Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 26 Apr 2015, 01:57

### 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≠0, is |x| <1? (1) x2<1 (2) |x| < 1/x

Author Message
TAGS:
Senior Manager
Joined: 25 Jun 2009
Posts: 309
Followers: 2

Kudos [?]: 86 [0], given: 6

If x≠0, is |x| <1? (1) x2<1 (2) |x| < 1/x [#permalink]  26 Jul 2009, 23:46
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
. If x≠0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

Thanks
Current Student
Affiliations: ?
Joined: 20 Jul 2009
Posts: 191
Location: Africa/Europe
Schools: Kellogg; Ross (); Tuck
Followers: 2

Kudos [?]: 31 [0], given: 6

Re: DS Question [#permalink]  27 Jul 2009, 00:35
IMO D

1) is sufficient
If x² <1 it means that x is between -1 and 1.
So |x| < 1

In statement 2, we can notice that x can not be negative. So for any positive x, |x| = x and then x<1/x means that (x²-1 )/x <0 ==> x²-1 < 0 same as statement 1) sufficient.

Director
Joined: 01 Apr 2008
Posts: 906
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 18

Kudos [?]: 302 [0], given: 18

Re: DS Question [#permalink]  27 Jul 2009, 00:52
nitishmahajan wrote:
. If x≠0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

D.
1) x^2<1 implies that -1<x<1 which is nothing but |x| < 1.
SUFF.
2)
If x>0, then the inequality becomes, x < 1/x. Is this possible for integers >= 1 ? No, because for integers >=1 the reciprocal is always smaller than the integer itself.

So we have to think about fractions between 0 and 1. Reciprocal of this fraction will an integer greater than 1. So all the fractions between 0 and 1 will satisfy the equality.

If x<0, then LHS becomes +ve and RHS becomes negative. Is this possible ? No, so anything <0 will not satisfy the inequality.

Hence, we can say for sure that if |x|<1, then |x|<1/x.
SUFF.
Senior Manager
Joined: 25 Jun 2009
Posts: 309
Followers: 2

Kudos [?]: 86 [0], given: 6

Re: DS Question [#permalink]  27 Jul 2009, 01:06
Economist wrote:
nitishmahajan wrote:
. If x≠0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

D.
1) x^2<1 implies that -1<x<1 which is nothing but |x| < 1.
SUFF.
2)
If x>0, then the inequality becomes, x < 1/x. Is this possible for integers >= 1 ? No, because for integers >=1 the reciprocal is always smaller than the integer itself.

So we have to think about fractions between 0 and 1. Reciprocal of this fraction will an integer greater than 1. So all the fractions between 0 and 1 will satisfy the equality.

If x<0, then LHS becomes +ve and RHS becomes negative. Is this possible ? No, so anything <0 will not satisfy the inequality.

Hence, we can say for sure that if |x|<1, then |x|<1/x.
SUFF.

Thanks for nice explanation ! It helped !
Senior Manager
Joined: 25 Jun 2009
Posts: 309
Followers: 2

Kudos [?]: 86 [0], given: 6

Re: DS Question [#permalink]  27 Jul 2009, 01:07
IMO D

1) is sufficient
If x² <1 it means that x is between -1 and 1.
So |x| < 1

In statement 2, we can notice that x can not be negative. So for any positive x, |x| = x and then x<1/x means that (x²-1 )/x <0 ==> x²-1 < 0 same as statement 1) sufficient.

OA is D only, thanks for he explanation !:)
Manager
Joined: 07 Apr 2009
Posts: 145
Followers: 1

Kudos [?]: 9 [0], given: 3

Re: DS Question [#permalink]  27 Jul 2009, 12:32
hey , I did not get this can you please elobarate ?

Quote:
If x<0, then LHS becomes +ve and RHS becomes negative. Is this possible ? No, so anything <0 will not satisfy the inequality.

Hence, we can say for sure that if |x|<1, then |x|<1/x.
SUFF.

if x < 0

then the inequation becomes -x < 1/x ==> $$-x^2 < 1$$==> $$x^2 > -1$$ how can we rule this out ?

please throw some light on this..
Director
Joined: 01 Apr 2008
Posts: 906
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 18

Kudos [?]: 302 [0], given: 18

Re: DS Question [#permalink]  27 Jul 2009, 14:03
skpMatcha wrote:
hey , I did not get this can you please elobarate ?
if x < 0

then the inequation becomes -x < 1/x ==> $$-x^2 < 1$$==> $$x^2 > -1$$ how can we rule this out ?

please throw some light on this..

There are two problems in what you have derived:

1. if x<0, then value of |x| is always x. Note that |x| is ALWAYS positive no matter whatever is the value of x. That is why we can say that when x<0, |x| = -x....because -(-x) will be positive x.

2. This is not related to your question because if x<0 your derivation -x < 1/x is wrong(as I explained above)..it should be x < -1/x.

Now, just analyzing your statement, If x<0 and -x < 1/x ==> [m]-x^2 < 1.
Here, if -x < 1/x then we can derive -x^2 < 1 only if x > 0. If x<0, then it becomes -x^2 > 1.
Manager
Joined: 21 Jun 2009
Posts: 156
Followers: 3

Kudos [?]: 8 [0], given: 1

Re: DS Question [#permalink]  27 Jul 2009, 15:20
Yep D.

Good explaination econ.
Re: DS Question   [#permalink] 27 Jul 2009, 15:20
Similar topics Replies Last post
Similar
Topics:
17 If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x 16 08 Apr 2012, 08:43
If x 0, is |x| <1? (1) x^2<1 (2) |x| < 1/2 3 26 Feb 2011, 15:23
1 If x ≠ 0, is x^2/|x|< 1? (1) x < 1 (2) x > −1 4 13 Oct 2008, 20:30
If x 0, is |x| <1? (1) x^2<1 (2) |x| < 1/x 6 03 Aug 2008, 18:53
If x≠0, is |x| <1? (1) x^2<1 (2) |x| < 1/x 7 14 Dec 2007, 21:37
Display posts from previous: Sort by

# If x≠0, is |x| <1? (1) x2<1 (2) |x| < 1/x

 Powered by phpBB © phpBB Group and phpBB SEO 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®.