AMC12 2001 A

AMC12 2001 A · Q4

AMC12 2001 A · Q4. It mainly tests Averages (mean).

The mean of three numbers is $10$ more than the least of the numbers and $15$ less than the greatest. The median of the three numbers is $5$. What is their sum?

三个数的平均数比其中最小的数大 $10$，且比其中最大的数小 $15$。这三个数的中位数是 $5$。它们的和是多少？

(A) 5 5

(B) 20 20

(D) 30 30

(E) 36 36

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $m$ be the mean of the three numbers. Then the least of the numbers is $m-10$ and the greatest is $m + 15$. The middle of the three numbers is the median, $5$. So $\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$, which can be solved to get $m=10$. Hence, the sum of the three numbers is $3\cdot 10 = \boxed{\textbf{(D) }30}$.

设 $m$ 为这三个数的平均数，则最小的数为 $m-10$，最大的数为 $m + 15$。三个数中间的数为中位数 $5$。因此 $\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$，解得 $m=10$。所以三数之和为 $3\cdot 10 = \boxed{\textbf{(D) }30}$。

Topics

AR.006 Averages (mean)