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Difference between revisions of "2000 AMC 8 Problems/Problem 23"

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<math> \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} </math>
 
<math> \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} </math>
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==Solution==
 
==Solution==

Revision as of 17:42, 21 December 2020

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 (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
All AJHSME/AMC 8 Problems and Solutions

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