It is currently 22 Oct 2017, 05:20

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If x is not equal to 0, is |x|<a? 1. x^2<1 2.

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Manager
Joined: 15 Aug 2003
Posts: 50

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

Location: Singapore
If x is not equal to 0, is |x|<a? 1. x^2<1 2. [#permalink]

### Show Tags

24 Jan 2004, 06:17
00:00

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.

If x is not equal to 0, is |x|<a?
1. x^2<1
2. |x|<1/x

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

Manager
Joined: 26 Dec 2003
Posts: 227

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

Location: India

### Show Tags

24 Jan 2004, 06:43
Definetely E as we dont have the information of variable a.

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

Manager
Joined: 15 Aug 2003
Posts: 50

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

Location: Singapore

### Show Tags

24 Jan 2004, 06:50
My apologies,,,the qn is

If x is not equal to 0, is |x|<1?
1. x^2<1
2. |x|<1/x

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

Manager
Joined: 26 Dec 2003
Posts: 227

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

Location: India

### Show Tags

24 Jan 2004, 16:45
Then A? becoz from (1) we can say that x is a fraction and so [x] < 1/x but from (2) we cant say

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

Director
Joined: 28 Oct 2003
Posts: 501

Kudos [?]: 34 [0], given: 0

Location: 55405

### Show Tags

24 Jan 2004, 17:46
I'm thinking that B is sufficient, too.

B) |x|<1/x

B tells us that for positive X values, any number between 0 and 1 works.
It does not work for any negative X values.

So b tells us that 0<x<1, thus making it sufficient.

I also agree that A is sufficient as well.

Kudos [?]: 34 [0], given: 0

Manager
Joined: 26 Dec 2003
Posts: 227

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

Location: India

### Show Tags

24 Jan 2004, 18:12
I think B is ambigious and we need a definite one which can be said from A, So I think its A.

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

Director
Joined: 28 Oct 2003
Posts: 501

Kudos [?]: 34 [0], given: 0

Location: 55405

### Show Tags

24 Jan 2004, 18:26
Quote:
I think B is ambigious

It's not an opinion question.

Disprove my contention.

Kudos [?]: 34 [0], given: 0

Manager
Joined: 26 Dec 2003
Posts: 227

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

Location: India

### Show Tags

24 Jan 2004, 18:42
Stoolfi u only proved that for +ve values it suffices but not for -ve values, so u can say that B is sufficient only if its true for both +ve and -ve values but in this case its not. So I think B is not sufficient. What do u say?

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

Director
Joined: 28 Oct 2003
Posts: 501

Kudos [?]: 34 [0], given: 0

Location: 55405

### Show Tags

24 Jan 2004, 18:56
is |x|<1 is the question.

B tells us, as I have shown, that x is a positive number that is less than 1.

1 will always be greater than all numbers less than 1, by definition.

You contend that I "only proved that for +ve values it suffices but not for -ve values", but B says that x is positive. In other words, there are no negative values of X that make B true.

Kudos [?]: 34 [0], given: 0

Manager
Joined: 26 Dec 2003
Posts: 227

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

Location: India

### Show Tags

24 Jan 2004, 19:12
Ur rite Stoolfi, I agree now.

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

Manager
Joined: 15 Aug 2003
Posts: 50

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

Location: Singapore

### Show Tags

24 Jan 2004, 20:45
U are right, Stoolfi

D is the answer. I had put A as my answer.

I just came up with one other method of answering, by drawing graphs.

the first eq, x^2<1, is a U shaped graph and for all values of x<1, y<1 too. So, A is correct.

For the second eq, |x|<1/x, draw two graphs, y=|x|, which is a 45 degree straight line hinged at (0.0) making a V.
y=1/x is an inverse of y=x.
For all values of graph 1 < graph 2, |x|<1. So, B is correct as well.

=>D is the answer

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

Manager
Joined: 26 Dec 2003
Posts: 227

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

Location: India

### Show Tags

24 Jan 2004, 21:18
Seems to be a nice method

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

24 Jan 2004, 21:18
Display posts from previous: Sort by

# If x is not equal to 0, is |x|<a? 1. x^2<1 2.

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 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®.