Is x an integer?

This is actually a much more difficult question (at least if you want a proper mathematical proof) then it might appear to be. It's probably easier to follow the solution below if you first look at some numerical examples. If Statement 2 is true, but the answer to the question is 'no' -- that is, if x is not an integer -- then 2x must be odd. So in those cases, 2x might equal something like 3 or 5 or 11, and x would equal something like 3/2 or 5/2 or 11/2. But notice in each case, x^2 will never be an integer -- if you square each of these, you'll get 2.25, or 6.25, or 30.25. In fact x^2 will always end in '.25' if you write it as a decimal.

So we can show that the two statements are sufficient if we start out by assuming we can get a 'no' answer, and then prove that no value of x can exist where both statements are true.

Look first at Statement 2. If the answer is 'no', so x is not an integer, then 2x is odd. Then x itself will be equal to some integer plus 1/2 (as a decimal, it will end in ".5"), because if 2x is odd, the remainder is 1 when we divide 2x by 2.

So if the answer to the question is 'no', then x must be equal to something like k + 1/2, where k is an integer. But if we also assume Statement 1 is true, we know x^2 is also an integer, so if the answer to the question is 'no', (k + 1/2)^2 would need to be an integer. But (k + 1/2)^2 = k^2 + k + (1/4), and since k is an integer, this is exactly 1/4 more than some integer (its decimal would end in ".25"), so it is impossible for this to be an integer.

That is, if you assume the answer to the question is 'no' and use both statements, you find it's impossible that both statements are true. That means the answer must be 'yes', so C is the answer (since neither statement is sufficient alone).