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A tank holds 30 liters of saline that is 25% 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% 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 tank starts with 30 liters that are 25% salt, so it holds 30 × 0.25 = 7.5 liters of salt. If the chemist adds s liters of pure salt, the salt rises to 7.5 + s and the volume rises to 30 + s, so the requirement is (7.5 + s)/(30 + s) ≥ 0.40.
Keep s in the denominator and multiply out: 7.5 + s ≥ 0.40(30 + s) = 12 + 0.40s. Subtract 0.40s from both sides: 0.60s ≥ 4.5, so s ≥ 7.5. Since s must be a whole number, check the boundary directly: adding 7 liters gives (7.5 + 7)/37 = 14.5/37, about 39.2%, which falls short of the floor. Adding 8 liters gives 15.5/38, about 40.8%, which clears it. So the least whole number of liters that works is s = 8, and the final volume of 38 liters still respects the 40-liter cap. That 40-liter cap sits in the stem to bait choice E, which fills the tank straight to it with 40 − 30 = 10 liters and never checks the concentration at all.
The least number of liters that must be added is 8.
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