Difference between revisions of "2004 AMC 10A Problems/Problem 13"
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| − | If each man danced with <math>3</math> 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 <math>2</math> men, so there must have been <math>\frac{36}2= | + | If each man danced with <math>3</math> 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 <math>2</math> men, so there must have been <math>\frac{36}2=\boxed{\mathrm{(D)}\ 18}</math> women. |
== See also == | == See also == | ||
Revision as of 15:11, 30 June 2015
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 were a total of 
pairs of a man and a woman. However, each woman only danced with 
men, so there must have been 
women.
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 | ||
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