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

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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) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = amount of alcohol, volume, liters, to add

Amount+of+Alcohol+=+12%2Ba
Amount+of+Mixture+to+Result+=+12%2B30%2Ba

%2812%2Ba%29%2F%2812%2B30%2Ba%29=0.60, for 60% alcohol result.
Solve the equation for a.

The first step should be at a%2B12=0.6%28a%2B42%29