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?

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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) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
x, amount of 30%
90-x, amount of 60%

%2830x%2B60%2890-x%29%29%2F90=50

-
3x%2B6%2890-x%29=9%2A50
x%2B2%2890-x%29=3%2A50
x%2B180-2x=150
-x=150-180
highlight%28x=30%29 ----------volume of the 30%
-
highlight%2890-x=60%29---------volume of the 60%