Difference between revisions of "2007 UNCO Math Contest II Problems/Problem 2"

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Let the number of men be equal to <math>m</math> and the number of women be equal to <math>w</math>, where <math>m</math> and <math>w</math> are positive whole numbers. Assuming a marriage consists of one man and one woman, we see that the number of married men is equal to the number of married women in the equation:
 
Let the number of men be equal to <math>m</math> and the number of women be equal to <math>w</math>, where <math>m</math> and <math>w</math> are positive whole numbers. Assuming a marriage consists of one man and one woman, we see that the number of married men is equal to the number of married women in the equation:
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<math>\frac{4}{5}\times m = \frac{3}{7}\times w</math>
 
<math>\frac{4}{5}\times m = \frac{3}{7}\times w</math>

Revision as of 08:05, 28 January 2018

Problem

In Grants Pass, Oregon $\frac{4}{5}$ of the men are married to $\frac{3}{7}$ of the women. What fraction of the adult population is married? Give a possible generalization.

Solution

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Let the number of men be equal to $m$ and the number of women be equal to $w$, where $m$ and $w$ are positive whole numbers. Assuming a marriage consists of one man and one woman, we see that the number of married men is equal to the number of married women in the equation:


$\frac{4}{5}\times m = \frac{3}{7}\times w$


$w = \frac{28}{15} m$

Dividing the the number of married persons by the entire adult population gives us:


$\%_{married} = \frac{\frac{3}{7}w + \frac{4}{5}m}{m + w}$


$\%_{married} = \frac{\frac{8}{5}m}{\frac{43}{15}m}$


$\%_{married} = \frac{24}{43} \approx 55.8\%$

See Also

2007 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions
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