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# If X is NOT equal to -Y, is (X-Y)/(X+Y) > 1? (1) X > 0

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Director
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If X is NOT equal to -Y, is (X-Y)/(X+Y) > 1? (1) X > 0 [#permalink]

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17 Jun 2007, 19:04
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If X is NOT equal to -Y, is (X-Y)/(X+Y) > 1?

(1) X > 0

(2) Y < 0

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Senior Manager
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19 Jun 2007, 13:22
A for me. Will explain if I am right

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19 Jun 2007, 13:29
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.

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VP
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19 Jun 2007, 13:53
this post was edited !!

the answer is (E)

since (X+Y) can be negative and if so we have to flip signs !!!

(X-Y) < 1*(X+Y)

but if (X+Y) is positive then:

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

X-Y > 1(X+Y)

X-Y > X+Y

0 > 2*Y

Statement 1

X > 0

can get any value depending on Y

insufficient

Statement 2

Y < 0

since 0 > 2*Y this will be false every time !

sufficent - but we can never know if (X+Y) is positive or negative !

Statement 1&2

still insufficient

the answer is (E) - see my above comment !

Last edited by KillerSquirrel on 19 Jun 2007, 14:12, edited 2 times in total.

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Intern
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19 Jun 2007, 14:02
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

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19 Jun 2007, 14:10
ridhimagupta wrote:
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

the answer for me is (E) - sorry I have edited my post !

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19 Jun 2007, 14:22
(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.

Last edited by Fig on 19 Jun 2007, 14:22, edited 1 time in total.

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Manager
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19 Jun 2007, 14:22
hey killersquirrel,

i got the same answer as you, but i did it by simply plugging in numbers and looking for counterexamples for each statement.

how do you normally approach DS inequality problems like this?

is there a logical process you go through in your head once you see a DS inequality? i tend to have a lot of difficulty with DS problems where it's really tough to see where to start. usually i can get the right answer given a fair amount of time, and most oft the time when i can't solve it i have an educated guess from PoE, but i'd like to know what your thought process is on the tough DS questions and if you have any tips.

thanks!

btw have you taken your gmat yet? you seem like you'd score just fine

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19 Jun 2007, 14:48
plaguerabbit wrote:
hey killersquirrel,

i got the same answer as you, but i did it by simply plugging in numbers and looking for counterexamples for each statement.

how do you normally approach DS inequality problems like this?

is there a logical process you go through in your head once you see a DS inequality? i tend to have a lot of difficulty with DS problems where it's really tough to see where to start. usually i can get the right answer given a fair amount of time, and most oft the time when i can't solve it i have an educated guess from PoE, but i'd like to know what your thought process is on the tough DS questions and if you have any tips.

thanks!

btw have you taken your gmat yet? you seem like you'd score just fine

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.

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Re: DS- number properties [#permalink]

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20 Jun 2007, 23:14
LM wrote:
If X is NOT equal to -Y, is (X-Y)/(X+Y) > 1?

(1) X > 0

(2) Y < 0

guys, i'm not sure if i comprehend your explanation

(X-Y)/(X+Y)

= (1-Y/X)/(1+Y/X)

now if Y is negative and X is positive, Y/X is negative.

so (1-Y/X)/(1+Y/X) = (1+positive real number)/(1-positive real number)

this indeed is greater than 1. so both statements together are sufficient.

answer should be C.
_________________

for every person who doesn't try because he is
afraid of loosing , there is another person who
keeps making mistakes and succeeds..

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Manager
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21 Jun 2007, 01:31
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...)

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21 Jun 2007, 06:34
ronron wrote:
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...)

Well, the stem of the question specifically says that X is not equal to -Y. This implies that X+Y is not equal to 0. If X is positive, dividing both numerator and denominator by X would not change their signs. So, this is not the special case as you just mentioned.
_________________

for every person who doesn't try because he is
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keeps making mistakes and succeeds..

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21 Jun 2007, 07:04
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.

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10 Sep 2007, 20:38
I agree that ST. 1 is insuff.
but how is ST. 2 insuff?
even after reading killersquirrel & fig's posts, i am not getting it

ST 2 says Y<0 & X-Y/X+Y>1
i.e., is X-Y> X+Y?

if Y<0, let Y= -1
if X is positive, let X=2
2+1>2-1
3>1 yes

if X is negative, let X= -2
-2-(-1)>-2-1
-1>-3 yes

ST. 2 seems sufficient to me, shouldn't answer be B?

Last edited by r019h on 10 Sep 2007, 21:23, edited 1 time in total.

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CEO
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Re: DS- number properties [#permalink]

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10 Sep 2007, 21:03
[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.

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Re: DS- number properties [#permalink]

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10 Sep 2007, 21:20
GMATBLACKBELT wrote:
LM wrote:
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.

when you try X= 1, Y= -2
it is 1+2>1-2
3> -1, how is it Not true???

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10 Sep 2007, 23:12
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

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Manager
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11 Sep 2007, 13:54
E for me too.
After long calculations i have reached this conclusion. It is a tricky q. Hope i find all the questions of such intellectual level.
Because this means i am probably scoring high.

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12 Sep 2007, 12:25
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?????

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Director
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Re: DS- number properties [#permalink]

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16 Sep 2007, 10:07
Given that x != (-y)

ST1:

x>0

Let's say x = 1 and y = 2
(we can not take y = (-1) since x != (-y))

Now, (x-y)/(x+y) = -(1/3) < Zero

Not sufficient.

ST2:

Given that y<0

Let us take x = (-1) and y = (-2)

Now, (x-y)/(x+y) = -(1/3) < Zero

Insufficient.

Let's take ST1 and ST2 together:

x>(0)

y<(0)

Let's say x = 1 and y = -2

So, (x-y)/(x+y) = (-3) hence Not Sufficient.

Answer should be E.

- Brajesh

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Re: DS- number properties   [#permalink] 16 Sep 2007, 10:07
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