SOLUTION: One solution of alcohol is 30% and a second is 60% alcohol. How much of each should be mixed in order to make 90 liters of a solution that is 50% alcohol?

Question 1073416: One solution of alcohol is 30% and a second is 60% alcohol. How much of each should be mixed in order to make 90 liters of a solution that is 50% alcohol?
Found 2 solutions by jorel1380, josgarithmetic:
Answer by jorel1380(3719) (Show Source): You can put this solution on YOUR website!
Let n be the amount of 60% alcohol. Then the amount of 30% alcohol would be 90-n. So:
.6n+.3(90-n)=.5(90)
.3n+27=45
.3n=18
n=60
60 liters of .6 alcohol and 30 liters of .3 alcohol will give you the desired mixture. ☺☺☺☺
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
x, amount of 30%
90-x, amount of 60%

-

----------volume of the 30%
-
---------volume of the 60%
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