There are 28 stomachs in 28 cows. However, since there are four chambers in each stomach, there would be 112 chambers amongst 28 cows.
28 depends how big the chickens are
chickens chickens
20 Cows 8 turkeys
There are 18 horses and 14 chickens. How? Horses= 4 legs 18*4=72 Chickens= 2 legs 14*2=28 72+28=100
28
Assuming each cow consumes the same amount per day. The farmer has enough for: 30 cows*28 days = 840 cow days or 840 cows for 1 day Therefore he can feed 840 cow*days/35days = 24 cows for 35 days. 30 cows - 24 cows = 6 cows The farmer must get rid of 6 cows.
Gemini Division - 2008 Smart Cows 1-28 was released on: USA: 9 October 2008
4 of each. Horses: 28 + 4 = 32 Cows: 36 + 4 = 40 Horses: cows = 32 : 40 = 4 x 8 : 5 x 8 = 4 : 5
28 days
Let's denote the number of chickens as C and the number of rabbits as R. We can write two equations based on the given information: C + R = 72 (total number of heads) and 2C + 4R = 200 (total number of feet). Solving these equations simultaneously, we find there are 50 chickens and 22 rabbits in the cage.
28 unless you consider a monkey's forepaws to be 'hands' instead of feet, in which case it would be 24.