DS - Numbers

yeah, so you should definitely rephrase the problem by getting rid of the fractions first. you can kill them all by multiplying through by 12, the common denominator of all the fractions:
2k + 3m = t
this is the base equation from which we'll be working to answer the problems.

you can also rephrase the question. if t and 12 have a common factor greater than 12, that's just another way of saying that something that goes into 12 (besides 1) also goes into t. so, here's a rephrase of the prompt question:
does 2, 3, 4, 6, or 12 go into t?
since 4, 6, and 12 are just compounds of the prime factors 2 and 3, it's actually redundant to ask whether those numbers are factors of t. therefore, we can rephrase the question down to its most distilled essence:
does 2 or 3 go into t?

-- statement (1) --
if k is a multiple of 3, then we can write k = 3n, where n is an integer. therefore, we have
2(3n) + 3m = t
6n + 3m = t
since 6n and 3m are both multiples of 3, it follows that 3 goes into t.
sufficient.

-- statement (2) --
if m is a multiple of 3, then we can write m = 3n, where n is an integer. therefore, we have
2k + 3(3n) = t
2k + 9n = t
these (2k and 9n) don't have a common factor, so this is insufficient.
if you want proof of insufficiency, go ahead and plug some easy numbers: if k = n = 1 (which would actually mean k = 1 and m = 3), then t = 11, which doesn't share a common factor with 12. if k = 3 and n = 1 (which would actually mean k = m = 3), then t = 15, which shares the common factor 3 with 12.
insufficient.

answer = a