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blueviper
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Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0 17 Jul 2018, 08:22
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

75% (hard)

Question Stats:

42% (01:44) correct 58% (01:40) wrong based on 99 sessions

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Is \(x<0\) ?


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

(2) \(x^2 - 1 < 0\)
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C
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tejyr
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Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0 17 Jul 2018, 08:27
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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 17 Jul 2018, 08:36
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blueviper
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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Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0 17 Jul 2018, 08:59
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blueviper
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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Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0 17 Jul 2018, 10:16
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rahul16singh28
blueviper
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:

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Answer is C
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Is x < 0 ? (1) (x - 1)(x)(x + 1) < 0 (2) x^2 - 1 < 0 11 Oct 2018, 01:14
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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 07 Nov 2019, 12:45
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blueviper
Is \(x<0\) ?

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

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

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

\(x(x^2-1)<0…x<0:x^2-1>0…x^2>1…|x|>1…(x<0):x<-1\)
\(x(x^2-1)<0…x>0:x^2-1<0…x^2<1…-1<x<1…(x>0):0<x<1\)

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

\(x^2-1<0…x^2<1…|x|<1…-1<x<1\)

(1 & 2) sufic.

\(since:-1<x<1…then:x<-1=invalid…and:0<x<1=valid\)

Ans. (C)
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