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:

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

14

00:00

A

B

C

D

E

Difficulty:

15% (low)

Question Stats:

78% (00:56) correct 22% (01:05) wrong based on 407 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.
_________________

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 >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Originally posted by VeritasKarishma on 04 Nov 2010, 20:14.
Last edited by VeritasKarishma on 04 Nov 2010, 20:18, edited 1 time in total.

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.

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

Show Tags

08 Jun 2012, 03:33

1

rggoel9 wrote:

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.

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

Show Tags

09 Jun 2012, 02:56

pavanpuneet wrote:

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...

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

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.
_________________

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

Show Tags

24 Feb 2017, 07:16

Top Contributor

jscott319 wrote:

Is x between 0 and 1 ?

(1) x² is less than x (2) x³ is positive

Target question:Is x between 0 and 1 ?

Statement 1:x² is less than x In other words, x² < x We can apply some inequality rules here. Since x² must be POSITIVE here, we can take x² < x and divide both sides by x² We get: 1 < 1/x Since 1/x is greater than 1, we can conclude that 1/x is positive, which means x is POSITIVE (i.e., x > 0) Since x is POSITIVE, we can take 1 < 1/x and multiply both sides by x to get: x < 1 When we combine our two inequalities, we get 0 < x < 1 In other words, x IS between 0 and 1 Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x³ is positive There are several values of x that satisfy statement 2. Here are two: Case a: x = 1/2 (so, x³ = (1/2)³ = 1/8). In this case, x IS between 0 and 1 Case b: x = 2 (so, x³ = 2³ = 8). In this case, x is NOT between 0 and 1 Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT