SOLUTION: A farmer bought 100 animals. He spent $100. Calves are $10, pigs are $3 and chickens are $.50. The farmer cannot remember what he bought and has lost his receipt. With this inf

Question 98937: A farmer bought 100 animals. He spent $100. Calves are $10, pigs are $3 and chickens are $.50. The farmer cannot remember what he bought and has lost his receipt. With this information, what did he buy?

Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=number of calves
y=number of pigs
And z=number of chickens
Now we are told that:
x+y+z=100---------------------eq1
10x+3y+0.5z=100-----------------eq2
Since we have three unknowns and only two equations, we will have to do a little trial and error to get at the correct answer or answers. I'm going to assume that there are fewer calves and pigs than chickens and that there are fewer calves than there are pigs or chickens, so I want to reduce eq1 and eq2 to one equation in two unknowns that relate the pigs to the calves. We can do this by multiplying eq1 by 0.5 and then subtracting it from eq2. When we do this we get:
eq2-0.5(eq1)=
9.5x+2.5y=50 subtract 9.5x from both sides
2.5y=50-9.5x divide both sides by 2.5
y=20-3.8x ------------------eq1a
now this tells us that:
the number of pigs=20-3.8 times the number of calves
What else do we know about this problem???
(1) We know that we cannot have a half a calf, a piece of pork or a chunk of chicken-------------------In other words, our solution must be whole numbers.
(2) Also, did you ever see a negative calf, pig or chicken???---No! ---So our solutions must be whole, positive numbers.
Armed with this information, we can quickly see that in eq1a, x cannot be greater than 5 or we will start getting negative pigs. So we know that:
0<=x<=5--------this means that we can only have between zero and five calves.
Let's just plug in values for the number of calves in eq1a and see what we get:
0 calves--------------20 pigs--------------------bingo
1 calf----------------piece of pork---no, no
2 calves--------------piece of pork---no, no
3 calves--------------piece of pork---no, no
4 calves--------------piece of pork---no, no
5 calves--------------1 pig---------------------bingo
This tells us that there are only two possible values for the pigs and calves:
(1) ----20 pigs and 0 calves
(2)------1 pig and---5 calves
Now let's go all the way back to eq1 and plug these bingo values in and see if we have whole, positive chickens. If so, then we have solved the problem:
x+y+z=100 substitute x=0 and y=20, we get:
0+20+z=100 subtract 20 from both sides:
z=80--------------------number of chickens
Answer #1
0 calves for---- $0
20 pigs for -----$60
80 chickens for -$40
100=100
x+y+z=100 substitute x=5 and y=1, we get:
5+1+z=100 collect like terms and subtract 6 from both sides
z=94-----------------------number of chickens

Answer #2
5 calves for -------------$50
1 pig for ----------------$3
94 chickens for-----------$47
100=100

Hope this helps-----ptaylor

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