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Is 1 > |x-1| ?

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Is 1 > |x-1| ? [#permalink] New post 24 Jun 2013, 23:18
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Is 1 > |x-1| ?

(1) (x-1)^2 > 1
(2) 0 > x
[Reveal] Spoiler: OA
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Re: Applying absolute value to an inequality? [#permalink] New post 24 Jun 2013, 23:33
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Is 1 > |x-1| ?
Rewrite the question as: is 0<x<2?

I. (x-1)^2 > 1
so x<0 or x>2, in both cases x is outside the interval 0-2.
Sufficient

II. 0 > x
x is less then 0, so it's outside the 0-2 interval.
Sufficient
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Re: Is 1 > |x-1| ? [#permalink] New post 25 Jun 2013, 01:32
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Is 1 > |x-1| ?

Is |x-1|<1? --> Is -1<x-1<1? --> add 1 to each side: is 0<x<2?

(1) (x-1)^2 > 1 --> since both sides are non-negative, then we can safely take the square root: |x-1|>1. So, this statement directly gives a NO answer to the question. Sufficient.

(2) 0 > x. Sufficient.

Answer: D.

Hope it's clear.
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Re: Is 1 > |x-1| ? [#permalink] New post 26 Jun 2013, 10:40
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I have a more intuitive approach to solve this.

The expression 1>|x-1| is true only when x=1. In all the other cases it will be false whether the value of x is +ve or negative. (try plugging different values).

Statement (1) (x-1)^2 > 1: if (x-1)^2 is greater 1 then clearly the value of x!=1. Hence the above expression in question stem is false and hence this statement is sufficient to answer.

Statement (2) 0 > x: since x!=1 again this statement is sufficient.

I would like to hear the reviews of this approach from the experts.

Thanks.
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Re: Is 1 > |x-1| ? [#permalink] New post 26 Jun 2013, 10:45
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vabhs192003 wrote:
I have a more intuitive approach to solve this.

The expression 1>|x-1| is true only when x=1. In all the other cases it will be false whether the value of x is +ve or negative. (try plugging different values).

Statement (1) (x-1)^2 > 1: if (x-1)^2 is greater 1 then clearly the value of x!=1. Hence the above expression in question stem is false and hence this statement is sufficient to answer.

Statement (2) 0 > x: since x!=1 again this statement is sufficient.

I would like to hear the reviews of this approach from the experts.

Thanks.


The red part is not correct, consider x=0.5; you get 1>0.5 still true. You are considering only integers values for x, but x can be a non-integer as well.
That expression is true if 0<x<2, all your method is based on that assumption, which is not correct (even if you get the correct answer).
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Re: Is 1 > |x-1| ? [#permalink] New post 26 Jun 2013, 11:00
Zarrolou wrote:
vabhs192003 wrote:
I have a more intuitive approach to solve this.

The expression 1>|x-1| is true only when x=1. In all the other cases it will be false whether the value of x is +ve or negative. (try plugging different values).

Statement (1) (x-1)^2 > 1: if (x-1)^2 is greater 1 then clearly the value of x!=1. Hence the above expression in question stem is false and hence this statement is sufficient to answer.

Statement (2) 0 > x: since x!=1 again this statement is sufficient.

I would like to hear the reviews of this approach from the experts.

Thanks.


The red part is not correct, consider x=0.5; you get 1>0.5 still true. You are considering only integers values for x, but x can be a non-integer as well.
That expression is true if 0<x<2, all your method is based on that assumption, which is not correct (even if you get the correct answer).


Yup. Thanks for correcting me on that. I guess 0<x<2 should be correct way to bound variable x.
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Re: Is 1 > |x-1| ? [#permalink] New post 26 Jun 2013, 11:47
Is 1 > |x-1| ?
x>1:
1>|x-1|
1>x-1
2>x
1<x<2
x<1
1>-(x-1)
1>-x+1
0>-x
x>0
0<x<1

(1) (x-1)^2 > 1
(x-1)^2 > 1
(x-1)>1
x>2
If x > 2 then it doesn't fall in either possible range (0<x<1 OR 1<x<2)
SUFFICIENT

(2) 0 > x
x<0
As with #1, the given value of x does not fall in either possible range (0<x<1 OR 1<x<2)
SUFFICIENT

(D)

Last edited by WholeLottaLove on 08 Jul 2013, 15:52, edited 2 times in total.
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Is |x – 1| < 1? [#permalink] New post 14 Nov 2013, 22:33
Is |x – 1| < 1?

(1) (x – 1)² > 1
(2) x < 0
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Re: Is |x – 1| < 1? (1) (x – 1)² > 1 (2) x < 0 [#permalink] New post 15 Nov 2013, 00:10
honchos wrote:
Is |x – 1| < 1?
(1) (x – 1)² > 1
(2) x < 0

A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
D. Either statement BY ITSELF is sufficient to answer the question.
E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.


From Stmt 1) (x-1)<-1 or (x-1)>1
then |x-1| is always >1
stmt 1 alone is sufficient

From stmt 2) x < 0
x-1<-1
|x-1|> 1
hence |x-1| is always >1
Stmt 2 alone is sufficient

Ans D ..hope its clear
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Re: Is |x – 1| < 1? [#permalink] New post 15 Nov 2013, 00:22
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Re: Is |x – 1| < 1?   [#permalink] 15 Nov 2013, 00:22
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