It is currently 20 Oct 2017, 20:47

### 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

# Is x negative ? (1) n3 (1-X2) < 0 (2) X2 - 1 < 0

Author Message
Director
Joined: 05 Jan 2005
Posts: 555

Kudos [?]: 23 [0], given: 0

Is x negative ? (1) n3 (1-X2) < 0 (2) X2 - 1 < 0 [#permalink]

### Show Tags

24 Jul 2005, 09:17
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.

Is x negative ?
(1) n3 (1-X2) < 0
(2) X2 - 1 < 0

Kudos [?]: 23 [0], given: 0

Director
Joined: 27 Dec 2004
Posts: 896

Kudos [?]: 54 [0], given: 0

### Show Tags

24 Jul 2005, 09:26
Hi Arsene, did you mean to write X^2 and n^3 or 2x and 3n? . thnaks

Kudos [?]: 54 [0], given: 0

Senior Manager
Joined: 29 Jun 2005
Posts: 403

Kudos [?]: 21 [0], given: 0

### Show Tags

24 Jul 2005, 11:37
If X2 = X^2, then it doesn't make any sense. The answer is E. X could be anything from -1 to 1.

Kudos [?]: 21 [0], given: 0

Director
Joined: 27 Dec 2004
Posts: 896

Kudos [?]: 54 [0], given: 0

### Show Tags

24 Jul 2005, 12:11
Answer is E and here is why

From stmt 1, n^3 (1 â€“ X^2) < 0
n^3 â€“ n^3 * x^2 < 0
n^3 < n^3*x^2
1 < x^2
Insuff cos x could be -2, or 2

From stmt 2, x^2 -1 < 0
X^2 < 1
Means x could be < 1 and -1

Combining both, they are insuff

Kudos [?]: 54 [0], given: 0

Senior Manager
Joined: 17 Apr 2005
Posts: 372

Kudos [?]: 30 [0], given: 0

Location: India

### Show Tags

25 Jul 2005, 11:24
Arsene_Wenger wrote:
Is x negative ?
(1) n3 (1-X2) < 0
(2) X2 - 1 < 0

E , both the inequalities would be valid for both "+ and - "x alike.

HMTG.

Kudos [?]: 30 [0], given: 0

Director
Joined: 05 Jan 2005
Posts: 555

Kudos [?]: 23 [0], given: 0

### Show Tags

27 Jul 2005, 00:14
sorry guys, that's not what the OA says.

Kudos [?]: 23 [0], given: 0

Senior Manager
Joined: 04 May 2005
Posts: 278

Kudos [?]: 79 [0], given: 0

Location: CA, USA

### Show Tags

27 Jul 2005, 09:57
E

x=-0.01 and x=0.01 both satisfy

n^3(1-x^2) < 0
and
x^2 -1 < 0

of course, n needs to be < 0

Not sure how the OA is not E

Kudos [?]: 79 [0], given: 0

Senior Manager
Joined: 29 Jun 2005
Posts: 403

Kudos [?]: 21 [0], given: 0

### Show Tags

28 Jul 2005, 08:09
Could you post OA and OE?
And please, check the question again. It seems that you missed an important detail.

Kudos [?]: 21 [0], given: 0

Director
Joined: 05 Jan 2005
Posts: 555

Kudos [?]: 23 [0], given: 0

### Show Tags

29 Jul 2005, 03:11
Yeah guys, OA is (E). I am still trying to convince myself on (E) b/c i do not have an OE.

i'll look at qpoo's post now.

Kudos [?]: 23 [0], given: 0

Senior Manager
Joined: 29 Nov 2004
Posts: 478

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

Location: Chicago

### Show Tags

29 Jul 2005, 04:34
Arsene_Wenger wrote:
Yeah guys, OA is (E). I am still trying to convince myself on (E) b/c i do not have an OE.

i'll look at qpoo's post now.

I am pretty sure you accept why COndition 2 is insufficient

As far as condition 1 is concerned i went like this

n^3 - n^3x^2 <0

subtract n^3 on both sides

- n^3x^2 < -n^3

n^3x^2 > n^3

Divide both sides ny n^3 which could be + or -

If it is positive then we get x^2 > 1 (or x > +/-1) so we cannot say
if it is negative then we get x^2 < 1 (or x < +/- 1) so again we cannot say

COmbining both conditions is useless because condition 1 is a subset of condition 2. Hence E
_________________

Fear Mediocrity, Respect Ignorance

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

29 Jul 2005, 04:34
Display posts from previous: Sort by