Well, the question can be phrased as "Is -1<x<1?". Let's look at the statements.

A. x / |x| < x

We have 2 cases here,
Case 1: If x < 0, then x / -x < x. x > -1.
Case 2: If x > 0, then x / x < x. x > 1.

Case 1 & 2 doesn't tell us whether -1<x<1. Insufficient.

B. |x| > x

Again, we have 2 cases here.
Case 1: If x < 0, then -x > x --> 2x < 0 --> x < 0
Case 2: If x > 0, then x > x. This solution is not possible.
So, we only know that x < 0. But we still do not know whether -1 < x < 1. Insufficient.

Combined,
From (A), we have 2 case (x > 0 and x < 0). From (B), we have only 1 case (x < 0)

From (A), we will take case (1) which is x > -1. And, we also know that x < 0. So, x lies between -1 and 0 (which answers the question -1 < x < 1).
Sufficient.

Hence, it should be C. Please share the OA

amazing reply papgust. Yes, OA is C.

Thank you very much for the detailed explanation.

Regards,
Shobha

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

(1) x/|x| < x

(2) |x| > x

is lxl < 1 ??
Now lxl is always positive and what positive number, not equal to zero, is less than 1
Now, I rephrased the question as:
Is x a negative fraction ('coz it's given that x is not equal to zero)

S1. lxl is either -x or x. Lets look at both cases:
if lxl = -x, then x/-x < x
or -1<x
However this means that x could be a negative fraction or positive number Not suff.

S2 lxl > x this is easy to translate.
This means that x is a negative number (negative integer or negative fraction) hence not suff.

S1 + S2 tells us that x could be -ve fraction/+ve number or -ve fraction/-ve integer
Intersection of both is negative fraction. Hence suff.

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

(1) x/|x| < x

(2) |x| > x

|x| < 1 => -1<x<1

1. x < x|x| => x(1-|x|) < 0 => x>0 1<|x| or x<0 1>|x| so two cases.
2. |x| > x => x<0,

Combining 1&2: x<0 and so |x|<1 so C