AMC8 2000

AMC8 2000 · Q23

AMC8 2000 · Q23. It mainly tests Averages (mean).

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

有一列七个数。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$6\frac{4}{7}$，则两个四数集合共有的那个数是

(A) $\frac{5}{7}$ $\frac{5}{7}$

(B) 6 6

(D) 7 7

(E) $\frac{7}{7}$ $\frac{7}{7}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Since the average of all seven numbers is $6\frac{4}{7}=\frac{46}{7}$, the sum of the seven numbers is $7\times\frac{46}{7}=46$. The sum of the first four numbers is $4\times5=20$ and the sum of the last four numbers is $4\times8=32$. Since the fourth number is used in each of these two sums, the fourth number must be $(20+32)-46=6$.

答案（B）：由于这七个数的平均数是 $6\frac{4}{7}=\frac{46}{7}$，因此这七个数的和为 $7\times\frac{46}{7}=46$。前四个数的和为 $4\times5=20$，后四个数的和为 $4\times8=32$。由于第四个数在这两个和中都被计算了一次，所以第四个数应为 $(20+32)-46=6$。

Topics

AR.006 Averages (mean)