Last visit was: 12 Jul 2025, 11:57 It is currently 12 Jul 2025, 11:57
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
705-805 Level|   Inequalities|            
User avatar
ccooley
User avatar
Manhattan Prep Instructor
Joined: 04 Dec 2015
Last visit: 06 Jun 2020
Posts: 931
Own Kudos:
1,620
 [1]
Given Kudos: 115
GMAT 1: 790 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 790 Q51 V49
GRE 1: Q170 V170
Posts: 931
Kudos: 1,620
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
avatar
OjhaShishir
Joined: 19 Dec 2019
Last visit: 10 Mar 2021
Posts: 31
Own Kudos:
Given Kudos: 6
Posts: 31
Kudos: 20
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 12 July 2025
Posts: 102,636
Own Kudos:
740,734
 [1]
Given Kudos: 98,172
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,636
Kudos: 740,734
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
PyjamaScientist
User avatar
Admitted - Which School Forum Moderator
Joined: 25 Oct 2020
Last visit: 15 May 2025
Posts: 1,118
Own Kudos:
Given Kudos: 633
GMAT 1: 740 Q49 V42 (Online)
Products:
GMAT 1: 740 Q49 V42 (Online)
Posts: 1,118
Kudos: 1,239
Kudos
Add Kudos
Bookmarks
Bookmark this Post
jusjmkol740
If \(x ≠ -y\), then is \(\frac{x-y}{x + y} > 1\)?


(1) \(x > 0\)

(2) \(y < 0\)
IanStewart chetan2u could you please see if this is right?
Is \(\frac{x-y}{x + y} > 1\) ?
Multiply both sides by \((x+y)^2\), although we don't know the sign of \(x \) and \(y\), we can be sure that \((x+y)^2\) is positive. So, it can be multiplied on both sides.
Is \(\frac{(x+y)^2*x-y}{x + y} > (x+y)^2*1\) ?
Is \((x+y)*(x-y) > (x+y)^2\) ?
Is \((x^2 - y^2) > (x^2 + y^2 + 2xy)\) ?
Is \((- y^2) > (y^2 + 2xy)\)
Is \((- 2y^2) > (2xy)\) ?
Is \((y^2) < (xy)\) ?
Multiply both sides by \(1/y{^2}\),
Is \( 1< (x/y)\) ?
User avatar
IanStewart
User avatar
GMAT Tutor
Joined: 24 Jun 2008
Last visit: 12 Jul 2025
Posts: 4,141
Own Kudos:
10,617
 [4]
Given Kudos: 97
 Q51  V47
Expert
Expert reply
Posts: 4,141
Kudos: 10,617
 [4]
4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PyjamaScientist
chetan2u could you please see if this is right?
Is \(\frac{x-y}{x + y} > 1\) ?
Multiply both sides by \((x+y)^2\), although we don't know the sign of \(x \) and \(y\), we can be sure that \((x+y)^2\) is positive. So, it can be multiplied on both sides.
Is \(\frac{(x+y)^2*x-y}{x + y} > (x+y)^2*1\) ?
Is \((x+y)*(x-y) > (x+y)^2\) ?
Is \((x^2 - y^2) > (x^2 + y^2 + 2xy)\) ?
Is \((- y^2) > (y^2 + 2xy)\)
Is \((- 2y^2) > (2xy)\) ?
Is \((y^2) < (xy)\) ?
Multiply both sides by \(1/y{^2}\),
Is \( 1< (x/y)\) ?

It's right up until the point I've highlighted. When you wrote the highlighted line, a negative sign disappeared (I think you were dividing by -2 on both sides, but then the right side should become negative). The other steps all look good -- you can certainly multiply or divide both sides of an inequality by a nonzero square, since squares are never negative.
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 12 Jul 2025
Posts: 11,295
Own Kudos:
41,699
 [1]
Given Kudos: 333
Status:Math and DI Expert
Products:
Expert
Expert reply
Posts: 11,295
Kudos: 41,699
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PyjamaScientist
jusjmkol740
If \(x ≠ -y\), then is \(\frac{x-y}{x + y} > 1\)?


(1) \(x > 0\)

(2) \(y < 0\)
IanStewart chetan2u could you please see if this is right?
Is \(\frac{x-y}{x + y} > 1\) ?
Multiply both sides by \((x+y)^2\), although we don't know the sign of \(x \) and \(y\), we can be sure that \((x+y)^2\) is positive. So, it can be multiplied on both sides.
Is \(\frac{(x+y)^2*x-y}{x + y} > (x+y)^2*1\) ?
Is \((x+y)*(x-y) > (x+y)^2\) ?
Is \((x^2 - y^2) > (x^2 + y^2 + 2xy)\) ?
Is \((- y^2) > (y^2 + 2xy)\)
Is \((- 2y^2) > (2xy)\) ?
Is \((y^2) < (xy)\) ?
Multiply both sides by \(1/y{^2}\),
Is \( 1< (x/y)\) ?

IanStewart has of course told you exactly where you have gone wrong.

Is \((- 2y^2) > (2xy)\) ?
Is \(-y^2>xy\) ? Or
Is \((y^2) <- (xy)\) ?
Multiply both sides by \(1/y{^2}\),
Is \( 1< -(x/y)\) ?

The two statements combined tell us that both are of opposite sign so x/y<0, hence -x/y>0.
Two possibilities
a) 1>x/y>0
b) x/y>1>0
But will it be between 0 and 1 or >1 cannot be told from given information
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,371
Own Kudos:
Posts: 37,371
Kudos: 1,010
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
   1   2 
Moderator:
Math Expert
102636 posts