2007 UNCO Math Contest II Problems/Problem 2

Revision as of 20:23, 28 January 2018 by Mitko pitko (talk | contribs) (Solution)

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:


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


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


$Fraction_{married} = \frac{24}{43}$


Using the above statements, we can derive a generalized formula. If $k$ is the fraction of married men and $p$ is the fraction of married women, then:


$Fraction_{married} = \frac{2 \times k m}{m + m \frac{k}{p}}$

$Fraction_{married} = \frac{2 p k}{p + k}$

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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