mod is always positive, |x|>0 so we can multiply here the inequalities
st(1) x/|x|<x, 1<|x|; two cases possible 1<x and 1<(-x), correspondingly x<-1 and x>1. We know for sure that x is not in the interval (-1;1), But we don't know the sign of x Not Sufficient, as we don't know if x is positive or negative

st(2) |x| > x, when multiplying both parts by positive x^2 we get x^3>x^3 which is only possible if the right-hand side is negative x<0 Not sufficient

combined st(1&2) must be sufficient as x<0 and x<-1, so |x|>1 we answer No and Sufficient

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

(1)x/|x| < x

(2)|x| > x

IMO C

Given x is not equal to 0
to find -> Is |x| less than 1
Now, |x| could could be less than 1 only for the possible set of values -1<x<1

Evaluating Statement 1
x/|x| < x


Testing following set of values{-10,-1,-1/2,1/2,1,10} to determine the range of x
-10/10< -10 False
-1/1 < -1 False
-0.5/0.5 < -0.5 True
0.5/0.5 < 1 False
1/1 < 1 False
10/10 < 10 True

So the range of X is ---> x>1 & -1<x<0

Now, we have two possible sets therefor this statement is insufficient because
if x = 10 then |x| is not less than 1
if x = -1/2 then |x| is less than 1

Evaluating Statement 2

|x|>x

This can only be possible when x<0

Therefore statement 2 is also insufficient
if x = -10 then |x| is not less than 1
if x = -1/2 then |x| is less than 1

Evaluating Statement 1&2 together


From 1 -> x>1 & -1<x<0
From 2 -> x<0

Now the only common range determined for x from the two statements is -1<x<0

For all values in the above determined range |x| is always less than 1
Therefor Statements 1 & 2 together are sufficient