Official Solution:We’ll go for ALTERNATIVE because we can use numbers to eliminate contradictory answers.

(1) The weights in R could be, say, {2, 4, 6}, or {100, 102, 104}, and the weights in S could be, say, {11, 13, 15}, so that the total weight of set S is greater than the first R set example, but lower than the second example. Insufficient. (A) and (D) eliminated.

(2) Median is the middle value, so if the weights in R are, say, {4, 5, 6}, then the median is 5. Say the weights in S are {1, 2, 100}, or {1, 2, 3} – in both cases, the median is 2, however in the first case the total weight of S is greater, and in the other case the total weight is lower. Insufficient. (B) is eliminated.

On combining the two statements, we can use the fact that for any set of consecutive numbers the median is also the average. So if the median and average are higher for one such set, and also both sets have the same number of elements (n), the set with the higher median weight would have a higher total weight. For example, if set R is {2, 4, 6} and set S is {1, 3, 5}. (E) is eliminated.

Answer: C

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