Last visit was: 25 Apr 2026, 17:21 It is currently 25 Apr 2026, 17:21
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
User avatar
Dek
Joined: 14 Mar 2007
Last visit: 14 Feb 2008
Posts: 52
Own Kudos:
Posts: 52
Kudos: 34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Dek
Joined: 14 Mar 2007
Last visit: 14 Feb 2008
Posts: 52
Own Kudos:
Posts: 52
Kudos: 34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
bsd_lover
Joined: 17 May 2007
Last visit: 15 Mar 2020
Posts: 2,432
Own Kudos:
Given Kudos: 210
Posts: 2,432
Kudos: 1,750
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
subhen
Joined: 20 Dec 2004
Last visit: 16 Feb 2009
Posts: 74
Own Kudos:
Posts: 74
Kudos: 28
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Answer should be E.

Check out the attachment
Attachments

Case.doc [28 KiB]
Downloaded 82 times

User avatar
Fig
Joined: 01 May 2006
Last visit: 02 Feb 2025
Posts: 1,031
Own Kudos:
Posts: 1,031
Kudos: 253
Kudos
Add Kudos
Bookmarks
Bookmark this Post
(E) for me too :)

Is |x-y|>|x| - |y| ?

From 1
y < x
<=> x - y > 0
So, |x-y| = x - y

Then, I will try to find all cases to eliminate the absolute values from the right side of the inequality.

If 0 < y < x then,
o |x| - |y| = x - y = |x-y|


If y < 0 < x then,
o |x| - |y| = x - (-y) = x + y
and
o x + y < x - y as y < 0

So,
o |x| - |y| = x + y < x - y = |x-y|

If y < x < 0 then,
o |x| - |y| = -x - (-y) = y - x
and
o y-x < 0 < x-y (Condition of stat 1)

Thus,
|x| - |y| < |x-y|

Finally,
o If 0 < y < x, then |x| - |y| = |x-y|
o If y < 0 and y < x, then |x| - |y| < |x-y|

INSUFF.

From 2
x*y > 0
Implies:
o x > 0 and y > 0
or
o x < 0 and y < 0

the cases analysed from 1 are still existing here:
o If 0 < y < x, then |x| - |y| = |x-y|
o If y < 0 and y < x, then |x| - |y| < |x-y|

INSUFF.

From (1) and (2):
We remain with the same cases.

INSUFF.
User avatar
Dek
Joined: 14 Mar 2007
Last visit: 14 Feb 2008
Posts: 52
Own Kudos:
Posts: 52
Kudos: 34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Wow! Thats a lot for 2 mins.
Maybe I'll request GMAT to let me do this during my essay time. :idea:
User avatar
Fig
Joined: 01 May 2006
Last visit: 02 Feb 2025
Posts: 1,031
Own Kudos:
Posts: 1,031
Kudos: 253
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Dek
Wow! Thats a lot for 2 mins.
Maybe I'll request GMAT to let me do this during my essay time. :idea:


I have detailed all here ;)...

Probabbly the best approach is to pick numbers : but u have to be sure to know what u do... In other words, to cover all relevant interval or conditions :)
User avatar
bsd_lover
Joined: 17 May 2007
Last visit: 15 Mar 2020
Posts: 2,432
Own Kudos:
Given Kudos: 210
Posts: 2,432
Kudos: 1,750
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Dek,

To be honest this is one of the lesser time consuming problems that I have seen on this forum. Look at my previous post, I didnt really need to check for too much to be able to prove that both the statements are insufficient.

Fig's reply is outstanding and his comprehensive method will work for pretty much all modulus / inequality type DS questions that most people have trouble with.

cheers,



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!