deviation = 74 3.5
6 74
The standard deviation can be roughly interpreted as the average distance
from the arithmetic mean for the n measurements. The standard deviation cannot
be negative, and when two sets of measurements are compared, the one with the
larger dispersion will have the larger standard deviation. 4.3 Frequency Distributions
For some sets of measurements, it is more convenient and informative to
display the measurements in a frequency distribution. For example, the following
values could represent the number of dependent children in each of 25 families
living on a particular street.
1, 2, 0, 4, 1, 3, 3, 1, 2, 0, 4, 5, 2,
3, 2, 3, 2, 4, 1, 2, 3, 0, 2, 3, 1
These data can be grouped into a frequency distribution by listing each different
value (x) and the frequency ( f ) of occurrence for each value.
Frequency Distribution
x
0 3 1 5 2 7 3 6 4 3 5 1 Total 52 f 25 The frequency distribution format not only provides a quick summary of the
data, but it also simplifies the calculations of the central location and dispersion
measures. For these data, the x’s can be summed by multiplying each x by its
frequency and then adding the products. So, the arithmetic mean is
(0 )(3) + (1)(5) + (2 )(7) + (3)(6) + ( 4)(3) + (5)(1)
= 2.16
25 The median is the middle (13th) x value in order of size. The f values show
that the 13th x value must be a 2. The range is 5 minus 0, or 5. The standard
deviation can also be calculated more easily from a frequency distribution,
although in practice it is likely that a programmable calculator would be
used to calculate both the mean and the standard deviation directly from the
25 measurements. 4.4 Counting
Some definitions and principles basic to counting are:
(a) If one task has n possible outcomes and a second task has m possible
outcomes, then the joint occurrence of the two tasks has (n)(m) possible
outcomes. For example, if Town A and Town B are joined by
3 different roads, and Town B and Town C are joined by 4 different
roads, then the number of different routes from Town A to Town C
through B is (3)(4), or 12. Each time a coin is flipped, there are
2 possible outcomes: heads or tails. Therefore, if a coin is flipped
4 times, then the number of possible outcomes is (2)(2)(2)(2), or 16.
(b) For any integer n greater than 1, the symbol n!, pronounced
“n factorial,” is defined as the product of all positive integers
less than or equal to n. Also, 0! = 1! = 1. Therefore,
0! = 1
1! = 1
2! = (2)(1) = 2
3! = (3)(2)(1) = 6
4! = (4)(3)(2)(1) = 24
and so on.
(c) The number of ways that n objects can be ordered is n!. For example,
the number of ways that the letters A, B, and C can be ordered is 3!,
or 6. The six orders are
ABC, ACB, BAC, BCA, CAB, and CBA 53 (d) The number of different subsets of r objects that can be selected from
n objects ( r n) , without regard to the order of selection, is n!
.
( n  r )! r !
For example, the number of different committees of 3 people that can
be selected from 5 people is
5!
5!
120
=
=
= 10.
2! 3!
(2 )(6)
(5  3)!3! These 10 subsets are called combinations of 5 objects selected 3 at
a time. 4.5 Probability
Everyday there are occasions in which decisions must be made in the face of
uncertainty. The decisionmaking process often involves the selection of a course
of action based on an analysis of possible outcomes. For situations in which the
possible outcomes are all equally likely, the probability that an event E occurs,
represented by “P(E)”, can be defined as P( E ) = The number of outcomes involving the occurrence of E
.
The total number of possible outcomes For example, if a committee of 11 students consists of 2 seniors, 5 juniors, and
4 sophomores, and one student is to be selected at random to chair the committee,
2
then the probability that the student selected will be a senior is .
11
In general, “P(E )” can be thought of as a number assigned to an event
E which expresses the likelihood that E occurs. If E cannot occur, then
P( E ) = 0, and if E must occur, then P( E ) = 1. If the occurrence of E is
uncertain, then 0 < P( E ) < 1. The probability that event E does NOT occur
is 1  P( E ). For example, if the probability is 0.75 that it will rain tomorrow,
then the probability that it will not rain tomorrow is 1  0.75, or 0.25.
The probability that events E and F both occur can be represented by
P(E and F )*, and the probability that at least one of the two events occurs can
be represented by P(E or F )**. One of the fundamental relationships among
probabilities is called the Addition Law:
P( E or F ) = P( E ) + P( F )  P( E and F ). _________ *Many texts use P( E ¬ F ).
**Many texts use P( E F ). 54 For example, if a card is to be selected randomly from a standard deck of
52 playing cards, E is the event that a heart is selected, and F is the event
that a 9 is selected, then P( E ) =
Therefore, P( E or F ) = 13
4
1
, P( F ) =
, and P( E and F ) =
.
52
52
52
13
4
1
16
4
+
=
=
.
52 52 52
52
13 Two events are said to be independent if the occurrence or nonoccurrence of
either one in no way affects the occurrence of the other. It follows that if events
E and F are independent events, then P( E and F ) = P( E ) P( F ). Two events
are said to be mutually exclusive if the occurrence of either one precludes the
occurrence of the other. In other words, if events E and F are mutually
exclusive, then P ( E and F ) = 0.
Example: If P( A) = 0.45 and P( B) = 0.20, and the two events are
independent, what is P( A or B) ?
According to the Addition Law:
P( A or B) =
=
=
= P( A) + P( B)  P( A and B)
P( A) + P( B)  P( A) P( B)
0 . 45 + 0 . 20  (0.45)(0.20)
0 . 56 If the two events in the example above had been mutually exclusive, then
P(A or B) would have been found as follows:
P( A or B) = P( A) + P( B)  P( A and B)
= 0 . 45 + 0 . 20  0
= 0 . 65 4.6 Data Representation and Interpretation
Data can be summarized and represented in various forms, including tables,
bar graphs, circle graphs, line graphs, and other diagrams. The following are
several examples of tables and graphs, each with questions that can be answered
by selecting the appropriate information and applying mathematical techniques. 55 Example 1. (a) For which year shown on the graph did exports exceed the previous
year’s exports by the greatest dollar amount?
(b) In 1973 the dollar value of imports was approximately what percent
of the dollar value of exports?
(c) If it were discovered that the import dollar amount shown for 1978 was
incorrect and should have been $3.1 billion instead, then the average
(arithmetic mean) import dollar amount per year for the 13 years would
be how much less?
Solutions:
(a) The greatest increase in exports from one year to the next is represented
by the dotted line segment with the steepest positive slope, which is
found between 1976 and 1977. The increase was approximately
$6 billion. Thus, the answer is 1977.
(b) In 1973, the dollar value of imports was approximately $3.3 billion,
and the dollar value of exports was $13 billion. Therefore, the answer
3.3
, or approximately 25%.
is
13
(c) If the import dollar amount in 1978 were $3.1 billion, rather than the
amount $7 billion from the graph, then the sum of the import amounts for the
13 years would be reduced by $3.9 billion. Therefore, the average
$3.9
billion, which is $0.3 billion, or
per year would be reduced by
13
$300 million. 56 Example 2. (a) In 1971, what was the ratio of the value of sensitized goods to the value of
stillpicture equipment produced in the United States?
(b) If the value of office copiers produced in 1971 was 30 percent higher than
the corresponding value in 1970, what was the value of office copiers
produced in 1970 ?
(c) If the areas of the sectors in the circle graph are drawn in proportion to
the percents shown, what is the measure, in degrees, of the central angle
of the sector representing the percent of prepared photochemicals
produced?
Solutions:
(a) The ratio of the value of sensitized goods to the value of stillpicture
equipment is equal to the ratio of the corresponding percents shown.
Therefore, the ratio is 47 to 12, or approximately 4 to 1.
(b) The value of office copiers produced in 1971 was (0.25)($3,980 million), or
$995 million. Therefore, if the corresponding value in 1970 was x,
then x(1.30) = $995 million, or x = $765 million.
(c) Since the sum of the central angles for the six sectors is 360, the
central angle for the sector representing prepared photochemicals is
(0.07)(360 ) , or 25.2 . 57 Example 3. (a) For which year was the ratio of parttime enrollment to total enrollment
the greatest?
(b) What was the fulltime enrollment in 1977 ?
(c) What was the percent increase in total enrollment from 1976 to 1980 ? 58 Solutions:
(a) It is visually apparent that the height of the shaded bar compared to the
total height of the bar is greatest in 1978 (about half the total height).
No calculations are necessary.
(b) In 1977 the total enrollment was approximately 450 students, and the
parttime enrollment was approximately 150 students. Thus, the fulltime
enrollment was 450  150, or 300 students.
(c) The total enrollments for 1976 and 1980 were approximately 400 and
750, respectively. Therefore, the percent increase from 1976 to 1980 was
750  400
350
=
= 0.875 = 87.5%.
400
400 Example 4.
CONSUMER COMPLAINTS RECEIVED
BY THE CIVIL AERONAUTICS BOARD
Category
Flight Problems........................................................
Baggage ...................................................................
Customer service......................................................
Oversales of seats.....................................................
Refund problems......................................................
Fares.........................................................................
Reservations and ticketing .......................................
Tours ........................................................................
Smoking...................................................................
Advertising...............................................................
Credit .......................................................................
Special passengers ...................................................
Other ........................................................................ 1980
(percent) 20.0%
18.3
13.1
10.5
10.1
6.4
5.8
3.3
3.2
1.2
1.0
0.9
6.2
100.0%
Total Number of Complaints ................................... 22,998 1981
(percent)
22.1%
21.8
11.3
11.8
8.1
6.0
5.6
2.3
2.9
1.1
0.8
0.9
5.3
100.0%
13,278 (a) Approximately how many complaints concerning credit were received by
the Civil Aeronautics Board in 1980 ?
(b) By approximately what percent did the total number of complaints
decrease from 1980 to 1981 ?
(c) Which of the following statements can be inferred from the table?
I. In 1980 and in 1981, complaints about flight problems, baggage,
and customer service together accounted for more than 50 percent
of all consumer complaints received by the Civil Aeronautics
Board. 59 II. The number of special passenger complaints was unchanged
from 1980 to 1981.
III. From 1980 to 1981, the number of flight problem complaints
increased by more than 2 percent.
Solutions:
(a) In 1980, 1 percent of the complaints concerned credit, so the number
of complaints was approximately (0.01)(22,998), or 230.
(b) The decrease in total complaints from 1980 to 1981 was
22,998  13,278, or 9,720. Therefore, the percent decrease
was 9,720 22,998, or 42 percent.
(c) Since 20.0 + 18.3 + 13.1 and 22.1 + 21.8 + 11.3 are both greater than
50, statement I is true. The percent of special passenger complaints did
remain the same for 1980 to 1981, but the number of special passenger
complaints decreased because the total number of complaints decreased.
Thus, statement II is false. The percents shown in the table for flight
problems do in fact increase more than 2 percentage points. However,
the number of flight problem complaints in 1980 was (0.2)(22,998), or
4,600, and the number in 1981 was (0.221)(13,278), or 2,934. So, the
number of flight problem complaints actually decreased from 1980 to
1981. Therefore, statement I is the only statement that can be inferred
from the table.
Example 5. In a survey of 250 European travelers, 93 have traveled to Africa, 155 have
traveled to Asia, and 70 have traveled to both of these continents, as illustrated
in the Venn diagram above.
(a) How many of the travelers surveyed have traveled to Africa but not to
Asia?
(b) How many of the travelers surveyed have traveled to at least one of the
two continents Africa and Asia?
(c) How many of the travelers surveyed have traveled neither to Africa nor
to Asia? 60 Solutions:
A Venn diagram is useful for sorting out various sets and subsets that may
overlap. The rectangular region represents the set of all travelers surveyed; the
two circular regions represent the two groups of travelers to Africa and Asia;
and the shaded region represents the subset of those who have traveled to both
continents.
(a) The set described here is represented by that part of the left circle that
is not shaded. This description suggests that the answer can be found by
taking the shaded part away from the first circle—in effect, subtracting
the 70 from the 93, to get 23 travelers who have traveled to Africa but not
to Asia.
(b) The set described here is represented by that part of the rectangle that
is in at least one of the two circles. This description suggests adding the
two numbers 93 and 155. But the 70 travelers who have traveled to both
continents would be counted twice in the sum 93 + 155. To correct the
double counting, subtract 70 from the sum so that these 70 travelers are
counted only once:
93 + 155  70 = 178.
(c) The set described here is represented by that part of the rectangle that is
not in either circle. Let N be the number of these travelers. Note that the
entire rectangular region has two main nonoverlapping parts: the part
outside the circles and the part inside the circles. The first part represents
N travelers and the second part represents 93 + 155  70 = 178 travelers
(from question (b)). Therefore,
250 = N + 178, and solving for N yields N = 250  178 = 72. 61 DATA ANALYSIS EXERCISES
(Answers on page 69)
1. The daily temperatures, in degrees Fahrenheit, for 10 days in May were
61, 62, 65, 65, 65, 68, 74, 74, 75, and 77.
(a) Find the mean, median, and mode for the temperatures.
(b) If each day had been 7 degrees warmer, what would have been the mean,
median, and mode for those 10 measurements?
2. The ages, in years, of the employees in a small company are 22, 33, 21, 28,
22, 31, 44, and 19.
(a) Find the mean, median, and mode for the 8 ages.
(b) Find the range and standard deviation for the 8 ages.
(c) If each of the employees had been 10 years older, what would have been
the range and standard deviation of their ages?
3. A group of 20 values has mean 85 and median 80. A different group of
30 values has mean 75 and median 72.
(a) What is the mean of the 50 values?
(b) What is the median of the 50 values?
4. Find the mean, median, mode, range, and standard deviation for x, given the
frequency distribution below.
x
0
1
2
3
4 62 f
2
6
3
2
4 5. In the frequency distribution below, y represents age on last birthday for
40 people. Find the mean, median, mode, and range for y.
y f 17
18
19
20
21
22
23 2
7
19
9
2
0
1 6. How many different ways can the letters in the word STUDY be ordered?
7. Martha invited 4 friends to go with her to the movies. There are 120 different
ways in which they can sit together in a row. In how many of those ways is
Martha sitting in the middle?
8. How many 3digit positive integers are odd and do not contain the digit “5”?
9. From a box of 10 light bulbs, 4 are to be removed. How many different sets
of 4 bulbs could be removed?
10. A talent contest has 8 contestants. Judges must award prizes for first, second,
and third places. If there are no ties, (a) in how many different ways can the
3 prizes be awarded, and (b) how many different groups of 3 people can get
prizes?
11. If the probability is 0.78 that Marshall will be late for work at least once next
week, what is the probability that he will not be late for work next week? 63 12. If an integer is randomly selected from all positive 2digit integers
(i.e., the integers 10, 11, 12, . . . , 99), find the probability that the
integer chosen has
(a) a “4” in the tens place
(b) at least one “4”
(c) no “4” in either place
13. In a box of 10 electrical parts, 2 are defective.
(a) If one part is chosen randomly from the box, what is the probability that
it is not defective?
(b) If two parts are randomly chosen from the box, without replacement,
what is the probability that both are defective?
14. The table shows the distribution of a group of 40 college students by gender
and class.
Sophomores
Males
Females Juniors Seniors 6 10 2 10 9 3 If one student is randomly selected from this group, find the probability
that the student chosen is
(a) not a junior
(b) a female or a sophomore
(c) a male sophomore or a female senior
15. P( A or B) = 0.60 and P( A) = 0.20.
(a) Find P(B) given that events A and B are mutually exclusive.
(b) Find P(B) given that events A and B are independent.
16. Lin and Mark each attempt independently to decode a message. If the
probability that Lin will decode the message is 0.80, and the probability
that Mark will decode the message is 0.70, find the probability that
(a) both will decode the message
(b) at least one of them will decode the message
(c) neither of them will decode the message 64 17. (a) Which station has a high wind speed that is the median of the high wind
speeds for all the stations listed?
(b) For those stations that have recorded hurricane winds at least once
during the 10year period, what is the arithmetic mean of their average
wind speeds?
(c) For how many of the stations is the ratio of high wind speed to average
wind speed greater than 10 to 1 ? 65 18. (a) In which year did total expenditures increase the most from the year
before?
(b) In 1979 private school expenditures were approximately what percent
of total expenditures? 66 19. (a) In 1981, how many categories each comprised more than 25 million
workers?
(b) What is the ratio of the number of workers in the Professional category
in 1981 to the projected number of such workers in 1995 ?
(c) From 1981 to 1995, there is a projected increase in the number of
workers in which of the following categories?
I. Sales
II. Service
III. Clerical 67 20. (a) In 1989 Family X used a total of 49 percent of its gross annual
income for two of the categories listed. What was the total amount
of Family X ’s income used for those same categories in 1990 ?
(b) Family X ’s gross income is the sum of Mr. X ’s income and Mrs. X ’s
income. In 1989 Mr. and Mrs. X each had an income of $25,000. If
Mr. X ’s income increased by 10 percent from 1989 to 1990, by what
percent did Mrs. X ’s income decrease for the same period? 68 ANSWERS TO DATA ANALYSIS EXERCISES
1. (a) mean = 68.6, median = 66.5, mode = 65
(b) Each measure would have been 7 degrees greater.
2. (a) mean = 27.5, median = 25, mode = 22
(b) range = 25, standard deviation 7.8
(c) range = 25, standard deviation 7.8
3. (a) mean = 79
(b) The median cannot be determined from the information given.
4. mean = 2, median = 2, mode = 1, range = 4, standard deviation 1.4
5. mean = 19.15, median = 19, mode = 19, range = 6
6. 120
7. 24
8. 288
9. 210
10. (a) 336 (b) 56 11. 0.22
12. (a) 1
9 (b) 13. (a) 4
5 (b) 21
40 (b) (c) 7
10 4
5 9
40 1
45 14. (a) (c) 1
5 15. (a) 0.40 (b) 0.50 16. (a) 0.56 (b) 0.94 17. (a) New York
18. (a) 1976 (c) 0.06 (b) 10.4 (c) Three (b) 19% 19. (a) Three
20. (a) $17,550 (b) 9 to 14, or 9
14 (c) I, II, and III (b) 30% 69 ...
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