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# M25-01

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Math Expert
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M25-01  [#permalink]

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16 Sep 2014, 01:22
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Question Stats:

74% (01:15) correct 26% (01:31) wrong based on 296 sessions

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Is $$|x - 6| \gt 5$$ ?

(1) $$x$$ is an integer

(2) $$x^2 \lt 1$$

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Re M25-01  [#permalink]

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16 Sep 2014, 01:22
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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
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Re: M25-01  [#permalink]

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30 Sep 2014, 13:43
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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06 Aug 2015, 08:57
GOOD
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Re: M25-01  [#permalink]

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08 Aug 2015, 13:37
1
this is a 700 level question?
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Re M25-01  [#permalink]

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15 Oct 2015, 01:03
Why are the questions containing $symbols and many other Questions have terms like "Frac" etc. Its very difficult to understand the Question itself. May I know if the real GMAT test will also have them? Math Expert Joined: 02 Sep 2009 Posts: 56244 Re: M25-01 [#permalink] ### Show Tags 15 Oct 2015, 21:38 WittyWitch wrote: Why are the questions containing$ 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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01 Dec 2015, 06:57
(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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01 Dec 2015, 06:58
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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01 Dec 2015, 07:33
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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13 Dec 2016, 01:10
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 cant 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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13 Dec 2016, 02:18
jahidhassan wrote:
Bunuel wrote:
Official Solution:

Let's work on the stem first. For which values of $$x$$ inequality $$|x - 6| &gt; 5$$ 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 cant 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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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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Re: M25-01  [#permalink]

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20 Feb 2017, 00:09
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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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: M25-01  [#permalink]

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05 Mar 2018, 07:46
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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Re: M25-01  [#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 (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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05 Mar 2018, 08:25
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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05 Mar 2018, 08:40
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  [#permalink]

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26 Apr 2019, 01:46
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

Can you please clarify this?
You said
So we have that inequality $$|x - 6| \gt 5$$ holds true for $$x \lt 1$$ and $$x \gt 11$$.

Shouldn't it be
$$|x - 6| \gt 5$$ holds true for $$x \lt 1$$ OR $$x \gt 11$$.

Thanks.
Re: M25-01   [#permalink] 26 Apr 2019, 01:46
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# M25-01

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