[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

[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

[AM_20151226EIQR] Looking for a Pea among Quadratic Roots?

Question

Introduction

     This question is about finding the parameter  p, and not about solving for the “unknown”  x.  It is heavy on algebra, one has to be patient, careful and meticulous.  Please refer to this article for a recapitulation of (Vieta’s) theory of Quadratic Roots.

Solution

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* other syllabuses that involve quadratic roots
* any learner who is interested

[AM_20151224QERI] Quadratic Roots and Use of Identities

Problem

Introduction

     Here is a fairly standard question on roots of quadratic equations, except that part (iii) is slightly more challenging.  To solve this question, one must know the square of sum identity well.

Recapitulation

     Please refer to this previous article  and  this article  for the theory on quadratic roots.

Solution

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Levels H2 Mathematics

* other syllabuses that involve roots of quadratic equations

* any learner who is willing to learn

[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

[H2_VJC2015PromoQ10_IAXS] Volume of a Doughnut

Problem

Introduction

     This is a problem involving the calculation of the volume of solid of revolution of an enclosed region.   The junior colleges (or senior high schools) like to set this type of question. 

Technique

     If the axis of revolution is the  y-axis, the basic formula is   ò px² dy   with the appropriate lower and upper limits.  Notice that this only works for the region between one curve and the axis and when rotated, this will generate a solid with no hollow parts.  An enclosed region, however, consists of two curves.  In our case, when we make  x  the subject, we find that we have two choices.  One of them leads to a curve that is further away from the axis of rotation.  I call that the outer curve.  The other curve is the inner curve, and this is nearer the axis of rotation.  We need to subtract the volume generated by the inner curve from that generated by the outer curve.

Solution

Remarks

     In this example, the curve on the right happens to be the outer curve.  If the equation were

(x + 93)² + y² = 15²,  the circular region would be on the left of the  y-axis and the outer curve would be on the left.

     For your information, the above solid of revolution is a torus.  This is the shape of a doughnut, (or hoopla-hoop, circular tube, or Polo mint perhaps?).  It is the inner curve that gives the hole in the “doughnut”.

     You can imagine in your mind’s eye that as the circular disk revolves around the  y-axis, its centre traces out a circular path of  93 units.  By the Second Centroid Theorem of Pappus,

     volume = length of path of centroid × area of cross section = 2p(93) × p(15)² =  41850p²

In general, the volume of a torus with major radius  R  and minor radius  r  is

     volume = 2pR × pr² =  2Rr²p²

If you know this fact, you use it to check your calculations.  Although this is not in the H2 Syllabus, but it is something interesting to explore.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards                                [e.g. making  x  the subject]

H09. Restate the problem in another way  [symmetry: volume is twice of upper half]

H10. Simplify the problem                         [integration by substitution]

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Levels H2 Mathematics

* IB Mathematics HL (Applications of Integration)

* Advanced Placement (AP) Calculus AB & BC

* University / College Calculus

* other syllabuses that involve Applications of Integration

* any precocious or independent learner who loves to learn

[OlymLS_20151220INEQ] An Egyptian Fraction Partition of Unity?

Problem

Introduction

     This question appeared in a “holiday homework” from an IP school.  It is not the usual type of question in exams and tests, but it is a good mental-stretching exercise.  It is asking us to split unity (the number “1”) into three unit fractions (a.k.a. Egyptian fractions).

     The first part can be solved by trial and error or “guess and check”.  The second part, proving that this is the only solution (are there any other solutions?), seems challenging.  We can solve this by making suppositions and using inequalities.

Solution

                                              

Heuristics Used

H03. Make a systematic list

H05. Work backwards

H07. Use guess and check

H08. Make suppositions  [ which may lead to negative conclusions ]

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Primary School / Lower Secondary Olympiad

* other syllabuses that involve fractions and inequalities

* any learner who is itching for a mental challenge

[AP_Calculus20151201] Integrate something with arcsecant by parts

Question

Introduction

     This question is taken from the Techniques of Integration chapter of Thomas’ Calculus, 12th edition.  It looks pretty nasty in that the arcsecant is just one of those things on the fringes of teachers’ and students’ minds.

Strategy

     We apply the “Integration by Parts” Formula                                                  

with the “d(etail)” heuristic.  The expression  t  is  algebraic whereas the arcsecant expression is of the “inverse” type.  Since “a” comes before  “i”  in “d(etail)”, we choose the algebraic   expression  t  to serve as our  ^dv/_dx.  We realise that we will need the derivative of the arcsecant                                                 

for  sec^-1 x  being a cute angle ... I mean, an acute angle.

Solution

Remarks

     A slight modification of this approach is to first re-express the arcsecant as  arcsec t = arccos(¹/_t).  One needs to work out the derivative of this arccosine expression when doing the integral.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* GCE ‘A’ Levels H2 Mathematics (challenge)

* IB Mathematics HL (challenge)

* Advanced Placement (AP) Calculus BC (challenge)

* University / College calculus

* other syllabuses that involve integration and inverse trigonometric functions

* any precocious learner who loves a challenge

[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

[EM20151129CGCB] Bisector of a Chord in a Circle

Problem

Introduction

     This “elementary” mathematics question poses a challenge because it actually testing Coordinate Geometry and Circle Geometry.  In the setting of tests and exams, there seems to be a trend of combining topics.  To solve this problem successfully, students need to know that when a chord is bisected (cut into two equal parts), the line segment joining its mid-point to the centre of the circle will be perpendicular to the chord itself.  Thereafter, we can proceed with Pythagoras’ Theorem.

Solution

Remarks

     There is actually no boundary between topics and even subjects.  Things to be learned are separated into topics only to facilitate teaching of the material.  Students are encourage to adopt a more wholistic view of knowledge.

Heuristics Used

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

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

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

* GCE ‘O’ Level Additional Mathematics (revision)

* other syllabuses that involve geometry and coordinate geometry / analytic geometry
* whoever is interested

[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

[AM_20151127DF2D] Differentiation with Chunking and Elimination

Question

Introduction

     Although this looks like a differential equation question, the student is not required to solve the differential equation.  The requirement is just to derive the equation.  This would be a challenging question for secondary 4 (~ grade 10) students taking Additional Mathematics or their counterparts in Integrated Programme schools.

Strategy

     One way to do this is to differentiate the given equation once and again and just verify the equation by substitution.  The problem is that when we repeatedly apply the Product rule

the terms tend to sprawl.  A way to keep things neat is to try to recognise chunks and also use elimination.

Solution

Remarks

     After differentiating once, we notice that  10xe^2x   is twice of  5xe^2x,  and this allows the simplification in [1].  The second differentiation yields  10e^2x   which, we notice, is twice of  5e^2x.  We can get rid of that term.   Multiplying equation [1] by 2 gives  10e^2x  in equation [3],  which matches nicely with the same term in  [2].  So we can eliminate that term via elimination.  After that, we just need to rearrange things to get the final equation.

Heuristics Used

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

* GCE ‘O’ Level Additional Mathematics, “Integrated Programme Mathematics”

* GCE ‘A’ Levels H2 Mathematics (revision)

* AP Calculus AB / BC (revision)

* University / College calculus (revision)

* other syllabuses that involve differentiation

* any learner interested in calculus

[H2_20151125DEST] Differential Equation via Substitution

Question

Introduction

     This question was taken from a Facebook group.  The suggested substitution was added in to make it accessible to students taking H2 Mathematics.  In the original question, no suggested substitution was given.

Strategy

     Observe that the given equation has  x  and  y.   After substitution, we should get an equation with  v  and  x only.  How to make  y  “disappear”?  One way is recognise patches that can be substituted for  v.  Another way is to make  y  the subject and differentiate that with respect to  x.  Replace the derivative ^dy/_dx  with your new expression.  Here, I do both at once!

Solution

Remarks

     Usually we try to express  y  in terms of  x,  but here, it is more convenient to express  x  in terms of  y.

     Since  A  is an arbitrary constant,  A+1  is still an arbitrary constant.  Also, whenever there is a “ln” appearing in the solution, it is good to introduce “ln” with an arbitrary constant.  So we can lump  A  and  1  together and call it  ln B.  Nothing is lost in this process.  ln B  is able to achieve all possible numbers.   ln B  is negative if  B  is a fraction between  0  and  1,  and is positive if  B  is more than  1  and is zero if  B = 1.

     This is not an applied differential equation, so there are no exogenic reasons or hints for us to suppose that  x + y + 2  is a positive quantity.  As such, we still need to leave it as | x + y + 2|.  To get rid of the absolute or modulus sign properly, we can replace  ±B  with  C.

Heuristics Used

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

H13. Use Equation / write a Mathematical Sentence

Heuristic that cannot be used here

H08. Make suppositions

Suitable Levels

*  GCE ‘A’ Level H2 Mathematics

*  IB HL Mathematics Calculus Option

*  Advanced Placement (AP) Calculus BC

*  university / college calculus

*  other syllabuses that involve differential equations

*  anyone who is game for a challenge