Question

一组五个互不相同的正整数的平均数是 $15$ ，中位数是 $18$ 。这五个整数中最大数的最大可能值是

The mean of a set of five different positive integers is $15.$ The median is $18.$ The maximum possible value of the largest of these five integers is

$19$

$24$

$32$

$35$

$40$

答案：D

知识点：平均数中位数（数据）最优化

难度评级：1550

解答：

中位数是第三大的数，即 $18$ 。有两个数小于 $18$ ，两个数大于它。

平均数为 $15$ 所以五个数的和为 $5 \cdot 15 = 75$ 要在这个总和下使最大数尽量大，其余数必须尽量小。

小于 $18$ 的两个数必须是不同的正整数，所以最小为 $1$ 和 $2$ 。

紧接在 $18$ 后的数也应尽量小，所以是 $19$ 。

因此剩下的数，也就是最大可能值，为 $75 - 1 - 2 - 18 - 19 = 35.$

所以正确答案是 D。

The median of the set of numbers is the third largest number, which is $18.$ There are two numbers less than $18$ and two numbers greater than it.

The mean of the set is $15,$ so the sum of all the numbers is $5 \cdot 15 = 75.$ In order to maximize the largest number with this sum, the other numbers must be as small as possible.

The two numbers less than $18$ must be positive and distinct, so they must be $1$ and $2.$

The number immediately after $18$ must also be as small as possible, so it must be $19.$

Therefore, the remaining number, the maximum possible value in the set, is $75 - 1 - 2 - 18 - 19 = 35.$

Thus, D is the correct answer.

2001 AMC 8 第 21 题

答案：D

难度评级：1550

解答：

其他年份的第 21 题