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A tank holds 30 liters of saline that is 25 percent salt. A chemist will add a whole number of liters of pure salt, each liter adding to the total volume, so that the mixture is at least 40 percent salt. The final volume may not exceed 40 liters. What is the least number of liters that must be added?
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Correct answer
D
Two constraints are live at once: the concentration floor drives the algebra, and the whole-number requirement decides the final liter. The 30 liters hold 30 × 0.25 = 7.5 liters of salt. Adding s liters of pure salt raises the salt to 7.5 + s and the volume to 30 + s, so the requirement is (7.5 + s)/(30 + s) at least 0.40.
The classic slip is dropping s from the denominator; keep it and multiply out: 7.5 + s at least 0.40(30 + s) = 12 + 0.40s. Subtract: 0.60s at least 4.5, so s at least 7.5. Now the integer step: 7 liters fails, since (7.5 + 7)/37 is about 39.2 percent, so the least whole number that clears the floor is s = 8, giving 15.5/38, about 40.8 percent. The final volume, 38 liters, respects the cap; that 40-liter cap sits in the stem to bait choice E, which fills straight to it with 10 liters and never touches the concentration.
Choice A drops s from the denominator, needing only 0.40 × 30 = 12 liters of salt and so 4.5 more, rounded up to 5. Choice B multiplies the 15-point gap by the 40-liter cap, 0.15 × 40 = 6, never modeling the growing volume. Choice C rounds 7.5 the wrong way to 7 and finishes under the floor. An at-least bound met in whole liters rounds up: the answer is choice D.
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