Is x^3 > x^2 ? : GMAT Data Sufficiency (DS)
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Is x^3 > x^2 ?

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Is x^3 > x^2 ? [#permalink]

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New post 16 Aug 2010, 13:18
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Is x^3 > x^2 ?

(1) x > 0
(2) x^2 > x
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Feb 2014, 12:21, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Is [m]x^3 > x^2[/m] ? [#permalink]

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New post 16 Aug 2010, 13:48
seekmba wrote:
Is \(x^3 > x^2\) ?

(1) \(x>0\)
(2) \(x^2>x\)


Is \(x^3>x^2\)? --> is \(x^3-x^2>0\)? --> is \(x^2(x-1)>0\)? --> is \(x>1\)?

(1) \(x>0\). Not sufficient.

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

(1)+(2) Intersection of the ranges from (1) and (2) is the range \(x>1\). Sufficient.

Answer: C.
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New post 17 Aug 2010, 05:43
Thanks so much Bunuel.

However I did not understand how you arrived at

\(x^2(x-1)>0\)--> is \(x>1?\)
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New post 17 Aug 2010, 06:19
seekmba wrote:
Thanks so much Bunuel.

However I did not understand how you arrived at

\(x^2(x-1)>0\)--> is \(x>1?\)


Is \(x^3>x^2\) --> is \(x^3-x^2>0\) --> is \(x^2(x-1)>0\) --> as \(x^2\) is non negative, for \(x^2(x-1)\) to be positive \(x\) must not be zero (first multiple) and \(x-1\) must be more than zero (second multiple) --> \(x\neq{0}\) and \(x-1>0\) --> \(x>1\).
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New post 17 Aug 2010, 06:49
Thanks a bunch. Makes sense now.
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Re: Is [m]x^3 > x^2[/m] ? [#permalink]

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New post 17 Aug 2010, 07:18
Another way of looking at the problem can be
X^2 > X means two things
1) X cannot be a fraction.
2) X can be a negative no.

Once it is ensured that X is non Negative number (as done by option A)
it will ensue that X^3 > X^2.

I Hope it helps :)
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Re: Is [m]x^3 > x^2[/m] ? [#permalink]

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New post 17 Aug 2010, 13:32
answer should be C, you need to have positive number that is > 1.

1) will fail for fractions like 1/2 , 1/4 ...

2) will fail for -ve numbers

combine both you will get the number that satisfy the base question
Re: Is [m]x^3 > x^2[/m] ?   [#permalink] 17 Aug 2010, 13:32
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