If x is not equal to 0, is |x| less than 1?
(1) x/|x|< x
(2) |x| > x
Manhattan Q
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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
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
Regards
Samir
Samir
- gabriel
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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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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
Samir