SOLUTION: A pharmacist has 2 liters of a solution containing 30% alcohol. If he wants to have a solution containing 44% alcohol, how much pure alcohol must be added?

Question 268964: A pharmacist has 2 liters of a solution containing 30% alcohol. If he wants to have a solution containing 44% alcohol, how much pure alcohol must be added?
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
This is a mixture problem. Here is a table based on the given information
solution . . . . . . . . .% . . . . . . . . .liter . . . . . . . . .%-Liter
30% . . . . . . . . . . . 30. . . . . . . . . 2. . . . . . . . . . . 60
pure . . . . . . . . . . 100 . . . . . . . . x . . . . . . . . . . . 100x
mixture . . . . . . . . 44 . . . . . . . . . 2+x . . . . . . . . 88 + 44x
Looking at the third column, we get
60 + 100x = 88 + 44x
subtract 44x and subtract 60 from both sides to get
56x = 28
divide by 56 to get
x = 1/2
You need 1/2 liter of pure alcohol.
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