SOLUTION: How many liters of a 15% acid solution should be mixed with 10 liters of a 36% acid solution to obtain a mixture that is 20%?

Question 1131385: How many liters of a 15% acid solution should be mixed with 10 liters of a 36% acid solution to obtain a mixture that is 20%?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!

x liters of 15%
10 liters of 36%

Solve for x

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(now fixed)
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Again tutor @josgarithmetic has shown a setup for a problem different than the one in the problem.

You want 10 liters of 36% acid solution and x liters of 15% acid solution to get (10+x) liters of 20% acid solution:

That equation is simply solved using basic algebra; you might want to first multiply the whole equation by 100 to get rid of the decimals.

Here is what I think is a faster and easier way to solve mixture problems like this, without the decimals and the formal algebra.

(1) The target percentage of 20% is 5% away from 15% (20-15 = 5) and 16% away from 36% (36-20 = 16).
(2) The ratio of those differences (5:16) is the ratio in which the two ingredients must be mixed; common sense says the larger portion needs to be the 15% solution, since 20% is closer to 15% than to 36%. So we need 16 parts of 15% acid solution to 5 parts of the 36% acid solution.
(3) Since we know there are 10 liters of the 36% acid solution, we solve the problem with a simple proportion: 10 liters of 36% acid solution and x liters of 15% acid solution need to be mixed in the ratio 5:16:

ANSWER: You need 32 liters of 15% acid solution to be mixed with the 10 liters of 36% acid solution to make a 20% acid solution.

All the words of explanation make it seem like more work than the traditional algebraic solution method. But without the words, you can see that the calculations are quick and easy:

36-20 = 16; 20-15 = 5
5:16 = 10:x --> x = 2*16 = 32
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