rahuljones1 wrote:If x is not equal to 0, is |x| less than 1?

(1)
x
--- < x
|x|



(2) |x| > x

IMO C

explantion if anwer is correct

IMO E
Equation reduces to -1<x<1? x not equal to zero

1) x/ lxl < x

x= 3, expression is true 1<3
x=-3, -1<-3 .....false
x=-1/2. -1<-1/2.......true
x=1/2...1 < 1/2 .....false

It is true for -1<x<0 and x>1
Insuff

2) lxl>x
This is true for x<0 but not for 0<x<1.
Insuff.

Together,
x<0 or x>1

Whats OA?

i get C

1) x/ lxl < x
=>x<x*|x| (|x|>0, given x not =0)
=>x-x*|x|<0
=>x(1-|x|)<0
either x<0, 1-|x|>0=>|x|<1........a or
x>0, 1-|x|<0=>|x|>1.........b

gives both yes and no answers, not sufficient

2) |x| > x
=>|x|-x>0
if x>0, x-x>0=> 0>0 not possible
so x must be negative. given that x is not 0
|x| may be <1 or >1 cant say
not sufficient

combining, we get x<0 from 2).
so |x|<1 from a in statement 1)

hence, C

What is wrong in my explantion?

schumi,
from 1) you get -1<x<0 or x>1
and from 2) x<0

when you combine, x>1 is ruled out. you get -1<x<0 which would make |x|<1 so C would be sufficient.

Thanks Scooby,

Is my assumption, lxl<1 ===> -1<x<1 correct?

I got it this way.

for x>0, x<1
for x<0, -x<1 i.e. x>-1

Hence -1<x<1

If this is correct, then the conditions does hold true for 0<x<1

If x is not equal to 0, is |x| less than 1?

(1)
x
--- < x
|x|


(2) |x| > x

=============================
Q is asking for whether : |x|<1 or -1<x<1

1). x<x|x|
if x<0, x<-x2 or x^2+x <0 or x(x+1)<0
as x<0 we know x+1 >0 or x>-1 so it says -1<x<0
here |x| <1
: for x>0 x<x^2 or x(x-1)<0 as x>0 x<1 so in range 0<x<1
so this case is ok

2. |x| > x
if x<0 -x>x or 0>2x or x<0 nothing new
if x>0 x>x or 0>0 not possible
so for sure we know x<0 as oter case is not possible here but less than 0 doesnt mean it cant be less than 1 it can well be -5 or -5000 so not sufficient

Combining 1 & 2 we know for x<0 and so we can take the range -1<x<0

so |x| <1 as it is between 0 and 1

i will go with CCCCCCCCCCCCCCC