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Cattle Selling (Posted on 2003-03-10) Difficulty: 3 of 5
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)

See The Solution Submitted by Ravi Raja    
Rating: 1.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
With the new rupee total--a solution or several | Comment 8 of 12 |
If c,p and s are the numbers of cows, pigs and sheep, then
c+p+s = 40k
17c+2p+2s = 285
subtracting twice the first equation from the second:
15c = 285 - 80k
Values of 1 or 2 for k yield fractional cows.
A value of 3 for k yields 3 cows.
Values of k higher than 3 yield negative cows.

So k is 3 and the total number of animals is 40*k=120, of which 3 are cows. That leaves a total of 117 sheep and pigs. As the sheep and pigs are sold at the same price, we can't tell how many of the 117 are sheep and how many are pigs.

None of the numbers is divisible by either 8 or 5, there can't be an even distribution in either droves or among purchasers, so that can't be used as a determinative for allocating the 117 to pigs vs. sheep.
  Posted by Charlie on 2003-03-11 03:23:07
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