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GMAT Club - m11#16

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Which of the following represents the range for all x which [#permalink] New post 02 Sep 2008, 22:11
Which of the following represents the range for all x which satisfy |1 - x| < 1 ?

A. (-1, 1)
B. (-1, 2)
C. (0, 1)
D. (0, 2)
E. (1, 2)
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Re: modulus [#permalink] New post 02 Sep 2008, 22:58
arjtryarjtry wrote:
Which of the following represents the range for all X which satisfy \(|1 - X| \lt 1\) ?


* (-1, 1)
* (-1, 2)
* (0, 1)
* (0, 2)
* (1, 2)

guys, show the working in detail


-1< 1-x < 1

x<2, x> 0

D.
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Re: modulus [#permalink] New post 03 Sep 2008, 05:46
arjtryarjtry wrote:
alpha_plus_gamma wrote:
arjtryarjtry wrote:
Which of the following represents the range for all X which satisfy \(|1 - X| \lt 1\) ?


* (-1, 1)
* (-1, 2)
* (0, 1)
* (0, 2)
* (1, 2)

guys, show the working in detail


-1< 1-x < 1........how????? i didnt get it

x<2, x> 0

D.


say 1-x =y

|y|<1 means -----> y values must be between -1 and 1
when y>0 --> y<1 --> 1-x<1 --> x>0
when y<0 --> -y<1 --> y>-1 --> 1-x>-1 --> x<2

(0,2)

is it clear
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Re: modulus [#permalink] New post 03 Sep 2008, 05:55
hmmm.... i went to the basics.. and it struck me suddenly

thanks suresh... cos ur explanation reinforced what i thought.

in maths when u discover something u feel like....

:band :band ..

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Re: modulus [#permalink] New post 03 Sep 2008, 08:51
When ever you see modulo you have two possibilities

(1-x) < 1 & -(1-x) < 1

If you solve both of them, 0<x<2 is what you will get.
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Re: modulus [#permalink] New post 03 Sep 2008, 10:10
Yep,

x> 2 and x < 0.

so it is in between 0 and 2. so, (0,2).

So IMO is D
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Re: modulus [#permalink] New post 03 Sep 2008, 11:00
Mathematicians usually think of absolute value as a measure of distance (more abstractly, a metric)

abs(a-b) is the distance from a to b (or vice versa)

abs(b) = abs(b-0) is the distance from b to zero

To find the values that satisfy the equation

abs(1-x) <1

we want to find all values whose distance from 1 is less than 1.

Think visually using the number line and visualize all the values that are less than 1 unit from the number 1

(0,2) is obviously those values
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GMAT Club - m11#16 [#permalink] New post 07 Sep 2010, 11:16
Which of the following represents the range for all \(X\) which satisfy \(|1 - X| \lt 1\) ?

(C) 2008 GMAT Club - m11#16

* (-1, 1)
* (-1, 2)
* (0, 1)
* (0, 2)
* (1, 2)

Answer:
[Reveal] Spoiler:
The correct answer is D.


Explanation:
[Reveal] Spoiler:
If \(X \ge 1\) , then the inequality turns into \(X - 1 \lt 1\) or \(X \lt 2\) . If \(X \lt 1\) , then the inequality turns into \(1 - X \lt 1\) or \(X \gt 0\) . Combine the intervals to get the answer.


My problem:
[Reveal] Spoiler:
|1-2|=1 not smaller than 1 ... ???
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Re: GMAT Club - m11#16 [#permalink] New post 07 Sep 2010, 11:30
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Expert's post
AndreG wrote:
Which of the following represents the range for all \(X\) which satisfy \(|1 - X| \lt 1\) ?

(C) 2008 GMAT Club - m11#16

* (-1, 1)
* (-1, 2)
* (0, 1)
* (0, 2)
* (1, 2)

Answer:
[Reveal] Spoiler:
The correct answer is D.


Explanation:
[Reveal] Spoiler:
If \(X \ge 1\) , then the inequality turns into \(X - 1 \lt 1\) or \(X \lt 2\) . If \(X \lt 1\) , then the inequality turns into \(1 - X \lt 1\) or \(X \gt 0\) . Combine the intervals to get the answer.


My problem:
[Reveal] Spoiler:
|1-2|=1 not smaller than 1 ... ???


\(|1-x|<{1}\) --> key point is \(x=1\) (key points are the values of \(x\) when absolute values equal to zero), thus two ranges to check:

\(x<1\) --> \(|1-x|=1-x\) and \(|1-x|<{1}\) becomes: \(1-x<{1}\) --> \(x>0\);
\(x\geq{1}\) --> \(|1-x|=-1+x\) and \(|1-x|<{1}\) becomes: \(-1+x<{1}\) --> \(x<2\);

So \(|1-x|<{1}\) holds true for \(0<x<2\).

Answer: D.

As for your question: \(x\) can not equal to 2, because \(0<x<2\) means that \(x\) MUST be less than 2 (and more than zero), for ANY \(x\) from this range given inequality will hold true.

I guess (0,2) should be changed to \(0<x<2\) (as well as all other options) to avoid confusion whether 0 and 2 are inclusive in the range.

Hope it helps.
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Re: GMAT Club - m11#16 [#permalink] New post 07 Sep 2010, 11:35
Bunuel wrote:
AndreG wrote:
Which of the following represents the range for all \(X\) which satisfy \(|1 - X| \lt 1\) ?

(C) 2008 GMAT Club - m11#16

* (-1, 1)
* (-1, 2)
* (0, 1)
* (0, 2)
* (1, 2)

Answer:
[Reveal] Spoiler:
The correct answer is D.


Explanation:
[Reveal] Spoiler:
If \(X \ge 1\) , then the inequality turns into \(X - 1 \lt 1\) or \(X \lt 2\) . If \(X \lt 1\) , then the inequality turns into \(1 - X \lt 1\) or \(X \gt 0\) . Combine the intervals to get the answer.


My problem:
[Reveal] Spoiler:
|1-2|=1 not smaller than 1 ... ???


\(|1-x|<{1}\) --> key point is \(x=1\) (key points are the values of \(x\) when absolute values equal to zero), thus two ranges to check:

\(x<1\) --> \(|1-x|=1-x\) and \(|1-x|<{1}\) becomes: \(1-x<{1}\) --> \(x>0\);
\(x\geq{1}\) --> \(|1-x|=-1+x\) and \(|1-x|<{1}\) becomes: \(-1+x<{1}\) --> \(x<2\);

So \(|1-x|<{1}\) holds true for \(0<x<2\).

Answer: D.

As for your question: \(x\) can not equal to 2, because \(0<x<2\) means that \(x\) MUST be less than 2 (and more than zero), for ANY \(x\) from this range given inequality will hold true.

Hope it helps.



Thanks for the quick reply!
Your explanation with the key-point value is very helpful!

Also, I was not aware, that if they talk about "range" the values given are excluded (hence \(0<x<2\) )

Thanks
André
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Re: GMAT Club - m11#16 [#permalink] New post 14 Jun 2011, 23:35
quoting bunuel :

" I guess (0,2) should be changed to 0<x<2 (as well as all other options) to avoid confusion whether 0 and 2 are inclusive in the range. "



This correction is needed IMO. when we say (0,2) we mean than the numbers are included.
Also another alternative would be to change the question to |1-x|<=1
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Re: modulus [#permalink] New post 26 May 2012, 11:32
I was able to get to the equation
2> x > 0

but it will be wrong to say (0,2) as this will include 0 and 2 as well where as x not equal to 0 or 2.

Please comment
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Re: modulus   [#permalink] 26 May 2012, 11:32
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