AMC8 1997

AMC8 1997 · Q14

AMC8 1997 · Q14. It mainly tests Averages (mean), Logic puzzles.

There is a set of five positive integers whose average (mean) is 5, whose median is 5, and whose only mode is 8. What is the difference between the largest and smallest integers in the set?

有一个五个正整数的集合，其平均数（均值）为 5，中位数为 5，且唯一众数为 8。求该集合中最大整数与最小整数之差。

(A) 3 3

(B) 5 5

(D) 7 7

(E) 8 8

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Since 5 is the median, there must be two integers greater than 5 and two less than 5. The five integers may be arranged this way: $\_\ \_\ 5\ 8\ 8$. Since the mean is 5, the sum of all five integers is 25. The numbers 5, 8, and 8 total 21, leaving 4 for the sum of the first two. They can't both be 2 since 8 is the only mode, so they are 1 and 3. The difference between 8 and 1 is 7.

答案（D）：由于 5 是中位数，必须有两个整数大于 5，两个小于 5。这五个整数可以按如下方式排列：$\_\ \_\ 5\ 8\ 8$。由于平均数是 5，这五个整数的总和是 25。数 5、8 和 8 的和为 21，因此前两个数的和为 4。它们不可能都是 2，因为 8 是唯一的众数，所以它们是 1 和 3。8 和 1 的差是 7。

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