2008 AMC 8 Problems/Problem 10

Problem

The average age of the $6$ people in Room A is $40$ . The average age of the $4$ people in Room B is $25$ . If the two groups are combined, what is the average age of all the people?

$\textbf{(A)}\ 32.5 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 33.5 \qquad \textbf{(D)}\ 34\qquad \textbf{(E)}\ 35$

Solution

The total of all their ages over the number of people is

$\frac{6 \cdot 40 + 4\cdot 25}{6+4} = \frac{340}{10} = \boxed{\textbf{(D)}\ 34}$

2008 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AJHSME/AMC 8 Problems and Solutions

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2008 AMC 8 Problems/Problem 10

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