Could You Have Been a Teacher in the 19th Century?

During my research into one of my relatives who was a teacher in the 19th century, I came across some math problems published in a newsletter for teachers. These questions were used during examinations of assistant teachers at elementary schools in 1878. People taking the assistant teachers’ exam were usually around 18 years old, had finished their own elementary school with good results and had worked for several years as a trainee, assisting already qualified teachers.

Children in class

Children using an abacus in school. Image credits: Spaarnestad Photo

Test questions

  • Barge versus man
    A walker, who walks 5 kilometers per hour, leaves half an hour before the departure of the draw barge from A to B. The barge progresses at 9 kilometers per hour and stops twice for five minutes, while the walker rests for fifteen minutes. If the barge arrives in B 15 minutes before the walker, how long is the road?
  • Filling a tub
    A tub is filled by three taps. If they are all opened together for half an hour, the strongest tap can fill the tub up in 3.75 hours. If they would be open together for a full hour, the same tap could fill the remainder in 2.5 hours. If the ratio of the strength of the second and third tap is 4:5, in how many hours can each tap fill the tub?
  • Selling potatoes
    Of a batch of potatoes, 10 percent is rotten and the rest lost 5% to dryness. They are sold for 5.50 guilders and the final profits is 6 7/8 percent. At what price were they purchased?
  • Playing with marbles
    Herman and Karel have some marbles. Herman has 16 less than Karel. If he gives 24 to Karel, Karel will have five times more than Herman. How many marbles does each boy have?

Can you solve these?

Personally, I think this is pretty advanced math for an elementary school teacher. I like how applied the problems are, so that you first have to translate them to abstract algebra before you can solve them (or maybe that’s just the result of my 20th century education). I’m sure they would have used similar types of problems in class, though probably less complicated. I like how these problems give us a nice glimpse into the level of math expected of these teachers and give us a taste of life in the 19th century to boot.

What do you think? Can you solve these? Kudos to anyone who posts the answers (including proof arguments, of course) in the comments 🙂


Source

Akte-examens,” De Wekker, 12 June 1878, p. 2, col. 4 – p. 3, col. 1; consulted as digital images, Koninklijke Bibliotheek, Delpher (http://www.delpher.nl : accessed 28 January 2016).

About Yvette Hoitink

Yvette Hoitink, CG®, QG™ is a professional genealogist in the Netherlands. She holds the Certified Genealogist credential from the Board for Certification of Genealogists and has a post-graduate diploma in Family and Local History from the University of Dundee. She has been doing genealogy for over 30 years and helps people from across the world find their ancestors in the Netherlands. Read about Yvette's professional genealogy services.

Comments

  1. Michaël Boers says

    Walker needs 30 + 10 + walking time +15 minutes.
    Boat needs 15 + sailing time minutes. Therefore walking time = sailing time + 40 minutes.
    Distance AB = walking time * 9/60 = sailing time * 5/60. Or in other words: walking time * 9 = sailing time * 5.
    Combine those two formulas and you get: walking time = 90. Therefore sailing time = 50, the distance AB is 7.5 km and the total travel time for the walker is 105 minutes.

  2. Michaël Boers says

    Karel – 16 marbles = Herman (1). 5 * (Herman – 24) = Karel + 24 (2).
    Substitute 1 in 2 gives: 5 * (Karel -16 -24) = Karel +24, or 5 * Karel – 200 = Karel + 24, or Karel = 56.
    And then Herman must be 40.

  3. michael boers says

    (Tap1 + Tap2 + Tap3) * 1/2 hour + Tap1 * 3.75 hour = Volume of tub (1).
    (Tap 1 + Tap2 + Tap3) * 1 hour + Tap1 * 2.5 hour = Volume of tub (2).
    Tap2 / Tap3 = 4 / 5 or Tap2 = 4 * Tap3 / 5 (3).
    Rewrite (1) gives: Tap1 + Tap2 + Tap3 + 7.5 * Tap1 = 2 * Volume tub or 8.5 * Tap1 + Tap2 + Tap3 = 2 * Volume tub (4).
    Rewrite (2) gives: 3.5 * Tap1 + Tap2 + Tap3 = Volume tub (5).
    Subtract formula (5) from (4) gives: 5 * Tap1 = Volume tub, or Tap1 fills the tub in 5 hours.
    This results in (5) being: 3.5 * Volume tub / 5 + Tap2 + Tap3 = Volume tub (6)
    Combine formulas (6) and (3): 3.5 * Volume tub / 5 + 4 * Tap3 / 5 + Tap3 = Volume tub. This leads to
    Tap3 * 9 / 5 = Volume tub * 3 / 10. Therefore Tap3 = 15 / 90 * Volume tub, or Tap3 fills the tub in 6 hours.
    And that leads to Tap2 filling the tub in 7.5 hours.

  4. Michaël Boers says

    I suspect there is an error in the text about the potatoes. If you know the total selling price and the profit then the buying price is a piece of cake. And you can ignore the losses you have. The calculating becomes much harder if the selling price of 5.5 guilders is not the total selling price, but the price per weight or per volume. In that case the losses to rot and dryness or significant.
    You were only able to sell 85.5% of the original amount or weight of potatoes. So the 6.875% profit would lead to a buying price per weight or amount of 5.5 * 0.855 / 1.06875 = 4.4 guilders.

    I am sure I used way more time to calculate all this than the original teachers had to do. So, hats off for those teachers who passed the test.

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