GMATinsight wrote:

ashwink wrote:

Is \(x^{3} > x^{2}\) ?

1. x > 0

2. x < 1

The solution given by Bunuel is THE BEST for this question but I am just mentioning a few rules for reduction for the help of readers

Rule 1: If there is something positive factor/variable on both sides of the Inequality sign then it can always be cancelled out like in this question x^2 can be cancelled from both sides as x^2 will always be positive

Rule 2: If there is something Negative factor/variable on both sides of the Inequality sign then it can always be cancelled out

BUT Inequality sign Flips Rule 3: If the sign of some factor/variable on both sides of the Inequality sign is UNKNOWN then NEVER cancel them out as one of the solutions of the variable may be "zero"

My Solution remains as mentioned by Bunuel (Saving my effort by copying the same

Reduce by x^2 (which is non-negative): is x > 1? (x≠0x≠0)

(1) x > 0. Not sufficient.

(2) x < 1. Sufficient.

Answer: B.

Ok, Great !! So according to this, please correct me if i am wrong,

1. As \(x^2\) is deemed to be always positive, we cancelled out \(x^2\). Had it been \(x^5>X^4\), we could have cancelled out \(X^4\) also.

2. Another situtation, not this question, if it is given "x<0 or x is negative", we can cancel \(x^3\) from \(x^6>x^3\) and it would result in \(x^3<1\).

and if nothing is given then \(x^3(x^3-1)>0\) ? because both of the factors could be greater than zero.

Am i right ? Thanks for the first reply though.

Also, i have a Gmat in 10Days, could use some Extra help on verbal. Current scores ranging (Q49V29)660. SC AND RC are areas to be concerned about.