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# M08-11

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Math Expert
Joined: 02 Sep 2009
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M08-11  [#permalink]

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16 Sep 2014, 00:37
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Difficulty:

75% (hard)

Question Stats:

46% (01:22) correct 54% (01:23) wrong based on 111 sessions

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Is $$|x - 1| \lt 1$$ ?

(1) $$(x - 1)^2 \le 1$$

(2) $$x^2 - 1 \gt 0$$

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16 Sep 2014, 00:37
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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20 Jan 2016, 06:54
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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02 Feb 2016, 11:14
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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02 Feb 2016, 11:18
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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02 Feb 2016, 11:21
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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02 Feb 2016, 11:29
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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20 Aug 2016, 04:57
I think this is a high-quality question and I agree with explanation.
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Joined: 07 Feb 2016
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Re: M08-11  [#permalink]

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07 May 2017, 02:06
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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Re: M08-11  [#permalink]

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07 May 2017, 02:19
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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07 May 2017, 02:21
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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30 Sep 2017, 07:15
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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30 Sep 2017, 07:19
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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21 Jul 2018, 06:18
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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21 Jul 2018, 06:29
Re: M08-11 &nbs [#permalink] 21 Jul 2018, 06:29
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# M08-11

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