My friend who owns a farm nearby, has five droves of animals on his farm consisting of cows, sheep and pigs with the same number of animals in each drove. One day he decided to sell them all and sold them to 8 dealers.

Each of the 8 dealers bought the same number of animals and paid at the rate of Rupees 17 for each cow, Rupees 2 for each sheep and Rupees 2 for each pig. My friend received from the dealer in total Rupees 285.

How many animals in all did he have and how many of each kind ?

(Given: 1 Rupee = 100 Paise)

c - cows; s - sheeps ; p - pigs.

17*c + 2*s + 2*p = 285

and (c + s + p) is a multiple of 8, say 8*k.

c = 8*k - s - p

17*(8*k - s - p) + 2*s + 2*p = 285

136*k - 15*s - 15*p = 285

15*(s + p) = 136*k - 285 

(s + p) = 136*k/15 - 19

k = 15---> (s + p) = 117------> c = 3----->120 animals.

(k = 30 ------>(s + p) = 253-----------> c < 0)

And this is all we can get with the given data. Since there is at least 1 of each animal, there are 116 solutions, named (c, s, p) = (3, 1, 116), (3, 2, 115), (3, 3, 114),...(3, 114, 3), (3, 115, 2) and (3, 116, 1).

Posted by pcbouhid on 2005-07-19 23:09:48