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:

The following question is located in the Equations, Inequalities & VICs Manhattan Guide book on page 135. I just need some clarification on the last part of their working out of the answer. I'll post this below in a spoiler section, so please click to find where I am a bit stuck.

MGMAT provides the following working out: "if we combine statements (1) and (2), we find that a and b can still have 2 different values. \(b-1=b^2-1\) \(b=b^2\) \(0=b^2-b\) \(0=b(b-1)\)

b= 0 or 1 a= -1 or 0

So \(b=0\) when \(a=-1\), and \(b=1\) when \(a=0\). However, in either case \(ab^2=0\). Therefore statements (1) and (2) combined are sufficient.

Please explain how b can equal 0 or 1? Could I deduce b=1 using this equation from above: \(0=b(b-1)\). Take out the b on the outside of the brackets and solve the remaining \(0=(b-1)\) equation, therefore \(b=1\)?

Likewise, how do I deduce that b=0 also? Is it from this equation \(0=b^2-b\)?

What I seem to find difficult is how would you know that b has two answers. THANKS IN ADVANCE!

What is the value of \(ab^2\)?

(1) \(a=b-1\). Clearly insufficient.

(2) \(a=b^2-1\). Clearly insufficient.

(1)+(2) \(b-1=b^2-1\) --> \(b^2=b\) --> \(b^2-b=0\) --> \(b(b-1)=0\). For the product of two numbers (\(b\) and \(b-1\)) to be zero one of them (or both) must be zero. So either: \(b=0\) --> \(a=b-1=-1\) --> \(ab^2=0\); or \(b-1=0\), \(b=1\) --> \(a=b-1=0\) --> \(ab^2=0\).

Re: Data sufficiency question on equations from MGMAT [#permalink]

Show Tags

09 Jun 2010, 13:12

Thanks for the quick reply Bunuel, kudos for the explanation.

What I am slightly concerned about is that if a similar question appears on the GMAT, I may fail to spot the fact that b can have two answers. If \(b^2-b=0\) then surely b must equal 0. Why do we then go and factorise to produce a new expression: \(b(b-1)=0\)? And when solving for b using this latter expression, why do we drop the first b and do \(b-1=0\) therefore \(b=1\) instead of \(b(b-1)=0\) and solve?

Sorry if this sounds convoluted but I'm probably overlooking a rule somewhere, which essentially needs clarification.

Thanks for the quick reply Bunuel, kudos for the explanation.

What I am slightly concerned about is that if a similar question appears on the GMAT, I may fail to spot the fact that b can have two answers. If \(b^2-b=0\) then surely b must equal 0. Why do we then go and factorise to produce a new expression: \(b(b-1)=0\)? And when solving for b using this latter expression, why do we drop the first b and do \(b-1=0\) therefore \(b=1\) instead of \(b(b-1)=0\) and solve?

Sorry if this sounds convoluted but I'm probably overlooking a rule somewhere, which essentially needs clarification.

First of all \(b^2-b=0\) is a quadratic equation and it can have 2 solutions. Next: \(b^2-b=0\) means that either \(b=0\) (0^2-0=0) or \(b=1\) (1^2-1=0) (so it's not necessary \(b\) to be zero). We did not drop \(b=0\) we just found the second solution \(b=1\).

Re: Data sufficiency question on equations from MGMAT [#permalink]

Show Tags

10 Jun 2010, 10:42

Ah got it now , silly mistake really. As it's been so long since I've touched algebra that I thought all quadratics looked like \(Ax^2+ Bx - C= 0\) so the \(b(b-1)=0\) threw me a little. I realize now that \(b(b-1)=0\) is the same as saying \((b) . (b-1)=0\) which follows the concept of what a quadratic looks like after its been factorised ie \((x+1) . (x-1)=0\). Thus, we can solve the two values for b by taking each expression individually (b & (b-1)) and equating them each to zero.

Re: Data sufficiency question on equations from MGMAT [#permalink]

Show Tags

12 Jun 2010, 23:50

Bunuel, Please can you explain my confusion? If I place the equation as b-1 = b^2-1 and solve it by thinking b-1 = (b-1) (b+1) ==> b = -1..... which is only one solution... is this method wrong?? or it is ok? Please kindly tell me if i am making some mistake.

Bunuel, Please can you explain my confusion? If I place the equation as b-1 = b^2-1 and solve it by thinking b-1 = (b-1) (b+1) ==> b = -1..... which is only one solution... is this method wrong?? or it is ok? Please kindly tell me if i am making some mistake.

First of all the roots of the equation \(b-1 = b^2-1\) are \(b=0\) and \(b=1\) (see the solution in my post), if you got these roots you solved the equation correctly and if you got different roots you solved incorrectly.

Next, I don't understand why aren't you cancelling (-1) from \(b-1=b^2-1\) --> \(b=b^2\).

But anyway if you don't and we proceed the way you are doing: \(b-1=b^2-1\) --> \(b-1=(b-1)(b+1)\). Now how is \(b=-1\) the root of this equation? \(b-1=-2\neq{(b-1)(b+1)}=0\).
_________________

Re: Data sufficiency question on equations from MGMAT [#permalink]

Show Tags

13 Jun 2010, 07:29

Sorry, it was an error from my side. I mean to say if I arrive with only one solution ==> b=0, accroding to the method i mentioned. Then why it is considered wrong? b-1 = b^2 - 1 b-1 = (b-1) (b+1) 1 = b+1 ==> b=0

Why I am not cancelling 1 on each side? I dont know... That is the first thought when i looked at the equation. I want to make a note in my mind why it is wrong so that i dont make these mistakes on the test day. Thanks for your ever woderful support.

Sorry, it was an error from my side. I mean to say if I arrive with only one solution ==> b=0, accroding to the method i mentioned. Then why it is considered wrong? b-1 = b^2 - 1 b-1 = (b-1) (b+1) 1 = b+1 ==> b=0

Why I am not cancelling 1 on each side? I dont know... That is the first thought when i looked at the equation. I want to make a note in my mind why it is wrong so that i dont make these mistakes on the test day. Thanks for your ever woderful support.

I see. When you write \(1 = b+1\) after \(b-1 = (b-1) (b+1)\) what you are actually doing is reducing (dividing) the equation by \(b-1\) but we can not do that as it can equal to zero.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

So basically when you divide by \(b-1\) you assume that it doesn't equal to zero, thus missing the second valid solution for \(b-1 = (b-1) (b+1)\) which is \(b-1=0\) --> \(b=1\).

Again if we proceed the way you are doing \(b-1 = (b-1) (b+1)\) --> \((b-1)-(b-1) (b+1)=0\) --> factor out \(b-1\) --> \((b-1)(1-b-1)=0\) --> \((b-1)b=0\) (the type of equation you'd receive right away if you'd cancel out \(-1\)) --> \(b=0\) or \(b=1\).

1)Doesn't give anything about a and b.So not sufficient.Eliminate A and D 2)Doesn't give anything about a and b.So not sufficient. Eliminate B 1+2) (b-1)=b^2-1 (b-1)=(b-1)(b+1) 1=b+1 b=0 So answer is C

1)Doesn't give anything about a and b.So not sufficient.Eliminate A and D 2)Doesn't give anything about a and b.So not sufficient. Eliminate B 1+2) (b-1)=b^2-1 (b-1)=(b-1)(b+1) 1=b+1 b=0 So answer is C

Just a caveat:Note the part that is highlighted,

You can't cancel out (b-1) on both the sides, as you don't know whether b=1 or not. Thus, either b=1 or b=0. In either case, you get ab=0.
_________________

1)Doesn't give anything about a and b.So not sufficient.Eliminate A and D 2)Doesn't give anything about a and b.So not sufficient. Eliminate B 1+2) (b-1)=b^2-1 (b-1)=(b-1)(b+1) 1=b+1 b=0 So answer is C

Just a caveat:Note the part that is highlighted,

You can't cancel out (b-1) on both the sides, as you don't know whether b=1 or not. Thus, either b=1 or b=0. In either case, you get ab=0.

Answer is C.

Just to add for St 2 we have

a = (b-1)(b+1) => a=a(b+1) { from St 1 we have a=b-1}

1)Doesn't give anything about a and b.So not sufficient.Eliminate A and D 2)Doesn't give anything about a and b.So not sufficient. Eliminate B 1+2) (b-1)=b^2-1 (b-1)=(b-1)(b+1) 1=b+1 b=0 So answer is C

Just a caveat:Note the part that is highlighted,

You can't cancel out (b-1) on both the sides, as you don't know whether b=1 or not. Thus, either b=1 or b=0. In either case, you get ab=0.

Answer is C.

Just to add for St 2 we have

a = (b-1)(b+1) => a=a(b+1) { from St 1 we have a=b-1}

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.
_________________

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...