AMC10 2002 B

AMC10 2002 B · Q25

AMC10 2002 B · Q25. It mainly tests Systems of equations, Averages (mean).

When 15 is appended to a list of integers, the mean is increased by 2. When 1 is appended to the enlarged list, the mean of the enlarged list is decreased by 1. How many integers were in the original list?

将 15 添加到整数列表后，均值增加 2。将 1 添加到扩大列表后，扩大列表的均值减少 1。原列表中有多少个整数？

(A) 4 4

(B) 5 5

(D) 7 7

(E) 8 8

Answer

Correct choice: (A)

正确答案：（A）

Solution

(A) Let $n$ denote the number of integers in the original list, and $m$ the original mean. Then the sum of the original numbers is $mn$. After $15$ is appended to the list, we have the sum $$(m+2)(n+1)=mn+15,\quad \text{so}\quad m+2n=13.$$ After $1$ is appended to the enlarged list, we have the sum $$(m+1)(n+2)=mn+16,\quad \text{so}\quad 2m+n=14.$$ Solving $m+2n=13$ and $2m+n=14$ gives $m=5$ and $n=4$.

（A）设 $n$ 表示原始列表中整数的个数，$m$ 表示原始平均数。则原始数的总和为 $mn$。将 $15$ 添加到列表后，总和为 $$(m+2)(n+1)=mn+15,\quad \text{因此}\quad m+2n=13.$$ 在扩大的列表中再添加 $1$ 后，总和为 $$(m+1)(n+2)=mn+16,\quad \text{因此}\quad 2m+n=14.$$ 解方程组 $m+2n=13$ 与 $2m+n=14$，得 $m=5$、$n=4$。

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