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16 Sep 2014, 01:22
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Re: M2501 [#permalink]
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30 Sep 2014, 13:43
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I use number line for Inequality problems as i find it more acurate x  6>5 or x  65>0 expression is 0 for 1 and 11 divides number line in 3 segments x<1, 1<x<6, x>11 now expression is positive for x>11 and x<1 while for 1<x<11 is negative.
(1) x is an integer not sufficient (2) x^2< 1 or 1<x<1 which is sufficient since expression is positive for all values for x<1



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Re: M2501 [#permalink]
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06 Aug 2015, 08:57
GOOD



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08 Aug 2015, 13:37
this is a 700 level question?



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15 Oct 2015, 01:03
Why are the questions containing $ symbols and many other Questions have terms like "Frac" etc. Its very difficult to understand the Question itself. May I know if the real GMAT test will also have them?



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15 Oct 2015, 21:38



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Re: M2501 [#permalink]
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01 Dec 2015, 06:57
(2) x2<1. So, −1<x<1, as all xes from this range are in the range x<1, hence inequality x−6>5 holds true. Sufficient.
I dont understand this statement. Because 1<x<1 means X=0 Or x=.25 and any other valu. But question inference say that x<1 or X>11. So 1<x<1 how x satisfy x>11. I am not clear about this. Would any one like to help me.



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Re M2501 [#permalink]
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01 Dec 2015, 06:58
I think this is a highquality question and the explanation isn't clear enough, please elaborate. (2) x2<1. So, −1<x<1, as all xes from this range are in the range x<1, hence inequality x−6>5 holds true. Sufficient.
I dont understand this statement. Because 1<x<1 means X=0 Or x=.25 and any other valu. But question inference say that x<1 or X>11. So 1<x<1 how x satisfy x>11. I am not clear about this. Would any one like to help me.



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Re: M2501 [#permalink]
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01 Dec 2015, 07:33
Bahadur wrote: I think this is a highquality question and the explanation isn't clear enough, please elaborate. (2) x2<1. So, −1<x<1, as all xes from this range are in the range x<1, hence inequality x−6>5 holds true. Sufficient.
I dont understand this statement. Because 1<x<1 means X=0 Or x=.25 and any other valu. But question inference say that x<1 or X>11. So 1<x<1 how x satisfy x>11. I am not clear about this. Would any one like to help me. The question asks whether x<1 and x>11. (2) says that −1<x<1, so we can say that yes, x is less than 1, which makes this statement sufficient.
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Re: M2501 [#permalink]
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13 Dec 2016, 01:10
Bunuel wrote: Official Solution:
Let's work on the stem first. For which values of \(x\) inequality \(x  6 > 5\) is true? If \(x \lt 6\), then \(x+6 \gt 5\) or \(x \lt 1\). If \(x \geq 6\), then \(x6 \gt 5\) or \(x \gt 11\). So we have that inequality \(x  6 \gt 5\) holds true for \(x \lt 1\) and \(x \gt 11\). (1) \(x\) is an integer. Clearly not sufficient. \(x\) can be 12 and the inequality holds true as we concluded OR \(x\) can be 5 and inequality doesn't hold true. (2) \(x^2 \lt 1\). So, \(1 \lt x \lt 1\), as all \(x\)es from this range are in the range \(x \lt 1\), hence inequality \(x  6 \gt 5\) holds true. Sufficient.
Answer: B I can`t understand why the option C is not the right answer. According to S2, x can only be anything within 0.99 to 0.99 as 1<x<1. While we take S1, we can say that there is only one integer which is O. Therefore, we can find a certain value for x. Please correct me if i am wrong. Thanks in advance.



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Re: M2501 [#permalink]
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13 Dec 2016, 02:18
jahidhassan wrote: Bunuel wrote: Official Solution:
Let's work on the stem first. For which values of \(x\) inequality \(x  6 > 5\) is true? If \(x \lt 6\), then \(x+6 \gt 5\) or \(x \lt 1\). If \(x \geq 6\), then \(x6 \gt 5\) or \(x \gt 11\). So we have that inequality \(x  6 \gt 5\) holds true for \(x \lt 1\) and \(x \gt 11\). (1) \(x\) is an integer. Clearly not sufficient. \(x\) can be 12 and the inequality holds true as we concluded OR \(x\) can be 5 and inequality doesn't hold true. (2) \(x^2 \lt 1\). So, \(1 \lt x \lt 1\), as all \(x\)es from this range are in the range \(x \lt 1\), hence inequality \(x  6 \gt 5\) holds true. Sufficient.
Answer: B I can`t understand why the option C is not the right answer. According to S2, x can only be anything within 0.99 to 0.99 as 1<x<1. While we take S1, we can say that there is only one integer which is O. Therefore, we can find a certain value for x. Please correct me if i am wrong. Thanks in advance. The question does not ask to find the value of x. The question asks whether x<1 and x>11. (2) says that −1<x<1, so we can say that yes, x is less than 1, which makes this statement sufficient.
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Re: M2501 [#permalink]
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19 Feb 2017, 18:05
Hi Bunuel, If the original stmt is x<1 and x>11
In Stmt II I think that the range 1<x<1 satisfies only one of the two conditions. In this case x<1. what about x>11? Are you implying that x<1<11? Kindly explain.



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20 Feb 2017, 00:09



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Re: M2501 [#permalink]
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21 Feb 2017, 05:03
Is \(x  6 \gt 5\) ?
(1) \(x\) is an integer
Put x =100 ....Answer is yes
Put x =1.........Answer is No
Insufficient
(2) \(x^2 \lt 1\)
1 < x <1 Put x=0.99...........5.01> 5..Answer is yes
Put x=0................6>5.......Answer is Yes
Any negative number will yield yes
Sufficient
Answer: B



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Re: M2501 [#permalink]
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05 Mar 2018, 07:46
HI chetan2u, Bunuel, While converting the mod x−6>5 If x≥6, then (x6)>5 ==> x>11 & If x<6, then (x6)>5 ==> x>1 I have taken ve sign towards 5, where Am'I going wrong? Do I need to change the sign(ve) if I take sign on 5.. Can you pls elaborate?
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Re: M2501 [#permalink]
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05 Mar 2018, 07:52
NandishSS wrote: HI chetan2u, Bunuel, While converting the mod x−6>5 If x≥6, then (x6)>5 ==> x>11 & If x5 ==> x>1I have taken ve sign towards 5, where Am'I going wrong? Do I need to change the sign(ve) if I take sign on 5.. Can you pls elaborate? when you take negative of mod.. when x<6, x6<0.. (x6)>5 or \(x6<5\)
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Re: M2501 [#permalink]
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05 Mar 2018, 08:25
Quote: when you take negative of mod.. when x<6, x6<0.. (x6)>5 or \(x6<5\) HI chetan2u, So If I add ve towards 5 than sign changes & if ve is towards (x6) >(sign) remains same Pls correct me if I'm wrong!!!
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Re: M2501 [#permalink]
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05 Mar 2018, 08:40
NandishSS wrote: Quote: when you take negative of mod.. when x5 or [m]x6(sign) remains same
Pls correct me if I'm wrong!!! Whenever you multiply both sides of INEQUALITY by  sign, the INEQUALITY sign reverses.. Reason.. 2<1.. (2)>(1)...2>1
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