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

2004 AMC 10A (Problems • Answer Key • Resources)
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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2004 AMC 10A Problems/Problem 13

Contents

Problem

Solution

Solution 2

See also