SOLUTION: How many liters of a 40%-alcohol solution must be mixed with 10 liters of a solution that is 80% alcohol to get a solution that is 60% alcohol?
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Question 323454: How many liters of a 40%-alcohol solution must be mixed with 10 liters of a solution that is 80% alcohol to get a solution that is 60% alcohol?
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
.80*10+.40x=.60(10+x)
8+.40x=6+.60x
.40x-.60x=6-8
-.2x=-2
x=-2/-.2
x=10 liters of 40% solution is used.
Proof:
.80*10+.40*10=.60(10+10)
8+4=.60*20
12=12
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