SOLUTION: A lab has a 20% acid solution and a 50% acid solution. How many liters of each are required to obtain 600 liters of a30% solution?

Question 363069: A lab has a 20% acid solution and a 50% acid solution. How many liters of each are required to obtain 600 liters of a30% solution?
Found 2 solutions by Fombitz, ewatrrr:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Let A be the amount of 20% solution, B the amount of 50% solution.
1.
.
.

2.
Multiply eq. 1 by (-2) and add to eq. 2,

liters
Then from eq. 1,

liters
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

Hi,
Let x represent the amount of the 50% solution, (600 - x) would then represent the 20% solution
.20(600L - x) + .50x = .30*600L
.20*600L - .20x + .50x = .30*600L
.30x = .10*600L
x = 200L of 50% solution and 400L of 20% solution are needed
checking our answer
.20*400L + .50*200L = 80L + 100L = 180L = .30*600L
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