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02 Sep 2025, 07:20
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There are at least three times as many boys as girls in a class. On a test, the boys’ average score is 78, and the girls’ average score is 94. Which of the following could be the average test score for the entire class?
Indicate all such values.
A. 88
B. 86
C. 84
D. 82
E. 80
Indicate all such values.
A. 88
B. 86
C. 84
D. 82
E. 80
21 Sep 2025, 04:12
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Official Explanation
If there were an equal number of boys and girls, then the class average would equal the average of 78 and 94. But since there are more boys in the class, the average is weighted toward their number, 78, and will be less than 86. So choices (A) and (B) may be eliminated. If there are at least 3 times as many boys, then the class average will be at least 3 times as close numerically to their average as to the girls’ average. To find this value, which will be the maximum average for the class, we find the number of the way from the boys’ average to the girls’ in order to obtain a 1:3 ratio of lengths. First, find the difference in the two averages, 94 - 78 = 16, and then divide this by 4 to obtain 16 ÷ 4 = 4. This number, 4, is added to 78, to obtain 82, the value 4 units from the boys’ average and 94 - 82 = 12 units from the girls’ average, which creates a 4:12 or 1:3 ratio of distances on the number line. 92
This means that “at least three times as many boys” implies that the class average is at most 82, and (D) and (E) are correct. (This is sometimes thought of as the leverage or “see-saw” principle. On a see-saw, a 100-pound boy must sit 3 feet from the fulcrum to balance a 150-pound boy sitting 2 feet from the fulcrum, because 100 × 3 = 150 × 2.
Answer: D,E
If there were an equal number of boys and girls, then the class average would equal the average of 78 and 94. But since there are more boys in the class, the average is weighted toward their number, 78, and will be less than 86. So choices (A) and (B) may be eliminated. If there are at least 3 times as many boys, then the class average will be at least 3 times as close numerically to their average as to the girls’ average. To find this value, which will be the maximum average for the class, we find the number of the way from the boys’ average to the girls’ in order to obtain a 1:3 ratio of lengths. First, find the difference in the two averages, 94 - 78 = 16, and then divide this by 4 to obtain 16 ÷ 4 = 4. This number, 4, is added to 78, to obtain 82, the value 4 units from the boys’ average and 94 - 82 = 12 units from the girls’ average, which creates a 4:12 or 1:3 ratio of distances on the number line. 92
This means that “at least three times as many boys” implies that the class average is at most 82, and (D) and (E) are correct. (This is sometimes thought of as the leverage or “see-saw” principle. On a see-saw, a 100-pound boy must sit 3 feet from the fulcrum to balance a 150-pound boy sitting 2 feet from the fulcrum, because 100 × 3 = 150 × 2.
Answer: D,E
