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Intern
Joined: 21 Jul 2009
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If x≠0, is x < 1?
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19 Oct 2009, 06:10
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If x≠0, is x < 1? (1) x^2 < 1 (2) x < 1/x
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Joined: 02 Sep 2009
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Re: Help with question involving x
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19 Oct 2009, 07:11
SMAbbas wrote: 1. Could someone solve the following DS prob If x≠0, is x <1? (1) x2 <1 == > (X Square)<1 (2) x < 1/x
Statement 1 is suff. I am able to solve this using statement 2.
2. Could someone share the strategy of solving questions involving x? This one is tricky: If \(x\neq{0}\), is \(x <1\)?Basically we are asked: is x in the range (1,1): is \(1<x<1\) true? (1) \(x^2<1\) > \(1<x<1\). Sufficient. (2) \(x < \frac{1}{x}\). LHS is absolute value, which means that it's positive (since \(x\neq{0}\)), thus RHS must also be positive, so \(x>0\). This on the other hand implies that \(x=x\). So, we have that \(x<\frac{1}{x}\). Multiply both parts by positive x: \(x^2<1\) > \(1<x<1\), but as \(x>0\), then the final range is \(0<x<1\). Sufficient. Answer D. Hope it's clear.
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Joined: 05 Jul 2009
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Re: Help with question involving x
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23 Oct 2009, 15:11
Yes, Thank you a lot!



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Re: Help with question involving x
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29 Nov 2013, 19:20
Please explain more why second condition is sufficient. I am not able to understand it Bunuel wrote: SMAbbas wrote: 1. Could someone solve the following DS prob If x≠0, is x <1? (1) x2 <1 == > (X Square)<1 (2) x < 1/x
Statement 1 is suff. I am able to solve this using statement 2.
2. Could someone share the strategy of solving questions involving x? This one is tricky: If \(x\neq{0}\), is \(x <1\)?Basically we are asked: is x in the range (1,1): is \(1<x<1\) true? (1) \(x^2<1\) > \(1<x<1\). Sufficient. (2) \(x < \frac{1}{x}\). LHS is absolute value, which means that it's positive (since \(x\neq{0}\)), thus RHS must also be positive, so \(x>0\). This on the other hand implies that \(x=x\). So, we have that \(x<\frac{1}{x}\). Multiply both parts by positive x: \(x^2<1\) > \(1<x<1\), but as \(x>0\), then the final range is \(0<x<1\). Sufficient. Answer D. Hope it's clear.



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Joined: 02 Sep 2009
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Re: Help with question involving x
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30 Nov 2013, 03:10
piyushmnit wrote: Please explain more why second condition is sufficient. I am not able to understand it Bunuel wrote: SMAbbas wrote: 1. Could someone solve the following DS prob If x≠0, is x <1? (1) x2 <1 == > (X Square)<1 (2) x < 1/x
Statement 1 is suff. I am able to solve this using statement 2.
2. Could someone share the strategy of solving questions involving x? This one is tricky: If \(x\neq{0}\), is \(x <1\)?Basically we are asked: is x in the range (1,1): is \(1<x<1\) true? (1) \(x^2<1\) > \(1<x<1\). Sufficient. (2) \(x < \frac{1}{x}\). LHS is absolute value, which means that it's positive (since \(x\neq{0}\)), thus RHS must also be positive, so \(x>0\). This on the other hand implies that \(x=x\). So, we have that \(x<\frac{1}{x}\). Multiply both parts by positive x: \(x^2<1\) > \(1<x<1\), but as \(x>0\), then the final range is \(0<x<1\). Sufficient. Answer D. Hope it's clear. You should specify what didn't you understand there. Thank you.
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New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Intern
Joined: 02 Sep 2011
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Re: Help with question involving x
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01 Dec 2013, 12:49
piyushmnit wrote: Please explain more why second condition is sufficient. I am not able to understand it Bunuel wrote: SMAbbas wrote: 1. Could someone solve the following DS prob If x≠0, is x <1? (1) x2 <1 == > (X Square)<1 (2) x < 1/x
Statement 1 is suff. I am able to solve this using statement 2.
2. Could someone share the strategy of solving questions involving x? This one is tricky: If \(x\neq{0}\), is \(x <1\)?Basically we are asked: is x in the range (1,1): is \(1<x<1\) true? (1) \(x^2<1\) > \(1<x<1\). Sufficient. (2) \(x < \frac{1}{x}\). LHS is absolute value, which means that it's positive (since \(x\neq{0}\)), thus RHS must also be positive, so \(x>0\). This on the other hand implies that \(x=x\). So, we have that \(x<\frac{1}{x}\). Multiply both parts by positive x: \(x^2<1\) > \(1<x<1\), but as \(x>0\), then the final range is \(0<x<1\). Sufficient. Answer D. Hope it's clear. Dear piyushmnit, the question asks whether 1<x<1 is true, so any range of x, that would fall into that one would be sufficient. The second statement tells that 0<x<1, which fall into the the aforementioned range. I hope it helps you.



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Re: If x≠0, is x < 1?
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27 Jul 2015, 11:57
Hello,
It would be helpful if someone explains why we don't open the mod under the second option. As well, what's wrong with my algebra here?
\(x < \frac{1}{x}\) if \(x<0\), then \(x*x<1\) and therefore, \(x\) could take all ve number.
However, picking any negative number surely invalidates that claim. Obviously, having trouble reconciling my algebra with picking numbers.
Thanks for your help,



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If x≠0, is x < 1?
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27 Jul 2015, 12:38
mejia401 wrote: Hello,
It would be helpful if someone explains why we don't open the mod under the second option. As well, what's wrong with my algebra here?
\(x < \frac{1}{x}\) if \(x<0\), then \(x*x<1\) and therefore, \(x\) could take all ve number.
However, picking any negative number surely invalidates that claim. Obviously, having trouble reconciling my algebra with picking numbers.
Thanks for your help, Question 1:We did not open the mod because the LHS of the inequality (=x) is a positive quantity for ALL x > If 1/x is > a positive quantity > x > 0 . This is the reason the mod was 'not opened'. \(1/x > a '+' quantity\) > x > 0 Question 2:If \(x<0\) > \(x = x\) > \(x<\frac{1}{x}\) > \(x^2>1\) (reversed the sign of inequality as x<0 and multiplying an inequality by a negative number reverses the sign of inequality!) > This is not possible as \(x^2 = +\), \(x^2 < 0\) and thus \(x^2\) can never be > 1. Thus , \(x<0\) is not a valid substitution. You did not reverse the sign of inequality when you multiplied the inequality by x as x<0



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Re: If x≠0, is x < 1?
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31 Aug 2016, 22:48
Statement 1: x^2<1, Only fractions in the range 1<0<1, will yield a lower value than 1 when squared. Given information is sufficient. Statement 2: When integers are considered, then modulus would be greater, given that 1/x is to be greater, fractions in the range1<0<1 would yield a higher value. Given information is sufficient.
Hence, option D



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Re: If x≠0, is x < 1?
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13 Sep 2018, 05:25
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Re: If x≠0, is x < 1? &nbs
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