Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 12 Feb 2016, 19:57

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Is x^2 greater than x ?

Author Message
TAGS:
Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1713
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Followers: 82

Kudos [?]: 627 [0], given: 109

Is x^2 greater than x ? [#permalink]  20 Jun 2012, 09:44
00:00

Difficulty:

75% (hard)

Question Stats:

47% (02:22) correct 53% (01:46) wrong based on 19 sessions
Is x^2 greater than x ?

(1) x^2 > x^3.
(2) x^2 > x^4.

I already solved it, but I tried to solve it by also the following method, and I don't know in which part I am wrong.

(1) $$x^2 > x^3$$.
$$x^2 - x^3 > 0$$
$$x^2 * (1-x) > 0$$
So, we get: x= 0 and x = 1

-----(+)------0----(-)-------1------(+)------

Therefore, the intervals are x<0 and x>1

However, if we solve it in this way:
$$x^2 > x^3$$
We divide both sides by $$x^2$$:
$$1 > x$$
The interval is not the same.

Please, tell me in which part of my first way I am wrong.
Thanks!
[Reveal] Spoiler: OA

_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: my-ir-logbook-diary-133264.html

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 28 Mar 2012
Posts: 287
Concentration: Entrepreneurship
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 23

Kudos [?]: 289 [0], given: 23

Re: Is x^2 greater than x ? [#permalink]  20 Jun 2012, 11:32
metallicafan wrote:
$$x^2 > x^3$$.
$$x^2 - x^3 > 0$$
$$x^2 * (1-x) > 0$$
So, we get: x= 0 and x = 1

-----(+)------0----(-)-------1------(+)------

Hi,

When you say,
$$x^2 (1-x) > 0$$
if $$x\neq 0$$ then,$$x^2 >0$$ for any real value of x.
thus, inequality reduces to $$1-x>0$$
or $$x < 1$$

Also, to use the number line method, you should change the coefficient of x as positive (for simplicity)
$$x^2 (x-1) < 0$$ (multiply by -1 and reverse the sign)
now, use the number line;
-----(-)-----1---(+)---------

Please go through the following post:
Solving inequalities

Regards,
Senior Manager
Joined: 28 Mar 2012
Posts: 287
Concentration: Entrepreneurship
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 23

Kudos [?]: 289 [0], given: 23

Re: Is x^2 greater than x ? [#permalink]  20 Jun 2012, 11:41
Hi,

Is $$x^2>x$$?
or $$x^2-x>0$$?
or $$x(x-1)>0$$?
or x<0 or x>1?

Detailed solution:
Using (1),
$$x^2>x^3$$
or $$x^2(1-x)>0$$
or x < 1, for x = 0.5
$$x^2=0.25<x$$
but for x=-1,
$$x^2=1>x$$. Thus, Insufficient.

Using (2),
$$x^2>x^4$$
or $$x^2(1-x^2)>0$$
or $$x^2(1-x)(1+x)>0$$
or $$x^2(x-1)(1+x)<0$$ (multiplied by -1)
-(+)-------(-1)---(-)-----(+1)-----(+)------
or -1<x<1
Again, for x = 0.5
$$x^2=0.25<x$$
but for x=-0.5,
$$x^2=0.25>x$$. Thus, Insufficient.

Even after using (1) & (2), we have -1 < x <1, Insufficient.

Regards,
Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1713
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Followers: 82

Kudos [?]: 627 [0], given: 109

Re: Is x^2 greater than x ? [#permalink]  20 Jun 2012, 12:32
Thank you both!

It seems that I just memorized the method and didn't analyze it well.

I think that I get it now. And also I think that I have found another way that also can be helpful to you. Please, confirm if I am Ok. Please, Bunuel, take a look too.

(1) $$x^2 > x^3$$.
$$x^2 - x^3 > 0$$
$$x^2 * (1-x) > 0$$
So, we get: x= 0 and x = 1

If $$x>1$$ -----> $$x^2 * (1-x) < 0$$
If $$0<x<1$$ -----> $$x^2 * (1-x) > 0$$
If $$x<0$$ -----> $$x^2 * (1-x) > 0$$

So, we get:

-----(+)------0----(+)-------1------(-)------

Therefore, the solution is x < 1.

The same solution using this method:
$$x^2 > x^3$$
We divide by $$x^2$$
$$1 > x$$
$$x < 1$$.

What do you think?
_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: my-ir-logbook-diary-133264.html

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 28 Mar 2012
Posts: 287
Concentration: Entrepreneurship
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 23

Kudos [?]: 289 [0], given: 23

Re: Is x^2 greater than x ? [#permalink]  20 Jun 2012, 12:43
metallicafan wrote:
If $$x>1$$ -----> $$x^2 * (1-x) < 0$$
If $$0<x<1$$ -----> $$x^2 * (1-x) > 0$$
If $$x<0$$ -----> $$x^2 * (1-x) > 0$$

Hi,

As you can see, the sign only depends on 1-x, so, even powers can be ignored here.

Regards,
Math Expert
Joined: 02 Sep 2009
Posts: 31303
Followers: 5364

Kudos [?]: 62513 [0], given: 9457

Re: Is x^2 greater than x ? [#permalink]  21 Jun 2012, 00:03
Expert's post
Is x^2 greater than x ?

Question: is $$x^2>x$$? --> is $$x(x-1)>0$$? is $$x<0$$ or $$x>1$$?

(1) x^2 > x^3. Now, from this statement we know that $$x\neq{0}$$, so $$x^2>0$$ and we can safely reduce by it to get $$1>x$$ ($$x<1$$). So, finally we have that given inequality holds true for $$x<0$$ and $$0<x<1$$ (remember we should exclude 0 from the range, since if $$x=0$$ then the given inequality doesn't hold true). Not sufficient.

(2) x^2 > x^4. Apply the same logic here: we know that $$x\neq{0}$$, so $$x^2>0$$ and we can safely reduce by it to get $$1>x^2$$ ($$x^2<1$$) --> $$-1<x<1$$. So, finally we have that given inequality holds true for $$-1<x<0$$ and $$0<x<1$$ (remember we should exclude 0 from the range, since if $$x=0$$ then the given inequality doesn't hold true). Not sufficient.Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) gives: $$-1<x<0$$ and $$0<x<1$$, so we can still have an YES answer (consider $$x=-0.5$$) as well as a NO answer (consider $$x=0.5$$).

Hope it's clear.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 31303
Followers: 5364

Kudos [?]: 62513 [0], given: 9457

Re: Is x^2 greater than x ? [#permalink]  21 Jun 2012, 00:07
Expert's post
Hi,

Is $$x^2>x$$?
or $$x^2-x>0$$?
or $$x(x-1)>0$$?
or x<0 or x>1?

Detailed solution:
Using (1),
$$x^2>x^3$$
or $$x^2(1-x)>0$$
or x < 1, for x = 0.5
$$x^2=0.25<x$$
but for x=-1,
$$x^2=1>x$$. Thus, Insufficient.

Using (2),
$$x^2>x^4$$
or $$x^2(1-x^2)>0$$
or $$x^2(1-x)(1+x)>0$$
or $$x^2(x-1)(1+x)<0$$ (multiplied by -1)
-(+)-------(-1)---(-)-----(+1)-----(+)------
or -1<x<1
Again, for x = 0.5
$$x^2=0.25<x$$
but for x=-0.5,
$$x^2=0.25>x$$. Thus, Insufficient.

Even after using (1) & (2), we have -1 < x <1, Insufficient.

Regards,

The red parts are not correct. You should exclude zero, from all the ranges, since if $$x=0$$, then neither of statements hold true.

Hope it's clear.
_________________
Senior Manager
Joined: 28 Mar 2012
Posts: 287
Concentration: Entrepreneurship
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Followers: 23

Kudos [?]: 289 [0], given: 23

Re: Is x^2 greater than x ? [#permalink]  21 Jun 2012, 00:14
Hi,

Thanks for pointing it out!

Regards,
Bunuel wrote:
Hi,

Is $$x^2>x$$?
or $$x^2-x>0$$?
or $$x(x-1)>0$$?
or x<0 or x>1?

Detailed solution:
Using (1),
$$x^2>x^3$$
or $$x^2(1-x)>0$$
or x < 1, for x = 0.5
$$x^2=0.25<x$$
but for x=-1,
$$x^2=1>x$$. Thus, Insufficient.

Using (2),
$$x^2>x^4$$
or $$x^2(1-x^2)>0$$
or $$x^2(1-x)(1+x)>0$$
or $$x^2(x-1)(1+x)<0$$ (multiplied by -1)
-(+)-------(-1)---(-)-----(+1)-----(+)------
or -1<x<1
Again, for x = 0.5
$$x^2=0.25<x$$
but for x=-0.5,
$$x^2=0.25>x$$. Thus, Insufficient.

Even after using (1) & (2), we have -1 < x <1, Insufficient.

Regards,

The red parts are not correct. You should exclude zero, from all the ranges, since if $$x=0$$, then neither of statements hold true.

Hope it's clear.
Re: Is x^2 greater than x ?   [#permalink] 21 Jun 2012, 00:14
Similar topics Replies Last post
Similar
Topics:
7 Is x^2 greater than x-y? 3 13 May 2014, 05:31
12 Is x^2 greater than x ? 16 09 Feb 2014, 23:24
7 Is x^2 greater than x ? 6 13 Feb 2013, 15:45
Is x^2 greater than x? 1) x^2 is greater than 1 2) x is 5 29 Oct 2011, 08:24
4 Is x^2 greater than x? 18 31 Oct 2010, 16:39
Display posts from previous: Sort by