SOLUTION: how many liters of a 60% acid solution must be mixed with a 10% acid solution to get 400 L of a 50% acid solution

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Question 674714: how many liters of a 60% acid solution must be mixed with a 10% acid solution to get 400 L of a 50% acid solution
Answer by checkley79(3341) About Me  (Show Source):
You can put this solution on YOUR website!
.60X+.10(400-X)=.50*400
.60X+40-.10X=200
.50X=200-40
.50X=160
X=160/.50
X=320L OF 60% ACID IS USED.
400-320=80L OF 10% ACID IS USED.
PROOF:
.60*320+.10*80=200
192+8=200
200=200