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A 24-liter tank holds a mixture that is 50 percent acid. A technician will remove a whole number of liters of the mixture and replace them with an equal volume of pure water so that the result is at most 30 percent acid. What is the least number of liters that must be removed?
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Correct answer
C
Two things must go right: modeling drain-and-replace, then rounding the resulting bound in the correct direction. The tank starts with 24 × 0.50 = 12 liters of acid. Removing r liters of the mixture takes 0.50r liters of acid with it, and refilling with water restores the volume to 24 liters, so the new fraction is (12 − 0.50r)/24.
The cap requires (12 − 0.50r)/24 ≤ 0.30. Multiply by 24: 12 − 0.50r ≤ 7.2. So 0.50r ≥ 4.8, and r ≥ 9.6. Choice B waits right here: the tempting move is rounding 9.6 to the nearest whole number, 9, but removing 9 liters leaves (12 − 4.5)/24 = 31.25 percent, above the cap. The at-most constraint forces the round to go up, so the least legal removal is 10 liters, which leaves 7/24, about 29.2 percent.
Choice A counts the 4.8 liters of departing acid instead of the removed mixture, D halves the tank by reflex, and E solves a different process, adding 16 liters of water without draining anything. When an inequality must hold at a whole number, round toward the side that keeps it true: the answer is choice C.
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