This multi-TNDOT is a little different, not as high a level or as old school as the previous ones, but the questions still seem pretty wild. We haven’t thought about the questions at any length, but conceivably they could also be used to challenge/tease younger students.
The TNDOT consists of two questions, each with two parts, from the 1911 Victorian Pass Algebra Matric exam. The exam paper instructed students to answer only one part from each question but we run a tough blog here and all readers are expected to answer all questions. As usual, commenters might refrain from posting answers too quickly, so that other readers have a chance to have a go. Have fun.
3(α) A certain quantity being divided into n equal parts, and also into n + 1 equal parts, the product of the n parts is n times that of the n + 1 parts. Find it.
(β) A farmer buys m sheep for £p and sells n of them at a profit of q per cent. At what price per head must he sell the rest to clear r per cent on his whole bargain?
6(α) if a men or b boys can dig m acres in n days, find the number of boys whose assistance will be required to enable a – p men to dig m + p acres in n – p days.
(β) To complete a certain work A requires a times as long as B and C would take if they worked conjointly ; B requires b times as long as C and A and would take if they worked conjointly ; and C requires c times as long as A and B would take if they worked conjointly. Prove that
and find the time which each would take separately to do the work, assuming that if they all worked together they would require h hours.
UPDATE (11/08/24)
OK, sorry, this one kind of fell off the truck. But here, finally, is an update with solutions. (We’ve tried to be careful, but we haven’t checked it all the way we would an ACARA-bashing post, so be aware there may be, and feel free to point out, errors and typos.)
For 3(α), the solution was provided by Red Five (noting the correction by Banacek Spaces). If we let the quantity in question be Q, then we are told,
Dividing both sides by n, and then moving the Q-stuff to the right and the n-stuff to the left, we find,
Or, as Banacek Spaces noted, we can write, 
, with obvious import.
For 3(β), Red Five started a solution, although I’m not convinced his algebra is correct. The main thing is to simply be methodical. To begin, the farmer sold m sheep at a profit of q%. Since he bought these sheep for £p each, he sold them for a total cost of
Similarly, suppose he sold the remaining m – n sheep at a profit of s%. Then these would be sold for a total cost of
Altogether, this is supposed to yield a profit of r% on the sale of the m sheep. Which leads to the equation,
Multiplying though by 100 and dividing by p,
Rearranging, this gives,
Then, finally, the price per head of the second batch of sheep is,
![\color{Plum}\boldsymbol{ p(1+\frac{s}{100}) = p\frac{100(m-n) + mr - nq}{100(m-n)}= p\left[1 + \frac{mr-nq}{100(m-n)}\right]}\,.](https://mathematicalcrap.com/wp-content/ql-cache/quicklatex.com-a6b4dc51d4884a3270639b7da8c778fa_l3.svg "Rendered by QuickLaTeX.com")
6(α) similarly seems to require care and perseverance but no real cleverness. Each man can dig m/na acres per day, and each boy can dig m/nb acres per day. So, if we have a – p men and, say, B boys, then in n – p days they will dig together
We require m + p acres to be dug. Equating with the above, and doing some obvious simplification, this gives,
We can then solve for B, although we can’t see any particular way to make the answer pretty:
Finally, 6(β) seems trickier, although not bad when you get the right angle on it. Let’s have A also stand for the rate (work-units/hour) at which A can work, and similarly for B and C. Then B and C working together work at the rate B + C, and so on. With this terminology, we are then told
Solving for a, b and c respectively in these equations, and then adding 1, we have,
Taking the reciprocals and summing gives the required equation.
For the second part, the rate at which A and B and C work together is A + B + C. From the above equations (or thinking about the initial information), this is 1 + a times the rate at which A works alone. So, if all of them working together takes h hours, then A alone will take (1 + a) times as long, and thus A will take (1 + a)h hours. Similarly, B and C will take (1 + b)h and (1 + c)h hours, respectively.
That Q6
is beautiful. The art of writing such questions seems long gone. Alas!
The art of writing a coherent sentence is long gone.
Long lost gems…
Where could we access similar questions for 7-10 kids nowadays?
It does require certain sophisticated level of maths and literacy skills to articulate such beautiful questions…sadly today we lack such training for teachers to craft similar questions!
You can order scans of old exam papers through SLV but it costs a bit and takes up to a month. In the 1900 to 1920ish era there were public Junior, Leaving and Senior examinations which roughly means Year 10, Year 11, Year 12 in today’s terms.
There were multiple Mathematics exams: Algebra, Arithmetic, Trigonometry (and much later Calculus and Applied Mathematics). Some were available at Pass as well as Honours level.
Also, not all papers are available and the SLV website is not the easiest to navigate.
Thanks RF.
I did try my luck with SLV old archives/collections.
Only managed to get a few earlier HSC/VCE math papers.
Different eras, different writing styles.
With more “Tasmania Jones” leaving us, not all of the good old treasures were successfully inherited…
I went to a bookstore at Kew a few years ago and purchased an Algebra textbook from 1960s. That’s indeed a rare gem from the dark, dusty shelfs … Only 10 bucks, can you believe it?
Lancelot, there’s a book, hopefully coming out by the end of the year, which will will be an excellent source of such questions. It’s not my book, but I’ll definitely be announcing and promoting it when it appears.
Great. Thanks Marty.
Will be happy to see this upcoming book.
The reading skills of students have also changed.
For better or worse? Questions are more wordy in 2024 than they were in 1924, so do we conclude this is because students are better readers than they were in 1924 or worse because they need more explanation of what the question is asking?
I don’t think either is a safe conclusion to be honest.
The reading skills of students are much worse, and the writing skills of examiners are JFC worse.
I am a little bit sad to admit it took me 3 seconds to decode JFC.
But on the second half of your point, I agree.
The first half… I don’t disagree, but I’m not sure I entirely agree either. Sample sizes are very different etc…
The evidence is overwhelmingly clear: The reading skills of students is much worse.
https://www.edweek.org/teaching-learning/reading-on-screens-worsens-comprehension-for-younger-students-what-can-teachers-do/2024/01
OK. Accepted. Now, back to the original questions.
I have found solutions for
and 
but in the case of 
I’m not totally satisfied with my answer (it feels like it should be able to be simplified).
Under the standard TNDOT rules I will refrain from posting for now, except to say no special tricks were required.
Thanks, RF. I deleted your duplicate comment.
I wouldn’t expect any particular tricks are required for these, just good care.
Indeed. Something else that has become short in supply.
Agree with all comments above, about abilities of students and many educators (or, I should say, exam-setters). It seems to me that we’ve lost a lot – not only in reading – and gained very little in return. Who nowadays can mount a reasonable argument, or tell a good argument from a specious one? Anyway, I had a brief look at the questions, and 6(beta) reminded me of this fine essay by the great Canadian humorist, Stephen Leacock: “The human element in mathematics”, at: https://en.wikisource.org/wiki/Literary_Lapses/A,_B,_and_C
Thanks, Alasdair. Ed Barbeau also pointed out that Leacock piece to me. It’s hilarious.
I think people can feel happy to start posting answers and solutions.
Is it part of the quantity divided into n equal parts and the rest into n+1 equal parts?
Is it clear from wording that there are two different ways to divide all ?
Can blame my English…
If i phrased it:
“When q is divided into n equal parts then the product of the parts is n times the product of n+1 parts when q is divided into n+1 equal parts.
Find q.” (These days: “in terms of n”)
Or am i in the woods…
Hi Banacek. I think your latter phrasing is intended: consider dividing q both into n parts and then into n+1 parts.
Thank you.
May be it “needs” “modernizing”, you know, taking every kid by the hand and walking through the stages…
Quantity q is split into n equal parts.
Find the algebraic expression for the product of all these parts.
Quantity q may be split into n+1 equal parts.
Find alg expr for the product…
Set up an equation, which translates the fact that the first product is n times the second product.
Find q ( in terms of n) by solving the above equation.
Because you know, kids cant do it without scaffldng
I hope i ruined this problem enough
I think there is an argument for (some) modernising. I wouldn’t write such questions in such a manner. But I admire the spareness and efficiency with which the above questions are written.
That is exactly what I assumed the question to be saying.
What I found interesting is that the answers were not “pretty” and required quite a bit of algebra.
That is assuming my answers are correct of course. I’ll start posting this week (assuming my other duties allow time for such distractions!)
Question
(my solution that has not been vetted, so let the criticism begin…)
Let the quantity be
.
The product of the
parts is then 
and the product of the 
parts is 
Since the second product is
times the first we have 
After some algebra, I get the result
…over n^(n+1) ?
Or alt. (1+1/n)^(n+1) which looks like an interesting one in the limit…
Hmmm… yes… I see… algebraic error perhaps, or LaTeX error when copying from my notes to here. Let me check my handwriting and update if needed.
RF, if you want me to correct the TeX (or maths) in your comment, just say.
Thanks. In my notebook, the final answer I have reads
.
So no changes needed, but I am willing to be convinced I am wrong/have made an error somewhere.
Thanks, RF. I think Banacek is correct. You gather the Q-bits on the RHS, and the n-bits end up in the denominator on the LHS.
Yes. Correction accepted and noted with thanks.
There being no objection to my answer to
, my answer to 
is
Working available on request.
Hi, RF. I don’t understand your answer, since the question asked for a price.
Feedback received with thanks.
I will attend to this shortly.
OK, I finally got around to updating this TNDOT. Solutions are above, although they’re not so thoroughly checked. So, please point out any errors and/or odditites.