Greek Anthology Book XIV: Metrodorus: 131

Arithmetical Epigram of Metrodorus
Open me and I, a spout with abundant flow, will fill the present cistern in four hours; the one on my right requires four more hours to fill it, and the third twice as much. But if you bid them both join me in pouring forth a stream of water, we will fill it in a small part of the day.

Solution
Let $t$ be the number of hours it takes to fill the cistern.

Let $a, b, c$ be the flow rate in numbers of cisterns per hour of (respectively) the first, second and third spouts.

In $t$ hours, the various contributions of each of the spouts is $a t$, $b t$ and $c t$ respectively.

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

We have:

and so:

So the cistern will be filled in $2 \frac 2 7$ hours.