AMC12 2007 A

AMC12 2007 A · Q15

AMC12 2007 A · Q15. It mainly tests Averages (mean), Arithmetic misc.

The set $\{3,6,9,10\}$ is augmented by a fifth element $n$, not equal to any of the other four. The median of the resulting set is equal to its mean. What is the sum of all possible values of $n$?

集合 $\{3,6,9,10\}$ 增加第五个元素 $n$，且 $n$ 不等于另外四个数。所得集合的中位数等于其平均数。求所有可能的 $n$ 的值之和。

(A) 7 7

(B) 9 9

(D) 24 24

(E) 26 26

Answer

Correct choice: (E)

正确答案：（E）

Solution

The median must either be $6, 9,$ or $n$. Casework: - Median is $6$: Then $n \le 6$ and $\frac{3+6+9+10+n}{5} = 6 \Longrightarrow n = 2$. - Median is $9$: Then $n \ge 9$ and $\frac{3+6+9+10+n}{5} = 9 \Longrightarrow n = 17$. - Median is $n$: Then $6 < n < 9$ and $\frac{3+6+9+10+n}{5} = n \Longrightarrow n = 7$. All three cases are valid, so our solution is $2 + 7 + 17 = 26 \Longrightarrow \mathrm{(E)}$.

中位数只能是 $6, 9,$ 或 $n$。分类讨论： - 中位数为 $6$：则 $n \le 6$ 且 $\frac{3+6+9+10+n}{5} = 6 \Longrightarrow n = 2$。 - 中位数为 $9$：则 $n \ge 9$ 且 $\frac{3+6+9+10+n}{5} = 9 \Longrightarrow n = 17$。 - 中位数为 $n$：则 $6 < n < 9$ 且 $\frac{3+6+9+10+n}{5} = n \Longrightarrow n = 7$。三种情况都成立，因此答案为 $2 + 7 + 17 = 26 \Longrightarrow \mathrm{(E)}$。

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