## NUS Math Ranked among Top in Asia

In the latest Quacquarelli Symonds (QS) World University Rankings by Subject (2014), NUS Math is ranked among the best mathematics departments in Asia.

Chinese students typically outperform U.S. students on international comparisons of mathematics competency. Paradoxically, Chinese teachers receive far less education than U.S. teachers–11 to 12 years of schooling versus 16 to 18 years of schooling.

Studies of U.S. teacher knowledge often document insufficient subject matter knowledge in mathematics. But, they give few examples of the knowledge teachers need to support teaching, particularly the kind of teaching demanded by recent reforms in mathematics education.

This book describes the nature and development of the “profound understanding of fundamental mathematics” that elementary teachers need to become accomplished mathematics teachers, and suggests why such teaching knowledge is much more common in China than the United States, despite the fact that Chinese teachers have less formal education than their U.S. counterparts.

## Make (y) the subject of the formulae (O Level Maths Tuition)

Question:

Make (y) the subject of the formulae

(1) $\displaystyle x =\frac{y}{y+1}$

Solution:

$\displaystyle \frac{x}{1}=\frac{y}{y+1}$

Cross multiply,

$xy+x=y$

$xy-y=-x$

$y(x-1)=-x$

$\displaystyle y=\frac{-x}{x-1}=\frac{x}{1-x}$

## Ten Year Series: How many questions or papers to practice for Maths O Levels / A Levels?

This is a question to ponder about, how many questions or papers to practice for Maths O Levels / A Levels for the Ten Year Series?

If you searched Google, you will find that there is no definitive answer of how many questions to practice for Maths O Levels/ A Levels anywhere on the web.

For O Level / A Level, practicing the Ten Year Series is really helpful, as it helps students to gain confidence in solving exam-type questions.

Here are some tips about how to practice the Ten Year Series (TYS):

1) Do a variety of questions from each topic. This will help you to gain familiarity with all the topics tested, and also revise the older topics.

2) Fully understand each question. If necessary, practice the same question again until you get it right. There is a sense of satisfaction when you finally master a tough question.

3) Quality is more important than quantity. It is better to do and understand 1 question completely than do many questions but not understanding them.

Back to the original query of how many questions or papers to practice for Maths O Levels / A Levels for the Ten Year Series, I will attempt to give a rough estimate here, based on personal experience.

5 Questions done (full questions worth more than 5 marks) will result in an improvement of roughly 1 mark in the final exam.

(The 5 Questions must be fully understood. )

So, if a student wants to improve from 40 marks to 70 marks, he/she should try to do 30×5=150 questions (around 7 years worth of past year papers). Repeated questions are counted too, so doing 75 questions (around 3 years worth of past year papers) twice will also count as doing 150 questions. In fact, that is better for students with weak foundation, as the repetition reinforces their understanding of the techniques used to solve the question.

If the student starts revision early, this may work out to just 1 question per day for 5 months. Of course, the 150 questions must be varied, and from different subject topics.

 Marks improved by Long Questions to be done Approx. Number of years of TYS OR (even better) 10 50 2 1 year TYS practice twice 20 100 4 2 year TYS practice twice 30 150 6 3 year TYS practice twice 40 200 8 4 year TYS practice twice 50 250 10 5 year TYS practice twice

This estimate only works up to a certain limit (obviously we can’t exceed 100 marks). To get the highest grade (A1 or A), mastery of the subject is needed, and the ability to solve creative questions and think out of the box.

When a student practices TYS questions, it is essential that he/she fully understands the question. This is where a tutor is helpful, to go through the doubts that the student has. Doing a question without understanding it is essentially of little use, as it does not help the student to solve similar questions should they come out in the exam.

## Challenging Geometry E Maths Question — St Andrew’s Sec 3 Maths Tuition Question

Question:

ABCD is a rectangle. M and N are points on AB and DC respectively. MC and BN meet at X. M is the midpoint of AB.

(a) Prove that $\Delta CXN$ and $\Delta MXB$ are similar.

(b) Given that area of $\triangle CXN$: area of $\triangle MXB$=9:4, find the ratio of,

(i) DN: NC

(ii) area of rectangle ABCD: area of $\triangle XBC$. (Challenging)

(ii) 20:3

Suggested Solutions:

(a)
$\angle MXB=\angle NXC$ (vert. opp. angles)

$\angle MBX = \angle XNC$ (alt. angles)

$\angle BMX = \angle XCN$ (alt. angles)

Therefore, $\Delta CXN$ and $\Delta MXB$ are similar (AAA).

(b) (i) $\displaystyle\frac{NC}{BM}=\sqrt{\frac{9}{4}}=\frac{3}{2}$

Let $BM=2u$ and $NC=3u$

Then $DC=2\times 2u=4u$

So $DN=4u-3u=u$

Thus, $DN:NC=1u:3u=1:3$

(ii)

We now have a shorter solution, thanks to a visitor to our site! (see comments below)

From part (a), since $\Delta CXN$ and $\Delta MXB$ are similar, we have $MX:XC=2:3$

This means  that $MC:XC=5:3$

Thus $\triangle MBC:\triangle XBC=5:3$ (the two triangles share a common height)

Now, note that $\displaystyle\frac{\text{area of }ABCD}{\triangle MBC}=\frac{BC\times AB}{0.5 \times BC \times MB}=\frac{AB}{0.5MB}=\frac{2MB}{0.5MB}=4$

Hence area of $ABCD=4\times\triangle MBC$

We conclude that area of rectangle ABCD: area of $\triangle XBC=4(5):3=20:3$

Here is a longer solution, for those who are interested:

Let area of $\triangle XBC =S$

Let area of $\triangle MXB=4u$

Let area of $\triangle CXN=9u$

We have $\displaystyle\frac{S+9u}{S+4u}=\frac{3}{2}$ since $\triangle NCB$ and $\triangle CMB$ have the same base BC and their heights have ratio 3:2.

Cross-multiplying, we get $2S+18u=3S+12u$

So $\boxed{S=6u}$

$\displaystyle\frac{\triangle BCN}{\triangle BDC}=\frac{3}{4}$ since $\triangle BCN$ and $\triangle BDC$ have the same base BC and their heights have ratio 3:4.

Hence,

$\begin{array}{rcl} \triangle BDC &=& \frac{4}{3} \triangle BCN\\ &=& \frac{4}{3} (9u+6u)\\ &=& 20u \end{array}$

Thus, area of $ABCD=2 \triangle BDC=40u$

area of rectangle ABCD: area of $\triangle XBC$=40:6=20:3