GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 24 Apr 2019, 01:47

### 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

# Is x greater than x^3 ?

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 25 Nov 2006
Posts: 327
Schools: St Gallen, Cambridge, HEC Montreal
Is x greater than x^3 ?  [#permalink]

### Show Tags

Updated on: 20 Apr 2014, 12:26
2
00:00

Difficulty:

75% (hard)

Question Stats:

52% (02:18) correct 48% (02:23) wrong based on 255 sessions

### HideShow timer Statistics

Is x greater than x^3 ?

(1) x is negative

(2) x^2 - x^3 > 2

Originally posted by lumone on 09 Feb 2009, 13:43.
Last edited by Bunuel on 20 Apr 2014, 12:26, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
SVP
Joined: 07 Nov 2007
Posts: 1632
Location: New York

### Show Tags

09 Feb 2009, 13:57
lumone wrote:
Is x greater than x^3

(1) x is negative

(2) x^2 - x^3 > 2

1)

x =-2 x^3=-8 x>x^3
x=-1/2 x^3 = -1/8 x<x^3

not sufficient

2)

x^2 (1 - x) > 2

x^2 is always postive.|.. so (1-x) need to be positve and x can't be fractional values between -1 and 1

x must be -ve value < -1 i.e

x > x^3

sufficient.

B.
_________________
Smiling wins more friends than frowning
Intern
Joined: 04 Aug 2011
Posts: 5

### Show Tags

20 Apr 2014, 10:15
x2suresh wrote:
lumone wrote:
Is x greater than x^3

(1) x is negative

(2) x^2 - x^3 > 2

1)

x =-2 x^3=-8 x>x^3
x=-1/2 x^3 = -1/8 x<x^3

not sufficient

2)

x^2 (1 - x) > 2

x^2 is always postive.|.. so (1-x) need to be positve and x can't be fractional values between -1 and 1

x must be -ve value < -1 i.e

x > x^3

sufficient.

B.

This is a really old question, and I found it recently in Jeff Sackmann's Data Sufficiency Challenge question bank. I am afraid I don't understand why Statement (2) above is sufficient, particularly the highlighted part above.

My interpretation is this: In order to keep x^2 > x^3, (1-x) > 0 (i.e. positive). This implies that 1 > x. x could therefore be negative integer, negative fraction or positive fraction. For instance:

x = -2 => x^2 = 4 and x^3 = -8 => x^2 > x^3 and 1-(-2) > 0. So, x > x^3

x = -1/2 => x^2 = 1/4 and x^3 = -1/8 => x^2 > x^3 and 1-(-1/2) > 0. However, x < x^3

x = 1/2 => x^2 = 1/4 and x^3 = 1/8 => x^2 > x^3 and 1-(1/2) > 0. So, x > x^3

Clearly, x CAN be fractional values between -1 and 1 for x^2 > x^3. But each time, the relation between x and x^3 may change. So B is not sufficient.

Can somebody please explain where I am going wrong? Or provide an alternative explanation. Thanks a lot!
Math Expert
Joined: 02 Sep 2009
Posts: 54493

### Show Tags

21 Apr 2014, 05:23
DrFawkes wrote:
x2suresh wrote:
lumone wrote:
Is x greater than x^3

(1) x is negative

(2) x^2 - x^3 > 2

1)

x =-2 x^3=-8 x>x^3
x=-1/2 x^3 = -1/8 x<x^3

not sufficient

2)

x^2 (1 - x) > 2

x^2 is always postive.|.. so (1-x) need to be positve and x can't be fractional values between -1 and 1

x must be -ve value < -1 i.e

x > x^3

sufficient.

B.

This is a really old question, and I found it recently in Jeff Sackmann's Data Sufficiency Challenge question bank. I am afraid I don't understand why Statement (2) above is sufficient, particularly the highlighted part above.

My interpretation is this: In order to keep x^2 > x^3, (1-x) > 0 (i.e. positive). This implies that 1 > x. x could therefore be negative integer, negative fraction or positive fraction. For instance:

x = -2 => x^2 = 4 and x^3 = -8 => x^2 > x^3 and 1-(-2) > 0. So, x > x^3

x = -1/2 => x^2 = 1/4 and x^3 = -1/8 => x^2 > x^3 and 1-(-1/2) > 0. However, x < x^3

x = 1/2 => x^2 = 1/4 and x^3 = 1/8 => x^2 > x^3 and 1-(1/2) > 0. So, x > x^3

Clearly, x CAN be fractional values between -1 and 1 for x^2 > x^3. But each time, the relation between x and x^3 may change. So B is not sufficient.

Can somebody please explain where I am going wrong? Or provide an alternative explanation. Thanks a lot!

I think you misread the second statement: it's x^2 - x^3 > 2 not (2) x^2 - x^3 > 0.
_________________
Manager
Joined: 12 Oct 2012
Posts: 112
WE: General Management (Other)
Is x greater than x^3 ?  [#permalink]

### Show Tags

19 Jul 2016, 09:49
Is x greater than $$x^3$$ ?

(1) x is negative.

(2) $$x^2- x^3 > 2$$

I find such questions challenging. Please explain.
Math Expert
Joined: 02 Sep 2009
Posts: 54493
Re: Is x greater than x^3 ?  [#permalink]

### Show Tags

19 Jul 2016, 09:54
Is x greater than $$x^3$$ ?

(1) x is negative.

(2) $$x^2- x^3 > 2$$

I find such questions challenging. Please explain.

Merging topics. please refer to the discussion above.
_________________
Retired Moderator
Joined: 26 Nov 2012
Posts: 591
Is x greater than x^3 ?  [#permalink]

### Show Tags

19 Jul 2016, 13:43
1
Is x greater than $$x^3$$ ?

(1) x is negative.

(2) $$x^2- x^3 > 2$$

Bunuel, can you explain Stat 2...I did half way ..$$x^2$$ ( 1 - x ) > 2 and taken $$x^2$$ > 0 as it is square root and tried with 1-x >2...and struck here.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9136
Location: Pune, India
Re: Is x greater than x^3 ?  [#permalink]

### Show Tags

19 Jul 2016, 22:44
2
1
lumone wrote:
Is x greater than x^3 ?

(1) x is negative

(2) x^2 - x^3 > 2

Is x > x^3?

Statement 1: x is negative

If -1 < x < 0, x is less than x^3.
If x < -1, x is greater than x^3.
Not sufficient alone.

Statement 2: $$x^2 - x^3 > 2$$
$$x^3 - x^2 + 2 < 0$$
$$(x + 1)(x^2 - 2x + 2) < 0$$

(x^2 - 2x + 2) has no real roots so it doesn't cut the x axis. It is positive for all real values of x.
So for (x+1)(x^2 - 2x + 2) to be negative, (x+1) must be negative.
So x + 1 < 0
x < -1
If x < -1, x is greater than x^3.

Sufficient

Note: Know the relation between x, x^2 and x^3 on the 4 sections of the number line: x > 1, 0 < x< 1, -1 < x < 0 and x < -1
_________________
Karishma
Veritas Prep GMAT Instructor

Director
Joined: 13 Mar 2017
Posts: 723
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Is x greater than x^3 ?  [#permalink]

### Show Tags

22 Aug 2017, 06:17
lumone wrote:
Is x greater than x^3 ?

(1) x is negative

(2) x^2 - x^3 > 2

DS : x>x^3?

Statement 1 : x<0
-1<x<0 , x^3 > x
x<-1, x<x^3

NOT SUFFICIENT

Statement 2 : x^2 - x^3 > 2
x^2> 2+x^3
For x>1, x^3 > x^2 So, x^3 +2 > x^2. This is not the case
For 0<x<1, x^2 <1 So, x^2 <2 +x^3 This is also not the case.
For -1<x<0, x^2<1 , -1<x^3<0 So, x^2 <2 +x^3 This is also not the case.
For x = -1 , x^2=2+x^3 This is also not the case.
For x<-1, x^2>2+x^3 This is the case.

Now if x<-1, Clearly x>x^3

_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler
UPSC Aspirants : Get my app UPSC Important News Reader from Play store.

MBA Social Network : WebMaggu

Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)

What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9136
Location: Pune, India
Re: Is x greater than x^3 ?  [#permalink]

### Show Tags

11 Sep 2017, 04:29
Responding to a pm:
Quote:
I didn't quite understand how you factored the cubic polynomial x^3 - x^2 + 2
Would appreciate it if you could walk me through the steps. Thanks in advance!

Here is how you will do it:
https://www.veritasprep.com/blog/2013/0 ... rd-degree/
_________________
Karishma
Veritas Prep GMAT Instructor

Re: Is x greater than x^3 ?   [#permalink] 11 Sep 2017, 04:29
Display posts from previous: Sort by