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

Answer: 85

Concepts:mixture percentage

Difficulty rating: 1950

Small Hint:

Imagine pouring all three jars together: the result is $10$ liters that must be $50$ % acid, which determines $k$

Big Hint:

With $k$ known, let $x$ be the amount added to jar $A$ and set jar $A$ ’s acid equal to half of its new volume

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

Answer: 85

Difficulty rating: 1950

Small Hint:

Big Hint:

Solution:

Problem 1 in Other Years