SOLUTION: How many liters of a 20% alcohol solution and how many liters of a 50% alcohol solution must be mixed to produce 9 liters of a 30% alcohol solution?

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Question 188486: How many liters of a 20% alcohol solution and how many liters of a 50% alcohol solution must be mixed to produce 9 liters of a 30% alcohol solution?
Found 2 solutions by stanbon, Mathtut:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
How many liters of a 20% alcohol solution and how many liters of a 50% alcohol solution must be mixed to produce 9 liters of a 30% alcohol solution?
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Equation:
alcohol + alcohol = alcohol
0.20x + 0.50(9-x) = 0.30*9
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20x + 50*9 - 50x = 30*9
-30x = -20*9
x = 6 (amt. of 20% solution in the mix)
9-x = 3 (amt of 50% solution in the mix)
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Cheers,
Stan H.

Answer by Mathtut(3670) (Show Source):
You can put this solution on YOUR website!
let x and y be the amounts of 20% and 50% solutions respectively
:
x+y=9........eq 1
.2x+.5y=.3(9).eq 2
:
rewrite eq 1 to x=9-y and plug this into eq 2
:
.2(9-y)+.5y=2.7
:
1.8-.2y+.5y=2.7
:
.3y=.9
:
y=3...liters of 50% solution
:
x=9-y=9-3=6 liters of 20% solution