Here is my step by step thoughts:

Reading question....Set S is composed of consecutive multiples of 3. Great! it is like {3,6,9,12...}. No restriction on the number of elements and we can shift it by 3 as far as we want. In other words, S could be {111, 114, 117,...}.

Set T is composed of consecutive multiples of 6.The same as about set S. But in set T the distance between the closest elements is twice as large as that in set S.

Each set contains more than one element.No essential information. Perhaps, we should return to the point latter.

Is the median of set S bigger than the median of set T?Median is a middle number or average between two for S and T with even number of elements. By the way, S and T are evenly distributed sets for which median = mean.

After reading....Each of sets is not fixed in the question. In other words, we can have {3, 6} S set and {111111111111, 111111111114}. And median vary in infinite range. So, our conditions should at least fix sets at some point.

Conditions....1. The least element in either set is 6Yeah, it fixes the beginning of both sets. But the ranges of sets could vary from 3 (for S) and 6 (for T) to infinity. So, insufficient.

2. Set T contains twice as many elements as set SIt does not fix positions of sets. So, insufficient.

1&2. So, our sets begin at 6. For example, S={6} and T={6,12}. If when S={6,9}, T={6,12,18,24}. We can see that median of T is increasing faster than median of S and it is larger. So, C is sufficient.

It is exactly the way I thought.

_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame