SOLUTION: When young MacDonald was starting his farm a long time ago, cows cost $10 each, pigs cost $3 each, and chickens cost $0.50 each. He purchased at least one of each kind of animal. H

Question 160772: When young MacDonald was starting his farm a long time ago, cows cost $10 each, pigs cost $3 each, and chickens cost $0.50 each. He purchased at least one of each kind of animal. He paid axactly $100 for exactly 100 animals. How many chickens did he buy?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=number of cows
y=number of pigs
z=number of chicks
Now we are told that:
x+y+z=100------------------------------eq1
10x+3y+0.5z=100--------------------------eq2 ($ understood)
Three unknowns and only two equations which means there will have to be some trial and error and maybe more than one solution.
Multiply eq2 by 2 (and we get 20x+6y+z=200) and then subtract eq1 from it and we get:
19x+5y=100 or
5y=100-19x divide each side by 5
y=(100-19x)/5--------------------------eq3
Now eq3 gives us a relationship between the number of pigs,y, and the number of cows, x.
What do we know about this problem:
First, we know that we cannot have a negative cow, pig or cluck--we deal in positive numbers.
Secondly, we know that we cannot have a fraction of a cow, pig or cluck--we deal in whole numbers.
Given the above constraints, we immediately see from eq3 that we cannot have more than 5 cows otherwise we start having negative pigs which is a no-no.
Now lets make a chart for cows and pics
cows 5--pigs=(100-19*5)/5=1------------------BINGO!!!!
NOW WE plug these values into eq1 and see if the chicks come out ok:
5+1+z=100
z=100-6=94 chicks
Tis gives us the 100 animals, next we wil see if the dollars work out by plugging these into eq2
10*5+3*1+94*0.5=100
50+3+47=100
100=100
So, we know that one answer is:
10 cows
1 pig
94 clucks
There may be other answers, you can check.
Hope this helps---ptaylor
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