m15 # 27

Set S is composed of consecutive multiples of 3. Set T is composed of consecutive multiples of 6. If each set contains more than one element, is the median of set S larger than the median of set T?

s1) The least element in either set is 6.
s2) Set T contains twice as many elements as set S.
(C) 2008 GMAT Club - m15#27

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient
Statement (1) by itself is insufficient. If set contains many more elements than set , its median can be greater than that of set .

Statement (2) by itself is insufficient. We don't know how the sets are positioned against each other.

Statements (1) and (2) combined are sufficient. The answer is "no".

The correct answer is C.

my question is...cant this still be proved ineffective?
it says consecutive multiples of 3 or 6, but you cannot just assume they start with small elements
to TEST C:
eg. S= (3,6,9,12,15,18) median = 10.5, T = (6,12,18) median = 12
median S < median T = therefore NO

BUT there is no restriction on where these consecutive multiples start:
what if S = (102,105,108, 111, 114, 117) median = 109.5 , T = (6, 12, 18) median = 12

so shouldnt the answer be E?

am i missing something here? thanks in advance..