Arithmetical Problems

Since I write for both adults and children, I often visit classrooms. And when I do, I usually bring  early 19th century school books with me, to show children what text books were like two hundred years ago, when students had to supply their own books, and often books were shared. (I have one Webster’s speller that was once re-covered in wood – which is now cracked and split – to protect it.)

Most of the books I have are concerned with reading, writing, and speaking. So a couple of weeks ago when I was at an antique show I was happy to pick up a copy of an arithmetic book. It was a little later than my early books — it was published in 1863 — but I love it. The title, Arithmetical Problems or Questions in Arithmetic, for the Use of Advanced Classes in Schools, says it all. It was written by W.H. Farrar, the principal of the Woonsocket (Rhode Island) High School. And it’s a collection of one thousand – yes — one thousand – “word problems. — all of which are actually practical applications of arithmetic.

At some time a student — perhaps the Charles T. Haynes who signed it in 1871 — started the book, and penciled in the answers to the problems. But he gave up at #99.

Since there’s a lot of talk today about our schools, and how our students are falling behind those in other countries, I thought today I’d share (without solutions) ten of the “arithmetical problems” in this little book. You might even want to share them with a young friend or relative.

Here then, for your edification and amusement ….

1.  What must I pay per hogshead for molasses, that I may keep it 8 months, when money is worth 6 percent, and then sell it at $42 per hogshead, and gain 12 per cent?

2.  The foot of a ladder, 60 feet long, remaining in the same place, the top will just reach a window 40 feet high on one side of the street and another 30 feet high on the other side. How wide is the street?

3. If a certain number be diminished by 7 and the remainder be divided by 8, and the quotient be multiplied by 5, that product increased by 4, the square root of the sum extracted, and 3/4 of that root be cubed, and that cube be divided by 9, the last quotient will be 24. What is that number?

4. Bought 200 yards of cambric for 90 pounds, but, it being damaged, I am willing to lose 7 pounds 10 shillings on it. What must I demand per ell English?

5. How many square yards of paper in a book of 672 pages, the leaves being 6 1/2 by 10 1/4 inches?

6. A house is 36 feet long and 28 feet wide, and the ridge is 12 feet above the beam. The roof projects one foot over the ends and eaves in all directions. How many shingles will be required to cover the roof, 6 shingles being allowed to the foot, and 10 per cent for waste?

7. A field is 72 chains long and 131 rods 10 feet wide. How many acres does it contain?

8.A stone weighs 120 pounds in the air and 100 pounds in water. What is the specific gravity of the stone, water being 1000?

9. If 63 men can build a wall, 45 1/3 feet long, 6 7/12 feet high and 3 1/8 feet thick, in 34 days of 11 1/3 hours each, in how many days of 8 1/2 hours each will 21 men build a wall 98 3/4 yards long, 2 1/2 yards high, and 1 1/4 yards thick?

10.  A man lost in a speculation 1/4 of his money. He then gained a sum equal to 1/3 of what he then had. Afterwards he lost 1/5 of what he then had, and then gained a sum equal to 1/4 on what he had left, when he found he had $1200. How much had he at first?

And they didn’t even have calculators ……

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5 Responses to Arithmetical Problems

  1. Okay, we’re talking cruel and unusual punishment here! I hated word problems in math. One of the hardest things for me to do as a principal was tutor some 4th graders in math problems. Oh, my goodness. I confess to not even reading all of these. They mess with my brain so bad. Of course some of that is the vocabulary, but even if I knew what some of those terms were, I’d be out of luck. LOL

  2. As someone who taught math for thirty years, I have to say that although these problems could not be solved by the average college student today, some of them are pedagogically unsound. Problem 2, for example, is an ingenious use of the Pythagorean theorem. The ladder forms two right triangles when it is leaned against the windows on each side of the street with the length of the ladder being the hypotenuse and the height of the windows being one other leg. Solve for the missing led in each triangle, add them together, and you get the width of the street. If a student can see that is the solution, she has mastered mathematical reasoning. But she may not get the answer because the actual calculation requires finding square roots, both of which are irrational numbers. The problem would be be so elegant if different lengths had been chosen. After all, the width of any actual street is always a rational number. One wonders whether the intent was to instruct or torment! Thanks for this post. It was fun.

  3. Judy Dee says:

    Do we get the answers too?

  4. These are problems I would have to pass on to my math-whiz children. What a fun post, Lea.

  5. Mike Ereksonjm says:

    So, anyone know what is meant by “when money is worth 6 percent”? A hogshead is a barrel or cask of liquid (about 63 gallons or 300L). An ell is 45 inches. A schilling is 0.05 pounds and there are 12 pence per schilling. A chain is 66 feet and there are 4 rods in a chain. A mile is 80 chains. A furlong is 10 chains. An acre is 1 chain by 10 chains or 4840 square yards or 43560 square feet. With that information these are all easily solvable without the use of a calculator. The square roots needed to solve problem two can be estimated knowing that 40 squared is 1600, 50 squared is 2500, and 60 squared is 3600. Hope that helps…I have answers to the first 5 and only used a calculator to check my work at the end. As an interesting side note, calculus had already been around for over 100 years at the date of publication of this book…so there was much more advanced math being done without the aid of calculators or computers at that time. I will work answers to the others when I have time in the next day or two.

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