At a farm, there are hens, cows and bullocks, and the kecpers to look after them. There are 69 heads less than legs, the number of cows is double of that of the bullocks, the number of cows and hens is the same and then is or keeper per ten birds and cattle. The total number of hens plus cows cows and bullocks, and their keepers does not exceed 50. How many cows are there?

Options

A. 10
B. 14
C. 12
D. 16

Answer

C

Explanation

Let the number of hens, cows, bullocks and the keepers be H, C, B and K respectively. From the given data, we get four equations as follows:
`C= 2B " " ...(i)`
`C=H " "... (ii)`
`K = (H + C + B)/10 " " ....(iii)`
`H + C + B + K lt 50` ... (iv)
Putting the values of (i) & (ii) in (iii), we get
`K = (C + C + B)/10 = (2C + B)/10 = (2(2B)+ B)/(10) " " [because C = 2B]`
or `K = (5B)/(10) " or " 10 K = 5B`
or `B = 2K`
So `B = 2K , C = 2B = 2 xx (2K ) = 4K , H = C = 4K`
According to the question,
Total number of heads =`H+C+B+K `
[`because` All have one head]
Total number of legs = `2H+4C + 4B + 2K `
[`because` Hen=2 legs, Cow & Bullock = 4 legs, Keeper = 2 legs]
According to the question
(Total no. of legs) - (Total no. of heads) = 69.
`(2H + 4C + 4B + 2K)-(H + C + B + K) = 69 `
or, `H + 3C + 3B + K = 69 `
By putting the values of H, C and B in terms of K we get,
`4K + 3 [4K] +3[2K] + K = 69 `
or, `23K = 69 `
or, `K=3` Therefore, number of cows `= C = 4K = (4 xx 3) = 12 `