M15-27

Official Solution:


Set \(S\) consists of consecutive multiples of 3, while set \(T\) consists of consecutive multiples of 6. Is the median of set \(S\) greater than the median of set \(T\)?

(1) The least element in both sets is 6.

If \(S=\{6, 9\}\) and \(T=\{6, 12\}\), then the answer is NO. However, if \(S=\{6, 9, 12, 15\}\) and \(T=\{6, 9\}\), then the answer is YES. Not sufficient.

(2) Set \(T\) has twice as many elements as set \(S\).

If \(S=\{6, 9\}\) and \(T=\{60, 66, 72, 78\}\), then the answer is NO. However, if \(S=\{6, 9\}\) and \(T=\{-78, -72, -66, -60\}\), then the answer is YES. Not sufficient.

(1)+(2) Given that set \(T\) contains twice as many elements as set \(S\), and the smallest element in both sets is 6, the median of set \(T\) will always be greater than the median of set \(S\). For example, we can have \(S=\{6\}\) and \(T=\{6, 12\}\), or \(S=\{6, 9\}\) and \(T=\{6, 12, 18, 24\}\), or \(S=\{6, 9, 12\}\) and \(T=\{6, 12, 18, 24, 30, 36\}\), and so on. In each case, the median of set \(T\) is greater than the median of set \(S\). Therefore, the answer to the question is NO. The combination of statements (1) and (2) is sufficient.


Answer: C