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?

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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}