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If x≠0, is |x| < 1?

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If x≠0, is |x| < 1?  [#permalink]

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New post 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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Re: Help with question involving |x|  [#permalink]

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New post 19 Oct 2009, 07:11
6
6
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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Re: Help with question involving |x|  [#permalink]

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New post 23 Oct 2009, 15:11
Yes, Thank you a lot!
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Re: Help with question involving |x|  [#permalink]

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New post 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|  [#permalink]

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New post 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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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

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What are GMAT Club Tests?
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Re: Help with question involving |x|  [#permalink]

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New post 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?  [#permalink]

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New post 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?  [#permalink]

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New post 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?  [#permalink]

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New post 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 range-1<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?  [#permalink]

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Re: If x≠0, is |x| < 1? &nbs [#permalink] 13 Sep 2018, 05:25
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