[Pri1_20161021DIV] Labelling as a strategy for Division

Introduction

          The Singapore mathematics syllabuses are very well designed, especially the primary school syllabus.  Fundamental concepts and skills are introduced before going on to complex calculations and problem solving.  At primary 1, pupils learn the idea of multiplication and division of small numbers by grouping (or partitioning).  They are not made to recite the times tables meaninglessly.

          Division is easy if the number of things in each group is known.  You just keep on circling the known number of objects until everything is circled.  However, if the number of groups is required but the number of things in each group is not given, and if the objects are not arranged in a convenient way, the task can be a bit more challenging.  Remember: they have not memorised the multiplication tables yet.

Problem / Question

Solution (Suggested)

          One way to solve this problem is to label the fish 1, 2, 3, 1, 2, 3, ... in a cyclic fashion, assigning fish to each of the three friends one at a time, thereby ensuring that each person gets the same number of fish.  Start with “1” somewhere on the left, “3” on the right and “2” somewhere in the middle.  Assign the next “1” close to the previous “1”, the next “2” close to the previous “2” and the next “3” close to the previous “3”.  So all the 1s are close together, the 2s are close together and the 3s are close together.  After all the fish have been labelled, the partitioning (or grouping) becomes obvious.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H09. Restate the problem in another way

Suitable Levels

* Primary School / Elementary School Mathematics

* any precocious or independent learner who is interested

[Maths Education] Mathematical Journalling

     Nowadays, I see some schools / textbooks asking students to search the internet and to write on a certain problem on their “mathematics journal”.  It's so very guided.  It's so artificial.  The questions should come from the learners themselves, out of their own curiosity.  The learn then seeks to answer their own questions.  The journal can serve as to document and summarise their process of learning.

     The best maths journals are self-initiated.  Great mathematician Karl Friedrich Gauss and renowned scientist Richard Feynman kept math journals on their own accord, not because some teacher told them to do it. 

     When I was a student, I borrowed books from the National Library on things out of the normal curriculum.  I kept notes of things I learned.  I also did my own investigations.  I accidently discovered quadratic equations when I was in Primary 4.  I read guidebooks, asked my friend's elder brothers and sisters, my Chinese teacher (!) and other people to find out more.  I did not like factorisation by trial-and-error.  Neither did I like completing the square nor using the quadratic formula.  So I did my own research to find a sure-fire way to factorise without trial-and-error.  I finally managed to find a way, but my method had an uncanny similarity to the quadratic formula.  It was a Pyrrhic victory, but it was fun!  I thoroughly enjoyed it.

     If students need to be told or goaded to write mathematics journals, then we as educators need to ask ourselves:  Why?  What is their conception of mathematics and education?  What experiences have they gone through that lead them to these beliefs?

     Some food for thought, eh?

[P6_20160217RTTU] Books on Bookshelves

Problem

Introduction

     Here we have a numerically challenging problem that involves ratios, and it ultimately reduces to an algebraic problem with two unknowns.  Nevertheless, we are spoilt for choice as regards to methods of solution:-

     (1)   Bar Diagram Modelling

     (2)   explicit letter-symbolic Algebra

     (3)   “p” and “u”  (parts and units)

     (4)   Distinguished Ratio Units

     Despite the fact that Bar Diagram Modelling made “Singapore mathematics” famous, let us remember that it is only one of the ways of solving problem by diagramming, which is just one of the eleven Primary School heuristics recommended by the Singapore Ministry of Education.
     The methods have a lot in common, and they differ mainly in the form of presentation.  However, standard Bar modelling is impractical under high-stakes high-stress examination conditions for this problem, not least because one would have to cut the bars into many pieces.  One should not cut off one’s feet just so as to fit the shoes (削足适履), as one Chinese saying goes.  We need to be flexible and open-minded.  I present a solution using my own Distinguished Ratio Units.

Solution

Ans:  735 books

Commentary

     First off, we need to equalise the numerators of  ²/₅  and  1¹/₄ = ⁵/₄  and put them ratio form.   This is because the  “2”  in the  ²/₅  represents the same quantity as the  “5”  in  ⁵/₄.

We do this adjustment by multiplying the former through by  5  and the latter through by  2.  Thus we deduce that the original number of books in A and in B are  25  and  8  “heart” units respectively. 
     Next, we add on the  2  and  3  “triangle” units.  By doing a comparison, we can figure out that  1  “triangle” unit must be  45  more than  17  “heart” units.  So  2  “triangle” units must be equal to  34  “heart” units plus  90.  Replacing the  2  “triangle” units (shown in yellow) with their equivalent, we now know that  59  “heart” units plus 90 gives  444.  This allows us to figure out that  1  “heart” is actually  6.  Thus, we can work out what  1  “triangle” unit, and then what  5 “triangle” units are worth.

Final Remarks

     Due to the difficulty of the numbers, the solution presented above is about as streamlined as I can make it to be.  

     There is another variation that can be used – equalising the “triangle” units (akin to the technique of elimination in standard algebra).  What we do is we multiply the group with total  444  by  3  and to multiply the group with total  489  by  2.  This would give  6  triangle units on each side.  Then we can compare the “heart” units and continue from there.  This way of proceeding is not for those who fear 4-digit numbers.

     If there are nicer or more elegant ways to tackle this question, I would definitely love to hear from you.

Heuristics Used

H01. Act it out

H04. Look for pattern(s)

H05. Work backwards

H06. Use before-after concept

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve whole numbers and ratios

* any problem solver who loves a challenge

Happy Chinese New Year 2016 (#LunarNewYear)

2016 = 12 × 168

     Wishing all my readers a Happy Chinese New Year ... or more accurately a Lunar New Year – many other Asians (e.g. the Japanese and the Koreans) also celebrate this festival.  This year  2016  is mathematically special, because it is the product of  12  and  168.  12  is a lucky number for Western people (e.g. 12 signs of the zodiac and 12 days of Christmas)  and “8” is especially auspicious in Chinese because it sounds like发 / 發  [fa in Mandarin,] which means to prosper or to grow.  168 sounds like 一路发 [yaad lou faat in Cantonese] which is to prosper all the way through life.

     I think this year will be challenging, because the Fire Monkey could be monkeying around even more with the economy and world peace.  Nevertheless if all human beings can unite together and learn to think critically, creatively and logically (and mathematics is about all these), the planet Earth can be a better place.  So best wishes to one and all!

[Maths_Education] Is Learning Mathematics a "Performance"?

Article
The Math-Class Paradox


My Comments
     This is an excellent article addressing the issue of mathematics education in schools today.  School "mathematics" barely touches even the tip of my iceberg.  Is the learning of mathematics merely a matter of performing under time pressure on tests and exams (and that puts many students off)?  What about exploration, creativity, asking questions and discovering connections ... etc?
     The thought provoking article is written by Jo Boaler, one of my favorite professors of mathematics education.  It is definitely worth your while reading it!

[S2_20151226EFQF] Factorisation without Trial and Error?

Problem

 

Introduction

     This problem was posed by a student going on to Secondary 1 (~ grade 7) next year.  This sort of problem is usually done at Secondary 2 or 3 (about grade 8 or 9).  This reminds me of my personal story.

     I accidentally discovered quadratic equations when I was in Primary 4.  I imagined a rectangle whose length is  2 cm  longer than the breadth.  If the breadth is  4 cm, the length is  6 cm and the area is obviously  24 cm².  But if I pretended that I knew the area but did not know the dimensions, I did not know how to solve it with the knowledge that I had at that time.  This started me on a quest to find out the answer.  I read secondary school guidebooks, asked my friend’s brothers and sisters, and even asked my Chinese teacher (who, after exams, offered to answer any question we had)!  Basically, I was offered two choices: (1) trial and error factorisation  and  (2) the quadratic formula.  I did not like guess and check (or hit and run?), and the quadratic formula looked formidable to me.

     So I started a quest to find a method of factorisation that did not require trial-and-error.  By secondary 1, after fiddling around with algebra, I managed to do it.  I reconstruct my derivation below.  And then I use my method to solve the above factorisation problem.

Derivation

Solution

Remark

     This looks like a Pyrrhic victory.  But like they say, it’s the journey and not the destination that matters.  Doing my own explorations prepared me for future learning and made me understand better.

     Nowadays, the new models of calculators give solutions to the associated equations and you can work backwards to get the factorisation.  Unfortunately, many students just blindly use this and forget to work backwards, giving the wrong factorisation.  If calculator gives 9 and ^-248/₂₉, and you write your factorisation as (x – 9)(x + ²⁴⁸/₂₉), your answer is wrong. Moral of the story: you still need to use your brain.

[S1_20151225ABEX] Apples and Cherries on Christmas?

Problem

The ratio of the mass of an apple to the mass of two cherries is  9 : 1.  The mass of the apple is  150 g.  What is the number of cherries that can be found in  y  kg?

Solution                

Remarks

     To obtain the answer, we made a simplifying assumption that all the apples and cherries are identical in mass.  The answer is an algebraic expression and it can be obtained by following the same procedure one would solve the problem if it were in concrete numbers.  Learning algebra is like learning a new but more powerful language.  It takes some time getting used to.  Since we do not know the value of  y,  we leave the answer in terms of  y.  But if we knew the value of  y,  we would know that the answer is  120 times that.  For example, with  3 kg,  we get (about)  360 cherries.

    

Heuristics Used

H02. Use a diagram / model

H05. Work backwards

H08. Make suppositions

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* Primary 6 Mathematics (challenge)

* Lower Secondary Mathematics (Secondary 1)

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve ratios and algebra

[S1_20151221ABEX] The Table as the Years go by ...

Problem

Simon is  15q  years old.  He is now  5  times as old as his son.  How old will he be when his son is  28  years old?

Introduction

     This is an introductory algebra problem, good for getting used to the language of algebra.  As we have seen in this previous article, tabulation is a good way to help organise our information.  Although no one will penalise you for not using tables, once you start using tables, you wonder how you could ever survive without them.

     I present two approaches.  One way is to consider the number of years passed by.  A second way is to observe that the age difference always remains the same as time goes by.

Solution 1

	Now	future
Simon	15q	?
Son	3q	28

The number of years passed is  28 – 3q.

Simon’s future age = 15q + (28 – 3q) = 12q + 28

Ans: When the son is  28  years old, Simon will be  (12q + 28)  years old.

Solution 2     (Refer to table as above)

Note that the age gap always remains the same.

Age difference = 15q  – 3q = 12q.

Simon’s future age = 28 + 12q.

Ans: When the son is  28  years old, Simon will be  (12q + 28)  years old.

Remark

     The answer required is an algebraic expression, in terms of  q.  Since we do not know the value of  q,  do not try to evaluate the expression, but just leave it as it is.  When learning algebra, one needs to get comfortable working with unknowns.

     Also remember to be mentally flexible.  There may be more than one way to “skin the cat”.

Heuristics Used

H02. Use a diagram / model  [tabulation]

H04. Look for pattern(s)

H05. Work backwards

H06. Use before-after concept

H11. Solve part of the problem

Suitable Levels

* Lower Secondary Mathematics (Secondary 1)

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve ratios and algebra

[Pri_20151223WNSV] Unravelling Four Whole Numbers

Problem

If  A,  B,  C  and  D  are whole numbers such that  A × B = 8,  B × C = 28, C × D = 63,  B × D = 36,  find the values of    A,  B,  C  and  D.

Introduction

     This question seems to be taken from a secondary school textbook from a chapter on linear equations.  However, I think a good  primary school pupil could attempt this.

Strategy

     The key to solving the above problem is to make observations.  When you multiply up the first two equations, you get an  A,  a  C  and two  Bs  in the product.  Hmmm ... This doesn’t look promising ...  Ah!  But when you multiply the second and the third equations together, you get an  B,  a  D  and two  Cs  in the product.  This can cancel (via division) with the fourth equation which has one B  and one  D  in the product.

Solution

Remark

     Always cancel as much as possible, to avoid large numbers and reduce chances of making careless mistakes.

     By the way, a whole number is a non-negative (zero or positive) integer that does not contain any fractional part.   As such, the set of whole numbers is {0, 1, 2, 3, 4, ...}.  Thus we do not need to consider the negative square roots.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

Suitable Levels

* Primary School Mathematics (Challenge)

* Lower Secondary School Mathematics (Challenge)

* other syllabuses that involve whole numbers

* anyone game for a challenge

[S1_20151216] Using a Table to Organise Information for Algebra

Problem

The average monthly salary of  m  male employees and  f  female employees of a company is  $2 000.  If the average monthly salary of the male employees is  $(b + 200), find the average monthly salary of the female employee.

Introduction

     This Secondary 1 (~ grade 7) problem in introductory algebra is challenging due to the multitude of pieces of information and their interrelationships.  Using tables is a good strategy to help us organise the information.  

Strategy

     What we do is to fill up each piece of given information in the table first (shown in green below).  Once that is done, proceed to figure out the other blank cells of the table.  The more you do that, the more you would be able to figure out the rest, until you get the solution.

Solution

 Final Remark

     I hope you enjoyed this tip!

Heuristics Used

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics (Secondary 1)

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve ratios or algebra

[PriOlym_20151130RTAC] Ratio with One Circle Overlapping Two

Question

Introduction

     This question is like this previous one, except it is of olympiad standard.  I illustrate the solution of this without algebra, by using Distinguised Ratio Units.  As before, I try to match parts to an equal number.  But here we have quite a mixture of different types of units.

Solution

Commentary

     Basically we make the triangle units to number 12 and do the same for the circle and square units.  It turns out that one triangle unit is the sum of one circle unit and square unit.  We deduce that 9 circle units (for the area of A) plus 6 circle units (for the area of B) is the same as 8 circle units and 8 circle units.  The reduction of circle units must be equally compensated by the increase in the circle units.  Thus one circle unit is the same as two square units.  From here, things become easy.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Primary School Olympiad Mathematics

* Primary School Mathematics (challenge)

* other syllabuses that involve areas and ratios

* anyone who is game for a challenge

[Pri20151129RTAO] Equalising Ratio Units for The Overlap

Question

 

Introduction

     This is a primary school ratio problem that is quite a favourite among question setters, but poses headaches for pupils and parents.  The trouble is that the ratios use different base units and this makes it difficult to compare the ratios.  Can we avoid using algebra or trial and error?  

Strategy

     Note [H04, H09] that the difference in the areas between the rectangle and the square (including the shaded overlapping part) is exactly the same as the difference between them without the overlapping part.  With this crucial observation, we can proceed to try to equalise the ratio units [H10] of the aforementioned differences.  This can be done by multiplying to get to the Lowest Common Multiple, which, in this example is 6.  Henceforth we can be sure of using the same ratio units, because the same number of units are used to refer to the same quantity.

Solution

Summary

     Ratio problems are solved by making sure that we use the same type of units.

Heuristics Used

H02. Use a diagram / model        [ table ]

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve areas and ratios

* anyone who wants to learn

[S1_20151124AESR] Slanted Rectangle does not need Pythagoras

Question

Introduction

     This is another “Bonus Question” at a secondary level from somewhere that the question poser did not mention, but I guess it is most likely an Integrated Programme school in Singapore.  It is a beautifully crafted question.  The presence of a slant line seems to necessitate the usage of Pythagoras’ Theorem.  However, we have seen that Pythagoras’ Theorem can actually be avoided even in Primary (Elementary) School problems.  So a 10 year old kid with a rudimentary knowledge of algebra could do this.  Can you spot a short cut?

Making Observations

     Stare at the diagram for a while.  What do you observe?

Solution

             area of  DDB_nC_n =  ½  of the area of  AB_nC_nD.

              area of  DDB_nC_n =  ½  of the area of  DB_nPQ.

        \  area of DB_nPQ  =  area of AB_nC_nD = n cm².

HeuristicsUsed

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Lower Secondary Mathematics

* challenge for Primary school Olympiad

* other syllabuses that involve areas and a tiny bit of algebra

* anyone game itching for a challenge

[U_Calculus_STKC58] Exploiting symmetry for a Complicated Integral

Question

     This problem appears as problem 58 from a Facebook group and it is set by Kunihiko Chikaya.  Ordinary integration problems are already challenging, but this one is tough on steroids. 

The Standard Approach

Solution 1

With this result, I realised that there is a short cut.  We can make use of symmetry.  Note that  sin(p – x) = sin x   and   cos(p – x) = - cos x.

Solution 2

Heuristics Used

H04. Look for pattern(s):         “onions”, exploit symmetry

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* University / College Mathematics

* challenge for ‘A’ Level H2 Mathematics

* challenge for IB HL Mathematics

* other syllabuses that involve trigonometry and integration 

[MathEdn 20151118] Abuse of The Equal Sign and Maths IQ Puzzles Debunked

Question

Introduction

     This puzzle is making its rounds on the Internet.  The “solution” is easy enough.

          3 ×   6 = 18                  

          4 ×   8 = 32                  

          5 × 10 = 50                  

          6 × 12 = 72                  

          7 × 14 = 98                  

Notice that the numbers in the second column is always twice the corresponding number in the first column.  For the number 10, we have

         10 × 20 = 200

Critique 1

     A few people might get tricked by taking 10 × 16 = 160,  since 16 is the next even number after  14.  But that is not my gripe.  My issue is with the abuse of the “=” sign.  This sign stands for “equal” which means, equal (Surprise!  Surprise!).  Equal means the same, having the same value.  3  is not equal to  18.  So we should not write “3 = 18”, because that is not true.  Writing “3 ×   6 = 18” is OK and correct, because it makes sense and it is a true statement.  Mathematics is not a jumble of nonsensical symbols, although to some people it seems like it.  The symbols have meanings.  And these symbols should not be abused.  If you want to say “corresponds to”, then you might want to use an arrow e.g. “3  ¾® 18”.  This is keeping with the modern concept of a function, in which a value is assigned unambiguously to another number.  Using a function notation we can write things like “f(3) = 18”.

Critique 2

     Another problem with puzzles such as this is that, using Lagrange Interpolation or Newton Interpolation and the like, it is always possible to invent a function that hits the given first few numbers and then any number you like (even a “wrong” number).  For the above puzzle, the Lagrange method allows us to cook up a function like this:-

This function looks complicated, but if you note carefully, when you substitute the numbers 3, 4, 5, 6, 7 and 10, one of the algebraic fractions with  x  becomes equal to  1  and the rest of them become  0.  Thus, we easily see that  f(3) = 18,  f(4) = 32,  f(5) = 50,  f(6) = 72  and  f(7) = 98.  For  f(10),  I could actually have chosen any value I like, but I chose the number 42.  So the correct answer does not have to be 200.  There is actually no single correct answer, since you can make it any answer you like!  This trick can be done on all similar puzzles, and hence these “IQ” puzzles have now been effectively debunked! 

[Pri20151117MSAS] MCQ tactic for the Area of a Hollow Square

Question

Solution 1

     Width of the smaller square  WX = (156 ¸ 4) cm   = 39 cm

     Width of the larger square     AB = (39 + 2´8) cm = 55 cm

     Area = (55² – 39²) cm² = 1504 cm² 

     Ans: (1)

Solution 2

     This is a Multiple Choice Question (MCQ).  Observe that the shaded area is an even number, because it is 8 cm width all around.  Since 1504 is the only even number among the options, (1) is the correct choice.

Remark

     No tedious calculation is needed!

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels
* Primary School Mathematics
* other syllabuses that involve areas and perimeters
* anyone who loves his/her brain tickled