X<0? 1) X^3<X^2 2) X^3<X^4 : DS Archive

thearch

Manager

Joined: 19 Feb 2005

Posts: 212

Own Kudos [?]: 69 [0]

Given Kudos: 0

Location: Milan Italy

Send PM

Re: X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]

13 Apr 2005, 04:04

for statement 2, is it right to simplify this way???
x^3<x^4 divide both by x^2 (it has to be positive and cant be 0)

so we have x<x^2 (which is far simpler)

awaiting your thoughts, inequalities are another weakness of mine

banerjeea_98

Director

Joined: 18 Nov 2004

Posts: 679

Own Kudos [?]: 198 [0]

Given Kudos: 0

Send PM

Re: X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]

13 Apr 2005, 08:27

thearch wrote:

for statement 2, is it right to simplify this way???
x^3<x^4 divide both by x^2 (it has to be positive and cant be 0)

so we have x<x^2 (which is far simpler)

awaiting your thoughts, inequalities are another weakness of mine

yes u can....u can do that for state 1 as well.....x^3 < x^2 ---> x < 1

rthothad

Senior Manager

Joined: 03 Nov 2004

Posts: 321

Own Kudos [?]: 108 [0]

Given Kudos: 0

Send PM

Re: X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]

13 Apr 2005, 09:38

I would go with C

1. x can be -ve or a fraction.
2. x can be -ve or a +ve

Both put together x will be -ve

DLMD

Manager

Joined: 07 Nov 2004

Posts: 218

Own Kudos [?]: 862 [0]

Given Kudos: 0

Send PM

Re: X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]

13 Apr 2005, 11:13

I got C too but the answer given is E.

for these who choose E, please elaborate why E is right.

otherwise, the answer given might be wrong.

Thanks

Folaa3

Senior Manager

Joined: 27 Dec 2004

Posts: 383

Own Kudos [?]: 108 [0]

Given Kudos: 0

Send PM

Re: X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]

13 Apr 2005, 12:38

DLMD wrote:

I got C too but the answer given is E.

for these who choose E, please elaborate why E is right.

otherwise, the answer given might be wrong.

Thanks

Well for each values of x, use Â½, -1/2, 2, -2, 3, -3 for trials and you will see how it is E. A little exhaustive but gives the right answer. .

banerjeea_98

Director

Joined: 18 Nov 2004

Posts: 679

Own Kudos [?]: 198 [0]

Given Kudos: 0

Send PM

Re: X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]

13 Apr 2005, 13:08

Folaa3 wrote:

DLMD wrote:

I got C too but the answer given is E.

for these who choose E, please elaborate why E is right.

otherwise, the answer given might be wrong.

Thanks

Well for each values of x, use Â½, -1/2, 2, -2, 3, -3 for trials and you will see how it is E. A little exhaustive but gives the right answer. .

OA can't be "E"...in ur values of X all ur +ve numbers do not satisfy both statements...1/2 doesn't satisfy state 2 and 2,3 don't satisfy state 1....only x < 0 satisfies both.

FN

Current Student

Joined: 28 Dec 2004

Posts: 1581

Own Kudos [?]: 642 [0]

Given Kudos: 2

Location: New York City

Concentration: Social Enterprise

Schools:Wharton'11 HBS'12

Send PM

Re: X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]

13 Apr 2005, 13:19

c for me...

1) statement 1 says x<0 or 0<x<1
two possibilites thus insuff!

2) statement 2 says
x<0 or X>1
again two possible answers...insuff

combine the two...

since both say X <0 is the only common piece...

gmat2me2

Senior Manager

Joined: 18 Feb 2005

Posts: 360

Own Kudos [?]: 20 [0]

Given Kudos: 0

Send PM

Re: X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]

13 Apr 2005, 15:20

DLMD wrote:

I got C too but the answer given is E.

for these who choose E, please elaborate why E is right.

otherwise, the answer given might be wrong.

Thanks

X<0?
1) X^3<X^2
2) X^3<X^4

From 1)

x^3 - x^2 <0
x^2(x-1)<0

==>> x^2<0 and (x-1)>0 (Since both of them cannot be negative)

or x^2>0 and (x-1)<0

So you cannot tell anything about x

from 2)

x^3(1-x)<0

Again the same logic

x^3<0 and (1-x)>0

or

x^3>0 and (1-x)<0

which again isnt sufficient.

Since(1) and (2) are giving multiple situations we cannot tell if x<0.

So I believe the answer is E.

MA

Director

Joined: 25 Nov 2004

Posts: 707

Own Kudos [?]: 448 [0]

Given Kudos: 0

Send PM

Re: X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]

13 Apr 2005, 21:47

from i, x = more than any negative value that is less than 1= -ve and less than 1>.
from ii, x=either -ve or +ve but more than 1.
to satisfy both, x must less than 0 but more than -1 = -1<x<0
suppose x = -0.5,
from i, X^3<X^2
(-0.5)^3<(-0.5)^2
-0.125<0.25

from ii, X^3<X^4
(-0.5)^3<(-0.5)^4
-0.125<0.0625. therefore it is C.

HongHu

Director

Joined: 03 Jan 2005

Posts: 971

Own Kudos [?]: 769 [0]

Given Kudos: 0

Send PM

Re: X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]

14 Apr 2005, 14:38

X<0?

1) X^3<X^2
x<1
Insufficient

2) X^3<X^4
1/x<1
x<0 or x>1
Insufficient

Combined:
x<1
x<0 or x>1
=>x<0

(C)

HongHu

Director

Joined: 03 Jan 2005

Posts: 971

Own Kudos [?]: 769 [0]

Given Kudos: 0

Send PM

Re: X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]

Updated on: 14 Apr 2005, 21:08

gmat2me2 wrote:

DLMD wrote:

I got C too but the answer given is E.

for these who choose E, please elaborate why E is right.

otherwise, the answer given might be wrong.

Thanks

X<0?
1) X^3<X^2
2) X^3<X^4

From 1)

x^3 - x^2 <0
x^2(x-1)<0

==>> x^2<0 and (x-1)>0 (Since both of them cannot be negative)

or x^2>0 and (x-1)<0

So you cannot tell anything about x

from 2)

x^3(1-x)<0

Again the same logic

x^3<0 and (1-x)>0

or

x^3>0 and (1-x)<0

which again isnt sufficient.

Since(1) and (2) are giving multiple situations we cannot tell if x<0.

So I believe the answer is E.

Right approach, but x^2 can never be negative. We are talking about real numbers.

Assume x>0, then you can multiple the two equations, and you get:
x^6<x^6 which is impossible. In other words it is impossible that x>0. You know that x can't be equal to zero either. So x must be less than zero. (C).

Originally posted by HongHu on 14 Apr 2005, 14:42.
Last edited by HongHu on 14 Apr 2005, 21:08, edited 1 time in total.

gmat2me2

Senior Manager

Joined: 18 Feb 2005

Posts: 360

Own Kudos [?]: 20 [0]

Given Kudos: 0

Send PM

Re: X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]

14 Apr 2005, 15:30

HongHu wrote:

gmat2me2 wrote:

DLMD wrote:

I got C too but the answer given is E.

for these who choose E, please elaborate why E is right.

otherwise, the answer given might be wrong.

Thanks

X> [color=red]x^20 (Since both of them cannot be negative)

or x^2>0 and (x-1)0

or

x^3>0 and (1-x)0, then you can multiple the two equations, and you get:
x^60. You know that x can be equal to zero either. So x must be less than zero. (C).

I would have cancelled x on both sides had this question came before I read your cameo on inqualities in this forum.Atleast the approach is right now.....Thanks HH

Archived Topic

Hi there,

This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.

Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum

Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.

Thank you for understanding, and happy exploring!

gmatclubot

Re: X<0? 1) X^3<X^2 2) X^3<X^4 [#permalink]

14 Apr 2005, 15:30