SOLUTION: Farmer wants to buy cows, pigs and chickens. He has $100 and wants to buy total of 100 combined animals. Cows are $10. Pigs are $3. Chickens are $.50. How many of each animal shoul

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Farmer wants to buy cows, pigs and chickens. He has $100 and wants to buy total of 100 combined animals. Cows are $10. Pigs are $3. Chickens are $.50. How many of each animal shoul      Log On


   



Question 281733: Farmer wants to buy cows, pigs and chickens. He has $100 and wants to buy total of 100 combined animals. Cows are $10. Pigs are $3. Chickens are $.50. How many of each animal should he buy to get 100 total animals for $100?
Found 2 solutions by josmiceli, richwmiller:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let c = number of cows
Let p = number of pigs
Let k = number of chickens
given:
10c+%2B+3p+%2B+%281%2F2%29%2Ak+=+100 dollars
c+%2B+p+%2B+k+=+100
There are 2 equations and 3 unknowns, so it
can't be solved directly. but you can get there
There can't be all cows- that would be
only 10 animals for $100
There can't be all pigs- that would be
33 animals for $99
Say there are 50 chickens and 50 pigs:
k+=+50 %281%2F2%29%2A50+=+25
p+=+50 3%2A50+=+150
That's $175 for 100 animals. There must be more
chickens than pigs
Let's say:
k+=+60 %281%2F2%29%2A60+=+30
p+=+40 3%2A40+=+120
That's $150 for 100 animals
I have to run, but keep going and you'll get it
You must have at least 1 cow, too

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
you know there are going to be a lot of chickens and few cows.
w+p+n=100
10w+3p+.5n=100
If there were all chickens he could buy 200 chickens
If there were all cows he could by 10 cows
If there were all pigs he could by 33 pigs and have a dollar left over.
So the numbers need to be between 0 and those upper limits.
0 < w < 11
0 < p < 34
0 < n < 201
chicken=n=94
cow=w=5
pig=p=1
5*10+94*.5+1*3=100
50+47+3=100
94+5+1=100