Difference between revisions of "2011 AIME I Problems/Problem 1"

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== Solution ==
 
== Solution ==
45/5
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There are <math>\frac{45}{100}(4)=\frac{9}{5}</math> L of acid in Jar A. There are <math>\frac{48}{100}(5)=\frac{12}{5}</math> L of acid in Jar B. And there are <math>\frac{k}{100}</math> L of acid in Jar C. After transfering the solutions from jar C, there will be
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<br> <math>4+\frac{m}{n}</math> L of solution in Jar A and <math>\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}</math> L of acid in Jar A.<br>
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<br> <math>6-\frac{m}{n}</math> L of solution in Jar B and <math>\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)</math> of acid in Jar B.
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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.

Revision as of 12:59, 19 March 2011

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 $\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)$ 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.

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