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Is x greater than x^3 ?

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Is x greater than x^3 ? [#permalink] New post 09 Feb 2009, 12:43
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

48% (01:53) correct 51% (01:29) wrong based on 29 sessions
Is x greater than x^3 ?

(1) x is negative

(2) x^2 - x^3 > 2
[Reveal] Spoiler: OA

Last edited by Bunuel on 20 Apr 2014, 11:26, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: DS - 97 [#permalink] New post 09 Feb 2009, 12:57
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.
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Re: DS - 97 [#permalink] New post 20 Apr 2014, 09:15
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!
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Re: DS - 97 [#permalink] New post 21 Apr 2014, 04:23
Expert's post
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.
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Re: DS - 97   [#permalink] 21 Apr 2014, 04:23
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