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) b^2-|b|-20=0. Solve quadratics for \(|b|\): \((|b|)^2-|b|-20=0\) --> \(|b|=-4\) or \(|b|=5\). Since absolute value cannot be negative then we have that \(|b|=5\) and \(|2b|=10\). Sufficient.

(2) |2b|=3b+25. Two cases:

If \(b\leq{0}\) then we would have that \(-2b=3b+25\) --> \(b=-5\). If \(b>0\) then we would have that \(2b=3b+25\) --> \(b=-25\), but since we are considering the rangeo when \(b>{0}\) then discard this solution.

So, we have that \(b=-5\), hence \(|2b|=10\). Sufficient.

(1) b^2-|b|-20=0. Solve quadratics for \(|b|\): \((|b|)^2-|b|-20=0\) --> \(|b|=-4\) or \(|b|=5\). Since absolute value cannot be negative then we have that \(|b|=5\) and \(|2b|=10\). Sufficient.

(2) |2b|=3b+25. Two cases: If \(b\leq{0}\) then we would have that \(-2b=3b+25\) --> \(b=-5\). If \(b>0\) then we would have that \(2b=3b+25\) --> \(b=-25\), but since we are considering the rangeo when \(b>{0}\) then discard this solution.

So, we have that \(b=-5\), hence \(|2b|=10\). Sufficient.

Answer: D.

Hope it's clear.

Hi, can you make the solution more clear?in case of (1) why you didn't consider the negative value of b? When b is positive, b^2 -b = 20 and when b is negative b^2 + b =20..then we will find 4 values of b...I am confused ..Can you explain? and in case of (2) how you discard the 2nd value? Thanks

hey farukqmul I also had the same doubt but then i tried to solve it properly and i found out that there is no need to do it the long way but i will post here to clear your doubt b^2-|b|-20=0 Considering b>0 eq. becomes b^2-b-20=0 which gives us solution b=-4 and 5 since we assumed b>0 so b=-4 is rejected Considering b<0 eq. becomes b^2+b-20=0 which gives us solution b=4 and -5 since we assumed b<0 so b=4 is rejected In both the cases |2b|=10 Now for the 2nd part (copying Buenel's explanation) |2b|=3b+25. Two cases: If b>0 then we would have that -2b=3b+25 --> b=-5. If b>0 then we would have that 2b=3b+25 --> b=-25, but since we are considering b>0 then discard this solution. Hope it's all clear now. _________________

(1) b^2-|b|-20=0. Solve quadratics for \(|b|\): \((|b|)^2-|b|-20=0\) --> \(|b|=-4\) or \(|b|=5\). Since absolute value cannot be negative then we have that \(|b|=5\) and \(|2b|=10\). Sufficient.

(2) |2b|=3b+25. Two cases: If \(b\leq{0}\) then we would have that \(-2b=3b+25\) --> \(b=-5\). If \(b>0\) then we would have that \(2b=3b+25\) --> \(b=-25\), but since we are considering the rangeo when \(b>{0}\) then discard this solution.

So, we have that \(b=-5\), hence \(|2b|=10\). Sufficient.

Answer: D.

Hope it's clear.

Hi, can you make the solution more clear?in case of (1) why you didn't consider the negative value of b? When b is positive, b^2 -b = 20 and when b is negative b^2 + b =20..then we will find 4 values of b...I am confused ..Can you explain? and in case of (2) how you discard the 2nd value? Thanks

OK.

Say \(x=|b|\), then we have that \(x^2-x-20=0\) --> \(x=-4\) or \(x=5\). Now, \(x=|b|=-4\) is not possible, since an absolute value of a number (\(|b|\)) cannot be negative. So, we have that \(|b|=5\) and \(|2b|=10\).

Now, you can solve this statement considering two ranges: \(b\leq{0}\) and \(b>0\), which will lead you to the same.

If \(b\leq{0}\) then we'll have \(b^2+b-20=0\) --> \(b=-5\) or \(b=4\) (not a valid solution since we are considering the range when \(b\leq{0}\)); If \(b>{0}\) then we'll have \(b^2-b-20=0\) --> \(b=-4\) (not a valid solution since we are considering the range when \(b>{0}\)) or \(b=5\);

So, only two valid solutions: \(b=-5\) or \(b=5\) --> \(|2b|=10\).

As for (2), it's explained in the post: If \(b>0\) then we would have that \(2b=3b+25\) --> \(b=-25\), but since we are considering the rangeo when \(b>{0}\) then discard this solution.

(1) If b >0, eqn reduces to (b-5)(b+4)=0 , since we have assumed b >0, this gives us a soln of b=5. If b <0, eqn reduces to (b+5)(b-4) =0 giving us a soln of b=-5 (since b <0). Since we are asked for the value of |2b| it doesnt matter if b is 5 or -5. Sufficient.

(2) If b >0, it gives us 2b=3b+25 or b =-25 but tht goes against our assumption. So b <0 and -2b=3b+25 or b=-5. Sufficient.

Hey Bunuel, the original question contained b^2-|b|-20=0 but you considered it to |b|^2-|b|-20=0 I know it does not matter that if it is |b|^2 or b^2 But it does matter especially when it is not mentioned that b is positive number... _________________

Hey Bunuel, the original question contained b^2-|b|-20=0 but you considered it to |b|^2-|b|-20=0 I know it does not matter that if it is |b|^2 or b^2 But it does matter especially when it is not mentioned that b is positive number...

I'm not sure I understand your question. Anyway:

We are asked to find the value of |2b|, so we don't really care for (1) whether b is positive or negative. _________________

(1) b^2-|b|-20=0. Solve quadratics for \(|b|\): \((|b|)^2-|b|-20=0\) --> \(|b|=-4\) or \(|b|=5\). Since absolute value cannot be negative then we have that \(|b|=5\) and \(|2b|=10\). Sufficient.

(2) |2b|=3b+25. Two cases: If \(b\leq{0}\) then we would have that \(-2b=3b+25\) --> \(b=-5\). If \(b>0\) then we would have that \(2b=3b+25\) --> \(b=-25\), but since we are considering the rangeo when \(b>{0}\) then discard this solution.

So, we have that \(b=-5\), hence \(|2b|=10\). Sufficient.

Answer: D.

Hope it's clear.

Hi, can you make the solution more clear?in case of (1) why you didn't consider the negative value of b? When b is positive, b^2 -b = 20 and when b is negative b^2 + b =20..then we will find 4 values of b...I am confused ..Can you explain? and in case of (2) how you discard the 2nd value? Thanks

Hi, I am wondering why |b| cannot be other numbers greater than 5. What if |b| was to equal 16? This would also allow us to solve the quadratics right? in this case (1) would have multiple answers, 5 and 6, therefore causing (1) to be insufficient?

Please let me know if I misunderstand something. Thank you

Solving the equation x^2 - x - 20 = 0 gives you two possible solution which are -4 and +5 In our case x = |b| so x cannot be negative. That let x = 5 the only possible solution

hi Arthur, I understand your point. However, I do not understand why "16" cannot be another possible answer. Since if we allow |b| to be 16, we can also solve the quadratics.

hi Arthur, I understand your point. However, I do not understand why "16" cannot be another possible answer. Since if we allow |b| to be 16, we can also solve the quadratics.

Frankly the above does not make ANY sense. How is |b|=16 satisfy any of the statements? _________________

(1) b^2-|b|-20=0. Solve quadratics for \(|b|\): \((|b|)^2-|b|-20=0\) --> \(|b|=-4\) or \(|b|=5\). Since absolute value cannot be negative then we have that \(|b|=5\) and \(|2b|=10\). Sufficient.

(2) |2b|=3b+25. Two cases: If \(b\leq{0}\) then we would have that \(-2b=3b+25\) --> \(b=-5\). If \(b>0\) then we would have that \(2b=3b+25\) --> \(b=-25\), but since we are considering the rangeo when \(b>{0}\) then discard this solution.

So, we have that \(b=-5\), hence \(|2b|=10\). Sufficient.

Answer: D.

Hope it's clear.

Hey Bunuel,

I have always been meaning to ask you. I know that (1) \(|x|=x\) if \(x>0\) and (2) \(|x|=-x\) if \(x <0\).

But the "=" of the (less than or equal to) or (greater than or equal to), where does it go? I noticed that you put it on equation 2, while others place it on equation (1). Its Driving me nuts!!

(1) b^2-|b|-20=0. Solve quadratics for \(|b|\): \((|b|)^2-|b|-20=0\) --> \(|b|=-4\) or \(|b|=5\). Since absolute value cannot be negative then we have that \(|b|=5\) and \(|2b|=10\). Sufficient.

(2) |2b|=3b+25. Two cases: If \(b\leq{0}\) then we would have that \(-2b=3b+25\) --> \(b=-5\). If \(b>0\) then we would have that \(2b=3b+25\) --> \(b=-25\), but since we are considering the rangeo when \(b>{0}\) then discard this solution.

So, we have that \(b=-5\), hence \(|2b|=10\). Sufficient.

Answer: D.

Hope it's clear.

Hey Bunuel,

I have always been meaning to ask you. I know that (1) \(|x|=x\) if \(x>0\) and (2) \(|x|=-x\) if \(x <0\).

But the "=" of the (less than or equal to) or (greater than or equal to), where does it go? I noticed that you put it on equation 2, while others place it on equation (1). Its Driving me nuts!!

First of all: \(|x|=-x\) when \(x\leq{0}\);

\(|x|=x\) when \(x\geq{0}\).

Next, as for "=" sign in the solution: you could include it either in the first case or in the second, it doesn't matter at all. _________________

(1) b^2-|b|-20=0 use quadratic formula to try and get a value for b. (|b|-5) (|b|+4) = 0 |b|=5 OR |b|=-4 Well, an absolute value cannot be equal to a negative number so |b|=5. Therefore, |2b|=10 SUFFICIENT

(2) |2b|=3b+25

Find positive/negative cases to try and isolate b. |2b|=3b+25 If x>= 0 2b=3b+25 -b=25 b=-25 Not valid as -2b is not greater than or equal to zero. x<0 -2b=3b+25 -5b=25 b=-5 Valid as -5 falls within the range of <0 SUFFICIENT (D)

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. _________________

1. b^2 - |b| = 20 Since both b^2 and absolute value of b cannot be negative,

b ( b - 1 ) = 20 and b= 5

2. I got two solutions following the same method as Bunuel did, but the part I don't understand is why we have to discard b = -25, where does it say that we are considering b>0 ??

Can someone please explain what exactly I missed or overlooked! Thank you

1. b^2 - |b| = 20 Since both b^2 and absolute value of b cannot be negative,

b ( b - 1 ) = 20 and b= 5

2. I got two solutions following the same method as Bunuel did, but the part I don't understand is why we have to discard b = -25, where does it say that we are considering b>0 ??

Can someone please explain what exactly I missed or overlooked! Thank you

For (1): b(b - 1) = b^2 - b not b^2 - |b|. Also, b(b - 1) = 20 has two roots: b = -4 and b = 5.

For (2): plug b = -25. Does it satisfy the equation? _________________

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. _________________

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...