Greek Anthology Book XIV: 49. - Problem

Problem

Make me a crown weighing sixty minae, mixing gold and brass, and with them tin and much-wrought iron.

Let the gold and bronze together form two-thirds,

the gold and tin together three-fourths,

and the gold and iron three-fifths.

Tell me how much gold you must put in, how much brass, how much tin, and how much iron, so as to make the whole crown weigh sixty minae.

Solution

Let $g, b, t, i$ be the quantities in minae of each of gold, brass, tin and iron.

We have:

\(\ds g + b + t + i\)

\(=\)

\(\ds 60\)

\(\ds g + b\)

\(=\)

\(\ds 60 \times \dfrac 2 3\)

\(\ds \)

\(=\)

\(\ds 40\)

\(\ds g + t\)

\(=\)

\(\ds 60 \times \dfrac 3 4\)

\(\ds \)

\(=\)

\(\ds 45\)

\(\ds g + i\)

\(=\)

\(\ds 60 \times \dfrac 3 5\)

\(\ds \)

\(=\)

\(\ds 36\)

\(\ds \leadsto \ \ \)

\(\ds b\)

\(=\)

\(\ds 40 - g\)

\(\ds t\)

\(=\)

\(\ds 45 - g\)

\(\ds i\)

\(=\)

\(\ds 36 - g\)

\(\ds \leadsto \ \ \)

\(\ds g + \paren {40 - g} + \paren {45 - g} + \paren {36 - g}\)

\(=\)

\(\ds 60\)

\(\ds \leadsto \ \ \)

\(\ds 121 - 2 g\)

\(=\)

\(\ds 60\)

\(\ds \leadsto \ \ \)

\(\ds 2 g\)

\(=\)

\(\ds 61\)

\(\ds \leadsto \ \ \)

\(\ds g\)

\(=\)

\(\ds 30 \frac 1 2\)

\(\ds \leadsto \ \ \)

\(\ds b\)

\(=\)

\(\ds 40 - 30 \frac 1 2\)

\(\ds \)

\(=\)

\(\ds 9 \frac 1 2\)

\(\ds t\)

\(=\)

\(\ds 45 - 30 \frac 1 2\)

\(\ds \)

\(=\)

\(\ds 14 \frac 1 2\)

\(\ds i\)

\(=\)

\(\ds 36 - 30 \frac 1 2\)

\(\ds \)

\(=\)

\(\ds 5 \frac 1 2\)

So the required quantities are:

gold: $30 \frac 1 2$ minae

brass: $9 \frac 1 2$ minae

tin: $14 \frac 1 2$ minae

iron: $5 \frac 1 2$ minae.

$\blacksquare$

Sources

• 1918: W.R. Paton: The Greek Anthology Book XIV ... (previous) ... (next): $49$. -- Problem