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(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.

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

(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) 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. _________________

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) 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.

(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. _________________

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