SOLUTION: How many liters of a solution that is 30% alcohol must be mixed with 80 liters of a solution that is 90% alcohol to get a solution that is 60% alcohol?

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Question 1196850: How many liters of a solution that is 30% alcohol must be mixed with 80 liters of a solution that is 90% alcohol to get a solution that is 60% alcohol?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


(1) An informal solution using common sense....

The desired concentration, 60% is halfway between 30% and 90%; that means equal amounts of the two ingredients are needed.

ANSWER: 80 liters

(2) A typical setup for solving the problem using formal algebra....

x liters of .30 alcohol, plus 80 liters of .90 alcohol, equals (80+x) liters of .60 alcohol:

.30%28x%29%2B.90%2880%29=.60%28x%2B80%29

Solve using basic algebra; I'll leave that to you.

Obviously your answer should be 80 liters....