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Math Help - Target Average Price

  1. #1
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    Target Average Price

    If I have 100 apples at different weights, how do I share them out between a given number of people to a given ratio, so that each has as close as possible the same average weight per apple.

    For example - 3 people require 20, 30 and 50 apples respectively. There are 100 apples of various known weights.

    The average weight of each persons apples needs to be as close as possible.

    I am afraid I don't know which branch of maths this falls under. Any help appreciated!
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  2. #2
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    I'm going with "applesauce".

    You need more information, I think. There are far too many solutions.
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  3. #3
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    Hi TKHunny

    Good thought - but the apples cannot be cut up! There are many permutations I know, but this is the only information you get:
    • The number of apples
    • The weight of each apple
    • The number of people
    • The number of apples each person requires
    When the variation in apple weight is realistic, I can get the correct result. However, if I have a couple of really (ridiculously) heavy apples the result can be bad.
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  4. #4
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    Quote Originally Posted by squelch View Post
    I can get the correct result.
    I continue to struggle with this concept. I think it highly unlikely that there is "the" correct result. I takes relatively few apples and people before there are many appropriate solutions. If a situation truly has an optimal solution, there must first be a definition of "optimal". Given a definition, the apples sizes also probably are derived to accomodate.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by squelch View Post
    Hi TKHunny


    Good thought - but the apples cannot be cut up! There are many permutations I know, but this is the only information you get:
    • The number of apples
    • The weight of each apple
    • The number of people
    • The number of apples each person requires
    When the variation in apple weight is realistic, I can get the correct result. However, if I have a couple of really (ridiculously) heavy apples the result can be bad.
    One algorithm might be:

    Assume there are sufficient apples.

    Give the first person the number of apples they require choosen to be the lightest apples.

    Now repeat for the second person etc.

    If you are feeling generous start with the heaviest apples.


    CB
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