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:

As I have said before, most mod questions are best tackled using a number line. You don't need to do many calculations then. |x| means the distance from 0. |x-3| means the distance from 3. etc. For details of this approach, check out:

Let's go on to this question now. Is |x| < 1 i.e. Is the distance of point x from 0 less than 1?

Statement 1: |x + 1| = 2|x – 1| This means 'distance of x from -1 is twice the distance of x from 1'. Draw the number line now. There will be 2 points where the distance from -1 will be twice the distance from 1.

Attachment:

Ques6.jpg [ 5.12 KiB | Viewed 2648 times ]

For one of these points, distance from 0 is less than 1, for the other it is more than 1. So not sufficient.

Statement 2: |x – 3| > 0 This statement tells us that distance of x from 3 is more than 0 i.e. x does not lie at 3. It can lie anywhere else. You can look at it in another way: Mods are always more than or equal to 0. All this statement tells us is that this mod is not equal to zero i.e. x is not equal to 3. For some of these points, distance from 0 will be less than 1, for the others it will be more than 1. So not sufficient.

Using both statements together, statement 1 says that x is either 3 or a point between 0 and 1 (which I don't really need to calculate). Statement 2 tells us that x is not 3. So together, x must be a point between 0 and 1 and its distance from 0 must be less than 1. Sufficient. Answer C. _________________

I always get confused how to approach the equality or inequality with absolute signs on both sides (statement 1). Any1 knows an efficient way to approach this ?

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.

Alternately we could just expand absolute values of RHS and LHS with same sign (for example both LHS and RHS positive: \(x + 1 = 2(x - 1)\)) and then with different sign (for example LHS positive and RHS negative \(x + 1 = 2(-x + 1)\)), solve for \(x\) both equations, and finally check whether the solutions satisfy \(|x + 1| = 2|x - 1|\).

OR: we can square given equation to get rid of the modulus: \((x + 1)^2 = 4(x - 1)^2\) --> \(3x^2-10x+3=0\) --> \(x=3\) or \(x=\frac{1}{3}\).

(2) \(|x - 3|>{0}\). Absolute value is always non-negative, more than or equal to zero: \(|some \ expression|\geq{0}\). We are told that absolute value of \(x-3\) is MORE than zero, so just it says that \(|x-3|\neq{0}\), which simply means 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.

Can you please explain why, in B above, we must NOT change the sign in |x+1| leaving it as x+1, and must change the sign in 2|x-1| making it 2(-x+1)?

Thank you.

Absolute value properties: When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\);

For B. if \(-1\leq{x}\leq{1}\) (green range) --> then \(x+1\geq{0}\) (try some value of \(x\) from the given range to check: for example \(x=0\)) so \(|x+1|=x+1\) BUT \(x-1\leq{0}\) ] (again try some value of \(x\) from the given range to check: for example \(x=0\)) so \(|x-1|=-(x-1)\) thus \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\).

Re: Inequality - absolute on both sides [#permalink]

Show Tags

04 Feb 2011, 12:19

2

This post received KUDOS

The question is

Is |x| < 1 ?

Asking whether x falls between 1 and -1, exclusive.

if |x|<1 x<1; 0.5, 0.2,0 and x>-1; -0.5,-0.2

or -1<x<1

Moment you write x=1 or x=-1; the |x|<1 becomes false.

(1) |x + 1| = 2|x – 1|

For modulus on both sides:

Case I. solve the equation with no sign change

so +(x+1) = +2(x-1) x+1=2x-2 x=3. Not between -1 and 1.

Case II. solve the equation with sign change on one side. Either RHS or LHS. Let's do the sign change on LHS

so -(x+1) = +2(x-1) -x-1=2x-2 3x=1. x=1/3 x is between -1 and 1.

CaseII(b): Even if we did sign change on RHS, we would have gotten the same result. Let's try +(x+1) = -2(x-1) x+1=-2x+2 3x=1 x=1/3. Same result as before.

So; sign change should be done for either LHS or RHS.

Now, we have two solutions for x; 1/3(between -1 and 1), 3(not between -1 and 1)

Not sufficient.

###Also please substitute these factors of x into the main equation and check whether the factors indeed satisfy the equation. because say if 3 didn't satisfy the equation and 1/3 does. The statement would be sufficient.###

(2) |x – 3| > 0

Here modulus only on LHS; So, try this with both signs;

+(x-3) > 0 x-3>0 x>3

and

-(x-3) > 0 -x+3>0 -x>-3 x<3

Here x can be either less than 3 or more than 3. Not 3.

However; this doesn't tell us definitively whether x lies between -1 and 1.

Not Sufficient.

Using both:

We know x can be either (1/3 or 3) by 1st statement. Second statement tells us that x can not be 3.

For B. if \(-1\leq{x}\leq{1}\) (green range) --> then \(x+1\geq{0}\) (try some value of \(x\) from the given range to check: for example \(x=0\)) so \(|x+1|=x+1\) BUT \(x-1\leq{0}\) ] (again try some value of \(x\) from the given range to check: for example \(x=0\)) so \(|x-1|=-(x-1)\) thus \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\).

So, as I understand, we need to plug the value from the available range into the modulus. If the expression in the modulus becomes negative of zero, we must change the sign. And if the expression becomes positive or zero, then we should leave it as it is. Am I correct?

Thanks again for your help, Bunuel.

Yes. You should really try the links I've provided in the above posts.

Matt1177 wrote:

fluke wrote:

Case II. solve the equation with sign change on one side. Either RHS or LHS. Let's do the sign change on LHS

So, we cannot change the signs on both LHS and RHS at once?

No that's not the point. When you have absolute values on both sides of the equation expansion can be either + + (or which is the same - - because \(x + 1 = 2(x - 1)\) is the same as \(-(x + 1) = -2(x - 1)\)) OR + - (or which is the same - + because \(x + 1 = -2(x - 1)\) is the same as \(-(x + 1)=2(x - 1)\)) so basically only two options.

So as I've written in my solution: "alternately you could just expand absolute values of RHS and LHS with same sign (for example both LHS and RHS positive: \(x + 1 = 2(x - 1)\)) and then with different sign (for example LHS positive and RHS negative \(x + 1 = 2(-x + 1)\)), solve for \(x\) both equations, and finally check whether the solutions satisfy \(|x + 1| = 2|x - 1|\)."

Bunuel...if the second statement was (2) |x – 3| = 0 then in that case could we say

x= 3.

If (2) were |x – 3| = 0 then yes, we would have that x=3 and in this case this statement would be sufficient, as we could answer NO to the question whether -1<x<1. _________________

Mustu, for (1) |x + 1| = 2|x – 1| when looking at X<0 case, how come you multiply only right-hand side by negative, but not both sides? thanks

We have to consider 4 cases totally..

1) + + 2) + - 3) - + 4) - -

If u analyse more closely, then u will find that case 1 and 4 ( - and - cancels out) are the same and cases 2 and 3 are the same( + and - 0r - and + are the same). Hope this is clear.

From Statement 1 square both sides: --> 2^2+2x+1=4x^2-8x+4 --> 0 = 3x^2-10x+3 --> 0=(3x-1)(x-3) x=3 or x=1/3 Hence, insufficient.

From Statement 2 values of x: any number larger or smaller than 3 (i.e. x not equals to 3) Hence, insufficient. From statement 1+2 Sufficient.

Answer: C _________________

"The best day of your life is the one on which you decide your life is your own. No apologies or excuses. No one to lean on, rely on, or blame. The gift is yours - it is an amazing journey - and you alone are responsible for the quality of it. This is the day your life really begins." - Bob Moawab

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

As you leave central, bustling Tokyo and head Southwest the scenery gradually changes from urban to farmland. You go through a tunnel and on the other side all semblance...