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PS: I found it tough

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PS: I found it tough [#permalink] New post 03 Feb 2009, 17:23
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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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Re: PS: I found it tough [#permalink] New post 03 Feb 2009, 19:11
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
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Re: PS: I found it tough [#permalink] New post 03 Feb 2009, 19:50
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
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Re: PS: I found it tough [#permalink] New post 04 Feb 2009, 01:36
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 :)
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Re: PS: I found it tough [#permalink] New post 04 Feb 2009, 03:55
Yes, I solved in exactly the same way as GMATTIGER (feeling a bit lazy in writing all those equations again :)
got answer x=2
I will categorize it in 600+ range
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Re: PS: I found it tough [#permalink] New post 04 Feb 2009, 08:31
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?

greatly appreciate your help!
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Re: PS: I found it tough [#permalink] New post 04 Feb 2009, 09:13
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..
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Math: new-to-the-math-forum-please-read-this-first-77764.html
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Re: PS: I found it tough   [#permalink] 04 Feb 2009, 09:13
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