Inequalities^(Exponents)

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

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: Beat The GMAT — Data Sufficiency | Fri Mar 18, 2011 3:23 am

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

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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MAAJ: Master | Next Rank: 500 Posts; Posts: 243; Joined: Sun Jul 12, 2009 7:12 amFollowed by: 2 members; Location: Dominican Republic; Thanked: 31 times; GMAT Score:480

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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HSPA: Legendary Member; Posts: 1101; Joined: Fri Jan 28, 2011 7:26 amFollowed by: 13 members; Thanked: 47 times; GMAT Score:640

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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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

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.

https://www.gmatpill.com

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