Question

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 $k + m + n.$

Answer

Solution:

If all three jars were combined, the result would be $10$ liters of $50$ % acid, since both final jars are $50$ % acid. The total acid is therefore $5$ liters, so $4(0.45) + 5(0.48) + 0.01k = 5,$ which gives $k = 80.$

Now let $x$ be the number of liters poured from jar C into jar A. Jar A then holds $4 + x$ liters containing $1.8 + 0.8x$ liters of acid, so $1.8 + 0.8x = 0.5(4 + x),$ giving $0.3x = 0.2,$ so $x = \frac{2}{3}.$

Thus $m + n = 2 + 3 = 5,$ and $k + m + n = 80 + 5 = 85.$

2011 AIME I Problem 1

Solution:

Problem 1 in Other Years