[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

[S2_20151107SLFC] Simultaneous Linear Equations with Fractional Coefficients

Question 

Introduction

     Somebody said, “Dear Algebra, please stop asking me about your  x.  She is not coming back.  Don’t ask me  y!”

     Here we have here linear equations that appear to be more tricky than the usual fanfare.  This is because the coefficients of the unknowns  x  and  y  are fractions.  In school, students learn to solve these using the method of substitution and the method of elimination.  They tend to prefer to do it by the former method, because psychologically it seems easier to accept.  However, the latter matter is generally more effective and yields a shorter solution.  This question is like a fly trap for those who prefer the method of substitution.  You make either  x  or  y  the subject from one of the equations, and then substitute that into the other equation.  This yields a complicated algebraic fractions within algebraic fractions.  The more complicated your solution is, the higher chance there is for making careless mistakes.  It is a good idea that students get out of their comfort zones and adopt a new skill.  So how do we do it?

The smart tactic

     First, we need to clear away the fractions by multiplying through with the LCM of the denominators appearing in each equation.  For the first equation, LCM(4, 8) = 8.  Since  4  is a factor of  8,  8  is like a giant that absorbs the number  4.  OK, so we multiply the first equation through by  8.  For the second equation, we multiply every term by LCM (3, 2) = 6

Solution

Remarks

     As you can see, equations [3] and [4] both contain – 3y  and this can be eliminated via subtraction.  So the value of  x  comes out easily.  Now once  x  is known, one can find the  y!

For another example of simultaneous equations, please refer to this article.

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)

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

* other syllabuses that involve algebra, linear simultaneous equations

[S2_20150520SERC] Simultaneous Equations with Reciprocals

Question

Introduction

     This is a secondary 2 (» grade 8) problem that involves linear simultaneous equations in two unknown.  It looks non-linear and indeed if you try to solve for  x  and  y  directly,  you might get into a big mess with extraneous solutions to boot.  What should our approach be?

Observations

     It is always a good idea to stare at the question for a little while longer before jumping in to try to solve it.  Let us (mentally) reformat the equations a little bit.

     Do you notice any repeated chunks?  Chunking is very useful.  Let use substitutions for those chunks to simplify the equations.

Solution

Final Remarks

     To recap: It is a good learn to make observations first before attempt to solve the problem.  If you see any repeated chunks, it is a good idea to use substitution to simplify the problem.  Once the reciprocals have been substituted, we try to use elimination (which is usually the more effective method).  Also, try as far as possible to avoid fractions.  By multiplying equation [2] by 5,  we get – 30Y, which can be cancelled if we multiply  15Y by 2.  Thus  Y  is eliminated.  The problem becomes easy from this point onwards.
     For another example of simultaneous equations, please refer to thisarticle.

HeuristicsUsed

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

* Lower Secondary Mathematics (Secondary 1)

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

* other syllabuses that involve algebra and simultaneous linear equations

[S2_20150402RLD] Fifty balls left behind

Question

This problem can be solved with Primary School knowledge using ratios.  The famous Singapore bar diagramming method can be used to model the situation, but I prefer my own Distinguished Ratio Units.  The former method is good for visualisation for beginners, while the latter is faster if you want to solve it quickly without fussing around drawing the perfect diagram.  My DRU method is also visual in another way, and it works with big numbers as well as small numbers.  Alternatively, this can be solved using algebra via simultaneous equations.

Solution 1 (Using my Distinguished Ratio Units method) [H02]

Explanation: Since the number of white balls is a multiple of  3,  I let “triangle” 3 represent the number of white balls.  I let “heart” 1 represent the number of red balls.  There are 50 more white balls than red balls.  [H04]  When the white balls are removed three at a time, the number of groups of three would be one-third of the number of white balls, i.e. 1 triangle unit.  [H04]  This number is less than the number of red balls (1 heart unit) by 50.  So 1 triangle unit plus 50 gives 1 heart unit.  [H04]  Following on from the heart to the “triangle” 3, one realises that 2 “triangle” units is the same as  100.  [H05]  So one triangle unit is  50.  [H11]  From here we can solve the rest of the problem.

Solution 2 (using Algebra)  [H13, H05]

                     w = k + 50 = 3h             –––––––––– [1]

                     r  = k         =   h + 50     –––––––––– [2]

for some unknown  k  and  h.  And then [1] – [2] gives  [H10]

                     w – r = 50 = 2h – 50

so                      2h = 50 + 50

                            h = 50

This quickly leads to

                           w = 150

and                       r = 100.

Heuristics Used

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

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Primary School Mathematics (“Ratio”)

* Lower Secondary School (“Simultaneous Linear Equations”)

* other syllabuses that involve ratio or algebra