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# Is x negative? 1) x^3(1-x^2)<0 2) x^2-1<0

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Retired Moderator
Joined: 05 Jul 2006
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Is x negative? 1) x^3(1-x^2)<0 2) x^2-1<0 [#permalink]

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16 Aug 2009, 00:43
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Question Stats:

50% (00:00) correct 50% (00:48) wrong based on 2 sessions

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Is x negative?

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

2) x^2-1<0

Kudos [?]: 443 [0], given: 49

Senior Manager
Joined: 17 Mar 2009
Posts: 302

Kudos [?]: 597 [0], given: 22

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16 Aug 2009, 21:54
IMO C

is x<0?
stmt1
since x^3(1-x^2)<0
either x^3 <0 or (1-x^2) <0,
so insufficient

stmt2
x^2-1<0
x2<1
then x lies between -1 and 1
insufficient

combining stmt1 and stmt2
stmt2 => -x^2+1>0 => (1-x^2)>0 ,apply this in stmt1
since (1-x^2)>0 , x^3 must be < o, so defintely x<o , so answer is C

Kudos [?]: 597 [0], given: 22

Retired Moderator
Joined: 05 Jul 2006
Posts: 1749

Kudos [?]: 443 [0], given: 49

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17 Aug 2009, 05:50
yezz wrote:
Is x negative?

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

2) x^2-1<0

from 1

x^3<x^5 either x >1 or -1<x<0....insuff

from 2

x^2<1 either 0<x<1 or -1<x<0... insuff

both

x belongs to interval -1<x<0 ie: -ve suff

C

Kudos [?]: 443 [0], given: 49

Intern
Joined: 30 Jul 2009
Posts: 17

Kudos [?]: 11 [0], given: 4

Location: Danbury CT
Schools: Wharton, Columbia , Cornell, CMU , Yale

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18 Aug 2009, 13:44
is x<0?
stmt1
x^3(1-x^2)<0

There are 3 cases a) x > 0 ( positive) then equation x^3(1-x^2)<0 holds true as (1-x^2) part of the equation will always be negative

b) x < 0 ( negative) then we are contradicting the equation x^3(1-x^2)<0 given as x^3 will be negative and 1-x^2) part of the equation will always be negative - So their multiplication will lead to a number greater than 0. ( Which is contradiction of the statement given)

Replace x with numbers ( integers, fractions ) it will reflect the results

According to me this statement ( Statement 1 ) is sufficient to deduce if x is negative or positive

Statement 2 x^2-1<0

then x^2<1 .. This case will be true only if 0<x<1 as for any other case square of a number is always positive. If x = 0 then statement holds true. So statement 2 is not sufficient to answer this question.

In my opinion , answer will be D.

I just started revisting my basics in maths after a long time and recently gearing up to start up my GMAT prep.

Thanks
_________________

Kudos [?]: 11 [0], given: 4

Retired Moderator
Joined: 05 Jul 2006
Posts: 1749

Kudos [?]: 443 [0], given: 49

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18 Aug 2009, 14:04
gulatin2 wrote:
is x<0?
stmt1
x^3(1-x^2)<0

There are 3 cases a) x > 0 ( positive) then equation x^3(1-x^2)<0 holds true as (1-x^2) part of the equation will always be negative

b) x < 0 ( negative) then we are contradicting the equation x^3(1-x^2)<0 given as x^3 will be negative and 1-x^2) part of the equation will always be negative - So their multiplication will lead to a number greater than 0. ( Which is contradiction of the statement given)

Replace x with numbers ( integers, fractions ) it will reflect the results

According to me this statement ( Statement 1 ) is sufficient to deduce if x is negative or positive

Statement 2 x^2-1<0

then x^2<1 .. This case will be true only if 0<x<1 as for any other case square of a number is always positive. If x = 0 then statement holds true. So statement 2 is not sufficient to answer this question.

In my opinion , answer will be D.

I just started revisting my basics in maths after a long time and recently gearing up to start up my GMAT prep.

Thanks

Hi, plz have a look at my solution above and try pluging in number to check.

Kudos [?]: 443 [0], given: 49

Re: Is x negative?   [#permalink] 18 Aug 2009, 14:04
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