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> | ||
| − | ==Solution== | + | ==Solution 1== |
Assume arbitrarily (and WLOG) 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 (and WLOG) 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>. | ||
| + | |||
| + | ==Solution 2 (Algebraic)== | ||
| + | Let single women be <math>S</math>, married wives to be <math>W</math>, and men to be <math>M</math>. As such, the number of men is equal to married woman since each husband has his own wife or <math>W=M</math>. We are asked to find the probability of men in the total group or <math>\frac{M}{S+W+M}</math>. | ||
| + | |||
| + | We are also given that out of all the women, 2/5 are single or <math>\frac{S}{S+W}=\frac{2}{5}</math>. Rewrite this as <math>S=\frac{2}{5}(S+W)</math>, <math>S=\frac{2S}{5}+\frac{2W}{5}</math>, isolating <math>S</math> gives us <math>\frac{3S}{5}=\frac{2W}{5}</math> or <math>S=\frac{2M}{3}</math> (because <math>W=M</math>). | ||
| + | |||
| + | Now, substituting gives us <math>\frac{M}{\frac{2M}{3}+M+M}=\frac{M}{\frac{2M}{3}+\frac{6M}{3}}=\frac{M}{\frac{8M}{3}}=\boxed{\textbf{(B)}\ \frac38}</math> | ||
| + | ~RandomMathGuy500 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2004|num-b=21|num-a=23}} | {{AMC8 box|year=2004|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 01:37, 26 December 2024
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?
Solution 1
Assume arbitrarily (and WLOG) 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 
.
Solution 2 (Algebraic)
Let single women be 
, married wives to be 
, and men to be 
. As such, the number of men is equal to married woman since each husband has his own wife or 
. We are asked to find the probability of men in the total group or 
.
We are also given that out of all the women, 2/5 are single or 
. Rewrite this as 
, 
, isolating gives us 
or 
(because ).
Now, substituting gives us 
~RandomMathGuy500
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 | ||
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