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Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0

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Manager
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Joined: 16 Jan 2018
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Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0  [#permalink]

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New post 17 Jul 2018, 07:22
1
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

39% (01:22) correct 61% (01:37) wrong based on 74 sessions

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Is \(x<0\) ?


(1) \((x-1)(x)(x+1) < 0\)

(2) \(x^2 - 1 < 0\)
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Joined: 26 Dec 2017
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Re: Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0  [#permalink]

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New post 17 Jul 2018, 07:27
IMO it is C.
On combining 1&2 we will get 0<x<1 which answers the question as no
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Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0  [#permalink]

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New post 17 Jul 2018, 07:36
blueviper wrote:
Is \(x<0\) ?


(1) \((x-1)(x)(x+1) < 0\)

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


Hi blueviper,

Can you please provide the OA also. The answer has to be C.

As, \(x^2 - 1 < 0\) for \((x-1)(x)(x+1)\) < 0, x > 0
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Re: Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0  [#permalink]

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New post 17 Jul 2018, 07:59
1
blueviper wrote:
Is \(x<0\) ?


(1) \((x-1)(x)(x+1) < 0\)

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


Imo C

From statement 1 we have range 0<x<1 and x<-1 in which x is negative. X can be negative or positive not sufficient

From statement 2 we have -1<x<1. X can be negative or positive not sufficient

Together there is a common range 0<x<1 so it is sufficient to answer.
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Re: Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0  [#permalink]

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New post 17 Jul 2018, 09:16
rahul16singh28 wrote:
blueviper wrote:
Is \(x<0\) ?


(1) \((x-1)(x)(x+1) < 0\)

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


Hi blueviper,

Can you please provide the OA also. The answer has to be C.

As, \(x^2 - 1 < 0\) for \((x-1)(x)(x+1)\) < 0, x > 0



Hi there,

1) Statement 1 gives \(x(x^2 - 1) < 0\) which means either \(x < 0 or x^2 - 1 < 0\) Insufficient

2) Statement 2 gives \(x^2 - 1 < 0\) which again means x can be both \(<0 or >0\) (between 1 & -1)

Combining 1) & 2) we can conclude \(x^2 - 1 < 0\) so x has to be greater than 0. Sufficient! :thumbup:

Answer is C
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Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0  [#permalink]

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New post 11 Oct 2018, 00:14
We can get the second equation \(x^2 -1<0\) from the first equation, so the second one seems unnecessary. Hence the answer should be either A or D? And since the first equation seems to be working for \(x<1\), \(x<0\) and \(x<-1\) then the answer should be D.
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Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0 &nbs [#permalink] 11 Oct 2018, 00:14
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