[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

[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

[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

[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

[Pri20150519WNRT] Girls in the Stadium

Question

     Teachers!  Do you ever realise how you set questions?  This must be a pretty small stadium.  Or else a very sparse one.  Which stadium would allow you to book it for its use when there are relatively so few people?  Anyway, let us ignore that and get on with the “problem”.

     I am going to illustrate my Distinguished Ratio Units method, as usual.  You can use bar models if you want.  There are many ways to skin the cat, as they say.

Solution

     We organise the given information by setting up a table.  We work out that there are  560  males in total.  I used “triangle” ratio units for the adults and “circle” units for the children.  You can use anything you like, as long as you make it clear they are different.

My favorite tactic is to equalise one of the ratio units.  It is particularly easy to use units that with the number  1.  Let us multiply the left column by  2,  as shown below.  Imagine what would happen if each male was cloned to have two copies of each person.

The circle units are now equalised.  This serves as a stepping stone or a bridge to connect 2  “triangle”  units with  6  “triangle” units.  By comparison and subtraction, we figure out that  4  “triangle” units correspond to  400.  And we can work out the rest easily.

Ans:  There are  520  girls.

Final Remarks

     Distinguished Ratio Units are easy to use.  The strategy is:

          (1)  use different types of units marked by differently-shaped outlines

          (2)  look for a unit with “1”, multiply to equalise the unit of that type

          (3)  compare the other type of unit and solve that

          (4)  solve the rest of the problem

With this method, you do not need to worry about drawing and redrawing bars, or cutting bars into many smaller bars.  You can just concentrate on the problem modelling and thinking.

HeuristicsUsed

H02. Use a diagram / model    (use an effective one J)

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 Mathematics

* other syllabuses that involve whole numbers and ratios

[Pri20150510PEP] Percentages of Erasers and Pens

Question

Introduction

     This is a question about percentages, which are really fractions based upon 100 as denominator.  For example,  60%  just means  ⁶⁰/₁₀₀.    It is possible to solve this using some sort of algebraic approach based on  100  units for a percentage.  However it is more convenient to use fractions in their lowest terms.  I present a solution based on my Distinguished Units Method, which is a proto-algebraic approach.

Solution

     Note that  60% = ⁶⁰/₁₀₀ = ³/₅  and  25% = ¹/₄.  The Lowest Common Multiple (LCM) of the denominators is  20.  I use  20 “square” units for the original number of erasers and  20  “circle” units for the original number of pens.  This makes the units easy to divide.  It does not matter what shape you use to envelop the different units, as long as different shapes are used for different types of units.

          Ans:  There were  240  pens at first.

Since the question asks for the original number of pens, it is a good idea to equalise the eraser’s “square” units.  Multiplying the first row numbers by  ⁸/₂₀  gives 8  “square” units for the third rows.  This serves as a stepping stone to connect the “circle” units.  See the part highlighted in yellow.  From  8  “circle” units to  15  “circle” units, the difference is  84.  This allows us to deduce the value of  1  “circle” unit.  The original number of pens is represented by  20  “circle” units corresponds to  240,  which is the answer we want.

Final Remarks

     It is a good idea to know the fractions of some of the more common percentages.  For example,

     25% = ¹/₄,   50% = ¹/₂,   75% = ³/₄

     20% = ¹/₅,   40% = ²/₅,   60% = ³/₅ ,   80% = ⁴/₅

The usage of  LCM of the denominators is very effective for making calculations easy.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H06. Use before-after concept

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 whole numbers and ratios

[Pri20150510OAC] An Oranges-to-Apples Comparison

Question

Introduction

     This primary school mathematics question is slightly tricky as the pieces of fruit* are packaged.  I present two solutions: the first uses the “assumption” or “supposition” method together with the concept of unit costs.  The second solution illustrates my Distinguished Ratio Units (DRU) method.

Solution 1 (using an Assumption / Supposition)

 Solution 2 (using my Distinguished Ratio Units method)

I use  3  “circle” units to denote the number of oranges (since I know it can be divided by 3) and

I use  2  “triangle” units to denote the number of apples (since the number is divisible by 2).  Then the cost of the oranges would is two “circle” units and the cost of the apples is  3  “triangles” units.  Let us equalise the “circle” units by multiplying the first row by  2  and the second row by  3.  With both “circle” units equal to  6,  the  6  “circle” units can be used as a stepping stone to connect  4  “triangle” units with  9  “triangle” units (highlighted in yellow).  We now know that  5  “triangle” units corresponds to  20.  It is easy now to work one  1  “triangle” unit and then  2  “triangle” units, which corresponds to  8.

Ans:  Abigail bought  8  Apples

Commentary

     Note that my DRU solution does not use any fractions!  The astute reader will note that my DRU method is actually the equivalent to the  u and  p (“units vs parts”) method used by many Singapore teachers/tutors.  Both are actually algebraic methods in disguise, just as the bar modelling (“the model method”) is.  In case you are wondering what the difference between units and parts is: units and parts actually have the same meaning, except that they refer to differently-sized unknowns and we need different names for different units.  Primary school mathematics in Singapore is actually rather challenging because many of these problems are equivalent to solving a pair of simultaneous equations in two unknowns.  It is possible to use one unknown unit, but you would kill a few brain cells in the process of formulating the problem in terms of only one unknown.

*A Note on Singlish and standard English

     Most Singaporeans would call, for example, 2 apples and 3 oranges as 5 “fruits”.  Actually, in standard English, “fruit” (as an uncountable noun) refers to general fleshy food that comes from flowering plants.  So if you eat “2 apples and 3 oranges”, you are eating fruit.  Yes, fruit is food.  The word “fruit” can also be used as a countable noun.  This refers to fruit coming from different botanical species.  So in the aforementioned example, “2 apples and 3 oranges” would be considered as 2 fruits (2 types of fruit) and  5  pieces of fruit.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H08. Make suppositions (assume, “what if”, imagine if ...)

H09. Restate the problem in another way

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve ratios, fractions and money

[Pri20150503RTB] Specky Men & Women

Question

Introduction

     This looks like a difficult problem because there are different fractions and different numbers.   There seems to be so much information.  How do we deal with that?

Solution

     One good way to organise information is to use a two-way table.  I have a row for the “speckies” (bespectacled people) and a row for the “non-speckies”.  I put one column for the ladies and one column for the men.  Put in column- and row- totals and the grand total.  Instead of spelling out the words, I use icons to represent the different groups.  Who says you cannot be creative in maths?

     I use “circle” units for the women and “square” units for the men.  This is my Distinguished Ratio Units method.  It is easy to work out the total for the speckies.  Just subtract  282  from  456.  After filling up the table, we get a diagram like this:-

Notice the  1  “circle” unit?  It is easy to multiply this by  5  so as to match the  5  “circle” units.  So I multiply everything from the speckies’ row by  5.  I am imagining what would happen if there were five times as many bespectacled men and women.  For then the numbers of bespectacled women would be the same as the number of the clear-sighted women.  The result is shown in green below.

With the “circle” units equalised to  5  units each, we can now compare the  25 “square” units with the  4  “square” units.  The difference of  21 “square” units must be due to the difference between 870 and 282, which is  588.  That allows us to work out the value of  1  “square” unit and then  5  “square” units (representing the number of male speckies).  Knowing two of the numbers in the speckies’ row, we can finally work out the remaining number, which is the number of female speckies.

Ans:   34 women wear spectacles.

Summary

     In this article, I have demonstrated the use of Distinguished Ratio Units and the use of tables for organising information.  I have also demonstrated the technique of equalising one type of units (in this case the “circle” units), so that we can compare the other type of units  (here the “square” units).  I hope you have found this article useful.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H08. Make suppositions

H11. Solve part of the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve whole numbers and ratios

[Pri20150504WNR] Emma and Francis’ Stickers

Question

Introduction

     This is another one of those Singapore primary school mathematics problems whose underlying algebraic structure is equivalent to that of a pair of simultaneous linear equations.  However, the primary school pupils are taught only simple algebra.  A very popular method is the use of bar diagrams a.k.a. “the model method”.  It is a useful tool to help pupils visualise the quantities involved.  However, some people are not comfortable with this method.  One needs to cut the bars into the correct number of sub-parts.  If the diagram is drawn wrongly, one would need to erase the diagram and redraw it.  Remember: the “model method” is just one of the many ways of drawing diagrams, which is only one of the many heuristics for solving mathematical problems.  It is worth the ship, but do not worship the “model method”.   If it works for you, go ahead.  If not, do not force it.  Try another method.  Do not cut your feet to fit the shoes (削足适履).

     Once again, I demonstrate my Distinguished Ratio Units (DRU) method as an alternative.    Let us say that a pupil who uses DRU actually drew the diagram in his/her mind, in a way.  Instead of drawing various sized multiple units, we now use numbers surrounded by different shapes to represent the different types units.  It is easy to read the passage sentence by sentence and translate them directly into the diagram without having to worry about whether the unit is drawn to the correct relative length.   To illustrate the facility of use, I shall actually present two ways to attack the problem using DRU. 

Solution 1  (Unifying units via LCM)

     I use one “circle” unit each for what Emma and Francis had at first.  Later on, after adding two “circle” units, Emma has  3  “circle” units.  After subtracting  8,  Francis ends up with, say, one “square” unit.  Emma has four times as much, so Emma has  4  “square” units, which is equal to  3  “circle” units.

     Now since the LCM of  3  and  4  is  12,  we can make Emma’s later holdings for stickers to be  12  “triangle” units, say.  We can express every of the original units in terms of this common “triangle” unit.  Just multiply the numbers in every “circle” unit by  4,  and multiply the numbers in every “square” unit by  12.  This is what we would get:-

From here, we easily see that the transition from  4  “triangle” units to  3  “triangle” units is a subtraction by  8.  Hence  1  “triangle” unit is  8  and therefore  4  “triangle” units (which represents Francis’ original number of stickers) represents  32.  So Francis had  32  stickers at first.

Solution 2  (Stepping stone)

     We model the situation in a way similar to the above solution, except that we reverse the arrow connecting  1  “circle”  unit to  1  “square” unit and replace the  “+8”  with  “–8”.  We have not changed the meaning by doing this.

Now, multiply everything in Francis’ column by  4  so as to obtain  4  square units.  This is the same as imagining what would have happened if Francis’ had  4  times his original number of stickers and he had given away  4  times (i.e. 32) the number of stickers.  Obviously, he would have  4  times the number left, as represented by  4  “square” units.  If we added  32  to the  4  “square” units,  we would get  4  “circle” units.

Notice that the  4  “square” units now serve as a stepping stone to connect the  3  “circle” units to  4  “circle” units, as highlighted in green.  Hence we can see straightway that  1  “circle” unit is the same as  32.  But this is the answer we want, because  Francis had the equivalent of  1  “circle” unit at the beginning!

Summary

     In the first method, I just use two different types of units.  Then I use the idea of Lowest Common Multiple (LCM) and then change all the quantities to a common type of unit.  In the second method, I multiply a relation by a certain number so that one type of units matches exactly and this serves as a stepping stone to find a link that solves the other unit.  I hope you like my Distinguished Ratio Units method.

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

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve whole numbers and ratios

[S2_20150402RLD] Fifty balls left behind

Question

This problem can be solved with Primary School knowledge using ratios.  The famous Singapore bar diagramming method can be used to model the situation, but I prefer my own Distinguished Ratio Units.  The former method is good for visualisation for beginners, while the latter is faster if you want to solve it quickly without fussing around drawing the perfect diagram.  My DRU method is also visual in another way, and it works with big numbers as well as small numbers.  Alternatively, this can be solved using algebra via simultaneous equations.

Solution 1 (Using my Distinguished Ratio Units method) [H02]

Explanation: Since the number of white balls is a multiple of  3,  I let “triangle” 3 represent the number of white balls.  I let “heart” 1 represent the number of red balls.  There are 50 more white balls than red balls.  [H04]  When the white balls are removed three at a time, the number of groups of three would be one-third of the number of white balls, i.e. 1 triangle unit.  [H04]  This number is less than the number of red balls (1 heart unit) by 50.  So 1 triangle unit plus 50 gives 1 heart unit.  [H04]  Following on from the heart to the “triangle” 3, one realises that 2 “triangle” units is the same as  100.  [H05]  So one triangle unit is  50.  [H11]  From here we can solve the rest of the problem.

Solution 2 (using Algebra)  [H13, H05]

                     w = k + 50 = 3h             –––––––––– [1]

                     r  = k         =   h + 50     –––––––––– [2]

for some unknown  k  and  h.  And then [1] – [2] gives  [H10]

                     w – r = 50 = 2h – 50

so                      2h = 50 + 50

                            h = 50

This quickly leads to

                           w = 150

and                       r = 100.

Heuristics Used

H02. Use a diagram / model

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

* Primary School Mathematics (“Ratio”)

* Lower Secondary School (“Simultaneous Linear Equations”)

* other syllabuses that involve ratio or algebra