Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 30 Aug 2015, 02:06

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0

Author Message
TAGS:
Math Expert
Joined: 02 Sep 2009
Posts: 29110
Followers: 4726

Kudos [?]: 49769 [0], given: 7403

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| > 0 [#permalink]  02 Nov 2012, 01:00
Expert's post
himanshuhpr wrote:
Is |x| < 1?

(1) |x + 1| = 2|x - 1|
(2) |x - 3| > 0

From 1st statement we know that x = 1/3 or x= 3
From 2nd statement we know that that x>3 or x<3

My question is can we combine these 2 results and say (c) is sufficient as x>3 in statement creates an ambiguity and therfore it should be E.

(2) |x - 3| > 0. Absolute value is always more than or equal to zero. Thus this statement just says that $$|x - 3|\neq{0}$$ --> $$x\neq{3}$$.

Hope it helps.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 29110
Followers: 4726

Kudos [?]: 49769 [0], given: 7403

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  07 Jul 2013, 23:53
Expert's post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

To find DS questions by Kudos, sort by Kudos here: gmat-data-sufficiency-ds-141/
To find PS questions by Kudos, sort by Kudos here: gmat-problem-solving-ps-140/

_________________
Moderator
Joined: 25 Apr 2012
Posts: 734
Location: India
GPA: 3.21
Followers: 31

Kudos [?]: 459 [1] , given: 723

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  08 Jul 2013, 00:24
1
KUDOS
yezz wrote:
Is |x| < 1?

(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

From St1 we can say that since both LHS and RHS is positive therefore squaring both sides we get

(x+1)^2= 4(x-1)^2
On simplifying we get 3x^2-10x-3= 0 -----> x=3 or x= 1/3

2 ans and hence st 1 is not sufficient. Therefore Option A and D ruled out

St 2 we have |x - 3| ≠ 0 and since Modulus of any no is greater than or equal to 0 (In this case greater than 0) which means x not equal to 3

Not sufficient.

Combining both equation we see that x=1/3 which is in the Question stem range

Ans C
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Senior Manager
Joined: 13 May 2013
Posts: 475
Followers: 1

Kudos [?]: 91 [1] , given: 134

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  08 Jul 2013, 15:27
1
KUDOS
Is |x| < 1?

(1) |x + 1| = 2|x - 1|
(My first problem was to decide how to solve this. I had recently tackled a problem to the effect of |x-z|=|x-y| where I squared both sides to help simplify. In this case, it might be better to find the check points of the LHS and RHS as we are given a variable and an integer as opposed to all variables like the problem I had to solve before)
|x + 1| = 2|x - 1|
x=-1, x=1
so, x<-1, -1<x<1, x>1
x<-1:
-(x+1) = 2 -(x-1)
-x-1 = 2(-x+1)
-x-1 = -2x+2
x=3 INVALID as 3 does not fall within the range of x<-1

-1<x<1
(x+1) = 2 -(x-1)
x+1 = -2x+2
3x=1
x=1/3 VALID as x falls within the range of -1<x<1

x>1
(x+1)=2(x-1)
x+1=2x-2
-x=-3
x=3 VALID as x falls within the range of x>1

We have two valid values for x and therefore we cannot determine a single, correct answer.
INSUFFICIENT

(2) |x - 3| ≠ 0
Tells us nothing about the sign of x. x could be -10 or +10 and |x - 3| ≠ 0 would be valid.
INSUFFICIENT

1+2) x=1/3 OR x=3 AND |x - 3| ≠ 0

Plug in both x values into the above equation. x=3 fails as it does = 0 when the equation says it does not. Therefore, the only remaining x value is 1/3.
SUFFICIENT

(C)
Intern
Joined: 02 Sep 2013
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  03 Sep 2013, 05:27
From (1) Taking sqaure of both sides ((X+1)^2)/((X-1)^2 = 4, solving this you get X = 3 or 1/3, which is not sufficient.

(2) says X is not = 3

Hence both together tell us the answer (C)
Manager
Joined: 28 Aug 2012
Posts: 52
Concentration: Operations, Marketing
GMAT 1: 510 Q36 V25
GPA: 4
WE: Information Technology (Other)
Followers: 0

Kudos [?]: 79 [0], given: 105

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  03 Sep 2013, 05:54
I m really sorry, even after reading all the above replies.. i just could'nt get it..

Could someone point out to what i m missing out here..
As per my understanding $$|x| < 1$$ means $$-1 < x < 1$$ .. which eventually means is x=0 ?
Why do we have to consider the cases of x <-1 etc etc.. ?
Math Expert
Joined: 02 Sep 2009
Posts: 29110
Followers: 4726

Kudos [?]: 49769 [0], given: 7403

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  03 Sep 2013, 05:58
Expert's post
thinktank wrote:
I m really sorry, even after reading all the above replies.. i just could'nt get it..

Could someone point out to what i m missing out here..
As per my understanding $$|x| < 1$$ means $$-1 < x < 1$$ .. which eventually means is x=0 ?
Why do we have to consider the cases of x <-1 etc etc.. ?

Why does "is $$-1 < x < 1$$?" means "is x=0"? We are not told that x is an integer.
_________________
SVP
Joined: 06 Sep 2013
Posts: 2046
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 32

Kudos [?]: 340 [0], given: 355

Re: Is |x| < 1? [#permalink]  30 Jan 2014, 06:18
Economist wrote:
C

1)
if x>1, solving the equality we get x=3.
if -1<x<1, we get x=1/3.
if x<-1, we get x=3.

Not suff
2)
we get x is not equal to 3.

Combining, x can only be 1/3
OA?

Actually your third range is wrong I believe. If x<-1 then x cannot be 3, it isn't a valid solution but the rest is OK

Cheers
J
Intern
Joined: 23 Dec 2011
Posts: 39
Location: United States
Concentration: Technology, General Management
GPA: 3.83
WE: Programming (Computer Software)
Followers: 1

Kudos [?]: 10 [0], given: 26

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  05 Jul 2014, 00:21
This is very good question
X<-1
-x-1=2(-x+1)

-1<=x<=1
x+1=2(-x+1)

X>1
x+1=2(x-1)
Intern
Joined: 05 Feb 2014
Posts: 24
Followers: 0

Kudos [?]: 2 [0], given: 65

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  02 Sep 2014, 03:17
Bunuel wrote:
tejal777 wrote:
I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

Is $$|x| < 1$$?

Is $$|x| < 1$$, means is $$x$$ in the range (-1,1) or is $$-1<x<1$$ true?

(1) $$|x + 1| = 2|x - 1|$$
Two key points: $$x=-1$$ and $$x=1$$ (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------

A. $$x<-1$$ (blue range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$-x-1=2(-x+1)$$ --> $$x=3$$, not OK, as this value is not in the range we are checking ($$x<-1$$);
B. $$-1\leq{x}\leq{1}$$ (green range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(-x+1)$$ --> $$x=\frac{1}{3}$$. OK, as this value is in the range we are checking ($$-1\leq{x}\leq{1}$$);
C. $$x>1$$ (red range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(x-1)$$ --> $$x=3$$. OK, as this value is in the range we are checking ($$x>1$$).

So we got TWO values of $$x$$ (two solutions): $$\frac{1}{3}$$ and $$3$$, first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) $$|x - 3|\neq{0}$$
Just says that $$x\neq{3}$$. But we don't know whether $$x$$ is in the range (-1,1) or not.

(1)+(2) $$x=\frac{1}{3}$$ or $$x=3$$ AND $$x\neq{3}$$ --> means $$x$$ can have only value $$\frac{1}{3}$$, which is in the range (-1,1). Sufficient.

Hope it helps.

Manager
Joined: 02 Jul 2012
Posts: 216
Location: India
GMAT 1: 720 Q50 V38
GPA: 2.3
WE: Consulting (Consulting)
Followers: 9

Kudos [?]: 114 [0], given: 82

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  20 Oct 2014, 22:57
yezz wrote:
Is |x| < 1?

(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

1 - On solving the equation, we get the following values of x - 3, 1/3
2 - There can be infinite values of x, except 3

Combining 1 & 2 = x = 1/3

Ans. C
_________________

Give KUDOS if the post helps you...

Manager
Joined: 06 Aug 2013
Posts: 86
Followers: 0

Kudos [?]: 2 [0], given: 16

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  21 Oct 2014, 03:46
Bunuel wrote:
tejal777 wrote:
I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

Is $$|x| < 1$$?

Is $$|x| < 1$$, means is $$x$$ in the range (-1,1) or is $$-1<x<1$$ true?

(1) $$|x + 1| = 2|x - 1|$$
Two key points: $$x=-1$$ and $$x=1$$ (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------

A. $$x<-1$$ (blue range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$-x-1=2(-x+1)$$ --> $$x=3$$, not OK, as this value is not in the range we are checking ($$x<-1$$);
B. $$-1\leq{x}\leq{1}$$ (green range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(-x+1)$$ --> $$x=\frac{1}{3}$$. OK, as this value is in the range we are checking ($$-1\leq{x}\leq{1}$$);
C. $$x>1$$ (red range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(x-1)$$ --> $$x=3$$. OK, as this value is in the range we are checking ($$x>1$$).

So we got TWO values of $$x$$ (two solutions): $$\frac{1}{3}$$ and $$3$$, first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) $$|x - 3|\neq{0}$$
Just says that $$x\neq{3}$$. But we don't know whether $$x$$ is in the range (-1,1) or not.

(1)+(2) $$x=\frac{1}{3}$$ or $$x=3$$ AND $$x\neq{3}$$ --> means $$x$$ can have only value $$\frac{1}{3}$$, which is in the range (-1,1). Sufficient.

Hope it helps.

Hi Bunuel,
I got most of it, just one doubt here
in case of |x|, we either consider x>0 or x<0,
why are we considering values between -1<x<1.
the reason i am asking this is because the question doesnt ask us to do that, right?
Math Expert
Joined: 02 Sep 2009
Posts: 29110
Followers: 4726

Kudos [?]: 49769 [0], given: 7403

Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]  21 Oct 2014, 05:25
Expert's post
arnabs wrote:
Bunuel wrote:
tejal777 wrote:
I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

Is $$|x| < 1$$?

Is $$|x| < 1$$, means is $$x$$ in the range (-1,1) or is $$-1<x<1$$ true?

(1) $$|x + 1| = 2|x - 1|$$
Two key points: $$x=-1$$ and $$x=1$$ (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------

A. $$x<-1$$ (blue range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$-x-1=2(-x+1)$$ --> $$x=3$$, not OK, as this value is not in the range we are checking ($$x<-1$$);
B. $$-1\leq{x}\leq{1}$$ (green range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(-x+1)$$ --> $$x=\frac{1}{3}$$. OK, as this value is in the range we are checking ($$-1\leq{x}\leq{1}$$);
C. $$x>1$$ (red range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(x-1)$$ --> $$x=3$$. OK, as this value is in the range we are checking ($$x>1$$).

So we got TWO values of $$x$$ (two solutions): $$\frac{1}{3}$$ and $$3$$, first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) $$|x - 3|\neq{0}$$
Just says that $$x\neq{3}$$. But we don't know whether $$x$$ is in the range (-1,1) or not.

(1)+(2) $$x=\frac{1}{3}$$ or $$x=3$$ AND $$x\neq{3}$$ --> means $$x$$ can have only value $$\frac{1}{3}$$, which is in the range (-1,1). Sufficient.

Hope it helps.

Hi Bunuel,
I got most of it, just one doubt here
in case of |x|, we either consider x>0 or x<0,
why are we considering values between -1<x<1.
the reason i am asking this is because the question doesnt ask us to do that, right?

It seems that you need to brush up fundamentals.

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1381430
_________________
Intern
Joined: 17 May 2015
Posts: 19
Followers: 0

Kudos [?]: 1 [0], given: 2

Is |x| < 1 ? (1) |x + 1| = 2|x – 1| (2) |x – 3| > 0 [#permalink]  13 Jul 2015, 05:31
Need help for a simple solution to get the answer as I m confuse with this question , many thankss
Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0

Really appreciate it , Many Thanks
Math Forum Moderator
Joined: 20 Mar 2014
Posts: 924
Concentration: Finance, Strategy
GMAT 1: 650 Q49 V30
GMAT 2: 690 Q49 V34
GMAT 3: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Followers: 20

Kudos [?]: 288 [0], given: 137

Re: Is |x| < 1 ? (1) |x + 1| = 2|x – 1| (2) |x – 3| > 0 [#permalink]  13 Jul 2015, 05:51
apple08 wrote:
Need help for a simple solution to get the answer as I m confuse with this question , many thankss
Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0

Really appreciate it , Many Thanks

Good question.

For absolute values, you need to take into consideration all possible values. For example, if I say |x| <1, then -1<x<1. We get this by looking at 2 cases :

1. When x>= 0, |x|=x ---> x<1. This value of x<1 does satisfy x>=0. So keep this.

2. When x<0 , |x| = -x ----> -x<1 ----> x>-1 . This value of x>-1 (think x= -0.7 or -0.5) still satisfies x<0. So keep this value as well.

Thus, from 1 and 2 , |x| <1 is nothing but a fancy way of saying -1<x<1. You can also visualise this from numberline:

|x| is the distance of x from 0 (as |x| = |x-0|) on the number line. Thus , when I say |x| <1 this means that distance of x from 0 is less than 1 on either side (as we can approach 0 from 2 directions, 1 from positive and 1 from the negative!).

So, the question actually asks is |x| <1 ----> is -1<x<1.

Now lets look at the 2 statements,

Per statement 1, |x + 1| = 2|x – 1|

Now, your points of analysis are x+1 =0 ----> x=-1 and x-1=0---> x=1. Thus you will have 3 regions to look at:

1. x<-1
2. -1<x<1
3. x>1

When you look at the above 3 regions, you will see that case (1) is not going to satisfy while case (2) will give you x=1/3 and case 3 will give you x = 3. So we see that -1<x<1 is true if x=1/3 but is false if x=3. Thus this statement is not sufficient.

Per statement 2, |x-3| >0 ----> this is true for all x as if you substitute x = -1 or -2 or -0.75 or 7 or 10,|x-3| will always greater than 0 as |x| is DISTANCE of x from 0 (as |x| = |x-0|). As such |x-3| is also distance of x from 3 and as distance is always > 0, this is true for all values of x. Note, that we will not take x = 3 as we have been given |x-3| > 0 and not |x-3| >= 0. Thus this statement is not sufficient.

Combining 1 and 2 we get,

x=1/3 or x=3

Now, |x-3| >0 will not be true if x=3 as |3-3| = 0 and NOT >0. Thus only 1 value of x is possible ----> x = 1/3. Hence both the statements together are sufficient. C is the correct answer.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
GMATCLUB Math Book: http://gmatclub.com/forum/gmat-math-book-87417.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 924
Concentration: Finance, Strategy
GMAT 1: 650 Q49 V30
GMAT 2: 690 Q49 V34
GMAT 3: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Followers: 20

Kudos [?]: 288 [0], given: 137

Is |x| < 1 ? (1) |x + 1| = 2|x – 1| (2) |x – 3| > 0 [#permalink]  13 Jul 2015, 05:55
apple08 wrote:
Need help for a simple solution to get the answer as I m confuse with this question , many thankss
Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0

Really appreciate it , Many Thanks

Please search for questions before posting. The same question has been discussed at is-x-1-1-x-1-2-x-1-2-x-84652.html#p635009

Theory on absolute values: absolute-value-tips-and-hints-175002.html , math-absolute-value-modulus-86462.html
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
GMATCLUB Math Book: http://gmatclub.com/forum/gmat-math-book-87417.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Is |x| < 1 ? (1) |x + 1| = 2|x – 1| (2) |x – 3| > 0   [#permalink] 13 Jul 2015, 05:55

Go to page   Previous    1   2   [ 36 posts ]

Similar topics Replies Last post
Similar
Topics:
Is |x| < 1 ? (1) |x + 1| = 2|x – 1| (2) |x – 3| > 0 0 13 Jul 2015, 05:55
x| < 1? 1) |x+1| = 2|x-1| 2) |x-3| does not equal 0 4 01 May 2011, 19:59
9 If x#0, is x^2/|x| < 1? 12 08 Sep 2010, 10:51
10 If x 0, is x^2/|x| < 1? 7 03 Jan 2010, 06:14
5 If x 0, is x^2 / |x| < 1? 12 14 Dec 2009, 00:03
Display posts from previous: Sort by