Statement 1: c*(d + 1) is even. Implies,
(1) c even, d even
or
(2) c even, d odd
or
(3) c odd, d odd

Not sufficient.

Statement 1: (c + 2)*(d + 4) is even. Implies,
(1) c even, d even
or
(2) c even, d odd
or
(3) c odd, d even

Not sufficient.

1 & 2 Together: Third possibility of statement 1 (c & d both odd) and third possibility of statement 2 ( c odd, d even) can't coexist. Only first two possibilities of both of the cases are possible. Thus d may be even or odd, but c is always even.

Sufficient.

The correct answer is C.

Night reader wrote:...
Rahul, you mention that statements 1 and 2 contain different sets of variations, then you continue with opting C as two statements are sufficient to solve the problem. What was reasoning behind C?

Try to understand the logic that when two statements are combined then we must treat both of them as true. That may or may not result in answering the question, but we must treat both of them as TRUE.

Now in this case, combining two statements we see that first two possibilities are identical. The third possibilities says that,

• (1) For statement 1, if c is odd then d must be odd.
  (2) For statement 2, if c is odd then d must be even.

This means for c being odd, d must be odd and even simultaneously. Which is not possible. So we are left with the first two cases which tells that whether d is even or odd, c is always even.

If this is still not clear, analyze the statements individually.
Case 1: c even

• (1) c*(d + 1) is always even for d even or odd
  (2) (c + 2)*(d + 4) is always even for d even or odd

Case 2: c odd

• (1) c*(d + 1) even if d is odd
  (2) (c + 2)*(d + 4) is even if d is even

Thus for any odd value of c, both the statements cannot be true simultaneously.
Hope this helps.