What is the value of ab^2 ?

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
\(b-1=0\), \(b=1\) --> \(a=b-1=0\) --> \(ab^2=0\).

The same answer for both cases. Sufficient.

Answer: C.

Hope it helps.
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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 \(b=1\).

You might need to revise basics of algebra.

Hope it helps.
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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.

Thanks for your help Bunuel, very kind!

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? \(b-1=-2\neq{(b-1)(b+1)}=0\).
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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 \(b=1\).

Hope it's clear.
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Here, we are asked to find the value of ab^2.

1. a = b-1. Clearly insufficient coz with this info it is impossible to find the value of ab^2.
2. a = b^2-1. Clearly insufficient due to reason stated above.

With equations 1 and 2, we could equate the value for a

b-1 = b^2-1
b^2 = b
b^2 - b = 0 (cannot divide by b on both sides since we don't know the value of b which could be 0)
b(b-1) = 0
Thus, b = 0 , or b = 1

If b = 0 , a = b-1 = 0 -1 = -1 and ab^2 = 0
If b = 1, a = b-1 = 1 - 0 = 0 and ab^2 = 0

Thus with 1 and 2 we get ab^2 as 0.

Hence C

hope it helps,
meshtrap

Hi Jinglander,

individually, the statements are insufficient, as meshtrap points out.

Notice that (2) gives us a difference of squares (commonly tested pattern on the GMAT):

a = b^2 - 1

can be rewritten as:

a = (b-1)*(b+1)

We know from (1) that a = b - 1. Thus, b + 1 must be equal to 1. Thus, we can compute b's value. Thus, we can compute a's value. Thus, we can compute a*b^2.

Choose 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.
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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}

a= a(b+1)
1=b+1 => b=0, so ab is 0, sufficient.

Merging similar topics. Please refer to the solutions above.


Discussed here: what-is-the-value-of-ab-143841.html
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You cannot reduce a=a(b+1) by a and write 1=b+1 because you exclude a possible solution a=0. Check here for more: what-is-the-value-of-ab-95543.html#p737359

Hope this helps.
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So actually this is a slightly tricky question

St 1

two variables not enough info

insuff

St 2

two variables not enough info

insuff

St 1 & 2

b^2-1= b-1=

b^2= b

b=, 0, 1

But since the question is what is the product of A(B)^2

In either scenario you would inevitably multiply by 0 - if 1 then A becomes 0 or if A is 1 then B is 0

C

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