[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

[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

[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