Is x negative ? (1) n^3(1-x^2) > 0 (2) x^2 - 1 > 0 : DS Archive
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# Is x negative ? (1) n^3(1-x^2) > 0 (2) x^2 - 1 > 0

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Manager
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Is x negative ? (1) n^3(1-x^2) > 0 (2) x^2 - 1 > 0 [#permalink]

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06 Mar 2006, 13:53
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Is x negative ?
(1) n^3(1-x^2) > 0
(2) x^2 - 1 > 0
Director
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06 Mar 2006, 14:03
E..

1st one is insufficient since we dont know the value of n..

2nd one is insufficient since -1 <x < 1..

Combining both it is still insufficient
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07 Mar 2006, 08:24
hellom3p wrote:
Is x negative ?
(1) n^3(1-x^2) > 0
(2) x^2 - 1 > 0

Yes I think the answer is E because
1) is insufficient as we do not know the value of n
2) is insufficient because it tells us that x^2 > 1 so x has to be either > 1 or < -1

I hope this makes sense.
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07 Mar 2006, 08:27
bewakoof wrote:
E..

1st one is insufficient since we dont know the value of n..

2nd one is insufficient since -1 <x < 1..

Combining both it is still insufficient

By the way Bewakoof can you explain me how you drived the inequality
-1<x<1?
In my opinion x has to be either > 1 or < -1. In short 1<x<-1

Please let me know if I am missing something.
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07 Mar 2006, 10:12
OA is C but I think there may be a typo in the question.
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07 Mar 2006, 10:17
MBAHopeful wrote:

Yes I think the answer is E because
1) is insufficient as we do not know the value of n
2) is insufficient because it tells us that x^2 > 1 so x has to be either > 1 or < -1

I hope this makes sense.

sorry my bad.. typo on my part.. I was too much in a rush to do that..

you are right.. x >1 or x < -1
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07 Mar 2006, 10:32
hellom3p wrote:
OA is C but I think there may be a typo in the question.

OA might be correct, IMO, if there is typo in S1. That is, if question reads:
Is x negative ?
(1) x^3(1-x^2) > 0
(2) x^2 - 1 > 0

S1:
case 1:
x^3 > 0 AND (1-x^2) > 0
or

x^3 > 0 => x > 0
(1-x^2) > 0 => -1 < x < 1
together: 0 < x < 1
ans: NO

case 2:
x^3 < 0 AND (1-x^2) < 0
or

x^3 < 0 => x < 0
(1-x^2) < 0 => x > 1 or x < -1
together: x < -1
ans: YES

INSUFFICIENT

S2:
x^2 - 1 > 0 or
x^2 > 1 => x > 1 or x < -1
INSUFFICIENT

together:

from S2:
x^2 - 1 > 0 or
(1 - x^2) < 0

applying this to S1 Case 2:
x^3 < 0 or x < 0
ans: YES
SUFFICIENT
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07 Mar 2006, 10:32
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