Understanding Maths requires one to be able to analyse the question and choose the most appropriate method to solve the question. To be able to decide on which method to use the child has to be able to identify the crux of the problem before he/she is able to slowly show the required workings to solve the problem.
Using the Model Method
From Primary 1 students are taught to draw models to solve Long Answer Questions. This is the method that is “promoted” as it is supposedly the easiest method to explain how you can derive the answer. ( And NO you are not allowed to even use Algebra to explain yourself)
Sammy and Gerald had an equal number of stamps at first. Sammy then gave away 20 of this stickers to his friend Winston, and Sammy Gerald bought another 12 stamps. In the end, Gerald had thrice as many stickers as Sammy. Find the number of stamps Sammy had at first.
Therefore 2 units → 20 + 12
1 unit → 32 2
1 unit → 16
Number of stamps Gerald had at first = 1 unit + 20
16 + 20
Gerald had 36 stamps at first.
If you notice the type of model drawn differs from the type of models drawn in lower primary. It is more complex and you are comparing more then two different items involving multiple steps to solve the question.
Looking at a Pattern
Identifying the Pattern
How do you solve this problem?
Firstly you would need to identify the pattern. Based on the diagram above you would notice that there are 3 different shapes – Square, Oval and Triangle. The pattern repeats itself after the 3rd shape.
Step 1- Find out the number of “groups of 3” which are possible from 1- 77
77 3 = 25.66 ~ 25 times ( 25 R 2)
25x 3 = 75
After calculating you should get “Oval” as your answer.
It does take a lot of practice to be able to solve a mathematical question easily. It’s fortunate that Maths is one of Monkey boy’s stronger subject and it doesn’t take him too much effort to understand the concepts which needs to be applied to the various questions.
How to you reinforce the learning of Maths for your kids?
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.
**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.
Square to rectangle
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
X (at 75%)
Y at ( +2)
Area of Rectangle
10-2 = 8
10 + 2 = 12
8 x 12 = 94 cm²
8- 2 = 6
8 +3 = 10
6 x 10= 60 cm²
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.
Multiple Part Long Answer 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
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.