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== | ||
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+ | Consider drawing out a diagram. Let a circle represent a | ||
== See also == | == See also == |
Revision as of 12:08, 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
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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