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Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]
04 Nov 2010, 18:49

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Difficulty:

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Question Stats:

91% (01:22) correct
9% (00:14) wrong based on 66 sessions

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?

Re: GMAT Quant Rev 2nd Ed - DS 76 [#permalink]
04 Nov 2010, 18:55

Expert's post

jscott319 wrote:

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?

Re: GMAT Quant Rev 2nd Ed - DS 76 [#permalink]
04 Nov 2010, 19:02

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 bex > 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 ?

Re: GMAT Quant Rev 2nd Ed - DS 76 [#permalink]
04 Nov 2010, 19:14

1

This post received KUDOS

Expert's post

jscott319 wrote:

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 bex > 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. _________________

Re: GMAT Quant Rev 2nd Ed - DS 76 [#permalink]
04 Nov 2010, 19:15

Expert's post

jscott319 wrote:

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?

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;

Re: GMAT Quant Rev 2nd Ed - DS 76 [#permalink]
04 Nov 2010, 19:27

1

This post received KUDOS

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]
07 Jun 2012, 17:44

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]
08 Jun 2012, 02:33

Expert's post

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]
08 Jun 2012, 21:14

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]
09 Jun 2012, 01:56

Expert's post

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.

Re: Is X between 0 and 1? [#permalink]
01 Mar 2013, 04:07

Expert's post

irfankool wrote:

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.

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