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Math Help - Puppy Power (Recursion)

  1. #1
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    Puppy Power (Recursion)

    The question is 10 female dogs average 16 puppies each year with 50% of those puppies being female. Every year the owners remove 60% of the female puppies and all of the male puppies.

    What will the growth be of female puppies left and their mothers over 15years?

    Only the original mothers and 20% of female puppies can have puppies (each year); the rest will be neutered.

    I know this is pretty easy to work out with a calculator if you keep entering the numbers, but is there a equation that will work it out?

    Thanks for any help working towards finding it.
    Last edited by Tally; March 10th 2008 at 06:23 PM.
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  2. #2
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    Let A(n+1)=A(n)+(10X16)/2X.1
    =10+(160)/2x.1
    =10+(16)/2
    =10+8
    =18 females after first year.

    A(n) being amount of present female dogs = 10

    Now I need to show A+2 is right (for next year), is that correct? If so thats where I start to struggle.
    Last edited by Tally; March 11th 2008 at 05:21 PM.
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  3. #3
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    I understand there is away to work this out in spreadsheet, could anyone please enlighten me on what would be the easiest way.

    Thanks.
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  4. #4
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    So would next year be A(n+2)=A(n+1)+(A(n+1)X16)/2X.1

    = 18+(18X16)/2X.1
    =18+288/2X.1
    =18+28.8/2
    =18+14.4
    =32.4 round down to 32
    Last edited by Tally; March 11th 2008 at 05:24 PM.
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  5. #5
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    Quote Originally Posted by Tally View Post
    The question is 10 female dogs average 16 puppies each year with 50% of those puppies being female. Every year the owners remove 60% of the female puppies and all of the male puppies.

    What will the growth be of female puppies left and their mothers over 15years?
    Let there be A_n femails in year n, then in year n+1 there are:

    A_{n+1}=A_n+A_n\times 8 \times 0.4=4.2A_n

    So:

    A_{15}=4.2A_{14}=4.2^2A_{13}= ... = 4.2^{15} A_0

    and A_0=10

    RonL
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  6. #6
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    Thank you, things are starting to become clearer now.
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