Greek Anthology Book XIV: Metrodorus: 130

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Arithmetical Epigram of Metrodorus

Of the four spouts one filled the whole tank in a day,
the second in two days,
the third in three days,
and the fourth in four days.
What time will all four take to fill it?


Solution

Let $t$ be the number of days it takes to fill the tank.

Let $a, b, c, d$ be the flow rate in numbers of tanks per day of (respectively) the first, second, third and fourth spouts.


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

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


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


We have:

\(\ds a\) \(=\) \(\ds 1\) that is, $1$ tank in $1$ day
\(\ds b\) \(=\) \(\ds \frac 1 2\) that is, $1$ tank in $2$ days
\(\ds c\) \(=\) \(\ds \frac 1 3\) that is, $1$ tank in $3$ days
\(\ds d\) \(=\) \(\ds \frac 1 4\) that is, $1$ tank in $4$ days


and so:

\(\ds t\) \(=\) \(\ds \dfrac 1 {a + b + c + d}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {1 + \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 4}\)
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds \dfrac {12} {12 + 6 + 4 + 3}\) multiplying top and bottom by $12$
\(\ds \) \(=\) \(\ds \dfrac {12} {25}\)

So the tank will be filled in $\dfrac {12} {25}$ of a day.

$\blacksquare$


Source of Name

This entry was named for Metrodorus.


Sources