whichscore wrote:

if x is different from -y, in (x-y)/(x+y) greater than 1 ?

1. x > 0

2. y< 0

Even though fluke has already provided the detailed solution, there are a couple of points I would like to reinforce here.

People often get confused when they read 'if x is different from -y'. Why do they give this information? They do that because you have (x+y) in the denominator and hence it cannot be 0. That is, x + y = 0 or x = -y is not valid. Hence, they are just clarifying that the fraction is indeed defined.

Also, we are used to having 0 on the right of an inequality. What do we do when we have a 1? We take the 1 to the left hand side and get a 0 on the right.

Is

\frac{(x-y)}{(x+y)} > 1? (Don't forget it is a question, not given information)

Is

\frac{(x-y)}{(x+y)} - 1 > 0?

which simplifies to: Is y/(x+y) < 0 ?

We know how to deal with this inequality! Note here that we have no information about x and y as yet (except that x is not equal to -y which is more of a technical issue rather than actual information)

1. x > 0

No information about y so not sufficient.

2. y< 0

No information about x so not sufficient.

Both together, we know that y is negative. We need the sign of (x+y) now. But (x+y) may be positive or negative depending on whether x or y has greater absolute value. Hence not sufficient.

Answer (E)

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