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M08-11

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M08-11  [#permalink]

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New post 15 Sep 2014, 23:37
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A
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E

Difficulty:

  75% (hard)

Question Stats:

45% (01:24) correct 55% (01:27) wrong based on 115 sessions

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Re M08-11  [#permalink]

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New post 15 Sep 2014, 23:37
Official Solution:


(1) \((x - 1)^2 \le 1\). Since both sides of the inequality are non-negative then we can take square root from both parts: \(|x-1| \le 1\), so \(|x-1|\) can be less than 1 (answer YES), as well as equal to 1, for \(x=2\) or \(x=0\) (answer NO). Not sufficient. Notice that \(|x-1| \le 1\), means \(0 \le x \le 2\).

(2) \(x^2 - 1 \gt 0\). Rearrange: \(x^2 \gt 1\). Again, since both sides of the inequality are non-negative then we can take square root from both parts: \(|x| \gt 1\). If \(x=1.5\) then the answer is YES but if \(x=2\) then the answer is NO. Not sufficient.

(1)+(2) \(x=1.5\) and \(x=2\) satisfy both statements and give different answer to the question. Not sufficient.


Answer: E
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M08-11  [#permalink]

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New post 20 Jan 2016, 05:54
Is |x−1|<1 ?
(1) (x−1)^2≤1
(2) x^2−1>0
First,though it sounds absurd, try to test the condition whether sufficient or not even without using (1) or (2),. Place 5, 4<1, answer is no. Again, place -2, 3<1..still no. Now, place 2,1<1..still no.Now place 1, now we get the answer yes for 0<1. again, place 0, we get 1<1, no. So, once yes, once no suggests we need more data!
Now take (1) whether we get constant yes or constant no. By rooting, we get |x−1|≤1, means 0≤x≤2. Now we have restricted freedom to choose value. We can't choose any value less than 0 or more than 2. Lets try within this boundary. Place 0 or 2, we get 1 but is 1 less than 1?,NO. Try 1, |x−1|=0 (<1), YES,now less than 1. So, insufficient because of both outcome. We need either yes or no, not both simultaneously.
Think solely on (2) now.x^2−1>0;|x|>1. so, we have to take such values either greater than 1 or less than -1. Any less than -1 value makes |x−1| more than 1, so NO. In addition, placing any value 2 or more than 2 gives us same NO. Whatif, we try placing values within 0 to 1;say,1.5. we get |x−1|=.5(<1),so YES. Once yes,once no suggests insufficient.[if x were integer, (2) would be sufficient alone]
Now combine (1) and (2). (1) restricts the freedom to choose within 0≤x≤2. (2) restricts it by |x|>1.Even after combining, nothing changes.Still we can put 1<x≤2. Try putting 2, answer NO. Try putting 1.5, YES. So, combining still fails to give a reliable or constant answer.
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Re M08-11  [#permalink]

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New post 02 Feb 2016, 10:14
Hi - For this question, Stmt1 solutions assumed inequality is less than equal to. Hence solutions states 0<= x <= 2 but it should be 0 < x < 2 which will make stmt 1 Sufficient. Please confirm?
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Re: M08-11  [#permalink]

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New post 02 Feb 2016, 10:18
fydpoctave wrote:
Hi - For this question, Stmt1 solutions assumed inequality is less than equal to. Hence solutions states 0<= x <= 2 but it should be 0 < x < 2 which will make stmt 1 Sufficient. Please confirm?


Why should it be 0 < x < 2 ? Both 0 and 2 satisfy \((x−1)^2≤1\).
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Re: M08-11  [#permalink]

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New post 02 Feb 2016, 10:21
Bunuel wrote:
fydpoctave wrote:
Hi - For this question, Stmt1 solutions assumed inequality is less than equal to. Hence solutions states 0<= x <= 2 but it should be 0 < x < 2 which will make stmt 1 Sufficient. Please confirm?


Why should it be 0 < x < 2 ? Both 0 and 2 satisfy \((x−1)^2≤1\).


Sorry should have clarified. I was referring to question stem (following part from the solution)

Notice that |x−1|≤1, means 0≤x≤2.

But question has |x−1|<1 only
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Re: M08-11  [#permalink]

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New post 02 Feb 2016, 10:29
fydpoctave wrote:
Bunuel wrote:
fydpoctave wrote:
Hi - For this question, Stmt1 solutions assumed inequality is less than equal to. Hence solutions states 0<= x <= 2 but it should be 0 < x < 2 which will make stmt 1 Sufficient. Please confirm?


Why should it be 0 < x < 2 ? Both 0 and 2 satisfy \((x−1)^2≤1\).


Sorry should have clarified. I was referring to question stem (following part from the solution)

Notice that |x−1|≤1, means 0≤x≤2.

But question has |x−1|<1 only


Please read the solution carefully:

Statement 1 - \((x−1)^2≤1\) is equivalent to \(|x−1|≤1\), which translates to \(0≤x≤2\).
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Re M08-11  [#permalink]

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New post 20 Aug 2016, 03:57
I think this is a high-quality question and I agree with explanation.
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Re: M08-11  [#permalink]

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New post 07 May 2017, 01:06
Am I correct, that if statement 1) mentions \((x-1)^2<1\) instead of \((x-1)^2≤1\) that both statements together would be sufficient?
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New post 07 May 2017, 01:19
ChrisL1988 wrote:
Am I correct, that if statement 1) mentions \((x-1)^2<1\) instead of \((x-1)^2≤1\) that both statements together would be sufficient?


In this case the first statement itself would be sufficient because \((x-1)^2<1\) is the same as \(|x-1|<1\) (by taking the square root).
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Re: M08-11  [#permalink]

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New post 07 May 2017, 01:21
Bunuel wrote:
ChrisL1988 wrote:
Am I correct, that if statement 1) mentions \((x-1)^2<1\) instead of \((x-1)^2≤1\) that both statements together would be sufficient?


In this case the first statement itself would be sufficient because \((x-1)^2<1\) is the same as \(|x-1|<1\) (by taking the square root).


Of course, you are right, my fault. Thank you for the quick reply!
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Re: M08-11  [#permalink]

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New post 30 Sep 2017, 06:15
why the negative value of x in case of 2nd statement i.e x < -1 not considered while solving?
for eg wen x > 1, it is not sufficient. but when x< -1, in every case the eq exceeds 1 and therefore it is sufficient to solve. By this logic optn 2 shud be sufficient right?
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Re: M08-11  [#permalink]

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New post 30 Sep 2017, 06:19
Kunal14792 wrote:
why the negative value of x in case of 2nd statement i.e x < -1 not considered while solving?
for eg wen x > 1, it is not sufficient. but when x< -1, in every case the eq exceeds 1 and therefore it is sufficient to solve. By this logic optn 2 shud be sufficient right?


In a Yes/No Data Sufficiency questions, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".

We can get both an YES and a NO answers for the second statement (even not considering negative numbers) hence the statement is not sufficient.
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Re: M08-11  [#permalink]

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New post 21 Jul 2018, 05:18
Hello,

Where can I get some theory and a lot of practice questions on absolute numbers.

Thank you.
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Re: M08-11  [#permalink]

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New post 21 Jul 2018, 05:29
vedanshimurarka wrote:
Hello,

Where can I get some theory and a lot of practice questions on absolute numbers.

Thank you.


10. Absolute Value



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Re: M08-11 &nbs [#permalink] 21 Jul 2018, 05:29
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