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: [tricky one] Is x > 0? [#permalink]
22 Feb 2012, 21:02

30

This post received KUDOS

Expert's post

6

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

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
25 Feb 2012, 09:45

1

This post received KUDOS

2

This post was BOOKMARKED

Bunnel, For statement one, I understand that you have not solved for x in your solution. I am still learning the abs value and in equalities. Can you share your thoughts on the following solution (solving for x in both statements)

a. |x+3| = 4x -3 |x+3| = x + 3 for x + 3 >= 0 ---> x > -3. Solving for x, x + 3 = 4x -3 ---> 6 = 3x ---> x = 2 ( solution accepted since x > -3)

| x + 3| = -(x +3 ) for x + 3 < 0 --> x < -3. Solving for x, -x - 3 = 4x -3 5x = 0 ---> x = 0 ( solution discarded as x is not < -3)

A is sufficient.

b. |x -3| = |2x -3| |x -3| = x -3 for x-3 >= 0 ---> x >= 3 |x -3| = -(x -3) for x-3 < 0 ---> x < 3

|2x -3 | = 2x -3 for 2x -3 >= 0 ----> x >=3/2 |2x -3 | = -(2x -3) for 2x -3 < 0 ----> x < 3/2

So we have 3 ranges

1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard 2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution ) 3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)

Let me know if I have solved the question correctly. I know the process is lengthy.

One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case 3/2 < x < 3), then both statements are sufficient independently and we get the answer as option (D).

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
25 Feb 2012, 09:52

1

This post received KUDOS

Expert's post

saxenaashi wrote:

Bunnel, For statement one, I understand that you have not solved for x in your solution. I am still learning the abs value and in equalities. Can you share your thoughts on the following solution (solving for x in both statements)

a. |x+3| = 4x -3 |x+3| = x + 3 for x + 3 >= 0 ---> x > -3. Solving for x, x + 3 = 4x -3 ---> 6 = 3x ---> x = 2 ( solution accepted since x > -3)

| x + 3| = -(x +3 ) for x + 3 < 0 --> x < -3. Solving for x, -x - 3 = 4x -3 5x = 0 ---> x = 0 ( solution discarded as x is not < -3)

A is sufficient.

b. |x -3| = |2x -3| |x -3| = x -3 for x-3 >= 0 ---> x >= 3 |x -3| = -(x -3) for x-3 < 0 ---> x < 3

|2x -3 | = 2x -3 for 2x -3 >= 0 ----> x >=3/2 |2x -3 | = -(2x -3) for 2x -3 < 0 ----> x < 3/2

So we have 3 ranges

1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard 2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution ) 3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)

Let me know if I have solved the question correctly. I know the process is lengthy.

One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case 3/2 < x < 3), then both statements are sufficient independently and we get the answer as option (D).

Please discuss.

Regards

It seems that you understand this method very well.

Everything is correct except the red parts: -(x-3) = -(2x -3) --> x=0 (not x=3), so it's also a valid solution. Hence for (2) you have two valid solutions x=2 and x=0 (just like in my post above), which makes this statement not sufficient.

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
25 Feb 2012, 10:07

Bunuel wrote:

saxenaashi wrote:

Bunnel, For statement one, I understand that you have not solved for x in your solution. I am still learning the abs value and in equalities. Can you share your thoughts on the following solution (solving for x in both statements)

a. |x+3| = 4x -3 |x+3| = x + 3 for x + 3 >= 0 ---> x > -3. Solving for x, x + 3 = 4x -3 ---> 6 = 3x ---> x = 2 ( solution accepted since x > -3)

| x + 3| = -(x +3 ) for x + 3 < 0 --> x < -3. Solving for x, -x - 3 = 4x -3 5x = 0 ---> x = 0 ( solution discarded as x is not < -3)

A is sufficient.

b. |x -3| = |2x -3| |x -3| = x -3 for x-3 >= 0 ---> x >= 3 |x -3| = -(x -3) for x-3 < 0 ---> x < 3

|2x -3 | = 2x -3 for 2x -3 >= 0 ----> x >=3/2 |2x -3 | = -(2x -3) for 2x -3 < 0 ----> x < 3/2

So we have 3 ranges

1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard 2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution ) 3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)

Let me know if I have solved the question correctly. I know the process is lengthy.

One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case 3/2 < x < 3), then both statements are sufficient independently and we get the answer as option (D).

Please discuss.

Regards

It seems that you understand this method very well.

Everything is correct except the red parts: -(x-3) = -(2x -3) --> x=0 (not x=3), so it's also a valid solution. Hence for (2) you have two valid solutions x=2 and x=0 (just like in my post above), which makes this statement not sufficient.

Hope it's clear.

Works. Thanks for pointing out the mistake and quick response too. I am just learning this so working ground up, hence I am not hesitant in taking long route initially. After next few questions, I would be tuned to extract the ranges I guess. Is there a condition on squaring the sides of the equation or inequation.

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
25 Feb 2012, 10:15

1

This post received KUDOS

Expert's post

saxenaashi wrote:

Works. Thanks for pointing out the mistake and quick response too. I am just learning this so working ground up, hence I am not hesitant in taking long route initially. After next few questions, I would be tuned to extract the ranges I guess. Is there a condition on squaring the sides of the equation or inequation.

Regards

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: \(2<4\) --> we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work. For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: \(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\); \(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

So in statement (2) since both parts of expression are non-negative we can safely apply squaring.

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
27 Feb 2012, 22:09

Well, I have a little bit another solution for this problem

Consider (1): if x>-3 then x+3=4x-3, so x=2, sufficient if x<-3, then -x-3=4x-3, which Leeds to answer x=0, which is wrong because we consider only x<-3 Sum up (1) is sufficient

The saim logic can be applied to the (2), and it is possible to derive that (2) is sufficient

So the answer is D either (1) is sufficient or (2) is sufficient

In your logic, you gave a mistake, because you've missed the root.

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
27 Feb 2012, 22:30

Expert's post

terance wrote:

Well, I have a little bit another solution for this problem

Consider (1): if x>-3 then x+3=4x-3, so x=2, sufficient if x<-3, then -x-3=4x-3, which Leeds to answer x=0, which is wrong because we consider only x<-3 Sum up (1) is sufficient

The saim logic can be applied to the (2), and it is possible to derive that (2) is sufficient

So the answer is D either (1) is sufficient or (2) is sufficient

In your logic, you gave a mistake, because you've missed the root.

Unfortunately your answer is not correct: OA for this question is A, not D (you can see it under the spoiler in the initial post).

(2) |x – 3| = |2x – 3| has two roots x=0 and x=2 (just substitute these values to see that they both satisfy the given equation), so you can not get the single numerical value of x, which makes this statement insufficient.

You can refer to above solutions for two different approaches of how to get these roots for (2). Please ask if anything remains unclear. _________________

A simple solution for this is solve for x in each statement, but remember when you solve for x with abs value you must make the term positive and negative. in both cases you will find x = +# and zero. plug the numbers back into the equation and you will find zero will not fit the equation so x must be > 0 _________________

Remember: When you have || on both sides of eqn you do not need to VERIFY the answer by putting them back in eqn BUT when you have || on only one side you MUST VERIFY the answer by putting them back in eqn

Re: [tricky one] Is x > 0? [#permalink]
18 Apr 2012, 06:46

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

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

and I get 0 & 2 for both statements, can u tell me what wrong am I doing?

Posted from my mobile device

If you substitute x=0 in |x + 3| = 4x – 3 you'll get: LHS=|x + 3|=3 and RHS=4x – 3=-3, thus \(LHS\neq{RHS}\), which means that x=0 is not the root of the given equation.

When expanding |x+3|:

When x<-3, then |x+3|=-(x+3), so in this case we'll have -(x+3)=4x-3 --> x=0 --> discard this value since 0 is not less than -3 (we consider the range when x<-3).

When x>=-3, then |x+3|=x+3, so in this case we'll have x+3=4x-3 --> x=2 --> this value of x is OK since 2>-3.

and I get 0 & 2 for both statements, can u tell me what wrong am I doing?

Posted from my mobile device

If you substitute x=0 in |x + 3| = 4x – 3 you'll get: LHS=|x + 3|=3 and RHS=4x – 3=-3, thus \(LHS\neq{RHS}\), which means that x=0 is not the root of the given equation.

When expanding |x+3|:

When x<-3, then |x+3|=-(x+3), so in this case we'll have -(x+3)=4x-3 --> x=0 --> discard this value since 0 is not less than -3 (we consider the range when x<-3).

When x>=-3, then |x+3|=x+3, so in this case we'll have x+3=4x-3 --> x=2 --> this value of x is OK since 2>-3.

So, |x + 3| = 4x - 3 has only one root, x=2.

Hope it's clear.

Thanks Bunuel, this is how I solve the Modulus questions, so as a rule i guess, I should check both values by putting them back in eq, I use to think that they both are the roots since I considered both scenarios, ie one with |X+3| and one with -(x+3). I guess I will have to add this additional step in my methodology. Thanks. _________________

Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back!

Re: [tricky one] Is x > 0? [#permalink]
04 Jun 2013, 06:56

Hi!

Why do we not find the positive and negative values for |x+3|=4x-3? Is it because this isn't a <, >, >= problem?

Thanks!

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

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Re: [tricky one] Is x > 0? [#permalink]
04 Jun 2013, 07:00

2

This post received KUDOS

Expert's post

WholeLottaLove wrote:

Hi!

Why do we not find the positive and negative values for |x+3|=4x-3? Is it because this isn't a <, >, >= problem?

Thanks!

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

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.

One can do this way too, but the way shown in my post is faster. _________________

Re: [tricky one] Is x > 0? [#permalink]
04 Jun 2013, 07:26

Hmmm...

When I solve for the pos. and neg. values of |x+3|=4x-3 I get:

I. x+3=4x-3 ==> -3x=-6 ==> x=2 II. x+3=-(4x-3) ==> x+3=-4x+3 ==> 5x=0 ==> x=0

So here in my presumably incorrect simplification, I have x=2 and x=0 in which case we can't be sure if x>0

For many abs. value questions it seems that you have to find the positive and negative cases each equation. I get that in this case, |x+3|=4x-3 means that 4x-3 is positive but why for other, similar questions, is solving for the positive and negative cases necessary?

Bunuel wrote:

WholeLottaLove wrote:

Hi!

Why do we not find the positive and negative values for |x+3|=4x-3? Is it because this isn't a <, >, >= problem?

Thanks!

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

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.

One can do this way too, but the way shown in my post is faster.

gmatclubot

Re: [tricky one] Is x > 0?
[#permalink]
04 Jun 2013, 07:26

Interested in applying for an MBA? In the fourth and final part of our live QA series with guest expert Chioma Isiadinso, co-founder of consultancy Expartus and former admissions...