Is x+y>0?

Question: Is x + y > 0
When will the sum of x and y be positive?
When either both x and y are positive or only one of them is positive with higher absolute value than the other one (which could be 0 or negative).
In other cases, x+y will be negative or 0.

(1) |x|>y
This could hold when both x and y are positive (x = 4, y = 2) or when both x and y are negative (x = -4, y = -2) or other cases.
Hence this statement alone is not sufficient.

(2) |y|>x
This statement is symmetrical to statement 2 with just the variables reversed. IF statement 1 is not sufficient, statement 2 will also not be sufficient.

Using both,
|x| > y
|y| > x

Can both x and y be positive? No, because then only one's absolute value will be greater than the other.
Is it possible that one of them is positive with higher absolute value than the other one? No. Say the numbers are x = 5 and y = -3.
Here |y| cannot be greater than x.
Both the conditions are not met. So, we can say for sure that x+y is not greater than 0

Answer (C)

Nice question. Thanks for posting. +1 kudos to you! Looks like a 700+ question to me.

(1) |x| > y

So, |x|-y > 0

The modulus can be opened in two ways:

So, x - y >0 ... (a1)

OR

-x - y > 0 ... (b1)

(b1) can be converted as x+y<0 and is sufficient.

However, consider what happens if we take (a1). From (a1), x could be 4 and y could be 2 (x+y > 0) OR x could be -2 and y could be -4 (in which case x+y < 0). Thus, (a1) is not sufficient, although (b1) is sufficient. So, overall, (1) alone is not sufficient as we don't know which of (a1) and (b1) will come into effect.

(2) |y| > x

Opening the modulus,

y > x
so, y-x > 0 ... (a2)

OR

-y > x
so, -x -y > 0
or x+y < 0 ...(b2)


Again, (b2) is sufficient, but (a2) is not (since (a2) is true in both cases viz. x=2, y=4 (in which case x+y is > 0) as well as x = -4 and y = -2 (in which case x+y<0)). So, we can't determine if x+y>0 uniquely from it. Therefore, overall, (2) is not sufficient.

Now, consider both (1) and (2) together.

For both (1) and (2) to hold, we cannot have these two together:
(a1) x-y>0
(a2) y-x>0

That means, the only options left are (b1) and (b2) which give us the same result viz. x+y < 0. This is sufficient to determine whether x+y is > 0. So, both (1) and (2) together are sufficient.

Hence, (C). BOTH TOGETHER ARE SUFFICIENT.

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