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 which 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 impossible to draw 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.
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 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
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.
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