AMC8 2001

AMC8 2001 · Q21

AMC8 2001 · Q21. It mainly tests Averages (mean), Arithmetic misc.

The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is

一个由五个不同正整数组成的集合的平均数是15。中位数是18。这五个整数中最大的数可能的最大值是

(A) 19 19

(B) 24 24

(D) 35 35

(E) 40 40

Answer

Correct choice: (D)

正确答案：（D）

Solution

The sum of all five numbers is 5(15)=75. Let the numbers be W, X, 18, Y and Z in increasing order. For Z to be as large as possible, make W, X and Y as small as possible. The smallest possible values are W = 1, X = 2 and Y = 19. Then the sum of W, X, 18 and Y is 40, and the difference, 75 −40 = 35, is the largest possible value of Z.

这五个数的和是$5\times15=75$。设这些数按升序为$W,X,18,Y,Z$。要使$Z$尽可能大，要使$W,X,Y$尽可能小。可能的最小值为$W=1,X=2,Y=19$。则$W+X+18+Y=40$，差值为$75-40=35$，即$Z$的最大可能值。

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