2004 AMC 8 Problems/Problem 9

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Problem

The average of the five numbers in a list is $54$. The average of the first two numbers is $48$. What is the average of the last three numbers?

$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 57 \qquad \textbf{(D)}\ 58 \qquad \textbf{(E)}\ 59$

Solution

Let the $5$ numbers be $a, b, c, d$, and $e$. Thus $\frac{a+b+c+d+e}{5}=54$ and $a+b+c+d+e=270$. Since $\frac{a+b}{2}=48$, $a+b=96$. Substituting back into our original equation, we have $96+c+d+e=270$ and $c+d+e=174$. Dividing by $3$ gives the average of $\boxed{\textbf{(D)}\ 58}$.

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AJHSME/AMC 8 Problems and Solutions

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