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If x≠0, is |x| < 1?

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If x≠0, is |x| < 1? [#permalink] 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
[Reveal] Spoiler: OA
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Re: Help with question involving |x| [#permalink] New post 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.

Answer D.

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

Answer D.

Hope it's clear.


You should specify what didn't you understand there. Thank you.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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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Re: Help with question involving |x| [#permalink] 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.
Re: Help with question involving |x|   [#permalink] 01 Dec 2013, 12:49
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