[Pri5 20170426FEM] Baking Éclairs and Macaroons

Problem / Question

     This problem for primary 5 from one of my acquaintances on Facebook, considered to be of intermediate level difficulty (in Singapore).  But it looks rather challenging to draw all those bar diagrams, doesn’t it?

     Here is my quickie solution without explicit algebra and without bar diagrams.

Solution

     For convenience, we use 6 circle units for Eclairs and 6 square units for Macaroons.

Suppose there were half as many Eclairs and Macaroons, then there would be 15 more Eclairs.  So 3 square units add 15 can be changed to 3 circle units.

     Add 15 to the 17 and change 3 square units to 2 circle units.  We deduce that 5 circle units is the same as 85.  From here we can easily figure out the rest.

Ans: 102 éclairs.

Comment

     The problem can be solved by bar diagrams.  However, there are many ways to skin the cat.  For more good stuff, please join my Facebook group “Effective and Elegant Mathematics”.

Heuristics Used

H02. Use a diagram / model

H05. Work backwards

H06. Use before-after concept

H08. Make suppositions

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Primary School / Elementary School Mathematics

* any precocious or independent learner who is interested

[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

[Pri20160223FPDM] Pernicious Portion Problem? Shift Happens!

Problem / Question

Strategy

     This seems to be a confounding question on decimals.  What shall we do with the triangles?  Is there a short cut?

     Yes!  What you can do is to imagine putting the two triangles together to form a rectangle.  And then the solution becomes easy!  This is because the area is unchanged and hence the proportion of the shaded area is unchanged, is the same as before.  We can make use of fractions and convert it to a decimal.

Solution

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve fractions and decimals

* any 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

[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

[Pri_20151224RAMN] A Coin Problem with Constant Difference

Problem

Danny saved some  50-cent coins and  $1-coins in his coin box.  The total value of the  50-cent coins to the total value of the  $1-coins he had was in the ratio  2 : 5.  After  $14  worth of  50-cent coins and an equal value of  $1-coins were added to the coin box, the ratio of the total value of  50-cent coins to the total value of  $1-coins became  5 : 9.  How many coins of each type did Danny have in the end?

Introduction

     Here is a “Singapore math” coin problem that can be befuddling for kids and even for adults.  To rub salt to the wound (or pour oil to the fire?), the value of a collection of coins is different than its number.  Whilst a $1-coin obviously has a value of one dollar, you would need two 50-cent coins to make up a dollar.

Strategy

     Notice that after adding  $14  worth of coins to both types of coins,  the difference in the total value of the two types of coins remains the same.  Some people call this a “constant difference” problem.  But how do we exploit this constant difference, when the type of ratio units used in  2 : 5  are most likely not the same as those used in  5 : 9?  Well, we need to bring them to a common unit! [H09]   How?  Read on!

Solution   [H02, H06]

Ans: Danny had  54  $1-coins and  60  50¢-coins in the end.

Commentary

     I am using Distinguished Ratio Units in my presentation.  This makes it clear that the ratio units are of different types.  In the the “before” stage [H06], the difference in the value of the two sets of coins is  3  circle units.  In the the “after” stage, the difference in the value of the two sets of coins is  4  square units.  But we know these two differences refer to the same numerical number.  The Lowest Common Multiple of  3  and  4  is  12.  So both of them must me equal to  12  common units (which I envelop with triangles).  We multiply the numbers inside the circle units by  4  and we multiply the numbers in square units by  3.  I put these multiplications in quotation marks because we are not really changing the numbers of coins.  We are merely changing the type of units used.  I am saying that each square unit is the same as  3  triangle units and each circle unit is the same as  4  triangle units.

     Once we bring everything to common units (triangle units), we can see the  $14  added corresponds to  7  triangle units.  Henceforth the whole problem unravels easily.  [H11, H05]

Heuristics Used

H02. Use a diagram / model

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 independent learner who is interested

[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

[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

[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

[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

[Pri20151116DNGC] Quadratic Plays Second Fiddle in Product-Difference Riddle

Question

What two numbers give a product of  21.5  and a difference of  6.1?

Introduction

     This question reminded me of a question that I set myself when I was in Primary 4 (» grade 4).  I imagined a rectangle with breadth 4  and  length 2 units longer then the breadth (i.e. 6) giving a area (product) of  24.  Then I pretended that I did not know the breadth and let it be  x.  This led to a quadratic equation  x(x + 2) which I did not know how to solve (if I did not know the answer).  So I accidentally discovered quadratic equations when I was in Primary 4.  This led me to a quest to learn the method of factorisation (by “trial and error” or “guess and check”) and the quadratic formula.  I never liked trial and error.  So I continued in my quest to invent a method of factorisation that did not require “guess and check”.  I finally succeeded doing that in secondary 1 (» grade 7).  This turned out to be a Pyrrhic victory.  The method I invented was quite similar to the quadratic formula.

     There is a place for “guess and check” in mathematics.  I present a simple solution to the above problem using just that.

Solution

smaller #	larger #	product
2	8.1	16.2	û
3	9.1	27.3	û
2.5	8.6	21.5	ü

Solved! J

Heuristics Used

H02. Use a diagram / model    (table)

H05. Work backwards             (if the smaller number is this, what is the bigger number?)

H07. Use guess and check

H09. Restate the problem in another way      (area = product)

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve decimal numbers

* anyone who loves to exercise their minds

[NumTh Expository] The Principle Behind Casting Out Nines

Introduction

Think of a number ... say 685932.  Divide by  9  and take the remainder.      685932 ¸ 9      = 76214 r 6  Add up the digits,  divide by  9  and take the remainder.      6+8+5+9+3+2 = 33 ¸ 9 = 3 r 6  What do you notice?  Try this with any other positive whole number.

Discussion

     Did you see that a number and its sum of digits always have the same remainder when divided by  9?  This is the principle behind the method of “casting out nines”, used in the past for checking arithmetical calculations.  Why does this work?  Where is its magic?

     The decimal number system that we use today is based on the number  10, which is just  1  larger than  9.  Observe that  9, 99, 999, 9 999, 99 999, ... etc are all divisible by  9.  Hence, the powers of 10, namely 10⁰ = 1,  10¹ = 10,  10² = 100,  10³ = 1 000,  10⁴ = 10 000,  10⁵ = 100 000,  etc  all leave a remainder of  1  when divided by  9.   Thus in our example,

     685932 = 6´10⁵ + 8´10⁴ + 5´10³ + 9´10² + 3´10¹ + 2´1

                  = 6´(99999+1) + 8´(9999+1) + 5´(999+1) + 9´(99+1) + 3´(9+1) + 2´1

                  = 6´99999+8´99995´999+9´993´9 + 6´1+8´1+5´1+9´1+3´1+2´1

                  = 9 ´ something + 6+8+5+9+3+2

As you can see, all the “´1” allow us to separate out the digits, and then the stuff with 9, 99, 999 etc can be lumped together as 9 ´ some whole number, but we do not need to care too much about this multiple of 9 as it would not make any difference to the remainder.  It is now obvious that  685932 and 6+8+5+9+3+2=33 will have the same number when divided by 9.

     Let us generalise the argument.   If two numbers  x  and  y  have the same remaider when divided by 9,  we say that  x  and  y  are congruent modulo 9, and we write     x  º  y  (mod 9).  Congruence is an equivalence relation and “º” behaves in many ways similar to “=”.

Theorem

For an arbitrary number  n   with digits  [dk...d3d2d1d0]                                         n º dk + ... + d3 + d2 + d1 + d0   (mod 9)

                 n = d_k ´10⁴ + ... + d₃´10³ + d₂´10² + d₁´10 + d₀.    

Since  10^k º 1 (mod 9)  for all integers  k > 0,  we have

                 n º d_k´1 + ... + d₃´1 + d₂´1 + d₁´1 + d₀

                 n º    d_k  + ... +   d₃   +    d₂  +   d₁    + d₀     (mod 9).   © (Q.E.D.)

As an example of application of this principle, please refer read thisarticle.

Suitable Levels

* Primary School Mathematics Olympiad

* syllabuses that involve congruences and Number Theory

* anybody who is interested