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DS

by sanjib » Mon Jun 22, 2009 6:13 am
. If x&#8800;0, is |x| <1?
(1) x2<1
(2) |x| < 1/x
OA is [[spoiler]spoiler]c[/spoiler][/spoiler]
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Source: — Data Sufficiency |
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Re: DS

by rahulg83 » Mon Jun 22, 2009 10:30 am
sanjib wrote:. If x&#8800;0, is |x| <1?
(1) x2<1
(2) |x| < 1/x
OA is [[spoiler]spoiler]c[/spoiler][/spoiler]
Did u intent to say x^2<1?
If it is correct, then A, IMO..
x is between -1 and 1, so |x|<1
Sufficient

Statement 2: if x>0 => x<1/x or x^2<1 or 0<x<1, which satisfies the inequality

if x<0 => -x<1/x or -x^2<1 or x^2>1 or x<-1, which doesn't satisfy the inequality. Insufficient

hence A
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by yogami » Mon Jun 22, 2009 3:53 pm
Why do i think the solution is D
?
(1) pretty much tells us that x can only be between -1 and 1 so mod x is always less than 1
(2) tells us that x can only be between 0 and 1 so mod x is always < 1.

SO D
Can someone give a thorough explanation please?
200 or 800. It don't matter no more.
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by joymukhi » Mon Jun 22, 2009 4:41 pm
Can someone please post a clear explanation of the problem? Thanks.
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by ssmiles08 » Mon Jun 22, 2009 5:05 pm
I dont know how the OA is C. I am getting D.

choice 1 tells us that x is a proper fraction. so -1<x<1 which is sufficient.

choice 2 tells us that x is a positive proper fraction. so 0<x<1 which is also sufficient.

Am I missing something?

IMO D.
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by pops » Mon Jun 22, 2009 10:04 pm
If x&#8800;0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

Asked: -1 < x < 1

statement 1. x^2 < 1
=> -1 < x < 1 ... hence sufficient

statement 2. |x| < 1/x
if x>0
x < 1/x
=> x^2 < 1
=> 0 < x < 1

if x<0
-x < 1/x
no solution

hence solution set for statement 2. is 0 < x < 1 which answers that |x| < 1

hence both statements can answer independently.
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Re: DS

by rahulg83 » Mon Jun 22, 2009 10:36 pm
if x<0 => -x<1/x or -x^2<1 or x^2>1 or x<-1, which doesn't satisfy the inequality.
should be x^2>-1

Yes, OA should be D..Silly mistakes are killing me :oops:
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