AMC 8 2000 Problem 23

AMC8 2000 年第 23

方程与不等式★★★★
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 $64/7$, then the number common to both sets of four numbers is
有一组 7 个数。前面 4 个数的平均值是 5,最后 4 个数的平均值是 8。若所有这 7 个数的平均值是 ,那么前面 4 个数和最后 4 个数共同拥有的那个数是
  1. A.537\displaystyle 5\frac{3}{7}
  2. B.6
  3. C.647\displaystyle 6\frac{4}{7}
  4. D.7
  5. E.737\displaystyle 7\frac{3}{7}

正确答案 · Correct Answer

B

解析 · Solution

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