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?

## Solution

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

$\begin{aligned} c+g &=& \frac{1}{40}, \\ c+w &=& \frac{1}{90}, \\ w+g &=& \frac{1}{60}. \end{aligned}$We add:

$2c + 2g + 2w = \frac{1}{40} + \frac{1}{90} + \frac{1}{60} = \frac{9 + 4 + 6}{360} = \frac{19}{360}.$

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

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