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

It is currently 17 Jun 2019, 08:45

Close

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
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.

Close

Request Expert Reply

Confirm Cancel

(16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55630
(16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?  [#permalink]

Show Tags

New post 05 Nov 2014, 08:35
2
6
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

68% (02:50) correct 32% (02:55) wrong based on 148 sessions

HideShow timer Statistics

Most Helpful Community Reply
SVP
SVP
User avatar
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1795
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: (16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?  [#permalink]

Show Tags

New post 10 Nov 2014, 01:58
5
3
\(\frac{(16x^4 – 81y^4)}{(2x + 3y)} = 12x^2 + 27y^2\)

\(\frac{(4x^2 + 9y^2)(4x^2 - 9y^2)}{(2x + 3y}) = 12x^2 + 27y^2\)

\(\frac{(4x^2 + 9y^2)(2x + 3y)(2x - 3y)}{(2x + 3y)} = 3*(4x^2 + 9y^2)\)

2x - 3y = 3

4x + 3y = 9 .......... Info already given in problem

6x = 12; x = 2

Answer = C
_________________
Kindly press "+1 Kudos" to appreciate :)
General Discussion
Intern
Intern
avatar
Joined: 22 Feb 2014
Posts: 29
Reviews Badge
Re: (16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?  [#permalink]

Show Tags

New post 05 Nov 2014, 18:07
2
Factorizing the given: ((2x)^2+(3y)^2)(2x-3y)(2x+3y)/(2x+3y) = 3(4x^2+9y^2)

Further: (4x^2+9y^2)(2x-3y) = 3(4x^2+9y^2) ---> 2x-3y = 3 and 4x+3y = 9 (given) ----> Adding up, 6x = 12 or x = 2; Answer: C
Manager
Manager
avatar
B
Joined: 23 Dec 2013
Posts: 142
Location: United States (CA)
GMAT 1: 710 Q45 V41
GMAT 2: 760 Q49 V44
GPA: 3.76
Reviews Badge
Re: (16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?  [#permalink]

Show Tags

New post 29 May 2017, 17:46
Bunuel wrote:

Tough and Tricky questions: Algebra.



(16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?

A. 1
B. 1.5
C. 2
D. 3
E. 4

Kudos for a correct solution.


It's important to work neatly here.

(16x^4-81y^4)/(2x+3y) = 12x^2 + 27y^2

(4x^2+9y^2)(4x^2-9y^2)/(2x+3y) = 12x^2 + 27y^2

(2x+3y)(2x-3y)(4x^2+9y^2)/(2x+3y) = 12x^2 + 27y^2

(2x-3y)(4x^2+9y^2) = 12x^2 + 27y^2

(2x-3y)(4x^2+9y^2) = 3(4x^2 + 9y^2)

2x-3y = 3
+4x+3y = 9

6x = 12
x = 2
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 11377
Re: (16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?  [#permalink]

Show Tags

New post 10 Sep 2018, 01:47
Hello from the GMAT Club BumpBot!

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.
_________________
GMAT Club Bot
Re: (16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?   [#permalink] 10 Sep 2018, 01:47
Display posts from previous: Sort by

(16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne