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
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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!
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?
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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
