AMC8 1991

AMC8 1991 · Q19

AMC8 1991 · Q19. It mainly tests Averages (mean), Optimization (basic).

The average (arithmetic mean) of 10 different positive whole numbers is 10. The largest possible value of any of these numbers is

10个不同的正整数的平均数（算术平均）是10。这些数中最大的可能值是

(A) 10 10

(B) 50 50

(D) 90 90

(E) 91 91

Answer

Correct choice: (C)

正确答案：（C）

Solution

Since the mean is 10, it follows that the sum of the numbers is $10 \times 10 = 100$. Taking the smallest possible value for the 9 smaller numbers would give the largest possible value of the tenth number. Thus, the largest possible number is $100 - (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 100 - 45 = 55$.

由于平均数是10，因此这些数的总和是$10 \times 10 = 100$。让9个较小的数取尽可能小的值，从而使第十个数最大。因此，最大的可能数是$100 - (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 100 - 45 = 55$。

Topics