2019/20 AHSMC Part 1 Problem 12

Problem 12 of 16

Four containers are weighed in pairs giving the values $215$ kg, $235$ kg, $260$ kg, $265$ kg, $290$ kg and $310$ kg. The weight (in kg) of the heaviest container is

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Solution

No two containers have the same weight otherwise we would have a duplicate weighing. Let $w_1 < w_2 < w_3 < w_4$ be the weights of the four containers. Then we must have $w_1 + w_2 = 215$ , $w_1 + w_3 = 235$ , $w_3 + w_4 = 310$ and $w_2 + w_4 = 290$ . There are two cases: $w_1 + w_4 = 260$ , $w_2 + w_3 = 265$ giving $w_4 = 167.5$ and $w_1 + w_4 = 265$ , $w_2 + w_3 = 260$ giving $w_4 = 170$ . The answer is (E).

University of Alberta

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