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
- 1918: W.R. Paton: The Greek Anthology Book XIV ... (previous) ... (next): Metrodorus' Arithmetical Epigrams: $130$