19th Century Schoolbooks

HORNBOOKS

Really not a book at all, the hornbook usually consisted of a single sheet of paper containing the alphabet, a shortened syllabary, the invocation, and the Lord's Prayer. It was pasted to a board or stiff card and covered with a translucent layer of horn (or varnished) to protect it.

BATTLEDORES

By the beginning of the 19th century, the hornbook had evolved into a cardboard structure folded into three. It is usually referred to as a "battledore," although in fact the term is better applied to the more substantial hornbook itself which could be and was used as a bat in a primitive form of badminton.

PRIMERS

The word Primer originally meant a book of prayers for the laity and was perhaps related to the monastic service of Prime. It came to mean an introduction to reading and later an introduction to any subject.

SPELLING BOOKS

These were introduced into the colonies at the turn of the 18th century. Their objective was to teach spelling, reading, religion, and morality. Until the time of the Declaration of Independence, all spellers printed in the colonies had been reprints of imported British works. The war for American independence made British texts much less acceptable. An early speller written by an American, published on an American press, was the work of a young patriot, Noah Webster.

READERS: OLD STYLE

In these early years, it was the speller that introduced a child to reading. A schoolbook called a "reader" was, until the 1830's, a book designed for children who could already read. Readers such as those by Noah Webster or by Caleb Bingham consisted of a compilation of essays originally written for adults on a variety of subjects.

Math problems from 19th century schoolbooks:

1.   A leaves a city at 9:15 and goes to another city going 4 mph. B leaves this second city at 9:30 and goes to the first city going 3 1/2 mph. The 2 cities are 21 miles apart. At what time are A and B going to meet?
Solution.
At 9:15 A is moving 4mph and B is not moving yet. We then found that at 9:30, the two people were 20 miles apart. At 10:30 they were 12 1/2 miles apart. We added 4 and 3 1/2 and got 7 1/2.   It was 20 miles at 9:30, which is 7 1/2.  So 20 mi / 7 1/2 mph = 2 2/3 hours = 2 hr. and 40 min. after 9:30, which equals 12:10. Therefore A and B will meet at 12:10.
 
2.   There is a garden with fruit trees, 1/2 of them are apple trees, 1/4 of them are peach trees, and 1/6 of them are plum trees. The remaining trees are 200 cherry trees. How many trees are there in the garden?
Solution.
We wrote out all of the fractions 1/2, 1/4, and 1/6 and found a common denominator which was 12. We then added up all the fractions to get 11/12. We saw that if you added 1/12, you would get 1. 1/12 is the number of cherry trees which is 200. Multiply 200 by 12 which equals 2400 which is the number of trees in the garden.
 
3.   A man who sells eggs, sells half his eggs and one more to the first buyer. To the second buyer he sells half of the remaining eggs and one more. The third buyer buys half of the remaining eggs and one more, and there are no more eggs left. How many eggs were there in the beginning?
Solution.
I did this problem by working backwards.
(((((0 + 1) x 2) + 1) x 2) + 1) x 2  = 14, the number of eggs he started with.
 
4.    If 30 men need 40 days to complete a job, how many men will it take to complete a job that is 5 times bigger in one-fifth the amount of time?
Solution.
There are 30 men, 40 days, and 1 project, (30,40,1). Since the project is 5 times as big, it becomes (30,40,5). Since it must take 1/5 of 40 days or 8 days, it then becomes (30,8,5). You multiply 5 (from the 1/5) by 5 ( the size of the new project) , which equals 25. You then multiply 25 by 30 (men) which equals the answer. 750 men are needed to finish 5 times the project in 8 days.
 
5.    A man left 38,000 dollars that was to be divided  among his three sons and his three daughters. Each son is going to get 33 1/3 % more than the oldest daughter; and the younger daughters are going to get 33 1/3 % less than the oldest daughter. What is everyone's portion?
Solution.
Let S = son, D = youngest daughter, X = oldest daughter. Then S = (1/3)X + X and D = X - (1/3)X.   The equation for the entire amount is 3S + 2D + X = 38,000. Substituting, 3((1/3)X + X) +2(X-(1/3)X) +X =38,000. Simplifying we got (6 1/3)X = 38,000. X= 6000 which is what the oldest daughter gets. We substituted and got S = 8000 and D = 4000.
 

In the midst of a meadow well stored with grass,
I took just an acre to tether my ass:
How long must the cord be, that feeding all round,
He may'nt graze less or more than acre of ground.
 
The answer is: 39.25073 ft

 
One ev'ning I chanc'd with a tinker to sit,
Whose tongue ran a great deal too fast for his wit:
He talked of his art with abundance of mettle;
So I ask'd him to make me a flat-bottom'd kettle.
Let the top and the bottom diameters be,
In just such proportion as five is to three:
Twelve inches the depth I propos'd, and no more;
And to hold in ale gallons seven less than a score.
He promis'd to do it, and straight to work went;
But when had done it he found it too scant.
He alter'd it then, but too big he had made it;
For though it held right, the diameters fail'd it;
Thus making if often too big and too little,
The tinker at last had quite spoiled his kettle;
But declares he will bring his said promise to pass,
Or else that he'll spoil every ounce of his brass.
Now to keep him from ruin, I pray find him out
The diameter's length, for he'll never do't, I doubt.
 
Ans. Bottom 14.44401, top 24.4002
 

Two porters agreed to drink off a quart of strong beer between them, at two pulls, or a draught each; now the first having given it a black eye, as it is called, or drank till the surface of the liquor touched the opposite edge of the bottom, gave the remaining part of it to the other:  what was the difference of their shares, supposing the pot was the frustum of a cone, the depth being 5.7 inches, the diameter at the top 3.7 inches, and that of the bottom 4.23 inches?

Ans. 7.05 cubic inches.