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by sanju09 » Fri Mar 18, 2011 3:23 am
Is x^3 > x^2?

(1) x > 0.

(2) x^2 > x.

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Source: — Data Sufficiency |

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by Anurag@Gurome » Fri Mar 18, 2011 4:29 am
sanju09 wrote:Is x^3 > x^2?

(1) x > 0.

(2) x^2 > x.

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(1) x > 0
Consider the following cases:
If x = 2, then 2^3 > 2^2 is true. So, x^3 > x^2.
If x = 1/2, then (1/2)^3 < (1/2)^2; here x^3 < x^2.
No unique answer.
NOT SUFFICIENT.

(2) x^2 > x
Consider the following cases:
If x = 2, then 2^3 > 2^2 is true. So, x^3 > x^2.
If x = -2, then (-2)^3 = -8 < (-2)^2 = 4; here x^3 < x^2.
No unique answer.
NOT SUFFICIENT.

Combining (1) and (2), we know that x has to be positive so that x^2 > x. So, x^3 > x^2 is always true.

The correct answer is C.
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by MAAJ » Fri Mar 18, 2011 5:48 am
Is x^3 > x^2 ? this is only possible if x > 1 so...

Is x > 1 ?

1) x > 0 Not sufficient

2) x^2 > x This is possible if x > 1 OR x < 0 Not sufficient

Combining 1) and 2) we get x > 1 which is sufficient to answer.

IMO C
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by HSPA » Fri Mar 18, 2011 8:16 am
If X= 0.5.. B will not hold

a) x > 0 and

x^2 > x since x>0 multiply on both sides with 1/x then we have X>1

For all values greater than 1 the requirement holds good

Yes C is the solution
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by Night reader » Fri Mar 18, 2011 9:06 am
x^3-x^2>0 OR x^2(x-1)>0. Since x^2 is always positive, Only one case is possible x>1 for our question to be true OR not true?

st(1) x>0 is Not Sufficient, as x can be greater OR less than 1;
st(2) x(x-1)>0, Two cases are possible x>0 and x>1 OR x<0 and x<1 Therefore Not Sufficient;

Combined st(1&2): st(1) specifies x>0 with st(2) specifies x>0 and x>1 Sufficient as x can be only greater than 1.

answer C
sanju09 wrote:Is x^3 > x^2?

(1) x > 0.

(2) x^2 > x.

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My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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