Weird homework question-can't figure it out?

Question

My pre-algebra teacher gave us this homework question,due next Friday.I took it to my Granddaddy and we worked on it until midnight-nothing.Here it is:

There are cows and chickens on a farm. There are 112 legs and 33 heads. How many chickens are there and how many cows?

My brain hurts.

Answer 1

23 cows and 10 chicken

Answer 2

so you have cows (a) and chickens (b)

there are 33 heads, meaning that there are 33 animals total.

a+b=33

cows have four legs and chicken have two, meaning that the total number of legs can be expressed as:

4a+2b=112

then you can solve it. ultimately, you should get 23 cows and 10 chickens

Answer 3

uhh... I'm really not good at math but I guess, 102 cows and 10 chickens? I might be wrong. c**p I did it wrong!!! Soo it's what the others say (?)

Answer 4

Trick Question  

Answer 5

I used algebra to solve this and got 23 cows and 10 chickens.  

I check my answer:

23+10=33 (the number of heads)

23(4)+10(2) =112 (the number of legs)

This is based on normal cows that have one head and four legs and regular chickens that have two legs and one head.

If you are in pre-algebra and don't know how to figure this out, I guess you would have to use a trial and error method.

Answer 6

Cows have 1 head and so do chickens, so there are 33 total animals.  But cows have 4 legs and chickens have two, so find two numbers that add up to 33, multiply one by 4 and the other by 2, and add the answers to see if you get 112.  If too low, you need more cows to get more legs; if too high you need more chickens.

Example:

10 cows = 40 legs

23 chickens = 46 legs, total 86 legs, too low so need more cows

Answer 7

Guess and Check Table.

Answer 8

you solve it by using substitution

a+b=33

4a+2b=112

now pick one equation and isolate the variable

b=33-a

now sub that in to the other equation

4a+2(33-a)=112

4a+66-2a=112

2a+66=112

2a=46

a=23

now sub a into one equation

23+b=33

b=10

there are 23 cows and 10 chickens