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# Is x > 0 ?

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Is x > 0 ? [#permalink]

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22 Jan 2011, 07:28
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Difficulty:

65% (hard)

Question Stats:

53% (02:26) correct 48% (00:59) wrong based on 80 sessions

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Is x > 0 ?

(1) x < x^2
(2) x < x^3
[Reveal] Spoiler: OA

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22 Jan 2011, 09:55
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rxs0005 wrote:
is x > 0

x < x^2

x < x^3

I did this algebraically i got A but OA is E can some one solve this algebraically

Is x > 0

(1) x < x^2 --> x*(x-1)>0 --> x<0 or x>1. Not sufficient.

(2) x < x^3 --> (x+1)*x*(x-1)>0 --> -1<x<0 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1<x<0 or x>1. Not sufficient.

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22 Jan 2011, 10:05
Bunuel in S1 why did u consider this as x< 0 i took this as x >0 and x-1 > 0
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22 Jan 2011, 10:17
rxs0005 wrote:
Bunuel in S1 why did u consider this as x< 0 i took this as x >0 and x-1 > 0

x*(x-1)>0 --> either both multiples are positive or both negative:
x>0 and x-1>0, x>1 --> so x>1;
x<0 and x-1<0, x<1 --> so x<0;

Thus x*(x-1)>0 holds true for x<0 and x>1.

Check for more here: x2-4x-94661.html#p731476 or here: inequalities-trick-91482.html

Similar questions:
gmat-quant-rev-2nd-ed-ds-104280.html
inequality-98674.html
ds-algebra-107401.html
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22 Jan 2011, 11:14
thanks for the prompt reply i now get it
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17 Mar 2011, 22:58
statement 1 says x can be either a negative fraction OR a positive integer
Statement 2 says x can be either a positive fraction OR a negative integer...
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18 Mar 2011, 00:24
(1)

x can be -ve (fraction or integer) or +ve (but not a +ve fraction), so not enough

(2)

x can be -ve fraction , say -1/2 or a +ve number > 1 so not enough.

Again, from (1) and (2), x can be a -ve fraction or +ve number, so not enough.

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19 Apr 2011, 07:39
Hi
How does one solve this question using Gupreet's method?
I was able to do the first statement but I dont know how to draw the root of x^2(x-1)>0. Thanks
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20 Apr 2011, 00:46
who is gurpeet?

the second should be

0>x^3-x

factor an x out

0> x(x^2-1)
0>x(x+1)(x-1)
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19 May 2011, 23:36
region on number scale

x>0 and -0.1<= x <= 0. satisfies both.

E
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Re: Is x > 0 ? [#permalink]

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09 Oct 2014, 09:15
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Re: Is x > 0 ? [#permalink]

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10 Oct 2014, 08:53
Bunuel : " x < x^3 --> (x+1)*x*(x-1)>0 --> -1<x<0 or x>1. Not sufficient."
i could not understand the concept of Roots of 3rd degree quad equation

cheers
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Re: Is x > 0 ? [#permalink]

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10 Oct 2014, 08:55
Bunuel : " x < x^3 --> (x+1)*x*(x-1)>0 --> -1<x<0 or x>1. Not sufficient."
i could not understand the concept of Roots of 3rd degree quad equation

cheers

This explained here:
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html
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Re: Is x > 0 ?   [#permalink] 10 Oct 2014, 08:55
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