Is x+y>0?

Answer: (C)

Question gives us some idea that the answer may be different for positive and negative integers. So let us think only along these lines.

Statement 1: INSUFFICIENT
Suppose x and y both are positive. Is it possible that |x| be greater than y? Yes, say x = 3 and y = 2. Is x+y>0? YES
If x and y both are negative, is it possible that |x| be greater than y? Yes again, say x = -3 and y = -2. Is x+y>0? NO

Statement 2: INSUFFICIENT
Same reasoning as before except that x and y are interchanged now.

Statements can also be analyzed as:
Statement 1 -> If magnitude of x is greater than y, then we have 2 possibilities:
y = positive: then distance of x from 0 is greater than the value of y.
y = negative: then x can be any integer.
Statement 2 -> If magnitude of y is greater than x, then the 2 possibilities are:
x = positive: then distance of y from 0 is greater than the value of x.
x = negative: then y can be any integer.

Hence after combining ->
x = negative and y = negative: consistent with both statements. Is x+y>0? NO
x = positive and y = negative: Since x is positive, so distance of y from 0 must be greater than the value of x (as per statement 2). Hence, y can be negative but will have a larger magnitude compared to x. Hence x + y is negative. Is x+y>0? NO
x = negative and y = positive: Since y is positive, so distance of x from 0 must be greater than the value of y (as per statement 1). Hence, x can be negative but will have a larger magnitude compared to y. Hence x + y is negative. Is x+y>0? NO
x = positive and y = positive: inconsistent with the 2 statements because in this case, either |x| > y or |y| > x. Both can't hold true together.

Hence, there is unique answer for the question asked after combining the 2 statements.

Correct Answer is C.

Try to manipulate by taking value x=1 and y=1 which holds insufficient condition in both cases.

Regards:-
Sumit kumar

Can you please check if the following method is fine?

To prove x + y >0 i.e ans should be a yes or a no

Statement a: |x|>y i.e LHS will always be +ve i.e y<0 or y = 0. Insufficient can't say anything about x
Statement b: |y|>x i.e LHS will always be +ve i.e x<0 or x = 0. Insufficient can't say anything about y

Statements combined. if y<0 i.e x also has to be less than 0 i.e x+y <0...similarly if y=0 i.e x also has to be 0 i.e x+y=0 in both the cases x+y is not greater 0. Hence C is the right answer