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You have exactly 100 dollars to buy 100 animals you must buy at least one of each of the following animals Chickens=50 cents each Pigs=3.00 each Cows=10.00 each

  • math -

    Don't you just love these 19th Century textbooks?
    A pig for $3 !!!!! WOW

    let number of cows be x
    let the number of pigs be y
    then the number of chickens = 100-x-y

    10x + 3y + .5(100-x-y) < 100
    times 2
    20x + 6y + 100 - x - y < 200
    19x + 5y < 100

    consider 19x + 5y = 100

    But , what is the question?

    Do you want the different combinations you can buy? Do we want as close as possible to $100?

    Remember x and y have to be positive integers.
    so x has to be between 1 and 5

    form a table with columns
    x, y, and 100-x-y, and cost

    1 16 83 99.50
    1 15 82 86.00
    ...
    5 1 94 100

    x=5, y=1 is the only combination that will use all the money

    He should buy 5 cows, 1 pig and 94 chickens

    5+1+94 = 100
    $50 + $3 + $47 = $100

  • math -

    A Golden Oldie

    If you had a $100.00 to spend and need to buy a 100 animals, and cows cost
    $10.00, pigs $3.00, chickens .50 cents each,how many of each can you buy?

    Let C, P, and F be the numbers of cows, pigs, and fowl.
    1--C + P + F = 100
    2--10C + 3P + .5F = 100 or 100C + 30P + 5F = 1000
    3--Multiplying (1) by 5 and subtracting from (2) yields 19C + 5P = 100
    4--Dividing through by 5 gives P + 3C + 4C/5 = 20
    5--4C/5 must be an integer as must be C/5
    6--Let C/5 = k making C = 5k
    7--Substituting (6) back into (3) gives 95k + 5P = 100 making P = 20 - 19k
    8--k can only be 1 making C = 5, P = 1, and F = 94
    Check: 10(5) + 3(1) + .5(94) = $100

    Alternatively

    Let C, P, and F be the numbers of cows, pigs, and fowl.
    1--C + P + F = 100
    2--10C + 3P + .5F = 100 or 100C + 30P + 5F = 1000
    3--Multiplying (1) by 5 and subtracting from (2) yields 19C + 5P = 100
    4--Solving for P, P = 20 - 19C/5
    5--19C/5 must be an integer meaning that C must be evenly divisible by 5.
    6--Thus, C must be 5 making P = 1, C = 5, and F = 94.

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