Arithmetical Reasoning 1.
At a farm , there are hens , cows and bullocks , and keepers 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 there is 1 keeper per 10 birds and cattle . The total number of hens plus cows and bullocks and their keepers does not exceed 50 . How many cows are there ?
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1 solution
Let no. of hens = h, no. of cows = c, no. of bullocks = b, no. of keepers = k
69 + ( h + c + b + k ) = 2h + 4c + 4b + 2k
Since c = 2b , and h = c. Then c = h = 2b
69 + ( 2b + 2b + b + k ) = 4b + 8b + 4b + 2k
69 + 5b + k = 16b + 2k
11b + k = 69
Since there is 1 keeper per 10 birds (hens) and cattle (cows and bullocks) , and h + c + b + k < 50. Then k< 4
k must equal to 3 in order to obtain an integer number for b.
Thus,
k = 3
b = 6
c = 12
h = 12