2011 AIME I Problems/Problem 1

Problem 1

Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is $k\%$ acid. From jar C, $\frac{m}{n}$ liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that $m$ and $n$ are relatively prime positive integers, find $k + m + n$ .

Solution

There are $\frac{45}{100}(4)=\frac{9}{5}$ L of acid in Jar A. There are $\frac{48}{100}(5)=\frac{12}{5}$ L of acid in Jar B. And there are $\frac{k}{100}$ L of acid in Jar C. After transfering the solutions from jar C, there will be
$4+\frac{m}{n}$ L of solution in Jar A and $\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}$ L of acid in Jar A.

$6-\frac{m}{n}$ L of solution in Jar B and $\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}\right$ (Error compiling LaTeX. Unknown error_msg) of acid in Jar B.
Since the solutions are 50% acid, we can multiply the amount of acid for each jar by 2, then equate them to the amount of solution.
$\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}$ $\frac{24}{5}-\frac{km}{50n}+\frac{k}{50}=6-\frac{m}{n}$ Add the equations to get $\frac{42}{5}+\frac{k}{50}=10$ Solving gives $k=80$ .
If we substitute back in the original equation we get $\frac{m}{n}=\frac{2}{3}$ so $3m=2n$ . Since $m$ and $n$ are relatively prime, $m=2$ and $n=3$ . Thus $k+m+n=80+2+3=\boxed{085}$ .

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2011 AIME I Problems/Problem 1

Problem 1

Solution