Difference between revisions of "2004 AMC 10A Problems/Problem 13"

 
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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?
 
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?
  
<math> \mathrm{(A) \ } 8 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 {2}\qquad \mathrm{(E) \ } 24  </math>
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<math> \mathrm{(A) \ } 8 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ } 24  </math>
  
 
==Solution==
 
==Solution==
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*[[2004 AMC 10A Problems/Problem 14|Next Problem]]
 
*[[2004 AMC 10A Problems/Problem 14|Next Problem]]
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[[Category:Introductory Algebra Problems]]

Revision as of 12:11, 5 November 2006

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 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