Difference between revisions of "1973 AHSME Problems/Problem 33"

Latest revision as of 13:04, 20 February 2020

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
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

@@ Line 20: / Line 20: @@
 ==See Also==
-{{AHSME 35p box|year=1973|num-b=32|num-a=34}}
+{{AHSME 30p box|year=1973|num-b=32|num-a=34}}
 [[Category:Introductory Algebra Problems]]

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Difference between revisions of "1973 AHSME Problems/Problem 33"

Latest revision as of 13:04, 20 February 2020

Problem

Solution

See Also