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# If x≠0, is |x| < 1?

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If x≠0, is |x| < 1? [#permalink]  19 Oct 2009, 06:10
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If x≠0, is |x| < 1?

(1) x^2 < 1
(2) |x| < 1/x
[Reveal] Spoiler: OA
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Re: Help with question involving |x| [#permalink]  19 Oct 2009, 07:11
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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.

Hope it's clear.
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Re: Help with question involving |x| [#permalink]  23 Oct 2009, 15:11
Yes, Thank you a lot!
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Re: Help with question involving |x| [#permalink]  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.

Hope it's clear.
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Posts: 28203
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Kudos [?]: 44878 [0], given: 6630

Re: Help with question involving |x| [#permalink]  30 Nov 2013, 03:10
Expert's post
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.

Hope it's clear.

You should specify what didn't you understand there. Thank you.
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Re: Help with question involving |x| [#permalink]  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.

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]  10 Mar 2015, 02:56
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Re: If x≠0, is |x| < 1?   [#permalink] 10 Mar 2015, 02:56
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