GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Jul 2018, 00:52

### 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

# What is x?

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 07 Nov 2009
Posts: 273

### Show Tags

Updated on: 11 Jul 2013, 00:52
5
27
00:00

Difficulty:

65% (hard)

Question Stats:

48% (00:40) correct 52% (00:42) wrong based on 951 sessions

### HideShow timer Statistics

What is x?

(1) |x| < 2
(2) |x| = 3x – 2

Originally posted by rohitgoel15 on 31 Jan 2012, 18:48.
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: 47161

### Show Tags

31 Jan 2012, 18:56
17
15
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.
_________________
##### General Discussion
Senior Manager
Joined: 12 Mar 2010
Posts: 315
Concentration: Marketing, Entrepreneurship
GMAT 1: 680 Q49 V34
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: 273

### Show Tags

01 Feb 2012, 03:42
2
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: 47161

### Show Tags

01 Feb 2012, 04:05
7
5
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.
_________________
Intern
Joined: 07 Jul 2013
Posts: 7
Re: What is x? (1) |x| < 2 (2) |x| = 3x – 2  [#permalink]

### Show Tags

11 Jul 2013, 00:51
1
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: 430

### Show Tags

11 Jul 2013, 11:22
4
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)
Manager
Joined: 13 Oct 2013
Posts: 131
Concentration: Strategy, Entrepreneurship

### 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: 153

### 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: 14
Location: India
Concentration: Marketing, Strategy
WE: Information Technology (Computer Software)

### 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
Manager
Joined: 29 Nov 2011
Posts: 109

### 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
Current Student
Joined: 20 Mar 2014
Posts: 2641
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)

### Show Tags

15 Jul 2015, 11:16
2
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.
Manager
Joined: 29 Nov 2011
Posts: 109

### 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.
Current Student
Joined: 12 Aug 2015
Posts: 291
Concentration: General Management, Operations
GMAT 1: 640 Q40 V37
GMAT 2: 650 Q43 V36
GMAT 3: 600 Q47 V27
GPA: 3.3
WE: Management Consulting (Consulting)

### Show Tags

23 Aug 2015, 10:07
OA:

(1) INSUFFICIENT: This expression provides only a range of possible values for x.

(2) SUFFICIENT: Absolute value problems often -- but not always -- have multiple solutions because the expression within the absolute value bars can be either positive or negative even though the absolute value of the expression is always positive. For example, if we consider the equation |2 + x| = 3, we have to consider the possibility that 2 + x is already positive and the possibility that 2 + x is negative. If 2 + x is positive, then the equation is the same as 2 + x = 3 and x = 1. But if 2 + x is negative, then it must equal -3 (since |-3| = 3) and so 2 + x = -3 and x = -5.

So in the present case, in order to determine the possible solutions for x, it is necessary to solve for x under both possible conditions.

For the case where x > 0:

x = 3x – 2
-2x = -2
x = 1

When dealing with absolute values that contain variables, you should always check that the solution is valid. Plug x = 1 into the original equation.

|1| = 3(1) – 2
1 = 3 – 2

The equation is true, so the solution is valid.

For the case when x < 0:

x = -1(3x – 2) We multiply by -1 to make x equal a negative quantity.
x = 2 – 3x
4x = 2
x = 1/2

Once again, we need to test whether the solution is valid. Plug x = 1/2 into the original equation.

|1/2| = 3(1/2) – 2
1/2 = 3/2 – 2
1/2 = -1/2

This equation is not true, so x = 1/2 is actually not a valid solution. That means that there is only one solution to the equation. x = 1. The statement is sufficient.

_________________

KUDO me plenty

Intern
Joined: 19 Feb 2016
Posts: 15

### Show Tags

07 May 2016, 08:06
OA:

(1) INSUFFICIENT: This expression provides only a range of possible values for x.

(2) SUFFICIENT: Absolute value problems often -- but not always -- have multiple solutions because the expression within the absolute value bars can be either positive or negative even though the absolute value of the expression is always positive. For example, if we consider the equation |2 + x| = 3, we have to consider the possibility that 2 + x is already positive and the possibility that 2 + x is negative. If 2 + x is positive, then the equation is the same as 2 + x = 3 and x = 1. But if 2 + x is negative, then it must equal -3 (since |-3| = 3) and so 2 + x = -3 and x = -5.

So in the present case, in order to determine the possible solutions for x, it is necessary to solve for x under both possible conditions.

For the case where x > 0:

x = 3x – 2
-2x = -2
x = 1

When dealing with absolute values that contain variables, you should always check that the solution is valid. Plug x = 1 into the original equation.

|1| = 3(1) – 2
1 = 3 – 2

The equation is true, so the solution is valid.

For the case when x < 0:

x = -1(3x – 2) We multiply by -1 to make x equal a negative quantity.
x = 2 – 3x
4x = 2
x = 1/2

Once again, we need to test whether the solution is valid. Plug x = 1/2 into the original equation.

|1/2| = 3(1/2) – 2
1/2 = 3/2 – 2
1/2 = -1/2

This equation is not true, so x = 1/2 is actually not a valid solution. That means that there is only one solution to the equation. x = 1. The statement is sufficient.

Shouldn't the statement 2 have a true definite answer since it consist of 2 answer x=1 and x=1/2
could you please provide a further explanation on this
Manager
Joined: 20 Mar 2015
Posts: 62

### Show Tags

Updated on: 07 May 2016, 08:57
satishchaudharygmat Statement 2 gives us two answers but, only one satisfies the equation. Whenever, you find answers in modulus questions, always check whether the value satisfies the equation. In this question 1/2 will not satisfy the equation from which 1/2 is derived.

Hope this helped.

Originally posted by bimalr9 on 07 May 2016, 08:17.
Last edited by bimalr9 on 07 May 2016, 08:57, edited 2 times in total.
Manager
Joined: 13 Apr 2016
Posts: 60
Location: India
GMAT 1: 640 Q50 V27
GPA: 3
WE: Operations (Hospitality and Tourism)

### Show Tags

07 May 2016, 08:20
rohitgoel15 wrote:
What is x?

(1) |x| < 2
(2) |x| = 3x – 2

statement 1 gives you value of x b/w 1 to - 1 not sufficient.
statement 2 gives you 2 values x=1 and 1/2. but you will 1/2 in the equation LHS will not be equal to RHS. so we can cancel this value. so X=1
Manager
Joined: 04 Jan 2014
Posts: 120
GMAT 1: 660 Q48 V32
GMAT 2: 630 Q48 V28
GMAT 3: 680 Q48 V35

### Show Tags

30 Aug 2016, 18:41
St1: |x| < 2

This implies that -2 < x < 2.
Insufficient.

St2: |x| = 3x – 2

Using one of Bunuel's legendary tricks:

Since absolute value is always >= 0, the expression on right hand side will be >= 0.

3x - 2 >= 0
x >= 2/3, meaning x is positive.

So, we can only check for one condition.

If x > 0

x = 3x - 2
2x = 2
x = 1 -> satisfies the condition that x is positive.
Sufficient.

The trick is legendary because it saves us from checking the condition x is negative. Let's see what will happen if we do check:

If x < 0

x = 2 - 3x
4x = 2
x = 1/2 -> does not satisfy the condition that x is negative.

As you can see, the trick saves time and can make a big difference in the actual exam.
SVP
Joined: 08 Jul 2010
Posts: 2120
Location: India
GMAT: INSIGHT
WE: Education (Education)

### Show Tags

30 Aug 2016, 20:30
neverplayd wrote:
What is x?

(1) |x| < 2

(2) |x| = 3x – 2

Question : x = ?

Statement 1: |x| < 2
i.e. -2 < x < 2
NOT SUFFICIENT

Statement 1: |x| = 3x – 2
i.e. x = 3x – 2 and -x = 3x – 2
i.e. x = 1 and x = 1/2
But 1/2 doesn't satisfy the given statement hence that is not acceptable. Therefore,
x=2 is the only acceptable solution

SUFFICIENT

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Current Student
Status: It`s Just a pirates life !
Joined: 21 Mar 2014
Posts: 235
Location: India
Concentration: Strategy, Operations
GMAT 1: 690 Q48 V36
GPA: 4
WE: Consulting (Manufacturing)

### Show Tags

30 Aug 2016, 22:45
neverplayd wrote:
What is x?

(1) |x| < 2

(2) |x| = 3x – 2

Spoiler: :: Solution
(1) INSUFFICIENT: This expression provides only a range of possible values for x.

(2) SUFFICIENT: Absolute value problems often -- but not always -- have multiple solutions because the expression within the absolute value bars can be either positive or negative even though the absolute value of the expression is always positive. For example, if we consider the equation |2 + x| = 3, we have to consider the possibility that 2 + x is already positive and the possibility that 2 + x is negative. If 2 + x is positive, then the equation is the same as 2 + x = 3 and x = 1. But if 2 + x is negative, then it must equal -3 (since |-3| = 3) and so 2 + x = -3 and x = -5.

So in the present case, in order to determine the possible solutions for x, it is necessary to solve for x under both possible conditions.

For the case where x > 0:

x = 3x – 2
-2x = -2
x = 1

When dealing with absolute values that contain variables, you should always check that the solution is valid. Plug x = 1 into the original equation.

|1| = 3(1) – 2
1 = 3 – 2

The equation is true, so the solution is valid.

For the case when x < 0:

x = -1(3x – 2) We multiply by -1 to make x equal a negative quantity.
x = 2 – 3x
4x = 2
x = 1/2

Once again, we need to test whether the solution is valid. Plug x = 1/2 into the original equation.

|1/2| = 3(1/2) – 2
1/2 = 3/2 – 2
1/2 = -1/2

This equation is not true, so x = 1/2 is actually not a valid solution. That means that there is only one solution to the equation. x = 1. The statement is sufficient.

As donnie mentioned, the primary condition for he problem to be solved is abs(x)>=0, so in the LHS the min value is 0. Applying x=2/3 results LHS as "0" , which implies that x>2/3 and for x=1 alone, the condition answers.
_________________

Aiming for a 3 digit number with 7 as hundredths Digit

Re: What is x? &nbs [#permalink] 30 Aug 2016, 22:45

Go to page    1   2    Next  [ 25 posts ]

Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.