C IMO.

Lets looks at the easier statement first.

(2) |x| > x

Implies x is negative (|x| could be less than or greater than 1- Insufficient)

(1) x/|x| can either be 1 or -1 ( 1 = when x is positive, -1 when x is negative)

So, 1<x (or) -1<x(<0) Between 0 and -1 because x should be -ve in this case) 2 cases, Insufficient

Together we know x can only take values from 0 to -1. Hence |x| < 1.

Prompt can be rephrased as Is -1<x<1 ?

Statement 1

x/|x| < x
|x| can have two solutions depending on whether x is positive on negative
When x is positive ; x>1
When x is negative ; x>-1 . Insufficient

Statement 2
|x|>x
When x is positive; x>x; This can't be true
When x is negative; x<0. Insufficient

Combining statement 1 and 2

From 1, we have x>1 or x>-1
From 2, we have x<0. Therefore only one solution fits,i.e, -1<x<0. This solution is a subset of -1<x<1. Sufficient

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

(1)x/|x| < x

(2) |x| > x

Question rephrased: Is x a positive or negative fraction?

Plug in the following:
One positive fraction
One positive integer greater than 1
One negative fraction
One negative integer less than -1.

Statement 1: x/|x| < x.
x = 1/2 does not work.
x = 2 works.
x = -1/2 works.
x = -2 does not work.
Since x = -1/2 is between -1 and 1 and x=2 is not, INSUFFICIENT.

Statement 2: |x| > x
x = 1/2 does not work.
x = 2 does not work.
x = -1/2 works.
x = -2 works.
Since x = -1/2 is between -1 and 1 and x=-2 is not, INSUFFICIENT.

Statements 1 and 2 combined:
Only x = -1/2 satisfies both statements.
It is not possible in statement 1 that x=1 or that x=-1.
Thus, we know that x must be a negative fraction.
SUFFICIENT.

The correct answer is C.

I plugged in -2, -0.5, 0.5, 2 into the 2.

And for
-2: N for 1 and Y for 2
-0.5: Y for 1 for Y for 2
0.5: N for 1 and N for 2
2: Y for 1 and N for 2.

Since more than 1 shows unique pattern (Y for 1 and N for 2), can I conclude it is C because of this?