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

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

I have to buy 100 animals, I have to use 100 dollars, I have to get at least 1 of each. horses(X) = 10 bucks. cows(Y) = 1 buck. chickens(Z) = 50 cents

Can you help me to find the answer

Poopy dulrp

Well, I have a clever solution for you. With 100 animals, you could go for the classic "chickens, pigs, and cows" combo. Let's do the math:

First, buy 48 chickens for 50 cents each, which will cost you $24.00 in total. Now you have $76.00 left.

Next, grab 33 pigs for $3.00 each, totaling $99.00. That leaves you with $1.00.

Finally, use that remaining dollar to buy yourself one cow for $10.00. You've reached your goal of purchasing 100 animals with your $100.00 budget!

So, there you have it: 48 chickens, 33 pigs, and 1 cow to create your very own farm animal collection.

To find out how many of each animal you can buy with 100 dollars, we can set up a system of equations. Let's call the number of chickens, pigs, and cows you can buy as x, y, and z respectively.

From the given information, we know the following prices:
- Chickens cost 50 cents each, which can be written as 0.50 dollars.
- Pigs cost 3.00 dollars each.
- Cows cost 10.00 dollars each.

Now we can set up the equations:
Equation 1: 0.50x + 3.00y + 10.00z = 100 (the total cost should be exactly 100 dollars)
Equation 2: x + y + z = 100 (the total number of animals should be 100)

We can use these two equations to solve for the values of x, y, and z.

To find the solution, we'll use the substitution method.

Step 1: Solve Equation 2 for x in terms of y and z.
x = 100 - y - z

Step 2: Substitute the value of x from Step 1 into Equation 1.
0.50(100 - y - z) + 3.00y + 10.00z = 100

Step 3: Simplify and solve for y.
50 - 0.50y - 0.50z + 3.00y + 10.00z = 100
2.50y + 9.50z = 50

Step 4: Simplify and solve for z.
5y + 19z = 100

To find the values of y and z that satisfy this equation, we can use trial and error or a solving method.

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.