[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

[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

[S1_Expository] Exploring the HCF and LCM with a calculator

How does it work?

     When we reduce a fraction to its lowest terms, we actually cancel out as many common factors as possible.  So eventually the numerator and the denominator of original fraction get cancelled by their Highest Common Factor (HCF), a.k.a. “Greatest Common Divisor (GCD) ” in America.  Just as England and the United States are divided by a common language^(*), we can divide  756  by  6  to get the HCF.  Why?  That is because  756  was divided by the HCF to get  6.  You may try a similar trick with the denominators.  You get the same conclusion.

     It is useful to know that the product of two numbers is equal to the product of their  HCF  and  LCM.  So   756 × 1386 = HCF × LCM.  Hence we have

Observe that the bracketed number  (⁷⁵⁶/₁₂₆)  is equal to  6  which is the numerator of the reduced fraction.  So we do not even need to make that calculation.  Just take the reduced numerator  6  and multiply that with the original denominator  1386  to get the LCM.  You may try a similar trick with  1386  and  126.  You end up multiplying  756  by  11,  which gives the same answer.

Further Exploration

     Try the above with different pairs of numbers.  How can you extend this to find the HCF and LCM of three or more numbers?

(*) OK, just kidding.  In Singapore, we sort of follow British English, but we are flexible.

Suitable Levels

* Lower Secondary Mathematics (Sec 1 ~ grade 7)

* other syllabuses that involve factors, HCF (GCD) and LCM

* any interested learner

[Pri20150522RPRW] More Hands make Light Work?

Question

Introduction

     The key to solving this is to find the total rate of work for Ahmad, Kumar and Calvin.  There is a relationship Work = Rate ´ Time, similar to the relationship Distance = Speed ´ Time, and you can use a similar triangle mnemonic for it.  For example if you cover ‘R’ with your finger, you get the relation Rate = Work / Time.  To organise your information, you may use a table to tabulate the given data.

     Do you believe “more hands make light work”?  Or is it “too many cooks spoil the broth”?  In real life, people may not work harmoniously together, or work at the same rate (never getting tired).  They may get distracted by Facebook, mobile phones or office gossip.  In school mathematics problems, we assume that when people work together, we can just add up their rates of work.  Yes, it is quite funny, but let us just assume.

Solution

Ans: 10 days

Commentary

     When we add up the given rates, we get  ¹/₅,  which is the rate that 2 Ahmads, 2 Kumars and 2 Calvins would work.  Unfortunately we do not have the clones.  We just have one Ahmad, one Kuman and one Calvin.  So we divide that by 2 to get  ¹/₁₀  and this is their combined rate.  This means that they can complete  1  house in  10 days.

     It is possble to do this question without the table, for example by taking the LCM of the denominators and considering how many houses can the guys build in  60  days.  However, the table is a good way of organising information and you can solve the problem by considering the total rate.

HeuristicsUsed

H02. Use a diagram / model   (including table)

H03. Make a systematic list

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 fractions 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

[AM_20150413RSD] Rationalising Denominators for #Surds

Question 

Introduction

     Expressions with surds in their denominators are cumbersome.  The good news is that we can make the denominators into rational numbers, which are nicer.  Rational numbers those that can be expressed as a ratio of integers i.e. they are (proper or improper) fractions or can be converted to fractions.  Whole numbers are also part of rational numbers because you can always put them upon a denominator of  1;  e.g. 2 = ²/₁,  so  2  is a rational number.

     The standard trick for simplifying expressions with surds in their denominators is to rationalise the denominator by mutiplying the numerator and the denominator with its conjugate surd.  For example, the conjugate surd of   Ö5 + Ö2   is   Ö5 – Ö2.   Just change the  +  to  –  or the  –  to  +.  Let us see how the magic works.

Solution

Remarks

     Note that in the first step, I pulled out 2 as the common factor of the denominator, so that I get a simpler surd to work with.  Always try to work with simpler expressions.  This not only shortens your working, it reduces your chances of making a careless mistake.
     In mathematics, “rationalising” does not mean you give some reason or excuse for something that you know you have done wrong.  It means “make it into a rational number”.  Why does rationalising the denominator work?  This is because on the bottom (denominator) we have a difference-of-squares expression of the form
                                           (a + b)(a – b)   which is equal to   a² – b².
Since squaring “gets rid” of square roots,  a²  and  b²  will give you rational numbers (whole numbers or fractions), you will end up with a nice number downstairs (on the denominator).  Pupils should make sure they have this technique in their repertoire of skills.

Suitable Levels
* GCE ‘O’ Level Additional Mathematics
* revision for GCE ‘A’ Level H2 Mathematics
* revision for IB Mathematics HL / SL
* other syllabuses that involve surds
* precocious kids who always want to learn more

[Pri20150330FBT] Open-ended Question on #Fractions

Question

Give a fraction with its value in between 2/5 and 1/2.  (Note: the denominator cannot be greater than 20)

Discussion

This is an open-ended question for primary (elementary) school, which ought to be easy because there are many possible answers and you just need to supply one that meets the requirements.  However, most people are used to closed-ended questions, which have only one correct answer.  This is what prompted a parent to pose this question on a parent-support forum on Facebook. 

Many people, including parents and even some private tutors, began to supply their answers to this question.  Much of the discussion clustered around the idea of the mid-point or the average of the two given numbers, namely ¹/₂ (²/₅ + ¹/₂) or  ⁹/₂₀, and why that gives an answer.

Then Dr Kho Tek Hong (retired curriculum specialist, the “father” of Singapore mathematics) chipped in.  He said that there are many possible answers e.g. ³/₇, ⁴/₉, ⁵/₁₁, ⁵/₁₂, ⁶/₁₃, etc.  He suggests that students be asked to justify their answers.  [ By the way, here is an interesting method to compare fractions. ]

And I thought that was a stroke of educational brilliance from The Guru!  It shows the spirit of the Singapore mathematics curriculum – to get students to think, to be open-minded, to use logic and to be able to communicate mathematically.  The Singapore curriculum is not meant to torture students, nor to get them to toil around a tortuous path around a high mountain seeking the elusive holy grail – although it often seems that way.  Teachers, tutors and parents would do well to help the learners it in the right spirit.  Mathematics is not always about calculations (although you need to do some calculation).  We also need to open our minds to multiple answers as well as to embrace various methods and concepts.

[Pri20150330CPF] How to #Compare #Fractions

Comparing Positive Fractions

     How do you compare positive fractions, which are taught in primary (elementary) school?  For example, which is bigger:  ⁵/₆  or  ³/₄ ?

The “orthodox” method is to put them both to a common denominator.  The Lowest Common Multiple (LCM) of  6  and  4  is  12.  Multiplying the left fraction by  ²/₂  and the right fraction with  ³/₃  gives, respectively,  ¹⁰/₁₂  and  ⁹/₁₂.

Since  ¹⁰/₁₂  >  ⁹/₁₂,  we conclude that  ⁵/₆  >  ³/₄.

Another Method

     Here is a “short-cut” that I learned from a schoolmate in primary school.  Basically you “cross-multiply”: multiply the left numerator with the right denominator, and multiply the right numerator with the left denominator, and then compare the products so formed.  That will give you the correct inequality or equality sign (viz. ‘<’, ‘=’ or ‘>’).

As we can see, since  20 > 18,  we conclude that  ⁵/₆  >  ³/₄. 

     Does this method work?  Yes, definitely.  You can try it out with a few pairs of fractions and you can see for yourself that it is so.  Is this method legit?  Why does it work?    I give a formal proof of the method below.

Further Discussion

     Use the above method with care.  Some school teachers may not accept the method not because it is not correct, but it sounds “dubious” to them because they have not heard of it or they are not able to prove it for themselves.  Pupils can use this short-cut to give them a quick look-ahead to certain questions, and as a back-up to check their answer after using the Lowest Common Denominator method.  In questions that ask pupils to arrange a few fractions in ascending and descending order, this “crossing method” may give some speed advantage if done carefully.

     In primary school, pupils focus on positive fractions.  In secondary school, negative numbers and fractions are introduced.  Does the above trick work for negative fractions?  Was my theorem and proof above carefully phrased enough to cover the negative fractions? 

     Note that this method works for comparing two fractions at a time only.  Sometimes this cross-multiplying gives rather big numbers.  In that case, it is better to multiply each fractions by the LCM of their denominators.  Essentially this is the same as the orthodox method, except that we do not write the denominators.  Can the above proof be extended to cover this new short-cut?  What do you think?