SOLUTION: A chemist wants to get a 50% acid solution by mixing 8 liters of 60% acid solution with 20% acid solution. How much 20% solution does the chemist need?

Question 1109268: A chemist wants to get a 50% acid solution by mixing 8 liters of 60% acid solution with 20% acid solution. How much 20% solution does the chemist need?
Found 2 solutions by addingup, greenestamps:
Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
0.6(8) + 0.2(x) = 0.5(8+x)
4.8 + 0.2x = 4 + 0.5x
-0.3x = -0.8
x = 2.67 liters of the 20% solution.
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!

The algebraic solution provided by the other tutor is fine.

Here is another way to solve these "mixture" problems.

The desired 50% is "3 times as close" to 60% as it is to 20%. (60-50 = 10; 50-20 = 30).

That means the amount of the 60% solution much be 3 times as much as the amount of the 20% solution.

Since the chemist is using 8 liters of the 60% solution, the amount of the 20% solution required is 1/3 of 8 liters, or 8/3 liters, or 2.67 liters (approximately).
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