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Re: If x is not equal to 0, is x less than 1? (1) x/x< x [#permalink]
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17 Mar 2017, 01:48
Hi, I'm new to this and my head is spining... so many details in all the posts, can't even begin to read them!
I thought in S2, x<0. So we know all the signs in S1 and therefore we can solve it like a simple inequality...and we find x<1. S1 &S2 together suff.
But this is too simple!? Am i wrong in my approach even if the result is right?... I m a newbie with no GMAT confidence...help



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Re: If x is not equal to 0, is x less than 1? (1) x/x< x [#permalink]
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13 May 2017, 08:25
Bunuel wrote: Hussain15 wrote: If x is not equal to 0, is x less than 1?
(1) x/x< x
(2) x > x
Will really appreciate if answer is supported by explanation. \(x\neq{0}\), is \(x<1\)? Which means is \(1<x<1\)? (\(x\neq{0}\)) (1) \(\frac{x}{x}< x\) Two cases: A. \(x<0\) > \(\frac{x}{x}<x\) > \(1<x\). But remember that \(x<0\), so \(1<x<0\) B. \(x>0\) > \(\frac{x}{x}<x\) > \(1<x\). Two ranges \(1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(1<x<1\). (For instance \(x\) can be \(0.5\) or \(3\)) Second approach: look at the fraction \(\frac{x}{x}\) it can take only two values: 1 for \(x>0\) > so we would have: \(1<x\); Or 1 for \(x<0\) > so we would have: \(1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(1<x<0\). The same two ranges: \(1<x<0\) or \(x>1\). (2) \(x > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(x\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(1<x<1\). (For instance \(x\) can be \(0.5\) or \(10\)) Or two cases again: \(x<0\)> \(x>x\)> \(x<0\). \(x>0\) > \(x>x\): never correct. (1)+(2) Intersection of the ranges from (1) and (2) is the range \(1<x<0\) (\(x<0\) (from 2) and \(1<x<0\) or \(x>1\) (from 1), hence \(1<x<0\)). Every \(x\) from this range is definitely in the range \(1<x<1\). Sufficient. Answer: C. Bunuel, please help me understand the following  Should the answer not be A? because in option A, we can simply take the mod to the other side  x<x*x  and the sign will remain the same since mod can only be positive; further on. because x multiplied by mod x is is larger than x, it tells us that x must be positive, because were it negative, x multiplied by mod x would have been smaller than x. Would really appreciate some light on this basic concept. Thanks.



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Re: If x is not equal to 0, is x less than 1? (1) x/x< x [#permalink]
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13 May 2017, 08:31
OreoShake wrote: Bunuel wrote: Hussain15 wrote: If x is not equal to 0, is x less than 1?
(1) x/x< x
(2) x > x
Will really appreciate if answer is supported by explanation. \(x\neq{0}\), is \(x<1\)? Which means is \(1<x<1\)? (\(x\neq{0}\)) (1) \(\frac{x}{x}< x\) Two cases: A. \(x<0\) > \(\frac{x}{x}<x\) > \(1<x\). But remember that \(x<0\), so \(1<x<0\) B. \(x>0\) > \(\frac{x}{x}<x\) > \(1<x\). Two ranges \(1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(1<x<1\). (For instance \(x\) can be \(0.5\) or \(3\)) Second approach: look at the fraction \(\frac{x}{x}\) it can take only two values: 1 for \(x>0\) > so we would have: \(1<x\); Or 1 for \(x<0\) > so we would have: \(1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(1<x<0\). The same two ranges: \(1<x<0\) or \(x>1\). (2) \(x > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(x\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(1<x<1\). (For instance \(x\) can be \(0.5\) or \(10\)) Or two cases again: \(x<0\)> \(x>x\)> \(x<0\). \(x>0\) > \(x>x\): never correct. (1)+(2) Intersection of the ranges from (1) and (2) is the range \(1<x<0\) (\(x<0\) (from 2) and \(1<x<0\) or \(x>1\) (from 1), hence \(1<x<0\)). Every \(x\) from this range is definitely in the range \(1<x<1\). Sufficient. Answer: C. Bunuel, please help me understand the following  Should the answer not be A? because in option A, we can simply take the mod to the other side  x<x*x  and the sign will remain the same since mod can only be positive; further on. because x multiplied by mod x is is larger than x, it tells us that x must be positive, because were it negative, x multiplied by mod x would have been smaller than x. Would really appreciate some light on this basic concept. Thanks. All I could say about this question is on previous 5 pages. In addition, there are many different solutions from different experts.
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Re: If x is not equal to 0, is x less than 1? (1) x/x< x [#permalink]
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15 Oct 2017, 14:02
For (1), wouldn't x<0 give you: x/x < x 1 < x 1 > x? When I plug in numbers for that, it doesn't work, but I don't see how we get x/x < x when plugging in x<0? Bunuel wrote: Hussain15 wrote: If x is not equal to 0, is x less than 1?
(1) x/x< x
(2) x > x
Will really appreciate if answer is supported by explanation. \(x\neq{0}\), is \(x<1\)? Which means is \(1<x<1\)? (\(x\neq{0}\)) (1) \(\frac{x}{x}< x\) Two cases: A. \(x<0\) > \(\frac{x}{x}<x\) > \(1<x\). But remember that \(x<0\), so \(1<x<0\) B. \(x>0\) > \(\frac{x}{x}<x\) > \(1<x\). Two ranges \(1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(1<x<1\). (For instance \(x\) can be \(0.5\) or \(3\)) Second approach: look at the fraction \(\frac{x}{x}\) it can take only two values: 1 for \(x>0\) > so we would have: \(1<x\); Or 1 for \(x<0\) > so we would have: \(1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(1<x<0\). The same two ranges: \(1<x<0\) or \(x>1\). (2) \(x > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(x\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(1<x<1\). (For instance \(x\) can be \(0.5\) or \(10\)) Or two cases again: \(x<0\)> \(x>x\)> \(x<0\). \(x>0\) > \(x>x\): never correct. (1)+(2) Intersection of the ranges from (1) and (2) is the range \(1<x<0\) (\(x<0\) (from 2) and \(1<x<0\) or \(x>1\) (from 1), hence \(1<x<0\)). Every \(x\) from this range is definitely in the range \(1<x<1\). Sufficient. Answer: C.



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Re: If x is not equal to 0, is x less than 1? (1) x/x< x [#permalink]
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15 Oct 2017, 21:17
cgarmestani wrote: For (1), wouldn't x<0 give you: x/x < x 1 < x 1 > x? When I plug in numbers for that, it doesn't work, but I don't see how we get x/x < x when plugging in x<0? Bunuel wrote: Hussain15 wrote: If x is not equal to 0, is x less than 1?
(1) x/x< x
(2) x > x
Will really appreciate if answer is supported by explanation. \(x\neq{0}\), is \(x<1\)? Which means is \(1<x<1\)? (\(x\neq{0}\)) (1) \(\frac{x}{x}< x\) Two cases: A. \(x<0\) > \(\frac{x}{x}<x\) > \(1<x\). But remember that \(x<0\), so \(1<x<0\) B. \(x>0\) > \(\frac{x}{x}<x\) > \(1<x\). Two ranges \(1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(1<x<1\). (For instance \(x\) can be \(0.5\) or \(3\)) Second approach: look at the fraction \(\frac{x}{x}\) it can take only two values: 1 for \(x>0\) > so we would have: \(1<x\); Or 1 for \(x<0\) > so we would have: \(1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(1<x<0\). The same two ranges: \(1<x<0\) or \(x>1\). (2) \(x > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(x\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(1<x<1\). (For instance \(x\) can be \(0.5\) or \(10\)) Or two cases again: \(x<0\)> \(x>x\)> \(x<0\). \(x>0\) > \(x>x\): never correct. (1)+(2) Intersection of the ranges from (1) and (2) is the range \(1<x<0\) (\(x<0\) (from 2) and \(1<x<0\) or \(x>1\) (from 1), hence \(1<x<0\)). Every \(x\) from this range is definitely in the range \(1<x<1\). Sufficient. Answer: C. If x < 0, then x = x. Substitute x by x in x/x< x to get x/(x) < x and then to get 1 < x. Since we consider the range when x < 0, then 1 < x < 0.
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Re: If x is not equal to 0, is x less than 1? (1) x/x< x [#permalink]
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29 Nov 2017, 07:09
Bunuel wrote: udaymathapati wrote: Hi Bunuel, Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen 1<x<1 since x>1 is area which will not fit into this equation. Can you explain? Range from (1): (1)(0)(1) \(1<x<0\) or \(x>1\), green area; Range from (2): (1)(0)(1) \(x<0\), blue area; From (1) and (2): (1)(0)(1) \(1<x<0\), common range of \(x\) from (1) and (2) (intersection of ranges from (1) and (2)), red area. Hope it's clear. the red zone indicates that the range is 1<x<0 , how do we arrive at 1<x<1?



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Re: If x is not equal to 0, is x less than 1? (1) x/x< x [#permalink]
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29 Nov 2017, 07:22
yousufkhan wrote: Bunuel wrote: udaymathapati wrote: Hi Bunuel, Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen 1<x<1 since x>1 is area which will not fit into this equation. Can you explain? Range from (1): (1)(0)(1) \(1<x<0\) or \(x>1\), green area; Range from (2): (1)(0)(1) \(x<0\), blue area; From (1) and (2): (1)(0)(1) \(1<x<0\), common range of \(x\) from (1) and (2) (intersection of ranges from (1) and (2)), red area. Hope it's clear. the red zone indicates that the range is 1<x<0 , how do we arrive at 1<x<1? The question asks whether \(1<x<1\) is true. We got that \(1<x<0\). Any, x from \(1<x<0\) IS in the range from 1 to 1, so we have an YES answer to the question.
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Re: If x is not equal to 0, is x less than 1? (1) x/x< x [#permalink]
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29 Nov 2017, 08:52
Hussain15 wrote: If x is not equal to 0, is x less than 1?
(1) x/x< x
(2) x > x
PLEASE READ THE WHOLE DISCUSSION Stmnt 2: x  x >0, which means x < 0... So, Insufficient Stmnt 1: x (11/x)>0, which means x < 0 , 11/x < 0 or x > 0, 11/x >0.. Insufficeint Combining 1 +2, We know x <0, 11/x < 0.. which is x < 1. Hope it helps.
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Re: If x is not equal to 0, is x less than 1? (1) x/x< x [#permalink]
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27 Dec 2017, 03:15



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Re: If x is not equal to 0, is x less than 1? (1) x/x< x [#permalink]
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17 Feb 2018, 11:51
(1) We dont have to remove the abs(x)...we can treat it as a variable that is always positive.
So if we multiply both sides by abs(x),we dont have to change the sign of the inequality and we get
abs(x)*x> x
if x>0 > abs(x) >1 if x<0> abs(x) <1 Not suf.
(2) abs(x) < x x must be <0, but we cant say that abs (x)<1, because this will be true for any x<0.
(1)(2): We are restricted to the inference: if x<0> abs(x) <1 so we its suf.




Re: If x is not equal to 0, is x less than 1? (1) x/x< x
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