Matsh Riddle?

Question

i know i should do my own homework but i need help. My teacher gave i to us for fun.

You have $100 to buy 100 animals
You must buy one of each

Sheep: $10
Pigs: $2
Chickens: $0.50

sorry i spelt maths wrong! :[

Answer 1

3 variables, 2 equations, a number of solutions are possible,

Suppose you have bought x no. of sheeps, y no. of pigs & z no. of chickens.
Hence, x + y + z = 100
cost of x sheeps = $10x
cost of y pigs = $2y
cost of z chickens = $0.5z
Hence, 10x+2y+0.5z = 100
or, 20x + 4y + z= 200
or, 20x + 4y + z - x - y - z = 200-100
or, 19x + 3y = 100
if x = 1 , 3y = 100 - 19 = 81
or, y = 27
hence, z = 100 - 27 -1 = 72,
if x = 2, 3y = 100 - 38 = 62
hence, in that case y is not an integerwhich is impossible (you can't buy a half of a pig)
for similar reason x is not equal to 3, 5
if x=4, 3y = 100 - 76 = 24
or, y = 8
hence z = 100 - 8 - 4= 88
x can't exceed 5, as in that case y becomes negative.

so the two possible solutions are,
You have to buy 1 sheep, 27 pigs & 72 chickens
or,
You have to buy 4 sheeps, 8 pigs & 88 chickens

In both the cases you have to spent $100 & you buy 100 animals.

Answer 2

s + p + c = 100
20s + 4p + c = 200
19s + 3p = 100
s ≤ 5, p ≤ 33
p = (100 - 19s)/3 must have integer solutions:
s = 1 or 4
.s.. p... c
1, 27, 72, (10 + 54 + 36 = 100)
4,.. 8, 88, (40 + 16 + 44 = 100)

There appear to be two valid solutions.

Answer 3

4 sheeps = $40
8 pigs = $16
88 chickens = $44
==============
$100

Answer 4

The answer is 4 sheep, 8 pigs, and 88 chickens.
I solved by trial and error and a bit of iteration, got it on my 8th guess.

Answer 5

I got a different correct answer (also by trial and error)
1 sheep
27 pigs
72 chickens