(2) alone is sufficient but (1) alone isn't... since y> 7, y has to be odd, and y+1 is even, so x(y+1) is even. However, for (1), if x = 7, y = 2, then x(y+1) isn't even.

If x and y are integers, is x(y+1) an even number?

1). x, and y are prime numbers.
2). y>7

1) e.g. x = 2, y = 5 --> 2(5+1) = 12 --> EVEN
x = 3, y = 2 --> 3(2+1) = 9 --> ODD.

2) y>7, y could 8, 9, 10, 11, 12...... and so on.
let's say y=8 and x = 2 --> 2(8+1) = 18 --> EVEN
OR y = 9 and x = 2 --> 2(9+1) -->20
BUT who's to say that x can't be odd? x =3, y = 8 --> 3(9) = 27 ODD.
OR x = 3, y = 9 --> 3(10) = 30 EVEN.

obviously the both alone are inefficient.

TOGETHER, if x and y are prime numbers AND y is > 7, then is x(y+1) even?
let's say x = 2 (even prime) and y = 11 (odd prime number > 7), then 2(11+1) = 24 EVEN.
OR say x = 3 (odd) and y = 13 (odd), then 3(13+1) = 3(14) = 42. EVEN.

SO we learned that: ODD * EVEN = EVEN, or EVEN*EVEN = EVEN

Answer should be C. Hope this helps.

answer should be - C

C for me as well. Excellent explanations.
I wanted to add one point.

stmt 2 could been written as a "y > anynumber that is greater than 2."