2004 AMC 8 Problems/Problem 22

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Problem

At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is $\frac25$. What fraction of the people in the room are married men?

$\textbf{(A)}\ \frac13\qquad \textbf{(B)}\ \frac38\qquad \textbf{(C)}\ \frac25\qquad \textbf{(D)}\ \frac{5}{12}\qquad \textbf{(E)}\ \frac35$

Solution

Assume arbitrarily (and WLOG) there are $5$ women in the room, of which $5 \cdot \frac25 =  2$ are single and $5-2=3$ are married. Each married woman came with her husband, so there are $3$ married men in the room as well for a total of $5+3=8$ people. The fraction of the people that are married men is $\boxed{\textbf{(B)}\ \frac38}$.

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AJHSME/AMC 8 Problems and Solutions

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