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) (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:
Solve using basic algebra; I'll leave that to you.
Obviously your answer should be 80 liters....
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