Difference between revisions of "2004 AMC 8 Problems/Problem 22"
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<math>\textbf{(A)}\ \frac13\qquad \textbf{(B)}\ \frac38\qquad \textbf{(C)}\ \frac25\qquad \textbf{(D)}\ \frac{5}{12}\qquad \textbf{(E)}\ \frac35</math> | <math>\textbf{(A)}\ \frac13\qquad \textbf{(B)}\ \frac38\qquad \textbf{(C)}\ \frac25\qquad \textbf{(D)}\ \frac{5}{12}\qquad \textbf{(E)}\ \frac35</math> | ||
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==Solution== | ==Solution== | ||
Assume arbitrarily there are <math>5</math> women in the room, of which <math>5 \cdot \frac25 = 2</math> are single and <math>5-2=3</math> are married. Each married woman came with her husband, so there are <math>3</math> married men in the room as well for a total of <math>5+3=8</math> people. The fraction of the people that are married men is <math>\boxed{\textbf{(B)}\ \frac38}</math>. | Assume arbitrarily there are <math>5</math> women in the room, of which <math>5 \cdot \frac25 = 2</math> are single and <math>5-2=3</math> are married. Each married woman came with her husband, so there are <math>3</math> married men in the room as well for a total of <math>5+3=8</math> people. The fraction of the people that are married men is <math>\boxed{\textbf{(B)}\ \frac38}</math>. | ||
Revision as of 00:00, 5 July 2013
Problem
At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is 
. What fraction of the people in the room are married men.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
Solution
Assume arbitrarily there are 
women in the room, of which 
are single and 
are married. Each married woman came with her husband, so there are 
married men in the room as well for a total of 
people. The fraction of the people that are married men is 
.
See Also
| 2004 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
