[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

[OlymPri20150526] The Smallest Angle in a Special Trapezium

Question

Introduction

     This looks challenging because we do not seem to be given much information.  For example, we do not know the individual angles of the trapezium (American: trapezoid).  Does this mean that this puzzle cannot be solved?  Are we trapped by this trapezium?  What is the secret key that unlocks the problem?

Solution

     First sketch trapezium  ABCD.  Introduce point  F,  the mid-point of  AD.  All contructed lines and points are shown in grey (American: gray).  Then  DF = FA = BC. 

Now draw a line parallel to  BC  passing through  A  and intersecting  DC  at  E, say.  Note that ÐAED = ÐBCD  (corresponding angles).  Then  ÐADE + ÐAED = ÐADC + ÐBCD = 120°.  Hence the remaining angle  in  DDAE,  ÐDAE = 180° – 120° = 60°.  We can deduce this, even though we do not (currently) know the individual angles  ÐADE  and  ÐAED.  This is the key step.

From here, things get easier.  DAFE  is an isosceles triangle with  ÐAFE = ÐAEF = (180° – 60°) ¸ 2 = 60°.  So  DAFE  is in fact an equilateral triangle.  That means  FE = FA = FD.  So  DFDE  is an isosceles triangle.

Note that  ÐAFE  is an exterior angle of the triangle  DFDE.  If you know that exterior angle of a triangle is the sum of the interior opposite angles,  then from  ÐAFE = ÐFED + ÐFDE,  we easily see that  ÐADE = ÐFDE = 60° ¸ 2 = 30°.  If you do not know the theorem about exterior angles, you can still quickly work out that  ÐDFE = 120°,  and then use  ÐADE = (180° – 120°) ¸ 2  and arrive at the same conclusion.  ©

Remarks

     We can now see that actually  ÐBCD  = ÐAED  = 90°,  but we do not need to rely on that (or on accurate drawing) to deduce the answer.  From solving this question, we learn that even though we do not know the individual angles, by construction and using the sum of angles in a triangle, it is possible to solve for an important angle, namely  ÐAFE.  The rest of the solving uses isosceles triangles and equilateral triangles, which are part of the common repertoire of tactics.

HeuristicsUsed

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

H12* Think of a related problem   (Draw construction lines)

Suitable Levels
* Primary School Mathematics

* other syllabuses that involve area of triangles and circles

[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

[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

[OlymUSec_20150412BDP] Guessing Cheryl’s Birthday

Question

Introduction

     This “Primary 5 mathematics” (actually an upper secondary Olympiad) logic puzzle has gone viral.  It has been making its rounds in various forums in Singapore and overseas, stumping adults and children alike.  It is actually a parody of an old puzzle.  Can it even be solved?  It seems that there is no information given by each parties that we can exploit.  Actually there is!  In a subtle way ...

Solution

     In the beginning, everybody knows that Albert knows only the month and Bernard knows only the numerical day of the month.

     When Albert tells us “I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.” he is leaking out information (from his knowledge of the month) that the day of the month appears more than once and cannot be (June 18 or May 19).  Actually, the original phrasing is more like “If I don’t know when Cheryl’s birthday is, then Bernard does not know too.”.  The person who set this question merely changed the names of the people and the dates, without appreciating the subtle but crucial difference between a statement of fact and an implication (an “if ... then ... ” statement). 

     Ruling out June 18 and May 19, we also know that Albert knows that the birthday month is neither June nor May.  Otherwise, how would he have been so confident in saying that he knows Bernard would not know Cheryl’s exact birthday?  So we can eliminate those months.

     Bernard acknowledges the above state of affairs and the embedded hint.  With the choice narrowed down and with his knowledge of the numerical date, he now knows Cheryl’s birthday.  Since we know that Bernard knows Cheryl’s birthday, we know that it cannot be a numerical date that appears more than once (otherwise he would not have been able to know).  So we can cross out July 14 and August 14.

     Now Albert would telepathically thank Bernard for this helpful hint.  Because now he is able to deduce Cheryl’s birthday with his knowledge of the month.  That would mean that this cannot be a month with two candidate dates.  We blot out the August dates and see for ourselves the only remaining possibility.

Conclusion: Cheryl’s birthday is  July 16.

Remarks

     This puzzle was solved using the process of elimination and analysing our knowledge of what each party knows and can know.  Thus we successively narrow down the possibilities until the answer becomes obvious.  Here we learn that

     knowledge of other people’s knowledge can itself give us knowledge. 

This principle is actually employed in cryptology (the use of secret codes) which finds applications in fields like banking, the military (cf. interesting story of how the German Enigma code was broken in WorldWar II) and communications.  As an example, radio communication can tell the enemy of troop positions and warn of an impending attack, and that is why radio silence is imployed as a precaution.  Sensitive information in certain organisations is restricted on a “need to know” basis.

Suitable Levels
* Upper Secondary Olympiad
* other syllabuses that involve knowledge or epistemology
* application of mathematical principles in real life
* for all people interested in logic puzzles

[S1_20150408DST] Funky Bus between Towns

Question

Introduction

     This is quite a convoluted word problem for a Secondary 1 (roughly equivalent of grade 7) pupil in Singapore.  There are a lot of data given and lots of unknowns.  How shall we cope with this?  One good way I recommend is to use a table to organise the information.  [H02]   Many primary school teachers teach the Triangle Mnemonic for the relationship between Distance, Speed and Time.  This continues to be useful in secondary school.

To get Distance, cover “Distance” with your finger, and you get Speed ´ Time.

To get Speed, cover “Speed” with your finger, and you get Distance ¸ Time.

To get Time, cover “Time” with your finger, and you get Distance ¸ Speed.

Solution

Remember: Tables are very useful to help to organise information before you try to solve the problem.  They also help you to formulate the algebraic equations correctly and quickly.

Heuristics Used

H02. Use a diagram / model

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels
* Secondary 1 Mathematics
* precocious primary school pupils
* anyone, young or old, who enjoys an algebra challenge