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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

Show Tags

04 Nov 2010, 19:49

2

This post received KUDOS

8

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

77% (00:41) correct
23% (00:53) wrong based on 300 sessions

HideShow timer Statistics

Is X between 0 and 1 ?

(1) x^2 is less than x (2) x^3 is positive

I am curious how to rephrase Statement 1 using inequalities. I rewrote it as \(x^2 - x < 0\) , which then gives me \(x(x-1) < 0\). If x < 0 and x < 1 then \(0>x<1\). Wouldnt this statement be insufficient? or am i writing that dual inequality incorrectly?

I am curious how to rephrase Statement 1 using inequalities. I rewrote it as \(x^2 - x < 0\) , which then gives me \(x(x-1) < 0\). If x < 0 and x < 1 then \(0>x<1\). Wouldnt this statement be insufficient? or am i writing that dual inequality incorrectly?

Hmm i think i am more confused after reading that the first time through....

I think i am missing the reason why the inequality sign for \(x < 0\) should actually be\(x > 0\). I determined x\(< 1\) because i set the inequality of \(x - 1 < 0\) and after subtracting from both sides give me \(x < 1\) . What is different about \(x < 0\) becoming \(x > 0\) ?

Hmm i think i am more confused after reading that the first time through....

I think i am missing the reason why the inequality sign for \(x < 0\) should actually be\(x > 0\). I determined x\(< 1\) because i set the inequality of \(x - 1 < 0\) and after subtracting from both sides give me \(x < 1\) . What is different about \(x < 0\) becoming \(x > 0\) ?

x(x - 1) < 0 is not the same as x <0 and (x - 1)< 0

When I multiply two terms, the result is negative if and only if one of them is negative and the other is positive. When I multiply x with (x - 1), the result x(x - 1) will be negative (less than 0) in two cases:

Case I: x < 0 (x is negative) but (x - 1) > 0 (x - 1 is positive) (x - 1) > 0 implies x > 1 But this is not possible. x cannot be less than 0 and greater than 1 at the same time.

Case II: x > 0 (x is positive) but (x - 1) < 0 (x - 1 is negative) (x - 1) < 0 implies x < 1 This will happen when x lies between 0 and 1. i.e. when 0 < x < 1.

The link gives you the shortcut of solving inequalities of this type.
_________________

I am curious how to rephrase Statement 1 using inequalities. I rewrote it as \(x^2 - x < 0\) , which then gives me \(x(x-1) < 0\). If x < 0 and x < 1 then \(0>x<1\). Wouldnt this statement be insufficient? or am i writing that dual inequality incorrectly?

Is x between 0 and 1?

Is \(0<x<1\)?

(1) x^2 is less than x --> \(x^2<x\) --> \(x(x-1)<0\):

Multiples must have opposite signs: \(x<0\) and \(x-1>0\), or \(x>1\) --> no solution (\(x\) can not be simultaneously less than zero and more than 1); \(x>0\) and \(x-1<0\), or \(x<1\) --> \(0<x<1\);

So \(x(x-1)<0\) holds true when \(0<x<1\). Sufficient.

Ok I see it now! I was not taking into consideration the 2 cases that you've just made clear for me. Now I see how x(x-1) < 0 must become 0<x<1 . Thanks guys!

Re: Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

Show Tags

07 Jun 2012, 18:44

I have a question is this. Why have we considered both the options.

x(x-1)<0:

Multiples must have opposite signs: 1. x<0 and x-1>0, or x>1 --> no solution (x can not be simultaneously less than zero and more than 1); 2. x>0 and x-1<0, or x<1 --> 0<x<1;

In the link to the post when we find the roots of the quadratic equation and if we know the sign is "<" we can directly right the roots as "root 1" < x < "root 2". The same way in this case also there are 2 roots 0 and 1 so we can directly write it this way. 0<x<1. Why do we have to consider case 1 also.

I have a question is this. Why have we considered both the options.

x(x-1)<0:

Multiples must have opposite signs: 1. x<0 and x-1>0, or x>1 --> no solution (x can not be simultaneously less than zero and more than 1); 2. x>0 and x-1<0, or x<1 --> 0<x<1;

In the link to the post when we find the roots of the quadratic equation and if we know the sign is "<" we can directly right the roots as "root 1" < x < "root 2". The same way in this case also there are 2 roots 0 and 1 so we can directly write it this way. 0<x<1. Why do we have to consider case 1 also.

Rahul

These are just two different approaches.
_________________

Re: Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

Show Tags

08 Jun 2012, 22:14

Hi Bunuel, Am I right in construing when I say that x(x-1)<0, which means the roots are 0, 1 and since it is "<" the solution must lie between 0 and 1 and hence, 0<x<1. Please confirm.

Hi Bunuel, Am I right in construing when I say that x(x-1)<0, which means the roots are 0, 1 and since it is "<" the solution must lie between 0 and 1 and hence, 0<x<1. Please confirm.

Is X between 0 and 1? 1. x^2 is less than x. 2. x^3 is positive

From F.S 1, we have \(x^2<x\)

or \(x*(x-1)<0\) . This is possible only if they have different signs. Thus, either x<0 AND (x-1)>0[ This is not possible as x can't be more than 1 and yet be negative] or x>0 AND (x-1)<0. This gives us that 0<x<1. Sufficient.

From F.S 2, we know that \(x^3\) >0. Thus, cancelling out x^2 from both sides, we have x>0. Insufficient.

Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

Show Tags

05 Nov 2014, 15:17

Could someone please explain why it's not possible:

x^2 < x try x=1/2 => 1/4 < 1/2 Yes, 0 < x < 1 try x=-1 => 1 > -1 No, x < 0 < 1

Why can x be only positive in this case since it can be negative and squared? It is not implied in"0 < x <1" that x must be a positive number? The question asks whether x is between 0 and 1, in case 1 x can be -1 and still satisfy the equation...

EDIT: Sorry, I realized that statement 1 must be correct in itself...

Re: Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

Show Tags

26 Dec 2015, 16:27

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.
_________________

Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

Show Tags

30 Jul 2016, 10:55

1

This post received KUDOS

nancy77 wrote:

Hi, when I first tackled this problem, I took the square root of both sides so that gave me the equation of x < sqrt(x).

Is it wrong to approach it this way? I now understand this is a positives and negatives problem based on the solutions above...

Even if you take x < sqrt(x), you know that :

1) x has to be positive because sqrt of -ve number is always imaginary. 2) for x to be less than its square root , it has to be less than 1 and greater than 0. because any number greater than 1 would have its square root less than itself.

Thus, your approach is also fine.
_________________

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...