SOLUTION: I have tried to solve to AW=L and so on... the answer is 150L but I cant figure out the formula. The question is "How many liters of a 20% alcohol solution should Maria mix with 50

Question 319752: I have tried to solve to AW=L and so on... the answer is 150L but I cant figure out the formula. The question is "How many liters of a 20% alcohol solution should Maria mix with 50 liters of a 60% alcohol solution to obtain a 30% solution?" the answer is 150L of the 20% solution. I think I'm doing too much. Ive figured out the following...

If L = Liters
LA = Liters of Alcohol
LW = Liters of Whatever
the 50 liter mix of the 60% solution = .012 to 1
if 60% of the solution = LA
then 40% = LW.
so(0.012LA)(50L)=0.6LA+0.4LW x 50= 50L?
20% of 1L = 0.2LA with 0.80LW
I still dont understand where I'm going wrong... my assignment is due Tuesday... wow, I'm super confused. Please help me!

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
How many liters of a 20% alcohol solution should Maria mix with 50 liters of a 60% alcohol solution to obtain a 30% solution?
---------------------
Add x liters of 20%
The total volume will be x + 50
0.2x + 0.6*50 = (x + 50)*0.3 is the amount of alcohol
0.2x + 30 = 0.3x + 15
x = 150 liters
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