Question

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

$5\frac{3}{7}$

$6$

$6\frac{4}{7}$

$7$

$7\frac{3}{7}$

Solution:

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 Problem 23

Solution:

Problem 23 in Other Years