Question
A 60-liter solution contains 30% alcohol. How much pure
alcohol must be added to make the solution 50% alcohol?Solution
Currently, the solution contains 30% of 60 liters = 18 liters of alcohol. Let x liters of pure alcohol be added. After adding x liters, total volume = 60 + x liters, Alcohol = 18 + x liters. We want: (18 + x) / (60 + x) = 0.5. Solving gives: 18 + x = 0.5(60 + x), 18 + x = 30 + 0.5x, 0.5x = 12, x = 24 liters. Answer: a) 24 liters
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