Henry Ernest Dudeney/Puzzles and Curious Problems/164 - Cow, Goat and Goose/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $164$

Cow, Goat and Goose
A farmer found
that his cow and goat would eat all the grass in a certain field in $45$ days,
that the cow and the goose would eat it in $60$ days,
but that it would take the goat and the goose $90$ days to eat it down.
Now, if he had turned cow, goat and goose into the field together, how long would it have taken them to eat all the grass?


Solution

$40$ days.


Proof

Let $t$ be the number of days it takes for them all to eat all the grass.

Let $a, b, c$ be the eating rate in numbers of fields per day of (respectively) cow, goat and goose.


In $t$ days, the various contributions of each of the animals to the eating process is $a t$, $b t$ and $c t$ respectively.

So for the total contribution to be $1$ field, we have:


\(\ds \paren {a + b + c} t\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \dfrac 1 {a + b + c}\)


We have:

\(\ds a + b\) \(=\) \(\ds \dfrac 1 {45}\) his cow and goat would eat all the grass in a certain field in $45$ days,
\(\ds a + c\) \(=\) \(\ds \dfrac 1 {60}\) that the cow and the goose would eat it in $60$ days,
\(\ds b + c\) \(=\) \(\ds \dfrac 1 {90}\) but that it would take the goat and the goose $90$ days to eat it down.


and so:

\(\ds 2 a + 2 b + 2 c\) \(=\) \(\ds \dfrac 1 {45} + \dfrac 1 {60} + \dfrac 1 {90}\)
\(\ds \) \(=\) \(\ds \dfrac {4 + 3 + 2} {180}\)
\(\ds \) \(=\) \(\ds \dfrac 9 {180}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {20}\)
\(\ds \leadsto \ \ \) \(\ds a + b + c\) \(=\) \(\ds \dfrac 1 {40}\)
\(\ds t\) \(=\) \(\ds \dfrac 1 {a + b + c}\)
\(\ds \) \(=\) \(\ds 40\)

So the field will be eaten bare in $40$ days.

$\blacksquare$


Sources

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