2022 AMC 10A 第 8 题

Question

一个数据集由

6

个不一定互异的正整数组成：

1

、

7

、

5

、

2

、

5

、

X

。这

6

个数的平均数等于数据集中的某个数。所有可能的

X

值之和是多少？

A data set consists of

6

(not distinct) positive integers:

1,

7,

5,

2,

5,

and

X.

The average (arithmetic mean) of the

6

numbers equals a value in the data set. What is the sum of all possible values of

X?

10

26

32

36

40

平均数必须等于所列数据中的某一个，或者等于 X

The average must equal one of the listed data values or X

分别令平均数等于每个可能的取值，只保留 X 为正的解

Set the average equal to each possible value and keep only positive X values

这

6

个数的平均数为

\dfrac{1 + 7 + \cdots + X}{6} = \dfrac{20 + X}{6}\text{。}

这个值可以等于数据集中的任一项，因此分别讨论。

\dfrac{20 + X}{6} = 1 \iff X = -14

\dfrac{20 + X}{6} = 7 \iff X = 22

\dfrac{20 + X}{6} = 5 \iff X = 10

\dfrac{20 + X}{6} = 2 \iff X = -8

\dfrac{20 + X}{6} = X \iff X = 4

所有正的

X

值之和为

36

。

所以正确答案是 D。

The average of the

6

numbers is

\dfrac{1 + 7 + \cdots + X}{6} = \dfrac{20 + X}{6}.

This value can equal any of the terms in the set, so we can case on what it equals.

\dfrac{20 + X}{6} = 1 \iff X = -14

\dfrac{20 + X}{6} = 7 \iff X = 22

\dfrac{20 + X}{6} = 5 \iff X = 10

\dfrac{20 + X}{6} = 2 \iff X = -8

\dfrac{20 + X}{6} = X \iff X = 4

Adding up all the positive values for

X,

we get

36.

Thus, D is the correct answer.

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