2004 AMC 10A Problems/Problem 13

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Problem

At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?

$\mathrm{(A) \ } 8 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 {2}\qquad \mathrm{(E) \ } 24$

Solution

If each man danced with 3 women, then there were a total of $3\times12=36$ pairs of a man and a women. However, each women only danced with 2 men, so there must have been $\frac{36}2=18$ women $\Rightarrow\mathrm{(D)}$.

See Also

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