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A 24-liter tank holds a mixture that is 50% 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% acid. What is the least number of liters that must be removed?
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Correct answer
C
Two things must go right here: modeling the drain-and-replace process correctly, 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 pure water restores the volume to 24 liters, so the new acid fraction is (12 − 0.50r)/24.
The cap requires (12 − 0.50r)/24 ≤ 0.30. Multiply both sides by 24: 12 − 0.50r ≤ 7.2. So 0.50r ≥ 4.8, and r ≥ 9.6. This is exactly where choice B waits: the tempting move is to round 9.6 to the nearest whole number, 9, but removing only 9 liters leaves (12 − 4.5)/24 = 31.25% acid, which is above the cap. Because the constraint is at most 30%, you must round up, not to the nearest integer, so the least number of liters that legally satisfies it is 10, which leaves 7/24, about 29.2% acid.
Choice A counts the 4.8 liters of departing acid instead of the removed mixture, choice D halves the tank on reflex, and choice E solves a different process entirely, adding 16 liters of water without draining any mixture at all. When an inequality must hold at a whole number, you round toward the side that keeps it true, so the answer is choice C.
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