2420.41 – The Cow, the Goat, and the Wildebeest

A cow and a goat can together eat clean farmer Hogan's pasture in 40 days, including the Edsel abandoned there in 1958. The cow and a wildebeest can do this in 90 days. The wildebeest and the goat would take 60 days.

How long should it take all three of them chomping away together?


We need the daily rate of consumption for each animal. Let c,g,wc, g, w be the daily rates of the cow, goat and wildebeest. What we are given is that

c+g=140,c+w=190,w+g=160. \begin{aligned} c+g &=& \frac{1}{40}, \\ c+w &=& \frac{1}{90}, \\ w+g &=& \frac{1}{60}. \end{aligned}

We add:

2c+2g+2w=140+190+160=9+4+6360=19360.2c + 2g + 2w = \frac{1}{40} + \frac{1}{90} + \frac{1}{60} = \frac{9 + 4 + 6}{360} = \frac{19}{360}.

If TT is the time all three animals will take, we can compute

T=1 whole field including carjoint rate of cow, goat and wildebeest=1c+g+w=11219360=7201937.8938 days. \begin{aligned} T &=& \frac{\text{1 whole field including car}}{\text{joint rate of cow, goat and wildebeest}} \\ &=& \frac{1}{c + g + w} = \frac{1}{\frac{1}{2} \frac{19}{360}} \\ &=& \frac{720}{19} \approx 37.89 \approx 38 \text{ days}. \end{aligned}

(Note: this problem doesn't take account of the growth of grass over the 38 days. See Stella 2500.62 for that.)

(And a query: How likely is it that a wildebeest and an abandoned Edsel would be in the same place on the globe?)