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# PS: I found it tough

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Joined: 28 Dec 2004
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PS: I found it tough [#permalink]  03 Feb 2009, 16:23
00:00

Difficulty:

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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
From MGMAT..they say this is a 600 level question..i beg to disagree..

If (16x^4 – 81y^4)/ (2x + 3y) = 12x^2 + 27y^2 and 4x + 3y = 9, what is x?
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SVP
Joined: 29 Aug 2007
Posts: 2497
Followers: 57

Kudos [?]: 555 [0], given: 19

Re: PS: I found it tough [#permalink]  03 Feb 2009, 18:11
FN wrote:
From MGMAT..they say this is a 600 level question..i beg to disagree..

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

(16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2
[(2x + 3y)(2x - 3y)(4x^2 + 9y^2)]/(2x + 3y) = 3 (4x^2 + 9y^2)
2x - 3y = 3 ...................................1
4x + 3y = 9 ...................................2

add 1 and 2: 6x = 12 or x = 2

agree that its 600 level question
_________________
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 180 [0], given: 2

Re: PS: I found it tough [#permalink]  03 Feb 2009, 18:50
this is at least 680 level.. recognizing x^2 - y^2 form requires greater skills

GMAT TIGER wrote:
FN wrote:
From MGMAT..they say this is a 600 level question..i beg to disagree..

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

(16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2
[(2x + 3y)(2x - 3y)(4x^2 + 9y^2)]/(2x + 3y) = 3 (4x^2 + 9y^2)
2x - 3y = 3 ...................................1
4x + 3y = 9 ...................................2

add 1 and 2: 6x = 12 or x = 6

agree that its 600 level question
Intern
Joined: 05 Oct 2006
Posts: 15
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Kudos [?]: 0 [0], given: 0

Re: PS: I found it tough [#permalink]  04 Feb 2009, 00:36
GMAT TIGER wrote:
FN wrote:
From MGMAT..they say this is a 600 level question..i beg to disagree..

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

(16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2
[(2x + 3y)(2x - 3y)(4x^2 + 9y^2)]/(2x + 3y) = 3 (4x^2 + 9y^2)
2x - 3y = 3 ...................................1
4x + 3y = 9 ...................................2

add 1 and 2: 6x = 12 or x = 6

agree that its 600 level question

Minor mistake in calculation though:
6x = 12 means x = 2 (not 6) .. not very imp
VP
Joined: 18 May 2008
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Re: PS: I found it tough [#permalink]  04 Feb 2009, 02:55
Yes, I solved in exactly the same way as GMATTIGER (feeling a bit lazy in writing all those equations again
I will categorize it in 600+ range
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 180 [0], given: 2

Re: PS: I found it tough [#permalink]  04 Feb 2009, 07:31
wow its unanimous.. somehow i felt this question was tough i spent over 4 min on it..

i guess i found one of my problem spots..i.e realizing the a^2 - b^2 form...

anyone have any questions similiar to this one that i can practice on?

SVP
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Re: PS: I found it tough [#permalink]  04 Feb 2009, 08:13
D-dude wrote:
GMAT TIGER wrote:
FN wrote:
From MGMAT..they say this is a 600 level question..i beg to disagree..

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

(16x^4 – 81y^4)/(2x + 3y) = 12x^2 + 27y^2
[(2x + 3y)(2x - 3y)(4x^2 + 9y^2)]/(2x + 3y) = 3 (4x^2 + 9y^2)
2x - 3y = 3 ...................................1
4x + 3y = 9 ...................................2

add 1 and 2: 6x = 12 or x = 6

agree that its 600 level question

Minor mistake in calculation though:
6x = 12 means x = 2 (not 6) .. not very imp

Thanks for the correction.
I edited the error..
_________________
Re: PS: I found it tough   [#permalink] 04 Feb 2009, 08:13
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