Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Friday Algebra DS [#permalink]
28 Jan 2011, 05:54

Expert's post

rxs0005 wrote:

Is x > x ^3

S1 x < 0

S2 x^2 - x^3 > 2

Is x> x^3?

Is x>x^3? --> is x^3-x<0? --> is (x+1)x(x-1)<0? is x<-1 or 0<x<1

(1) x<0. Not sufficient.

(2) x^2-x^3>2 --> x^2(1-x)>2 --> only true for x<-1 (note that if x>1 then x^2(1-x) is negative so this range is not good and if -1\leq{x}\leq{1} then x^2(1-x)\leq{2} so this range is also not good). Sufficient.

Re: Friday Algebra DS [#permalink]
17 Mar 2011, 21:49

1

This post received KUDOS

I goofed on this because I tricked myself but this is easy...think about it as always before putting pen to paper.. When is x>x^3? ONLY when x = negative integer OR a positive fraction. 1) x= -ve, ok but is it a fraction? Insuff 2) x^2 - x^3 >2 x^2 (x-1)<2 Use gurpreets method to draw the number line. You will see that Statement is positive for all: x>1 = positive, x>0 = positive Therefore between 0 and 1 the statement is negative..hence the original statement holds true. Because X is a positive fraction, B is sufficient.

Re: Friday Algebra DS [#permalink]
05 Mar 2013, 05:49

Bunuel wrote:

rxs0005 wrote:

Is x > x ^3

(2) x^2-x^3>2 --> x^2(1-x)>2 --> only true for x<-1 (note that if x>1 then x^2(1-x) is negative so this range is not good and if -1\leq{x}\leq{1} then x^2(1-x)\leq{2} so this range is also not good). Sufficient.

Answer: B.

Hi Bunuel,

For x^2-x^3>2 how did you directly arrive at the intervals <-1 between -1 and 1 and >1. When we deal with (1-x) >2 we get x <-1. But how did you chose the other point 1?

Re: Friday Algebra DS [#permalink]
06 Mar 2013, 01:55

Expert's post

Jaisri wrote:

Bunuel wrote:

rxs0005 wrote:

Is x > x ^3

(2) x^2-x^3>2 --> x^2(1-x)>2 --> only true for x<-1 (note that if x>1 then x^2(1-x) is negative so this range is not good and if -1\leq{x}\leq{1} then x^2(1-x)\leq{2} so this range is also not good). Sufficient.

Answer: B.

Hi Bunuel,

For x^2-x^3>2 how did you directly arrive at the intervals <-1 between -1 and 1 and >1. When we deal with (1-x) >2 we get x <-1. But how did you chose the other point 1?

Re: Friday Algebra DS [#permalink]
06 Mar 2013, 23:08

Bunuel wrote:

Check here:

Hi Bunuel, Thanks for your reply and reference to some wonderful materials around quadratic inequalities.

May I ask you one more question - going back to the basics now. What would be the roots of the eqn we have in hand in this post: x^2 - x^3 >2. I solved the roots to be 0 and -1 by following the below steps: x^2(1-x)>2 X^2 implies 0 is a root. 1-x>2 implies x<-1, so -1 is a root. From your explanations the roots seem to be 1 and -1. Where am I going wrong? (I also saw another post of yours where I was not able to solve the correct roost when x^3 was involved. Please help.)

Last edited by Jaisri on 07 Mar 2013, 05:28, edited 1 time in total.

Nothing to add after Bunuel's explanation. But just writing down how to get the inequality x<-1 from F.S 2 for those who couldn't get it.

F.S 1 clearly not sufficient. Take x= -1,you get a NO for stem, but for -0.5, you get a YES.

F.S 2, states x^2-x^3-2>0

or( x^2-1)-(x^3-1)>0

or(x-1)[(x+1) - (x^2+1+x)]>0

or(x-1)(-x^2)>0. Thus, (x-1) has to be negative

or x-1<0

or x<-1.

Sufficient.

B.

+1 Kudos. Thanks for showing how to solve this inequality! There is a minor change though... x-1 >0 gives x<1 and not x <-1. Bunuel seems to have considered -1 and 1 as the roots. Any reasons for that?

+1 Kudos. Thanks for showing how to solve this inequality! There is a minor change though... x-1 >0 gives x<1 and not x <-1. Bunuel seems to have considered -1 and 1 as the roots. Any reasons for that?

thanks for pointing out the mistake. It was a splendidly foolish mistake. Apologies. _________________