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A chemist has 30 liters of a solution that is 20 percent acid by volume. How many liters of pure acid must be added so that the resulting mixture is 50 percent acid by volume?
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Correct answer
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The starting acid is 0.20 × 30 = 6 liters. Let x be the liters of pure acid added. Pure acid adds to both the acid amount and the total volume, so the mixture condition is (6 + x) / (30 + x) = 0.50. Cross-multiply: 6 + x = 0.5(30 + x) = 15 + 0.5x. Then 0.5x = 9, so x = 18.
Check: acid = 6 + 18 = 24, total = 30 + 18 = 48, and 24 / 48 = 0.50. The key translation point is that pure acid changes the denominator too; holding the total at 30 or stopping at the intermediate 0.5x = 9 are the static-denominator and stopped-early traps.
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