[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

[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

[Pri20150308CSR] Charitable Savings Ratios?

Question

Introduction

     Here is another one of those Singapore Mathematics problems that are two-variable simultaneous equations in disguise.  The key to solving this question quickly is to exploit the fact that the amount of donations are the same in this case.

Solution

     First, read the question and translate the information into a diagram [H02. Use a diagram / model].  I use different shapes (e.g. circle and square) to envelop the different types of units.

     Since we have ‘-80’  for both Sharon and Ryan [H04. Look for pattern(s)], we may deduce that 1 ‘circle’ unit  (5 minus 4 ‘circle’ units) is equal to 3 ‘square’ units (10 minus 7 ‘square’ units).  That means 5 ‘circle’ units is 15 ‘square units’.  [ H10. Simplify the problem]

     By comparison again, we realise that 5 square units is 80 [H05. Work backwards] and hence 15 ‘square’ units is 240.

Ans: Sharon’s savings was $240 at first.

Check

Before    $240   $192      5 : 4

After      $160   $112     10 : 7

Solution makes sense.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

[Pri20150306APC] Apples-with-Pears Comparison

Question

Introduction

     This question involves money and looks rather challenging, because there are a few unknown quantities.  To solve it, we use the concept of unit costs, so that we can make an “apples-to-apples” ... er ... I mean “pears-to-apples” J comparison of the prices.

Solution

     Since absolute dollar amounts are given, we can quickly solve [H11. Solve part of the problem] for the total costs of pears and apples as follows [H02. Use a diagram / model]:-

     Now we know that the total cost (in dollars) of pears is 45 and for the apples it is 50 (5 more than 45).  Although we do not know the absolute numbers of pears and apples, we know their ratio.  So let us write these down as, say, ‘square’ units. 

     Dividing the total costs by the numbers gives the unit costs, which we know only in ratio terms.  So let us use, say, ‘circle’ units to denote these.  But we know that the unit cost (in $) of an apple is 0.50 less than that of a pear.  And that is equivalent to 5 ‘circle’ units.  From here [H05. Work backwards], we quickly work out the cost of a pear (15 ‘circle’ units) as $1.50.

     Ta da!

Heuristics Used

H02. Use a diagram / model

H05. Work backwards

H11. Solve part of the problem

[Pri20150306COH] The Cards of Hearts?

Question

Introduction

     For this question, I shall illustrate my technique of Distinguished Ratio Units to model the situation.  Read the question carefully and translate the information into a diagram [ using heuristic H02 ] like below:-

     I used three different shapes (circle, triangle and square) to envelop the numerical counts of the different kinds ratio units. Although we do not know how many circle units’ worth of cards Kelly had at first, we quickly notice that  9 circle units are equivalent to 3 triangle units, so that one triangle unit is the same as 3 circle units.  So Kelly had 2 circle units’ worth of cards [ heuristic H05 ], as depicted below:-

 

This allows us to answer part (a) of the question already, namely that the required ratio is 10 : 2 i.e. 5 : 1.  With different types of units, it is difficult to compare things.  However, note that in the exchange of cards, the total number of cards remains constant.  Taking the LCM of 12, 4 and 11 which is 132, we can change all the ratio units to a common type of unit [ H09], say ‘heart’ unit, based on the total being 132 ‘heart’ units.  To do that, we can multiply the numerical counts in columns 1 & 2 by 11, multiply column 3 by 33 and multiply column 4 by 12.  This is what we would get:-

     To answer part (b) of the question, we actually do not need to bother about columns 2 and 3.  Just focus on columns 1 and 4.  From column 4 we observe that 84 minus 48 which is 36 ‘heart’ units gives 72, so one ‘heart’ unit corresponds to 2.  The number of cards won by Kelly can be found by comparing the 84 ‘heart’ units and 22 ‘heart’ units highlighted in yellow.


  That means 62 ‘heart’ units and that corresponds to 134.  And we are done!

Answer (a)   5 : 1

              (b)   134

Heuristics Used

H02. Use a diagram / model

H05. Work backwards

H09. Restate the problem in another way

Thinking Back

     In this question, we have used Distinguished Ratio Units to model the given situation.  We worked backwards to find that Kelly’s initial holdings were worth 2 circle units.  Then we converted everything to a common unit (‘heart’ unit) based on the constant total of 132 heart units.  Once again, I © hearts!