Question

有一个包含七个数的列表。前四个数的平均数是 $5$ ，后四个数的平均数是 $8$ 。如果全部七个数的平均数是 $6\frac{4}{7}$ ，那么两组四个数中共同的那个数是

There is a list of seven numbers. The average of the first four numbers is $5,$ and the average of the last four numbers is $8.$ If the average of all seven numbers is $6\frac{4}{7},$ then the number common to both sets of four numbers is

$5\frac{3}{7}$

$6$

$6\frac{4}{7}$

$7$

$7\frac{3}{7}$

答案：B

知识点：平均数双重计数

难度评级：1610

解答：

前四个数的和是 $4 \cdot 5 = 20$ 。后四个数的和是 $4 \cdot 8 = 32$ 。

全部七个数的和是 $7 \cdot 6\frac{4}{7} = 46$ 。共同的数包含在前两个和中。

也就是说，前两个和相加会把每个数计算一次，但共同的数计算两次。

而全部七个数的和只把每个数计算一次。因此用前两个和的总和减去七个数的总和，就得到共同的数。

所以共同的数为 $20 + 32 - 46 = 52 - 46 = 6.$

所以正确答案是 B。

The sum of the first four numbers is $4 \cdot 5 = 20.$ The sum of the last four numbers is $4 \cdot 8 = 32.$

The sum of all seven numbers is $7 \cdot 6\frac{4}{7} = 46.$ We know that the number common to both sets is included in both of first two sums.

This means that the sum of the first two sums includes every number once, except for the common number which is included twice.

The third sum, however, only includes every number once. This means that the sum of the first two sums minus the third sum yields our desired number.

Therefore, the common number is $20 + 32 - 46 = 52 - 46 = 6.$

Thus, B is the correct answer.

2000 AMC 8 第 23 题

答案：B

难度评级：1610

解答：

其他年份的第 23 题