# math

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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.