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)

I don't think this problem has a solution unless we can have fractions of animals. Here are my thoughts:

Each of the 5 droves contains the same number of animals, therefore the total number of animals is a multiple of 5.
Each of the dealers bought the same number of animals, therefore the total number of animals is a multiple of 8.
From the above two statements it follows that the total number of animals must be a multiple of 40.

The least expensive animals (sheep and pig) cost 2 Rupees, therefore the most animals there could be is 301/2 = 150 animals. Since the total must be a multiple of 40 then there are 40, 80 or 120 animals.

Let C be the number of cows, let N be the number of sheep/pigs (since sheep and pigs both cost 2).
Then 17C + 2N = 301

Assume C + N = 40, ie N = 40 - C
This gives: 17C + 2*(40 - C) = 301 -> C = 221/15 = 14.7

Assume C + N = 80, ie N = 80 - C
This gives: 17C + 2*(80 - C) = 301 -> C = 141/15 = 9.4

Assume C + N = 120, ie N = 120 - C
This gives: 17C + 2*(120 - C) = 301 -> C = 61/15 = 4.1

So none of the possibilities give an integer number of cows.

Are the values of the sheep and pig correct? As they are both the same it is going to be impossible to work out how many of each animal there are unless there are only cows, which there can't be since 301 isn't a multiple of 17.

Posted by fwaff on 2003-03-10 02:32:54