Mathematical Reasoning Workbook for the GED Test

32. The equation relating the number of Chapter 12 ~ Graphing Equations   89
customized cell phones produced and the 34. Graph the equation 5x – 4y = –20.
profit per cell phone is p = –2.50n2 + 21n,
where n is in 100,000s. Plot the resulting y
graph. Be sure to label and number the axes
appropriately and indicate the maximum x
value of p.

y

x 35. The coordinates of four points are given in
absolute terms (all values positive). Plot and
33. Graph the equation of a straight line that label each point in the quadrant requested.
passes through (–1, 3) and has a slope of –3. Different quadrants will require changing the
signs of some or all the coordinates.
y Point A: Plot the point with absolute
coordinates of (3, 6) in the third quadrant.
Point B: Plot the point with absolute
coordinates of (5, 5) in the second quadrant.
Point C: Plot the point with absolute
coordinates of (2, 7) in the third quadrant.
Point D: Plot the point with absolute
coordinates of (5, 9) in the fourth quadrant.

y

x

x

90    McGraw-Hill Education  Mathematical Reasoning Workbook for the GED® Test

The following question contains a set of choices 37. Which graph shows the proper end behavior
marked Select . . . ▾ . Indicate the choice that is for the equation y = 4x2 – 2x as x becomes a
correct. (Note: On the real GED® test, the choices will large positive or negative number?
appear as a “drop-down” menu. When you click on a
choice, it will appear in the blank.) A. y

36. Two rockets are launched. Rocket A follows the x
trajectory shown here. Rocket B follows the
equation e(t) = – 100t2 + 1200t, where e equals
feet of elevation and t equals seconds in flight.
Which rocket climbs to a greater height?

y
3000

2000

Feet

1000

0 12345678 x B. y
Seconds x

Select . . . ▾
Rocket A
Rocket B

y Chapter 12 ~ Graphing Equations   91

C. 38. Sketch the function f(x) = x2 + 6x + 8, and list
the intercepts, if any, with the x and f(x) axes.

y

x
x

D. y

x

92    McGraw-Hill Education  Mathematical Reasoning Workbook for the GED® Test

39. Which graph(s) that follow show symmetry C. y
with the x-axis?

A. y

x

x

D. y
x
B. y

x

40. What are the coordinates of a point symmetric
to the origin with the point (3, –9)?
(,)

41. What are the end behaviors of
y = 3x3 + 15x2 – 6x + 5?
A. +∞ for x < 0 and –∞ for x > 0
B. –∞ for x < 0 and –∞ for x > 0
C. –∞ for x < 0 and +∞ for x > 0
D. +∞ for x < 0 and +∞ for x > 0

42. Which of the following graph(s) does NOT C. Chapter 12 ~ Graphing Equations   93
show symmetry with the y-axis?
y
y

A.

x
x

D. y
x
B. y

x

94    McGraw-Hill Education  Mathematical Reasoning Workbook for the GED® Test

43. What is the slope of line B in the diagram 46. Given the graph below, place the letter of the
below? correct equation in the appropriate box.

y A. y = 2x + 11

A B. y = 1 x + 10
C 2

D C. y = x3 + 5
B
D. y = 3 x + 1
4 2

Distance E. y = – 2 x + 1
3 2

y

x 10

Time 5

A. 3 x
5
3
B. 4 –10 –5 5 10
–5
C. 3
D. 9

44. Are two lines, one passing through (6, 9)

and (2, 4) and another with slope 5 and –10
4
passing through the point (2, 1), parallel? 47. Given the equation y = –2x2 + 5 , graph the
equation so that the maxima or minima
Check Yes or No. intercepts with the x-axis and end behaviors
are evident.
45. Place an X next to the two of the four
lines below that definitely form part of a y
parallelogram.

Line A:  3x – 2y = 10

Line B:  y = – 2 x – 5
3

Line C:  3x + 2y = 10

Line D:  y = 3 x – 15
2

x

48. What is the equation of a line passing through B. Chapter 12 ~ Graphing Equations   95

(4, 5) with slope – 2 ? y
3 x

Fill in the boxes to complete the equation. y
x
y = x +

49. Graph the intersection of a line

perpendicular to y = – 2 x + 5 and passing
3
through (–2, –2) and a second line parallel to
y = –x + 1 and passing through (1, 5).

y

C.

x

50. Which graph shows the proper end behaviors D. y
for the equation y = –x2 + 7x – 7?

A. y

x
x

This page intentionally left blank

CHAPTER 13

Functions

Directions: Answer the following questions. For multiple-choice questions, choose the best answer. For
other questions, follow the directions provided. Answers begin on page 144.

1. Graph y = 3 x. The following question contains a set of choices
2 marked Select . . . ▾ . Indicate the choice that is
correct. (Note: On the real GED® test, the choices will
y appear as a “drop-down” menu. When you click on a
10 choice, it will appear in the blank.)

8 2 4 6 8 10 x 3. Which of the functions has the larger positive
6 x-intercept: Function A, the function graphed
4 here, or Function B, the function given by the
2 equation y = –7x + 24?

–10 –8 –6 –4 –2 y
–2
–4 10
–6 8
–8 6
4
–10 2

–10 –8 –6 –4 –2 2 4 6 8 10 x
–2
2. Which of the tables of input-output pairs,
where x represents the input and y the output, –4
does NOT represent a function?
–6

A. x 3 2 7 3 8 2 –8
y941924 –10

B. x 1 2 4 7 8 9 Select . . . ▾
y482715 Function A
Function B
C. x 1 4 6 4 3 1
y238762 4. Which of the following implicitly defines a
linear function?
D. x 5 8 2 4 2 5
y000300 A. x2 + y2 = 25

B. 3x – 5y = 9
1
C. y = 3x −2

D. xy = 8

97

98    McGraw-Hill Education  Mathematical Reasoning Workbook for the GED® Test

5. Where is the function in the graph increasing? The following question contains a blank marked
Select . . . ▾ . Beneath the blank is a set of choices.
y Indicate the choice that is correct and belongs in the
blank. (Note: On the real GED® test, the choices will
5 appear as a “drop-down” menu. When you click on a
4 choice, it will appear in the blank.)
3
2 9. Function A is given by f(x) = –2x + 10. Function
1 B is graphed below. has the
larger initial value.
–5 –4 –3 –2 –1 x
–1 12345 Select . . . ▾
–2 Function A
–3
–4 Function B
–5
y

A. –2 < x < 1 and x > 3 10
B. x < –2 and x > 3 9
C. x < –2 and 1 < x < 3 8
D. everywhere 7
6
6. Which function assigns the domain value 12 to 5
the range value 7? 4
3
A. f(x) = 2x – 2 2
B. f(x) = 3x – 4 1
1 x
C. f( x ) = 2 x + 1 1 2 3 4 5 6 7 8 9 10

D. f( x ) = 3 x + 12 10. Describe the symmetry displayed by the graph.
7
y

7. For f( x ) = 2 x + 4 , find f(–7). 10
5 5 8
6
A. –14 4
B. –10 2
C. –2
D. 5 –10 –8 –6 –4 –2
–2
–4 2 4 6 8 10 x
–6
8. A function is being used by the manager of a –8
factory producing lawn mowers. The output of
the function tells the manager how many lawn –10
mowers will be made, using the number of
workers for the input. What types of numbers A. symmetric about the x-axis
are acceptable values for the input? B. symmetric about the y-axis
A. Any number will do. C. symmetric about the origin
B. Only positive numbers can be used. D. no symmetry
C. Only integers can be used.
D. Only non-negative integers can be used.

Chapter 13 ~ Functions   99

11. Which equation does NOT implicitly define a 14. Which of the following sets of ordered pairs
linear function? (x, y) illustrate why the equation y2 = x does
A. y = x2 + 5 not represent a function?
B. y = 3x + 2
C. 4x – 6y = 11 A. (4, 2) and (4, –2)
D. y + 3 = –2(x – 5) B. (9, 3) and (4, 2)
C. (16, 4) and (9, 3)
12. A function d(t) is being used to predict the D. (4, 2) and (16, 4)
distance a bird flies t hours after release from
a wildlife rehab program. What numbers are 15. Graph f( x ) = − 3 x + 8.
acceptable domain values? 2
A. any number will do
B. only positive numbers y
C. only integers
D. only non-negative numbers 10 2 4 6 8 10 x
8
The following question contains a blank marked 6
Select . . . ▾ . Beneath the blank is a set of choices. 4
Indicate the choice that is correct and belongs in the 2
blank. (Note: On the real GED® test, the choices will
appear as a “drop-down” menu. When you click on a –10 –8 –6 –4 –2
choice, it will appear in the blank.) –2
13. The graph below shows the distance of a –4
–6
freight train moving away from the center –8
of a large city. A taxi is moving away from
the center of the same city, with distance –10
determined by the function f(t) = 40t + 5.
The  is moving faster. 16. Find f(–3) for f(x) = –2x2 – 7x + 9.
A. –30
Select . . . ▾ B. –18
taxi C. 12
train D. 48

d The following two questions each contain a set of
choices marked Select . . . ▾ . Indicate the choice
350 that is correct. (Note: On the real GED® test, the
325 (10, 320) choices will appear as a “drop-down” menu. When
300 you click on a choice, it will appear in the blank.)
275 (9, 290) 17. A linear function f(x) is specified by the values
250 (8, 260)
225 (7, 230) in the table. Another is given by g(x) = –3x + 5.
200 (6, 200) Which function has the larger slope?
175 (5, 170)
150 x –2 0 3 5 9
125 (4, 140) f(x) –7 –3 3 7 15
100 (3, 110)
Select . . . ▾
75 (2, 80) f(x)
50 (1, 50) g(x)
25 (0, 20)

t
1 2 3 4 5 6 7 8 9 10 11 12

100    McGraw-Hill Education  Mathematical Reasoning Workbook for the GED® Test

18. Which of the functions, g(x) given in the graph 20. Sketch the quadratic function that has a
or f(x) = x2 – 8x + 8, has the larger minimum maximum at (3, 6) and passes through (7, 2)
value? and (–3, –3).

y y

10 2 4 6 8 10 x 10 2 4 6 8 10 x
8 8
6 6
4 4
2 2

–10 –8 –6 –4 –2 –10 –8 –6 –4 –2
–2 –2
–4 –4
–6 –6
–8 –8

–10 –10

Select . . . ▾ 21. What is the domain of the function in the
f(x) graph?
g(x)
y

19. Which type of symmetry is NOT exhibited by 10
this graph? 8
6
y 4
2
6 2 4 6 8 10 x
5 –10 –8 –6 –4 –2
4 123456 x –2
3 –4
2 –6
1 –8

–6 –5 –4 –3 –2 –1 –10
–1
–2 A. x > –5
–3 B. x ≥ –5
–4 C. x ≥ 0
–5 D. x > 4.5
–6
22. Fill in the missing value in the table so that the
A. symmetry about the x-axis pairs represent a function.
B. symmetry about the y-axis x 7 –3 –2 4 –3
C. symmetry about the origin y 4 8 –1 –4
D. symmetry about the diagonal y = x

Chapter 13 ~ Functions   101

23. An accountant uses the function 25. The graph of f(x), as shown here, has two
2000 x-intercepts. The graph of g( x) = x 2 − 2x + 1
R(v) = v + 100 to predict the pattern of has only one. Which of the following
statements is true?
return of a particular investment, where R is
the return, expressed as a percentage, and v is y
the dollar value invested. What return can she
expect from an investment of $400?
10
A. 2% 8
B. 4% 6
C. 5% 4
D. 10% 2

The following question contains a blank marked –10 –8 –6 –4 –2 2 4 6 8 10 x
Select . . . ▾ . Beneath the blank is a set of choices. –2
Indicate the choice that is correct and belongs in the –4
blank. (Note: On the real GED® test, the choices will –6
appear as a “drop-down” menu. When you click on a –8
choice, it will appear in the blank.)
–10

24. Landon and Molly have a footrace. Molly’s A. The x-intercept of g(x) lies between those
distance in feet from the starting line after of f(x).
t seconds is given by M(t) = 3.95t. Landon’s
distance in feet from the starting line after B. The x-intercept of g(x) lies to the right of
t seconds during the race is in the table below. those of f(x).
Assuming they both keep a constant pace,
wins the race. C. The x-intercept of g(x) lies to the left of
those of f(x).
Select . . . ▾
Landon D. The x-intercept of g(x) coincides with one
of those of f(x).

Molly 26. Evaluate f( x) = 5 x 2 − 2 x − 7 for x = −6.
9 3
t 5 10 15 20 25 30 35 40 45 A. 7
L(t) 19.5 39 58.5 78 97.5 117 136.5 156 175.5 B. 13
C. 17
D. 20

The following two questions each contain a set of
choices marked Select . . . ▾ . Indicate the choice
that is correct. (Note: On the real GED® test, the
choices will appear as a “drop-down” menu. When
you click on a choice, it will appear in the blank.)

102    McGraw-Hill Education  Mathematical Reasoning Workbook for the GED® Test

27. A sand hopper is emptied through a chute. 29. Graph the function given in the following table.
The amount w of sand in kilograms x –7 –6 –3 –1 0 2 5 6 8
t seconds after the chute is opened is given f(x) 4 –1 0 7 –2 6 2 –4 1
by w(t) = 1000 − 5t. The hopper next to it is
being filled from a dump truck. The truck’s y
entire load of 125 kilograms of sand is dumped
in 30 seconds. Which is moving sand faster, the 10
open chute or the truck while dumping?

Select . . . ▾ 8
the open chute 6
the dump truck 4

2

28. A tool factory manager is evaluating two 2 4 6 8 10 x
machines that stamp parts for hammers. The –10 –8 –6 –4 –2
result of his testing is displayed in the graph. –2

Which machine should the factory buy? –4

100 –6
90
80 The following que–s8tion contains a set of choices
70 marked Select ▾ . Indicate the choice that is
60 Machine A correct. (Note: . . . real GED® test, the choices will
50 Machine B O–n1t0he
40 appear as a “drop-down” menu. When you click on a
Hammer parts 30 choice, it will appear in the blank.)
20
10 30. Line A passes through the point (0, 17) on the
y-axis and continues to the right, dropping
3 units for every unit to the right. Line B is
given by 5x − 3y = 30. Which line has the
larger x-intercept?

2 4 6 8 10 12 14 16 18 20 Select . . . ▾
Minutes Line A
Line B
Select . . . ▾
Machine A
Machine B 31. What word best describes the function
−143 7
f( x ) = x − 5 ?

A. negative
B. increasing
C. linear
D. symmetric

Chapter 13 ~ Functions   103

32. A set D has 5 elements, and a set R has 35. Mark the area in the graph below where the
3 elements. How many functions can be function is constant.
defined with domain D and range R?
A. 15 y
B. 125
C. 150 10
D. 243 8
6
33. What is the value of the function 4 2 4 6 8 10 x
f(x) = – 143 x – 57 when x = 0? 2
A. – 16151
B. – 75 –10 –8 –6 –4 –2
C. 0 –2
7 –4
D. 5 –6
–8

–10

34. The function I(n) = 28, 000 + 4, 000n The following two questions each contain a set of
represents the average annual income in choices marked Select . . . ▾ . Indicate the choice
dollars for a person with n years of college that is correct. (Note: On the real GED® test, the
education. What is the best interpretation for choices will appear as a “drop-down” menu. When
the equation I(4) = 44, 000? you click on a choice, it will appear in the blank.)

A. A person with 4 years of college should 36. Which is steeper, the line g(x) whose
request an annual average salary of x-intercept is 7 and y-intercept is –5, or the
$44,000 when interviewing. linear function f( x) = −3x + 9?

B. A person with 4 years of college will earn Select . . . ▾
$44,000 more each year on average than if f(x)
they didn’t attend college. g(x)

C. A person with 4 years of college will earn 37. Luz built a function h(t) = −16t 2 + 320t
$44,000 annually, on average. to estimate the height in feet of an object
launched vertically from the ground t seconds
D. A person with 4 years of college should after launch. The object she launches carries
look for a position that starts at $44,000 a miniaturized telemetry broadcaster, from
annually. which she records the data in the table below
during a test launch. Does the function predict
the object’s falling to the ground sooner or
later than is indicated by the telemetry?

t 3 6 9 12 15
T(t) 576 864 864 576 0

Select . . . ▾
sooner
later

104    McGraw-Hill Education  Mathematical Reasoning Workbook for the GED® Test

38. The equation f(a) = b implies that what point 41. The following table displays values in a
is on the graph of f(x)? proportional relationship. The graph that
follows displays a different proportional
A. (a, b) relationship. Which relationship has the larger
B. (b, a) rate of change?
C. (–a, b) x  3  5   8  12
D. (–b, a) P(x) 54 90 144 216

39. Where is the function f(x) = 5 x − 25 y
negative? 8 16
200 (10, 200)
A. x< 5 160
8 120
5
B. x< 2 80
40
C. x> 5
4 2 4 6 8 10 x
D. x = 0

40. Mark the area in the graph below where the Select . . . ▾
function is positive. table
graph
y

10 42. Is the following graph that of a function?
8
6 y
4
2 x 10
2 4 6 8 10 8
–10 –8 –6 –4 –2 6
–2 4 2 4 6 8 10 x
–4 2
–6
–8 –10 –8 –6 –4 –2
–2
–10 –4
–6
The following two questions each contain a set of –8
choices marked Select . . . ▾ . Indicate the choice
that is correct. (Note: On the real GED® test, the –10
choices will appear as a “drop-down” menu. When
you click on a choice, it will appear in the blank.)

Select . . . ▾
yes
no

Chapter 13 ~ Functions   105

43. Evaluate f −23  for f( x) = 4 x 2 − 6x + 3. 45. Which quadratic function is negative over a
A. 3 larger subset of its domain, f(x), with values
B. 12 given in the following table, or g(x), graphed
C. 18 below?
D. 21
x –5 –4 –1  0  3  4 7 8
f(x)  3  0 –6 –7 –7 –6 0 3

44. What is the domain of the function in the y
following graph?
10
y 8
6
10 4 2 4 6 8 10 x
8 2
6 2 4 6 8 10 x
4 –10 –8 –6 –4 –2
2 –2
–4
–10 –8 –6 –4 –2 –6
–2 –8
–4
–6 –10
–8
Select . . . ▾
–10 f(x)
g(x)
and

46. Is the following graph that of a function?

y

The following three questions each contain a set 2
of choices marked Select . . . ▾ . Indicate the 1
choice that is correct. (Note: On the real GED® test, the
choices will appear as a “drop-down” menu. When
you click on a choice, it will appear in the blank.)

–2 –1 x
12

–1

–2

Select . . . ▾
yes
no

106    McGraw-Hill Education  Mathematical Reasoning Workbook for the GED® Test

47. Which of the intercepts of f(x) = −190 x + 9 is Use the following for questions 49–50.
closer to the origin? A baseball is hit by the batter; its height in
feet t seconds after being hit is given by
Select . . . ▾ h(t) = −16t 2 + 128t + 4 .
x-intercept
y-intercept 49. What is the maximum height of the ball,
reached 4 seconds after being hit?
48. Graph f(x) = x 2 − 4 x − 5. A. 16 ft
B. 260 ft
y C. 512 ft
D. 516 ft
10 2 4 6 8 10 x
8 50. What is the domain of the baseball height
6 function?
4 Fill in the boxes to complete the inequality.
2
≤ t ≤
–10 –8 –6 –4 –2
–2
–4
–6
–8

–10

ANSWERS AND SOLUTIONS

Chapter 1 Whole Numbers and Integers

1. 19. B Death Valley is not as far below sea level
as the Dead Sea is.
–80 –60 –40 –20 0 20 40 60 80
20. D The absolute value of a number is the
47 is closer to 50 than to 40. distance of the number from 0.
2.
21. B All statements are true, but in geometric
–80 –60 –40 –20 0 20 40 60 80 terms, “<” means “is to the left of.”

−25 is negative, so it is to the left of 0. 22. D They all have a debt equal to the absolute
3. < Any negative number is less than any value of their balance.

positive number. 23. 6 factors of 42: 1, 2, 3, 6, 7, 14, 21, 42
4. < −47 is farther to the left than −44. factors of 36: 1, 2, 3, 4, 6 , 9, 12, 18, 36
5. > Any positive number is greater than any common factors: 1, 2, 3, 6 
largest of these: 6
negative number.
6. D −(−5) = 5 24. B multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, . . .
7. B The opposite of the opposite of a number multiples of 9: 9, 18, 27, 36, 45, . . .
common multiples: 18, 36, . . . 
is the same number. smallest of these: 18
8. A The vertical bars indicate absolute value.
9. D Absolute value is the distance from 0, 25. B from the friend’s point of view, –85 + 5(17) =
−85 + 85 = 0
which is always positive.
10. D Positive x values are to the right; negative 26. D Adding a negative is a move to the left on
the number line.
y values are below the x-axis.
11. C Quadrants are named counterclockwise, 27. B Number + opposite starts at zero, goes to
the number, and then returns.
starting in the upper right. Quadrant III is to
the left of the y-axis and below the x-axis. 28. B 3 − 7 = 3 + (−7) = −4
12. A Reflections in the x-axis change the sign
of the y value. 29. D 5 − (−6) = 5 + 6 = 11
13. B Reflections in the x-axis change the sign
of the y value. 30. −7 7 + (−5) + (−9) = 2 + (−9) = −7
14. D Reflections in the y-axis change the sign
of the x value. 31. C −3 − (−8) + (−4) − 5 = −3 + 8 + (−4) + (−5) =
15. C C hanging the sign of 0 does not affect its 5 + (−4) + (−5) = 1 + (−5) = −4
value.
16. S (−3, 2): 3 units left and 2 units up 32. C −9 + 3 + (−4) = −6 + (−4) = −10
17. X (4, –5): 4 units right, 5 units down
18. B −3 is a larger number than −7 and so is to 33. A − 8 − (−10) − 5 = −8 + 10 + (−5) =
its right. 2 + (−5) = −3

34. C The product of two numbers with
identical signs is positive.

35. B 2(−4)(−1) = (−8)(−1) = 8

36. C The quotient of two numbers with
different signs is negative.

107

108    ANSWERS AND SOLUTIONS: Exponents, Roots, and Properties of Numbers
ANSWERS AND SOLUTIONS

37. A The quotient of two numbers with 42. C −17 − (−52) = −17 + 52 = 35
different signs is negative. 43. B D ivision by 0 is undefined; there is no

38. A The quotient of two numbers with need to compute any of the numerators.
identical signs is positive. 44. −20, −15, −8, −2, 0, 6, 13
The negative number with largest absolute
39. C −20 ÷ (−4) = 5
value is the smallest in the group, so it is first.
40. D The distance between two numbers is 45. D |−7 − 6| = |−7 + (−6)| = |−13| = 13 
the absolute value of their difference Distance must be positive.
(as opposed to the difference of their 46. neither  Z ero is the only number that is
absolute values, as in C, or any sum, as in
A and B). neither positive nor negative.
47. C Start at 0, follow arrow to 5, back up
41. y
(negative) 8, end up at −3.
6 123456 x 48. A Start at 0, go left 6, then right 2,
5 A
4 gives the addition problem −6 + 2.
3 As a subtraction problem, this would
2 have been changed from −6 − (−2),
C1 ending up at −4.
49. B −40(9) = −360; −360 + 160 = −200;
–6 –5 –4 –3 –2 –1 −200 ÷ 5 = −40
–1 50. D −7(−5) − 4(−8 − 6) ÷ (−2) =
–2 35 − 4(−8+ (−6)) ÷ (−2) =
35 − 4(−14) ÷ (−2) =
B –3 35 − (−56) ÷ (−2) =
–4 35 − 28 = 7
–5
–6

Negative x values are left of the y-axis.
Positive x values are right of the y-axis.
Negative y values are below the x-axis.
Positive y values are above the x-axis.

Chapter 2 Exponents, Roots, and Properties of Numbers

1. C T he number of factors is the power. 5. A T he base is the quantity to which the
2. B T he power is the number of factors. exponent is attached.
3. A The number of factors is the power. 6. 125 53 = 5 • 5 • 5 = 25 • 5 = 125
4. C 43 = 4 • 4 • 4 = 16 • 4 = 64 7. D  9 5 • 93 = 95 + 3 = 98

ANSWERS AND SOLUTIONS: Exponents, Roots, and Properties of Numbers   109
ANSWERS AND SOLUTIONS

8. D 4 6 • 26 = (4 • 2)6 = 86 30. C 2 6 • 56 = (2 • 5)6 = 106 = 1,000,000

9. A 25 • 35 • 67 = (2 • 3)5 • 67 = 65 • 67 = 65 + 7 = 612 31. D (54 • 79)3 = (54)3 • (79)3 = 54 • 3 • 79 • 3 = 512 • 727

10. B 73 • 72 • 75 = 73 + 2 + 5 = 710 1

11. C 23 • 43 • 53 = (2 • 4 • 5)3 = 403 32. B 1 002 = 100 = 10

12. B Any ­non-​z­ ero number to the zero power 1
is 1.
33. A 83 = 3 8 = 2

13. D (53)4 = 53 • 4 = 512 34. C 25– 1 = 1 = 1 = 1
2 25 5
1
252

14. C Any number to the first power is that 35. B 1 000–31 = 1 = 1 = 1
number. 3 1000 10
1
1000 3
15. B 3 • 33 = 31 • 33 = 31 + 3 = 34
1
16. A 65(62)3 = 65 • 62 • 3 = 65 • 66 = 65 + 6 = 611 36. A 42–1 = = 4 =2
42

17. D ( 35)2(34)3 = 35 • 2 • 34 • 3 = 310 • 312 = 310 + 12 = 322 37. D b y definition

18. A 5−2 = 1 = 1 38. B by definition
52 25
39. C by definition
19. C (2­ –3​­ )2 = 2­ –​­3 • 2 = ­2–­​6 = 1 = 1
26 64 40. A b y definition

20. B ( ­5–​2­ )–4 = 5­ –2​­ • (–4) = 58 41. C 92 = 9 • 9 = 81

21. 100,000 10 to the 5th is 1 followed by 42. 6 36 = 6 x 6 = 6
5 zeroes.
43. D 4 3 = 4 • 4 • 4 = 16 • 4 = 64
22. D 1 to any power is 1.
1 1 44. A 33 = 3 • 3 • 3 = 9 • 3 = 27
23. A 2 4 • ­2–­​9 = 24 + (–9) = 2­ –​5­ = 25 = 32

24. A 3375 = 37 – 5 = 32 = 3 • 3 = 9 45. D 5 4 – 30 = 6 • 9 – 6 • 5 = 6(9 – 5)

46. A 5(7 + 2) = 5 • 7 + 5 • 2 = 35 + 10

25. C 53 = 53 – 6 = 53 + (–6) = ­5–3​­ = 1 = 1 47. C 3(9 – 5) = 3 • 9 – 3 • 5 = 27 – 15
56 53 125
48. B 6 4 + 28 = 4 • 16 + 4 • 7 = 4(16 + 7)
1 1
26. positive ­1 0–2​­ = 102 = 100 49. D 2(3 + 5 – 4) = 2 • 3 + 2 • 5 – 2 • 4 = 6 +
10 – 8
27. A 0 to any positive power is 0.

28. D  78 5 = 75 50. A –6(8 + 3) = –6 • 8 + (–6) • 3 = –48 + (–18) =
85 –48 – 18

29. B  23 4 = 24 = 16
34 81

110    ANSWERS AND SOLUTIONS: Fractions and Operations
ANSWERS AND SOLUTIONS

Chapter 3 Fractions and Operations

1. 6 Cuts that give whole numbers of 8. D 185 = 12450 ; 4 4 = 4 32 ;
2 1 5 40
slices for 3 and 2 are 6, 12, 18, etc.
12450 + 4 32 = 5 57 = 6 17
6 is the lowest common 40 40 40

denominator AND gives the correct
answer.
9. C

1 + 2 = 3 + 4 = 161 1 5 3 = 15 6 ; 15 6 +12 3 = 27 9 = 28 1
2 3 6 6 4 8 8 8 8 8

2. $147.92 1 × 1 = 1 ; 1 × $7100 = $147.92
8 6 48 48
10. A
3. C 14040 = 2151. Numerator and
denominator are divided by their 23 1 = 23 5 = 22 2205; 7 3 = 7 12 ;
4 20 5 20
greatest common factor, 4.
22 25 – 7 12 = 15 13
20 20 20
4. B

1 000 × 3 = 750; 750 × 2 = 500; 500 × 1 = 250 11. D Yes, he did not add 1 when he
4 3 2 converted the mixed number to an
3
5. D 31 ÷ 53 = 31 × 3 = 1 improper fraction (he multiplied 4
5 5
by 5 rather than by 6 ).
5 5
slTehufeet.sdd43ida–y4141tho=efrte21hweoawfsthh43oeloef
6. C On Monday 12. B 22255 = 455 = 1
backlog. On 9
the backlog

backlog. Looked at another way, 13. A 6 21 = 6 48 = 5182 ; 5182 – 3 7 = 2 5
8 8
1 1 1
she did 4 + 4 = 2 of the backlog,

leaving 1 to be done. 14. C
2
7 43 = 31 ; 55 × 4 = 220 ; 220 ÷ 31 = 220 × 4 = 220 = 7 3
7. C After first reduction, each item costs 4 4 4 4 4 4 31 31 31

2 of the original price. The second 220 ÷ 34 1 = 220 × 4 = 220 = 7 3 = 7 full glasses
3 4 4 31 31 31
2
reduction leaves 3 of the reduced

price, so the final price equals

2 × 2 = 4 of the original price. The
3 3 9
5
reduction is 9 of the original price.

ANSWERS AND SOLUTIONS: Fractions and Operations   111
ANSWERS AND SOLUTIONS

15. A 24. B  92 10 = 912300 – 45 15 = 46 15 = 46 3
20 20 20 4

1 1196 × 5 = 55 45 ; 1 ×5 = 5 = 10 ; 25. 33 71 = 1700 ; 3 = 42 ; 2 = 14 ;
16 8 8 16 35 5 70 10 70

55 45 + 10 = 55 55 = 58 7 ; 10 + 42 + 14 = 66 = 33
16 16 16 16 70 70 70 70 35

8 × 12 = 96 ; 96 – 58 7 = 37 9 26. B Overtime pay =
16 16

$9 × 121 × 8 = $9 × 3 × 8 = $108 ;
2
1 20 14 6 3
16. A 5 = 100 – 100 = 100 = 50 Regular pay = $9 × 40 = $360;

17. 1 by definition, any number multiplied Total = 360 + 108 = $468; $108 = 3
by its reciprocal equals 1. $ 468 13

3 16 48 27. > The lower the denominator, the larger
4 16 64
18. 48 × = the part, so 3 is slightly bigger than
54
3
19. C Any ­non-z​­ ero number divided by 56 .
itself is 1.

20. B Greatest common factor is 72. 28. < 5 = 141 < 121
4

21. 76 7 3 = 38 ÷ 5 = 38 × 2 = 76 29. > 7 > 2 ; 13 < 2
5 5 2 5 5 25 30. = 4 8

22. D 2 is a whole number, so it may be Both reduce to 1 .
ignored. 18

8 : 2 × 2 × 2   31. 5 1= 4 ; 4 + 1 = 5
24 : 2 × 2 × 2 × 3 4 4 4 4 4
36 : 2 × 2 × 3 × 3
32. 585 6 = 48 ; 48 + 7 = 55
LCD has the maximum 8 8 8 8
number of each distinct factor:
2 × 2 × 2 × 3 × 3 = 72. 127 120 7 127
10 10 10 10
23. C 33. 12 = 120 ; + =
10

C ommon denominator: 34. 4 1 17 ÷ 4 = 4 1
3 : 3 4 4
5 : 5
8 : 2 × 2 × 2 35. 11131 124 ÷ 11 = 11131

Maximum number of each factor 36. 1158 92 ÷ 72 = 17220 = 1158
2 × 2 × 2 × 3 × 5 = 120.

31 = 40 ; 1 = 24 ; 3 = 45 ; 37. 3 2 = H
120 5 120 8 120 3

14200 + 24 + 45 = 109 38. –1 1 = D
120 120 120 2

112    ANSWERS AND SOLUTIONS: Decimal Numbers and Operations
ANSWERS AND SOLUTIONS

39. 1 1 =G 46. Point G = 2
2

40. 6 4 = I 47. A 1
5
643 = 3 64 = 4

41. – 3 = E 16– 1 1 1 1
4 2 16 4
3 48. B = 1 = =
4 49. C
42. Point A = –3 50. D 16 2

43. Point C = –2 1  – 1 2 = – 1 × – 1 = 1
4  3  3 3 9

44. Point D = – 1  23 = 2 2 2 = 8
4  5  5 5 5 125
× ×

45. Point F = 3
4

Chapter 4 Decimal Numbers and Operations

1. > 14. A 14.00 – 8.37 = 5.63 gallons
2. <
3. = 5.63 gallons × 27.6 miles =
4. < gallon
5. >
155.388 miles
( ) ( ) 6. C –6 × 106 × –2 × 102 = 12 × 108 =
15. B
1.2 × 109
7. B 1.235 + (–1.235) = 0 $ 132.50 + $675.00 + $512.50 = $1320
8. 16.784 12.389 + 4.3950 = 16.7840 = 16.784 $3300 ÷ $1320 = 2.5
9. 21.33
10. 17.1442 16. B 63.55 cents = $0.6355
11. 17.85
12. 8.88 235,500 ÷ 1000 = 235.5
13. 683.25
$0.6355 × 235.5 = $149.66025

17. C

2.25 × 108 = 0.75 × 103 = 7.5 × 102
3.0 × 105

= 750 seconds

18. tens
19. C |–1.45 –(–8.34)| = |–1.45 + 8.34| =

|6.89| = 6.89

ANSWERS AND SOLUTIONS: Decimal Numbers and Operations   113
ANSWERS AND SOLUTIONS

20. A |2.735 – (–1.989)| = |2.735 + 1.989| = 36. 3rd
|4.724| = 4.724 The decimal is moved 3 places to the left, so

21. B The second decimal place to the the exponent will be positive.
right of the decimal expresses
hundredths. 37. C $0.0125 × 150,000 = $1875

22. D In D, the digit 5 represents units. 38. D 8 × $10.50 = $84.00

23. A 0.460 × $5.50 = $2.53 $1.17 × 85 = $99.45
24. A 4 × $1.35 = $5.40
$99.45 – $84.00 = $15.45

$5.40 + $3.29 + $2.10 = $10.69 39. B $3.98 × 3.45 = $13.731 ≈ $13.73

$20.00 – $10.69 = $9.21 40. C 9 3,000,000 = 9.3 × 107 in scientific
notation
25. D $0.25 × 7 = $1.75
$20.00 – $1.75 = $18.25 41. D 1 .496 × 109 × 0.39 = 0.58422 × 109 = 5.8422 ×
42. C
26. C 32.75 × $17.50 = $573.125

27. A $0.95 × 15 = $14.25 $19,500 ÷ $421.55 = 46.26 months
She will make her last payment in the
28. D 1 50 × 6 × 104 = 900 × 104 = 9.0 × 106
47th month.

29. C 11.74 × $3.459 = $40.608 ≈ $40.61 43. A 1÷ 7.05 × 10–3 ≈ 1.418 × 102
1.418 × 102 × $1025 ≈ $145,390
30. B $ 20.00 ÷ $4.599 ≈ 4.34877. But
we can’t round up because that 44. 47.28 100 – 52.72 = 100.00 – 52.72 =
would cost more than $20.00. 47.28
So 4.34 is the correct answer.

31. B $0.265 × 1050 = $278.25 . 45. B $17.50 × 40 = $700

32. C $44.94 + $10.50 + 7.00 + 79.50 = $141.94

9.45 × 12.15 = 114.8175 ≈ 114.82 square feet$44.94 + $1 0 .50 + 7.00 + 79.50 = $141.94

33. B 0.625 $700 – $141.94 = $558.06
46. B
8)5.000

34. C 0.060 12.35 + 123.56 + 111.23 + 73.4 + 45.65 = 366.19
366.19 – 7.50 = 358.69
)50 3.000

35. A 0.109375

)64 7.000000

114    ANSWERS AND SOLUTIONS: Ratios, Rates, and Proportions
ANSWERS AND SOLUTIONS

47. C 50. C 5.7 + 8.6 = 14.3 miles each way, or
7.36 × 1022 = 0.0736 × 1024 28.6 miles per round trip
13.5 gallons × 32.5 miles per gallon =
0.0736 × 1024 + 5.9742 × 1024 = 438.75 miles
6.0478 × 1024= 438.75 miles ÷ 28.6 miles per trip =
6.05 × 1024 15.34 trips
He would run out of gas on the 16th trip,
48. 0.00006022 so he can make 15 trips without running
out of gas.
49. A 17.35 – 11.23 = 6.12 km

Chapter 5 Ratios, Rates, and Proportions

1. C 10 feet to 12 feet = 7. A 70 words in 6 sentences =

10 feet = 10 = 5 70 words = 35 words =
12 feet 12 6 6 sentences 3 sentences
3
2. 4 21 students to 28 students = = 1123 words
8. B sentence
21 students = 21 = 3
28 students 28 4

3. D 24 miles to 36 minutes = $750 for 40 hours = $750 = 4 $75 = $18.75 = $18.75 / ho
40 hours hours 1 hour
24 miles = 32mmini4ule0t$seh7s5o0urs h$7o5ur s = $11h8o.7u5r = $18.75/hour
36 minutes =
4

4. 5 leaves 20 leaves to 8 twigs = 9. C 3 model 6 model 3
2 twigs 4 x week 4
20 leaves 5 leaves = →  x = 3 → x = 4
8 twigs = 2 twigs 1
2 week

5. B 200 miles per 4 hours = 10. C

200 miles = 50 miles = 50 miles Packets of drink mix 24 ×18 36
4 hours 1 hour hour Quarts of water 36 54
×18
6. 8.4 ounces 42 ounces for 5 mugs = 7
mug 10.50 89
42 ounces 8.4 ounces ounces 11. 5 6 12.00 13.50
5 mug = 1 mug = 8.4 mug 7.50 9.00

425omunucge s = 8 .41 ounces = 8.4 ounces Each hot dog costs $1.50.
mug mug

ANSWERS AND SOLUTIONS: Ratios, Rates, and Proportions   115
ANSWERS AND SOLUTIONS

12. B 7 = x → 28 • 7 = 28 • x → 28 • 7 = x → 19. C 5 picture = 5 • 12 picture = 1 • 3 picture
12 28 12 28 12 8 8 5 hour 2 1 hour
5
12 hour

x → 28 • 7 = 2 8 • 2x8 → 28 • 7 = x → 7 •7 = x → 43985152=pihcxotu→urre1=6 5831 •=152x p hicot uurre = 21 • 3 picture = 3 picture = 121 pictures/hour
8 12 7 3• 7 = x → 12 3 1 hour 2 hour

13. C 49 = x → 16 1 = x 35 x 35 x
3 3 2 14 2 14
20. D = → 14 = 14 → 7 • 35 = x → 245 = x
• •

132 persons = 132 3p25er=so1nx4s p→er1s4qu• a32r5e =m1il 4e • 1x4 → 7 • 35 = x → 245 = x
40 square miles 40

4013sq2upaerre s om n isles = 132 persons per square mile = 21. D 6 = x → 3 1 • 6 = 3 1 • x → 10 • 6 = x → 10 • 2 =
40 1 3 1 3 3 1
3 1 3 1
= 3.3 persons per square mile 3 3

3.3 6 = x → 3 1 • 6 = 3 1 • x → 130 • 61 = x → 10 • 2 = x → 20 = x
40 132.0 1 3 1 3 =129 x • 61 = x →61 =1 99x•213 →= 9x 21→•
)→ 3 1 3 1
10 6• 3 3 6 1 x 19 6
1 1 2 2 1
= 9 • 9 1 → • = x → 19 •3=
x 2 • x → 54 =x
14. D 3 = x → 3 =9x2110→• 19x021→• 6114=5 9=21x•→9x213→43
8 10 8= 57 = x

10 • 3 = 10 • 1x0 → 145 = x → 3 3 = 9 x 9 x
8 4 16 96 16 96
22. B = → 96 = 96 → 6 9=
• •

15. C m196ile=s x → 96 • 9 = 96 • x → 6 • 9 = x → 54 = x
234.50hmoiulerss p9e6r hour 16 96
= 68.5

) 68.5 23. A 2.5 = x → 90 • 2.5 = 90 • x → 9 • 2.5 = x → 11.2
20 90 20 90 2
35 2400.0 ≈ 69 miles per hour.
28x2.05→= 190x5016→=9x0 2.5 = 9 0 • 9x0 9 • 2.5 = x 11.25 = x
• 20 → 2 → x

16. B 127 = x → 8 • 127 = 8 • → 203.2 = x • 10
5 8 5
3 1 x 3 1 5 7 35
x 127 = 8 • 8x → 10516 = x → 203.2 = x 24. D 2 = 10 → 10 • 2 = 10 → 2 • 2 =x → 4 = x
= 8 8 5 4 4
→ •

17. A 2520cmg 3 = 4 g/3c421m=3 =1x0.8→g1/c0m•33421 = 10 x → 25 • 72 = x → 35 = x → 8 3 = x
5 • 10 4 4

18. A 25. B 298.9 miles = 24.1 miles per gallon
12.4 gallons
$30 $5 $1.25
24 bottles = 4 bottles = 1 bottle = $1.25 24.1

$1.25 per bottle )12.4 298.9

116    ANSWERS AND SOLUTIONS: Ratios, Rates, and Proportions
ANSWERS AND SOLUTIONS

26. C 1 = x → 864 • 1 = 864 • x →3= x 32. B 1$23b.6a0rs = $0.30 per bar
288 864 288 864

x → 864 • 281 8 = 864 • x → 3 = x ) 0.30
864 864 12 3.60

27. A 2475 = x → 120 • 27 = 120 x → 120 • 3 = 3x3. C 49.95 = x → 16 • 99 = 16 x → 1584 = x → 352 = x
120 45 • 120 5 16 4.5 • 16 4.5

x 12 0 • 24 75 = x 120 3 = x49.9→5 =2114x6• 3
120 → 120 5 →1
)= → 120 • → • =16x •→49.9572= =16 x • 1x6 → 1584 = x → 352 = x
4.5
24 3 = x 72 =x
1 • 1 → 14 48
21 72
28. D The r­ eal-​w­ orld square is 6 feet. 34. Yes Write = and ­cross-​m­ ultiply:

1 = x → 6 • 1 = 6 • x → 6 = x 14 • 72 = 1008 = 21 • 48
5 6 5 6 5
35. D The unit cost of gasoline should be
1.2 multiplied by the quantity of gasoline
5 6.0 get the total cost. to

29. B 23 = x → 8 3 = 8 x → 4• 3 = x → 12 = x 36. B Each crate weighs 40 pounds, so the unit
8 2 8
• • rate is 40 pounds per crate. ( 1 crates
40
→ 8 • 3 = 8 • 8x → 4 • 3 = x → 12 = x
2 per pound conveys the same relationship,

30. C 12 = x → 3 1 12 = 3 1 • x → 13 • 4 12 but unit rates are typically concerned with
3 41 • 1 412 4 4 4 1 •1 = x quantity per item, not items per quantity.)
1 3 1 •1 3 1 37. 200; 320.
4 4 4 4 Since the unit rate is 40 pounds per crate,

= x → = 3 1 • x → 13 • 4 • 12 = x 5 crates weighs 200 pound and 8 crates
4 4 1 1
3 1 3 1
4 4

→ 13 • 12 = x → 156 = x weighs 320 pounds.

156 minutes – 2 hours = 156 minutes – 38. 3 is the smaller ratio
120 minutes = 36 minutes 5

x = 2 hours 36 minutes Number 3 24 Number 5 25

31. A 264 feet = 264 • 1 mile = 264 1 60 mile = Price 5 40 Price 8 40
1 min 5280 1 5280 1 hour 60
1 • •
60 hour

= 264 • 1 mile = 264 1 60 mile = ) 39. D 5 300
5280 1 5280 1 hour
1 ho ur • •
60
Dennis’s unit rate is 60 miles .
15, 840 miles per hour = hour
5280 Multiplying this by time in hours will cancel
the hours and leave miles, or distance.
3 miles per hour

3 D = 60t

) 5280 15,840 18 34
81 154
40. No Write = and ­cross-m​­ ultiply:

18 • 154 = 2772, but 81 • 34 = 2754

ANSWERS AND SOLUTIONS: Percents and Applications   117
ANSWERS AND SOLUTIONS

41. Nick $48.75 = $9.75/hour 47. Poplar Hills
5 hours
362 persons
$58.32 = $9.72/hour Poplar Hills: 0.02 square mile = 18,100
6 hours
42. D 1720.7..90375262s=qp1ue8ar.r5seo0nm•s1ile8 p.5= 0 1→S8a,1n101F0.r1ap0nec=rissocpnos: per square mile
1 8.5 0 • 172.7.975 7.77 = p → 18.50 •
12.95 18.50 805,000 persons
47 square miles = 17,128
p p
77 = 18.50 → = 18.50 • 18.50 → 11.10 = p 84075s,0q0u0arpeemrsiolen ss = 17,128 persons per square mile
.95

43. C The unit price = 60 $1620 = 48. A 72 = 144 = 192 = 288 = 231434,6th=e 360 = 24
baseballs 3 6 8 12 15

$27 per baseball mileage in miles per gallon, so D = 24f

44. B 22 = d → 3• 22 = 3• d → 11 = d → 3 2 49. 8 1
18 3 18 3 3 3 =d 3

22 d 3 • 128 2 = 3 • d 11 2 93,000,000 miles = 93,000 miles • 1 second = 500 seco
18 3 3 3 3 1 186 miles
= → → = d → 3 = d = 3.67 feet 186, 000 miles
second

141 75 miles 1 hour 1 m931i1n,80u60t,0e0,00=000s5mecmmiole4nsiidllee ss = = 19413, 000 miles 1 second = 500 seconds
hour 60 minutes 1 • 186 miles
45. miles • •

les 601mhionuurtes • 1 m i1nute = 5 miles = 141 miles 500 seconds • 1 minute = 25 minutes = 8 1 minutes
r 4 60 seconds 3 3
• 1

46. C 250q5bu0oa0wrtsslse1c=on9dqsuna• r6ts10→mseinc9ouqntuedasrt=s •232550mqbuionawruttsles 5s0==. 989q.31u7amrtsin•u9teqsunarts

250qbuoawrtsls = 9 q unarts → 9 quarts • 20 bowls = 9 quarts • n $17.37 = $17.37 1 pound = 17.37 pounds = 9.7 pounds
5 quarts 9 quarts 1 $1.79 1.79
$1.79 •

n → 9 quart s • 25 0q buoawrtsls = 9 quarts • 9 q=unnar→ts$p1$3→o17u6.7.n39db97o•=w4l$sb1o=71w.3nl7s = n → 36 bowpolsun=d n
quarts → 9 • 4 bowls • 1$p1o.u7n9 d 17.37
= 1.79 pounds = 9.7 pounds

36 bowls – 20 bowls = 16 bowls

Chapter 6 Percents and Applications

1. 17.5% 2. 80% 3. 660.5% Questions 1–5 are solved by multiplying by
4. 1520% 5. 0.17% 100, moving the decimal two places right,
and adding the percent sign.

118    ANSWERS AND SOLUTIONS: Percents and Applications
ANSWERS AND SOLUTIONS

6. E 7. D 8. A 27. C $1147 × 0.0575 = $65.95

9. C 10. B 28. B $2175 = $12, 428.571 ≅ $12, 428.57
Questions 6–10 are solved by converting the 0.175
fraction into a decimal, moving the decimal $12,428.57
two places right, and adding the percent sign.
29. D $11.03652.250 = $127.25, which is
11. 1430 12. 53 13. 530 the sale price.
14. 11270 15. 20700
$149.70 – $127.25 = $22.45
Questions 11–15 are solved by dividing discount
the percents by 100, clearing decimals by
additional multiplications as necessary, and $22.45 = 14.99% ≅ 15%
reducing the fractions to lowest terms. $149.70
16. 0.11 17. 0.04 18. 27.56
30. C 172 pages × 0.285 = 49 pages of
ads. 172 × 0.25 gives 43 pages
as a maximum for 25% ads. The
difference is 6 pages.

19. 0.071875 20. 0.000076 ptho1ei5n0t ,2000506,00 301 .= A0. 0000137 = 01.5000,2100305706%,00≅00=.0001.040%00137 = 0.00137% ≅ 0.0014%
Questions 16–20 are solved by
percents by 100 or moving the dividing
two places to the left. decimal

21. B $17.89 × 1.085 ≅ $19.41 32. D This month’s sales are 115% of

22. A $159.95 × 0.85 (net from last month’s, so $3036 = $2640.
1.15
discount) × 1.0525 (sales tax) = Last month’s sales were 110% of
$143.10 the month before, and

23. C $36.99 × 0.8 × 0.9 = $26.63 $2640 = $2400.
1.10

24. C 1106582 = 16% = boxes 33. B 150,000,000 × 0.57 = 85,500,000

unloaded by Susan 85, 500, 000 = 0.041 = 41%
207, 634, 000
100% – 16% = 84% = remaining
boxes to unload 34. C $9.99 × .06 = 0.5994 ≈ .60, so the
total for the watch will be $10.59.
25. A E ither $9250 × 0.68, or Since he has $10.00, he will need
$9250 – ($9250 × 0.32) = $6290. $0.59 more.

26. B $1150 × 0.06 = $69, but the 35. B T his year’s total is 83% of last
year’s.
loan is only for 1 of a year,
3
475 × 0.83 = 394
so interest is 1 that, or $23. 36. B
3 422, 000
$1150 + 23 = $1173 987, 000 0.4275 43%
= ≅

ANSWERS AND SOLUTIONS: Statistics   119
ANSWERS AND SOLUTIONS

37. C $225 – $180 = $45 45. A $104.8.449 = $17.25

$45 = 0.20 = 20% 46. loss Suppose Sal invested $100.
$225 100 × 1.10 × 0.95 × 1.02 × 0.93 ≅
99.13, so Sal lost $0.87, an overall
38. C T he total, $63.70, is 107.95% of decrease of 0.87%.
the cost.

$63.70 = $59 47. D 6.2% + 1.45% + 14%= 21.65%
1.0795 reduction

39. B $3720 × 0.015 = $55.80 100% – 21.65% = 78.35% net
t­ ake-​h­ ome
40. C A 15% discount means she pays
only 85% of the cost. 0.7835 × $718.15 ≅ $562.67
48. D $555.00 × 0.75 × 0.85 = $353.81
$170.00 = $200
0.85 49. A T he total weight of a batch is
112 pounds.
41. C Let us say Tom invests $100. He
loses 50%, which leaves $50. A 9 = 0.080 = 8%
50% increase of $50 is only $25, 112
which raises his investment value
to $­75—​a­ 25% loss overall. 50. D This month is 120% of last
90, 000
42. A 2 69 × 1.075 ≅ $289.18, or month. 1.20 = $75, 000 =
269 × 0.075 + $269 ≅ $289.18
last month’s sales. $75,000 is
43. C 1,544,400 – 1,320,000 = 224,400 80% of the previous month’s
increase.
sales. $75, 000 = $93,750 =
0.80
224, 000 0.169 ≅ 17% sales two months ago.
1, 320, 000 =

44. D $ 76 for one month means $76 ×
12, or $912, interest per year.

$912 = 0.1824 ≅ 18.2%
$5000

Chapter 7 Statistics

1. B The numbers are roughly proportional, 4. B   A straight line passing through the
with a proportionality constant of 7. origin represents a relationship that
never needs a constant.
2. D Y = 7X
3. C The most probable source of the “noise” 5. C   Put the data in order. The median is
the average of the two middle terms,
in the data is random effects. 28 and 29. The median is 28.5.

120    ANSWERS AND SOLUTIONS: Statistics
ANSWERS AND SOLUTIONS

6. Histogram

6Frequency
5
4
3
2
1

1–10
11–20
21–30
31–40
41–50
51–60
61–70
71–80
81–90

Groups

Group ­1–1​­ 0 ­11–​2­ 0 2­ 1–​3­ 0 ­31–4​­ 0 ­41–5​­ 0 5­ 1–6​­ 0 6­ 1–7​­ 0 7­ 1–8​­ 0 8­ 1–9​­ 0
Frequency 356532002

7. 86, 87 These data seem fairly far removed 16. C The largest number of people made sales
from the rest of the data set. in this range, so that is the mode.

8. For this distribution, yes. 17. B There are 21 salespersons. The median
will be salesperson #11, who is in the
9. D There is no mode. Each term appears $101–$150 range.
only once in the set.
18. B Probabilities are calculated for every
10. 86 The range is the difference between terminal event. When that is done,
the largest and smallest terms. the average gas well has the highest
87 – 1 = 86. probability.

11. C The sum of the numbers is 99, and there

are 11 numbers. Drilling Product Amount Probability

The mean is 99 = 9. Abo0v.e3a0v.e4rage 0.0132
11 Average 0.0176

12. C The sorted data set is 1, 2, 6, 7, 8, 9, 10, 11,
11, 14, 20.
0.4Oil Below ave0r.a3ge
The median is the number in the middle, 9. Ga0s.6 0.0132

13. D 11 appears more than any other number. Eco0n.1o1mic well Ab0o.0v5e average 0.0033
furtDhreyr hao0ctl.8eiv9(intyo) 0.8 0.0528
14. D The highest data value minus the lowest Average 0.0099
data value = 20 – 1 = 19. Below av0e.r1a5ge

15. A If x is equal to or larger than 10, then 19. C Probabilities are calculated for every
it falls to the right of the median and terminal event. When that is done, the
increases the number of items in the data ­above-a​­ verage gas well has the lowest
set to 12. The new median will be the probability.
average of the sixth and seventh terms
(9 and 10), so the median is 9.5.

ANSWERS AND SOLUTIONS: Statistics   121
ANSWERS AND SOLUTIONS

20. B 26. C 75 + 87 + 96 + 76 + 88 = 422. Divide
Expected number of producing wells = 11, by 5 = 84.4 average.

expected revenue = 11 × $23 million = $253 27. B Range is the highest minus the
million; expected number of dry holes = 89, lowest, so 96 – 75 – 21.
expected cost = 89 × $1.2 million = $106.8
million; profit = revenue – cost = 28. D With the five scores she already has,
$253 million – $106.8 million = $146.2 million. she has a total of 422 points. For an
21. average of 90, she needs a minimum of
90 × 6 = 540 points. 540 − 422 = 118. It
Cats Dogs is not possible for her to make an A.
187° 158°
29. B A double grade of 69 gives a
weighted average of 80.

Guinea pigs 30. B If the final counts as three grades, she
14° would have 3 + 5 = 8 grades. For an
average of 90, she would need 8 ×
There are 100 items, so each item gets 90 = 720 points. She has 422 points
3.6 degrees. before the final, so she needs 298
more. Divide 298 by 3 (the weight of
the final) to see what she must score
on the final. 298 ÷ 3 = 99.3.

31. C 78 + 83 + 81 + 85 + 89 + 94 + 96 = 606.
606 ÷ 7 = 86.57 = 87.

22. A 52 out of 100 own a cat, and 44 out of 32. D T here is no mode. All the scores
100 own a dog, so 52 + 44 = 96 out of appear only once.
100 own either a cat or dog.

23. 11 : 13 44 dog owners = 11 33. 18 The range is the highest score minus
52 cat owners 13 the lowest score. 96 − 78 = 18.

24. D She has 16 credit hours of 4 points 34. A Put the scores in order: 78, 81, 83, 85,
each for As, so 16 × 4 = 64. She has 89, 96, 96. The median is the middle
27 credit hours of 3 points each for Bs, number, 85.
so 27 × 3 = 81 and 64 + 81 = 145.

25. B 35. C In question 31, you found that she has
606 total points. Subtract her lowest
Grades ABCD F Total score, 78, to get 528 total points.
1 70 Divide that by 6 scores to find her
Credit hours 16 27 18 8 0 average, 88.
0 189
Grade points 4 3 2 1

Total grade 64 81 36 8 36. B In question 31, you found that she has
points 606 total points. Add another 96 so
that the final counts twice. Now she
189 = 2.70 has 702 points. Divide that by 8 scores
70 to get 87.75.

122    ANSWERS AND SOLUTIONS: Statistics
ANSWERS AND SOLUTIONS

37. A Multiply each score in the chart by its 39.
percentage weight.
Favorite Sports

Elena’s Math Scores Weight Weighted

Test 1 78 10% 7.8 football hockey
Test 2 83 10% 8.3 basketball soccer
Test 3 81 10% 8.1 baseball
Midterm soccer baseball football
exam 85 20% 17 hockey
Test 4 basketball
Test 5 89 10% 8.9
Final Exam 94 10% 9.4
96 30% 28.8

Add up the weighted scores to get her weighted 40. B 4 = 2 = 0.20 = 20%.
average. 20 10
7.8 + 8.3 + 8.1 + 17 + 8.9 + 9.4 + 28.8 = 88.3.
38. 41. 4:1 Eight football fans to two soccer fans =
8:2 = 4:1.
100
90 42. A Only 1 out of 20 students favors hockey.
80
70 43. 78% Now there are seven soccer fans out
60
50 of 25 total students. 7 = 0.28 = 28%.
40 25
30
20 44. A There were 5 out of 20 before: 5 =
10 20
0 Test 1 Test 2 Test 3 Midterm Test 4 Test 5 Final 1 , and there are 5 out of 25 now.
54 210.
Elena's Math Scores = 1 . 1 − 1 =
5 4 5
45. 25

2% 34% 34% 2%
71 14% 14% 99

78 85 92

ANSWERS AND SOLUTIONS: Probability and Counting   123
ANSWERS AND SOLUTIONS

46. B 1 standard deviation from the mean is 49. B The center of the box is the median so
78 to 92. That range is 92 – 78 = 14. the median is 180.

47. D Because 2% is below 71, 98% passed. 50. C The highest is 200 and the lowest is
160. 200 – 160 = 40.
48. B 14% + 2% are above 92, so 16% of the
25 students scored above 92. 25 ×
0.16 = 4.

Chapter 8 Probability and Counting

1. C The probability of having a boy 4. G GGG
B GGB
is 21, independent of the number G G G GBG
of boys already born. The B B GBB
probabilities multiply:

1 × 1 × 1 = 1 = 0.125 = 12.5%
2 2 2 8

2. B The events “having three girls” and G G BGG
“having three boys” are mutually B B BGB
exclusive, so the probabilities for G BBG
each event are added: B B BBB

12.5% + 12.5% = 25%

3. C The only way not to have a girl is
for there to be all boys. Subtract
that probability from 1 to get the
probability of at least one girl:

1− 1 = 7 = 0.875 = 87.5% Of the eight possible combinations only three
8 8 have two girls and one boy: BBB, BBG, BGB,
BGG, GBB, GBG, GGB, GGG. Each of the three

desired combinations has probability 81, and
they are mutually exclusive, so the probabilities

are added:

1 + 1 + 1 = 3 = 0.375 = 37.5%
8 8 8 8

124    ANSWERS AND SOLUTIONS: Probability and Counting
ANSWERS AND SOLUTIONS

5. C  H ere the order of the problems does not First Second Third Fourth Outcomes
matter, so we use a combination. Toss Toss Toss
Toss
10! 10 × 9 8×7×6 H
 0C5 = (10 = 5× 4 × 3×2×1 = 2 × 3×2×7× 3 = 252 H HHHH
− 5)!5! × T H T HHHT
10 × 9 × 8 ×× 2 7 ×× 6
5× 4 × 3 1 = 2 × 3 × 2 × 7 × 3 = 252 H H HHTH
T T HHTT
6. B There are 20 socks (8 red and 12 black) H T
that, if picked, will not make the pair. H HTHH
Picking one of these will require at least a H
third pick to get a pair. The probability of
T HTHT
picking one of those is 20 or 4 .
35 7 H HTTH
T T HTTT
7. A Having picked two socks, there are now
H THHH
34 left in the drawer. The only way to H T THHT
avoid making a pair: pick a red sock. The
H THTH
probability of that is 8 . Subtracting from T T T THTT
34
8 26 1173. H TTHH
1 gives a probability of 1 – 34 = 34 = H

T TTHT
H TTTH
T T TTTT

8. D The next sock must be one of the already
drawn three colors. It is impossible to 13. Given a fair coin, the most likely  result 
avoid a pair. Impossible events have a
zero probability. of six separate flips is  three  heads
and  three   tails  , which is also the  mean 
9. A 4 salespersons out of 21 did this, so the and the  median  of the  random  data set.
probability is 241.

10. C 10 salespeople did that, so the probability 14. A The trials are independent and
probabilities are multiplied:
is 1201.
1 × 1 × 1 × 1 = 1
11. D   There are 6! or 720 arrangements. Only 2 2 2 2 16

one is in alphabetical order. The 15. B Of the ­three-​t­ oss combinations that
include one head and two tails, only one
 probability is 1 .
720 is in the desired order.

12. B  D raw a probability tree. There are six 1 × 1 × 1 = 1
outcomes that have exactly two tails 2 2 2 8
6 3
 out of 16 total outcomes. 16 = 8 . 16. D  T he only way no heads are tossed is if all

tails are tossed. That probability

 is 1 × 1 × 1 × 1 × 1 × 1 = 1 . Since
2 2 2 2 2 2 64
 t he sum of probabilities equals one,

1− 1 = 63 is the probability of at
64 64
  least one head.

ANSWERS AND SOLUTIONS: Probability and Counting   125
ANSWERS AND SOLUTIONS

17. D The probability of throwing four tails 24. D  W e add the probabilities of the
two types but must subtract the
in a row with a fair coin is 1 . This is number of people who fit both
16
not a rare enough occurrence categories at the same time to avoid
to support a decision about the ­double-​c­ ounting them.
fairness of the coin. Since each throw
is independent of all others, the   14 + 13 − 11 = 16
probability of throwing another tail 23 23 23 23

with a fair coin is still 21, so whatever 25. A 7P7 = 7! = 5040

happens on the next throw proves 26. A There are 10 multiples of 3 out of 30
10 1
nothing. total possibilities. 30 = 3 .

18. D He has eight choices of shirt, four 27. C    The trials are not run with
choices of pants, and three choices of replacement. There are only
shoes. 8 × 4 × 3 = 96. 27 numbers left. Only one of those
papers has the number 1 on it, so the
19. C If he chooses two shirts, he has 8 × 7 = probability is 1 out of 27.
2 × 1
56 = 28 choices for the shirts.
2
Multiply that times the number of 28. D T he order in which the instructors sit
on a committee does not matter, so we
choices for pants and shoes. 28 × 4 × use a combination. Six women selected
3 = 336.
6!
20. B For the first shirt, he can pick any two at a time = 6C2 = (6 = 15;
− 2)!2!
1 eight men selected two at a time =
of the 8, so 8 . The second shirt chosen
8!
must match the first, and there are 8C2 = (8 = 28; 15 × 28 = 420.
− 2)!2!
s81o×1771. T=he516pr.obability
seven shirts left, 29. D 6 women out of 14 people gives
for both those is
6 = 73 .
21. C There are a total of nine men out of a 14

total of 23 people. 30. B 6 × 5 × 4 × 3 = 3 × 5 × 1 × 3 = 15 = 0.0
B oThf aerteoatarel osfe2v3enpemo1ap64lele×a. s1s53em×b14l2er×s o13u1t= 15 3 × 31 × 14 13 12 11 7 13 3 11 1001
22. 3 3 15
7 11 1001 0.0145 = 1.45%
× = =

23. C There are 14 females, so the is 2143. 31. B There are two green spaces out of
probability of choosing a female
2 119.
The probability of choosing a second 38 total spaces. 38 =
ofefmthaoleseisis12322134. T×he2132pr=ob15a08b62il=ity29f51o3r. both
32. A T here are 18 winning numbers in the
wheel out of 38, so the probability is

18 = 0.4736 = 47.36%.
38

126    ANSWERS AND SOLUTIONS: Probability and Counting
ANSWERS AND SOLUTIONS

33. C The only way to avoid a n­ on-​ 42. C There are four queens, but only
C­ alifornian is to pick three 51 total cards because a king has
Californians. That works out to already been chosen.

6 × 5 × 4 = 1 × 5 × 1 = 5 = 1 43. D There is a 4 chance of drawing a 7
30 29 28 5 29 7 1015 203 52

5 × 4 = 51 × 2 59 × 1 = 5 = 1 . and then a 4 chance of drawing a 2.
29 28 7 1015 203 51

Subtracting from 1 gives the 4 + 4 = 412 = 103 .
probability that the group will have at 52 51 2652 663

least one ­non-​C­ alifornian: 202 . 44. A She has drawn four cards already, so
34. C 203 there are 48 cards left. There are still
four aces, so the probability of
The combinations give 5 × 2 × 3 = 30.

35. 42 There are now 7 × 3 × 2 = drawing an ace is 4 = 112.
42 possibilities. 48
45. D
The probability of rolling a 2 is 1 , so
6
36. D You still have your 5 tickets but now the probability of not rolling a 2 is 56.
there are only 98 tickets total, so the 46. C

probability is 5 . The probability of not rolling a 6 is
98
47. A 5
37. A There are 31 × 31 × 31 possible 6 , so the probability of doing it twice

combinations but only 31 that have all is 5 × 5 = 25 .
6 6 36
three digits the same.
The probability of rolling a 3 is 1 ,
31 = 1 6
31 × 31 × 31 961 and the probability of rolling a 4 is

38. 26,970 If each digit must be different, there 1 . The probability of doing both is
6
are 31 × 30 × 29 = 26,970 possibilities. 1 1 316.
6 × 6 =
number of outcomes 165
39. 66% probability = number of trials = 250 = 480.. 66B = 6 6 % There are not a lot of combinations

that are less than 4, so you could just
list them: (1, 1) or (2, 1) or (1, 2). That
number of oouf ttrci oamls es = 165 = 0.66 = 66% is 3 out of 36 possible combinations.
number 250
You could also draw a grid to count
40. D 4 cards are kings, so 48 cards are them:
52 52
48 12
not kings. 52 = 13 .

Die #2

41. B 4 cards are 3s and 13 cards are 123456
52 52
12 3 4 5 6 7
Die #1 23 4 5 6 7 8

hearts. 4 + 13 = 17 . However, one 34 5 6 7 8 9
52 52 52
card is the 3 of hearts and was counted 4 5 6 7 8 9 10
in both categories, so subtract it once 5 6 7 8 9 10 11
6 7 8 9 10 11 12

from the total. 17 − 1 = 16 = 4 .
52 52 52 13

ANSWERS AND SOLUTIONS: Geometry   127
ANSWERS AND SOLUTIONS

49. C There are three odd numbers on each die,

so 3 × 3 = 9 = 14 . 1 2 3456
50. C 6 6 36 1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
“At least” problems are easier to do by 3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
finding the opposite and subtracting 4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
from 1. The probability of not rolling a 2 5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
or a 3 is 64 , so 4 × 4 = 16 = 4 . 1− 4 = 5 .
6 6 36 9 9 9
You could also draw a grid to count them.
Twenty out of the 36 combinations have a

2 or a 3 or both. 20 = 5 .
36 9

Chapter 9 Geometry

1. D 3. doubled

Working from the area, the largest plot is Doubling the sides a and b gives us
40 feet high. The square is half this height
because both plots are 20 feet wide. This (2a)2 + (2b)2 = 4a2 + 4b2 =

makes the dimensions of the triangle 20 feet 4(a2 + b2 ) = 2 a2 + b2 = 2c

on each side. The area of a triangle is 1 bh. 4. C For a triangle, A = 1 bh. Doubling
2 2
The triangle has an area of 200 square
feet. (It is also half the area of the small the lengths gives

square.) Total area is 800 + 400 + 200, or A = 1 (2b)(2h) = 1 • 4bh = 4  1 bh , or
1400 square feet. 2 2  2

2. A The major difficulty here is the length four times the area.

of the hypotenuse of the triangle. Each 5. B Each sphere has a radius r = 6
of the triangle’s legs is 20 feet, so the inches. The surface area is given by
hypotenuse is 4πr2 = 4π • 62 = 4π • 36 = 144π ≈ 144 • 3.14 = 452.1

202 + 202 = 400 +44π0r02 = 4π80•062≈ =284.3π • 36 = 14 4π ≈ 144 • 3.14 = 452.16. Each sphere has an
area of 452.16 square inches. Together,
28.3 feet. Adding the sides gives all three require 1356.48 square inches
20 + 40 + 40 + 20 + 28.3 = 148.3 feet, but of paint. At 288 square inches a can, this
the problem asks for a whole number of means she needs 1356.48 ÷ 288 =
feet, which means the answer is 149 feet. 4.71 cans. She must buy 5 whole cans.

6. B A square 4 feet on a side allows a circle of
radius r = 2 feet. The area of this square
will be A = πr2 = π22 = 4π ≈ 4 • 3.14 =
12.56 square feet.

128    ANSWERS AND SOLUTIONS: Geometry
ANSWERS AND SOLUTIONS

7. 3.44 square feet 20. A

The area of the square with a side of length The volume of the cube is V = s3 = 13 = 1.
s = 4 is given by A = s2 = 42 = 16 square The sphere, with radius r = 21, has volume
feet. Subtracting the area of the circle gives 3
16 – 12.56 = 3.44 square feet. 4 4  1  4 1 π 3.14
V = 3 πr 3 = 3 π  2  = 3 π • 8 = 6 ≈ 6 = 0.523
8. D If the area of the circle is equal to the area
outside the circle, tthheenartehaeoafrethaeocfitrhc=lee43.
rectangle is twice π • 1 = π6 ≈ 3.614 = 0.523. The difference is 1 – 0.523 =
8
The circle, with radius r = 3, has an area 0.477 cubic centimeters.
A = π • 32 = 9π ≈ 9 • 3.14 = 28.26 square
feet; twice that is 56.52. Dividing this by 6 21. C The half sphere has the area
gives the result. 4 πr 2
Per the formula for the area of a A = 2 = 2π • 12 ≈ 2 • 3.14 • 1 =

9. C b1 b2 6.28. The lateral area of a right cone

trapezoid, A = + h, the area of each is A = πrl, where l is the slant height.
2 Use the Pythagorean theorem on a right
end is 432 square inches. Double this
and get 864 square inches for the two triangle with a 1‑foot base and a 6‑foot
height to find the slant height:
ends. Note that the bottom of the hood
is open; calculate: The top is 24 × 36, or 62 + 12 = 36 + 1 = 37 ≈ 6.08. The
864 square inches. The back is 16 × 36, lateral area is then A ≈ 3.14 • 1 • 6.08 ≈
or 576 square inches. The sloping front is 19.1. Adding these two areas yields
the hypotenuse of a right triangle whose 25.4 square feet of paintable surface.

legs are 6 and 16 inches. Its length is 22. D
17.1 inches. The area of this piece is then
17.1 × 36 or 615.6 square inches. The If one side is x and the other 2x, the area of the
total area is 864 + 864 + 576 + 615.6 =
2919.6 square inches. triangle is A = 1 bh = 21( x)(2x) = x2 = 72.25.
2
Taking the square root gives x = 8.5. The two
10. H sides of the triangle are 8.5 and 17 inches. Using

11. C 8.52 + 172 = the Pythagorean theorem, you find that the
12. A hypotenuse is 8.52 + 172 = 72.25 + 289 = 361.25 ≈
13. F
72.25 + 28 9 = 361.25 ≈ 19. The perimeter is the sum of the
three numbers: 8.5 + 17 + 19 = 44.5 inches.

14. B 23. A A regular hexagon inscribed in a circle
15. D has a side equal in length to the radius
16. G of the circle, since segments drawn from
17. J the center to the vertices of the hexagon
18. E form six equilateral triangles. An apothem
drawn in one of these triangles forms a
right triangle with hypotenuse 3 and half

19. I

ANSWERS AND SOLUTIONS: Geometry   129
ANSWERS AND SOLUTIONS

a side, 1.5, for a base. Use the Pythagorean 30. B Use C = 2πr to find the radius of each
theorem to find the apothem: 4
ball and V = 3 πr 3 to find each volume.
32 − 1.52 = 9 − 2.25 = 6.75 ≈ 2.6.
The perimeter is P = 6 • 3 = 18. The area is Dividing the circumference of the soccer
21(2.6)(18) = 23.4 square centimeters. ball by 2π gives a radius of 10.8 and a
volume of 5310. Doing the same for the

24. B basketball gives a radius of 12 and a
volume of 7268 cubic centimeters. The
difference is 1958 cubic centimeters.
A 1 aP 1 4.13 • 30 61.95
= 2 = 2 • = 31. A

25. B The area of the end is 10 square feet The perimeter of the field is 450 feet. The
equilateral triangle has three equal sides, so
4 + 6 • 2 . The volume is the area of the each will be 150 feet.
 2 
 32. C
end times the length of the trough:
10 • 11 = 110 cubic feet.
The rectangle has an area of 12,500. Use the
26. C Pythagorean theorem to find the height of the
triangle. The hypotenuse is known, and one leg
( ) The volume of a right cylinder (a can) is πr2 h, is half the side of the triangle, so the other leg is

the height times the area of one end. With a 1502 − 752 = 22,500 − 5625 = 16,875 ≈130.
3.75-centimeter radius, the area of each 1 1
The area of the triangle is 2 bh = 2 (150)(130) ==
circular end is 44 square centimeters. Dividing
this into 296 cubic centimeters tells us the 9750. The difference is 2750 square feet.
height will be 6.7 centimeters high.
33. D If he takes down the fences, the farmer
27. B The circumference of a can can irrigate a circle 330 feet in radius for
7.5 centimeters in diameter is an area of 342,119 square feet. If he leaves
πd ≈ 3.14 • 7.5 = 23.55 centimeters. his fences, he gets four circles with radii of
Adding 1 centimeter for gluing overlap, 165 feet. Each small circle has an area of
this becomes 24.55 centimeters. The 85,530 square feet, or 342,119 total square
label area is 24.55 • 6.7 = 164.5 square feet. There is no difference.
centimeters.
34. A
28. C The volume of a package is 6 • 12 • 3 =
216 cubic inches. Dividing 13,842 by 216 The square has an area of 15,625 square feet.
tells us that there will be 64 packages per The semicircle has a radius of 62.5 feet. The
box. area of a circle is πr2, so an entire circle would
have an area of 12,272 feet. However, we need
29. D to cover only half that, or 6136 square feet. The
sum of 15,625 + 6136 is 21,761 square feet.
The surface area of the box is
A = 2hl + 2hw + 2lw = 35. D
2 • 3 • 12 + 2 • 3 • 6 + 2 • 12 • 6 =
72 + 36 + 144 = 252 square inches Since the area of the triangle is 1 bh, the
2
triangle’s other side is 8 inches long. Using the
The volume of plastic for one box is Pythagorean theorem, a2 + b2 = c2, we find the

252 • 1 = 15.75 cubic inches per package. hypotenuse, c, to be 10 inches long.
16

130    ANSWERS AND SOLUTIONS: Geometry
ANSWERS AND SOLUTIONS

36. A The radius of both the cylinder and the 42. C The original box has panels of 7 • 12 = 84;
half spheres is 6. The volume of the two 9 • 12 = 108, and 9 • 7 = 63, which add up
half spheres is the same as the volume of to 255 square inches for half the panels.
one whole sphere,
The new box has panels of 3 1 • h, 4 1 • h,
4 3 4 • 63 2 2
V = 3 πr ≈ 3 • 3.14 ≈ 905. 1 1 3
and 4 2 3 2 15 4 square inches for half
The volume of the cylinder is • =

V = πr2h ≈ 3.14 • 62 • 50 = 5652. The the surface area. Adding the new box’s

sum is 6557 cubic feet. dimensions means that 8h + 15 3 = 255.
4
37. A The surface area of the side of a cylinder
is needed, A = 2πrh ≈ 2 • 3.14 • 6 • 50 = 1884 Then, 8h = 239 1 square inches, and
4
h ≈ 2 • 3.14 • 6 • 50 = 1884 square feet. The half spheres have
the surface area of a whole sphere, h = 29 29 inches.
A = 4πr2 ≈ 4 • 3.14 • 62 ≈ 452 square feet. 32
The sum of the areas is 2336 square feet.
43. D
38. C The radius is proportional to the cube of The area of the square is 2500. The
the volume, so the volume is proportional
to the cube root of the radius. Doubling the circles each have r = 25, and total area
volume increases the radius by 3 2 ≈ 1.26. 2πr2 ≈ 2 • 3.14 • 252 == 3925 square feet. Total
area is 6425 square feet.

39. 262.5 44. D
The total perimeter is that around two circles
A right triangle is formed, with one leg =
250 feet along the ground and the other leg = with r = 25: 2(2πr ) ≈ 2 • 2 • 3.14 • 25 ==314 feet.
85 – 5 feet (the change in height). Using the
Pythagorean theorem, we find that the wire is 45. A
The cube root of the volume is the length of
2502 + 802 = 62,500 + 6400 = 68,900 ≈
262.5. one edge, since V = l3. This gives an edge
length of 144 inches.

40. D The area of the base of the pyramid is 36. 46. B Using the Pythagorean theorem, we find
that the length of the other leg is
The volume is 31(36)(20) = 240. The
11.42 − 72 = 129.96 − 49 = 80.96 ≈ 9.
cone has a circular base with r = 3. Its area 1 1
The area of the triangle is A = 2 bh = 2 • 7 • 9 =
is πr2 = π (3)2 ≈ 28.27. The volume of = 21 bh = 21 • 7 • 9
A = 31.5. The volume of the prism
the cone is V = 31(28.27)h ≈ 9.42h = 240.
Dividing 240 by 9.42 gives the height of is V = Bh = 31.5h = 787.5. Dividing 787.5
by 31.5 gives a height of 25 units.
the cone, 25.5 feet.
1 1 1 1
41. A 47. D The area of one end is A = 2 (b1 + b2 )h = 2 •  11 + 7 2  • 3 2

The top is a cone with r = 1.5 and h = 6,
1 1  1 1 3
so V = 1 πr 2h ≈ 1 • 3.14 • 1.52 • 6 = 14.13. A= 2 (b 1 + b2 )h = 2 •  11 + 7 2  • 3 2 • = 32 8 . Multiplying
3 3
Half of that is 7.065.
by the height of the box gives the

volume: V = Bh = 32 3 • 7 = 226 5 cubic
8 8
centimeters.

ANSWERS AND SOLUTIONS: Polynomial and Rational Expressions   131
ANSWERS AND SOLUTIONS

48. A The volume of the space station is V = Bh, where
B is the area of the base, which is the same as
Using the formula, A = 3 s2 = 3 • 152 = 3 • 225 the area of the hexagonal ­cross-​s­ ection. Each
4 4 4
side of the hexagon must be 60 feet long. Split
3 s2 = 3 • 15 2 = 3 • 225 ≈ 1.732 • 225 ≈ 97.4. the hexagon into six equilateral triangles 60 feet
4 4 4 4 on a side. The area of the

49. A Multiplying the given area by 4 and then triangle is A = 3 s2 = 3 • 602 = 3 • 3600 ≈ 1.732 • 36
dividing by 3 gives the square of the 4 4 4 4
length of the side: s2 = 200. Take the
square root to get the length of a side: = 3 • 3600 ≈ 1.732 • 3600 = 1558.8 .
s = 200 ≈ 14. 4 4
The hexagon’s area is 6 times this:
1558.8 × 6 = 9352.8. The volume of the station is
50. 4,676,400 V = Bh = 9352.8 • 500 = 4,676,400 cubic feet.

Chapter 10 Polynomial and Rational Expressions

1. C 2a + 5b − 7 + a − 9b − 6 = 8. A 6x and 9y are added, so the expression is
2a + 1a + 5b − 9b − 7 − 6 = a sum.
3a − 4b − 13
9. A The coefficient is the numerical factor in a
2. B 5(3x − 2y + 4) = variable term.
5 • 3x − 5 • 2y + 5 • 4 =
15x − 10y + 20 10. D Sharon interviews 2s households. Seven
fewer than 2s is 2s − 7.
3. A (2x + 5) − (5x − 7) =
2x + 5 − 5x + 7 = 11. B (3x2 + x − 2) + (x2 − 4x + 7) =
2x − 5x + 5 + 7 = 3x2 + x2 + x − 4x − 2 + 7 =
−3x + 12 4x2 − 3x + 5

4. 5 + 2x 12. A (3x + 2y) − (2x + 3y) =
3x + 2y − 2x − 3y =
“five more than twice a number” = “5 more 3x − 2x + 2y − 3y = x − y
than 2x” = 5 + 2x
13. C (6x2 + 2x − 4) − (2x2 − 5x + 1) =
5. C 7 and (y − 1) are multiplied, so the 6x2 + 2x − 4 − 2x2 + 5x − 1 =
expression is a product. 6x2 − 2x2 + 2x + 5x − 4 − 1 =
4x2 + 7x − 5
6. x3 − 8
14. A Add the exponents: x3 • x 6 • x2 = x3+6+2 = x11
“eight less than the cube of a number” = x 3 • x 6 • x 2 = x 3+6+2 = x11
“8 less than x3” = x3 − 8
15. B 2x 4 • 4x5 = (2 • 4)x 4+5 = 8x9
7. B 95 C 32 9 (−40) 32 9(−8) 32 −72 32 = −40
+ = 5 + = + = + Subtract exponents: x8 x 8−2 x6
16. D x2
(−40) + 32 = 9(−8 ) + 3 2 = −72 + 32 = −40 = =

132    ANSWERS AND SOLUTIONS: Polynomial and Rational Expressions
ANSWERS AND SOLUTIONS

17. D 25x9 y 4 = 5 • 5x9−6 = 5x3 27. x2 + x − 3
15x 6 y12 5 • 3y12− 4 3y8 “the sum of the square of a number and three

18. C 5x3y(3xy2 + 2x2y3) = less than the number” = “the sum of x2 and
5x3y • 3xy2 + 5x3y • 3x2y3 = x – 3” = x2 + x − 3

15x4y3 + 10x5y4 28. C −16t2 + 350 = −16 • 42 + 350 =

19. D (3x + 4)(2x − 5) = −16 • 16 + 350 = −256 + 350 = 94 feet

3x • 2x − 3x • 5 + 4 • 2x − 4 • 5 = 29. A The leading coefficient is the number in
6x2 − 15x + 8x − 20 =
6x2 − 7x − 20 front of the term with the largest exponent.

20. B (x − 2y)(2x − y) = 30. B 12x4y + 9x3y2 − 6x2y2 =
x • 2x − x • y − 2y • 2x + 2y • y =
2x2 − xy − 4xy + 2y2 = 3x2y • 4x2 + 3x2y • 3xy − 3x2y • 2y =
3x2y(4x2 + 3xy − 2y)

2x2 − 5xy + 2y2 31. D 3x2 − 8x + 4 =

21. C 12p3q − 16p5q2 + 10p4q4 = 12p3q − 16 p5q2 + 10 p 4 q 4 3x2 − 6x − 2x + 4 =
8p2q3 8p2q3 8p2q3 8p2q3 3x(x − 2) − 2(x − 2) =
(x − 2)(3x − 2)

16 p5q2 + 1 0p 4q 4 = 12p3q − 16 p5q2 + 10 p 4 q 4 = 32. C 2x2 − xy − y2 =
8p2q3 8p2q3 8p2q3 8p2q3 2x2 − 2xy + xy − y2 =
2x(x − y) + y(x − y) =
= 3p − 2p3 + 5p2q (x − y)(2x + y)
2q2 q 4 33. A 12x2y + 40xy − 32y =
4y[3x2 + 10x − 8] =
22. A 21x3 − 14 x2 = 7x2(3x − 2) = 3x − 2 4y[3x2 − 2x + 12x − 8] =
14 x3 + 21x2 7x2(2x + 3) 2x + 3 4y[x(3x − 2) + 4(3x − 2)] =
4y(3x − 2)(x + 4)
23. A 9s3t + 6st2 = 3st(3s2 + 2t) = 3 34. (4x – 9y)(4x + 9y) 
6s3t + 4st2 2st(3s2 + 2t) 2

24. C 2x − 3 16x2 − 81y2 = (4x)2 − (9y)2 = (4x − 9y)(4x + 9y)
35. 2x2 = x + 5
)x + 2 2x2 + x − 6 “twice the square of a number is five more

2x2 + 4x than the number” → “twice the square of
− 3x − 6 a number = five more than the number” →
−3x − 6 “twice x2 = five more than x” → 2x2 = x + 5
36. A The degree of a polynomial in a single
25. B 2x2 − 4xy − 3y2 =
2 • 52 − 4 • 5(−1) + 3(−1)2 = variable is the largest exponent.
2 • 25 + 4 • 5 + 3 • 1 =
50 + 20 + 3 = 73
26. x4 − 2x3 − 7x2 + 6x + 5 37. D t d = 390 = 390 = 65 miles per hour
+4 2+4 6

Write the highest-degree term, then the next
highest, then the next, etc.

ANSWERS AND SOLUTIONS: Polynomial and Rational Expressions   133
ANSWERS AND SOLUTIONS

38. C 2 + 5 = 2 • 2 + x • 5 = 4 + 5x = 4 46+2x.5 2 xC x +1 + x −1 − 2x 4 =
3x2 6x 2 3x2 x 6x 6x2 6x2 2x − 4 2x + 4 x2 −

2 2 + 6 5x = 2 • 2 + x • 5 = 4 + 5x = 4 + 5x = x +1 + x −1 − (x 2x =
3x 2 3x2 x 6x 6x2 6x2 6x2 2( x − 2) 2( x + 2) − 2)( x + 2)

5 = 2 • 2 + xx • 6 5x = 4 + 5x = 4 + 5x = x +2 • x +1 + x −2 x −1 − 2 • (x 2x
6x 2 3x2 6x2 6x2 6x2 x +2 2( x − 2) x −2• 2( x + 2) 2 − 2)( x + 2)

39. B 2x − 5 + x +1 ==52(xxxx+++−2225)• +2(3xx(x+x−+1+21)2+) x − 2 • 2 (xx −+ 12) − 2 • (x − 2x + 2) == x2 + 3x + 2 + x2 − 3x + 2 −
5x + 10 3x + 6 x − 2 2 2)( x 2( x − 2)( x + 2) 2( x − 2)( x + 2)

2x − 5 + 3xx ++ 16 = 2x − 5 + x +1 = = 2(xx2−+23) (xx ++2 2) + x2 − 3x + 2 − 2( x − 4x + 2) =
5x + 10 5( x + 2) 3( x + 2) 2( x − 2)( x + 2) 2)( x

= 3 • 2x −5 + 5 • x +1 = 6x − 15 + 5x ++52 ) = 1151(x=x−2+2(12xx0)2−−24)(xx + 4 = 2( x2 − 2x + 2) =
3 5( x + 2) 5 3( x + 2) 15( x + 2) 15( x + 2) 2( x − 2)( x + 2)

2x − 5 + 5 • 3( xx ++1 2) = 6x − 15 + 5x + 5 = 11x − 10 = x2 − 2x +2
5( x + 2) 5 15( x + 2) 15( x + 2) 15( x + 2) ( x − 2)( x + 2)

3x 4y 3x 91•4x51y2x30−xy=8−3y90222xxyy238yab•−x314y4538y0x•yx926=yab336=90xyx953x382yaxb=3x23−0y4−4xx38y4•8y0y 4y96•x232yaba .36 x5=xyy 5A5339 =x323480x−abxy4x8y3•y4ya2•x5 y5963ab63=xy 5 x4 • x5 = x4x5 = x9
10 y 15x 3x 30 xy 4y a5 y 3 4a5 yy 3 4a5 y 4
40. D − = y 3=
y• x5 x9
130xy − 145 yx = 3x 3x − 2 x4 yy 3 = 4a5 y 4
= 3x 10 y 2 = 4a5
•

3x 3x 2 yy • 145 yx 9x2 8y2 = x4x5 = x9
3x 10 y 2 30 xy 30 xy 4a5 yy 3 4a5 y 4
= • − −

3x x 3x x 44. B x2 − 2x − 3 • x2 − 9 = (x − 3)( x + 1) • ( x − 3)( x +
2x − 10 2x + 2( x − x2 + 3x x2 + 2x +1 x( x + 3) ( x + 1)( x +
41. D − 6 = 5) −x22( x x3)−
2 x x+ 6 3x − − +2 x2 − 9
3x 2( x − 5) x 3 • + 2x +1 = ( x − 3)( x + 1) • (x − 3)( x + 3) =
2x − 10 x+ = x( x + 3) (x + 1)( x + 1)
− = = x2 + 3x x 2 =
=
= 2( 3) x − 3 x −3 ( x − 3)2
x2 +=9xx −x−3x •2 x+x −5+ x31 x x +1 x( x + 1)
x +3 3x x − 5 x = 2( x − 5)( x + 3) • =
x +3 2( x − x − 5 + 3) 3
• 5) − • 2( x 3) ==

3x2 + 9x 3) − x2 − 5x ( x − 3)2
2( x − 5)( x + 2( x − 5)( x + x( x + 1)

9x − x2 5−) (5xx + 3) = 3x2 + 9x − x2 + 5x = 45. D 20x2 y5 ÷ 10b2 x 6 = 20x2 y5 9a8 y 3 = 2y5 • a2 y 3
x+ − 2( x − 5)( x + 3) 27a6b 9a8 y 3 27a6b • 10b2 x 6 3b b2x 4
3) 2( x
( x220−x7(x5ax2)6(y+bx57+÷) 319 )0ab 82yx 36 20x2 y5 9a8 y 3 2y5 a2 y 3 2a2 y 8
= 2x2 + 14 x = 2x(x + 7) = = 27a6b • 10b2 x 6 = 3b • b2x 4 = 3b3 x 4
2( x − 5)( x + 3) 2( x − 5)( x + 3)
=220(7xxa2−x6y(b5x5)(+÷x71+9)0a3b8)2yx36 20x2 y5 • 19 0ab 82yx 36 2y5 a2 y 3 2a2 y 8
= 2 x2 5+) (1x4 +x 3) = 2x(x + 7) = 27a6b = 3b • b2x 4 = 3b3 x 4
x − 2( x − 5)( x + 3)
2(

14 x = 2( x2−x(5x) (+x7 +) 3) = (x x(x + 7) 3)
x + 3) − 5)( x +

134    ANSWERS AND SOLUTIONS: Solving Equations and Inequalities
ANSWERS AND SOLUTIONS

46. A 4x + 8 ÷ x2 − 4 = 4x + 8 • x2 + x − 6 48. C 3x2 + 7x − 2 = 3(−4)2 + 7(−4) − 2 = 3 • 16 + 7(−4) − 2
x2 + 3x x2 + x − 6 x2 + 3x x2 − 4 2x2 − 7x + 3 2(−4)2 − 7(−4) + 3 2 • 16 − 7(−4) + 3

+8 ÷ x 2x+2 −x −4 6 = 4x + 8 • x2 + x − 6 = 3x2 + 7x − 2 = 3(−4)2 + 7(−4) − 2 = 3 • 16 + 7(−4) − 2
+ 3x x2 + 3x x2 − 4 2x2 − 7x + 3 2(−4)2 − 7(−4) + 3 2 • 16 − 7(−4) + 3

= 4( x + 2) • (x + 3)( x −−2322xx))22=+−4x77 x −1 2 =4 3(−4)2 + 7(− 4 ) − 2 = 3 • 16 + 7(−4) − 2 =
x( x + 3) (x + 2)( x x• 1+ =3 x 2(−4)2 − 7(−4) + 3 2 • 16 − 7(−4) + 3

2) ( x + 23)) (( xx −− 22)) 4 1 4 = 48 − 28 − 2 = 18 = 2
3) ( x + x 1 x = 32 + 28 + 3 63 7
• = • =

x2 − 5x x2 − 7x + 12 = 48 − 2288 +− 32 18 = 2
4x2 • x2 − 16 32 + 63 7
2x − 10
47. D ÷ x2 + 2x − 8 =

x2 − 5x x2 − 7x + 12 x2 + 2x − 8 49. A
4x2 x2 − 16 • 2x − 10
= • = “the quotient of 5 more than a number and 5

x( x − 5) ( x − 3)( x − 4) ( x − 2)( x + 4) less than the number” = “the quotient of x + 5
4 x2 • ( x − 4)( x + 4) • 2( x − 5)
= = and x – 5” = x+5
x −5

= 1 • x − 3 • x −2 = (x − 3)( x − 2)
4x 1 2 8x
50. B

1 • x − 3 • x 2− 2 = (x − 3)( x − 2) After the donation to charity, there are
4x 1 8x
p – 1000 dollars left. This is split evenly
between three people, so each person gets

p − 1000 dollars.
3

Chapter 11 Solving Equations and Inequalities

1. p = 17 p – 5 + 5 = 12 + 5 → p + 0 = 17 → p = 157. 0 5x + 2( x – 9) = 7x + 10 → 5x + 2x – 18 = 7x + 10 →
– 5 + 5 = 12 + 5 → p + 0 = 17 → p = 17 5x + 2( x – 9) = 7x + 10 →  5x + 2x – 18 = 7x + 10 →
 7x – 18 = 7x + 10 → 7x – 18 – 7x = 7x + 10 – 7x
2. s = –8 7s = –56 x11s–=18–8= s =7x–8– 18 = 7 x + 10 →  7x – 18 – 7x = 7x + 10 – 7x
7 7 → → + 10 → 7x – 18 – 7x = 7x + 10 – 7x –18 = 10. This equation is
7 7x
3. x = 4 –3x + 17 = 5 → –3x + 17 – 17 = 5 – 17 → true for no value of x.
– 3 x + 17 = 5 –3x +==1–7–1–1–2321→→7 =––x353x=– =17––→132
→ –3x 4 → x=4 6. B   7(b + 11) = 245 → 7(b + 11) = 245 → b + 11 = 35 →
– 3x = –12 → –3x 7(b + 11) 7 7
–3 += = 822744=55–→→7 +      bb78(++b→117+111–a=11)=31=51=→237455
4. a = 1 –7 a=–2845=→–77→(b + 11) → b+ 11 = 35 →
–46 2 → a – 8 = 7a –8 – 11 → b=
6(a – 8) =7–(b462+ 11) 24
6 →
6(a – 8)
6 = → a – 8 + 8 = –7 + 8 → a = 1

ANSWERS AND SOLUTIONS: Solving Equations and Inequalities   135
ANSWERS AND SOLUTIONS

7. A 95 = 9 C + 32 → 95 – 32 = 9 C + 32 – 3 123→. A6 3 = 9 C → 3k –7+7 ≥ 17 + 7 → 3k ≥ 24 → 3k ≥ 24 →
95 5 5 5 3k k≥8 3 3
9 9 + 32 – 32 3→k –637 += 795≥C 1→7 + 7 → 3k ≥ 24 → 3 ≥ 24
9 5 = 5 C + 32 → – 32 = 5 C 3 →
– 32 →
+ 32 → 95 – 32 = 95 C + 32 63 = 9 C → 14. C –2x – 3 + 3 < 3 + 3 → –2x < 6 → –2 x > 6
5 → 5 –2 –2
9 = 3 < 3 + 3 → –2 x 6
5 63 C 5 9 C 5 7 –2 x –3 + = 35 –2 x < 6 → –2 > –2 → x > –3
9 9 5 =C → C
• • → •

• 6 3 = 5 • 9 C 5•7= → C = 35 15. A Time for 1st train: t; distance: 120t.
9 5 Time for 2nd train: t – 2; distance:
80(t – 2).
8. C y – b < mx + b – b → y – b < mx →
y – b < mx +b –b y – b < mx → → 3x + 16. (2, –1) 3x + y = 5 → 3x + y – 3x = 5 – 3x → y = 5
→ 3 x + y = 5 3x + y – 3x = 5 – 3x → y = 5 – 3x;
y – 3 x = 5 – 3x → y = 5 – 3x;
y –b < mx → y m–3bx <+ yx = 5 →
m m
15 → 2 x – 3y = 7 2x – 3y = 7 → 2x – 3(5 – 3x) = 7 → 2x – 15
9. 5:00 p.m. 5c + 15 = 40 → 5c + 15 – 15 = 40 – 2x – 3 (5 – 3x) = 7 2x – 3(5 – 3x) = 7 → 2x – 15 + 9x = 7 →
5c + 15 = 40 → 5c555ccc=+==512h2555o5–→u→1r5s5a5c=cft=4e=0r151n2–5x5o1o–25→nx1→5–c=3=y75=→71→1x → 2x – 15 + 9x = 7 →
→ 11x – 15 = 7 → 11x – 15 + 15 = 7 + 15 → 11x
11 x – 15 = 7
5c = 25 → – 1 5 + 1 5 = 7 + 15 → 11x – 15 + 15 = 7 + 15 → 11x = 22 →
25 → 55c 25 → 11x = 22 →
5c = = 5 → 11x 22
11 = 11 x = 2;
10. D →

2(w + 5) + 2w = 90 → 2w + 10 + 2w = 90 → 4w + 10 = 9 y0 =→5 – 3x y = 5 – 3x →y = 5 – 3 •2→ y =5 − 6 → y
y = 5 – 3• 2→ y = 5 − 6 → y = –1
w + 5) + 2w = 90 → 2w + 10 + 2w = 90 → 4w + 10 = 90 → →
w + 10 + 2w = y = 5 – 3x → y = 5 – 3 • 2 → y = 5 − 6 → y = –1
819840000 →→→→ w44l444www=w=2++==201108+080040–5=→1→=90024w=45w→9=0=2–804100→→w4wy= == 580– →3x 4→w y==8450 –→3 •w 2= →2 0y = 5 − 6 → y = –1
10 – 10 = 90 – 20 4 17. D With r milliliters of solution added

– 10 → 4w = to the cylinder each minute,
after 60 minutes 60r milliliters
= 80 → 4w = will be added to the 72 milliliters
4 already in the cylinder. The total

11. B Total cost of m bags of milo = amount 60r + 72 cannot exceed
50m; total cost of s bags of the capacity of the cylinder, so it
soybean = 40s; total seed cost = must be ≤ 500.

= 50m + 40s = 2000. 18. B 4 x – 2(3x + 7) = 6 + 5( x – 3) → 4 x – 6x – 14

12. 5 When the rock strikes the4 x – 2(3x + 7 ) = 6 + 5( x – 3) → 4 x – 6x – 14 = 6 + 5x – 15 →
ground, h = 0; –2x – 14 = 5x – 9 → –2x – 14 – 5x = 5x – 9
80t – 16t2 = 0 → –2x – 1 4 = 5x – 9 → –2x – 14 – 5x = 5x – 9 – 5x →
16t(5 – t) = 0 → 16t = 0 or 5 – t = 0 → –7x – 14 = –9 → –7x – 14 + 14 = –9 + 14 →
1 6t (5 – t) = 0 → 16t = 0 or 5 – t = 0 → – 7x – 14 = –9 → –7x – 14 + 14 = –9 + 14 → –7x = 5 →
16t 0 –7 x 5 5
16 = 16 or–57–xt–+1t4 == 0–9+ →t →–7t x=–014o r+t 1=4 5= –9 + 14 → –7x = 5 → –7 = –7 → x = – 7
= 1 06 or 5 – t + t
16t = 0+t →t = 0 or t = 5 ––77x = 5 → x = – 5
16 –7 7

136    ANSWERS AND SOLUTIONS: Solving Equations and Inequalities
ANSWERS AND SOLUTIONS

→ 5 y + 3825 –1925• 3. 25y8 By+≥ ≥348825≥–• 1341088 →→→ Iy225525n≥yyyy,+y≥2ien31•ac8r02h, Ce–→→sa3.r8o25y25ly≥y≥n• +4254w83yil–8l≥g3≥r825o4w→•8121→02521→yy25,≥oyyr1+≥032→8•– 2234→8. ≥Cy 4≥8 4–  38 → 25 y0taTraaq0ehmmnh.u.≥55medea0oo01craaeuut0o•i•msnnnnn5→5odsttocfi=oocequnpoffnung0unsptat.orwrs2uroaetlt0ruitafsteai(nthopo5niaotnuft.n+0nirptft.er.i2wuimefSa0rret)een(e,az5etsaresiztf+nit,iernhnetwciegeferz)ewteehiizteshe
2

20. C p pounds of peanuts have a total   0.50 • 5 = 0.20(5 + w ) → 2.5 = 1+ 0.2w →
value of 2p dollars. 0.50 • 5 = 0.20(5 + w ) → 2.5 = 1+ 0.2w →
c pounds of cashews have a total  2.5 – 1 = 1+ 0.2w – 1 → 1.5 = 0.2w → 1.5 = 0.2w
0.2 0.2
value of 7c dollars. 2.5 – 1 = 1 + 0.2 w – 1 → 1.5 = 0.2w 1.5 = 0.2w 7.5 = w
10 pounds of mixture has a total 0.2w → 10 ..52 = 00.2.2w → 7.5 → 0.2 0.2 →

value2o.5f 5– 1• 1=01=+500.2dwol–la1rs→. 1.5 = =w

2p + 7c = 50; also, p + c = 10 25. B   The area of the base is given

c =10 → p → p + c =10 → p + c – p =10 – p → c = 10 – p by B = lw = 6w; the 6 + 2w = 12 + 2w
+ c – p =10 – p → c = 10 – p perimeter of the base is
2p + 7c = 50 → 2p + 7(10 – p) = 50 → 2p +p7=0 –2l7+p2=w5=0 given by p = 2l + 2w = 2 •
2 p + 7c = 50 2p + 7(10 – p) = 50 → 2p + 70 – 7p = 50 → →2 • 6 + 2w = 12 + 2w. The
50 → 2p + 7( 10 – p) = 50 2p + 70 – 7p = 50 → surface area is given by  
→ –5p 50 → ==6w–520=0–1 →720h→+ 2(–h15w2p ++=21w–22w)  0h =+→1S2A2•h=6+wp1h2=w+122+Bh2+=hw2(1h2=w+8+621w.2)wh + 2 • 6w = 12h + 2hw
–5 p + 70 = 70 –5p + 70 = 50 → –5p + 70 – 70 = 12h + 12w + 2hw =
→ –5 p + 70 – 70 =(1520 +– 720w +–52p•
+ 70 – 70 = 50 – → –5 p = –20 → →)h

–5p = –20 → p = 4 26. x = –2 or   x2 – 10x – 24 = 0 → ( x + 2)( x – 12) = 0
–5 –5 x2 – 10xx=–1224 = 0 → ( x + 2)( x – 12) = 0

21. infinite  x + 2 = 0 or x – 12 = 0
 x = –2 or x = 12
3x – 2( x – 1) = 2( x + 1) – x →
3x – 2x + 2 = 2x + 2 – x → x + 2 = x + 2 →
= 2x + 2 – x → x + 2 = x + 2 → x + 2 – x = x + 2 – x → 2 = 2, 27. x = –3; y = –2   3 • 6x – 5y = –8 → 18 x – 15y = –24 →
5 • 4 x + 3y = –18 20 x + 15y = –90
x + 2 – x = x + 2 – x → 2 = 2, which is true for all x.

22. D 17 red blocks weigh 17r ounces; 3 • 64xx –+53yy == ––818 → 1280xx – 15y = –24 →
23. 13 blue blocks weigh 13b ounces. 5 • + 15y = –90
Total weight is 17r + 13b = 99.
–114 →4 x33+8 8x3y== –13–18148   →→ 34x8(–=x3–=) 3+–→31y14=4→x–1+83338→8yx ==––113818→4 →4(–x3=) +–33y→= –18 →
A The weight of any object must
be a positive number, so r >380x. =

 –12 + 3y = –18 → –12 + 3y + 12 = –18 + 12 →
–12 + 33yy= =–1–86   →→ 3–331yy2==+––3636y→+→1323yy===––1–2386 + 12 →
→y= –2

ANSWERS AND SOLUTIONS: Solving Equations and Inequalities   137
ANSWERS AND SOLUTIONS

28. C Shading to the right of –2 34. A – 2 x ≥ 4 → – 3  – 2 x  ≤ – 3 • 4 → x ≤ –6
indicates sense of “greater 3 2  3  2

than” and the open end point – 2 x ≥ 4 → – 3  – 2 x  ≤ – 3 • 4 → x ≤ –6
indicates “strict” inequality. 3 2  3  2

x + 9 2x – 4x + 29. D → –––2––2222x2xxxx–++><499–x–1––>+16296–9→>7→>→––47–x2x2–x<–9<87→–––142–6x2(w→x→+>x–4–2)<1wx68+=→94>–5 33–265→2–x. .7 <w(DC→Aw 2–=–+1+2l6w44)→ww= x2 = ± 25 → x = ±5 (w + 9)(w – 5) = 0
–9 > 9 > 4 x – 7 – 4 x x<8
–2 x → 45; l = w + 4
–7 – + 9 – 9 > –7 – 9 → = 45 → w2 + 4w – 45 = 0 →
– 45 = 0 → (w + 9)(w – 5) = 0
9 → –2x > –16

30. 0   2 •  3x – 5 y = 2  6x – 10 y = 4 w + 9 = 0 or w – 5 = 0 → w = –9 or w = 5, + 4 = 5 + 4
 3x – 5 –6x + 10 y = 7 → –6x + 10 y = 7 → 0 =bu1t1width cannot be negative, so w = 5., l = w
–6x + 10 0 = 11 37. 28 gallons of the strong detergent,
2 • y = 2  6x – 10y = 4
y = 7 → –6x + 10y = 7 → 32 gallons of the weak
The amount of pure detergent in a solution
is the concentration times the amount of
This statement is not true, so solution. There are 0.90s gallons of pure
there are no solutions for this detergent in s gallons of the strong solution,
system of equations. and 0.15w gallons of pure detergent in

31. B At $40 per lawn, mowing x w gallons of the weak solution. These are
added together to get 0.50 • 60 = 30 gallons
lawns will bring in $40x. This is of pure detergent in the mild solution:
added to the $420 already saved.
The total cannot be smaller 0.90s + 0.15w = 30. Also, s + w = 60.
s + w = 60 → s + w – s = 60 – s → w = 60 – s
than $1500, so a “greater than” 0.90s + 0.15w = 30 → 0.90s + 0.15(60 – s) = 30 →
inequality is used.
0.90s + 0.15w = 30 → 0.90s + 0.15(60 – s) = 30 →
32. A 0.90s + 9– →90.1=5003.s7.075=5–s39=0→→02.71005.7.7→55ss s=+=2912=→8 3000.→7.755s
0.75s + 9– = 21 s
2x2 – 7x + 4 = 0 cannot be facto0r.7e5ds; u+s9e –th9e = 9 → 0.75s = 21 0.75 →

b2 30 –
2a
quadratic formula x = –b ± – 4ac , with w = 60 – s = 60 – 28 = 32

a = 2, b = –7, and c = 4: : 7x 38. A 7x – 6 – 2x ≤ 2x + 4 – 2x → 5x – 6 ≤4→ 5
– 6 – 2x ≤ 2x + 4 – 2x 5x – 6 ≤ 4 → 5x – 6 + 6 ≤ 4 + 6 → 5x ≤ 10 5
(–7)2 → →
x = –(–7) ± 2·2 – 4·2·4 5x – 6 + 6 ≤ 4 + 6 →
= 5x ≤ 10 → 5x ≤ 10 → x ≤ 2
5 5

7 ± 49 – 32 = 7 ± 17 39. B The “less than” part of the inequality
4 4 indicates shading to the left; the “or
equal to” part indicates that the end
33. B point is a solid dot.

For a square, A = s2 , so

64 = s2 → ± 64 = s2 → ±8 = s.

Lengths are positive, so the negative solution
is discarded.

138    ANSWERS AND SOLUTIONS: Solving Equations and Inequalities
ANSWERS AND SOLUTIONS

40. A 0.35m + 0.65 ≥ 4.50 → 0.35m + 0.65 – 0.65 ≥464. .5C0 – 0.65 → Buying g gallons of gas at $3.15
0.35m + 0. 65 ≥ 4.50 0.35m +≥≥0330..6..8835555–→→0.60m50.3.≥35≥5m41.15≥0 30–..83055.65→→m per gallon incurs a cost of
→ 0.35m ≥ 11 $3.15g. This is added to $1.65 to
→ 0.35m get the total, $40.71.
0.35 m ≥ 3.85 0.35 47. A 5f + 150 = 580 → 5f + 150 – 150 = 580 – 150 →

41. B Replacing x with 11 in each of the 5f + 150 = 580 → 5f + 150 – 150 = 580 – 150 →
inequalities and evaluating produces
the following results: 5f = 430 → 5f = 430 → f = 86
43 0 → 55f = 430 → 5 5
5
A. 28 < –18 B. 74 ≥ –24 5f = f = 86

C. 45 ≤ 36 D. 12 > 27 48. D 5( x – 6) < 2( x – 9) → 5x – 30 < 2x – 18 →

42. A → ax + by =c→ ax + by – ax = c – a–axx52(→x→x <–b 62y)x =<– 2c(–x a–x9→) → 5x – 30 < 2x – 18 → < –18 →
a x + by = c → ax + by – ax = c – ax → b5yx=–c3–0 18 – 2x → 5x – 30 – 2x < 2x – 18 – 2x → 3x – 30
by – a x = c – ax by 3x – 30 < –18 →
c ax + = c – ax → 3x – 30 + 30 < –18 + 30 → 3x < 12 → 3x < 12
→ by 3x 3 3 →
b
= c – ax → y = c – ax 3x – 30 + 30 < –18 + 30 → < 12 → 3x < 12 → x < 4
b b 3 3

bby = c – ax → y = c – ax 49. x = 2 or 2x2 – 7x + 6 = 0 → (2x + 3)( x – 2) = 0 →
b b
420x0 2– – 17 3x5x =+→–623=
43. C 5p + 135 ≤ 400 → 5p + 135 – 135 ≤ 0 → (2x + 3)( x – 2) = 0 →
5 p + 135 2x + 3 = 0 or x – 2 =
≤ 400 → 5p +≤≤1223665555–→→13555pp≤≤≤450230655– 135 → 0 →
≤ 265 → 5p → p≤ the
5p 53 2x = –3 or x = 2 →
5p 3
5 x = – 2 or x = 2

50. D When the rock reaches
bottom of the cliff, D = 1024:
44. D x + 19 < 15 → x + 19 – 19 < 15 – 19 → x < –4
x + 19 < 15 → x + 19 – 19 < 15 – 19 → x < –4
15 → x + 19 – 19 < 15 – 19 → x < –4 1024 = 16t 2 → 1024 = 16t 2 → 64 = t2 → ± 64
16 16
≥4→ x161– 01–1 26 29 4x≥=+4 19 +16x61t→22 →– 9D6x±48i→sr==egtt2a. r→din±g
– 4( x 45. B → 2x – 4( x + 3) ≥ 4 + 3(4 –130x2)4→= 126xt–2 64 = t2 →
+ 3) ≥ 4 + 3(4 – 3x) → 2x – 4 x – 12 ≥ 4 + 12 – 9x → the negative time
–2 x – 12 ≥ 16 – 9x → –2x – 12 + 9x
–2 x – 12 + 9x ≥ 16 – 9x + 9x →
–2x – 12 ≥ 16 – 9x 7x – 12 ≥ 16 → 7x – 12 + 12 ≥ 16 + 12 → 7x ≥ 28 → value, t = 8.

7x – 12 ≥ 16 → 7x – 12 + 12 ≥ 16 + 12 → 7x ≥ 28 →
6 → 7x – 12 + 12 ≥ 16 + 12 → 7x ≥ 28 →
7x 28
7 x 28 7 ≥ 7 → x ≥ 4
7 7
≥ → x≥4