[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

[Enrich20170101SQT] Calculating Square Roots by Hand

Introduction

          Happy New Year to our readers!  I wish this year will be a fruitful one for everybody.

          Today, I will illustrate how to calculate square roots by hand, using  54 756  as an example.  It is similar to long division, but has some modifications.

Solution

          Starting from the right, pair up the digits.

          2×2 = 4  is the nearest perfect square to  5.  Subtract and bring down the next two digits, giving  147.

          Double the digit  2  to get  4.  Think:  ? × 4?  gives  147  or nearest possible value.  We have 3×43 = 129.

          Subtracting and bringing down the next two digits gives  1856.  Replicate the digit  4  on the left and double the digit  3,  giving  46.

          Now think:  ? × 46?  gives  1856  or nearest possible value.  It turns out that  4 × 464 gives exactly  1856.  We are done!  The square root of  54 756  is  234.

How does it work?

          This relies on the algebraic identity  (10a + b)² = 100a² + 20ab + b², the right-hand expression is equal to   100a² + (20a + b)b.  For example, at stage 4, we have  a = 23,  b = 4  and  (20a + b) = 464.

          Did you learn something today?

[S1_20160227FZCK] Factorisation by Chunking

Problem / Question

 

Solution

Commentary

     Here I illustrate the usefulness of chunking to factorise (AmE: factor) an algebraic expression.  Observe that  3a – 2b  is a repeated part of the expression.  I call it a “chunk”.  To make it clear, I rewrite  (3a – 2b)²  as   (3a – 2b)(3a – 2b)  so that you can see it as two copies of the same chunk.  I highlight in yellow one copy of  (3a – 2b)  from each of 
(3a – 2b) (3a – 2b)   and  -3(3a – 2b).  The remaining stuff are highlighted in blue and green.  Take out the yellow chunk as common factor by writing it out on the left in the third line, shown in yellow.  You can pull out the common factor by writing it out to the right if you want, but here I chose to put it on the left.  The result would be equivalent anyway.  Once you have written out the common factor,  you write out the other stuff (shown highlighted in blue and green) into another other bracket.

     Once you understand how it works, you can actually do the second line mentally and write down the answer straightaway.  Chunking is a very useful technique in mathematics.  Here are some more examples of the technique of chunking: (1), (2), (3).

Heuristics Used

H04. Look for pattern(s)

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* Lower Secondary Mathematics (Sec 1 ~ grade 7)

* GCE ‘O’ Level “Elementary” Mathematics

* other syllabuses that involve algebra and factorisation (factoring)

* 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_20160117AFNS] Much Ado About Nothing?

Problem

 

Introduction

     Assuming no typing errors, this is a tricky Secondary 2 question involving an equation with algebraic fractions.  How to solve it?  How to present the solution?

Solution

Remarks

[1]   Once the LHS expression has no meaning, it would not even make sense to continue.

[2]   This is a proof by contradiction type of argument.

It turns out that not all algebraic equations are soluble (or solvable).  This problem is one case in point.  The “unknown”  m  cannot be ⁵/₂ because that would make the expression undefined.  But if you substitute any other value, you always end up with nonsense like “15 = 0”.  So no matter what, there is no solution.  In other words, there is no value of  m  that you can substitute into the equation that makes it a true statement.

Heuristics Used

H05. Work backwards

H08. Make suppositions

Suitable Levels

* Lower Secondary Mathematics (Sec 2 ~ grade 8)

* GCE ‘O’ Level “Elementary” Mathematics

* other syllabuses that involve algebra

* any learner who is interested in algebra

[OlymLSec_20160118PPPC] A Square Proof by Contradiction

Question

Explanation

     If  a + b = 11,  then  2ab = (a + b)² – (a² + b²) = 121 – 100 = 21.  But  2ab  is an even number, whereas  21  is odd.  This is a contradiction.  So (B) is impossible.  ©

Remarks

     Short and sweet isn’t it?  This uses the square-of-sum identity   (a + b)² = a² + 2ab + b².  I used the tactic of assuming the answer is correct  [H08]  and showing that this leads to something nonsensical [H05].  So the original assumption must be wrong.  This is called “proof by contradiction” or reductio ad absurdum (in Latin).

     By the way, the correct answer option is (E) from the Pythagorean Triplet   8² + 6² = 10²  with  {a, b} = {8, 6}.  The question seems to be taken from some Kangaroo mathematics competition.

Heuristics Used

H05. Work backwards

H08. Make suppositions

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics competition

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

* other syllabuses that involve whole numbers and Pythagorean triplets

* any precocious or independent learner who loves a challenge

[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

[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

[S2_20151122IXBQ] A Bonus for your Index Fun?

Question

Introduction

     This is a “bonus” question which most likely came from an Integrated Programme (IP) school.  It really tests the wits of students’ knowledge of indices and problem solving tactics.  I present two solutions without the use of logarithms.  The first solution uses reciprocal indices and the matching of the base ^a/_b.  In the second solution, we match up the indices to  xy.

Review of Important Laws of Indices

Solution 1

 

Solution 2

Remarks

     No logs from any forest were harmed in the process of making this blog post.  J

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* challenge for Lower Secondary Mathematics (Secondary 2)

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

* other syllabuses that involve algebra and indices

[AM_20151119ISSD] Differences of Squares Hiding under Square Roots

Question

Introduction

     Here is another question involving surds.  As we know, surds are literally absurd, because they are irrational.  How to we do this one?

Strategy

Solution

Remark

     Reflect: What did you learn from solving this question?

     For another example of using the difference of squares formula, please look at this article.

Heuristics Used

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* GCE ‘O’ Level Additional Mathematics (indices / surds)

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

* revision for IB Mathematics HL / SL

* other syllabuses that involve indices and/or surds
* any precocious or independent learner who is interested

[AM_20151117ISCP] Estimating a Crazy “Prosperous” surd to the Nearest Integer

Question

Introduction

     To the Chinese, the number  8  (八)  is considered to be auspicious, because it sounds like “prosper” (发) in the various Chinese languages/dialects.  But the LHS expression featured above seems too prosperous for comfort.  There is an explosion of  8s  coupled with eighth roots.  How to even handle that?

Strategy

     As usual, often one good tactic is to look for patterns or chunks.  [H04]  Can you see the sub-expressions containing the eighth roots (highlighted in green and pink)?

Do you notice any similarities between the two chunks?  Do you notice any difference(s)?

If we call the green chunk  a  and the pink chunk  b,  we can remove the roots by taking the eighth powers.  Then we get whole numbers, which are less complicated.

Solution

Remark

     The crux of the problem is the factorisation of  a⁸ – b⁸.  It is based on repeated application of the difference of squares    X² – Y² = (X + Y) (X – Y)  formula which schools expect students to know.
     For another example of using the difference of squares formula, please look at thisarticle.

Heuristics Used

H04. Look for pattern(s)

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
* “IP Mathematics” so called
* challenge for students taking GCE ‘O’ Level Additional Mathematics
* other syllabuses that involve surds
* precocious kids who like to test themselves

[S2_20151112CSSR] Change of Subject and Square Roots

Question

Introduction

     In this secondary 2 (approx. grade 8~9) algebra question, we are essentially asked to make  k  the subject.  This means that through a series of algebraic manipulation, we arrive at a final equation in which  k  appears alone on one side (conventionally the LHS) and all the other “stuff” on the other side.  This question looks challenging firstly because k  appears in more than one place and then also we need to deal with the square root.

The Square Root

     Note that the principal square root (or simply “the square root”) is by the modern definition non-negative i.e. zero or positive.  Of course, what goes under the square root must also be non-negative, otherwise it would not even make sense as a real number. 

     So observe that in the given equation, the RHS is non-negative.  Hence the LHS which is just  k,  must be non-negative.  It is tacitly understood that  3a – k² > 0  for the square root to make sense.  

     To get rid of the square root, we can square both sides of the equation.  After that we bring all the terms with  k  to the LHS.  Finally, we need to “unsquare” both sides by taking square roots.  The solution takes only about 5 steps, as shown below.

Solution

Remarks

     There is no need for  ±  in the final line because we already know that  k  is non-negative (k  is zero or positive).  Here is something that students and even teachers / tutors can get confused over.  Modern mathematics tends to take a “function” approach in which each expression can take only a single unambiguous value.  “ Ö ”  may be regarded as a function with the non-negative reals as domain and the non-negative reals as range.  Although in traditional parlance, we say things like the “square roots” of  9 are 3 and -3, once you see the  “ Ö ”  symbol (apart from the “±”), it is the result of a calculation and the result is by definition non-negative.

     Another thing that people get confused over is: What about the ± symbol ?  Note that  ±  by itself is actually meaningless!  Something like  ±3  is just a short-cut for lazy people to say “the answer is 3 or -3”  (and we tend to be lazy, don’t we?).  But this is the result of solving an equation like “x² = 9” when  x  is a real number with no other restrictions.  This equation has two roots: 3 and -3.  If  x  is known to be non-negative, then  x = 3  is the only solution.  Of course solving an equation involves calculation.

     So what is the difference between solving and mere calculation?  Solving is a process of finding values for unknowns and it usually involves a more than one step and it may include calculation.  When you see something like  Ö9   you are just calculating, and there is only one answer.  But when you see something like “Find the values of  x  such that  x² = 9” you are solving.  There is an unknown (e.g.  x)  and you are supposed to find number(s) that you can plug into  x  to satisfy that equation.  After you calculate  Ö9 = 3, you still need to write “x = 3 or x = -3” or its short form “x = ±3”.  I hope this clears the confusion.

     Remember “Ö something non-negative” Þ one non-negative answer.  “find / solve something” Þ maybe more than one answer.

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

* Lower Secondary Mathematics (Secondary 2 » Grade 8/9)

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

* other syllabuses that involve algebra