AMC12 2018 A

AMC12 2018 A · Q6

AMC12 2018 A · Q6. It mainly tests Averages (mean), Area & perimeter.

For positive integers $m$ and $n$ such that $m + 10 < n + 1$, both the mean and the median of the set $\{m, m + 4, m + 10, n + 1, n + 2, 2n\}$ are equal to $n$. What is $m + n$?

对于正整数 $m$ 和 $n$，满足 $m + 10 < n + 1$，集合 $\{m, m + 4, m + 10, n + 1, n + 2, 2n\}$ 的平均数和中位数均为 $n$。求 $m + n$ 的值。

(A) 20 20

(B) 21 21

(D) 23 23

(E) 24 24

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Note that the given conditions imply that the 6 values are listed in increasing order. Because the median of the these 6 values is $n$, the mean of the middle two values must be $n$, so $$\frac{(m+10)+(n+1)}{2}=n,$$ which implies $m=n-11$. Because the mean of the set is also $n$, $$\frac{(n-11)+(n-7)+(n-1)+(n+1)+(n+2)+2n}{6}=n,$$ so $7n-16=6n$ and $n=16$. Then $m=16-11=5$, and the requested sum is $5+16=21$.

答案（B）：注意给定条件意味着这6个数按递增顺序排列。由于这6个数的中位数是$n$，中间两个数的平均数必须是$n$，因此 $$\frac{(m+10)+(n+1)}{2}=n,$$ 由此得到$m=n-11$。因为这组数的平均数也为$n$， $$\frac{(n-11)+(n-7)+(n-1)+(n+1)+(n+2)+2n}{6}=n,$$ 所以$7n-16=6n$，从而$n=16$。于是$m=16-11=5$，所求的和为$5+16=21$。

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