Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 16 Jul 2019, 19:39 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # M25-01

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 56244

### Show Tags

2
3 00:00

Difficulty:   15% (low)

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

### HideShow timer Statistics Is $$|x - 6| \gt 5$$ ?

(1) $$x$$ is an integer

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

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 56244

### Show Tags

2
2
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.

_________________
Intern  Joined: 24 Sep 2014
Posts: 1
GMAT Date: 05-02-2015
GPA: 3.75
WE: Human Resources (Retail)

### Show Tags

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
Manager  S
Joined: 22 Mar 2014
Posts: 120
Location: United States
Concentration: Finance, Operations
GMAT 1: 530 Q45 V20 GPA: 3.91
WE: Information Technology (Computer Software)

### Show Tags

GOOD
Intern  Joined: 17 Jul 2015
Posts: 5

### Show Tags

1
this is a 700 level question?
Intern  Joined: 14 Oct 2015
Posts: 1

Why are the questions containing $symbols and many other Questions have terms like "Frac" etc. Its very difficult to understand the Question itself. May I know if the real GMAT test will also have them? Math Expert V Joined: 02 Sep 2009 Posts: 56244 Re: M25-01 [#permalink] ### Show Tags WittyWitch wrote: Why are the questions containing$ 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.
_________________
Intern  Joined: 18 Nov 2015
Posts: 5

### Show Tags

(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.
Intern  Joined: 18 Nov 2015
Posts: 5

### Show Tags

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 56244

### Show Tags

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.
_________________
Intern  B
Joined: 23 Jun 2016
Posts: 14

### Show Tags

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.

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.

Math Expert V
Joined: 02 Sep 2009
Posts: 56244

### Show Tags

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.

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.

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.
_________________
Intern  B
Joined: 03 Apr 2016
Posts: 16

### Show Tags

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 56244

### Show Tags

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.
_________________
SVP  V
Joined: 26 Mar 2013
Posts: 2283

### Show Tags

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

(1) $$x$$ is an integer

Put x =100 ....Answer is yes

Insufficient

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

-1 < x <1

Any negative number will yield yes

Sufficient

Director  V
Joined: 06 Jan 2015
Posts: 680
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE: Information Technology (Computer Software)

### Show Tags

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?
_________________
आत्मनॊ मोक्षार्थम् जगद्धिताय च

Resource: GMATPrep RCs With Solution
Math Expert V
Joined: 02 Aug 2009
Posts: 7764

### Show Tags

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 $$x-6<-5$$
_________________
Director  V
Joined: 06 Jan 2015
Posts: 680
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE: Information Technology (Computer Software)

### Show Tags

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!!!
_________________
आत्मनॊ मोक्षार्थम् जगद्धिताय च

Resource: GMATPrep RCs With Solution
Math Expert V
Joined: 02 Aug 2009
Posts: 7764

### Show Tags

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
_________________
Intern  B
Joined: 07 Jul 2018
Posts: 47

### Show Tags

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.

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
Display posts from previous: Sort by

# M25-01

Moderators: chetan2u, Bunuel  