# 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 1

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. ! Missing delimiter (. inserted).) 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 $k+m+n=80+2+3=\boxed{085}$.

## Solution 2

One might cleverly change the content of both Jars.

Since the end result of both Jars are $50\%$ acid, we can turn Jar A into a 1 gallon liquid with $50\%-4(5\%) = 30\%$ acid

and Jar B into 1 gallon liquid with $50\%-5(2\%) =40\%$ acid.

Now, since Jar A and Jar B contain the same amount of liquid, twice as much liquid will be pour into Jar A than Jar B, so $\dfrac{2}{3}$ of Jar C will be pour into Jar A.

Thus, $m=2$ and $n=3$.

$\dfrac{30\% + \frac{2}{3} \cdot k\%}{\frac{5}{3}} = 50\%$

Solving for $k$ yields $k=80$

So the answer is $80+2+3 = \boxed{85}$