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 Shepherd's Puzzle (2) (Posted on 2004-10-12)
Farmer Joe owns a cow, a goat, and a sheep. The animals each eat grass at a constant rate, and the grass grows at a constant rate. And Farmer Joe occasionally lets them eat the grass on a small pasture of his.
• If the cow and the goat graze together, the pasture is bare after 45 days.
• If the cow and the sheep graze together, the pasture is bare after 60 days.
• If the cow grazes alone, the pasture is bare after 90 days.
• If the goat and the sheep graze together, the pasture is bare after 90 days, also.
How long will it take for the pasture to be bare if all three animals graze together?

 See The Solution Submitted by SilverKnight Rating: 3.7500 (8 votes)

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 Puzzle Solution Comment 30 of 30 |

Let c = daily ration of the cow.
g = daily ration of the goat.
s = daily ration of the sheep.
r = daily rate of growth of grass in the pasture.
I = initial amount of grass in the pasture.

Then, by the problem,

I + 45r = 45(c+g).....(i)
I + 60r = 60(c+s).....(ii)
I + 90r = 90c...........(iii)
I + 90r = 90(g+s).....(iv)

From (i), (ii) and (iii), we obtain:
c+g = I/45 + r
c+s = I/60 + r
g+s = I/40 + r

Or, (c+g+s) = I/40 + 3r/2......(v)
Or, c = (I/40 + 3r/2) - (I/90 + r) = I/72 + r/2

But, from (iii), we observe that: c = I/90 + r
Thus, I/72 + r/2 = I/90 + r
Or, I/360 = r/2
Or, I = 180r

Let the number of days that the pasture will support all the three
animals be d.

Then, I + dr = d(c+g+s)
Or, 180r + dr = d(c+g+s)
Or, 180r/d + r = c+g+s
Or, 180r/d +r = 180r/40 + 3r/2 ( from (v))
Or, 180r/d = 5r
Or, d = 36

Consequently, it will take precisely 36 days for the pasture to be
bare if all the three animals graze together.

Edited on October 23, 2007, 6:03 am
 Posted by K Sengupta on 2007-10-23 06:02:41

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