Question

有一组五个正整数，它们的平均数是 $5$ ，中位数是 $5$ ，且唯一的众数是 $8$ 。这组数中最大整数与最小整数的差是多少？

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?

$3$

$5$

$6$

$7$

$8$

答案：D

知识点：平均数中位数（数据）众数

难度评级：1270

解答：

五个数的和是 $5 \cdot 5 = 25$ 。

按从小到大排列，中间的数是 $5$ 。因为 $8$ 是唯一的众数，最后两个数都必须是 $8$ ；但不能有三个 $8$ ，否则中位数也会是 $8$ 。

这三个已知数的和是 $5 + 8 + 8 = 21$ ，所以两个较小的数之和必须是 $25 - 21 = 4$ 。它们必须是不同的正整数，否则另一个数也会与 $8$ 并列为众数。因此它们是 $1$ 和 $3$ 。

因此所求的差为 $8 - 1 = 7.$

所以正确答案是 D。

The sum of all the numbers in the list is $5 \cdot 5 = 25.$

In increasing order, the middle number is $5$ . Because $8$ is the only mode, the last two numbers must both be $8$ ; there cannot be three $8$ s, since then the median would be $8$ .

These three known numbers sum to $5 + 8 + 8 = 21,$ so the two smaller numbers must add to $25 - 21 = 4.$ They must be distinct positive integers; otherwise a second value would tie $8$ as a mode. Thus they are $1$ and $3$ .

The desired difference is then $8 - 1 = 7.$

Thus, D is the correct answer.

1997 AMC 8 第 14 题

答案：D

难度评级：1270

解答：

其他年份的第 14 题