Liber Abaci/Problems/Lion, Leopard and Bear

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Classic Problem

A lion would take $4$ hours to eat $1$ sheep.
A leopard would take $5$ hours.
A bear would take $6$.
If a single sheep were to be thrown to them, how many hours would it take to devour it?


The three together would consume the sheep in $1 \frac {23} {37}$ hours.


Using the Method of False Position:

For $4$ hours, in which the lion eats a sheep, put $\dfrac 1 4$.

For the $5$ hours the leopard takes, put $\dfrac 1 5$.

For the $6$ hours the bear takes, put $\dfrac 1 6$.

Because $\dfrac 1 6$, $\dfrac 1 5$ and $\dfrac 1 4$ are found exactly in $60$, suppose that in $60$ hours they devour the sheep.

Then consider how many sheep a lion can eat in $60$ hours;

since in four hours it can devour one sheep, it can consume $15$ sheep in $60$ hours

and the leopard would eat $12$ as a fifth of $60$ is $12$.

Similarly the bear would eat $10$, as $\dfrac 1 6$ of $60$ is $10$.

Therefore in $60$ hours, all $3$ together would eat $15 + 12 + 10 = 37$ sheep.

So if takes them $60$ hours to eat $37$ sheep, it takes then $\dfrac {37} {60}$ hours to consume $1$ sheep.

Hence the result.


Historical Note

As pointed out by John Fauvel and Jeremy Gray in their The History of Mathematics: A Reader of $1987$, this is a "more gory version" of the cistern problem, as seen for example in the Greek Anthology Book $\text {XIV}$, no. $132$.

Instead of $3$ pipes pouring water into a pool at given rates, $3$ animals remove flesh from a sheep at different rates.