It is currently 20 Sep 2017, 22:17

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

DS mod

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Director
Director
avatar
Joined: 20 Sep 2006
Posts: 653

Kudos [?]: 132 [0], given: 7

DS mod [#permalink]

Show Tags

New post 29 Sep 2008, 20:07
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) 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.

Attachment:
m_2.JPG
m_2.JPG [ 29.98 KiB | Viewed 958 times ]

Kudos [?]: 132 [0], given: 7

SVP
SVP
avatar
Joined: 17 Jun 2008
Posts: 1540

Kudos [?]: 278 [0], given: 0

Re: DS mod [#permalink]

Show Tags

New post 29 Sep 2008, 22:02
C.

What the two statements are really saying is that mod of (x-3) will be absolute positive and that positive number is greater than y and less than -y. This is possible only when y = 0 .

Now, if we re-write the two statements then
|x-3| >= 0 and |x-3| <=0

Again, both the statements will be true only if |x-3| = 0 and that means x = 3.

Kudos [?]: 278 [0], given: 0

Intern
Intern
avatar
Joined: 29 Sep 2008
Posts: 46

Kudos [?]: 9 [0], given: 0

Re: DS mod [#permalink]

Show Tags

New post 30 Sep 2008, 16:38
rao, can you post the OA please.

I think it should be B.
Statement I is saying |x-3| is greater than or equal to a positive or zero number - y. Since the mod of any number is always positive or zero ,this leaves a lot of possibilities open.

statement II is saying that |x-3| is less than or equal to a number that is either negative or zero (remember that y is greater than or equal to zero). Since |x-3| can never be negative, it can never be less than a negative number, thus it has to be equal to zero. |x-3| = 0 => x = 3

Kudos [?]: 9 [0], given: 0

VP
VP
User avatar
Joined: 18 May 2008
Posts: 1262

Kudos [?]: 510 [0], given: 0

Re: DS mod [#permalink]

Show Tags

New post 30 Sep 2008, 17:33
I think we dnt need 2 combine the two equations.
S1 says that |x-3| >= 0 which is can give either x=3 or x>3 Insufficient
S2 says that |x-3| <=0 which is only possible when x=0 because modulus cn nev be less than 0
So I think B should be the ans

scthakur wrote:
C.

What the two statements are really saying is that mod of (x-3) will be absolute positive and that positive number is greater than y and less than -y. This is possible only when y = 0 .

Now, if we re-write the two statements then
|x-3| >= 0 and |x-3| <=0

Again, both the statements will be true only if |x-3| = 0 and that means x = 3.

Last edited by ritula on 01 Oct 2008, 16:44, edited 1 time in total.

Kudos [?]: 510 [0], given: 0

SVP
SVP
avatar
Joined: 17 Jun 2008
Posts: 1540

Kudos [?]: 278 [0], given: 0

Re: DS mod [#permalink]

Show Tags

New post 01 Oct 2008, 12:01
I got what you are saying. Question already gives clue that y >=0 and hence B should be sufficient.

aim2010 wrote:
rao, can you post the OA please.

I think it should be B.
Statement I is saying |x-3| is greater than or equal to a positive or zero number - y. Since the mod of any number is always positive or zero ,this leaves a lot of possibilities open.

statement II is saying that |x-3| is less than or equal to a number that is either negative or zero (remember that y is greater than or equal to zero). Since |x-3| can never be negative, it can never be less than a negative number, thus it has to be equal to zero. |x-3| = 0 => x = 3

Kudos [?]: 278 [0], given: 0

SVP
SVP
User avatar
Joined: 07 Nov 2007
Posts: 1795

Kudos [?]: 1032 [0], given: 5

Location: New York
Re: DS mod [#permalink]

Show Tags

New post 01 Oct 2008, 12:48
rao_1857 wrote:
Attachment:
m_2.JPG


|x-3|<= -y (Y is positve or zero)

|x-3| --> must be postive or zero.
-y is always -ve

so only possibility is |x-3|=0 -->x=3

B is the answer.
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

Kudos [?]: 1032 [0], given: 5

Director
Director
avatar
Joined: 20 Sep 2006
Posts: 653

Kudos [?]: 132 [0], given: 7

Re: DS mod [#permalink]

Show Tags

New post 02 Oct 2008, 06:46
aim2010 wrote:
rao, can you post the OA please.

I think it should be B.
Statement I is saying |x-3| is greater than or equal to a positive or zero number - y. Since the mod of any number is always positive or zero ,this leaves a lot of possibilities open.

statement II is saying that |x-3| is less than or equal to a number that is either negative or zero (remember that y is greater than or equal to zero). Since |x-3| can never be negative, it can never be less than a negative number, thus it has to be equal to zero. |x-3| = 0 => x = 3


Thanks aim2010 thanks for clarifying ... OA is B

Kudos [?]: 132 [0], given: 7

Re: DS mod   [#permalink] 02 Oct 2008, 06:46
Display posts from previous: Sort by

DS mod

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.