It is currently 24 Nov 2017, 06:20

### 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+y)^2<x^2, which of the following must be true? I.

Author Message
Current Student
Joined: 11 May 2008
Posts: 554

Kudos [?]: 226 [0], given: 0

If (x+y)^2<x^2, which of the following must be true? I. [#permalink]

### Show Tags

29 Aug 2008, 04:11
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If (x+y)^2<x^2, which of the following must be true?
I. x*y<0
II. y<x
III. y*(y+2x)<0
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III

Kudos [?]: 226 [0], given: 0

Manager
Joined: 15 Jul 2008
Posts: 205

Kudos [?]: 71 [0], given: 0

### Show Tags

29 Aug 2008, 04:56
arjtryarjtry wrote:
If (x+y)^2<x^2, which of the following must be true?
I. x*y<0
II. y<x
III. y*(y+2x)<0
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III- this one for me

expand the LHS and cancel the x^2 terms on both sides. 2xy + y^2 < 0 here y^2 is +ve. so xy <0.
also y(2x+y) <0
1 and 3.

Kudos [?]: 71 [0], given: 0

Director
Joined: 01 Jan 2008
Posts: 617

Kudos [?]: 204 [0], given: 1

### Show Tags

29 Aug 2008, 05:32
i second E. Similar logic: 2xy+y^2 < 0 -> 2xy < - y^2 -> xy < 0 and y*(2x+y) < 0

Kudos [?]: 204 [0], given: 1

Director
Joined: 01 Aug 2008
Posts: 727

Kudos [?]: 868 [0], given: 99

### Show Tags

29 Aug 2008, 08:18
I am getting even option II also right which is y < x...

(x+y)^2 < x^2

Case 1):
x+y < x -----> y < 0
Case 2):
x+y < -x ------> 2x+ y < 0
Case 3):
-(x+y) < x = 2x>y ------>y<x

Am I wrong with my case 3?

Kudos [?]: 868 [0], given: 99

SVP
Joined: 07 Nov 2007
Posts: 1790

Kudos [?]: 1091 [0], given: 5

Location: New York

### Show Tags

29 Aug 2008, 08:25
ugimba wrote:
I am getting even option II also right which is y < x...

(x+y)^2 < x^2

Case 1):
x+y < x -----> y < 0
Case 2):
x+y < -x ------> 2x+ y < 0
Case 3):
-(x+y) < x = 2x>y ------>y<x

Am I wrong with my case 3?

do it this way.

(x+y)^2 < x^2
--> x^2+y^2+2xy <x^2
--> y^2+2xy<0
--> y(y+2x)<0
_________________

Smiling wins more friends than frowning

Kudos [?]: 1091 [0], given: 5

Director
Joined: 01 Aug 2008
Posts: 727

Kudos [?]: 868 [0], given: 99

### Show Tags

29 Aug 2008, 08:30
x2suresh thanks for your response. I agree choice E will be the correct answer using your way of solving the queation.

If I use , sqrt ign both sides, I am getting option II which is (y<x)
Shouldn't I do a sqroot to solve the equations?

Kudos [?]: 868 [0], given: 99

SVP
Joined: 07 Nov 2007
Posts: 1790

Kudos [?]: 1091 [0], given: 5

Location: New York

### Show Tags

29 Aug 2008, 09:23
ugimba wrote:
I am getting even option II also right which is y < x...

(x+y)^2 < x^2

Case 1):
x+y < x -----> y < 0
Case 2):
x+y < -x ------> 2x+ y < 0
Case 3):
-(x+y) < x = 2x>y ------>y<x

Am I wrong with my case 3?

Sorry I thought you are talking about III.

(x+y)^2 < x^2 --> means |x+y| <|x|

here you will get the following four solutions
x+y>0 and x>0 --> x+y <x --> y<0
x+y>0 and x<0 --> x+y < -x --> x+x+y<0 --> x<0
x+y<0 and x>0 --> -( x+y) <x --> y<2x
we can say --> y<x (This is only when X>0 AND X+Y <0)
this is one possible soltuions what if X<0.. it leads to other solutions..

Question stem says which of the following "MUST BE TRUE" -- this is ruled out.
_________________

Smiling wins more friends than frowning

Last edited by x2suresh on 29 Aug 2008, 10:26, edited 1 time in total.

Kudos [?]: 1091 [0], given: 5

Director
Joined: 01 Aug 2008
Posts: 727

Kudos [?]: 868 [0], given: 99

### Show Tags

29 Aug 2008, 10:19
Thanks x2suresh for the explanation.

Kudos [?]: 868 [0], given: 99

Re: which is ...?   [#permalink] 29 Aug 2008, 10:19
Display posts from previous: Sort by

# If (x+y)^2<x^2, which of the following must be true? I.

Moderator: chetan2u

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.