1) X < 1
Let P = x^2/|x|
X=0 P < 1
X=1/2 P < 1
X=-2 P > 1
Not sufficient

2) X > -1
x=0 P < 1
x=-1/2 P < 1
X=2 P > 1
Not sufficient

1 and 2 combined,
-1 < X < 1
sufficient P < 1

I believe the answer is (C) because when x takes any value between -1 and 1 the inequality is always false. Thereby ensuring that the question can be answered

vipulgoyal wrote:Is x^2/|x| > 1 ?
1.x < 1
2.x > -1

ansc

x² = |x||x|.

Thus, the question stem can be rephrased as follows:
x²/|x| > 1
|x||x| / |x| > 1
|x| > 1.

Question stem rephrased: Is the distance between x and 0 greater than 1?

Statement 1: x < 1
If x=1/2, then the distance between x and 0 is less than 1.
If x=-2, then the distance between x and 0 is greater than 1.
INSUFFICIENT.

Statement 2: x > -1
If x=-1/2, then the distance between x and 0 is less than 1.
If x=2, then the distance between x and 0 is greater than 1.
INSUFFICIENT.

Statements combined: -1 < x < 1.
Thus, the distance between x and 0 is not greater than 1.
SUFFICIENT.

The correct answer is C.

Quick Qn: What if x = 1/2 (then x^2/|x| < 1). So how is -1 < x < 1 ?

Thanks.

I still juggling with why this approach experts please sugggest whats wrong

Is x^2/|x| > 1

opening mode first time with positive

x^2/x > 1
x^2 > x
x^2 - x >0
x(x-1)>0
x>0, x>1 ....1

opening mode with negitive

x^2 + x >0
x(x + 1) >0
x>0,x>-1 ....2

so the ineqaliti becomes for 1 and 2
is x>-1

so B could be the ans