Greek Anthology Book XIV: Metrodorus: 134

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Arithmetical Epigram of Metrodorus

O woman, how hast thou forgotten Poverty?
But she presses hard on thee, goading thee ever by force to labour.
Thou didst use to spin a mina's weight of wool in a day,
but thy eldest daughter spun a mina and one-third of thread,
while thy younger daughter contributed a half-mina's weight.
Now thou providest them all with supper, weighing out one mina only of wool.


Solution

The exercise appears to be to calculate the contribution of each of the women during the time (less than a day) they spent spinning.


Let $a$, $b$ and $c$ (respectively) be the contribution of the mother, the older daughter and the younger daughter of the $1$ mina of wool.

Let $w$ be the weight of wool in minae they would produce together if all three were to spin all day.


Then:

\(\ds w\) \(=\) \(\ds 1 + 1 \frac 1 3 + \frac 1 2\)
\(\ds \) \(=\) \(\ds 1 + \frac 4 3 + \frac 1 2\) clearing the mixed fraction
\(\ds \) \(=\) \(\ds \dfrac {6 + 8 + 3} 6\) putting over a common denominator
\(\ds \) \(=\) \(\ds \dfrac {17} 6\)


Thus if all $3$ worked all day, they would produce $\dfrac {17} 6$ minae.

It follows to produce $1$ mina, they worked for $\dfrac 6 {17}$ of a day.

Thus their individual contributions are

\(\ds a\) \(=\) \(\ds 1 \times \dfrac 6 {17}\)
\(\ds \) \(=\) \(\ds \dfrac 6 {17}\)
\(\ds b\) \(=\) \(\ds \dfrac 4 3 \times \dfrac 6 {17}\)
\(\ds \) \(=\) \(\ds \dfrac 8 {17}\)
\(\ds c\) \(=\) \(\ds \dfrac 1 2 \times \dfrac 6 {17}\)
\(\ds \) \(=\) \(\ds \dfrac 3 {17}\)


So:

the mother contributed $\dfrac 6 {17}$ of a mina of wool
the older daughter contributed $\dfrac 8 {17}$ of a mina of wool
the younger daughter contributed $\dfrac 3 {17}$ of a mina of wool.

$\blacksquare$


Source of Name

This entry was named for Metrodorus.


Sources