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Re: Is 1 > |x-1| ? [#permalink]
25 Jun 2013, 01:32

1

This post received KUDOS

Expert's post

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.

Re: Is 1 > |x-1| ? [#permalink]
26 Jun 2013, 10:40

4

This post received KUDOS

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.

Re: Is 1 > |x-1| ? [#permalink]
26 Jun 2013, 10:45

1

This post received KUDOS

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). _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Is 1 > |x-1| ? [#permalink]
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. _________________

Re: Is |x – 1| < 1? (1) (x – 1)² > 1 (2) x < 0 [#permalink]
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