The problem here is that, when you take stmt 2, you get that y can be both 1 and 2. Since x is a multiple of y and x + 2 is also a multiple of y, then x divides x + 2 - x = 2. There are two options for this, 1 and 2. So 2 by itself is insufficient.
But combine it with stmt 1 and you eliminate 1, so 2 will be the answer indeed.

Naga,
Excellent trap question. Whats ths source?

Regards,
CR

1000 DS cramya.

(1) y ≠ 1
y could be {2, 3, 4, 5, ...}. INSUFF

(2) x + 2 is a multiple of y <==> (x+2 = ny)
x is a multiple of y <==> (x = my)
So, my + 2 = ny
2 = ny - my
2 = (n-m)y
y = 2 / (n-m)

Since y must be positive integer. (n-m) could only be 1 or 2.
If (n-m) = 1, y = 2.
If (n-m) = 2, y = 1.
y could be either 1 or 2. INSUFF.

(1) + (2)
Since y &#8800; 1, we know that y must be equal to 2. SUFF.

C. is the answer.

C is the answer. Lets take it this way :

1. Y can be anything , not a solution

2. Lets take x = 4 multiple of 2 (y). Then x+2 is not a multiple of y. Same for other values of y .. But if we club both the conditions , it satisfies

nice solution Bidisha. kudos