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 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 $6\frac{4}{7}$, 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 $\boxed{(B) 6}$

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 $d = \boxed{(B) 6}$, and $d$ is the number that appeared twice.

Yay!  :D

See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
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
All AJHSME/AMC 8 Problems and Solutions

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