SOLUTION: How much pure alcohol must be added to a mixture of 12 liters of alcohol and 30 liters of water to produce a 60% alcohol solution.

Question 951767: How much pure alcohol must be added to a mixture of 12 liters of alcohol and 30 liters of water to produce a 60% alcohol solution.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the current ratio of alcohol to total beverage is 12/42.
that's because the amount of alcohol is 12 liters and the amount of water is 30 liters for a total of 42 liters of beverage.
you will be adding x liters of alcohol to the mixture to get 60% alcohol.
the formula becomes:
(x + 12) / (42 + x) = .6
multiply both sides of this equation by (42 + x) to get:
x + 12 = .6 * (42 + x)
distribute the multiplication to get:
x + 12 = .6 * 42 + .6 * x
subtract .6 * x from both sides of the equation and subtract 12 from both sides of the equation to get:
x - .6 * x = .6 * 42 - 12
simplify the left side of the equation to get:
.4 * x = .6 * 42 - 12
divide both sides of the equation by .4 to get:
x = (.6 * 42 - 12) / .4
distribute the multiplication on the right side of the equation to get:
x = .6 * 42 / .4 - 12 / .4
perform the operations indicated to get:
x = 63 - 30 = 33.
you would need to add 33 liters of alcohol to the mixture to get a mixture that is 60% alcohol.
12 + 33 = 45
42 + 33 = 75
45/75 = .6 * 100 = 60% alcohol.

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