SOLUTION: How many liters of a 10% alcohol solution must be mixed wih 40L of a 50% solution to get a 40% solution
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Question 439571: How many liters of a 10% alcohol solution must be mixed wih 40L of a 50% solution to get a 40% solution
Answer by jorel1380(3719) (Show Source): You can put this solution on YOUR website!
40L of a 50% solution equals 20L of pure alcohol.
.10x+.50(40)=.40(x+40)
.10x+20=.4x+16
4=.3x
4/.3=x
13.33=x
You need 13 1/3 liters of a 10% solution to mix up a 40% solution..
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