SOLUTION: Suppose that one solution contains 20% alcohol and another solution contains 50% alcohol. How many liter of each solution should be mixed to make 12 liters of a 40% solution?

SOLUTION: Suppose that one solution contains 20% alcohol and another solution contains 50% alcohol. How many liter of each solution should be mixed to make 12 liters of a 40% solution?

Algebra.Com

Question 732069:
Suppose that one solution contains 20% alcohol and another solution contains 50% alcohol. How many liter of each solution should be mixed to make 12 liters of a 40% solution?
Answer by checkley79(3341) (Show Source): You can put this solution on YOUR website!
.50X+.20(12-X)=.40*12
.50X+2.4-.20X=4.8
.30X=4.8-2.4
.30X=2.4
X=2.4/.30
X=8 LITERS OF 50% SOLUTION IS USED.
12-8=4 LITERS OF 20% SOLUTION IS USED.
PROOF:
.50*8+.20*4=.40*12
4+.8=4.8
4.8=4.8
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