Manhattan Q

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Manhattan Q

by [email protected] » Thu Sep 20, 2007 7:37 pm
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

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by samirpandeyit62 » Fri Sep 21, 2007 12:38 am
stmt 1: x/|x|< x

say x =-5 then we have here -1 < -5 (not valid)

say x= -1 then -1 < -1 (not valid)

sat x = 0.99 the we have here 1 < 0.99 (again not valid)

now if x= 1 then 1 < 1 (not valid)

so x needs to be greater than 1 ie.e 2 ...

if x=2 then 1<2 (valid)
i.e |2| =2 & so on

so SUFF


(2) |x| > x

this is possible only when x is -ve coz for all +ve values of x
|x| = x

hence x is less than 0 (-ve)
so SUFF (|x| is less than 1 )

so ans should be D
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Samir

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Re: Manhattan Q

by gabriel » Fri Sep 21, 2007 2:43 am
[email protected] wrote:If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x
Samir ur solution is wrong ..

the question says x not equal to 0 , so x can be any real number other than 0 , it then asks if mod (x) is less than 1 .. this is possible only when -1<x<1 ..

take a look at the second statement ...

over here if x = -0.1 ( x will be negative ) then mod(x)=0.1 and mod(x)> x .. also x can be -5... mod(x) =5 and > x .. so second statement is not sufficient as mod (x) can be greater than or less than 1 ..

now, take a look at the first statement ..

it says x/mod(x)<x .. now there are 2 possibilites over here

1.)x is a real number greater than 1 .. for eg x =2 in which case x/mod(x) is < x .. in this case mod(x)>1

2.) x is a negative number between -1 and 0 .. in this case too x/mod(x) < x .. if x is < -1 then x/mod(x) > x .. in this case mod(x)<1 ..

So the first statement is also insufficient ..

Now, combine the 2 statements .. from the 2nd statement w know x has to be negative and from the first statement we know that if x is negative then it has to be between -1 and 0 .. so the answer is C ..

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by samirpandeyit62 » Fri Sep 21, 2007 3:07 am
Thanks Gabreil for correcting, did this one a little hastily (plus some good work load) forgot to evaluate the range between 0 & -1 (which is IMO a danger zone in all abs val questions), thanks buddy for pointing out, I really expect this from u :) .
Regards
Samir