1973 AHSME Problems/Problem 33

Problem

When one ounce of water is added to a mixture of acid and water, the new mixture is $20\%$ acid. When one ounce of acid is added to the new mixture, the result is $33\frac13\%$ acid. The percentage of acid in the original mixture is

$\textbf{(A)}\ 22\% \qquad \textbf{(B)}\ 24\% \qquad \textbf{(C)}\ 25\% \qquad \textbf{(D)}\ 30\% \qquad \textbf{(E)}\ 33\frac13 \%$

Solution

Let $a$ be the original number of ounces of acid and $w$ be the original number of ounces of water. We can write two equations since we know the percentage of acid after some water and acid. $\frac{a}{a+w+1} = \frac{1}{5}$ $\frac{a+1}{a+w+2} = \frac{1}{3}$ Cross-multiply to get rid of the fractions. $5a = a+w+1$ $3a+3=a+w+1$ Solve the system to get $a=1$ and $w=3$ . The percentage of acid in the original mixture is $\tfrac{1}{1+3} = \boxed{\textbf{(C)}\ 25\%}$ .

1973 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 32	Followed by Problem 34
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1973 AHSME Problems/Problem 33

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