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:

I've understood everything about Absolute Values in these posts except for Example #1 in walker's first post. I was hoping someone could help me better understand the logic behind the solution.

First, I think what i'm struggling with most is the fact that we have absolute values on both the left and right hand side of the equation. How do you deal with those?

Second, I noticed that the three key points mentioned here are -8, -3, 4. I get that these three values for \(x\) equate the value inside each modulus to \(0\). But why do we need the value of each modulus to be \(0\)?

Finally, I don't understand how we derived the 4 conditions. With the first one, for example, if \(x<-8\), let's say \(x=-10\), then \(|8+(-10)|=2\) not \(0\). I'm completely lost here.

I've understood everything about Absolute Values in these posts except for Example #1 in walker's first post. I was hoping someone could help me better understand the logic behind the solution.

First, I think what i'm struggling with most is the fact that we have absolute values on both the left and right hand side of the equation. How do you deal with those?

Second, I noticed that the three key points mentioned here are -8, -3, 4. I get that these three values for \(x\) equate the value inside each modulus to \(0\). But why do we need the value of each modulus to be \(0\)?

Finally, I don't understand how we derived the 4 conditions. With the first one, for example, if \(x<-8\), let's say \(x=-10\), then \(|8+(-10)|=2\) not \(0\). I'm completely lost here.

I am assuming that you have understood that mod is nothing but distance on the number line. If so, then you can easily solve the question without equations. I re arrange the eg to get |x + 3| = |x + 8| + |x - 4| (Inside the mod, 4-x is same as x-4 since mod will always be non-negative) What I get from this question is: "I want the point on the number line whose distance from -3 is equal to the sum of its distances from -8 and from 4." i.e. the green distance is equal to red distance + blue distance.

Attachment:

Ques5.jpg [ 3.7 KiB | Viewed 3668 times ]

Is there any such point possible? If x is between -8 and 4, you see above that it is not possible. If x < -8 or x > 4, see the diagrams below and it will be obvious that there is no such point. In the first case, the blue distance itself is greater than the green distance and in the second case, the red distance itself is greater than the green distance. Since there is no such point on the number line, no such value of x exists.

Of course it can be done using algebra as well. It doesn't matter how many mods there are. you always deal with them in the same way. |x|= x when x is >= 0, |x|= -x when x < 0

|x - 2|= (x - 2) when x - 2 >= 0 (or x >= 2), |x - 2|= -(x - 2) when (x-2) < 0 (or x < 2)

Then you solve the equations using both conditions given above. That is the importance of the points. So if you have: |x - 2|= |x + 3|

You say, |x - 2|= (x - 2) when x >= 2. |x - 2|= -(x - 2) when x < 2 |x + 3| = (x + 3) when x >= -3 |x + 3| = -(x + 3) when x < -3

Now, x can be either greater than 2, between -3 and 2 or less than -3. So you solve for these 3 cases: Case 1: x >= 2 (x - 2) = (x + 3) -2 = 3 No solution

Case 2: -3 <= x < 2 -(x - 2) = (x + 3) x = -1/2 which lies between -3 and 2 So this is a solution to the equation

Case 3: x <= -3 -(x - 2) = -(x + 3) 2 = -3 No solution

Similarly you can solve for as many terms as you want.
_________________

Of course it can be done using algebra as well. It doesn't matter how many mods there are. you always deal with them in the same way. |x|= x when x is >= 0, |x|= -x when x < 0

|x - 2|= (x - 2) when x - 2 >= 0 (or x >= 2), |x - 2|= -(x - 2) when (x-2) < 0 (or x < 2)

Then you solve the equations using both conditions given above. That is the importance of the points. So if you have: |x - 2|= |x + 3|

You say, |x - 2|= (x - 2) when x >= 2. |x - 2|= -(x - 2) when x < 2 |x + 3| = (x + 3) when x >= -3 |x + 3| = -(x + 3) when x < -3

Now, x can be either greater than 2, between -3 and 2 or less than -3. So you solve for these 3 cases: Case 1: x >= 2 (x - 2) = (x + 3) -2 = 3 No solution

Case 2: -3 <= x < 2 -(x - 2) = (x + 3) x = -1/2 which lies between -3 and 2 So this is a solution to the equation

Case 3: x <= -3 -(x - 2) = -(x + 3) 2 = -3 No solution

Similarly you can solve for as many terms as you want.

-3 \leq x < 4 x \geq 4. if we take /4-x/ than we have 4-x >=0 so x >=4 and 4-x<0 4<=0 i see in the solution that you switched x-4 and wrote x-4>= 0 and x-4 <o i understand that it is perfectly ok to do so. my question: when do we have to know when to switch ? because if not switched than we shall have x >=4 if switched x<=4 i hope i made myself clear thanks

-3 \leq x < 4 x \geq 4. if we take /4-x/ than we have 4-x >=0 so x >=4 and 4-x<0 4<=0 i see in the solution that you switched x-4 and wrote x-4>= 0 and x-4 <o i understand that it is perfectly ok to do so. my question: when do we have to know when to switch ? because if not switched than we shall have x >=4 if switched x<=4 i hope i made myself clear thanks

You can switch inside the mod when you like! When you feel switching makes it easier for you to handle. We are used to getting terms in the form |x-4|, not |4-x|... so you switch.. it doesn't matter at all... let me show with a simpler example:

Question: |4-x| > 5

Case 1: When 4 - x >= 0 i.e. when x <= 4 4 - x > 5 or x < -1 So solution is x < -1

Case 2: When 4 - x <= 0 i.e. when x >= 4 -(4-x) > 5 x > 9 So solution is x > 9

Answer: x is either less than -1 or greater than 9.

Now switch: Question: |x-4| > 5

Case 1: When x - 4 >= 0 i.e. when x >= 4 x - 4 > 5 or x > 9 Solution is x > 9

Case 2: When x - 4 <= 0 i.e. when x <= 4 -(x - 4) > 5 x < -1 So solution is x < -1

Answer: x is either less than -1 or greater than 9.

The same two cases in both the questions.. same answer in both...
_________________

-3 \leq x < 4 x \geq 4. if we take /4-x/ than we have 4-x >=0 so x >=4 and 4-x<0 4<=0 i see in the solution that you switched x-4 and wrote x-4>= 0 and x-4 <o i understand that it is perfectly ok to do so. my question: when do we have to know when to switch ? because if not switched than we shall have x >=4 if switched x<=4 i hope i made myself clear thanks

You can switch inside the mod when you like! When you feel switching makes it easier for you to handle. We are used to getting terms in the form |x-4|, not |4-x|... so you switch.. it doesn't matter at all... let me show with a simpler example:

Question: |4-x| > 5

Case 1: When 4 - x >= 0 i.e. when x <= 4 4 - x > 5 or x < -1 So solution is x < -1

Case 2: When 4 - x <= 0 i.e. when x >= 4 -(4-x) > 5 x > 9 So solution is x > 9

Answer: x is either less than -1 or greater than 9.

Now switch: Question: |x-4| > 5

Case 1: When x - 4 >= 0 i.e. when x >= 4 x - 4 > 5 or x > 9 Solution is x > 9

Case 2: When x - 4 <= 0 i.e. when x <= 4 -(x - 4) > 5 x < -1 So solution is x < -1

Answer: x is either less than -1 or greater than 9.

The same two cases in both the questions.. same answer in both...

III. Thinking about absolute values as distance between points at the number line.

For example, Problem: A<X<Y<B. Is |A-X| <|X-B|? 1) |Y-A|<|B-Y| Solution: Image We can think about absolute values here as distance between points. Statement 1 means than distance between Y and A is less than Y and B. Because X is between A and Y, distance between |X-A| < |Y-A| and at the same time distance between X and B will be larger than that between Y and B (|B-Y|<|B-X|). Therefore, statement 1 is sufficient.
_________________

The proof of understanding is the ability to explain it.

when 0 is neither positive nor negative but neutral.

I am not sure I got your question.. |x|>=0 implies |x| is positive(when x is positive or negative) or 0 (when x = 0). |x| is of course never negative.
_________________

Hi, In the original post, in the section GMAC books, there are some prob numbers stated. Can you please explain how they relate to the post? How one will use those numbers is what I am asking.

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...