If X is NOT equal to -Y, is (X-Y)/(X+Y) > 1?

(1) X > 0

(2) Y < 0

A for me too.

(i) If X=0
then for any -ve no. Y the equation (X-Y)/(X+Y) should result in -1.
-1<1 hence sufficient

(ii) Only gives us Y<0 which we already know from the question stem. X could be any no. hence insufficient.

I think it is time for me to go home and sleep.

I don't know why I read X>0 as X=0. And yes it should be B....thanks KillerSquirrel

(E) for me too

(X-Y)/(X+Y) > 1 ?
<=> (X-Y)/(X+Y) - 1 > 0 ?
<=> (X-Y)/(X+Y) - (X+Y)/(X+Y) > 0 ?
<=> -2* Y/(X+Y) > 0 ?
<=> Y/(X+Y) < 0 ?
<=> Sign(Y) = - Sign(X+Y) ?

From 1
X > 0 : No information on Y, and so on the sign of X+Y...

INSUFF.

From 2
Y < 0 : No information on X, and so on the sign of X+Y...

INSUFF.

Both (1) and (2)
Y < 0 and X > 0.... We can not conclude.

Because we have 2 cases:
o |Y| < |X| : X + Y > 0 then yes, Sign(Y) = - Sign(X+Y).
o |Y| > |X| : X + Y < 0 then no, Sign(Y) = Sign(X+Y).

INSUFF.

hello plaguerabbit

This is the method I use when solveing DS problems:

The DS Approach:

1. The principles in DS problems are very basic - what usualy throw you off is the need to find fast those basic principles hiding in the stem.

2. Always write something down from the stem. Play with the information given in the stem. See if you can make an equation out of the information or if an equation is already provided to you then see if you can reduce it or change it to something else. This will help you to better understand the principles you are looking for.

3. Then, see if Statement I helps (if you need cover statement II).

4. Then, see if Statement II helps (if you need cover statement I).

5. Don't do the mistake of using info from statement I in II or the other way around !!!

6. Then see if both Statements combined help.

7. Im some cases the answer will be (E) but you might be tempted to think that the answer is (C) and that you don't know enough Quant to get to the answer. Resist this temptation to choose (C) and go with (E) - more often than not you will be right.

8. But beware ! In the real GMAT the answer will rarely be (E) - I think they assume this will be the obvious pick for those who don't know how to solve the problem given.

As for myself , DS was always a big problem, I'm doing very well on PS and have very good quantitative skills - but when I first started solveing DS problems, my success rate was something like 4 out of 10 for me - 'practice makes perfect' - and after a while you start to see the beauty in DS problems (yes ! there is such beauty).

If you have the time I strongly recommend that you try and write your own DS and PS problems (just for practice & fun). write one PS problem and one DS problem, then you will see that PS problems are kind of lame in comparison to DS problems.

TIP: I would use the number picking strategy for this kind of problem.

Check the special cases if in doubt (a special case would be when you pick a number that makes the nominator equal to zero and positive or negative etc...)

One more E.

1) INSUFF we don't know Y

2) INSUFF it can be both + and -

1) + 2) INSUFF we don't know the magnitude of X and Y so still, it can be both + and -


Hence E.

[quote="LM"]If X is NOT equal to -Y, is (X-Y)/(X+Y) > 1?

(1) X > 0

(2) Y <0>:-X

Just try a bunch of numbers. Say X=4 Y=5.

4-5/4+5>1 Not true.

Say X=16 Y=-4.

16+4/16-4=20/12 which is >1.

Insuff.

S2:
Y is negative. X=4, Y=-2

4+2/4-2=3>1

X=1, Y=-2.

3/-1-->-3>1 Not true.


So E.

St1:
Case1:
x = 2, y = 3, then (X-Y)/(X+Y) <1> 1
Insufficient.

St2:
Case1:
x = 2, y = -3, then (X-Y)/(X+Y) <1> 1.
Insufficient.

Using both st1 and st2:
Relate to the argument in st2. X is positive, y is negative, but (X-Y)/(X+Y) may or may not be less than 1. Insufficient.

Ans E

I go for E

1) definitely Insufficient

2) Insufficient since it will depend on x (which can be virtually any number)

together NOT Sufficient

check by once |y|>|x| than |y|<|x| you will see that they are different

SO, ANS: E

I DON"T UNDERSTAND WHAT ARE YOU DOUBTING?????