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OG 154 Is x negative?

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OG 154 Is x negative? [#permalink] New post 12 Jan 2007, 10:43
Data Sufficiency Question -
Is x negative?

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

I read the explanation in the OG but found it very confusing. Any easier way to solve it?

Please let me know.

Thanks
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 [#permalink] New post 12 Jan 2007, 10:48
The statment 2 is impossible in the scope of GMAT with real numbers.

Are u sure about it x^2 < 0 ? :)
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 [#permalink] New post 12 Jan 2007, 10:50
Sorry my bad... 2nd statement is

Data Sufficiency Question -
Is x negative?

1) x^3(1-x^2) < 0
2) x^2 - 1 < 0
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 [#permalink] New post 12 Jan 2007, 10:58
(C) for me :)

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

implies that:
x^3 > 0 and 1-x^2 < 0 <=> x > 0 and x^2 > 1 <=> x > 1
or
x^3 < 0 and 1-x^2 > 0 <=> x < 0 and x^2 < 1 <=> -1 < x < 0

INSUFF.

Stat 2
x^2 - 1 < 0
<=> x^2 < 1
<=> -1 < x < 1

INSUFF.

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

We finally have : -1 < x < 0

SUFF.
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 [#permalink] New post 12 Jan 2007, 12:12
Data Sufficiency Question -
Is x negative?

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

from one

either x^3 is -ve or 1-x^2 is

if x^3 is -ve then x is -ve

and if 1-x^2 is -ve then x^2 >1 and x is -ve or +ve ....insuff

from two

x^2<1 x could be either +ve or -ve...insuff

both together

x^2 <1 thus 1-x^2 is sure +ve

and from one thus x^3 has to be -ve and so is x

my answer is C, I BACK UP MY FRIEND FIG ALL THE WAY :wink:

CHEERS MATE
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 [#permalink] New post 12 Jan 2007, 12:23
Thanks guys... The OG explanation put me to sleep... Your explanations made more sense to me! :)
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 [#permalink] New post 12 Jan 2007, 14:39
axl_oz wrote:
Thanks guys... The OG explanation put me to sleep... Your explanations made more sense to me! :)


Well ;).... Sometimes a rest is good as well :D :)

By the way, u are welcome :)
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 [#permalink] New post 12 Jan 2007, 14:44
Welcome back Yezz :)

I hope u to achieve your goal :).... to crack this GMAT :)
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Re: [#permalink] New post 10 Jul 2011, 01:09
Is this correct way to solve for statement 1??


St1 x^3(1-x^2)<0

therefore, x^3 < 0 or 1-x^2 < 0


therefore, x < 0 or 1 < x^2


which yields => x<0 OR x> 1 OR x < 1??
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Re: Re: [#permalink] New post 10 Jul 2011, 01:22
siddhans wrote:
Is this correct way to solve for statement 1??


St1 x^3(1-x^2)<0

therefore, x^3 < 0 or 1-x^2 < 0


therefore, x < 0 or 1 < x^2


which yields => x<0 OR x> 1 OR x < 1??


X^3*(1-x^2)<0
it mean either X^3<0 or 1-x^2<0
so u have 2 scenarios;

first : X^3>0 = X>0
1-x^2 <0 = X^2>1= x> +/-1... combining this two we get X>1

second X^3 <0 = X<0
1-X^2 >0 = -X^2>-1; X^2<1 X<+/-1.. combining this two we get x<-1
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Re: Re: [#permalink] New post 10 Jul 2011, 01:29
sudhir18n wrote:
siddhans wrote:
Is this correct way to solve for statement 1??


St1 x^3(1-x^2)<0

therefore, x^3 < 0 or 1-x^2 < 0


therefore, x < 0 or 1 < x^2


which yields => x<0 OR x> 1 OR x < 1??


X^3*(1-x^2)<0
it mean either X^3<0 or 1-x^2<0
so u have 2 scenarios;

first : X^3>0 = X>0
1-x^2 <0 = X^2>1= x> +/-1... combining this two we get X>1

second X^3 <0 = X<0
1-X^2 >0 = -X^2>-1; X^2<1 X<+/-1.. combining this two we get x<-1


can this problem be solved in a easy manner?
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Re: [#permalink] New post 10 Jul 2011, 01:33
Fig wrote:
(C) for me :)

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

implies that:
x^3 > 0 and 1-x^2 < 0 <=> x > 0 and x^2 > 1 <=> x > 1
or
x^3 < 0 and 1-x^2 > 0 <=> x < 0 and x^2 < 1 <=> -1 < x < 0

INSUFF.

Stat 2
x^2 - 1 < 0
<=> x^2 < 1
<=> -1 < x < 1

INSUFF.

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

We finally have : -1 < x < 0

SUFF.



From this statement : x^3 < 0 and 1-x^2 > 0 <=> x < 0 and x^2 < 1 <=> -1 < x < 0

How do we have the lower bound -1 < x??? I dont understand ... All it says is x < 0 and x < 1 So lower bound could be even less than -1 ???
Re:   [#permalink] 10 Jul 2011, 01:33
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