1. Percent of Change pages 207–209 Exercises 12. 14.4%; increase
ALGEBRA 1 LESSON 4-4
pages 207–209  Exercises
1. 50%; increase
%; decrease
3. 25%; increase
4. 20%; decrease
%; increase
6. 25%; decrease
7. 25%; increase
8. 20%; increase
%; increase
%; increase
%; decrease
%; increase
13. 39%
14. 60% ft cm g
in.
cm2, cm2
mi2, mi2
in.2, in.2
km2, km2
in.2, in.2
km2; km2
25. 25%
26. 25% % %
29. a. 48 cm3
b cm3
c cm3
d cm3
e. 55%
30. 23%; decrease
31. 22%; decrease 1 3 1 3
4-4

2. 44. Answers may vary. Sample: Joan bought shoes for $10.
Percent of Change
ALGEBRA 1 LESSON 4-4
%; increase
%; increase
34. 4%; increase
35. 3%; decrease
36. 56%; decrease
37. 9%; decrease
38. 17%; increase
39. 2%
40. 19%
41. 1 mm
42. no; 16% increase
but a 14% decrease
43. no; increases to $70.40
but decreases to $63.36
44. Answers may vary. Sample:
Joan bought shoes for $10.
Sarah bought the same shoes
3 days later for $7. What was
the percent change?
30% decrease
cm2, 25.5 cm2
mi2, 59.6 mi2
in.2, 54.3 in.2
48. a. 100%
b. 100%
c. 50%
d. 50%
49. 11%
50. 34%
51. Answers may vary.
Sample: Use the
greatest possible error
to calculate the
maximum, minimum,
and measured areas.
Find the amounts by
which the maximum
and minimum differ
from the measured
area. Divide the greater
difference by the
measured area.
4-4

3. 52. Jorge found the change of $5 but divided by the final price
Percent of Change
ALGEBRA 1 LESSON 4-4
52. Jorge found the change of $5
but divided by the final price
instead of the original price.
53. a. 9%, 3%
b. Answers may vary. Sample:
The larger a measure,
the smaller is the percent error.
54. Yes; > 3 (48.7) = 146.1,
and %.
55. a. 21%
b. 21%
c. 21%; answers may vary.
Sample: 1.1a • 1.1a = 1.21a2,
which is 21% greater than a • a = a2.
Relationship between % increase of
side and area of the square doesn’t
depend on the side length.
56. C
57. I
58. A
59. [2] perimeter of softball diamond:
4(60) = 240, perimeter 240 ft,
side of baseball diamond:
1.5(60) = 90, side 90 ft, perimeter
of baseball diamond: 4(90) = 360,
perimeter 360 ft, % of increase
= = 50%; area of softball
diamond: 60(60) = 3600,
area 3600 ft2, area of baseball
diamond: 90(90) = 8100,
area 8100 ft2, percent of increase
= = 125%
OR computation that gives
same results
148.3 – 48.7
48.7
360 – 240
240
8100 – 3600
3600
4-4

4. [1] appropriate methods, but with one
Percent of Change
ALGEBRA 1 LESSON 4-4
67. q –17
68. x –1
[1] appropriate methods, but with one
computational error OR finds only
one % of increase
60–65. Equations may vary.
= , 7%
= , 87%
= , 179.5
63. = , 300%
= , 1.7
= , 37.2
66. n <
> –
> – x
100 5 67 x
100 13 15 44
100 79 x x
100 96 32
0.2
100 x
834
266
100 x 14 1 15
4-4