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fattty
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(|x| - 1)/(x - 1) =? 13 Jan 2016, 23:36
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25% (medium)

Question Stats:

75% (01:30) correct 25% (01:25) wrong based on 1195 sessions

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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) =? 14 Jan 2016, 00:08
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fattty
(|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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(|x| - 1)/(x - 1) =? 13 Jul 2017, 03:37
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fattty
(|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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(|x| - 1)/(x - 1) =? 13 Jul 2017, 04:10
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fattty
(|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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(|x| - 1)/(x - 1) =? 17 Sep 2018, 07:39
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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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(|x| - 1)/(x - 1) =? 17 Sep 2018, 08:06
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Sidbendale1
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) =? 14 Nov 2018, 09:45
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chetan2u
fattty
(|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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(|x| - 1)/(x - 1) =? 13 Oct 2020, 19:20
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chetan2u Didn't understand at all. Why are the statements insufficient?

chetan2u
Sidbendale1
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) =? 13 Oct 2020, 20:15
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revamoghe
chetan2u Didn't understand at all. Why are the statements insufficient?

chetan2u
Sidbendale1
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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(|x| - 1)/(x - 1) =? 24 Oct 2020, 18:35
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JessicaSheoran
chetan2u
fattty
(|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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(|x| - 1)/(x - 1) =? 24 Oct 2020, 20:11
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CEdward
JessicaSheoran
chetan2u

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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(|x| - 1)/(x - 1) =? 25 Feb 2022, 10:29
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Doesn't option B imply that x=0 ,and therefore, B is sufficient?
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(|x| - 1)/(x - 1) =? 26 Apr 2024, 22:51
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