Posted by Max on Tuesday, February 15, 2011 at 12:18am.
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
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.