A math question don't worry this is not for a test/quiz/homework please help?

Question

A farmer buys feed...
chicken feed cost .47 per ton
pig feed cost 1.23 per ton
cattle feed cost 1.44 per ton

the farmer can't buy any partial tons
how much of each would he have, if he bought 100 tons of feed and paid 100 dollars???

please help me i need it

Answer 1

Lets convert the problem to cents.

In equation form, with c for chicken feed in tons, p for pig feed in tons, and b for cattle feed in tons, we have
47c+123p+144b=10000
c+p+b = 100

Eliminating c, we arrive at
76p+97b = 5300.

So p = (5300-97b)/76 = 69-b+(56-21b)/76
Let z = (56-21b)/76 then
76z + 21b = 56
So b = (56-76z)/21 = 2-3z+(14-13z)/21
Let x = (14-13z)/21
Then 21x+13z=14
So z=(14-21x)/13 = 1-x+(1-8x)/13
Let y = (1-8x)/13 then
13y+8x=1, x=(1-13y)/8 = -y+(1-5y)/8
Because this can go on for quite some time, let y=5, then x = -8, z = 14, b=-48, and p=131. Shifting answers by the original values (76 and 97), b=-48+76, p=131-97, so b=28, p=34; therefore c=38. This is a long winded solution, but it illustrates the general rule for solving linear diophantine equations.

Steve

Answer 2

Let c be chicken feed
Let p be pig feed
Let m be cattle feed

0.47c + 1.23p + 1.44m = 100
c + p + m = 100

Two equations and three unknown, this will take some guessing to find the answer.

Let's start by multiplying the top equation by 100 to get rid of decimals:
47c + 123p + 144m = 10000

Let's multiply the bottom by 47:
47c + 47p + 47m = 4700

Subtract the two:
76p + 97m = 5300

Now time for some guessing... if p was 0, then m couldn't be bigger than 55. So let's try values of m between 0 and 55.

I put the formula p = (5300-97m)/76 into a spreadsheet and fed it the values of m = 0 to 54. It came out with an integer result when m = 28.

m = 28, p = 34

That would make c = 38

Double-checking:
38 tons of chicken feed = $17.86
34 tons of pig feed = $41.82
28 tons of cattle feed = $40.32

Adding that up you get exactly $100.00