Sunday, 19 October 2014

Australian Mathematics Competition 2014

Dear S1-02

Here's the summary of your class' performance:

DistinctionMoon Gijoo
Fyodor Lee
Owen Tee Hao Wei
Janice Tan Sze Hui
Wong Kang
Javier Yeo
Joel Tio Jin Hon
Joshua Yew Jek Yong
Hong Dominic
Oon Chwen Yueh Celest
Kay Sandy Maung
Joshua Chua Tze Ern
Lew Jiajun
Low Tjun Lym
Zhu Zhanyan
Elgin Ng
Liew Yong Jun Javier
Khoo Bo Si
Christopher Yong Wei Jie
Justin Lim
Yoke Xuan Woon

Tuesday, 7 October 2014

Concluding Year 2014...

Dear S1-02

It has been a pleasure to be with you on your maiden learning journey in SST :) 
The closure of this academic year would a good time for us to reflect and plan for the following year.

We would like to hear your experience as the feedback would be taken into account when we shape your learning experience, as well as your juniors'.

Click HERE to access the Feedback Form.

Saturday, 27 September 2014

2-to-3 Quiz (Revision) Construction

Question 1

To construct the figure:
(i) Measure and draw line AB using ruler.

(ii) Measure and draw angle ABC using protractor.

(iii) Stretch the 2 arms of the compass to measure 5.2 cm and mark an arc on the line BC. Label it point C.

Here are 2 ways of drawing the parallel line DC
(iv) Use a protractor to draw angle BCD
(iv) Use a set-square and ruler to draw line CD

(v) Stretch the 2 arms of the compass to measure 3.8 cm, place the sharp end at "C" and mark an arc on the line DC. Label it point D.

(vi) Join A and D to complete the figure.

Question 2

Three properties to take note of:

  • Diagonals of the rhombus bisect each other
  • Diagonals of the rhombus are perpendicular to each other
  • Opposite angles are equal

[Someone asked...] The difference between Approximation and Estimation...

Estimation involves calculation, whereas Approximation is about rounding off a number.
  • The rounding off of a number can be carried out before or after the calculation.

Consider the following scenario:
The length of the door is 1.95 m and its height is 2.06 m (assuming that the dimensions given are exact).

(a) Find the area of the door and round the answer off to 2 significant figures.
The actual area of the door is 1.95 m x 2.06 m = 4.017 m^2.
Answer = 4.0 m^2 (to 2 sig fig) [Ans]

The step to round off a number (to present the final answer) is an approximation.
There is no mention of estimation in question (a), so we use the given values to compute the area first, then round off as required.

(b) Estimate the area of the door to 2 significant figures.

The purpose of doing 'estimate' or 'estimation' is to enable us carry out calculation quickly and when we don't need an exact value. A rough figure should give us a sense of the actual value. It should not be too far off from the actual value. 

To estimate, we will round the numbers off to the required "degree of accuracy" (in this case, 2 significant figures) first before doing any calculation. In other words, we are using the approximated numbers to do calculation.

Area of the door ≈ 2.0 m x 2.0 m <<<<< This is the 'estimation' step when the approximated values are used for calculation
= 4.0 m^2 [Ans]

Since 1.95 m ≈ 2.0 m (2 sig fig) and  2.06 m ≈ 2.0 (2 sig fig)

When doing Estimation, we round off the values based on the degree of accuracy given.

However, there is an exception when dealing with roots because by rounding it off to 1 SF, 2 SF, etc. may not land itself nicely to be rooted.

To estimate square root of 26.77, it does not help us to figure out what number it's 'near' to if we round it off to 1 SF (i.e. 30) or 2 SF (i.e. 27). 
However, we know that 25 (which is a perfect square) is quite close to it; so, we will estimate it to be square root of 25, and get the answer "5". 
If you key square root 26.77 in the calculator, your answer should be quite close to 5.

To extend/ combine the above.... 
If the question asks: Estimate 150.5677 x "square root of 8.343" to 2 sig fig
Then we'll have 150.5677 x "square root of 8.343" ≈ 150 x "square root 9" 
= 150 x 3
= 450 [Ans]

150.5677 when rounded off to 2 sig fig = 150
square root of 8.343 will be square root of 9 (which is the closest perfect square).

Wednesday, 24 September 2014

Revision (2-to-3 Quiz): Linear Graph Question 1

Refer to Handout 2-to-3 Quiz (10) Question 1
For discussion...

Revision (2-to-3 Quiz): Linear Graph Question 1 - It should be...


(a) Recall, on the graph paper, each 'big' square with the 'darkened' green link measures 2 cm x 2 cm.

(b) Substitute the given values of x and y to find the unknowns.

(c) Remember to check the domain for the plot.

(d) Things that must be present when plotting the graph:

  • label x and y axes
  • markings on the axes must be clearly indicated/ written (including the origin)
  • label the line with the given equation
  • plot the points with crosses clearly


(e) To read points, the working is represented by the dotted lines.

Sunday, 21 September 2014

Content Pages - 2nd Semester

Content Page for the individual topics (2nd semester) are now available via the GoogleSite.
  • Click HERE to view Summary [hardcopy will be given]
  • Click HERE to access the individual content page

[Someone asked...] When simplifying expressions... do we factorise?

When we are asked to simplify expressions, we would usually expand to work on the like terms. However, there are occasions when we need to look out for "patterns".

Here are some possible scenarios:

1. If no variables are involved in the denominator of any algebraic fractions, we can just expand to simply.

2. If there is variable in the denominator, we'll need to see how complex it is.

In the following example, you notice that the denominator is very simple - it has only a numerical value multiplied to "x".
  • Now, we simply need to simplify the numerator like what we did (usually). 
  • When we reach Line 2, then check if there's any common factor amongst the terms in the numerator. 
  • If there is, 'take out' the common factor, just in case it can be 'cancelled' with the denominator.
  • In this case, you notice that in Line 3, we have "2x" in the numerator, which we can reduce further with the denominator "6x", resulting in what we see in Line 4.

3. In the following, you notice that the denominator seems a bit 'more complex', and there seems to be a common factor between the terms.
  • In addition, the numerator has 2 'brackets', hinting a possibility of 'reducing' with some terms in the denominator. So, do not expand the 'brackets' in the numerator yet. 
  • In Line 2, we 'take out' the common factor of the denominator.
  • Now check if there's any common factor between the numerator and the denominator., You would notice (x+2) is the common factor. 'Cancel' them and we'll get Line 3. 
  • Now, simplify and the final answer is as shown in Line 4.

[Someone asked...] How to form that INEQUALITY?

Someone asked: How to form the inequality of the following problem
(Fairfield Methodist Secondary End-Of-Year 2013 Paper 2)

Here's how you could approach the problem step-by-step:

End-of-Year Revision - Suggested Solutions

Please refer to the GoogleSite for the suggested solution/ working for the revision papers.

Friday, 19 September 2014

Term 4 Week 2/3 Consultation

Dear S1-02

As you should have moved into the final stage of the exam preparation in Term 4 Week 2, instead of conducting remedial lessons, I've created several consultation slots so that you can see me to go areas that you are not sure, e.g. questions in the revision papers.

  • 22 Sep (Monday) 2.30 pm to 4.30 pm
  • 24 Sep (Wednesday) 1.45 pm to 2.15 pm
  • 29 Sep (Monday) 11 am to 12 noon
  • 30 Sep (Tuesday) 12 noon to 1 pm
You may give me a call (@419, outside the staff room) or send me an email alert.
Venue: Outside Staff Room

Tuesday, 9 September 2014

[Someone asked....] Number of sides of a polygon if the exterior angle is a reflex angle...?

How to find "The number of sides a regular polygon has if its exterior angle is ≥180º?

In the syllabus, we focus on convex polygons.
One of the properties that all convex polygons have would be the sum of Interior and Exterior angles is 180º.
Hence, the size of each exterior angle cannot be more than 180º.

Below are 2 examples of polygons with exterior angles being acute angle and obtuse angle.

Thursday, 4 September 2014

Support Programme during Sep Holidays

Hi Everybody

I'll be in the school (during the Sep Holidays) to run the Maths Support Programme for some of your classmates on
  • 8 Sep 2014 (Monday)
  • 9.30 am to 11.30 am
  • Venue: Info Hub (Level 4/5)
  • Attire: School uniform
You may turn up if you wish to... 

What are we doing:
Attempt a Practice Paper from another school; consultation if you would like to clarify your doubts face-to-face.

If you intend to turn up, things to bring:
1. Long Ruler, Eraser & Sharp pencil - for graph plotting; Graph paper
2. Calculator
3. Writing materials 

Do let me know if you are unable to turn up so that I would prepare enough materials for you.

Who's attending the support programme:


Monday, 1 September 2014

Pythagoras' Theorem: How would Romeo meet Juliet?

[Post activity discussion]

Some of you enquired the "complete" set of solution (after the class discussion) - i.e. the various possibilities that arise from various assumptions.

Here are the possibilities and assumptions that we need to be clear before solving the problem:

In scenario 2, some of you suggested using nets (something you learnt in primary school).
Here, it's good to draw the net and write the dimensions clearly to ensure the correct numbers are used for calculation.

Look, there are 2 possible 'nets' that could give the shortest path.
Do NOT assume the 2 nets give the same "diagonal" length.
[Do not apply "seeing is believing"]
Always compute the numbers to check.

The 3rd scenario was the one that we spent a significant amount of time to discuss in class - Refer to the "Romeo & Juliet" box that I brought to the class to help you visualise better.
Here, you would have to apply Pythagoras Theorem twice.