1979 AHSME Problems/Problem 15

Problem 15

Two identical jars are filled with alcohol solutions, the ratio of the volume of alcohol to the volume of water being $p : 1$ in one jar and $q : 1$ in the other jar. If the entire contents of the two jars are mixed together, the ratio of the volume of alcohol to the volume of water in the mixture is

$\textbf{(A) }\frac{p+q}{2}\qquad \textbf{(B) }\frac{p^2+q^2}{p+q}\qquad \textbf{(C) }\frac{2pq}{p+q}\qquad \textbf{(D) }\frac{2(p^2+pq+q^2)}{3(p+q)}\qquad \textbf{(E) }\frac{p+q+2pq}{p+q+2}$

Solution

Solution by e_power_pi_times_i

The amount of alcohol in the jars are $\frac{p}{p+1}$ and $\frac{q}{q+1}$ , and the amount of water in the jars are $\frac{1}{p+1}$ and $\frac{1}{q+1}$ . Then the total amount of alcohol is $\frac{p}{p+1} + \frac{q}{q+1} = \frac{p+q+2pq}{(p+1)(q+1)}$ , and the total amount of water is $\frac{1}{p+1} and \frac{1}{q+1} = \frac{p+q+2}{(p+1)(q+1)}$ . The ratio of the volume of alcohol to the volume of water in the mixture is $\frac{\frac{p+q+2pq}{(p+1)(q+1}}{\frac{p+q+2}{(p+1)(q+1)}} = \boxed{\textbf{(E) } \frac{p+q+2pq}{p+q+2}$ (Error compiling LaTeX. Unknown error_msg).

1979 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1979 AHSME Problems/Problem 15

Problem 15

Solution

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