Mixture Problem

Hi prachi18oct,

This is an example of a 'Weighted Average' question.

We're told that we're going to mix a certain amount of a 50% alcohol solution with a certain amount of a 25% alcohol solution and end up with a 30% alcohol solution. Here's how we can set up that calculation using Algebra:

A = # of ounces of 50% solution
B = # of ounces of 25% solution
A+B = total ounces of the mixed solution

(.5A + .25B)/(A+B) = .3

.5A + .25B = .3A + .3B
.2A = .05B
20A = 5B
4A = B
A/B = 1/4

This means that for every 1 ounce of solution A, we have 4 ounces of solution B.

To answer the question that's asked, imagine that you have 5 ounces of solution A. We're told to REPLACE 4 ounces of it with 4 ounces of solution B (thus, we'll end up with the 30% mixture that we're after).

Since 4 ounces of the 5 ounces were replaced, 4/5 of the original alcohol and 4/5 of the original water was replaced.

4/5 = 80%

Final Answer: E

GMAT assassins aren't born, they're made,
Rich