2004 AMC 8 Problems/Problem 9

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 $\boxed{\textbf{(D)}\ 58}$ .

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2004 AMC 8 Problems/Problem 9

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