GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Oct 2018, 09:36

### 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

# Is x < y?

Author Message
TAGS:

### Hide Tags

Intern
Joined: 19 Oct 2013
Posts: 9
Location: United States
Concentration: Finance, Technology
GMAT Date: 11-06-2013
GPA: 3.5
WE: Engineering (Investment Banking)

### Show Tags

Updated on: 01 Nov 2013, 01:01
1
3
00:00

Difficulty:

55% (hard)

Question Stats:

61% (02:02) correct 39% (02:04) wrong based on 152 sessions

### HideShow timer Statistics

Is x < y?

(1) x^2 - y^2 < 0
(2) -x - y < 0

Originally posted by Puneethrao on 31 Oct 2013, 20:02.
Last edited by Bunuel on 01 Nov 2013, 01:01, edited 2 times in total.
Edited the question.
Intern
Joined: 29 Oct 2013
Posts: 18
Re: Is x < y?  [#permalink]

### Show Tags

31 Oct 2013, 22:15
2
Puneethrao wrote:
Is x < y?

x2 - y2 < 0
-x - y <0

St.1- can be rearranged as x^2 If x=1, y=2, the answer would be YES.
If x=1, y=-2, Its a NO. So 1 is insufficient

St.2- Again it can be (2, -1) or (-1, 2) or such values. Insufficient

(1)+(2)--> (x^2-y^2) and -(x+y) are both negative so their division must be positive.

(x^2-y^2)/-(x+y)> 0
(x+y)(x-y)/-(x+y)> 0
-(x-y)> 0
x-y <0
So x Therefore (1)+(2) is sufficient.

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 49915
Re: Is x < y?  [#permalink]

### Show Tags

01 Nov 2013, 01:01
Puneethrao wrote:
Is x < y?

(1) x^2 - y^2 < 0
(2) -x - y <0

Is x < y?

Is x < y? --> is x - y < 0?

(1) x^2 - y^2 < 0 --> x^2<y^2 --> |x|<|y|. Not sufficient. Consider: (x, y)=(1, -2) and (x, y)=(1, 2).

(2) -x - y < 0 --> x+y>0. The sum of two numbers is greater than 0, from this we cannot say which one is greater. Consider (x, y)=(1, 2) and (x, y)=(2, 1). Not sufficient.

(1)+(2) From (1) we have that (x-y)(x+y)<0 and from (2) we have that x+y>0, thus x-y<0. Sufficient.

_________________
Manager
Joined: 24 Oct 2013
Posts: 146
Schools: LBS '18
GMAT 1: 720 Q49 V38
WE: Design (Transportation)
Re: Is x < y?  [#permalink]

### Show Tags

18 Apr 2014, 23:14
1
x^2 - y^2 < 0 --> x^2>y^2

is that a typo or did you mean that?
Math Expert
Joined: 02 Sep 2009
Posts: 49915
Re: Is x < y?  [#permalink]

### Show Tags

19 Apr 2014, 03:25
gauravkaushik8591 wrote:
x^2 - y^2 < 0 --> x^2>y^2

is that a typo or did you mean that?

Yes, it was a typo. Edited. Thank you.
_________________
Current Student
Joined: 23 May 2013
Posts: 188
Location: United States
Concentration: Technology, Healthcare
Schools: Stanford '19 (M)
GMAT 1: 760 Q49 V45
GPA: 3.5
Re: Is x < y?  [#permalink]

### Show Tags

23 Sep 2014, 07:42
1
This question is also very easily solved algebraically.

1) $$x^2 - y^2 <0. (x+y)(x-y) <0$$. One of these is negative, so either$$x+y<0$$, which means that $$x<-y$$, or $$x<y$$. Two options; not sufficient.

2) $$-x-y<0$$. Not sufficient.

Using number 2), we can multiply by a negative to see that $$x+y>0$$. Now using option 1, this means that x-y must be less than 0. Therefore, $$x-y<0$$ and $$x<y$$. Sufficient.

Non-Human User
Joined: 09 Sep 2013
Posts: 8415
Re: Is x < y?  [#permalink]

### Show Tags

02 Apr 2018, 05:54
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Is x < y? &nbs [#permalink] 02 Apr 2018, 05:54
Display posts from previous: Sort by