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Cow, Pig and Sheep

June 29, 2008

“Many years ago, a farmer had 200 dinars with which to purchase 100 animals. Cows cost 20 dinars apiece, pigs cost 6 dinars apiece, while a sheep could be bought for a single dinar. Assuming that he had to purchase at least one of each of the three animals, and assuing that he had to use all of his money, how many of each kind of animal did he purchase?”

I came across this question when I went to Pustaka Mukmin to purchase some reference book. The author claims that no advanced maths is needed…but I seriously don’t have clue for this question. I will update this once I found the answer. (or if any reader manage to solve this)

3 Comments leave one →
  1. June 29, 2008 7:14 pm

    Pretty simple.. Author was right.. No complex mathematics required..

    Let the number of cows be a;
    the number of pigs be b;
    the number of sheep be c;

    You will realise that in order to satisfy the above conditions, two laws (equations) have to be met.

    1. 20a + 5b + c = 200
    2. a + b + c = 100

    Subtract equation 2 from equation 1, you get :

    3. 19a + 5b = 100 (A line on the Cartesan Plane)

    Since you can never has decimals of animals, solutions for a, b, and c are obviously integers.

    From equation 3 above, when you multiply 19 by any number other than 5 (or multiples of 5), it is not possible to obtain a zero for the last digit. Hence, the solution for a is a multiple of 5.

    Substituting a = 5 into equation 3 generates the answer, where:

    a = 5 b = 1 c = 94

    So, 5 cows, one pig and ninety-four sheep were bought.

  2. June 30, 2008 3:20 pm

    Thanks Fate…I feel like banging my head on the wall when I found the solution…

  3. Howard permalink
    July 3, 2008 4:12 pm

    Ok la, this is solvable using that method coz there is only 1 answer. However, if you change the question to something like minimizing and maximizing cost, you might get more answers. Look for assignment problem and simplex method for details. hehe. MA1508

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