GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 07 Dec 2019, 19:32

Close

GMAT Club Daily Prep

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.

Close

Request Expert Reply

Confirm Cancel

Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Director
Director
avatar
Joined: 07 Jun 2004
Posts: 552
Location: PA
Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t  [#permalink]

Show Tags

New post Updated on: 04 Feb 2012, 09:30
3
4
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

66% (01:35) correct 34% (01:48) wrong based on 182 sessions

HideShow timer Statistics

Is x > x^2 ?

(1) x is greater than x^3

(2) x is greater than x^4


hi can you solve this Agebraically instead of picking numbers

Originally posted by rxs0005 on 04 Feb 2012, 09:23.
Last edited by Bunuel on 04 Feb 2012, 09:30, edited 1 time in total.
Edited the question
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59588
Re: x > x^2  [#permalink]

Show Tags

New post 04 Feb 2012, 09:29
1
2
rxs0005 wrote:
hi can you solve this Agebraically instead of picking numbers

Is x > x^2 ?

S1 x is greater than x^3

S2 x is greater than x^4


Is \(x > x^2\) --> is \(x(x-1)<0\)? --> is \(0<x<1\)?

(1) x > x^3 --> \(x*(x^2-1)<0\) --> \((x+1)*x*(x-1)<0\) --> \(0<x<1\) or \(x<-1\). Not sufficient.

(2) x > x^4 --> \(0<x<1\) (only in this range x will be more than x^4). Sufficient.

Answer: B.

Similar questions (by the way all posted by you):
is-x-0-x-x-2-x-x-3-i-did-this-algebraically-i-108110.html
is-x-between-0-and-1-1-x-x-3-2-x-x-107401.html
is-x-0-s1-x-x-2-s2-x-2-x-108207.html
is-x-x-3-s1-x-0-s2-x-2-x-108395.html

Check this for more on solving this kind of inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.
_________________
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59588
Re: x > x^2  [#permalink]

Show Tags

New post 04 Feb 2012, 09:58
1
rxs0005 wrote:
B

what I have been realizing ( i feel its wrong now ) is that

x ( x - 1 ) < 0 Means

either x< 0 or x - 1 < 0 or x < 1

so x < 0 or x < 1 how did u get 0 < x < 1 ?


\(x(x-1)<0\) --> roots are \(x=0\) and \(x=1\) --> "<" sign indicates that the solution lies between the roots: \(0<x<1\):
Attachment:
1.gif
1.gif [ 891 Bytes | Viewed 3187 times ]


If it were: \(x(x-1)>0\) --> the same roots: \(x=0\) and \(x=1\) --> ">" sign indicates that the solution lies to the left of the smaller root and to the right of the larger root: x<0 or x>1:
Attachment:
2.gif
2.gif [ 997 Bytes | Viewed 3192 times ]


Follow the links in my previous post for more on this subject.

Hope it helps.
_________________
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 9850
Location: Pune, India
Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t  [#permalink]

Show Tags

New post 24 Apr 2014, 22:31
1
jlgdr wrote:
Good that you bumped onto this thread Bunuel

Here's my question

On second statement how do we do critical points?

So we have x>x^4

Therefore x^4-x<0

Now then factorizing we have: x^2 (x-1)(x+1) , we only need to take care of (x-1)(x+1) since x^2 is always >=0

But then I get the range -1<x<1 which is not correct given that for instance when x=0 clearly inequality does not hold

Would you mind showing me where I went wrong here?

Thanks!
Cheers
J :)


Note that the inequality is
\(x^4 - x < 0\)

So on factoring you get
\(x (x^3 - 1) < 0\)
\(x(x - 1)(x^2 + x + 1) < 0\)

Now, (x^2 + x + 1) is always positive. It has no real roots. Hence we might as well assume it to be a positive number and ignore it for the sake of getting our transition points.
Transition points are 0 and 1. Since we want the negative region, the inequality will be negative between 0 and 1.

0 < x < 1

Though, you should try to understand the relation between x, x^2 and x^3 in the ranges < -1, -1 < x < 0, 0 < x <1 and x > 1 so that you don't need to use much algebra for such questions.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
SVP
SVP
avatar
B
Joined: 06 Nov 2014
Posts: 1870
Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t  [#permalink]

Show Tags

New post 22 Jun 2015, 13:53
1
Is x > x^2 ?

This would hold true if x<1

(1) x is greater than x^3
x can be a negative integer. For example x = -2. -2^3 = -8. But -2 is not greater than -2^2 = 4. Not sufficient.
(2) x is greater than x^4 For this to hold true, x would have to be less than 1. sufficient
Veritas Prep GMAT Instructor
User avatar
V
Joined: 16 Oct 2010
Posts: 9850
Location: Pune, India
Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t  [#permalink]

Show Tags

New post 10 Jul 2017, 00:04
1
gmatexam439 wrote:
VeritasPrepKarishma wrote:
jlgdr wrote:
Good that you bumped onto this thread Bunuel

Here's my question

On second statement how do we do critical points?

So we have x>x^4

Therefore x^4-x<0

Now then factorizing we have: x^2 (x-1)(x+1) , we only need to take care of (x-1)(x+1) since x^2 is always >=0

But then I get the range -1<x<1 which is not correct given that for instance when x=0 clearly inequality does not hold

Would you mind showing me where I went wrong here?

Thanks!
Cheers
J :)


Note that the inequality is
\(x^4 - x < 0\)

So on factoring you get
\(x (x^3 - 1) < 0\)
\(x(x - 1)(x^2 + x + 1) < 0\)

Now, (x^2 + x + 1) is always positive. It has no real roots. Hence we might as well assume it to be a positive number and ignore it for the sake of getting our transition points.
Transition points are 0 and 1. Since we want the negative region, the inequality will be negative between 0 and 1.

0 < x < 1

Though, you should try to understand the relation between x, x^2 and x^3 in the ranges < -1, -1 < x < 0, 0 < x <1 and x > 1 so that you don't need to use much algebra for such questions.


Hello VeritasPrepKarishma, Its been a while since i studies all such formulaes like (a+1)^3 or a^3-1 or (a+b+c)^2 etc, so should I revisit each and every of these formulaes?

Coming back to the question, Can I say that since x>x^4 and since x^4>0 always therefore x MUST be >0
Thus, x>0 and x^4-x<0
=> x(x^3-1)<0
=> x^3-1<0
=> x^3<1
Now, since x >0 then x^3 will be less than 1 only when 0<x<1; since for any real number greater than 1, the cube is greater than the number itself.

Thus i can say x<0<1. Hence sufficient.

Is my reasoning correct?

Regards


Some important algebraic identities:

\((x + y)^2 = x^2 + y^2 + 2xy\)

\((x - y)^2 = x^2 + y^2 - 2xy\)

\(x^2 - y^2 = (x + y)*(x - y)\)

Others, if needed, you can calculate there and then.

\((x + y)^3 = (x + y)(x + y)(x + y) = (x^2 + y^2 + 2xy)(x + y) = x^3 + xy^2 + 2x^2y ...\)

As for this question, yes you can arrive at 0 < x< 1 through that method too. No problem in the reasoning though I would suggest you to look at the other methods discussed above too. They are more generic.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Director
Director
avatar
Joined: 07 Jun 2004
Posts: 552
Location: PA
Re: x > x^2  [#permalink]

Show Tags

New post 04 Feb 2012, 09:30
thanks B

nice to have u back
Director
Director
avatar
Joined: 07 Jun 2004
Posts: 552
Location: PA
Re: x > x^2  [#permalink]

Show Tags

New post 04 Feb 2012, 09:39
B

what I have been realizing ( i feel its wrong now ) is that

x ( x - 1 ) < 0 Means

either x< 0 or x - 1 < 0 or x < 1

so x < 0 or x < 1 how did u get 0 < x < 1 ?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59588
Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t  [#permalink]

Show Tags

New post 23 Feb 2014, 05:57
SVP
SVP
User avatar
Joined: 06 Sep 2013
Posts: 1545
Concentration: Finance
GMAT ToolKit User
Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t  [#permalink]

Show Tags

New post 24 Apr 2014, 06:54
Good that you bumped onto this thread Bunuel

Here's my question

On second statement how do we do critical points?

So we have x>x^4

Therefore x^4-x<0

Now then factorizing we have: x^2 (x-1)(x+1) , we only need to take care of (x-1)(x+1) since x^2 is always >=0

But then I get the range -1<x<1 which is not correct given that for instance when x=0 clearly inequality does not hold

Would you mind showing me where I went wrong here?

Thanks!
Cheers
J :)
Current Student
User avatar
D
Joined: 12 Aug 2015
Posts: 2549
Schools: Boston U '20 (M)
GRE 1: Q169 V154
GMAT ToolKit User
Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t  [#permalink]

Show Tags

New post 13 Mar 2016, 08:31
Retired Moderator
User avatar
V
Joined: 28 Mar 2017
Posts: 1192
Location: India
GMAT 1: 730 Q49 V41
GPA: 4
Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t  [#permalink]

Show Tags

New post 08 Jul 2017, 03:45
VeritasPrepKarishma wrote:
jlgdr wrote:
Good that you bumped onto this thread Bunuel

Here's my question

On second statement how do we do critical points?

So we have x>x^4

Therefore x^4-x<0

Now then factorizing we have: x^2 (x-1)(x+1) , we only need to take care of (x-1)(x+1) since x^2 is always >=0

But then I get the range -1<x<1 which is not correct given that for instance when x=0 clearly inequality does not hold

Would you mind showing me where I went wrong here?

Thanks!
Cheers
J :)


Note that the inequality is
\(x^4 - x < 0\)

So on factoring you get
\(x (x^3 - 1) < 0\)
\(x(x - 1)(x^2 + x + 1) < 0\)

Now, (x^2 + x + 1) is always positive. It has no real roots. Hence we might as well assume it to be a positive number and ignore it for the sake of getting our transition points.
Transition points are 0 and 1. Since we want the negative region, the inequality will be negative between 0 and 1.

0 < x < 1

Though, you should try to understand the relation between x, x^2 and x^3 in the ranges < -1, -1 < x < 0, 0 < x <1 and x > 1 so that you don't need to use much algebra for such questions.


Hello VeritasPrepKarishma, Its been a while since i studies all such formulaes like (a+1)^3 or a^3-1 or (a+b+c)^2 etc, so should I revisit each and every of these formulaes?

Coming back to the question, Can I say that since x>x^4 and since x^4>0 always therefore x MUST be >0
Thus, x>0 and x^4-x<0
=> x(x^3-1)<0
=> x^3-1<0
=> x^3<1
Now, since x >0 then x^3 will be less than 1 only when 0<x<1; since for any real number greater than 1, the cube is greater than the number itself.

Thus i can say x<0<1. Hence sufficient.

Is my reasoning correct?

Regards
Retired Moderator
User avatar
V
Joined: 28 Mar 2017
Posts: 1192
Location: India
GMAT 1: 730 Q49 V41
GPA: 4
Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t  [#permalink]

Show Tags

New post 10 Jul 2017, 07:09
VeritasPrepKarishma wrote:
Some important algebraic identities:

\((x + y)^2 = x^2 + y^2 + 2xy\)

\((x - y)^2 = x^2 + y^2 - 2xy\)

\(x^2 - y^2 = (x + y)*(x - y)\)

Others, if needed, you can calculate there and then.

\((x + y)^3 = (x + y)(x + y)(x + y) = (x^2 + y^2 + 2xy)(x + y) = x^3 + xy^2 + 2x^2y ...\)

As for this question, yes you can arrive at 0 < x< 1 through that method too. No problem in the reasoning though I would suggest you to look at the other methods discussed above too. They are more generic.


Thank you Karishma :) You guyz are doing a very good job. Will look at the other methods described above as well.
Regards
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 13723
Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t  [#permalink]

Show Tags

New post 12 Sep 2019, 20:32
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Bot
Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t   [#permalink] 12 Sep 2019, 20:32
Display posts from previous: Sort by

Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne