IMO C

I did it this way -

xy+z=z
xy = 0
i.e x=0 or y = 0

so a. x!0 then y = 0, so |x-y| will be greater than 0
frm b. y=0 so x!0, so |x-y| again > 0.

so D but ans is A .. ??

OA is A?

or is it this way !!

from b . y=0 so even x could be x=0 . in tht case x-y =0 is not greater than 0.

hence the OA is A.

Hussh!! self solve ..

I didn't get the answer until you put the OA up... but it's a good question.

From the statement xy + z = z, we can deduce that xy = 0

This gives us 2 conditions:
CASE A: x or y is 0
CASE B: x and y both are 0

The problem is knowing whether we have CASE A or B.

With Case A |x-y| > 0 for sure. With Case B |x-y| is 0 which is not greater than 0.

Statement I helps us identify that only one of the numbers is 0... So |x-y| is definitely greater than 0. So I is sufficient

Statement II states that y is 0... but that doesn't help as we don't know if x is 0 or x != 0 .. so II is insufficient. Answer is A

Rajani wrote:if xy+z=z, is |x-y|>0?

a. x!=0
b. y=0

Will be A.