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?

Solution

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=140c+w=190w+g=160\begin{aligned}c+g &= \dfrac{1}{40} \\[1em]c+w &= \dfrac{1}{90} \\[1em]w+g &= \dfrac{1}{60}\end{aligned}

We add:

2c+2g+2w=140+190+160=9+4+6360=19360\begin{aligned}2c + 2g + 2w &= \dfrac{1}{40} + \dfrac{1}{90} + \dfrac{1}{60} \\[1em]&= \dfrac{9 + 4 + 6}{360} \\[1em]&= \dfrac{19}{360}\end{aligned}

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=1(12⋅19360)=72019≈37.89≈38 days.\begin{aligned}T &= \dfrac{\text{1 whole field including car}}{\text{joint rate of cow, goat and wildebeest}} \\[1em]&= \dfrac{1}{c + g + w} = \dfrac{1}{\left(\dfrac{1}{2} \cdot \dfrac{19}{360}\right)} \\[1em]&= \dfrac{720}{19} \approx 37.89 \approx 38 \text{ days}.\end{aligned}