# Bridging the Gap- Modeling and Methods in Maths(Part 1)

Maths is a subject which requires a lot of practice in order to be familiar with it. There are many different concepts which one need to master and formulas to memorise in order to complete the array of questions within the given time.

In the last post I gave an overview of the components tested in a Math’s paper.  In this post I will be showing how Long Answer Questions can be solved using different methods.

**These are actual questions which are taken from the homework worksheets that Monkey boy has brought  home and  I have worked through together with him.

My first inclination was to see if it was possible to draw a model to analyse the question.  After pondering on it for a while both of us found it more complicated to explain if we drew a model for it and eliminated it from the ” potential way to solve the question” list.

We decided to use the Guess/List and check method as there were two different unknowns which had to be identified.

We worked through the possible  chance of  what the two different lengths of the rectangle could be given ( X and Y) given that the square was of a certain length for each side.

 Length of Side Area X (at 75%) Y at ( +2) Area of Rectangle 10 100cm² 10-2 = 8 10 + 2 = 12 8 x 12 = 94 cm² 8 64cm² 8- 2 = 6 8 +3 = 10 6 x 10= 60 cm² 6 36cm² 6-2.5=4.5 6=2= 8 4.5 x 8 = 36cm²

As shown above I assumed that each side of the square was 10cm so the area of the square would be 100 cm2.

Using the projections that ” one side had to be decreased by 25%”  the possibility of 1 side of the rectangle would  be 10-2 = 8 cm.

Following through on the part ” increase the length of the other side by 2cm” I added 2 to 10 to get 12cm.  However when you multiplied both sides to find the area of the rectangle I got 8 x 12 = 94 cm2.

This is not correct as there has been a change in the total area from 100cm2.    to  94 cm2.

I slowly moved down the list of possible numbers till I got to 6 cm as the length of the side of the square  and that the area of the square was equal to the area of the rectangle.

That was how we found the answer for this particular question.

If you are anything like me you too may draw a blank after reading through the whole question. I had to re-read through it a couple of times just to process what the questions was asking.  I found the question really complicated.

I seriously don’t remember having to attempt such complex questions when I was in Primary 5 back then. The difficultly level of questions attempted in Maths has certainly increased a lot!!

To solve this sum we had to use the Working backward method.

We start off from the sentence

” David paid \$3 back to Cali”

If we divide the amount David should have paid in 2 parts

The amount Cali paid : The amount  Aini and Bala paid = 2u: 5u

Therefore if all the kids paid  and equal proportion of the present  each

2 u (x4) : 5 u (x4)

8 u : 20 u

Cost of the present = 28u

Each child would have to pay 7 u. However Calia paid  1u more as she paid part of David share so

8u- 7u = 1 u

1 u = \$3

28 u = \$3 x 28

= \$84

a) The price of the present is \$84

The amount of money Aini and Bala paid in total = 20 u (\$60)

Based on the fact that Aini paid \$10 more then Bala we find the amount the Aini paid by doing the following.  (60 + 10) / 2 = \$35

Bala pays \$10 lesser

\$35 – \$10 = \$25

(Checking – \$25 + \$35 = \$60)

Since each person originally needed to pay 7 u = \$21

David had to return \$25-\$21 = \$4 to Bala  and \$35-\$21 = \$14 to Aini

b) David returns \$4 to Bala

c) David returns \$14 to Aini.

In the next post I will go through the remaining two methods- Modelling and Looking at a pattern to show how they can be applied in problem solving.

Post in this series

9) Bridging the Gap- Upper and Lower Primary Science

10) Bridging the Gap-  Moving from Lower to Upper Primary

11) Bridging the Gap- Moving from Lower to Upper Primary ( Part 2)

# Sakamoto Method

Last week I wrote a post about Abacus. This week I am reviewing another Maths method that is for more advance and complicated as compared to Abacus. This method is more relevant for  Singapore Primary Maths as it teaches the child the concepts of how to interpret the questions and breaking it down into smaller and more manageable steps which are easier to solve.

Dr. H. Sakamoto the founder of the method bases his method on 3 simple steps

1. Grasping the limitation ( Understanding the data in a systematic way through analysis and tabulation)
2. Diagram ( drawing a pictorial representation of the variables in question simplifies and provides the solution easily)
[This is already currently being taught in all primary/elementary schools]

3. Writing (providing a systematic way of writing numerical equations to derive at the given answer)
[Also  already currently being taught in all primary/elementary schools]

Certain schools in Singapore , like Poi Ching a top notch school in my district,  implemented this method as an enrichment class for their P4 students in 2007.

I personally feel that his method is very comprehensive and will definitely aid a student in scoring and excelling in mathematics