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Difficult Math Question #19 - Difficult Alegbra

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Question #19 - Difficult Alegbra

Post Fri Sep 22, 2006 10:03 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    OA coming after a few people answer. see image below..
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    gamemaster Senior | Next Rank: 100 Posts Default Avatar
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    Post Fri Sep 22, 2006 1:05 pm
    The expression in the multiple brackets equals eventually:

    S = x+x^2+x^3+x^4+x^5

    according to the question, An/S = x^5

    => An = x^6+x^7+x^8+x^9+x^10

    so its easy to see n = 7

    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Tue Sep 26, 2006 5:53 pm
    here's the OA:

    The method I followed was to reduce the Q to (x^6/ x ) * ( Y/Y)
    the eqn An= (x ^ n -1)(1+ x + x^2 + x^3 +.... ) ----------------------------(1)

    and the eqn x(1+x(1+x(1+x...))))
    which I call Z can be reduced to x( 1+x+x^2+x^3 ..) --------(2)

    from (1) and (2) we get An / Z = x^(n-1) / x
    therefore for getting answer x^5 (n-1) = 6
    therefore n=7

    Ans: B

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