# A math question I need help with

Discussion in 'Random Ramblings' started by ArizonaNessa, Jan 3, 2010.

1,798
19
171
Apr 7, 2009
Texas
I was playing a game and to open a chest I needed to answer this question. Can anyone figure it out and explain how you got the answer?

Question:

There is a farm with chickens and horses. There are 120 heads and 314 legs. How many chickens are on the farm?

2. ### BarkerChickensMicrobrewing Chickenologist

Nov 25, 2007
High Desert, CA
Horses have 1 head and 4 legs. Let's let Horses = X
Chickens have 1 head and 2 legs. Let's let Chickens = Y

So, 4X + 2Y = 314 legs
Similarly, 1X + 1Y = 120 heads

Let's set each one equal to Y (Chickens (because 314 is divisible by 2 easier than 4))

2Y = 314 - 4X (then, divide my two to get rid of the 2Y)
Y = 157 - 2X
and
Y = 120 - X

Now that they both equal Y, they must be equal to each other, right? So,

157 - 2X = 120 - X

Set the equation to equal X:

157 - 120 = -X + 2X, which equals

X = 37

So, if X equals 37, then plug it back in to see if it works.

4 (37) + 2Y = 314 legs
2Y = 314 legs - 148 horses
Y = 83

Similarly,

1 (37) + 1Y = 120 heads
Y = 120 heads - 37 horses
Y = 83

So, since Y is the same on both, the answer should be correct.

37 horses and 83 chickens

Last edited: Jan 3, 2010

1,798
19
171
Apr 7, 2009
Texas

Not in this lifetime or the next would I have ever figured that out!!!

4. ### bheilaChillin' With My Peeps

Feb 8, 2008
Kent, Wa
My 11yr old son has those exact same problems every night for homework, I hate them because he always needs help from me I love math but these are ridiculous problems

5. ### BarkerChickensMicrobrewing Chickenologist

Nov 25, 2007
High Desert, CA
They are complicated, but if you think of it as 2 separate problems it is a little easier. There are legs and heads. Horses and chickens. Each horse and chicken has a head and legs.

So, # Chicken (2 legs) + # Horse (4 Legs) = XYZ total legs

They have horses and chickens in common.

If you make each one equal the same thing, then you can take that common one away.
For example,
Y = 1 + 5
Y = 2 + 4
therefore, 1 + 5 = 2 + 4

This lets you solve 2 separate equations with a common variable. Once you know how many Xs or Ys is there, then you can go back to your original equation and solve for the other variable. They should have the same answer.

It's a pain, but if you set back and make it two different problems, it is easier to handle.

Let add that I am not a math geek. In fact, I hated math growing up...I just happen to do well in the subject (well, up to trig anyway. Calculus and I don't get along to well).

Edited because my brain was having some shorting of wires!

Last edited: Jan 3, 2010
6. ### WhiteWingedDoveOut Of The Brooder

93
0
29
Nov 25, 2009
Montana
I had a chicken with three legs once.

.. Just sayin'.

Jun 29, 2009

8. ### BarkerChickensMicrobrewing Chickenologist

Nov 25, 2007
High Desert, CA
Quote:

In that case, it cannot be solved due to the unknown number of 3 legged chickens and 3 legged horses.

9. ### Eggs4SaleChillin' With My Peeps

Jun 29, 2009
Quote:

In that case, it cannot be solved due to the unknown number of 3 legged chickens and 3 legged horses.

Wouldn't that be Z?

Last edited: Jan 3, 2010