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?

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

Solution

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

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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