Oh for the mathematics of yesteryear
Problem
If the transport of $60$ cwt. (hundredweights) of bread for $20$ miles costs the garrison $ £14 \; 10\text{s}$. (i.e. Fourteen pounds ten shillings), what weight can be carried $30$ miles for $ £5 \; 8\text{s } 9 \text{d}$?
n.b. in those days:
$12$d(pennies) = $1$s (shilling)
$20$s = $ £1$ (pound)
$16$ ounces = $1$ lb (pound - "weight")
$112$ lb= $1$ cwt
Getting Started
For part one:
How much bread is there altogether?
For part two:
I worked in pennies - how many pennies (d) in the pound (Â £)
Student Solutions
Correct solutions were received from Mary (Birchwood Community High School) and Chor Kiang. Well done.
Part 1
First, find the total ounces of bread.
$600 \times35 \times24$ ounces $= 504000$ ounces
Now, find the new number of men-shares needed.
$4800 \times45 = 216000$
Divide $504000$ ounces by $216000$ to get $2\frac{1}{3}$ ounces a day.
Hence, each man must only eat $2\frac{1}{3}$ ounces of bread a day.
Part 2
$ £14 10\text{s} = 290\text{s}$
$ £5 \; 8\text{s } 9 \text{d} = 108.75$ shillings
| Weight (cwt) | Distance (miles) | Amount (shillings) | |
|---|---|---|---|
| $60$ | can be transported for | $20$ | for $290$ |
| $1$ | can be transported for | $20$ | for $\frac{290}{60} = \frac{29}{6}$ |
| $1$ | can be transported for | $30$ | $\frac{29}{6}\times\frac{30}{20}= 7.25$ |
| $\frac{1}{7.25}$ | can be transported for | $30$ | $1$ |
| $\frac{1}{7.25}\times 108.75$ | can be transported for | $30$ | $108.75$ |
Since $\frac{108.75}{7.25} = 15$
$15$ cwt can be carried for $30$ miles at a cost of $ £5 \; 8\text{s } 9 \text{d}$