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

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Manager
Joined: 16 Jan 2018
Posts: 94
Location: New Zealand
Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0  [#permalink]

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17 Jul 2018, 08:22
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Difficulty:

75% (hard)

Question Stats:

42% (01:27) correct 58% (01:35) wrong based on 36 sessions

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Is $$x<0$$ ?

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

(2) $$x^2 - 1 < 0$$
Manager
Joined: 26 Dec 2017
Posts: 156
Re: Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0  [#permalink]

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17 Jul 2018, 08:27
IMO it is C.
On combining 1&2 we will get 0<x<1 which answers the question as no
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Senior Manager
Joined: 31 Jul 2017
Posts: 464
Location: Malaysia
GMAT 1: 700 Q50 V33
GPA: 3.95
WE: Consulting (Energy and Utilities)
Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0  [#permalink]

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17 Jul 2018, 08:36
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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VP
Status: Learning
Joined: 20 Dec 2015
Posts: 1167
Location: India
Concentration: Operations, Marketing
GMAT 1: 670 Q48 V36
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WE: Engineering (Manufacturing)
Re: Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0  [#permalink]

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17 Jul 2018, 08:59
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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Manager
Joined: 16 Jan 2018
Posts: 94
Location: New Zealand
Re: Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0  [#permalink]

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17 Jul 2018, 10:16
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!

Re: Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0 &nbs [#permalink] 17 Jul 2018, 10:16
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