GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Sep 2019, 19:52

### GMAT Club Daily Prep

#### 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

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

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58098
If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

05 Nov 2014, 08:46
5
44
00:00

Difficulty:

95% (hard)

Question Stats:

28% (02:58) correct 72% (02:58) wrong based on 538 sessions

### HideShow timer Statistics

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.

_________________
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1768
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

09 Nov 2014, 20:45
21
5
$$\frac{3(ab)^3 + 9(ab)^2 - 54ab}{(a-1)(a + 2)} = 0$$

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

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)

_________________
Kindly press "+1 Kudos" to appreciate
##### General Discussion
Senior Manager
Joined: 13 Jun 2013
Posts: 269
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

06 Nov 2014, 02:19
4
2
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
Intern
Status: Miles to go....before i sleep
Joined: 26 May 2013
Posts: 14
Location: India
Concentration: Finance, Marketing
GMAT 1: 670 Q47 V35
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

06 Nov 2014, 11:43
4
1
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
Intern
Joined: 31 May 2013
Posts: 11
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

06 Nov 2014, 11:56
3
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
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9644
Location: Pune, India
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

08 Jun 2016, 01:37
3
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.

_________________
Karishma
Veritas Prep GMAT Instructor

Intern
Joined: 06 May 2014
Posts: 4
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

06 Nov 2014, 23:37
2
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
Senior Manager
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 407
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

14 Jan 2015, 04:06
1
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)

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?
Senior Manager
Joined: 04 Aug 2010
Posts: 466
Schools: Dartmouth College
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

20 Aug 2018, 11:05
1
dwhitmer21 wrote:
Why does the 3ab which is factored out not translate to ab=0 as a possible answer?

The prompt indicates that a and b are NONZERO INTEGERS.
Thus, ab=0 is not a valid case.
_________________
GMAT and GRE Tutor
Over 1800 followers
GMATGuruNY@gmail.com
New York, NY
If you find one of my posts helpful, please take a moment to click on the "Kudos" icon.
Available for tutoring in NYC and long-distance.
Manager
Joined: 05 Jun 2014
Posts: 60
GMAT 1: 630 Q42 V35
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

10 Nov 2014, 23:46
I just have this question, why cant be a=1 or -2. Why cant the denominator equal zero..
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1768
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

10 Nov 2014, 23:54
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, $$\frac{5}{0} = Infinite$$
_________________
Kindly press "+1 Kudos" to appreciate
Math Expert
Joined: 02 Sep 2009
Posts: 58098
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

11 Nov 2014, 03:51
I just have this question, why cant be a=1 or -2. Why cant the denominator equal zero..

Division by 0 is not allowed: anything/0 is undefined.
_________________
Manager
Status: Final Call! Will Achieve Target ANyHow This Tym! :)
Joined: 05 Jan 2016
Posts: 81
Location: India
GMAT 1: 620 Q49 V25
GPA: 3.8
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

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.
_________________
Regards,
Varun

Trying my best..... will succeed definitely! :)

The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long.
+1 Kudos if you find this post helpful. :)

Do Check OG 2017 SC Solutions - http://gmatwithcj.com/solutions-gmat-official-guide-2017-sentence-correction-questions/
Manager
Status: Final Call! Will Achieve Target ANyHow This Tym! :)
Joined: 05 Jan 2016
Posts: 81
Location: India
GMAT 1: 620 Q49 V25
GPA: 3.8
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

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

Now the question is crystal clear.
_________________
Regards,
Varun

Trying my best..... will succeed definitely! :)

The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long.
+1 Kudos if you find this post helpful. :)

Do Check OG 2017 SC Solutions - http://gmatwithcj.com/solutions-gmat-official-guide-2017-sentence-correction-questions/
Intern
Joined: 08 Nov 2016
Posts: 2
GMAT 1: 710 Q47 V40
GPA: 3.93
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

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.
Manager
Joined: 19 Aug 2016
Posts: 79
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

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)

which denominator are u talking about?
Intern
Joined: 13 May 2017
Posts: 1
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

07 Aug 2017, 18:37

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.
Intern
Joined: 08 Feb 2017
Posts: 1
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

20 Aug 2018, 10:36
Why does the 3ab which is factored out not translate to ab=0 as a possible answer?
Manager
Joined: 17 Jul 2017
Posts: 164
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

17 Nov 2018, 10:21
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.

3(ab)^3 + 9 (ab)^2=54 ab
3(ab)^2[ab+3]=54 ab
div by 3 ab b/s
ab[ab+3]=18
let ab =z
z[z+3]=18
z[z+3]=3*6
hence, ab=3
so a cant b 1 so b cnt b 3
but m not getting rest of soltuins?i know i some how messed up by not making quadratic equation ,can u please guide me where i m going wrong?
Math Expert
Joined: 02 Aug 2009
Posts: 7844
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo  [#permalink]

### Show Tags

17 Nov 2018, 20:03
vanam52923 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.

3(ab)^3 + 9 (ab)^2=54 ab
3(ab)^2[ab+3]=54 ab
div by 3 ab b/s
ab[ab+3]=18
let ab =z
z[z+3]=18
z[z+3]=3*6
hence, ab=3
so a cant b 1 so b cnt b 3
but m not getting rest of soltuins?i know i some how messed up by not making quadratic equation ,can u please guide me where i m going wrong?

You are missing out on other value of z...
z(z+3)=18..
Here z can be -6 too.
-6(-6+3)=-6*-3=18
So ab=-6..
when a is -3, b is 2.... So I is valid..

Now for b as 4, not possible, otherwise a will become fraction.
_________________
Re: If [3(ab)^3 + 9(ab)^2 – 54ab] / [(a-1)(a + 2)] = 0, and a and b are bo   [#permalink] 17 Nov 2018, 20:03

Go to page    1   2    Next  [ 22 posts ]

Display posts from previous: Sort by