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M24 Q 30

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Senior Manager
Senior Manager
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Joined: 18 Aug 2009
Posts: 416

Kudos [?]: 144 [0], given: 16

Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0
M24 Q 30 [#permalink]

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New post 18 Jan 2010, 19:54
Please explain, how to solve this problem...
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_________________

Never give up,,,

Kudos [?]: 144 [0], given: 16

Senior Manager
Senior Manager
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Joined: 30 Aug 2009
Posts: 283

Kudos [?]: 188 [0], given: 5

Location: India
Concentration: General Management
Re: M24 Q 30 [#permalink]

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New post 19 Jan 2010, 03:43
mirzohidjon wrote:
Please explain, how to solve this problem...


taking different values for x

let x = -2 then we have -2/ |-2| = -1 < -2. NO

let x = -1 then we have -1/ |-1| = -1 < -1. NO

Let x = 0 then we have 0/|0| = 0<0. NO

Let x= 1 then we have 1/|1| = 1< 1. NO

let x = -1/3 then we have (-1/3)/ |(-1/3)| = -1 < (-1/3) YES

let x = 1/2 then we have (1/2)/|(1/2)| = 1 < 1/2 NO

let x = 2 then we have 2/|2| = 1 < 2 YES

so answer b expresses this correctly.

Kudos [?]: 188 [0], given: 5

Intern
Intern
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Status: Focus
Joined: 02 Jan 2010
Posts: 1

Kudos [?]: [0], given: 1

Schools: Stanford, HAAS, and Kellogs
Re: M24 Q 30 (SHORTER SOLUTION) [#permalink]

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New post 19 Jan 2010, 08:17
x/|x|<x =sgn(x)<x (sgn is signum function)

=>x/|x|<x ={ 1<x; x>=0 OR -1<x ; x<0 }
=> x>1 OR -1<x<0
So, B is right.

Kudos [?]: [0], given: 1

Senior Manager
Senior Manager
avatar
Joined: 18 Aug 2009
Posts: 416

Kudos [?]: 144 [0], given: 16

Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0
Re: M24 Q 30 [#permalink]

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New post 21 Jan 2010, 09:00
Yeah, the problem looks like will require a bit of time...
_________________

Never give up,,,

Kudos [?]: 144 [0], given: 16

Re: M24 Q 30   [#permalink] 21 Jan 2010, 09:00
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M24 Q 30

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