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 350,000 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:

Re: Absolute value DS [#permalink]
03 Sep 2010, 13:29

4

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

Is x > 0?

(1) \(|x+3|=4x-3\) --> LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient.

(2) \(|x+1|=2x-1\) --> the same here --> \(2x-1\geq{0}\) --> \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient.

So you see that you don't even need to find exact value(s) of \(x\) to answer the question.

Re: Absolute value DS [#permalink]
03 Sep 2010, 13:43

1

This post received KUDOS

thanks that helped your explanation is more simpler and fundamental!

the below explanation i did not get especially the last part where we need to substitute and check

(1) SUFFICIENT: Here, we are told that |x + 3| = 4x – 3. When dealing with equations containing variables and absolute values, we generally need to consider the possibility that there may be two values for the unknown that could make the equation work. We can find the two possible solutions by setting the expression inside the absolute value equal to ± the expression on the right side of the equation. In this case, our two solutions are:

x + 3 = +(4x – 3) and x + 3 = –(4x – 3) x + 3 = 4x – 3 x + 3 = –4x + 3 6 = 3x 5x = 0 2 = x x = 0

However, you need to be careful. When there is a variable outside the absolute value, both solutions are not always valid. We need to plug both x = 2 and x = 0 back into the original equation and test them.

Re: Absolute value DS [#permalink]
03 Sep 2010, 15:15

Expert's post

rxs0005 wrote:

thanks that helped your explanation is more simpler and fundamental!

the below explanation i did not get especially the last part where we need to substitute and check

(1) SUFFICIENT: Here, we are told that |x + 3| = 4x – 3. When dealing with equations containing variables and absolute values, we generally need to consider the possibility that there may be two values for the unknown that could make the equation work. We can find the two possible solutions by setting the expression inside the absolute value equal to ± the expression on the right side of the equation. In this case, our two solutions are:

x + 3 = +(4x – 3) and x + 3 = –(4x – 3) x + 3 = 4x – 3 x + 3 = –4x + 3 6 = 3x 5x = 0 2 = x x = 0

However, you need to be careful. When there is a variable outside the absolute value, both solutions are not always valid. We need to plug both x = 2 and x = 0 back into the original equation and test them.

Re: Absolute value DS [#permalink]
07 Sep 2010, 22:33

Whenever i see this type of question i start to solve for X. Now in this case, from st-1 i got X = 2 or 0 From st-2 also the same answer and i ticked E. Can someone explain how i should start to think for this type of questions?? however i forgot to put back tha values and check.

Re: Absolute value DS [#permalink]
19 Sep 2010, 23:42

rxs0005 wrote:

thanks that helped your explanation is more simpler and fundamental!

the below explanation i did not get especially the last part where we need to substitute and check

(1) SUFFICIENT: Here, we are told that |x + 3| = 4x – 3. When dealing with equations containing variables and absolute values, we generally need to consider the possibility that there may be two values for the unknown that could make the equation work. We can find the two possible solutions by setting the expression inside the absolute value equal to ± the expression on the right side of the equation. In this case, our two solutions are:

x + 3 = +(4x – 3) and x + 3 = –(4x – 3) x + 3 = 4x – 3 x + 3 = –4x + 3 6 = 3x 5x = 0 2 = x x = 0

However, you need to be careful. When there is a variable outside the absolute value, both solutions are not always valid. We need to plug both x = 2 and x = 0 back into the original equation and test them.

Therefore, x = 0 is not a valid solution and we know that 2 is the only possible solution. x is definitely positive.

This appears every so often on absolute value questions. It relies on you not testing the possible solution. You must always verify the solutions you get in your first step. I've made this mistake several times

Re: Absolute value DS [#permalink]
20 Sep 2010, 18:44

1

This post received KUDOS

amitjash wrote:

Whenever i see this type of question i start to solve for X. Now in this case, from st-1 i got X = 2 or 0 From st-2 also the same answer and i ticked E. Can someone explain how i should start to think for this type of questions?? however i forgot to put back tha values and check.

I did same and got tricked. Can someone explain why we are getting x=0 and it doesn't fit the equation. Are we doing anything wrong on absolute problems? _________________

If you like my post, consider giving me some KUDOS !!!!! Like you I need them

Re: Absolute value DS [#permalink]
20 Sep 2010, 19:10

saxenashobhit wrote:

amitjash wrote:

Whenever i see this type of question i start to solve for X. Now in this case, from st-1 i got X = 2 or 0 From st-2 also the same answer and i ticked E. Can someone explain how i should start to think for this type of questions?? however i forgot to put back tha values and check.

I did same and got tricked. Can someone explain why we are getting x=0 and it doesn't fit the equation. Are we doing anything wrong on absolute problems?

When you are dealing with absolute values, and you solve it in 2 cases assuming first that expression >= 0 and then that expression < 0, you must verify the solution you get to make sure you don't violate the extra assumption you made in the beginning

Eg.When you assume x+3>0, it means you assume x>-3. You get an answer 2, which is valid as no assumptions are violated When you assume x+3<0, it means you assume x<-3. You get an answer 0, which is invalid given your assumption

You do not need to necessarily "plug the solution back in", it is sufficient to check against the assumption you made while deciding the sign of the absolute value _________________

Re: Absolute value DS [#permalink]
20 Sep 2010, 20:54

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

saxenashobhit wrote:

amitjash wrote:

Whenever i see this type of question i start to solve for X. Now in this case, from st-1 i got X = 2 or 0 From st-2 also the same answer and i ticked E. Can someone explain how i should start to think for this type of questions?? however i forgot to put back tha values and check.

I did same and got tricked. Can someone explain why we are getting x=0 and it doesn't fit the equation. Are we doing anything wrong on absolute problems?

Solving for \(x\) is not the best way to deal with this problem.

But if you do this way you should know that checking validity of the solutions is very important for absolute value questions.

When you solve these kind of questions with check point method (discussed at: some-inequalities-questions-93760.html?hilit=substitute%20method), you test validity of solution on the stage of obtaining these values (by checking whether the value is in the range you are testing at the moment) and if the value obtained IS in the range you are testing at the moment, you don't need to substitute it afterwards to check, you've already done the checking and if the value obtained IS NOT in the range you are testing at the moment you also don't need to substitute it afterwards to check, you've already done the checking.

When you just expand the absolute value once with negative sign and once with positive sign then you should check whether obtained solution(s) satisfy equation be substituting solution(s) back to the equation.

Re: Is x > 0? (1) |x + 3| = 4x 3 (2) |x + 1| = 2x 1 Can [#permalink]
08 Oct 2013, 16:17

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. _________________

Re: Absolute value DS [#permalink]
09 Oct 2013, 06:28

Bunuel wrote:

Is x > 0?

(1) \(|x+3|=4x-3\) --> LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(4x-3\geq{0}\) --> \(x\geq{\frac{3}{4}}\), hence \(x>0\). Sufficient.

(2) \(|x+1|=2x-1\) --> the same here --> \(2x-1\geq{0}\) --> \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient.

So you see that you don't even need to find exact value(s) of \(x\) to answer the question.

Answer: D.

Hope it helps.

For those more inclined to solving equations, here is a different approach: (1) |x+3|=4x-3

If x+3>0 then: a) x > -3 AND b) x+3 = 4x-3 => 3x=6 => x=2. Since x=2 does NOT contradict x>-3, x=2 is a solution and x is positive

However if x+3<0 then: a) x<-3 AND b) x+3=-4x+3 => x=0. BUT x must be less than -3, so x=0 is NOT a solution.

Therefore only the case where x+3>0 can be possible being x=2 the only possible solution, therefore always positive.

(2) |x+1|=2x-1 Works exactly the same:

If x+1>0 then: a) x > -1 AND b) x+1 = 2x-1 => x = 2 Since x=2 does NOT contradict x>-1, x=2 is a solution and x is positive

However if x+1<0 then: a) x<-1 AND b) x+1 = -2x+1 => x = 0. BUT x must be less than -1, so x=0 is NOT a solution.

Therefore only the case where x+1>0 can be possible being x=2 the only possible solution, therefore always positive.

In BOTH cases the only possible value for x is always positive. Therefore D is the answer.

Re: Is x > 0? (1) |x + 3| = 4x 3 (2) |x + 1| = 2x 1 Can [#permalink]
02 Jan 2015, 06:07

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. _________________

How the growth of emerging markets will strain global finance : Emerging economies need access to capital (i.e., finance) in order to fund the projects necessary for...

Good news for globetrotting MBAs: travel can make you a better leader. A recent article I read espoused the benefits of traveling from a managerial perspective, stating that it...