SOLUTION: How many liters of pure alcohol should be mixed with 20 liters of a 15% alcohol solution to obtain a 20% mixture?

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Question 1131251: How many liters of pure alcohol should be mixed with 20 liters of a 15% alcohol solution to obtain a 20% mixture?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39618) About Me  (Show Source):
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Let x be the number of liters of pure alcohol you are adding. Then the amount of alcohol in the mixture is 15% of the original 20 liters plus 100% of the x liters. That amount of alcohol is to be 20% of the whole mixture, which is now (20+x) liters. So

.15%2820%29%2B1%28x%29+=+.20%2820%2Bx%29

I'll leave it to you to solve that equation; you might want to start by multiplying both sides by 100 to get rid of the decimals....

But here is a much easier and faster way to solve mixture problems like this, without the formal algebra.

Here is the key concept behind this method:

The ratio in which the two ingredients have to be mixed is exactly equal to the ratio of how far the percentage of the mixture is from the percentages of the two ingredients

The words may seem confusing; but the calculations are simple.

In this problem, the percentage of the mixture (20%) is 5% away from the original 15% solution(20-15=5); it is 80% away from the percentage of the pure alcohol (100-20=80).

The ratio of those difference is 5:80 or 1:16. That means the ingredients must be mixed in the ratio 1:16.

And since 20% is closer to 15% than it is to 100%, the larger portion of the mixture must be the 15% ingredient.

So the amount of pure alcohol that must be added is 1/16 of 20 liters, or 1.25 liters.

CHECK:
.15(20)+1(1.25) = 3+1.25 = 4.25
.20(21.25) = 4.25