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AMC10 2001 A

AMC10 2001 A · Q16

AMC10 2001 A · Q16. It mainly tests Linear equations, Averages (mean).

The mean of three numbers is 10 more than the least of the numbers and less than the greatest. The median of the three numbers is 5. What is their sum?
三个数的平均数比最小的数大10,并且小于最大的数。这三个数的中间数是5。它们的和是多少?
(A) 5 5
(B) 20 20
(C) 25 25
(D) 30 30
(E) 36 36
Answer
Correct choice: (D)
正确答案:(D)
Solution
(D) Since the median is 5, we can write the three numbers as $x$, 5, and $y$, where $\frac{1}{3}(x+5+y)=x+10$ and $\frac{1}{3}(x+5+y)+15=y$. If we add these equations, we get $\frac{2}{3}(x+5+y)+15=x+y+10$ and solving for $x+y$ gives $x+y=25$. Hence the sum of the numbers $x+y+5=30$.
(D)由于中位数是 5,我们可以把这三个数写成 $x$、5 和 $y$,其中 $\frac{1}{3}(x+5+y)=x+10$ 且 $\frac{1}{3}(x+5+y)+15=y$。 把这两个方程相加,得到 $\frac{2}{3}(x+5+y)+15=x+y+10$ 解出 $x+y$ 得 $x+y=25$。因此这三个数的和为 $x+y+5=30$。
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