SOLUTION: Mix a solution that is 50%alcohol with a solution that is 10% alcohol to make 300 L of a solution that is 30% alcohol. How much of each should you use?

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Question 66435: Mix a solution that is 50%alcohol with a solution that is 10% alcohol to make 300 L of a solution that is 30% alcohol. How much of each should you use?
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
Mix a solution that is 50%alcohol with a solution that is 10% alcohol to make 300 L of a solution that is 30% alcohol. How much of each should you use?
Let x=amount of 50% alcohol
Then 300-x=amount of 10% alcohol
Now we know that the amount of pure alcohol in the 50% solution (.50)(x) plus the amount of pure alcohol in the 10% solution (.10)(300-x) has to equal the amount of pure alcohol in the final solution (300)(.30). So our equation to solve is:
(.50)(x)+(.10)(300-x)=(300)(.30) simplifying, we get:
.5x+30-.1x=90 subtract 30 from both sides:
.4x=90-30=60
x=150 Liters of 50% alcohol
300-x=300-150=150 liters of 10% alcohol
Ck
150(.1)+150(.5)=300(.3)
15+75=90
90=90
Hope this helps----ptaylor