# farm mathcoolhistory

(3 comments)June 30, 2008

So back in high school my calc teacher (Mr. Pawlowski! Yay old yearbooks.) gave me an interesting way of solving this one classic-sounding algebra problem.

Like many math problems, the premise is a bit absurd: Farmer Brown knows he has, I dunno, 30 animals, cows and chickens. He doesn't know how many of each he has, but he

*does*know that among them they have, say, 74 legs. (Why Farmer Brown is able to count legs but not animals, and not distinguish a chicken leg from a cow egg, is not made clear...)

Now there's the fancy-pants school-larnin' way of solving this: (let c be the number of chickens, m (for moo) the number of cows)

we know c+m=30

thus c = 30 - m

plus the legs means (c*2)+(m*4)=74

((30-m)*2)+4m=74

60-2m+4m=74

60+2m=74

30+m=37

m=7

and c = 23

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That math seems a lot easier to do in your head! I'm not sure what the equations for it look like though... let me see...

c + m = 30So that's a lot of steps that seemed easier to manage when you chunked things the right way. Maybe it's more like

he quickly figured 2 * (c + m) = 60

but he knows 2c + 4m = 74 I guess he was able to tell that

(2c + 4m) - (2c + 2m) = 74 - 60

2m = 14

m = 7

c = 23

Let a be the number of animalsI'm not quite sure what the takeaway math lesson from this is... maybe it's the use of more variables when you're trying to do stuff in your head?

t be the total number legs (2c + 4m), 74

let d be the number of legs down, 2*a, 60

d - t = 14

2m = 14 (I think that's the smart bit)

m = 7

c = a - m = 23

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