AMC 8 2000 Problem 23

AMC8 2000 年第 23 题

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

\displaystyle 6\frac{4}{7}

, then the number common to both sets of four numbers is

有一组 7 个数。前 4 个数的平均值是 5，后 4 个数的平均值是 8。若这 7 个数的平均值是

\displaystyle 6\frac{4}{7}

，那么同时属于前 4 个数和后 4 个数的那个数是多少？

A. $\displaystyle 5\frac{3}{7}$
B.6
C. $\displaystyle 6\frac{4}{7}$
D.7
E. $\displaystyle 7\frac{3}{7}$

正确答案 · Correct Answer

解析 · Solution

Solution 1

Remember that if a list of

n

numbers has an average of

k

, then the sum

S

of all the numbers on the list is S = nk. So if the average of the first 4 numbers is 5, then the first four numbers total

4 \cdot 5 = 20

. If the average of the last 4 numbers is 8, then the last four numbers total

4 \cdot 8 = 32

. If the average of all 7 numbers is

\displaystyle 6\frac{4}{7}

, then the total of all seven numbers is

\displaystyle 7 \cdot 6\frac{4}{7} = 7\cdot 6 + 4 = 46

. If the first four numbers are 20, and the last four numbers are 32, then all "eight" numbers are 20 + 32 = 52. But that's counting one number twice. Since the sum of all seven numbers is 46, then the number that was counted twice is 52 - 46 = 6, and the answer is 6 Algebraically, if a + b + c + d = 20, and d + e + f + g = 32, you can add both equations to get a + b + c + 2d + e + f + g = 52. You know that a + b + c + d + e + f + g = 46, so you can subtract that from the last equation to get d = 6, and

d

is the number that appeared twice. Yay!:D

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