Is x negative? 1. x^3(1-x^2)<0 2. x^2-1<0 : DS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 22 Jan 2017, 10:49

### 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. x^3(1-x^2)<0 2. x^2-1<0

Author Message
Senior Manager
Joined: 10 Mar 2008
Posts: 371
Followers: 5

Kudos [?]: 259 [0], given: 0

Is x negative? 1. x^3(1-x^2)<0 2. x^2-1<0 [#permalink]

### Show Tags

17 Jun 2008, 18:05
1
This post was
BOOKMARKED
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. x^3(1-x^2)<0
2. x^2-1<0
Director
Joined: 01 Jan 2008
Posts: 629
Followers: 5

Kudos [?]: 175 [0], given: 1

### Show Tags

17 Jun 2008, 18:09
vksunder wrote:
Is x negative?

1. x^3(1-x^2)<0
2. x^2-1<0

1) equivalent to x*(1-x^2)=x*(1-x)*(1+x) < 0 -> x < -1 or x between 0 and 1 //insufficient
2) (x-1)*(x+1) < 0 -> x is between -1 and 1 -> insufficient

1&2: x is between 0 and 1 -> x negative -> sufficient -> C
Manager
Joined: 06 Feb 2008
Posts: 89
Followers: 1

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

### Show Tags

18 Jun 2008, 03:53
Answer should be E. Will explain if it's right..actually i have typed in full explain but some problem occured in n/w...
Current Student
Joined: 28 Dec 2004
Posts: 3384
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 15

Kudos [?]: 282 [0], given: 2

### Show Tags

18 Jun 2008, 04:50
vksunder wrote:
Is x negative?

1. x^3(1-x^2)<0
2. x^2-1<0

lets see..

1)
x^3<0 or (1-x^2)<0
if X^3<0 then x is negative.. if (1-x^2)<0 then X is a positive integer..Insuff

2) x^2<1 again x=0.5 or -2 Insuff

together..

we know that x^2<1 then it means (1-x^2) >0

which leaves x^3<0...if x^3 is less then zero then X<0

C it is
Director
Joined: 01 Jan 2008
Posts: 629
Followers: 5

Kudos [?]: 175 [0], given: 1

### Show Tags

18 Jun 2008, 05:12
fresinha12 wrote:
1)
x^3<0 or (1-x^2)<0
if X^3<0 then x is negative.. if (1-x^2)<0 then X is a positive integer..Insuff

that's not correct. it's either
a) x^3 < 0 and (1-x^2) > 0 //not just x^3 < 0
or
b) x^3 > 0 and (1-x^2) < 0 //not just 1-x^2 < 0
Current Student
Joined: 28 Dec 2004
Posts: 3384
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 15

Kudos [?]: 282 [0], given: 2

### Show Tags

18 Jun 2008, 05:20
maratikus wrote:
fresinha12 wrote:
1)
x^3<0 or (1-x^2)<0
if X^3<0 then x is negative.. if (1-x^2)<0 then X is a positive integer..Insuff

that's not correct. it's either
a) x^3 < 0 and (1-x^2) > 0 //not just x^3 < 0
or
b) x^3 > 0 and (1-x^2) < 0 //not just 1-x^2 < 0

what i mean is that Either X^3 can be neagtive or (1-x^2) can be negative but not both!
Senior Manager
Joined: 07 Jan 2008
Posts: 297
Followers: 1

Kudos [?]: 42 [0], given: 0

### Show Tags

18 Jun 2008, 06:15
Is x negative?

1. x^3(1-x^2)<0
2. x^2-1<0

1. x (1-x) (1+x) < 0 - insufficient.

2. x^2 -1 < 0 ==> 1-x^2 > 0 => (1-x)(1+x)>0

Combining 1 & 2: x<0

Guys I'm doubtful of my math skills so any response on this approach will help me get that confidence back!
Intern
Joined: 18 Jun 2008
Posts: 42
Followers: 0

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

### Show Tags

19 Jun 2008, 06:25
I will take from what was said above but come to the E conclusion:

1) equivalent to x*(1-x^2)=x*(1-x)*(1+x) < 0 -> *** Here is where I differ. This conclusion gives us -1 < x < 1 but x /= 0. So still insufficient, but it changes the final answer.
2) (x-1)*(x+1) < 0 -> x is between -1 and 1 -> insufficient

The key to this problem is seeing that 1 and 2 are essentially the same information except 1 excludes 0.

So now we combine the two. Both answers allow for a positive or negative result. Try -1/2 or +1/2. So E is the correct answer.
Manager
Joined: 11 Apr 2008
Posts: 128
Location: Chicago
Followers: 1

Kudos [?]: 49 [1] , given: 0

### Show Tags

19 Jun 2008, 07:11
1
KUDOS
jmaynardj wrote:
I will take from what was said above but come to the E conclusion:

1) equivalent to x*(1-x^2)=x*(1-x)*(1+x) < 0 -> *** Here is where I differ. This conclusion gives us -1 < x < 1 but x /= 0. So still insufficient, but it changes the final answer.
2) (x-1)*(x+1) < 0 -> x is between -1 and 1 -> insufficient

The key to this problem is seeing that 1 and 2 are essentially the same information except 1 excludes 0.

So now we combine the two. Both answers allow for a positive or negative result. Try -1/2 or +1/2. So E is the correct answer.

"So now we combine the two. Both answers allow for a positive or negative result. Try -1/2 or +1/2. So E is the correct answer."

If you plug in +1/2 for x in statement (1) the inequality would be x^3 > x^5, not x^3 < x^5. Thus the answer is C
_________________

Factorials were someone's attempt to make math look exciting!!!

Intern
Joined: 18 Jun 2008
Posts: 42
Followers: 0

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

### Show Tags

19 Jun 2008, 07:36
jmaynardj wrote:
I will take from what was said above but come to the E conclusion:

1) equivalent to x*(1-x^2)=x*(1-x)*(1+x) < 0 -> *** Here is where I differ. This conclusion gives us -1 < x < 1 but x /= 0. So still insufficient, but it changes the final answer.
2) (x-1)*(x+1) < 0 -> x is between -1 and 1 -> insufficient

The key to this problem is seeing that 1 and 2 are essentially the same information except 1 excludes 0.

So now we combine the two. Both answers allow for a positive or negative result. Try -1/2 or +1/2. So E is the correct answer.

Ok so 1) equivalent to x*(1-x^2)=x*(1-x)*(1+x) < 0 -> *** Here is where I differ. This conclusion gives us -1 < x < 1 but x /= 0. So still insufficient, but it changes the final answer.

has to be changed to -1 < x < 0 or x > 1. So by combining the two ranges from 1 & 2 you have only -1 < x < 0. Therefore C. Great problem. Hope I don't see anything quite so hard on the real test
Re: DS: Inequalities   [#permalink] 19 Jun 2008, 07:36
Display posts from previous: Sort by