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A chemist has 20 liters of a solution that is 30% acid. How many liters of pure acid must be added so that the resulting mixture is 50% acid? Assume the added acid mixes completely.
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Correct answer
B
Track the actual liters of acid and the total volume. The original 20 liters at 30% contain 0.30 × 20 = 6 liters of acid. Let x be the liters of pure acid added, so the acid becomes 6 + x and the total becomes 20 + x, and this must be 50% acid: (6 + x) ÷ (20 + x) = 0.5.
Multiply both sides by (20 + x) to get 6 + x = 10 + 0.5x. Subtract 0.5x and 6 from both sides: 0.5x = 4, so x = 8. So the answer is B. Check: acid is 6 + 8 = 14, total is 20 + 8 = 28, and 14 ÷ 28 = 0.5.
Choice A takes a flat 20% of the original volume. Choice C aims for half the original volume and ignores the starting 6 liters and the changing total. Choice D takes half of an assumed 24-liter total and reports that 12 as the amount to add, forgetting the 6 liters of acid already present. Choice E sums partial pieces, overcounting the acid needed.
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