Free Exam Paper: CHIJ Katong Convent Sec 3 Express Additional Mathematics

Katong Convent Exam Paper

(Sec 3 A Maths, with answers)

 

 

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.

Hope this information will help your revision.

Exam Time Management and Speed in Maths (Primary, O Level, A Level)

Time management is a common problem for Maths, along with careless mistakes.

For Exam Time management, here are some useful tips:
1) If stuck at a question for some time, it is better to skip it and go back to it later, rather than spend too much time on it. I recall for PSLE one year, there was a question about adding 1+2+…+100 early in the paper, and some children unfortunately spent a lot of time adding it manually.
2) Use a exam half-time strategy. At the half-time mark of the exam, one should finish at least half of the paper. If no, then need to speed up and skip hard questions if necessary.

To improve speed:
1) Practice. It is really important to practice as practice increases speed and accuracy.
2) Learn the faster methods for each type of question. For example, guess and check is considered a slower method, as most questions are designed to make guess and check difficult.

Sincerely hope it helps.
For dealing with careless mistakes (more for O and A levels), you may read my post on How to avoid Careless Mistakes for O-Level / A-Level Maths?

English: an animated clock
English: an animated clock (Photo credit: Wikipedia)

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.

recommended maths tuition geometry

(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)

[Answer Key] (b) (i) 1:3

(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

O Level Logarithm Question (Challenging)

Question:

Given \displaystyle\log_9{a} = \log_{12}{b} =\log_{16}{(a+b)}, find the value of  \displaystyle\frac{a}{b}.

Solution:

Working with logarithm is tricky, we try to transform the question to an exponential question.

Let \displaystyle y=\log_9{a} = \log_{12}{b} =\log_{16}{(a+b)}

Then, we have a=9^y=3^{2y}

b=12^y=3^y\cdot 2^{2y}

a+b=16^y=2^{4y}.

Here comes the critical observation:

Observe that \boxed{a(a+b)=b^2}.

Divide throughout by b^2, we get \displaystyle (\frac{a}{b})^2+\frac{a}{b}=1.

Hence, \displaystyle (\frac{a}{b})^2+\frac{a}{b}-1=0.

Solving using quadratic formula (and reject the negative value since a and b has to be positive for their logarithm to exist),

We get \displaystyle\frac{a}{b}=\frac{-1+\sqrt{5}}{2}.

If you have any questions, please feel free to ask me by posting a comment, or emailing me.

(I will usually explain in much more detail if I teach in person, than when I type the solution)

Hwa Chong IP Sec 2 Maths Question – Equation of Parabola

Question:

Given that a parabola intersects the x-axis at x=-4 and x=2, and intersects the y-axis at y=-16, find the equation of the parabola.

Solution:

Sketch of graph:

maths-tutor-parabola

Now, there is a fast and slow method to this question. The slower method is to let y=ax^2+bx+c, and solve 3 simultaneous equations.

The faster method is to let y=k(x+4)(x-2).

Why? We know that x=4 is a root of the polynomial, so it has a factor of (x-4). Similarly, the polynomial has a factor of (x-2). The constant k (to be determined) is added to scale the graph, so that the graph will satisfy y=-16 when x=0.

So, we just substitute in y=-16, x=0 into our new equation.

-16=k(4)(-2).

-16=-8k.

So k=2.

In conclusion, the equation of the parabola is y=2(x+4)(x-2).

Sec 2 IP (HCI) Revision 1: Expansion and Factorisation

This is a nice worksheet on Expansion and Factorisation by Hwa Chong Institution (HCI).

There are no solutions, but if you have any questions you are welcome to ask me, by leaving a comment, or by email.

Hope you enjoy practising Expansion and Factorisation.

The worksheet may be downloaded here:

Expansion and Factorisation by Hwa Chong Institution (HCI)

 

If your child needs help in Maths, please feel free to contact us for Maths Tuition. 🙂

Why is e^(ln x)=x? (O Level Math/ A Level Math Tuition)

Why is \boxed{e^{\ln x}=x}?

This formula will be useful for some questions in O Level Additional Maths, or A Level H2 Maths.

There are two ways to show or prove this, first we can let

y=e^{\ln x}

Taking natural logarithm (ln) on both sides, we get

\ln y=\ln x\ln e=\ln x

So y=x. Substitute the very first equation and we get e^{\ln x}=x. 🙂

Alternatively, we can view e^x and \ln x as inverse functions of each other. So, we can let f(x)=e^x and f^{-1}(x)=\ln x. Then, e^{\ln x}=f(f^{-1})(x)=x by definition of inverse functions. This may be a better way to remember the result. 🙂

The above method of inverse functions can be used to remember \ln (e^x)=x too.

The mass of particles of a certain radioactive chemical element (O Level Math Tuition Question)

Question from http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=29&t=8315&start=780

The mass of particles of a certain radioactive chemical element is halved every 10 months.  During a chemical experiment, the initial mass of particles of the chemical element is 3mg.

(i) write down an expression, in terms of t, for the mass of particles after t years.
(ii) Hence, find the value of t, if the mass is reduced to 0.046875 mg after t years.

Solution:

(i)

1\: \text{year} = 12\: \text{months}

Therefore, t\: \text{years}=12t\: \text{months}.

How many 10 months are there in t\: \text{years}? (Ans: \frac{12t}{10}=1.2t)

Hence, the mass of particles after t years is 3\times(0.5)^{1.2t} mg.

(ii)
We need to solve 3(0.5)^{1.2t}=0.046875.

Dividing by 3, we have (0.5)^{1.2t}=0.015625.

Ln both sides, we have 1.2t\ln{0.5}=\ln{0.015625}.

Hence, t=\frac{\ln{0.015625}}{1.2\ln{0.5}}=5.

 

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