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Re: Data sufficiency question on equations from MGMAT [#permalink]
09 Jun 2010, 05:44
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Expert's post
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BigBrad wrote:
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]
09 Jun 2010, 12:12
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.
Re: Data sufficiency question on equations from MGMAT [#permalink]
09 Jun 2010, 12:45
Expert's post
BigBrad wrote:
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]
10 Jun 2010, 09:42
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]
12 Jun 2010, 22:50
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.
Re: Data sufficiency question on equations from MGMAT [#permalink]
13 Jun 2010, 03:55
Expert's post
amitjash wrote:
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]
13 Jun 2010, 06:29
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.
Re: Data sufficiency question on equations from MGMAT [#permalink]
13 Jun 2010, 06:43
Expert's post
amitjash wrote:
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 \(b=1\).
Re: what is the value of ab? [#permalink]
20 Nov 2013, 11:06
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
Re: what is the value of ab? [#permalink]
20 Nov 2013, 11:12
Expert's post
vikrantgulia wrote:
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. _________________
Re: what is the value of ab? [#permalink]
20 Nov 2013, 12:36
mau5 wrote:
vikrantgulia wrote:
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}
Re: what is the value of ab? [#permalink]
21 Nov 2013, 02:11
Expert's post
viksingh15 wrote:
mau5 wrote:
vikrantgulia wrote:
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}
Re: What is the value of ab^2 ? [#permalink]
30 Jun 2015, 22:07
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