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If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo

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If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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Tough and Tricky questions: Algebra.



If \(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\), and a and b are both non-zero integers, which of the following could be the value of b?

I. 2
II. 3
III. 4

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III

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[Reveal] Spoiler: OA

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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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New post 06 Nov 2014, 02:19
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Bunuel wrote:

Tough and Tricky questions: Algebra.



If \(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\), and a and b are both non-zero integers, which of the following could be the value of b?

I. 2
II. 3
III. 4

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III

Kudos for a correct solution.


\(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\)

we cannot have a=1, and a=-2. because this will make denominator zero.

now, consider the numerator. let ab=k
thus we have k(3k^2 +9k-54)
=k(3k^2+18k-9k-54)
=k(3k(k+6)-9(k+6))
=k(3k-9)(k+6)

if 3k-9=0
3k=9
k=3
i.e. ab=3
now if b=3, then a=1, which is not acceptable as a=1, will make the denominator equal to zero.
if k+6=0
k=-6
or ab=-6

again, if b=3, then a=-2, which is not acceptable as a=-2, will make the denominator equal to zero.

thus except b=3, all other values could be possible. hence answer should be D

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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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manpreetsingh86 wrote:
Bunuel wrote:

Tough and Tricky questions: Algebra.



If \(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\), and a and b are both non-zero integers, which of the following could be the value of b?

I. 2
II. 3
III. 4

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III

Kudos for a correct solution.


\(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\)

we cannot have a=1, and a=-2. because this will make denominator zero.

now, consider the numerator. let ab=k
thus we have k(3k^2 +9k-54)
=k(3k^2+18k-9k-54)
=k(3k(k+6)-9(k+6))
=k(3k-9)(k+6)

if 3k-9=0
3k=9
k=3
i.e. ab=3
now if b=3, then a=1, which is not acceptable as a=1, will make the denominator equal to zero.
if k+6=0
k=-6
or ab=-6

again, if b=3, then a=-2, which is not acceptable as a=-2, will make the denominator equal to zero.

thus except b=3, all other values could be possible. hence answer should be D



As ab=-6 or ab=3, so if we consider value of b as 4, a becomes Fraction in both the cases, but a is a non-zero integer(Given in the question). Hence b cannot be 4, so the answer should be (A) I only

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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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New post 06 Nov 2014, 11:56
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IMO ans shud be A

using numerator and taking ab= x
we get
3x(x^2+3x-18)=0
gives 2 equations
3x=0 .i.e 3 ab =0 i.e a=0 or b=0 # but questions has given both are non zero so it can be true
therefore
x^2-3x+18=0 , gives
can be factorized as
(x-6)(x+3)=0
gives
x=6 or ab=6 it has soloutions as (1,6),(6,1),(2,3),3,2)
a=1 and a=2 is not possible as questions will become non mathematical.
Therfore solutions possible are (6,1) and (3,2)
Therfore ans is A b=2

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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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New post 06 Nov 2014, 23:37
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manpreetsingh86 wrote:
Bunuel wrote:

Tough and Tricky questions: Algebra.



If \(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\), and a and b are both non-zero integers, which of the following could be the value of b?

I. 2
II. 3
III. 4

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III

Kudos for a correct solution.


\(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\)

we cannot have a=1, and a=-2. because this will make denominator zero.

now, consider the numerator. let ab=k
thus we have k(3k^2 +9k-54)
=k(3k^2+18k-9k-54)
=k(3k(k+6)-9(k+6))
=k(3k-9)(k+6)

if 3k-9=0
3k=9
k=3
i.e. ab=3
now if b=3, then a=1, which is not acceptable as a=1, will make the denominator equal to zero.
if k+6=0
k=-6
or ab=-6

again, if b=3, then a=-2, which is not acceptable as a=-2, will make the denominator equal to zero.

thus except b=3, all other values could be possible. hence answer should be D


ab=-6
consider b=4, then a=-3/2 which is not a integer. But according to the question a,b are non zero integer. So, answer is A

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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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\(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\)

\(3(ab)^3 + 9(ab)^2 - 54ab = 0\)

Divide the complete equation by 3ab

\((ab)^2 + 3(ab) - 18 = 0\)

ab = -6 OR ab = 3

For ab = -6

if b = 2, then a = -3 (Satisfies denominator)

if b = 3, then a = -2 (Does not satisfy denominator)

For ab = 3

if b = 3, then a = 1 (Does not satisfy denominator)

Answer = A
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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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New post 10 Nov 2014, 23:46
I just have this question, why cant be a=1 or -2. Why cant the denominator equal zero..

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2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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New post 10 Nov 2014, 23:54
sunaimshadmani wrote:
I just have this question, why cant be a=1 or -2. Why cant the denominator equal zero..


If the denominator is zero, then the equation fails. Its value turn Infinite

For example, \(\frac{5}{0} = Infinite\)
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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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New post 14 Jan 2015, 04:06
PareshGmat wrote:
\(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\)

\(3(ab)^3 + 9(ab)^2 - 54ab = 0\)

Divide the complete equation by 3ab

\((ab)^2 + 3(ab) - 18 = 0\)

ab = -6 OR ab = 3

For ab = -6

if b = 2, then a = -3 (Satisfies denominator)

if b = 3, then a = -2 (Does not satisfy denominator)

For ab = 3

if b = 3, then a = 1 (Does not satisfy denominator)

Answer = A


One question here.

For For ab = 3, why cant we have:
ab=3 --> ab=1(3), cannot be
ab=3 --> ab=3(1), couldn't this one also be? Or it is not included in the solution because there is not "1" as an answer option?

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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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New post 07 Jun 2016, 11:23
Hi Bunuel,

Can you please provide a proper solution to this question? I am not very clear from the solutions provided above.

Thanks.
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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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New post 08 Jun 2016, 01:37
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Bunuel wrote:

Tough and Tricky questions: Algebra.



If \(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\), and a and b are both non-zero integers, which of the following could be the value of b?

I. 2
II. 3
III. 4

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III

Kudos for a correct solution.



\(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\)

Take 3ab common in the numerator.

\(\frac{3(ab) * [(ab)^2 + 3(ab) - 18]}{(a-1)(a + 2)} = 0\)

Note that \((ab)^2 + 3ab - 18\) is a quadratic which we can split into factors just like we do for \(x^2 + 3x - 18\).
\(x^2 + 3x - 18 = (x + 6)(x - 3)\) so \((ab)^2 + 3ab - 18 = (ab + 6)*(ab - 3)\)


\(\frac{3(ab) * [(ab + 6)*(ab - 3)]}{(a-1)(a + 2)} = 0\)

Now, when will this fraction be 0? When one of the factors in the numerator is 0. (Note that denominator cannot be 0. So a cannot the 1 or -2.)

So either ab = 0 (a and b are non zero so not possible)
or (ab + 6) = 0 i.e. ab = -6
or (ab - 3) = 0 i.e. ab = 3

So, we get that one of these two must hold. Either ab = -6 or ab = 3.

Now let's look at the possible values of b.
Can b be 2? If b = 2, a = -3 (in this case, ab = -6. Satisfies)
Can b be 3? If b = 3, a is -2 or 1. Both values are not possible so b cannot be 3.
Can b be 4? If b = 4, a is not an integer in either case. So b cannot be 4.

Answer (A)
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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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New post 08 Jun 2016, 03:20
Thanks for the quick reply Karishma! :)

Now the question is crystal clear.
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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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New post 08 Jun 2016, 04:49
vnigam21 wrote:
Hi Bunuel,

Can you please provide a proper solution to this question? I am not very clear from the solutions provided above.

Thanks.



HI,

If someone has a problem in factoring an euatio, we can apply logic and substitute to get to our answer..

Quote:
If \(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\), and a and b are both non-zero integers, which of the following could be the value of b?

I. 2
II. 3
III. 4

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III


Now \(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\) to be TRUE, \(3(ab)^3 + 9(ab)^2 - 54ab\) or \(3ab((ab)^2 + 9(ab) - 54\) should be 0 but with following restrictions...
a)\(a\neq{0}\)...........................\(b\neq{0}\)...... It is given that a and b are non-zero
b) \(a\neq{1}\)...........................\(a\neq{-2}\).........WHY? because the denominator will become 0 and the equation undefined..

so first substitute value of b given in choices abd check if substituting a as 1 or -2 gives us value of that equation as 0..

I. 2
\((ab)^2 + 3(ab) - 18\) ..................\((a*2)^2 + 3(a*2) - 18.....................4a^2+6a-18.................\)
when a= 1, ans is 4+6-18 = -8 OK
when a =-2, ans is 4*4-6*2-18=16-12-18 = 14..OK

so 2 is a valid value..


II. 3
\((ab)^2 + 3(ab) - 18\) ..................\((a*3)^2 + 3(a*3) - 18.....................9a^2+9a-18.................\)
when a= 1, ans is 9+9-18 = 0 NO

so 3 is NOT a valid value..


III. 4
\((ab)^2 + 3(ab) - 18\) ..................\((a*4)^2 + 3(a*4) - 18.....................16a^2+12a-18.................\)
when a= 1, ans is 16+12-18 = 10 OK
when a =-2, ans is 16*4-12*2-18=64-24-18 = 22..OK

so 4 is a valid value..

Thus I and III are valid
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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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New post 24 Jul 2017, 09:40
Given (a-1) and (a+2) in the denominator, a is not equal to 1 or -2
Simplifying the numerator we get:
[3(ab)^3 + 9(ab)^2 – 54ab]
= 3ab[(ab)^2+3ab-18]
= 3ab[(ab)^2+6ab-3ab-18]
= 3ab(ab+6)(ab-3)
Since the whole equation is equal to 0, we can say the numerator is also equal to 0.
Therefore, (ignoring 3ab=0)
ab+6=0 which implies b = -6/a -------say equ(1)
ab-3=0 which implies b = 3/a --------say equ(2)
Since a and b are both non-zero integers (note, they could be negative)
from equ (2) we can say a = {-3,-1,1,3}
However, applying the constraint set by the denominator, a = {-3,-1,3}
Substituting these values of 'a' in the equ(1) and equ(2), we get b = {-3,-2,-1,1,2,6}
Hence, the answer is option A.

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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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New post 05 Aug 2017, 18:14
PareshGmat wrote:
\(\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0\)

\(3(ab)^3 + 9(ab)^2 - 54ab = 0\)

Divide the complete equation by 3ab

\((ab)^2 + 3(ab) - 18 = 0\)

ab = -6 OR ab = 3

For ab = -6

if b = 2, then a = -3 (Satisfies denominator)

if b = 3, then a = -2 (Does not satisfy denominator)

For ab = 3

if b = 3, then a = 1 (Does not satisfy denominator)

Answer = A


which denominator are u talking about?

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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo [#permalink]

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New post 07 Aug 2017, 18:37
Start with initial analysis of the problem:

a and b are integers
ab cannot = 0
We cannot have a = 1 or a = -2, since denominator cannot be 0
Since we want the solution = 0 , then the numerator must = 0

So, focusing on the numerator:

Factoring 3(ab)^3 + 9(ab)^2 - 54ab
=3ab(ab)(ab) + 3ab(3)(ab) - 3ab(18)
=3ab(ab^2 + 3ab - 18)

I know that 3ab cannot be 0, so focusing on:

(ab^2 + 3ab - 18) = 0
=(ab + 6)(ab - 3)

So,
ab = -6
ab = 3 are my only solutions

I know that a cannot = 1 or -2, so b cannot = -6 or 3

b could be -3, -2, -1, 1, 2, 6

2 is only choice above provided in the answer.
Answer is A

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Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo   [#permalink] 07 Aug 2017, 18:37
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