Difference between revisions of "2004 AMC 10A Problems/Problem 13"
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| − | If each man danced with 3 women, then there | + | 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. |
| − | ==See | + | == See also == |
| − | + | {{AMC10 box|year=2004|ab=A|num-b=12|num-a=14}} | |
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 20:53, 5 January 2016
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.
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. 
