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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, 18:49
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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 \(x(x1) < 0\). 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, 18:55
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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 \(x(x1) < 0\). If x < 0 and x < 1 then \(0>x<1\). Wouldnt this statement be insufficient? or am i writing that dual inequality incorrectly? \(x(x1) < 0\) gives you the solution 0 < x < 1. check out the link below for the explanation: http://gmatclub.com/forum/inequalitiestrick91482.html#p804990
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Re: GMAT Quant Rev 2nd Ed  DS 76 [#permalink]
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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 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 \(x > 0\) ?



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Re: GMAT Quant Rev 2nd Ed  DS 76 [#permalink]
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04 Nov 2010, 19:14
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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.
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Re: GMAT Quant Rev 2nd Ed  DS 76 [#permalink]
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04 Nov 2010, 19:15
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 \(x(x1) < 0\). 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 > \(x^2<x\) > \(x(x1)<0\): Multiples must have opposite signs: \(x<0\) and \(x1>0\), or \(x>1\) > no solution (\(x\) can not be simultaneously less than zero and more than 1); \(x>0\) and \(x1<0\), or \(x<1\) > \(0<x<1\); So \(x(x1)<0\) holds true when \(0<x<1\). Sufficient. For alternate approach check "How to solve quadratic inequalities": x24x94661.html#p731476(2) x^3 is positive > \(x^3>0\) just tells us that x is positive. Not sufficient. Answer: A.
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Re: GMAT Quant Rev 2nd Ed  DS 76 [#permalink]
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04 Nov 2010, 19:27
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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(x1) < 0 must become 0<x<1 . Thanks guys!



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Arithmetic operation [#permalink]
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21 Feb 2011, 10:12
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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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Re: Arithmetic operation [#permalink]
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Re: GMAT Quant Rev 2nd Ed  DS 76 [#permalink]
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21 Feb 2011, 10:22
Q: Is 0<x<1? 1. x^2<x x^2x<0 x(x1)<0 Means; case I: x<0 and x1>0=>x>1 OR case II: x>0 and x1<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, 17:44
I have a question is this. Why have we considered both the options.
x(x1)<0:
Multiples must have opposite signs: 1. x<0 and x1>0, or x>1 > no solution (x can not be simultaneously less than zero and more than 1); 2. x>0 and x1<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, 02:33



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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, 21:14
Hi Bunuel, Am I right in construing when I say that x(x1)<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, 01:56



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Re: Is X between 0 and 1? [#permalink]
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01 Mar 2013, 04:07
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 \(x*(x1)<0\) . This is possible only if they have different signs. Thus, either x<0 AND (x1)>0[ This is not possible as x can't be more than 1 and yet be negative] or x>0 AND (x1)<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, 04:08
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, 14:17
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, 09:45
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, 09:55
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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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Need advice on number property [#permalink]
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24 Feb 2017, 05:48
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 Ans please with explanation Sent from my Lenovo A7020a48 using GMAT Club Forum mobile app




Need advice on number property
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