Difference between revisions of "2004 AMC 10A Problems/Problem 13"
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==Solution== | ==Solution== | ||
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. | 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== | ||
| + | |||
| + | 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. | ||
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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 == | ||
Latest revision as of 11:10, 17 August 2021
Contents
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?
Solution
If each man danced with 
women, then there will be a total of 
pairs of men and women. However, each woman only danced with 
men, so there must have been 
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:
~yk2007
See also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
