cramya wrote:Is x negative?

1) x^3(1-x^2) < 0
2) x^2-1 <0

OA:C

I get statement 2's insufficiency

Can someone share their thoughts on statement I in particular and also the solution as a whole? I coudnlt quite get the explanations in the older post for this.

Statement I

x^3(1-x^2) < 0

x^3 (1+x) (1-x) < 0

if x^3 is negative then (1-x^2) is positive or vice versa
if x^3 is positive, then (1+x) can be negative, and (1-x) is positive.
if x^3 is positive, then (1+x) can be positive, and (1-x) is negative.

if x^3 is negative, x is negative
if 1-x^2 is negative, x could be positive or negative
Insufficient.

Statement II
x^2-1 <0

you can simplify this in 2 ways

either
(x+1) (x-1) < 0
or
x^2 < 1

either way you can get x as positive and negative, but only in fractions not integers.

Insufficient.

Combining I & II

(x+1) (x-1) < 0

change the sign
- (x+1) (x-1) < 0
(1+x) (1-x) > 0 ------------I

x^3 (1+x) (1-x) < 0 --------II


if (1+x) (1-x)>0, then x^3 will always be less than 0, therefore, x must be negative.

Hence C

Hope this helps.

Thanks PC.

(x+1) (x-1) < 0

change the sign
- (x+1) (x-1) < 0

Thats the missing link on relating I and II.... Appreciate it!

Statement 1: distributed you get x^3-x^5<0 or x^3<x^5 (by adding x^5 to both sides). This statement gives you these values of x: x>1 or -1<x<0. So x can be positive or negative. INSUFFICIENT.

Statement 2: implies -1<x<1 so x can be positive or negative. INSUFFICIENT.

Taken together, the only values that satisfy both equations are -1<x<0, so x must be negative. SUFFICIENT.

Hence, C.