Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 24 May 2017, 10:05

### 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

# What is x?

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 07 Nov 2009
Posts: 306
Followers: 9

Kudos [?]: 592 [2] , given: 20

### Show Tags

31 Jan 2012, 18:48
2
KUDOS
20
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

47% (01:43) correct 53% (00:46) wrong based on 674 sessions

### HideShow timer Statistics

What is x?

(1) |x| < 2
(2) |x| = 3x – 2
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Jul 2013, 00:52, edited 1 time in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 38856
Followers: 7724

Kudos [?]: 106011 [15] , given: 11602

### Show Tags

31 Jan 2012, 18:56
15
KUDOS
Expert's post
8
This post was
BOOKMARKED
rohitgoel15 wrote:
What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

What is x?

(1) |x| < 2 --> $$-2<x<2$$. Not sufficient.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: $$3x-2\geq{0}$$ --> $$x\geq{\frac{2}{3}}$$ --> $$x$$ positive, so $$|x|=x$$ --> $$x=3x-2$$ --> $$x=1$$. Sufficient

Hope it's clear.
_________________
Senior Manager
Joined: 12 Mar 2010
Posts: 361
Concentration: Marketing, Entrepreneurship
GMAT 1: 680 Q49 V34
Followers: 2

Kudos [?]: 193 [0], given: 87

Re: What is x? (1) |x| < 2 (2) |x| = 3x – 2 [#permalink]

### Show Tags

31 Jan 2012, 20:38
Vow! that is a very good solution It opened my eyes on how to read modulus based problems...
Senior Manager
Joined: 07 Nov 2009
Posts: 306
Followers: 9

Kudos [?]: 592 [0], given: 20

### Show Tags

01 Feb 2012, 03:42
Bunuel wrote:
rohitgoel15 wrote:
What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

What is x?

(1) |x| < 2 --> $$-2<x<2$$. Not sufficient.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: $$3x-2\geq{0}$$ --> $$x\geq{\frac{2}{3}}$$ --> $$x$$ positive, so $$|x|=x$$ --> $$x=3x-2$$ --> $$x=1$$. Sufficient

Hope it's clear.

Hi Bunuel,

Thanks for reply .. But this is not clear to me.
Generally in modulus questions we do as below:
(1) |x| < 2
I agree with u. -2<x<2
(2) |x| = 3x – 2
In this case i did
a. x is positive so x = 3x - 2 , upon solving x = 1
a. x is negative so -x = 3x - 2 , upon solving x = 1/2

Why have you considered the bold part in this particular question. for eg. if |x| = 2 we always consider x = +2/-2.
I am confused !
Math Expert
Joined: 02 Sep 2009
Posts: 38856
Followers: 7724

Kudos [?]: 106011 [5] , given: 11602

### Show Tags

01 Feb 2012, 04:05
5
KUDOS
Expert's post
1
This post was
BOOKMARKED
rohitgoel15 wrote:
Bunuel wrote:
rohitgoel15 wrote:
What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

What is x?

(1) |x| < 2 --> $$-2<x<2$$. Not sufficient.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: $$3x-2\geq{0}$$ --> $$x\geq{\frac{2}{3}}$$ --> $$x$$ positive, so $$|x|=x$$ --> $$x=3x-2$$ --> $$x=1$$. Sufficient

Hope it's clear.

Hi Bunuel,

Thanks for reply .. But this is not clear to me.
Generally in modulus questions we do as below:
(1) |x| < 2
I agree with u. -2<x<2
(2) |x| = 3x – 2
In this case i did
a. x is positive so x = 3x - 2 , upon solving x = 1
a. x is negative so -x = 3x - 2 , upon solving x = 1/2

Why have you considered the bold part in this particular question. for eg. if |x| = 2 we always consider x = +2/-2.
I am confused !

Your way is more common, orthodox. Though you should discard x=1/2 as you get this value when considering x<0 and 1/2 is not less than 0. Also, if x=1/2 then 3x-2<0 and you get that |x|=negative, which is never true.

Next, the way I'm doing is based on the fact that absolute value is always non negative, thus 3x-2 which equals to some absolute value must also be non-negative: $$3x-2\geq{0}$$ --> $$x\geq{\frac{2}{3}}$$ --> $$x$$ positive, so $$|x|=x$$ and no need to consider $$|x|=-x$$ --> $$x=3x-2$$ --> $$x=1$$.

Check Absolute Values chapter of Math Book for more: math-absolute-value-modulus-86462.html

Hope it helps.
_________________
Manager
Joined: 23 Aug 2011
Posts: 81
Followers: 3

Kudos [?]: 222 [0], given: 13

what is the value of x? [#permalink]

### Show Tags

27 Aug 2012, 10:38
What is the value of x?

(1) |x| < 2

(2) |x| = 3x – 2

I marked E but i was wrong, my reasoning was as follows:

statement one has a range (-2<x<2) of values so it is insufficient.

Statement 2: i solved for two cases

when x>0
x=3x-2
x=1

when x<0
-x=3x-2 ( |x|=-x if x is -ve)
x=1/2

again 2 values not sufficient.
Using 1 and 2 it also does not help either. SO I marked E. But somewhere i went wrong.
Can some one point out my mistake and also let me know how to best deal with combo of inequalities ad absolute values such as "|x| = 3x – 2" in DS problems.
_________________

Whatever one does in life is a repetition of what one has done several times in one's life!
If my post was worth it, then i deserve kudos

Last edited by conty911 on 27 Aug 2012, 11:20, edited 2 times in total.
Math Expert
Joined: 02 Sep 2009
Posts: 38856
Followers: 7724

Kudos [?]: 106011 [2] , given: 11602

### Show Tags

27 Aug 2012, 10:42
2
KUDOS
Expert's post
conty911 wrote:
Sorry, if this question has already been posted earlier, i tried searching the forum but it never came up.

What is x?

(1) |x| < 2

(2) |x| = 3x – 2

statement one has a range (-2<x<2) of values so it is insufficient.

Statement 2: i solved for two cases

when x>0
x=3x-2
x=1

when x<0
-x=3x-2 ( |x|=-x if x is -ve)
x=1/2

again 2 values not sufficient.
Using 1 and 2 it also does not help either. SO I marked E. But somewhere i went wrong.
Can some one point out my mistake and also let me know how to best deal with combo of inequalities ad absolute values such as "|x| = 3x – 2" in DS problems.

Merging similar topics. Please refer to the solution above.
_________________
Manager
Joined: 23 Aug 2011
Posts: 81
Followers: 3

Kudos [?]: 222 [1] , given: 13

### Show Tags

27 Aug 2012, 11:24
1
KUDOS
Bunuel wrote:
conty911 wrote:
Sorry, if this question has already been posted earlier, i tried searching the forum but it never came up.

What is x?

(1) |x| < 2

(2) |x| = 3x – 2

statement one has a range (-2<x<2) of values so it is insufficient.

Statement 2: i solved for two cases

when x>0
x=3x-2
x=1

when x<0
-x=3x-2 ( |x|=-x if x is -ve)
x=1/2

again 2 values not sufficient.
Using 1 and 2 it also does not help either. SO I marked E. But somewhere i went wrong.
Can some one point out my mistake and also let me know how to best deal with combo of inequalities ad absolute values such as "|x| = 3x – 2" in DS problems.

Merging similar topics. Please refer to the solution above.

Thanks for merging the topic .
By the way the post is tagged in the wrong section , it should be in DS rather than PS.section?
_________________

Whatever one does in life is a repetition of what one has done several times in one's life!
If my post was worth it, then i deserve kudos

Math Expert
Joined: 02 Sep 2009
Posts: 38856
Followers: 7724

Kudos [?]: 106011 [0], given: 11602

### Show Tags

27 Aug 2012, 11:48
conty911 wrote:
Thanks for merging the topic .
By the way the post is tagged in the wrong section , it should be in DS rather than PS.section?

Topic moved to the DS forum. Thank you.
_________________
Intern
Joined: 07 Jul 2013
Posts: 7
Followers: 0

Kudos [?]: 6 [1] , given: 2

Re: What is x? (1) |x| < 2 (2) |x| = 3x – 2 [#permalink]

### Show Tags

11 Jul 2013, 00:51
1
KUDOS
Think of statement 2 this way:

If we consider x to be negative, then -x=3x-2 which translates to x=1/2. Now when we test this situation 1/2=3*1/2-2 the outcome is =3/2-2= -1/2. The RHS is not equal to the LHS. Hence x has to be 1, because when we test x=3x-2 we get x=1.
and the RHS and LHS are equal to each other
Senior Manager
Joined: 13 May 2013
Posts: 468
Followers: 3

Kudos [?]: 172 [3] , given: 134

### Show Tags

11 Jul 2013, 11:22
3
KUDOS
What is x?

(1) |x| < 2

x<2 x>-2
Invalid, x could be any number greater than 2 or less than 2.
INSUFFICIENT

(2) |x| = 3x – 2
x>=0
|x| = 3x – 2
x = 3x - 2
-2x = -2
x = 1 Valid
OR
x<0
-x = 3x - 2
-4x = -2
x = (1/2) Invalid as 1/2 is not less than zero.
SUFFICIENT

(B)
Intern
Joined: 12 Mar 2013
Posts: 4
Concentration: Marketing, General Management
Schools: HBS '16 (A)
GMAT 1: 750 Q50 V41
Followers: 1

Kudos [?]: 7 [0], given: 22

### Show Tags

21 Jul 2013, 06:29
What is x?

(1) |x| < 2

(2) |x| = 3x – 2

[Reveal] Spoiler: Variable inside absolute value
I solved equation 2 and got two values. Hence I thought 2 was insufficient. However, mgmat says "When dealing with absolute values that contain variables, you should always check that the solution is valid".They solve the equation, and then test the value once again in the equation and one the solutions is now invalid. Why is such a test required? Usually when you solve a equation, what you get is the SOLUTION right? Can someone please clarify why one solution will be 'invalid' in case of 'variables with absolute values'? I am unable to visualize that.
Math Expert
Joined: 02 Sep 2009
Posts: 38856
Followers: 7724

Kudos [?]: 106011 [0], given: 11602

### Show Tags

21 Jul 2013, 06:32
chk86 wrote:
What is x?

(1) |x| < 2

(2) |x| = 3x – 2

[Reveal] Spoiler: Variable inside absolute value
I solved equation 2 and got two values. Hence I thought 2 was insufficient. However, mgmat says "When dealing with absolute values that contain variables, you should always check that the solution is valid".They solve the equation, and then test the value once again in the equation and one the solutions is now invalid. Why is such a test required? Usually when you solve a equation, what you get is the SOLUTION right? Can someone please clarify why one solution will be 'invalid' in case of 'variables with absolute values'? I am unable to visualize that.

Merging similar topics. Please refer to the solution above.

_________________
Manager
Joined: 13 Oct 2013
Posts: 136
Concentration: Strategy, Entrepreneurship
Followers: 2

Kudos [?]: 47 [0], given: 125

### Show Tags

28 Nov 2013, 18:35
Bunuel wrote:
rohitgoel15 wrote:
What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

What is x?

(1) |x| < 2 --> $$-2<x<2$$. Not sufficient.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: $$3x-2\geq{0}$$ --> $$x\geq{\frac{2}{3}}$$ --> $$x$$ positive, so $$|x|=x$$ --> $$x=3x-2$$ --> $$x=1$$. Sufficient

Hope it's clear.

Hi Bunuel,
I have a question.
(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: 3x-2\geq{0} --> x\geq{\frac{2}{3}} --> x positive, so |x|=x --> x=3x-2 --> x=1. Sufficient

If it is true as above then why we are taking as -2<X<2 for |x| ,i mean why we are considering negative value (<0) for |X|?

_________________

---------------------------------------------------------------------------------------------
Kindly press +1 Kudos if my post helped you in any way

Manager
Joined: 05 Nov 2012
Posts: 170
Followers: 1

Kudos [?]: 34 [0], given: 57

### Show Tags

28 Nov 2013, 19:53
sunita123 wrote:
Bunuel wrote:
rohitgoel15 wrote:
What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

What is x?

(1) |x| < 2 --> $$-2<x<2$$. Not sufficient.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: $$3x-2\geq{0}$$ --> $$x\geq{\frac{2}{3}}$$ --> $$x$$ positive, so $$|x|=x$$ --> $$x=3x-2$$ --> $$x=1$$. Sufficient

Hope it's clear.

Hi Bunuel,
I have a question.
(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: 3x-2\geq{0} --> x\geq{\frac{2}{3}} --> x positive, so |x|=x --> x=3x-2 --> x=1. Sufficient

If it is true as above then why we are taking as -2<X<2 for |x| ,i mean why we are considering negative value (<0) for |X|?

We never considered negative value for |x|.... we considered negative value for just x...... By definition, mod never has a negative value....
Intern
Joined: 20 Feb 2013
Posts: 16
Location: India
Concentration: Marketing, Strategy
WE: Information Technology (Computer Software)
Followers: 0

Kudos [?]: 5 [0], given: 20

### Show Tags

25 Mar 2014, 21:38
rohitgoel15 wrote:
What is x?

(1) |x| < 2
(2) |x| = 3x – 2

When absolute values are given I guess it's better to go for number substitution.

1) |x| < 2 implies x can be either 1 or -1. 0 won't come here as it is neither positive nor negative and there is no need to take it here.
2) |x| = 3x - 2 is possible only when x = 1.

So, the final answer will be B
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15427
Followers: 649

Kudos [?]: 207 [0], given: 0

### Show Tags

31 Mar 2015, 01:29
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Manager
Joined: 29 Nov 2011
Posts: 113
Followers: 0

Kudos [?]: 16 [0], given: 367

### Show Tags

15 Jul 2015, 10:53
Hi Bunuel

What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

Your way is more common, orthodox. Though you should discard x=1/2 as you get this value when considering x<0 and 1/2 is not less than 0. Also, if x=1/2 then 3x-2<0 and you get that |x|=negative, which is never true.

Next, the way I'm ding is based on the fact that absolute value is always non negative, thus 3x-2 which equals to some absolute value must also be non-negative: 3x-2>=0 --> x>=2/3 --> x positive, so |x|=x and no need to consider |x|=-x --> x=3x-2 --> x=1.

ok my concern is if by cosidering |x| as positive and solving equation , if we get x as -negative value then , will we put |x| = -x and solve equation as we did in case of +x. please clarify
Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2644
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Followers: 128

Kudos [?]: 1470 [0], given: 789

### Show Tags

15 Jul 2015, 11:16
Hi Bunuel

What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

Your way is more common, orthodox. Though you should discard x=1/2 as you get this value when considering x<0 and 1/2 is not less than 0. Also, if x=1/2 then 3x-2<0 and you get that |x|=negative, which is never true.

Next, the way I'm ding is based on the fact that absolute value is always non negative, thus 3x-2 which equals to some absolute value must also be non-negative: 3x-2>=0 --> x>=2/3 --> x positive, so |x|=x and no need to consider |x|=-x --> x=3x-2 --> x=1.

ok my concern is if by cosidering |x| as positive and solving equation , if we get x as -negative value then , will we put |x| = -x and solve equation as we did in case of +x. please clarify

I dont think so. You need to follow the following 2 step procedure for ALL absolute value questions:

Treat |x| as following:

1. Let x $$\geq$$0 ----> |x| = x ---> x=3x-2 ---> x=1 and as x=1 satisfies x $$\geq$$0, this is a valid answer.

2. Let |x| < 0 ----> |x| = -x ----> -x=3x+2 ---> x = 1/2 and as x= 1/2 does not satisfy x<0, this solution is rejected. Thus we only 1 value, x = 1 and thus combining the 2 statements is sufficient. Thus C is the correct answer.

ok my concern is if by cosidering |x| as positive and solving equation , if we get x as -negative value then , will we put |x| = -x and solve equation as we did in case of +x. please clarify

If you consider |x| as positive (or >0) and you get x <0 , then this solution is not possible as you have 2 conflicting scenarios.

Hope this helps.
_________________

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
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Manager
Joined: 29 Nov 2011
Posts: 113
Followers: 0

Kudos [?]: 16 [0], given: 367

### Show Tags

15 Jul 2015, 22:48
HI Enge2012
I completely agree what you have mentioned, and that is way we normally solve mod equation, but if you see the Bunuel solution, he consider |x| as positive first and put equation >= 0 through which he gets +ve X and gain he solve the equation. Now my concern is over above equation what if we get -ve value of X example
instead of 3x-2>=0 if we have equation as 3x +2 >=0 now we get x as negative. hope i am cleared with doubt.
Re: What is x?   [#permalink] 15 Jul 2015, 22:48

Go to page    1   2    Next  [ 31 posts ]

Similar topics Replies Last post
Similar
Topics:
1 What is X? 16 06 Apr 2015, 12:05
2 What is x? 4 23 Feb 2016, 15:45
1 What is x? 2 21 Feb 2017, 20:20
What is x ? 2 09 Sep 2010, 15:59
12 What is x? 16 22 Nov 2016, 08:56
Display posts from previous: Sort by