# 2004 AMC 10A Problems/Problem 13

## 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.

## Solution 2

Consider drawing out a diagram. Let a circle represent a man, and let a shaded circle represent a woman.

Then, we know that for every 2 men, there will be 3 woman using our diagram. Therefore, the ratio between the number of men and women is 2:3.

Hence, we know that: $\frac{2}{3} = \frac{12}{x} \implies x = 18 \implies \boxed{D}.$

~yk2007

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