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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
Re: Is 1 > |x-1| ? [#permalink]
19 May 2015, 21:31
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Re: Is 1 > |x-1| ? [#permalink]
20 May 2015, 04:46
Expert's post
For those who have difficulty dealing with modulus in inequality, you can use the definition of modulus to interpret the range of the variable inside the modulus.
In this question we are asked if |x -1 | < 1. We know that modulus denotes the magnitude of the distance of a number from a particular point. So |x - 1| means the distance of x from 1. Since | x - 1| < 1 it means that x lies at a distance of less than 1 unit on either side of 1.
Hence 1 - 1 < x < 1 + 1 i.e. 0 < x < 2 which is what the question is asking us.
Re: Is 1 > |x-1| ? [#permalink]
20 May 2015, 18:38
Expert's post
Hi All,
When absolute values 'interact' with inequalities, it can sometimes be difficult to automatically spot all of the possible values that "fit." In these situations, you can usually TEST VALUES to figure out which values "fit" and which do not.
Here, we're asked if 1 > |X-1|. This is a YES/NO question.
Before dealing with the two Facts, we can think about what would gives us a "YES" answer and what would give us a "NO" answer...
X = 1 is an obvious YES answer, since 1 > 0. X = 2 gives us a NO answer, since 1 is NOT > 1 X = 0 gives us a NO answer too.
So 0 and 2 are the "borders"; with a bit more 'playing around', you'll see that any value BETWEEN 0 and 2 will give us a YES answer. All other values, including 0 and 2, will give us a NO answer.
Fact 1: (X-1)^2 > 1
Neither 0 nor 2 will fit this inequality (since 1 is NOT > 1). Any value BETWEEN 0 and 2 will turn the parentheses into a positive fraction, which also doesn't fit (since a positive fraction is NOT > 1). By eliminating all of the values that would lead to a NO answer, the only values that fit will create a YES answer. The answer will ALWAYS be YES. Fact 1 is SUFFICIENT.
Fact 2: 0 > X
This Fact also eliminates all of the values that would lead to a NO answer, so all that is left are YES answers. Fact 2 is SUFFICIENT
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