It is currently 18 Nov 2017, 01:51

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If x not equal to -y, is (x-y)/(x+y) > 1

Author Message
Manager
Joined: 10 Apr 2008
Posts: 53

Kudos [?]: 7 [1], given: 4

If x not equal to -y, is (x-y)/(x+y) > 1 [#permalink]

### Show Tags

19 Apr 2009, 21:19
1
KUDOS
00:00

Difficulty:

(N/A)

Question Stats:

50% (01:25) correct 50% (01:25) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x not equal to -y, is $$(x-y)/(x+y) > 1$$ ?

1) x>0
2) y<0

The answer explanation uses "number picking" to solve. Is there another way?

Last edited by thinkblue on 19 Apr 2009, 21:35, edited 1 time in total.

Kudos [?]: 7 [1], given: 4

Senior Manager
Joined: 01 Mar 2009
Posts: 367

Kudos [?]: 96 [0], given: 24

Location: PDX
Re: Q128 OG 11th Ed [#permalink]

### Show Tags

19 Apr 2009, 21:24
If (x-y)/(x+y) > 1 Then

x-y > x+y
=> 2y>0 or y>0

From B You have y<0 which answers the question. I haven't looked at the answer but is this correct ?

_________________

In the land of the night, the chariot of the sun is drawn by the grateful dead

Kudos [?]: 96 [0], given: 24

Manager
Joined: 10 Apr 2008
Posts: 53

Kudos [?]: 7 [0], given: 4

Re: Q128 OG 11th Ed [#permalink]

### Show Tags

19 Apr 2009, 21:37
pbanavara wrote:
If (x-y)/(x+y) > 1 Then

x-y > x+y
=> 2y>0 or y>0

From B You have y<0 which answers the question. I haven't looked at the answer but is this correct ?

Nope that's not it. You have to consider (x+y) could be negative. In which case the inequality changes - not completely sure how.
Even after I did that I didn't get the answer though

Kudos [?]: 7 [0], given: 4

Intern
Joined: 13 Apr 2009
Posts: 12

Kudos [?]: 2 [0], given: 0

Re: Q139 OG 11th Ed [#permalink]

### Show Tags

20 Apr 2009, 05:40
so what is the right answer?

I think it is E

Kudos [?]: 2 [0], given: 0

Intern
Joined: 18 Feb 2009
Posts: 5

Kudos [?]: 1 [0], given: 0

Re: Q139 OG 11th Ed [#permalink]

### Show Tags

20 Apr 2009, 07:02
even I got E.. but again by number picking.

For S1.. X>0
Case 1: x=4 y=8 Fraction = -1/3 i.e. <1

Case 2: x=4 y=-6 Fraction = -5 <1

Case 3: x=4 y=-2 Fraction = 3 > 1

For S2.. y< 0
Cas2 and Case 3 proves that S2 is not sufficient

Case 1 & Case 2 also proves that S1 and S2 together are also not sufficient.

But will be good to know if this can be done without number picking ... more than individual signs of X and Y, sign of (X+Y) is more vital.

Thanks
Akshay

Kudos [?]: 1 [0], given: 0

Manager
Joined: 22 Feb 2009
Posts: 136

Kudos [?]: 149 [1], given: 10

Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Re: Q139 OG 11th Ed [#permalink]

### Show Tags

20 Apr 2009, 12:27
1
KUDOS
Is x-y/x+y>1 ?
or we can say is
=> (x-y/x+y)-1>0
=> -2y/x+y >0
=> 2y/x+y<0?

St 1: x>0
Not sufficient to tell whether 2y/x+y < 0

St 2: 2) y<0

Not sufficient to tell whether 2y/x+y < 0

Combining also we cannot say 2y/x+y < 0, therefore E

Kudos [?]: 149 [1], given: 10

Intern
Joined: 14 Apr 2009
Posts: 4

Kudos [?]: 3 [0], given: 0

Re: Q139 OG 11th Ed [#permalink]

### Show Tags

21 Apr 2009, 15:24
bandit wrote:
Is x-y/x+y>1 ?
or we can say is
=> (x-y/x+y)-1>0
=> -2y/x+y >0
=> 2y/x+y<0?

St 1: x>0
Not sufficient to tell whether 2y/x+y < 0

St 2: 2) y<0

Not sufficient to tell whether 2y/x+y < 0

Combining also we cannot say 2y/x+y < 0, therefore E

@bandit : Great technique to look at the equation whether it is less than or equal to zero.
But, When we combine then, then we need to use the numbers to confirm that the answer is truly E.

Kudos [?]: 3 [0], given: 0

Manager
Joined: 19 Aug 2006
Posts: 238

Kudos [?]: 13 [0], given: 0

Re: Q139 OG 11th Ed [#permalink]

### Show Tags

22 Apr 2009, 21:22
I could not see how to solve it other than by picking numbers.
However, I got E.

Kudos [?]: 13 [0], given: 0

Re: Q139 OG 11th Ed   [#permalink] 22 Apr 2009, 21:22
Display posts from previous: Sort by