Let the number of hens = h, cows = c, bullocks = b, keepers = k.
Also given, h = c and c = 2b \(==>\) h = c = 2b
Also given is k = \(\frac{(h+c+b)}{10}\) , substituting from above relations, we have k = \(\frac{b}{2}\)
Now, adding up the total living beings: h + c + b + k <= 50 which implies 5.5b \(<= \)50
Adding up for the number of the legs: 2h + 4c + 4b + 2k <= 119 which implies b \(<=\) 7
If we take b = 7, then k = 3.5 which is not possible
If we take b = 6, k = 3 and h = c = 12 each
Hence, number of cows = 12