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 $k + m + n$.
Jar A contains $\frac{11}{5}$ liters of water, and $\frac{9}{5}$ liters of acid; jar B contains $\frac{13}{5}$ liters of water and $\frac{12}{5}$ liters of acid.
The gap between the amount of water and acid in the first jar, $\frac{2}{5}$, is double that of the gap in the second jar, $\frac{1}{5}$. Therefore, we must add twice as much of jar C into the jar $A$ over jar $B$. So, we must add $\frac{2}{3}$ of jar C into jar $A$, so $m = 2, n=3$.
Since jar C contains $1$ liter of solution, we are adding $\frac{2}{3}$ of a liter of solution to jar $A$. In order to close the gap between water and acid, there must be $\frac{2}{5}$ more liters of acid than liters of water in these $\frac{2}{3}$ liters of solution. So, in the $\frac{2}{3}$ liters of solution, there are $\frac{2}{15}$ liters of water, and $\frac{8}{15}$ liters of acid.  So, 80% of the $\frac{2}{3}$ sample is acid, so overall, in jar C, 80% of the sample is acid.
Therefore, our answer is $80 + 2 + 3 = \boxed{85}$.
~ ihatemath123