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

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==Solution==
 
==Solution==
If each man danced with 3 women, then there were a total of <math>3\times12=36</math> pairs of a man and a woman.  However, each woman only danced with 2 men, so there must have been <math>\frac{36}2=18</math> women <math>\Rightarrow\mathrm{(D)}</math>.
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If each man danced with <math>3</math> women, then there will be a total of <math>3\times12=36</math> pairs of men and women.  However, each woman only danced with <math>2</math> men, so there must have been <math>\frac{36}2 \Longrightarrow \boxed{\mathrm{(D)}\ 18}</math> women.
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==Solution 2==
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Consider drawing out a diagram. Let a circle represent a man, and let a shaded circle represent a woman.
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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.
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Hence, we know that:
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<math>\frac{2}{3} = \frac{12}{x} \implies x = 18 \implies \boxed{D}.</math>
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~yk2007
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 11:10, 17 August 2021

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

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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