SOLUTION: How many liters of 80% alcohol solution and 20% alcohol solution must be mixed to obtain 12 liters of 30% alcohol solution?

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Question 906813: How many liters of 80% alcohol solution and 20% alcohol solution must be mixed to obtain 12 liters of 30% alcohol solution?
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
a+b=12,
0.8*a+0.2*b=0.3*12
a=12-b
0.8*(12-b)+0.2*b=3.6
9.6-0.8b+0.2*b=3.6
-0.6*b=-6
b=10
a=12-b
a=2 liter at 80%
b=10 liter at 20%
check
0.8*2+0.2*10=0.3*12
1.6+2=3.6
3.6=3.6
ok