AMC8 2015

AMC8 2015 · Q13

AMC8 2015 · Q13. It mainly tests Averages (mean), Combinations.

How many subsets of two elements can be removed from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ so that the mean (average) of the nine remaining numbers is 6?

从集合 $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ 中移除多少个两元素子集，使得剩余九个数的平均数（均值）为6？

(A) 1 1

(B) 2 2

(D) 5 5

(E) 6 6

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): If the average of the remaining $9$ numbers is $6$, then their sum is $54$. Because the sum of the numbers in the original set is $66$, the sum of the two numbers removed must be $12$. There are five such subsets: $\{1,11\}$, $\{2,10\}$, $\{3,9\}$, $\{4,8\}$, and $\{5,7\}$.

答案（D）：如果剩下的 $9$ 个数的平均数是 $6$，那么它们的和是 $54$。由于原集合中所有数的和是 $66$，因此被移除的两个数之和必须是 $12$。满足条件的子集有五个：$\{1,11\}$, $\{2,10\}$, $\{3,9\}$, $\{4,8\}$, and $\{5,7\}$。

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