Mod

[ketkoag]

Please explain the answer.. IMO E..
OA given : c
second statement is "x is less than equal to 1"
in the figure the second line after 1 is the cursor..please ignore it.

[gmatmachoman]

When X lies between 0 & 1,we can certainly say that the exp is equal to 1

ex: let x= 0.2
x-1= -0.8
|x-1|=0.8

|x| +|x-1|=0.2+|-0.8|
=1.

So it proves that any value between 0& 1 will be eual to 1 for |x| +|x-1|

[ketkoag]

thanks for pointing it out..
kinda silly mistake.. isn't it.

[qwe12]

Please explain the answer.. IMO E..
OA given : c
second statement is "x is less than equal to 1"
in the figure the second line after 1 is the cursor..please ignore it.

where did you get this question?