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Math Help - Required a mathematical solution of the problem

  1. #1
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    Cool Required a mathematical solution of the problem

    Hello Friend .............

    Here is a problem stated as:

    Let there is a milkman who requires 40 liter of milk a day. constraint is that he can have exactly 19 animals. and the animals he can choose are as below with their capability of giving milk.

    Cow can give 2 liters a day.
    Buffalo can give 5 liters a day. and
    Goat can give (1/4) liter a day.

    This problem's solution is:
    No of Cows =12 milk produced = 12 * 2 = 24
    No of Buffaloes = 3 milk produced = 5*3 = 15
    No of Goats = 4 milk produced =(1/4) * 4 = 1
    No of animals =19 milk produced =40 liters

    Now Problem is that is there any system of liner equations that can be used to get answer of such type of problems, Trial and error methods is not needed.


    Thank you all........
    Last edited by shailesh1987; February 24th 2009 at 07:37 PM.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by shailesh1987 View Post
    Hello Friend .............

    Here is a problem stated as:

    Let there is a milkman who requires 40 liter of milk a day. constraint is that he can have exactly 19 animals. and the animals he can choose are as below with their capability of giving milk.

    Cow can give 2 liters a day.
    Buffalo can give 5 liters a day. and
    Goat can give (1/4) liter a day.

    This problem's solution is:
    No of Cows =12 milk produced = 12 * 2 = 24
    No of Buffaloes = 3 milk produced = 5*3 = 15
    No of Goats = 4 milk produced =(1/4) * 4 = 1
    No of animals =19 milk produced =40 liters

    Now Problem is that is there any system of liner equations that can be used to get answer of such type of problems, Trial and error methods is not needed.


    Thank you all........


    As far as I can think I don't see any method devoid of trial and error "completely", hope to see it

    Lets take "m" buffaloes , n cows and 4k(Why!!----total milk produced is 19 , an integer) goats

    where "m" "n" and "k" are integers

    2n+ 52 +k = 40

    4k+m+n = 19

    Try bringing last equation in terms of "k" and any one of "m" and "n"

    This is what I got

    k= \frac{3m-2}{7}

    Now since m<8(Why!! ---milk=40) , its better if you try to put the values till the first time you get an integer ...that's your answer
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  3. #3
    Like a stone-audioslave ADARSH's Avatar
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    A slight typo
    .....................
    Lets take "m" buffaloes , n cows and 4k(Why!!----total milk produced is 40 , an integer) goats................
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