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December 20, 2014

December 20, 2014

Posted by **amanda** on Thursday, October 29, 2009 at 7:41pm.

- math -
**Reiny**, Thursday, October 29, 2009 at 7:57pmFirst of all let me predict that there is no unique answer.

Here is why...

Let the number of horses be x

let the number of cattle by y

let the number of sheep be 100-x-y

then 10x + y + (1/2)(100-x-y) = 100

multiply 2 to get rid of fractions.

20x + 2y + 100 - x - y = 200

y = 100 - 19x

Of course we know that x and y must be natural numbers, (can't have 1/2 a cow)

so the largest value that x can have is 5

x=5, then y= 5 and 100-5-5 = 90

so 5 horses, 5 cattle and 90 sheep

let x = 4, y = 24, 100-4-24 = 72

so 4 horses, 24 cattle and 72 sheep

let x = 3, y = 43 100-3-43 = 54

3 horses, 43 cattle and 54 sheep

etc.

each of the above combinations will cost $100

(looks like this is from a book printed in the 1800's. Time for the schoolboard to upgrade)

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