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 Your Progress

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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

(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

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.. ?

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. _________________

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.

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.

Answer: C.

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?

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.

Answer: C.

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?

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. _________________

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. _________________

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...

Rishabh from Gyan one services, India had a one to one interview with me where I shared my experience at IMD till now. http://www.gyanone.com/blog/life-at-imd-interview-with-imd-mba/ ...