Note: Method by Khong is basically algebra and can be used in most of the P6/S1 questions for solution. Not sure about others, but feedback from my child is that P6 Math teacher discourages use of algebra because it usually presents “one-liner” workings with very little method marks that can be awarded if answer turns out to be wrong. Of course, full marks will be awarded for correct ending answer. Although the decision depends on the proficiency of individual students applying algebraic equations, there are definitely instances where the more advanced students are penalised completely for carelessness in a “one-liner” equation forming. So the discouragement.

Maybe that is why algebra for solving problem sums was not the intention of PSLE syllabus. Although I must admit the many Heuristics methods (units and parts) are like the basics of algebra and the lines between the two start to blur. As kids/parents/teachers/tutors get more used to and become proficient with “u”, and begin to apply “u” like the “x” we are used to in algebra, that is when it has unwittingly evolved from P6 Heuristics to solving by algebraic equations.

So much for my long-windedness…  I feel the intention of the question was that each bundle is not separable (should mention), so rather than divide the cost to each piece, the purpose is to know LCM is 12 for bundle of 4 and bundle of 3, and to understand the concept that there is minimum of 12 pcs for equal quantity sets and work on from there. (like Ace Starling)

[just for interest: Suppose the difference is $15.30 instead $37.40. That is when the “bundle” is broken up (not likely intention of setter in PSLE) and best solved with the above solution by Khong. See below.]

Method A (continuing from Ace Starling)

$15.30/$3.40 = 4.5 (times of 12 pcs set)

54 pcs each (there are 13.5 sets of pen bundle of 4)

Method B (continuing from Khong)

0.95u = 2u / 3 + 37.40 15.30

2.85u = 2u +45.90

u=54 pcs each  (there are 13.5 sets of pen bundle of 4)

SAHMom & Khong,

Thank you for sharing with us your view. I tends to agree that many instances the model approach, units and parts methods are quite “closely related” to the basics of algebra and at times, the lines between them are blur. 🙄  Sometimes they can look quite “similar” if you look at both from different perspectives.

Usually there is more than one way to solve each question…. be it model approach, branching method, using tables, algebra . A student will usually choose the method he is most familiar with, even though it may not be the most “efficient” method. 😉  

Hi SAHMom,

Thank you for your feedback. Your comment is very true for Primary school pupils.

Most Primary school teachers are not teaching Algebraic method, in my opinion, may be the following reasons.

1.  They have to follow the MOE instruction method, Heuristic method. 
2. The marking scheme is only based on the Heuristic method. No mark will be given for those sentences not in the marking scheme, unless you get the right answer. But no guarantee that marks will be given, base on teacher to teacher.
3.  One method is enough to teach all pupils. Other methods will be a burden to them. In fact there are 1001 methods in the market.
4. Not every child understand Algebra. Our algebra only be taught in P6. Only basic substitution in taught. Other skill like moving across the equal sign, Negative numbers, are not taught.

I use this method for those who are weak in Heuristic Method. To them, that method is too abstract.

By writing in table form, they can ‘see’  the pieces of information. Also, they save the time to draw the Modals which they are very weak in. Of course basic skill practice are needed before they can use this method.

Not every child learn in the same way. Methods are just assisting tools. To learn a skill that can help them to solve the problems is our main target. In short, with no good mark means no good secondary school.

Just a piece of my opinion. Not to offence any Teacher in the world.

Hi,

This is a fractional question.

We shall first list out in table.

0.95u = 2u / 3 + 37.40

2.85u = 2u + 112.20

2.85u – 2u = 112.20

0.85u = 112.20

1u = 132

0.95u = 132 × 0.95 = 125.40

2u/3 = 2/3 × 132 = 88

125.40 + 88 = $213.40