Here's a holiday maths challenge from the 1790s

[mohill]

This question comes from Algebraical Problems by Miles Bland. (My edition dated in the early 1800s).
A city barge, with chairs for the company and benches for the rowers, went a summer excursion, with two bargemen on every bench. The number of gentlemen on board was equal to the square of the number of bargemen; and the number of ladies was equal to the number of gentlemen, twice the number of bargemen, and one over.
Among other provisions there were a number of (soused) turtles equal to the square root of the number of ladies; and a number of bottles of wine less than the cube of the number of turtles by 361.
The turtles in dressing consumed a great quantity of wine, and the party having stayed out until the turtles were all eaten, and the wine all gone, it was computed, that supposing them all to have consumed an equal quantity, (viz. gentlemen, ladies, bargemen and turtles,) each individual would have consumed as many bottles as there were benches in the barge.
Required the number of turtles

Cheers from the antipodes!

[crofter]

The archaic language is a little difficult to follow, but I make it 19 turtles.
If there are b bargemen, g gentlemen, l ladies and t turtles, then
g = b^2 : l = (b+1)^2 and t = (b +1)
Hence the barge carries a total of 2(b+1)^2 bodies
each drinking b/2 bottles. But the total wine drunk is given by t^3 - 361.
So we need to solve 2(b+1)^2 x b/2 = (b+1)^3 -361
Therefore b = 18, so t = 19 (I reckon)