x2suresh wrote:

lumone wrote:

Is x greater than x^3

(1) x is negative

(2) x^2 - x^3 > 2

1)

x =-2 x^3=-8 x>x^3

x=-1/2 x^3 = -1/8 x<x^3

not sufficient

2)

x^2 (1 - x) > 2

x^2 is always postive.|.. so (1-x) need to be positve and

x can't be fractional values between -1 and 1x must be -ve value < -1 i.e

x > x^3

sufficient.

B.

This is a really old question, and I found it recently in Jeff Sackmann's Data Sufficiency Challenge question bank. I am afraid I don't understand why Statement (2) above is sufficient, particularly the highlighted part above.

My interpretation is this: In order to keep x^2 > x^3, (1-x) > 0 (i.e. positive). This implies that 1 > x. x could therefore be negative integer, negative fraction or positive fraction. For instance:

x = -2 => x^2 = 4 and x^3 = -8 =>

x^2 > x^3 and 1-(-2) > 0. So, x > x^3

x = -1/2 => x^2 = 1/4 and x^3 = -1/8 =>

x^2 > x^3 and 1-(-1/2) > 0.

However, x < x^3x = 1/2 => x^2 = 1/4 and x^3 = 1/8 =>

x^2 > x^3 and 1-(1/2) > 0. So, x > x^3

Clearly, x CAN be fractional values between -1 and 1 for x^2 > x^3. But each time, the relation between x and x^3 may change. So B is not sufficient.

Can somebody please explain where I am going wrong? Or provide an alternative explanation. Thanks a lot!