O Level E Maths Tuition: Statistics Question

statistics-olevel-tuition-graph

Solution:

From the graph,

Median = 50th percentile = $22,000 (approximately)

The mean is lower than $22000 because from the graph, there is a large number of people with income less than $22000, and fewer with income more than $22000. (From the wording of the question, calculation does not seem necessary)

Hence, the median is higher.

The mean is a better measure of central tendency, as it is a better representative of the gross annual income of the people. This is because more people have an income closer to the mean, rather than the median.

Secondary Maths Tuition: Kinematics Question

kinematics-question-o-levels

Solution:

acceleration of car =\frac{12}{6}=2m/s^2

\frac{v}{15-5}=2

v=2\times 10=20

Let T be the time (in seconds) when the car overtakes the truck.

Total distance travelled by car at T seconds = area under graph = \frac{1}{2}(T-5)(2(T-5))

(2(T-5) is the velocity of car at T seconds, it is obtained in the same way as we calculated v.)

Total distance travelled by truck at T seconds = \frac{1}{2}(6)(12)+\frac{1}{2}(15-6)(12+16)+16(T-15)

Equating the two distances will lead to a quadratic equation T^2-26T+103=0

Solving that gives T=21.12s or T=4.876s (rejected as car only starts at t=5)

21.12-5=16.1s (3 s.f.)

O Level 2007 E Maths Paper 2 Q3 Solution (Geometry Question)

(a)
\angle DAB=\angle DCB
(opposite angles of parallelogram)

\displaystyle\angle PAD=\frac{1}{2}\angle DAB=\frac{1}{2}\angle DCB=\angle RCB
(shown)

(b)
AD=BC

\angle PAD=\angle RCB
(from part a)

\angle ADC=\angle ABC
(opposite angles of parallelogram)

\displaystyle\angle ADP=\frac{1}{2}\angle ADC=\frac{1}{2} ABC=\angle RBC

Thus, triangles ADP and CBR are congruent (ASA).

(c)(i)
\angle ADC+\angle DAB=180^\circ
\angle DAP+\angle ADP=90^\circ

Considering the triangle ADP,
\angle DPA=180^\circ-90^\circ=90^\circ
(shown)

(ii)
\angle DAB+\angle ABC=180^\circ

\angle BAQ+\angle ABQ=90^\circ

Considering the triangle ABQ,
\angle PQR=180^\circ-90^\circ=90^\circ
(shown)

Sec 3 Hwa Chong Institution Maths Test Papers and Resources

Site: http://anglc.wiki.hci.edu.sg/space/content

Description: Includes Sec 3 Hwa Chong Institution Maths Test Papers and Resources

 

Please visit our site https://mathtuition88.com/free-exam-papers for updates on more Free Exam Papers and Maths Tips.

Maths Tuition Free Exam Papers (Primary, Secondary, O Levels, A Levels) Links

The Free Exam Papers Database has moved to:

https://mathtuition88.com/free-exam-papers

(Click here to go to Free Exam Papers Database)

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.

A Maths Tuition: Trigonometry Formulas

Many students find Trigonometry in A Maths challenging.

This is a list of Trigonometry Formulas that I compiled for A Maths. Students in my A Maths tuition class will get a copy of this, neatly formatted into one A4 size page for easy viewing.

A Maths: Trigonometry Formulas

\mathit{cosec}x=\frac{1}{\sin x}\mathit{sec}x=\frac{1}{\cos x}

\cot x=\frac{1}{\tan x}\tan x=\frac{\sin x}{\cos x}

all science teachers crazy maths tuition(All Science Teachers Crazy)

y=\sin x sin curve maths tuition

y=\cos x cosine curve maths tuition

y=\tan x tangent curve maths tuition

\frac{d}{\mathit{dx}}(\sin x)=\cos x

\frac{d}{\mathit{dx}}(\cos x)=-\sin x

\frac{d}{\mathit{dx}}(\tan x)=\mathit{sec}^{2}x

\int {\sin x\mathit{dx}}=-\cos x+c

\int \cos x\mathit{dx}=\sin x+c

\int \mathit{sec}^{2}x\mathit{dx}=\tan x+c

Special Angles:

\cos 45^\circ=\frac{1}{\sqrt{2}}

\cos 60^\circ=\frac{1}{2}

\cos 30^\circ=\frac{\sqrt{3}}{2}

\sin 45^\circ=\frac{1}{\sqrt{2}}

\sin 60^\circ=\frac{\sqrt{3}}{2}

\sin 30^\circ=\frac{1}{2}

\tan 45^\circ=1

\tan 60^\circ=\sqrt{3}

\tan 30^\circ=\frac{1}{\sqrt{3}}

right angle maths tuitionequilateral triangle maths tuition

y=a\sin (\mathit{bx})+c Amplitude: a; Period: \frac{2\pi }{b}

y=a\cos (\mathit{bx})+c Amplitude: a; Period: \frac{2\pi }{b}

y=a\tan (\mathit{bx})+c Period: \frac{\pi }{b}

\pi \mathit{rad}=180^\circ

Area of  \triangle \mathit{ABC}=\frac{1}{2}\mathit{ab}\sin C

Sine Rule:  \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}

Cosine Rule:  c^{2}=a^{2}+b^{2}-2\mathit{ab}\cos C

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. 🙂

Is d/dx (a^x)=x a^{x-1}? (a conceptual error in O/A Level Math)

In O Level, students are taught that \boxed{\frac{d}{dx}(x^{n})=nx^{n-1}}

So naturally, students may think that \displaystyle\frac{d}{dx}(a^{x})=xa^{x-1}?? (a is a constant)

Well, actually that is good pattern spotting, but unfortunately it is incorrect. Do not be too disheartened if you make this mistake, it is a common mistake.

The above is a conceptual error as  \boxed{\frac{d}{dx}(x^{n})=nx^{n-1}} only holds when n is a constant.

Fortunately, this question is rarely tested, though it is quite possible that it can come up in A Levels.

To fully understand the following steps, it would help read my other post (Why is e^(ln x)=x?) first.

First, we write \displaystyle a^x=e^{\ln a^x}=e^{x\ln a}.

Hence
\displaystyle    \begin{array}{rcl}    \frac{d}{dx} (a^x)&=&\frac{d}{dx} (e^{x\ln a})\\    &=&e^{x\ln a}(\ln a)\\    &=&e^{\ln a^x}(\ln a)\\    &=&a^x(\ln a)    \end{array}

After fully understanding the above steps, you may memorize the formula if you wish:

\boxed{\frac{d}{dx} (a^x)=a^x(\ln a)}

Memory Tip: If you let a=e, you should get \boxed{\frac{d}{dx} (e^x)=e^x(\ln e)=e^x}

The above steps involve the chain rule, which I will cover in a subsequent post.

How to avoid Careless Mistakes for Maths?

Many parents have feedback to me that their child often makes careless mistakes in Maths, at all levels, from Primary, Secondary, to JC Level. I truly empathize with them, as it often leads to marks being lost unnecessarily. Not to mention, it is discouraging for the child.

Also, making careless mistakes is most common in the subject of mathematics, it is rare to hear of students making careless mistakes in say, History or English.
Fortunately, it is possible to prevent careless mistakes for mathematics, or at least reduce the rates of careless mistakes.

From experience, the ways to prevent careless mistakes for mathematics can be classified into 3 categories, Common Sense, Psychological, and Math Tips.

Common Sense

  1. Firstly, write as neatly as possible. Often, students write their “5” like “6”. Mathematics in Singapore is highly computational in nature, such errors may lead to loss of marks. Also, for Algebra, it is recommended that students write l (for length) in a cursive manner, like \ell to prevent confusion with 1. Also, in Complex Numbers in H2 Math, write z with a line in the middle, like Ƶ, to avoid confusion with 2.
  2. Leave some time for checking. This is easier said than done, as speed requires practice. But leaving some time, at least 5-10 minutes to check the entire paper is a good strategy. It can spot obvious errors, like leaving out an entire question.

Psychological

  1. Look at the number of marks. If the question is 5 marks, and your solution is very short, something may be wrong. Also if the question is just 1 mark, and it took a long time to solve it, that may ring a bell.
  2. See if the final answer is a “nice number“. For questions that are about whole numbers, like number of apples, the answer should clearly be a whole number. At higher levels, especially with questions that require answers in 3 significant figures, the number may not be so nice though. However, from experience, some questions even in A Levels, like vectors where one is suppose to solve for a constant \lambda, it turns out that the constant is a “nice number”.

Mathematical Tips

Mathematical Tips are harder to apply, unlike the above which are straightforward. Usually students will have to be taught and guided by a teacher or tutor.

  1. Substitute back the final answer into the equations. For example, when solving simultaneous equations like x+y=3, x+2y=4, after getting the solution x=2, y=1, you should substitute back into the original two equations to check it.
  2. Substitute in certain values. For example, after finding the partial fraction \displaystyle\frac{1}{x^2-1} = \frac{1}{2 (x-1)}-\frac{1}{2 (x+1)}, you should substitute back a certain value for x, like x=2. Then check if both the left-hand-side and right-hand-side gives the same answer. (LHS=1/3, RHS=1/2-1/6=1/3) This usually gives a very high chance that you are correct.

Thanks for reading this long article! Hope it helps! 🙂

I will add more tips in the future.

Recommended Maths Book:

Math Doesn’t Suck: How to Survive Middle School Math Without Losing Your Mind or Breaking a Nail

This book is a New York Times Bestseller by actress Danica McKellar, who is also an internationally recognized mathematician and advocate for math education. It should be available in the library. Hope it can inspire all to like Maths!


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.

 

If you liked our solution above, please consider signing up for Maths Tuition with us! 🙂