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M25-01

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M25-01  [#permalink]

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New post 16 Sep 2014, 00:22
2
3
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

75% (00:48) correct 25% (00:59) wrong based on 470 sessions

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Re M25-01  [#permalink]

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New post 16 Sep 2014, 00:22
2
2
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 \(x-6 \gt 5\) 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
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Re: M25-01  [#permalink]

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New post 30 Sep 2014, 12:43
1
I use number line for Inequality problems as i find it more acurate
|x - 6|>5 or |x - 6|-5>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: M25-01  [#permalink]

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Re: M25-01  [#permalink]

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New post 08 Aug 2015, 12:37
1
this is a 700 level question?
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Re M25-01  [#permalink]

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New post 15 Oct 2015, 00:03
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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Re: M25-01  [#permalink]

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New post 15 Oct 2015, 20:38
WittyWitch wrote:
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?


Can you please post a screenshot of a question with these symbols? You should not get them.
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Re: M25-01  [#permalink]

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New post 01 Dec 2015, 05:57
(2) x2<1. So, −1<x<1, as all x-es 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 M25-01  [#permalink]

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New post 01 Dec 2015, 05:58
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. (2) x2<1. So, −1<x<1, as all x-es 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: M25-01  [#permalink]

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New post 01 Dec 2015, 06:33
Bahadur wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. (2) x2<1. So, −1<x<1, as all x-es 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: M25-01  [#permalink]

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New post 13 Dec 2016, 00:10
Bunuel wrote:
Official Solution:


Let's work on the stem first. For which values of \(x\) inequality \(|x - 6| &gt; 5\) is true?

If \(x \lt 6\), then \(-x+6 \gt 5\) or \(x \lt 1\).

If \(x \geq 6\), then \(x-6 \gt 5\) 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: M25-01  [#permalink]

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New post 13 Dec 2016, 01:18
jahidhassan wrote:
Bunuel wrote:
Official Solution:


Let's work on the stem first. For which values of \(x\) inequality \(|x - 6| &gt; 5\) is true?

If \(x \lt 6\), then \(-x+6 \gt 5\) or \(x \lt 1\).

If \(x \geq 6\), then \(x-6 \gt 5\) 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: M25-01  [#permalink]

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New post 19 Feb 2017, 17:05
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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New post 19 Feb 2017, 23:09
alados14 wrote:
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.


\(|x - 6| \gt 5\) is true if x<1 or x>11.

(2) says that −1<x<1, so we can say that yes, x is less than 1, thus \(|x - 6| \gt 5\) holds true.
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Re: M25-01  [#permalink]

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New post 21 Feb 2017, 04:03
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: M25-01  [#permalink]

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New post 05 Mar 2018, 06:46
HI chetan2u, Bunuel,

While converting the mod |x−6|>5

If x≥6, then (x-6)>5 ==> x>11

& If x<6, then (x-6)>-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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New post 05 Mar 2018, 06:52
NandishSS wrote:
HI chetan2u, Bunuel,

While converting the mod |x−6|>5

If x≥6, then (x-6)>5 ==> x>11

& If x-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?


when you take negative of mod..
when x<6, x-6<0..
-(x-6)>5 or \(x-6<-5\)
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Re: M25-01  [#permalink]

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New post 05 Mar 2018, 07:25
Quote:
when you take negative of mod..
when x<6, x-6<0..
-(x-6)>5 or \(x-6<-5\)


HI chetan2u,

So If I add -ve towards 5 than sign changes & if -ve is towards (x-6) >(sign) remains same

Pls correct me if I'm wrong!!!
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Re: M25-01  [#permalink]

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New post 05 Mar 2018, 07:40
NandishSS wrote:
Quote:
when you take negative of mod..
when x5 or [m]x-6(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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Re: M25-01 &nbs [#permalink] 05 Mar 2018, 07:40
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