SOLUTION: Old Mcdonald was buying animals for his farm. Cows cost $10, Pigs cost $3, and chickens cost $.50. He only bought 100 animals and bought at least one of each animal. He only spen

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Old Mcdonald was buying animals for his farm. Cows cost $10, Pigs cost $3, and chickens cost $.50. He only bought 100 animals and bought at least one of each animal. He only spen      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 206376: Old Mcdonald was buying animals for his farm. Cows cost $10, Pigs cost $3, and chickens cost $.50. He only bought 100 animals and bought at least one of each animal. He only spent $100. How many of each animal did he buy?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Cows cost $10, Pigs cost $3, and chickens cost $.50.
He only bought 100 animals and bought at least one of each animal.
He only spent $100. How many of each animal did he buy?
:
Let c = no. of cows; p = no. of pigs; s = no. chickens
:
No. of animals equation:
c + p + s = 100
:
Cows cost $10, Pigs cost $3, and chickens cost $.50.
the $$ equation:
10c + 3p + .5s = 100
:
We have two equations but 3 unknowns, however, we know there can't very many cows,
After trying a couple values for no. of cows, came up with 5 cows,
they're worth $50, our two equations then are:
:
p + s = 100 - 5
p + s = 95
and
3p + .5s = 100 - 50
3p + .5s = 50
:
Multiply the 1st equation by 3 and subtract the 2nd equation:
3p + 3s = 285
3p + .5s = 50
---------------- subtracting eliminates p
0p + 2.5s = 235
s = 235%2F2.5
s = 94 chickens
:
Find pigs
p + 94 = 95
p = 95 - 94
p = 1 pig
:
We have then: 5 cows, 1 pig, 94 chicken
;
;
Check in the cost equation
10(5) + 3(1) + 94(.5) =
50 + 3 + 47 = $100