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