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# Is x^3 > x^2 ?

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Is x^3 > x^2 ? [#permalink]

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16 Aug 2010, 14:18
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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, 13:21, 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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16 Aug 2010, 14:48
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.

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Re: Is [m]x^3 > x^2[/m] ? [#permalink]

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17 Aug 2010, 06:43
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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Re: Is [m]x^3 > x^2[/m] ? [#permalink]

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17 Aug 2010, 07:19
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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Re: Is [m]x^3 > x^2[/m] ? [#permalink]

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17 Aug 2010, 07:49
Thanks a bunch. Makes sense now.
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Re: Is [m]x^3 > x^2[/m] ? [#permalink]

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17 Aug 2010, 08:18
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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17 Aug 2010, 14:32
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, 14:32
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