formula for the farm question

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formula for the farm question

Postby nocebo » Thu Nov 19, 2009 1:55 am

is it alright for me to post here the formula i've discovered?
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Re: formula for the farm question

Postby admin » Thu Nov 19, 2009 2:32 am

I don't see why not. As long as you don't write a program and then sell it to other players, you can share any information you discovered in the game here.
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Re: formula for the farm question

Postby nocebo » Thu Nov 19, 2009 3:41 am

it's not really a formula. more of a tip on what to do to solve the problem

just some basic math problem solving.

[*]let's say you are given the variables:
59 heads
150 legs
x chickens
x 3legged cows


step1. choose a value for # of 3legged cows(between 1 and total # of heads -1)

let 20 be the # of 3legged cows heads.
and 39 be the # chicken heads.

20x3=60 39x2=78

step 2. now add the total of legs that you came up with.

60+78=138. that's wrong, isn't it?

(with a ratio of heads to legs. 59:138)

BUT

since the ratio of legs is 3:2

step 3. get the difference between # of legs from the "trial" and the actual(150-138). and that's 12. now simple add 12 to the creature with more legs. and subtract the same amount with the other.

ANSWER based on above:

20+12=32 3legged cows
39-12=27 chickens

32x3=96 legs
27x2=54 legs

96+54=150

...

note. for horses and chicken problems, during step 3, after calculate the difference, divide it by 2 before adding and subtracting.

for horses and 3legged cows, i'll be back to redo it. it seems i made a mistake
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Re: formula for the farm question

Postby menot » Thu Nov 19, 2009 5:05 am

i kind of do something similar just without making a real "guess"-

X = number of heads
y= number of legs
a= number of animals that you're trying to find out (lets say chickens)
b= number of the other animal (lets say 3 legged cows)
lets say theres 52 heads, 118 legs..

what I do is i straight away assume ALL animals are the 2nd one (cows in this case)
so times the number of heads, by the number of legs that animal has.
that makes 52x3 =156
now work out how far off that is... 38
each chicken has 1 less leg than a cow, so there must be 38 chickens to correct this!
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Re: formula for the farm question

Postby JRottschaefer » Thu Nov 19, 2009 3:27 pm

Here are the easy formulas:

x = total # of cows, y = total # of chickens, H = total # of heads, L = total # of legs
When it wants you to figure out how many chickens: y = 3H - L
When it wants you to figure out how many cows: x = L - 2H

For those of you who want to know the math used to get these, here goes:
3x + 2y = L and x + y = H

solving for chickens
x + y = H
x = H - y

substitute in
3x + 2y = L
3(H - y) + 2y = L
3H - 3y + 2y = L
3H - y = L
-y = L - 3H
y = 3H - L

solving for cows
x + y = H
y = H - x

substitute in
3x + 2y = L
3x + 2(H - x) = L
3x + 2H - 2x = L
x + 2H = L
x = L - 2H
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Re: formula for the farm question

Postby nocebo » Thu Nov 19, 2009 8:43 pm

menot i tried your method, and i must say i find it quicker to solve than using mine.
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Re: formula for the farm question

Postby Dac » Mon Nov 23, 2009 11:05 pm

Come on people this is simply 9th grade math, you are making it to complicated.

Lets say 3-legged cows and horse here.

3x + 4y = #legs
x + y = #heads

Solve for x and y:
3x + 4y = #legs
3 {x + y = #heads) = 3x + 3y = 3*#heads
Therefore
3x + 4y = legs
-3x + 3y = 3*heads
=================
y = legs - 3*heads, and there you have the number of horses

This is simple algebra
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Re: formula for the farm question

Postby Xiizhan » Fri Nov 27, 2009 5:12 pm

For those struggling with the heads and feet math, use an online linear equation solver. There's one at:
http://wims.unice.fr/wims/wims.cgi?modu ... nsolver.en
It works pretty well. fill it out like this:

x+y= (number of animal heads)
(feet of animal x)x + (feet of animal y)y = (total number of feet)

so chickens and 3-legged cows with a combined total of 92 heads and 201 feet would look like:

x+y=92
2x+3y=201

Solver spits back:

{ x = 75, y = 17 }

Which means that you have 75 chickens and 17 3-legged cows.

Good luck!

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Friend me if you want, make sure you say "Dream World" or "DW" in the personal message of the request. And feel free to send me any crystals that this advice helps you win... ;-)
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Re: formula for the farm question

Postby Smokey » Fri Nov 27, 2009 5:44 pm

good grief, here....

if its horses and 3 legged cows....

take number of heads X 4(amount of horses legs)... take that result and minus the number of legs on the farm.

Now take that number and subtract it from the amount of heads on the farm. The number you got from first subtraction is the number of 3 legged cows, so if thats what you need, you dont have to subtract the second time.

If its 3 legged cows and chickens, the math is done the same except you multiply amount of cows legs(3) instead of horses(4).

if its horses and chickens, the only thing different is after you subtract the legs from your multiplied number, divide it by 2 and you get the number of chickens on the farm, for the number of horses, simply subtract the number of chickens from the number of heads. Easy as pie, simple math, quicker than all that other mumbo jumbo.
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