This is a classic case for picking numbers on DS inequalities.

1.
x = 3, y = 2 -> 1/5
x = 3, y = -2 -> 5/1

2.
x = 3, y = -2 -> 5/1 (REUSE!)
x = -3, y = -2 -> -1/-5 = 1/5

Combined:
x = 3, y = -2 -> 5/1 (REUSE!)
x = 3, y = -4 -> 7/-1

The keys to this strategy:
Pick positives and negatives
Don't pick 1
Pick numbers that are a little bigger and a little smaller than the other numbers you are picking. Notice my use of 2, 3, 4, -2, -3, -4.
Reuse examples whenever you can.
Keep picking different types of numbers. For example, don't pick 3 and -2, and then pick 3 and -1. Both of those give you x + y > 0. We want to mix things up! That's why I chose 3 and -4.

jainrahul1985 wrote:If x not equal to -y, is (x-y)/(x+y) > 1 ?
(1) x > 0
(2) y < 0

OA E

Mate dont mention answer..The fun goes if you read answer before reading ques's.

gmatboost... is there any strategy for doing such questions. i find picking numbers very less convincing. Its a risk a lot of times.

Almost any question can be approached both by picking numbers and using algebra. The specific algebra varies by question, so it's tough to generalize.

Here, when you learn that x > 0 and y < 0, you should be able to figure out a few things:
x - y > 0
x + y could be positive, negative or zero
x - y > x + y

The "cases" to consider here are probably
1. x + y > 0
2. x + y = 0
3. x + y < 0

1. (x-y)/(x+y) is > 1
2. Impossible, cannot divide by 0
3. (x-y)/(x+y) is negative

Every question is a little different, but in general you are often trying to determine:
1. What's positive? What's negative? What could be either?
2. Which is greater?
3. Are we talking about a quantity that is > 1, between 0 and 1, between -1 and 0, or < -1.

These same questions should be in your mind when you choose numbers to plug in.

Hope this is at least a little helpful. Don't hesitate to ask follow up questions!

Hi Greg,

The answer would be C if we were given another condition from the stem. |x| > |y|.
Correct ?

That's correct.
If |x| > |y|, we would know that x + y > 0.

And since we know that (x-y)>(x+y), we could conclude that
(x-y)/(x+y) > 1

option 2 is not useful at all.

E is the choice.
The key clue should be whether x>Y or x<Y