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# Is |x| < 1 ? 1. x^4 - 1 > 0 2.

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Manager
Joined: 20 Apr 2010
Posts: 151

Kudos [?]: 19 [0], given: 16

Location: I N D I A
Is |x| < 1 ? 1. x^4 - 1 > 0 2. [#permalink]

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27 Jul 2010, 06:27
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Question Stats:

50% (01:25) correct 50% (01:29) wrong based on 10 sessions

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Is |x| < 1 ?
1. $$x^4$$ - 1 > 0
2. $$(1/(1-|x|))$$ > 0

One more basic doubt i ve regarding DS ?

Say If statement 1 . gives values of y as 0 , 1 , 2 , 3
Say If statement 2 . gives values of y as 1 , 2 , 3

Then while checking for C do we have to include 0 OR we just have to take common values i.e. 1,2,3 and not 0.. i hope i am able to make my Q clear.. I am missing somewhere..

Thanks

Kudos [?]: 19 [0], given: 16

Manager
Joined: 09 Jan 2010
Posts: 123

Kudos [?]: 42 [0], given: 12

Re: Dont have the official Ans for this.. [#permalink]

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27 Jul 2010, 07:20
as per cond 1

x^4 -1 > 0
-> X^4 > 1

i.e x can be -2, -3 .... ,2 , 3 , 4.....
mod x is always greater than 1
sufficient

as per cond 2

1/ (1-MOD X ) >0
i.e 1-mod x> 0
i.e 1> mod x

hence sufficient

Kudos [?]: 42 [0], given: 12

Math Expert
Joined: 02 Sep 2009
Posts: 42575

Kudos [?]: 135405 [0], given: 12692

Re: Dont have the official Ans for this.. [#permalink]

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27 Jul 2010, 07:49
sag wrote:
Is |x| < 1 ?
1. $$x^4$$ - 1 > 0
2. $$(1/(1-|x|))$$ > 0

Thanks

Not a good question.

Is $$|x| < 1$$? --> is $$-1<x<1$$?

(1) $$x^4-1>0$$ --> $$x^4>1$$ --> $$x<-1$$ or $$x>1$$. So $$x$$ is not in the range (-1,1). Sufficient.

(2) $$\frac{1}{1-|x|}>0$$ --> nominator is positive thus denominator must also be positive for fraction to be positive --> $$1-|x|>0$$ --> $$|x|<1$$. Sufficient.

But: From (1) we have that $$x$$ is NOT in the range (-1,1) and from (2) that $$x$$ is in the range (-1,1). Two statements contradict each other.

This will never occur on GMAT as: on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.

sag wrote:
One more basic doubt i ve regarding DS ?

Say If statement 1 . gives values of y as 0 , 1 , 2 , 3
Say If statement 2 . gives values of y as 1 , 2 , 3

Then while checking for C do we have to include 0 OR we just have to take common values i.e. 1,2,3 and not 0.. i hope i am able to make my Q clear.. I am missing somewhere..

Thanks

Consider the following question (I just made it up):

If $$y$$ is an integer, is $$|y+1|<3$$?

$$|y+1|<3$$ means is $$-4<y<2$$ (-3, -2, -1, 0, 1)?

(1) $$-3<y^3<10$$ --> $$y$$ can be: -1, 0, 1, or 2. Not sufficient.

(2) $$(y^2+4y)(y-1)=0$$ --> $$y$$ can be: -4, 0, or 1. Not sufficient.

(1)+(2) Intersection of the values from (1) and (2) are $$y=0$$ and $$y=1$$, both these values satisfy inequality $$|y+1|<3$$. Sufficient.

So if statement (1) gives one set of values for x and statement (2) gives another set of values for x, then when considering statements together we should take only the values which satisfy both statements.

Hope it helps.
_________________

Kudos [?]: 135405 [0], given: 12692

Manager
Joined: 20 Apr 2010
Posts: 151

Kudos [?]: 19 [0], given: 16

Location: I N D I A
Re: Dont have the official Ans for this.. [#permalink]

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27 Jul 2010, 23:24
Bunuel wrote:
sag wrote:
Is |x| < 1 ?
1. $$x^4$$ - 1 > 0
2. $$(1/(1-|x|))$$ > 0

Thanks

Not a good question.

Is $$|x| < 1$$? --> is $$-1<x<1$$?

(1) $$x^4-1>0$$ --> $$x^4>1$$ --> $$x<-1$$ or $$x>1$$. So $$x$$ is not in the range (-1,1). Sufficient.

(2) $$\frac{1}{1-|x|}>0$$ --> nominator is positive thus denominator must also be positive for fraction to be positive --> $$1-|x|>0$$ --> $$|x|<1$$. Sufficient.

But: From (1) we have that $$x$$ is NOT in the range (-1,1) and from (2) that $$x$$ is in the range (-1,1). Two statements contradict each other.

This will never occur on GMAT as: on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.

sag wrote:
One more basic doubt i ve regarding DS ?

Say If statement 1 . gives values of y as 0 , 1 , 2 , 3
Say If statement 2 . gives values of y as 1 , 2 , 3

Then while checking for C do we have to include 0 OR we just have to take common values i.e. 1,2,3 and not 0.. i hope i am able to make my Q clear.. I am missing somewhere..

Thanks

Consider the following question (I just made it up):

If $$y$$ is an integer, is $$|y+1|<3$$?

$$|y+1|<3$$ means is $$-4<y<2$$ (-3, -2, -1, 0, 1)?

(1) $$-3<y^3<10$$ --> $$y$$ can be: -1, 0, 1, or 2. Not sufficient.

(2) $$(y^2+4y)(y-1)=0$$ --> $$y$$ can be: -4, 0, or 1. Not sufficient.

(1)+(2) Intersection of the values from (1) and (2) are $$y=0$$ and $$y=1$$, both these values satisfy inequality $$|y+1|<3$$. Sufficient.

So if statement (1) gives one set of values for x and statement (2) gives another set of values for x, then when considering statements together we should take only the values which satisfy both statements.

Hope it helps.

Thanks Bunuel once again for both the explanations.. u rock...

Kudos [?]: 19 [0], given: 16

Re: Dont have the official Ans for this..   [#permalink] 27 Jul 2010, 23:24
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