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 825498: 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
Found 2 solutions by TimothyLamb, josgarithmetic:
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
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12/30 * 100 = 40% alcohol solution
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given:
42 L of 40% alcohol solution
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40*42 + 100x = 60(x + 42)
1680 + 100x = 60x + 2520
40x = 840
x = 840/40
x = 21
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answer:
add 21 liters of pure alcohol
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Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
Let a = amount of alcohol, volume, liters, to add

, for 60% alcohol result.
Solve the equation for a.

The first step should be at
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