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(|x| - 1)/(x - 1) =?

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fattty
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(|x| - 1)/(x - 1) =? [#permalink] New post   13 Jan 2016, 23:36
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(|x| - 1)/(x - 1) =?

(1) x < 0
(2) |x| = -x
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E
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chetan2u
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(|x| - 1)/(x - 1) =? [#permalink] New post   14 Jan 2016, 00:08
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fattty wrote:
(|x| - 1)/(x - 1) =?

(1) x < 0
(2) |x| = -x


Hi,
if we look at the eq..
(|x| - 1)/(x - 1) ..

the value depends on :-
1) if x is 0 or +ive, ans is 1, irrespective of value of x..
2) if x is -ive, the value of denominator and numerator changes, and the equations value will depend on value of x..

lets see the statements now..
(1) x < 0
here x is negative , so the value depends on value of x.. insuff

(2) |x| = -x
again value of x is 0 or -ive.. insuff..

combined .nothing new.. insuff
E..

just for knowledge...
Why x is -ive in statement 2..
|x| = -x..
The LHS is positive because it is absolute mod, so RHS will also be positive..
RHS=-x... now -x will become positive only when x is -ive
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Bunuel
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Re: (|x| - 1)/(x - 1) =? [#permalink] New post   13 Jul 2017, 03:37
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fattty wrote:
(|x| - 1)/(x - 1) =?

(1) x < 0
(2) |x| = -x


Similar question to practice: https://gmatclub.com/forum/is-x-134652.html
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Re: (|x| - 1)/(x - 1) =? [#permalink] New post   13 Jul 2017, 04:10
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fattty wrote:
(|x| - 1)/(x - 1) =?

(1) x < 0
(2) |x| = -x



Given : Nothing
DS : (|x| - 1)/(x - 1) =?

Option 1: x < 0 ,
So, |x| = -x
(|x| - 1)/(x - 1) = (-x-1)/(x-1)
NOT SUFFICIENT

Option 2: |x| = -x
(|x| - 1)/(x - 1) = (-x-1)/(x-1)
NOT SUFFICIENT

Answer E
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chetan2u
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Re: (|x| - 1)/(x - 1) =? [#permalink] New post   24 Oct 2020, 20:11
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CEdward wrote:
JessicaSheoran wrote:
chetan2u wrote:

Hi,
if we look at the eq..
(|x| - 1)/(x - 1) ..

the value depends on :-
1) if x is 0 or +ive, ans is 1, irrespective of value of x..
2) if x is -ive, the value of denominator and numerator changes, and the equations value will depend on value of x..

lets see the statements now..
(1) x < 0
here x is negative , so the value depends on value of x.. insuff

(2) |x| = -x
again value of x is -ive.. insuff..

combined .nothing new.. insuff
E..

just for knowledge...
Why x is -ive in statement 2..
|x| = -x..
The LHS is positive because it is absolute mod, so RHS will also be positive..
RHS=-x... now -x will become positive only when x is -ive




Firstly this statement isn't correct: "1) if x is 0 or +ive, ans is 1, irrespective of value of x.."
If x is equal to 1 then ans is 0.
Secondly shouldn't 0 be the only solution for this: |x| = -x since |x| can never be negative.


Maybe I missed it, but I concur with Jessica on this one.

For statement 2, we can set x = 0 which would give us a unique value for the expression. No other number would do that as chetah explains. So why isn't this statement sufficient?



1) The statement that “ if x is 0 or +ive,..” is perfectly fine. x cannot be 1, since then the denominator x-1 will be come 0, making the term undefined.
2) All values of x as a -ive number will fit in. We are not saying |x| is negative, but that x is negative. -x, that is negative of negative will also become positive.
For example
Let x=-2, then |x|=|-2|=2, and -x=-(-2)=2. Thus |x|=-x for all values of x where \(x\leq 0\)
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Sidbendale1
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Re: (|x| - 1)/(x - 1) =? [#permalink] New post   17 Sep 2018, 07:39
chetan2u what is mod of 0? Is it zero or undefined?
If its undefined, why do we say that |x| >=0. I think i am getting confused between 2 things.
Please advise
Thanks in advance
Siddharth
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chetan2u
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Re: (|x| - 1)/(x - 1) =? [#permalink] New post   17 Sep 2018, 08:06
Expert Reply
Sidbendale1 wrote:
chetan2u what is mod of 0? Is it zero or undefined?
If its undefined, why do we say that |x| >=0. I think i am getting confused between 2 things.
Please advise
Thanks in advance
Siddharth



Siddharth,

|0|=0, it is not undefined..
Modulus just removes the negative sign
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(|x| - 1)/(x - 1) =? [#permalink] New post   14 Nov 2018, 09:45
chetan2u wrote:
fattty wrote:
(|x| - 1)/(x - 1) =?

(1) x < 0
(2) |x| = -x


Hi,
if we look at the eq..
(|x| - 1)/(x - 1) ..

the value depends on :-
1) if x is 0 or +ive, ans is 1, irrespective of value of x..
2) if x is -ive, the value of denominator and numerator changes, and the equations value will depend on value of x..

lets see the statements now..
(1) x < 0
here x is negative , so the value depends on value of x.. insuff

(2) |x| = -x
again value of x is -ive.. insuff..

combined .nothing new.. insuff
E..

just for knowledge...
Why x is -ive in statement 2..
|x| = -x..
The LHS is positive because it is absolute mod, so RHS will also be positive..
RHS=-x... now -x will become positive only when x is -ive




Firstly this statement isn't correct: "1) if x is 0 or +ive, ans is 1, irrespective of value of x.."
If x is equal to 1 then ans is 0.
Secondly shouldn't 0 be the only solution for this: |x| = -x since |x| can never be negative.
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revamoghe
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Re: (|x| - 1)/(x - 1) =? [#permalink] New post   13 Oct 2020, 19:20
chetan2u Didn't understand at all. Why are the statements insufficient?

chetan2u wrote:
Sidbendale1 wrote:
chetan2u what is mod of 0? Is it zero or undefined?
If its undefined, why do we say that |x| >=0. I think i am getting confused between 2 things.
Please advise
Thanks in advance
Siddharth



Siddharth,

|0|=0, it is not undefined..
Modulus just removes the negative sign
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chetan2u
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Re: (|x| - 1)/(x - 1) =? [#permalink] New post   13 Oct 2020, 20:15
Expert Reply
revamoghe wrote:
chetan2u Didn't understand at all. Why are the statements insufficient?

chetan2u wrote:
Sidbendale1 wrote:
chetan2u what is mod of 0? Is it zero or undefined?
If its undefined, why do we say that |x| >=0. I think i am getting confused between 2 things.
Please advise
Thanks in advance
Siddharth



Siddharth,

|0|=0, it is not undefined..
Modulus just removes the negative sign



If x is positive, then |x|=x, and hence the numerator and denominator will be the same. Value will be 1.

If x is negative say -a, then (|x|-1)/(x-1)=(|-a|-1)/(-a-1)=(a-1)/-(a+1)=(1-a)/(a+1).
Thus answer will depend on value of a, that is the value of x, which is not given by any of the statements
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Re: (|x| - 1)/(x - 1) =? [#permalink] New post   24 Oct 2020, 18:35
JessicaSheoran wrote:
chetan2u wrote:
fattty wrote:
(|x| - 1)/(x - 1) =?

(1) x < 0
(2) |x| = -x


Hi,
if we look at the eq..
(|x| - 1)/(x - 1) ..

the value depends on :-
1) if x is 0 or +ive, ans is 1, irrespective of value of x..
2) if x is -ive, the value of denominator and numerator changes, and the equations value will depend on value of x..

lets see the statements now..
(1) x < 0
here x is negative , so the value depends on value of x.. insuff

(2) |x| = -x
again value of x is -ive.. insuff..

combined .nothing new.. insuff
E..

just for knowledge...
Why x is -ive in statement 2..
|x| = -x..
The LHS is positive because it is absolute mod, so RHS will also be positive..
RHS=-x... now -x will become positive only when x is -ive




Firstly this statement isn't correct: "1) if x is 0 or +ive, ans is 1, irrespective of value of x.."
If x is equal to 1 then ans is 0.
Secondly shouldn't 0 be the only solution for this: |x| = -x since |x| can never be negative.


Maybe I missed it, but I concur with Jessica on this one.

For statement 2, we can set x = 0 which would give us a unique value for the expression. No other number would do that as chetah explains. So why isn't this statement sufficient?
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Vaishali2004
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Re: (|x| - 1)/(x - 1) =? [#permalink] New post   25 Feb 2022, 10:29
Doesn't option B imply that x=0 ,and therefore, B is sufficient?
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Re: (|x| - 1)/(x - 1) =? [#permalink] New post   09 Apr 2023, 07:29
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Re: (|x| - 1)/(x - 1) =? [#permalink]
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