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

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

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04 Nov 2010, 19:49
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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?
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Re: GMAT Quant Rev 2nd Ed - DS 76 [#permalink]

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04 Nov 2010, 19:55
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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?

$$x(x-1) < 0$$ gives you the solution 0 < x < 1.

check out the link below for the explanation:
http://gmatclub.com/forum/inequalities-trick-91482.html#p804990
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 31 Aug 2010 Posts: 44 Followers: 0 Kudos [?]: 27 [0], given: 2 Re: GMAT Quant Rev 2nd Ed - DS 76 [#permalink] ### Show Tags 04 Nov 2010, 20:02 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 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 $$x > 0$$ ? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7370 Location: Pune, India Followers: 2285 Kudos [?]: 15093 [1] , given: 224 Re: GMAT Quant Rev 2nd Ed - DS 76 [#permalink] ### Show Tags 04 Nov 2010, 20:14 1 This post received KUDOS Expert's post 2 This post was BOOKMARKED 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 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 My Blog Get started with Veritas Prep GMAT On Demand for$199

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Last edited by VeritasPrepKarishma on 04 Nov 2010, 20:18, edited 1 time in total.
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Re: GMAT Quant Rev 2nd Ed - DS 76 [#permalink]

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04 Nov 2010, 20:15
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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 $$x<1$$ --> $$0<x<1$$;

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

For alternate approach check "How to solve quadratic inequalities": x2-4x-94661.html#p731476

(2) x^3 is positive --> $$x^3>0$$ just tells us that x is positive. Not sufficient.

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Re: GMAT Quant Rev 2nd Ed - DS 76 [#permalink]

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04 Nov 2010, 20:27
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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!
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21 Feb 2011, 11:12
is x between 0 and 1?
1. x^2 is less than x
2. x^3 is positive

I answer C considering x could be negative or positive but option 2 ensures x is positive. Please help what is the wrong with me.
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21 Feb 2011, 11:19
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Merging similar topics.

Baten80 wrote:
is x between 0 and 1?
1. x^2 is less than x
2. x^3 is positive

I answer C considering x could be negative or positive but option 2 ensures x is positive. Please help what is the wrong with me.

In (1) as x^2<x then x can not be negative, because if it is then we would have x^2>0>x.
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Re: GMAT Quant Rev 2nd Ed - DS 76 [#permalink]

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21 Feb 2011, 11:22
Q: Is 0<x<1?

1.
x^2<x
x^2-x<0
x(x-1)<0

Means;
case I:
x<0 and x-1>0=>x>1
OR
case II:
x>0 and x-1<0=>x<1
case I is impossible. x can't be greater than 1 and less than 0 at the same time.

Thus; only case II is valid and x>0 and x<1
In other words; 0<x<1
Sufficient.

2.x^3 is +ve.
if x=0.1; x^3=.001(a positive value); 0<x<1
if x=2; x^3=8(a positive value); but x>1
Not sufficient.

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

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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.

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

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08 Jun 2012, 03:33
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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.
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Re: Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

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

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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.

Yes, that's correct: x(x-1)<0 --> 0<x<1.

Explained here: x2-4x-94661.html#p731476
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Re: Is X between 0 and 1? [#permalink]

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01 Mar 2013, 05:07
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.

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

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26 Mar 2013, 05:08
One of my favorite number property questions. Really good approach and you need to come to inferences fast on this one.
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Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

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

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

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30 Jul 2016, 10:45
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...
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Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

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30 Jul 2016, 10:55
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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.
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24 Feb 2017, 06:48
Refer OG 16 quant review
Question no 89

Is x between 0 & 1
1. X square is less than X
2. X cube is positive

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