Propositiones ad Acuendos Juvenes/Problems/7 - De Disco Pesante Libras XXX

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Propositiones ad Acuendos Juvenes by Alcuin of York: Problem $7$

De Disco Pesante Libras $\text {XXX}$
A Dish weighing $30$ Pounds
A dish weighing $30$ pounds is made of gold, silver, brass, and lead.
It contains $3$ times as much silver as gold;
$3$ times as much brass as silver;
and $3$ times as much lead as brass.
How much is there of each?


Solution

Gold: $9$ ounces
Silver: $2$ (troy) pounds and $3$ ounces
Brass: $6$ (troy) pounds and $9$ ounces
Lead: $20$ (troy) pounds and $3$ ounces.


Proof

Let $x$ be the amount in (troy) pounds of gold in the dish.

Then:

the amount of silver in the dish is $3 x$
the amount of brass in the dish is $3 \times 3 x = 9 x$
the amount of lead in the dish is $3 \times 9 x = 27 x$


Thus, knowing the total quantity of metal in the dish:

\(\ds \paren {1 + 3 + 9 + 27} x\) \(=\) \(\ds 30\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \tfrac 3 4\) after algebra
\(\ds \leadsto \ \ \) \(\ds 3 x\) \(=\) \(\ds \tfrac 9 4\) \(\ds = 2 \tfrac 1 4\)
\(\ds \leadsto \ \ \) \(\ds 9 x\) \(=\) \(\ds \tfrac {27} 4\) \(\ds = 6 \tfrac 3 4\)
\(\ds \leadsto \ \ \) \(\ds 27 x\) \(=\) \(\ds \tfrac {81} 4\) \(\ds = 20 \tfrac 1 4\)


Given that there are $12$ (troy) ounces in one (troy) pound, we can express this as:

\(\ds x\) \(=\) \(\ds \tfrac 9 {12}\)
\(\ds 3 x\) \(=\) \(\ds 2 \tfrac 3 {12}\)
\(\ds 9 x\) \(=\) \(\ds 6 \tfrac 9 {12}\)
\(\ds 27 x\) \(=\) \(\ds 20 \tfrac 3 {12}\)

$\blacksquare$


Sources

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