McGraw-Hill Education Conquering GRE Math by Robert E. Moyer

Author:Robert E. Moyer
Language: eng
Format: epub
Publisher: McGraw-Hill Education
Published: 2020-06-15T00:00:00+00:00

9.38 SOLUTIONS

1. A. SUM = 144 N = 10

AVE = 144 ÷ 10 = 14.4

B. SUM = 135 N = 7

AVE = 135 ÷ 7 = 19.2857 = 19.3

C. SUM = 306 N = 8

AVE = 306 ÷ 8 = 38.25 = 38.3

D. SUM = 213 N = 9

AVE = 213 ÷ 9 = 23.666 = 23.7

E. SUM = 460 N = 5

AVE = 460 ÷ 5 = 92

2. A. Class A: 40, AVE = 78 Class B: 25, AVE = 72

SUM = 40(78) + 25(72) = 3,120 + 1,800 = 4,920

N = 40 + 25 = 65

AVE = 4,920 ÷ 65 = 75.6923 = 75.7

B. January: 32, AVE = $97.40 February: 48, AVE = $87.20

SUM = 32($97.40) + 48($87.20) = $3,116.80 + $4,185.60 = $7,302.40

N = 32 + 48 = 80

AVE = $7,302.40 ÷ 80 = $91.28

C. Last month: 5, AVE = $52.70 This month: 4, AVE = $67.50

SUM = 5($52.70) + 4($67.50) = $263.50 + $270.00 = $533.50

N = 5 + 4 = 9

AVE = $533.50 ÷ 9 = $59.27777 = $59.28

3. A. Ordered values: 7, 10, 18, 26, 26, 40, 43, 76 The middle two values of the 8 values are 26 and 26.

The median is 26.

B. Ordered values: 9, 24, 36, 72, 85, 500

The middle two values are 36 and 72. (36 + 72) ÷ 2 = 54

The median is 54.

C. Ordered values: 13, 14, 24, 29, 37, 65, 71

The middle value of the 7 values is 29.

The median is 29.

D. Ordered values: 3, 5, 10, 17, 46, 48, 81, 97, 140

The middle value of the 9 values is 46.

The median is 46.

4. A. AVE = 300 N = 4

SUM = 300(4) = 1, 200. Given values: 306 + 284 + 292 = 882

missing value = 1, 200 − 882 = 318

John needs to travel 318 miles on day 4.

B. AVE = 240 N = 5

SUM = 240(5) = 1, 200. Given values: 260 + 210 + 250 + 230 = 950

missing value = 1, 200 − 950 = 250

Larry needs to bowl a 250 on his next game.

C. AVE = 90 N = 7

SUM = 90(7) = 630. Given values: 90 + 78 + 96 + 94 + 88 + 100 = 546

missing value = 630 − 546 = 84

She needs to score an 84 on the last test.

5. A. AVE = (1 + 9) ÷ 2 = 5

B. AVE = (501 + 506) ÷ 2 = 503.5

C. AVE = (298 + 306) ÷ 2 = 302

D. AVE = (37 + 46) ÷ 2 = 41.5

E. AVE = (632 + 638) ÷ 2 = 635

6. A. Because 6 occurs twice and 9 occurs twice, the modes are 6 and 9.

B. Because 4 occurs three times, the mode is 4.

C. Because 11 occurs twice, the mode is 11.

D. Because every value occurs the same number of times, there is no mode.

7. A. MAX = 17 and MIN = 3, so R = MAX − MIN = 17 − 3 = 14.

B. MAX = 91 and MIN = 12, so R = MAX − MIN = 91 − 12 = 79.

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