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Exponents [#permalink] New post 08 May 2006, 14:01
How do you handle a question like this?


(x^2001 + x^2002)/(x^2002 - x^2000)
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 [#permalink] New post 08 May 2006, 14:17
just collect

x^2001(x+1) for the numerator
x^2000(x^2-1) for the denominator

simplify to get x/(x-1)
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Re: Exponents [#permalink] New post 08 May 2006, 16:36
consultinghokie wrote:
How do you handle a question like this?


(x^2001 + x^2002)/(x^2002 - x^2000)


Use componendo/dividendo

c/d = c+d/c-d

using the same principle above

(x^2001 + 2*x^2002 - x^2000)/(x^2001 + x^2000)

x^2000(x+2*x^2 -1)/x^2000(x+1)

(2x-1)(x+1)/(x+1) = 2x-1
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 [#permalink] New post 08 May 2006, 20:03
Solve by taking out common factors:

(x^2001 + x^2002)/(x^2002 - x^2000)

= (x^2001)(1+x)/(x^2000)(x^2-1)

= x^2(1+x)/(x+1)(x-1)

= x^2/(x-1)
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 [#permalink] New post 08 May 2006, 22:55
so we have three different answers. maybe this wasn't such an easy problem.
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 [#permalink] New post 09 May 2006, 00:11
taking x^2001 common from numerator leaves, (x+1)
taking x^2000 from denominator leaves, (x^2-1) i.e. (x+1)(x-1)

cancelling x^2000 & (x+1) from numerator & denominator, it leaves us with
x/(x-1)
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 [#permalink] New post 09 May 2006, 02:45
I am getting x/(x-1)
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 [#permalink] New post 09 May 2006, 06:41
what do you think the difficulty rating of a question like this is?
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 [#permalink] New post 09 May 2006, 07:37
consultinghokie wrote:
what do you think the difficulty rating of a question like this is?


I'd say this is a harder question. Maybe not the hardest but i woauld say Quant 46+ level...
Agree? Disagree?
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 [#permalink] New post 09 May 2006, 07:40
kook44 wrote:
consultinghokie wrote:
what do you think the difficulty rating of a question like this is?


I'd say this is a harder question. Maybe not the hardest but i woauld say Quant 46+ level...
Agree? Disagree?


I doubt it. I think it is easy/medium group.
I know there are different answers to this qtn. But that is only bcoz of carelessness. Not bcoz the qn is difficult.
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 [#permalink] New post 09 May 2006, 07:43
i guess difficulty is all relative - who knows how ETS assigns "value" to the questions
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 [#permalink] New post 09 May 2006, 08:25
gettig x(x-1), nice question.
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 [#permalink] New post 09 May 2006, 08:56
agree
x^2001(1+x)/x^2000(x^2-1)=x/(x-1)
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 [#permalink] New post 09 May 2006, 10:25
(x^2001 + x^2002)/(x^2002 - x^2000)=
x^2001(X+1)/x^2000(X^2-1)=x(x+1)/(x-1)*(x+1)=x/(x-1)
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 [#permalink] New post 09 May 2006, 10:27
ywilfred wrote:
Solve by taking out common factors:

(x^2001 + x^2002)/(x^2002 - x^2000)

= (x^2001)(1+x)/(x^2000)(x^2-1)

= x^2(1+x)/(x+1)(x-1)

= x^2/(x-1)
should be x(1+x)/(x+1)*(x-1)
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 [#permalink] New post 09 May 2006, 13:42
consultinghokie wrote:
what do you think the difficulty rating of a question like this is?


simple easy bin question 34-40 level ..just my opinion . who knows what they think .
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 [#permalink] New post 09 May 2006, 14:00
One more for x/x-1 using factorization
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 [#permalink] New post 10 May 2006, 00:52
thearch wrote:
just collect

x^2001(x+1) for the numerator
x^2000(x^2-1) for the denominator

simplify to get x/(x-1)



:good
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 [#permalink] New post 10 May 2006, 02:31
Definitely the answer is x/(x-1).
  [#permalink] 10 May 2006, 02:31
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