A math question I need help with

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

  1. ArizonaNessa

    ArizonaNessa Joyfully Addicted

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    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? [​IMG]
     
  2. BarkerChickens

    BarkerChickens Microbrewing Chickenologist

    Nov 25, 2007
    High Desert, CA
    Alright...120 heads and 314 legs
    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

    edited to add...if you need me to explain anything, just ask! [​IMG]
     
    Last edited: Jan 3, 2010
  3. ArizonaNessa

    ArizonaNessa Joyfully Addicted

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    [​IMG]
    [​IMG]
    Not in this lifetime or the next would I have ever figured that out!!!
     
  4. bheila

    bheila Chillin' 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 [​IMG] I love math but these are ridiculous problems [​IMG]
     
  5. BarkerChickens

    BarkerChickens Microbrewing 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
    and # Chicken (1 head) + # Horse (1 head) = XYZ total heads

    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. [​IMG]

    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! [​IMG]
     
    Last edited: Jan 3, 2010
  6. WhiteWingedDove

    WhiteWingedDove Out Of The Brooder

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


    .. Just sayin'.
     
  7. Eggs4Sale

    Eggs4Sale Chillin' With My Peeps

    Jun 29, 2009
    The answer is BLUE!!!!
     
  8. BarkerChickens

    BarkerChickens Microbrewing Chickenologist

    Nov 25, 2007
    High Desert, CA
    Quote:[​IMG]

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

    [​IMG]
     
  9. Eggs4Sale

    Eggs4Sale Chillin' With My Peeps

    Jun 29, 2009
    Quote:[​IMG]

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

    [​IMG]

    Wouldn't that be Z?
     
    Last edited: Jan 3, 2010
  10. ArizonaNessa

    ArizonaNessa Joyfully Addicted

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    In school I was the geometry girl. I can do most math things that I can see but algebra is beyond me. My 7th grade teacher told me it is because I ask "Why?" too much. She said I would not just accept things as they are. Heck I don't know but I just don't get algebra but I can find you the volume, circumference, diameter, radius, and perimeter to just about anything. I need to be able to see it or it gets me so stressed and confused I eventually give up. I don't have a problem using formulas as long as you don't hold back information.

    Edited to say: I also can't do those questions where you have two trains leaving on different days in different cities going different speeds with the same person on both trains and you want to know when they will get to Backwoods, Nowhere, USA!!! [​IMG]
     
    Last edited: Jan 3, 2010

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