There are 13 animals in the barn. Some are chickens and some are pigs. There are 40 legs in all. How many of each animal are there?

Question

There are 13 animals in the barn. Some are chickens and some are pigs. There are 40 legs in all. How many of each animal are there?

Answer 1

Hi Cameron--

 

chickens + pigs = 13  

 

so...

 

c + p = 13, so...

 

p = (13 - c)

 

2 legs times the number of chickens (2c) + 4 legs times the number of pigs (4p) = 40 total legs. So...

 

2c + 4(13-c) = 40, so

 

2c + 52 - 4c = 40, so

 

-2c = -12, so

 

2c = 12, so

 

c = 6.

 

There are 6 chickens, and therefore 7 pigs.

 

Hope this is clear!

 

Matt

Answer 2

Hi, Bob, let's turn the word problem into a set of equations:

 

Let's say that x = # of chickens, and y = # of pigs

 

If there are 13 animals altogether, that means that

 

    x + y = 13

 

If there are 40 legs together, we know that chickens have 2 legs, while pigs have 4, so the total number of legs would be:

 

    2x + 4y = 40

 

We now have two different equations, with the same two variables, so we can rearrange and substitute one equation into the other to solve. We can rearrange the first equation as:

 

    x + y = 13

 

    x = 13 - y

 

Let's plug this expression for "x" into the second equation:

 

    2(13-y) + 4y = 40

 

Now we have an equation with only one variable, so we can use algebra to solve for y:

 

    26 - 2y + 4y = 40

 

    2y = 14

 

    y = 7, there must be 7 pigs.

 

From the first equation that we rearranged,

 

    x = 13 - y

 

Now that we know the numerical value of "y", we can plug that in:

 

    x = 13 - 7

 

    x = 6, there must be 6 chickens