Yes, yes, all very interesting about the economics of early 20thC publishing, but worrabout the 'Rithmetic? You'd think that arithmetic would be the same as it ever was, since the Egyptians used a 3-4-5 triangle to to make their pyramids square but the past is a different country in ways that transcend the fact that pocket calculators only came on the scene in about 1975. Consider the gratituous difficulty of adding in units which are related by different multipliers. In sensible = decimal = "metric" addition "carries" numbers when they exceed ten. In summing weights, as in the top pictured example:
- 16 oz = 1 lb [ounces ~ 30 g; pound ~ 450 g)
- 14 lb = 1 st [stone]
- 2 st = 1 qr [quarter]
- 4 qr = 1 cwt [hundredweight which thus weighs 112 lb)
- 20 cwt = 1 [long] ton [which thus weighs 2240 lb or as close a dammit to 1 tonne = 1,000 kg]
That's not the only difference in a 100 year old arithmetic book. The interests and obsessions are quite different too. In the Problems at the end of the book, Pendlebury tries to relate the theory to practice in real life.
- 83. A besieged garrison have sufficient provisions to last them for 23 weeks at the rate of 18 oz per man per diem; but receiving a reinforcement of 40% on their original number, this allowance is reduced to 15 oz per diem. How many days will they be able to hold out?
- 99. A ship 600 miles from shore springs a leak which admits 6 tons of water in 20 minutes. 60 tons of water would suffice to sink her but pumps can throw out 70 tons in 4 hours. Find her average rate of sailing that she may reach shore just as she begins to sink
- 157. The proprietor of a boarding school having already 30 pupils, finds that an addition of 5 increases his gross yearly expenditure by £300, but diminishes the average cost per head by £1. What did his annual expense originally amount to?
- 194. A rectangular fold is to be made of hurdles 6 ft long, to contain at least 1,000 sheep allowing 8sq.ft per sheep. Find the number of hurdles needed for the cases where one side of the fold consists of 10, 11, . . . 20 hurdles. Draw a graph shewing the relation between the total number of hurdles and the number on one side. Hence find the smallest number that will suffice.