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January 20, 2017
Posted by **Max** on Tuesday, February 15, 2011 at 12:18am.

- math -
**Reiny**, Tuesday, February 15, 2011 at 12:48amDon'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 -
**tchrwill**, Tuesday, February 15, 2011 at 8:25amA 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.