There need to be a mistake
X=2, Y=1/2 => X*Y=1
(5/2)^2)/(3/2)^2=(5/3)^2=25/9
and several other results...
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Stuart@KaplanGMAT
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Definitely a typo.CIAO1982 wrote:If X*Y = 1
what is the value of
2 ( X+Y )^2
__________ =?
2 ( X-Y )^2
A) 2 B) 4 C) 8 D) 16 E) 32
I hope that the question is clear, thans you...
The fraction reduces to:
(x^2 + 2xy + y^2)/(x^2 - 2xy + y^2)
(x^2 + 2 + y^2)/(x^2 - 2 + y^2)
Which has an infinite number of solutions.
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cramya
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Stuart,
U r correct; ther is a typo. The actual question if I am not
mistaken is 2 ^ (x+y)^2 / 2 ^ (x-y)^2
given xy=1
CIAO1982,
Here's the solution:
2 ^ (x+y)^2 / 2 ^ (x-y)^2
{
((x+y)^2 = x^2+y^2+2xy
(x-y)^2 = x^2+y^2-2xy)
}
= 2 ^ (x^2+y^2+2xy) / 2^ (x^2+y^2- 2xy)
(a^m/a^n = a ^ (m-n) where a=2 m = (x+y)^2 n = (x-y)^2)
= 2 ^ (x^2+y^2+2xy - ( x^2+y^2-2xy)
= 2 ^ (x^2+y^2+2xy-x^2-y^2+2xy)
= 2^4xy
= 2 ^4 (since xy=1)
= 16
Friendly request: Please post the correct question so that the experts/other users can provide u the correct solution. Also include the official answer using spoiler function if possible. I know u r getting started so its ok. Please take this as nothing but a friendly request
OA: whatever the oa is
Regards,
Cramya
U r correct; ther is a typo. The actual question if I am not
mistaken is 2 ^ (x+y)^2 / 2 ^ (x-y)^2
given xy=1
CIAO1982,
Here's the solution:
2 ^ (x+y)^2 / 2 ^ (x-y)^2
{
((x+y)^2 = x^2+y^2+2xy
(x-y)^2 = x^2+y^2-2xy)
}
= 2 ^ (x^2+y^2+2xy) / 2^ (x^2+y^2- 2xy)
(a^m/a^n = a ^ (m-n) where a=2 m = (x+y)^2 n = (x-y)^2)
= 2 ^ (x^2+y^2+2xy - ( x^2+y^2-2xy)
= 2 ^ (x^2+y^2+2xy-x^2-y^2+2xy)
= 2^4xy
= 2 ^4 (since xy=1)
= 16
Friendly request: Please post the correct question so that the experts/other users can provide u the correct solution. Also include the official answer using spoiler function if possible. I know u r getting started so its ok. Please take this as nothing but a friendly request
OA: whatever the oa is
Regards,
Cramya
