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?

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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) About Me  (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.A%2BB=600
.
.
20A%2B50B=30%28600%29
2.2A%2B5B=1800
Multiply eq. 1 by (-2) and add to eq. 2,
-2A-2B%2B2A%2B5B=-1200%2B1800
3B=600
highlight%28B=200%29liters
Then from eq. 1,
A%2B200=600
highlight%28A=400%29liters

Answer by ewatrrr(24785) About Me  (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