2000 AMC 8 Problems/Problem 23

Problem

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

$\text{(A)}\ 5\frac{3}{7}\qquad\text{(B)}\ 6\qquad\text{(C)}\ 6\frac{4}{7}\qquad\text{(D)}\ 7\qquad\text{(E)}\ 7\frac{3}{7}$

Solution

Remember that if a list of $n$ numbers has an average of $k$ , then then 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 $6\frac{4}{7}$ , then the total of all seven numbers is $7 \cdot 6\frac{4}{7} = 7\cdot 6 + 4 = 46$ .

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 $\boxed{B}$

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.

2000 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2000 AMC 8 Problems/Problem 23

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