## Maths, Science and

## Chinese specialist

Breeze education offers tuition covering these subjects for PSLE and all levels in secondary school leading to the ‘O’ level.

## The Breeze!

## Style

Exam preparation and studying techniques are key to good performance in the exam. At Breeze, tutors adopt the philosophy of producing more with less.

## Excellent

## tutors

Breeze education is neither a tutor centre nor a tuition agency. All tutors are recruited via invitation.

Breeze education have tutors who are competent in teaching the below subjects at primary and secondary

level. In the near future, we will also be covering subjects at advanced level.

## Primary Level (P5 & P6)

- Maths
- Science

## Secondary Level

We taught the below subjects at both lower secondary levels and ‘O’ levels.

### Lower secondary:

- Chinese
- Maths
- Science

### ‘O’ Level:

- A. Maths
- E. Maths
- Chinese
- Biology
- Chemistry
- Physics
- Science (Biology/Chemistry)
- Science (Chemistry/Physics)

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- testimonials 1
I was failing in class, but thanks to Koon Yeow, I was able to do well for my exams. He has always been helpful and willing to sacrifice his time for me when I need it. He is patient as well and is the best tuition teacher I ever had!

##### Jonathan Long

- testimonials 3
I have been Koon Yeow’s student for about 3 years since the year 2010. He had been a great help to me in Mathematics and Science as I prepared for my GCE ‘O’ and ‘A’ Levels examinations. Koon Yeow is a wonderful tutor who is fully committed to the students he tutors, understanding them and their preferred methods of learning and adapting his teachings accordingly to suit their needs, to maximise their learning and understanding. Koon Yeow had also kept me well motivated by giving genuine encouragement and rewarding me at times for my studying. Personally, he has also been a good friend to me, relating to my interests and this had at the same time served as more motivation for me to work hard in tuition. Because of his excellent tutoring, my parents have also arranged for Koon Yeow to continue to help by tutoring my younger sibling. In conclusion, Koon Yeow’s commitment and great knowledge of what he teaches are some good points he has that can make him an effective tutor and also a good personal friend to his students.

##### Bryan Soh Jing Yu

- testimonials 5
During the years that I’ve been taught by Koon Yeow, I felt that he has been a patient,friendly and approachable tutor. He had creative methods to address my areas of weaknessin the various subjects that he taught me, such as Elementary and Additional Mathematics,as well as Chemistry and Physics. Apart from being excellent at his tutoring, Koon Yeow isalso a good listener and encourager.

##### Jonathan Loh

- Testimonial 7
Lorem ipsum dolor sit amet, cu agam zril prodesset has, ei quo autem fugit prompta. At mel augue everti accusam. Ne sea enim summo everti. Nusquam mnesarchum sea cu, option utroque vivendum ius ad. Mei quando nemore lobortis.

##### Ema Token, Chemistry student

- testimonials 2
7 years ago, Koon Yeowhelped me ace my A and E math for O levels. He was always patient and explained each topic that I was weak in in ways that I could understand well. He always knew which topics to work on, which really helped me focus my studying for the exams. He genuinely cared about my grades and progress and always stayed past the hours that he was paid for if I needed help, or if there was an upcoming test. Lessons with him were never boring and he instilled an appreciation for mathematics in me. He is a truly great tuition teacher and you will be hard pressed to find a better one.

##### Perlyn Long

- testimonials 4
As a parent, I found Koon Yeow to be a dedicated and patient tutor. He is good in helpinghis students build strong foundations in the respective subjects he taught. Moreover, hehas innovative methods of addressing their weak areas in the various subjects. Duringthe 8 years that he tutored my two children, he has proven to be capable of motivatingthem to achieve their personal best, in both Secondary School and Junior College. Myson and daughter have secured places in local universities, majoring in Engineering andAccountancy respectively.

##### Angela Wong

- testimonials 6
Lorem ipsum dolor sit amet, cu agam zril prodesset has, ei quo autem fugit prompta. At mel augue everti accusam. Ne sea enim summo everti. Nusquam mnesarchum sea cu, option utroque vivendum ius ad. Mei quando nemore lobortis.

##### Ema Token, Chemistry student

- Testimonials 9
I took ‘O’ Level Chemistry lessons under Francis. Before I met him, my chemistry grade was a D7, and I was approached by one of my school teachers who advised me to drop the subject unless I could improve my grade. That was when my mum suggested that I took tuition under him. I was at first skeptical but after a few lessons I was convinced that Francis was committed to seeing me not only to pass the subject, but excel in it as well. In the next exam, I scored better and though it was only a C6, I was praised by my teacher who was pleased that I had passed. Francis is great at explaining concepts and ensuring essential information is not easily forgotten.

##### Samuel, 2014, Chemistry, Presbyterian High School

## ‘O’ Level Chemistry

- Chemistry
### Testing Subject

cccccc

- Chemical calculation
### Chemical calculation

(A): Writing & Balancing Chemical Equations

(1): Sodium + Water Sodium hydroxide + Hydrogen gas

Step 1: Convert the word equation into a chemical equations by replacing all the reactants & products with their corresponding chemical symbols.

Step 2: Balance the equation by ensuring the number of each type of atoms is the same on both sides of the equation.

2Na + 2HO 2NaOH + H

(2): Sodium chloride + Fluorine Sodium fluoride + Chlorine gas

NaCl + F NaF + Cl (Unbalanced)

2NaCl + F 2NaF + Cl (Final equation)

Practise Questions:

Write the chemical equations for the following equations:

(A): Calcium hydroxide + Hydrochloric acid ? Calcium chloride + water

(B): Sodium + Zinc fluoride ? Sodium fluoride + Zinc

(C): Iron + Copper (II) sulphate ? Copper + Iron (II) Sulphate

(D): CH+ O ? CO+ HO

(E): Ca(OH) + HSO4 ? CaSO + HO

Relative Atomic Mass is the average mass of 1 atom of an element as compared with 1/12 of the mass of 1 atom of carbon-12 .

Relative Molecular Mass is the average mass of one molecule of the substance compared with 1/12 of the mass of 1 atom of carbon-12 .

Calculate the relative atomic mass & relative molecular mass of the following:

(i) Sodium Chloride

(ii) FeCl (Iron (III) chloride)

(iii) NaCO (Sodium carbonate)

(iv) CaF (Calcium fluoride)

The mole is defined as the amount of substance that contains as many particles (atoms/molecules/ions) as the number of carbon atoms, in 12 grams of Carbon-12.

12 grams of carbon atoms (1 mole of carbon) has a total of 6.0 X 10 23} oxygen molecules. 1 mole of water has a total of 6.0 X 10 23}_{2} molecules of CaCl.

(3) Calculate the number of hydrogen atoms present in 2 moles of water.

Molar mass

The mass in grams of 1 mole of a substance is called the molar mass .

mole = mass of sample molar mass

The molar mass of an element is the equivalent of relative atomic mass but with an unit (g).

Note: Relative atomic mass (or relative molecular mass) does not have units while molar mass has a unit of g/mol.

Questions:

(1) Calculate the mass of 2 moles of sulphur dioxide.

(2) Calculate the mass of 1.5 moles of oxygen gas.

(3) Calculate the number of moles present in 60g of hydrogen fluoride.

(4) Calculate the mass of 18 X 10 23}$ molecules of chlorine gas.

(5) Calculate the number of moles of oxygen atoms present in 2 moles of sulphur trioxide.

% Composition

## ‘O’ Level Chemistry Practice Questions

- Practise Questions (Basic) – Ammonia
### Practise Questions (Basic) – Ammonia

Practise questions (Basic) are designed to ensure that students are aware of the basic concepts of the topic. This should be the first step that students undertake while revising for their chemistry examination. For the O Level examination, these basic concepts will be tested although questions will not be straight forward and students will need to be able to apply these concepts to solve the questions. For the chapter of ammonia, there are fewer questions as this is a relatively small chapter and the scope is much narrower.

(1): State the optimal conditions for the Haber process.

(2): Describe how the raw materials of the Haber process, hydrogen and nitrogen, are obtained.

(3): Other than the Haber process, describe another method in which ammonia can be produced.

(4): Describe a test for ammonium ions.

About Breeze Education:

Breeze Education offers chemistry tuition to O Level Students. Tuition classes are conducted in small group with class size of not more than 3 students to facilitate active discussion and consultation. Chemistry is taught in a way to facilitate easy understanding of the topic. Students are also honed on their revision techniques so that they can improve faster over a shorter period of time. Tuition sessions are planned and taught by Dr Francis who has a PhD in Biomedical Science from National University of Singapore.

## ‘O’ Level Physics

- Physics test
### Energy test

cccccc

- Physics Chapter 1
### Chemical Bonds and the Structure of Substances

**Electronic Structure of Noble Gases**Noble gases or Group 0 (or Group VIII) elements

__exist as atoms__and are__unreactive (inert)__due to their stable electronic structures.The outermost (valence) electron shell of these elements are fully-filled

e.g. Helium:**2**

Neon: 2.**8**

Argon: 2.8.**8**Atoms of other elements tend to

__lose, gain or share electrons__in order to obtain the stable electronic structure of a noble gas. They do this by forming chemical bonds.There are 3 main types of bonding –

: Very strong attraction between the nuclei of two atoms and the shared electrons. 2 characteristics:*Covalent bond*Covalent bond is formed between

__atoms of the same element__or atoms of__different elements__.It is only formed between atoms of

.__non-metals__Examples of compounds with covalent bond: CH (methane), SO (sulphur dioxide), CO (carbon dioxide)

:Strong electrostatic attraction between positive and negative ions formed from the transfer of electrons from one atom to another.*Ionic bond*Metals always form positive ions (via losing electrons eg. )

If non-metals form ions, they will produce negative ions (via gaining electrons).

**IONIC BOND IS THUS FORMED BETWEEN A METAL \& NON-METAL.**Examples of compounds with ionic bond: NaCl (sodium chloride), NaO (sodium oxide), MgF (magnesium flouride)

:Strong attractive forces between the*Metallic bond*__positive metal____cations__and the__‘sea of delocalised negative electrons’__.The number of electrons used to form chemical bonds is called the

.__valency__**Covalent Bonds – Electron Sharing**Atoms share electrons in order to attain the electronic structure of a noble gas. When they share electrons, very strong covalent bonds are formed between the atoms.

There are 2 ways to represent a covalent bond: a dot-cross diagram or a line-bond diagram. In a dot-cross diagram, the electrons of one atom are represented by ‘

**x**‘ while the electrons of another atom are represented by ‘●’. There are 2 electrons in each covalent bond with each atom contributing 1 electron to the bond.A dative (coordinate) bond is a covalent bond where both electrons in the bond come from one of the atom and the other atom contributing no electrons.

**Ionic Bonds – Electron Transfer**Positive ions (cations) are formed by

__removing electrons from atoms__. In a cation, the__number of electrons is less than the number of protons__.Negative anions (ions) are formed

__by adding electrons to atoms__. In an anion, the__number of electrons is more than the number of protons__.Metals form cations and non-metals form anions, so ionic compounds are formed when a metal reacts with a non-metal.

Ions are formed when electrons are transferred from a metal atom (which becomes a positive ion) to a non-metal atom (which becomes an a negative ion)

**Chemical Bonds \& the Periodic Table**Elements in the same group have the

__same number of outer (valence) electrons__, which means they have the same valency. Thus they form compounds (ions and molecules) of similar formulas with other atoms.Form ions with similar formula e.g. Group I:Na, K

Group VII:Cl, Br

Form similar covalent molecules e.g.Group IV:CH, SiH

Form compounds with similar formula e.g.Group I with Group VII

NaCl, NaBr, KBr

**(Simple) Molecular Substances**Most non-metallic elements (except noble gases) and covalent compounds have simple molecular structures consisting of small simple molecules e.g. water, nitrogen gas and glucose.

Between the two atoms

**within a simple molecule**, there exist very strong intramolecular covalent bonds.Between two molecules, there exist

__weak intermolecular forces__of attraction.Molecular structures have

__low melting, boiling points__(usually below 200C)**due to the weak intermolecular forces**, generally do not conduct electricity in any state**due to the non-charged molecules**present, usually insoluble (or low solubility) in water but soluble in organic solvents.**Ionic Compounds**Ionic compounds are made up of positive metal cations and negative non-metal anions.

These ions are held together by strong ionic bonds or electrostatic forces of attraction.

Large amounts of energy are required to break these bonds, so ionic compounds have

.__high melting, boiling points__Ionic compounds

They only__cannot conduct electricity in solid state.____conduct electricity in the liquid (molten) state and aqueous (dissolved in water) state because the ions can move in these states.__Many of them dissolve in water but not in organic solvents.

**Metals (Metallic Bonding)**In a metal, each atom gives up some electrons to become a positive ion (cation)

**due to weak attraction of the nucleus for the outermost electrons**.Metallic bonding is the force of attraction between the positive metal cations and the ‘sea of negative electrons’

These small electrons are mobile, and go into the spaces in between the cations. The cations are big and cannot move (only vibrations).

Metals can conduct electricity in

, because the electrons can move throughout the metal.__both solid \& aqueous state__Metallic bonds are very strong, so metals usually have very high melting and boiling points.

The strength of the metallic bonds depends on the charge on the metal cation (or the number of valence electrons contributed to the sea of electrons per atom) and the ionic radius (size) of the cation.

**Macromolecular (Giant Covalent) Structures**In a macromolecule, all the atoms are joined together by

There are__strong covalent bonds.__within a macromolecule.__no intermolecular forces of attraction__Macromolecular structures have high melting, boiling points

**due to the strong covalent bond**s, are insoluble in all solvents and do not conduct electricity (except for graphite)**due to the absence of charged particles**. Polymers like plastics are also macromolecules.Examples of compound with macromolecular structures: Diamond, silicon, silicon dioxide (sand), graphite.

__Diamond and graphite same atomic constituent but different physical properties__Diamond and graphite are allotropes of carbon.

In allotropes, atoms of the same element are arranged (bonded) in different ways.

Both of them have macromolecular structures. In diamond, each carbon atom is joined covalently to 4 other carbon atoms in a tetrahedral arrangement.

In graphite, each carbon atom is joined covalently to 3 other carbon atoms in flat, hexagonal layers. Between the layers are weak intermolecular (Van der Waals) attractive forces. The weak forces between layer account for the soft and slippery properties of graphite.

In addition, in graphite, one outermost electron of the carbon atom is not involved in covalent bonding. This delocalized electron is mobile and thus graphite can conduct electricity.

*Silicon*and*silicon dioxide*have macromolecular structure too. Silicon has a similar structure to diamond where each silicon atom is joined covalently to 4 other silicon atoms in a tetrahedral arrangement.

## ‘A’ Level Chemistry

- ‘A’ Level Chemistry
### ‘A’ Level Chemistry

**Atomic Structure & the Periodic table***Key points:*(1) Atom – the most basic building block of elements/compounds

(2) An atom is very small. 12 g of carbon contains 6 X 10

^{23}atoms(3) In fundamental sense, the three important subatomic particles (components) of an atom are protons, neutrons and electrons.

(4)

Protons Neutrons Electrons Relative Mass 1 1 0 Relative Charge +1 0 -1 (5) In essence,

(a) The**mass**of an atom is contributed by the**total**number of**protons**and**neutrons**.

(b) Both the protons and neutrons are located at the centre of the atom,**nucleus**.

(c) In an*uncharged atom*, the number of protons and electrons are the same.

(d) During a chemical reaction, electrons might be gained or lost by an atom,**BUT**the number**of protons and neutrons remained the same**.(6) The periodic table is arranged in vertical column called group, and horizontal row called period.

(7) Atoms in the same group

**have similar chemical properties**, though they might**differ in their relative reactivity**in a chemical reaction.(8) Atoms in the same group have the same number of

**outermost electrons**.*Dissecting the periodic table*Atom Proton Number Neutron Number Electron number Atomic Mass Lithium 3 4 3 7 Magnesium Potassium Carbon Aluminium Fluorine Oxygen Sulphur Hydrogen Helium Argon *Atom, molecule, ion and isotopes***Atom**: Sodium atom, Carbon atom, Hydrogen atomA

**molecule**is made up of 2 or more atoms chemically combined together.

Eg. H_{2}(2 hydrogen atoms), CO_{2}(1 carbon atom & 2 oxygen atoms), O_{2}(2 oxygen atoms), NaCl (1 sodium atom & 1 chlorine atom)An

**ion**is a**charged atom**. It contains an unequal number of*protons*and*electrons*.

Cl^{–}(A chlorine atom gains one more electron. Proton number = 17, Electron number = 18)**Isotopes**are atoms of the same element with the**same number of protons**but**different number of neutrons**. Since the number of electrons is the same, isotopes have the same chemical properties.*Writing electronic configuration*Note: At your level, it is sufficient to know that electrons are arranged in shell radiating out from the nucleus. The first shell can contain 2 electrons and subsequent shells can contain 8 electrons.

Hence, the electronic configuration of,

Electronic configuration Lithium Oxygen Fluorine Sulphur Magnesium Aluminium Nitrogen Neon *Metals, Non-metals & Nobel gases*Metals have 3 or less outermost electrons (Group 1-3).

Non-metals have outermost electrons between 4 to 7 (Group 4-7).

Nobel gases are highly unreactive as they have a complete shell of outermost electrons (either 2 or 8).

## ‘A’ Level Biology

- ‘A’ Level Biology
### ‘A’ Level Biology

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## ‘O’ Level EMaths

- Maths
### Area of circle

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- ‘O’ Level Emaths
### ‘O’ Level Emaths

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## ‘O’ Level AMaths

- Integration: Revision Exercise
### Integration: Revision Exercise

- Vectors
### Vectors

- AMath Example 1
### Differentiation & Applications

(1). A particle travels in a straight line, starting from rest at point A, passing through point B and coming to rest again at point C. The particle takes 5s to travel from A to B with constant acceleration. The motion of the particle from B to C is such that its speed, v t seconds after leaving A, is given by

(i) Find the speed of the particle at B and the value of T.

(ii) Find the acceleration of the particle when t = 14.

(iii) Sketch the velocity-time curve for

(iv) Calculate the distance AC. (J02/P2/Q12)(2) A particle moves in a straight line so that, t seconds after leaving a fixed point O, its displacement, s metres from O, is given by s Find

(i) the positive value of t for which the particle is instantaneously at rest,

(ii) the total distance travelled by the particle from t = 0 to t = 4,

(iii) the acceleration of the particle when t = 1. (D00/P1/Q7)(3) A particle moves in a straight line so that, at time t seconds after leaving a fixed point O, its velocity, is given by v =

(i) Sketch the velocity-time curve.

(ii) Find the value of t when v = 10.

(iii) Find the acceleration of the particle when v = 10.

(iv) Obtain an expression, in terms of t, for the displacement from O of the particle at time t seconds. (D98/P2/Q5)(4) A particle moves in a straight line so that at time t seconds after passing through a fixed point O, its displacement s m is given by Find

(i) the times when the particle is momentarily at rest,

(ii) the total distance traveled in the first 5 seconds,

(iii) the interval of time during which the velocity is negative and sketch the velocity-time graph for the first 5 seconds.(5) A particle travels in a straight line so that at time t seconds, its distance s metres from the origin O is given by

(a) Show that the velocity is always positive.

(b) Find the least velocity attained.

(c) Find the range of values of t for which the particle is decelerating.

(d) Find the distance of the particle from O when the acceleration of the particle is instantaneously zero.

(e) Find the average velocity of the particle during the first 2 seconds.(6) The diagram below shows a vertical cross-swection of a container in the form of an inverted cone of height 60cm and base radius 20cm. The circular base is held horizontal and uppermost. Water is poured into the container at a constant rate of 40

(i) Show that, when the depth of water in the container is x cm, the volume of water in the container is

(ii) Find the rate of increase of x at the instant when x = 2. (D98/P1/Q6)(7) The diagram shows a quadrilateral ABCD in which are right angles, AB = 8 cm, BD = 10cm and

(i) Show that the area S of the quadrilateral ABCD is given by

S = 24 +

(ii) Hence find the maximum value of S and the value of when S is a maximum.

(8a) The tangent to the curve y = is parallel to the normal to the curve y =

(b) An open rectangular fish tank of capacity 1152 is to be constructed using materials of negligible thickness. If the length of the fish tank is 3x cm while its width is x cm, show that the amount of material needed to build the tank is given by A where A = Find the value of x for which A is a minimum.

(9) In the figure, PQRS is a square plastic plate of side 4 cm and ABCD is a square whose centre coincides with that of PQRS. The shaded regions are to be cut off an the remaining plastic is folded to form a right pyramid with base ABCD. Let AB = 2x cm and let V be the volume of the pyramid.

(i) Show that the height of the pyramid is

(ii) Show that V = .

(iii) Find the value of x such that V is maximum.(10i). Find the equation of the normal to the curve at the point where the curve meets the x-axis.

(ii) Given that where A and k are constants, find an expression for Hence find the value of k and of A for which (D01/P2/Q3)

(11) The gradient of a curve at any point is given by The curve intersects the x-axis at the point P. Given that the gradient of the curve at P is 1, find the equation of the curve.

(12) A curve has the equation y = Find

(i) an expression for the gradient of the curve,

(ii) the x-coordinate of each of the stationary points of the curve for which 0 x radians. (J00/P2/Q6)(13) Given that y =

(a) find

(b) find the value of k for which x + y = k is a tangent to the curve,

(c) show that y decreases as x increases.(14) Two variables, x and y, are related by the equation Given that both x and y vary with time, find the value of y when the rate of change of y is 12 times the rate of change of x. (D01/P1/Q16)

(15) Variables p and q are connected by the equation Find an expression, in terms of q, for and hence find the approximate change in p as q increases from 6 to 6 + k, where k is small. (J98/P1/Q2)

(16) The variable y is given in terms of x by y = Given that x is increasing at 0.5 radians per second, find the rate of change of y with respect to time when x = (J96/P2/Q6ac)

(17) The time T taken by a planet to revolve around the sun and its mean distance r from the sun are related by T = k where k is a constant. Obtain If the planet’s mean distance from the sun was to be increase by 2\%, estimate the approximate percentage increase in the period T.

(18) A vessel is in the shape of an inverted right circular cone whose base-radius is equal to its height and whose axis is vertical. Liquid is poured into the vessel at a constant rate of The volume of liquid in the vessel is when the depth of liquid is x cm. Calculate, at the instant when the depth of liquid is 10 cm, the rate of increase of

(i) the depth of the liquid,(ii) the area of the horizontal surface of the liquid.

(b) Given that y = use calculus to find, in terms of p, the approximate percentage increase in y when x increases from 2 by p\%, where p is small. (D95/P1/Q13)

(19) The diagram below shows a circle of radius r cm inside an equilateral triangle of side x cm. It is given that x has an initial value of 10 cm and r has an initial value of 3 cm. Both x and r are increasing at the rate of 0.2 cm/s.

(i) Express x in terms of r and show that the shaded area, A , is given by

A =

(ii) Hence, find the rate of change of the shaded area after 20 seconds.

(20) In the diagram,PQ is a straight line. A is on PQ and triangle APB is right-angled at P. BP = 10 cm and PA = x cm. AB is the diameter of the semi-circle. Find

(i) Find in terms of x, the area A, of the semi-circle,

(ii) Given that A moves along PQ such that x is increasing at 0.8 cm per second, find the rate of increase in the area of the semi-circle at the instant when x = 12 cm.(21) A container is such that when the depth of liquid in it is x cm, the volume is V

(a) A small increase in the depth of the liquid in the container leads to a small increase The depth is initially 10 cm, and then the volume is increased by 10 Calculate the approximate increase in the depth.(b) Show that the percentage increase in V is always approximately 2.5 times the percentage increase in x.

(c) Given that the volume is increasing at a rate of 20 per second. Calculate the rate at which the depth of the liquid in the container is increasing when x = 15.

(22) The curve whose equation is y = where k is a constant, has a turning point where x = -1.

(i) Calculate the value of k.

(ii) Calculate the value of x at the other turning point on the curve.

(iii) Draw a rough sketch of the curve and find the set of value of x for which y 0.

(D00/P2/Q1a)(23) For the curve y = calculate the coordinates of

(i) the points of intersection with the axes,

(ii) the turning points. Hence sketch the curve. (D96/P2/Q1c)(24) The variables and t are related by the equation where and k are constants. When t = 30,

(i) Show that the value of k, correct to 4 decimal places, is 0.0231. When t = 40,

(ii) Calculate the value of When t = 50, calculate

(iii)

(iv) .Find the average rate of change of with respect to t over the interval

(D00/P2/Q3b)

(25) Differentiate the following expressions with respect to x.

(i)

(ii) (D96/P2/Q5ac)

- Where is tuition conducted
## 1. Where is tuition conducted?

For one-to-one tuition, it is conducted at student’s house. For small group tuition, it can be conducted at one of the student’s house or another alternative location that is conducive for study.

- How does your small size group tuition work?
## 2. How does your small size group tuition work?

For group tuition, we have either group size of 2 or 3. At the moment, we do not actively organize group tuition and we encourage students to form their own group. This is because different schools have different pace of studying and that students might need to prepare for test from time to time. We believe that it will work more effectively if the students are from the same school and this explains why it will work better if the group is started by the students.

- What subjects are taught by tutors of Breeze Education?
## 3. What subjects are taught by tutors of Breeze Education?

At the moment, we offered secondary level Mathematics (both A. Maths and E.Maths) and Science (Physics, Chemistry and Biology). We have also on board a Chinese tutor who can teach all levels of Chinese, however, please do check with us on her availability.

- How long does it take for students to produce results?
## 4. How long does it take for students to produce results?

The speed and extent of improvement will differ from students to students. In general, it should take about half a year to see students improving.

- How long is a session?
## 5. How long is a session?

Each session is two hours and generally students just need to have one session a week per subject.

- Does Breeze Education offer tuition for PSLE or ‘A’ Level?
## 6. Does Breeze Education offer tuition for PSLE or ‘A’ Level?

For PSLE, we do have tutors who are adept at teaching PSLE. Please feel free to contact Dr Leow who can arrange a suitable tutor for you.

’A’ Level tuition will probably be launched in 2014. - Can we request for a particular tutor?
## 7. Can we request for a particular tutor?

Tutor usually makes the decision on whether they would like to accept the assignment. If the location is a problem for the tutor, we can discuss on other possible arrangement. We will also be launching tuition using web conference platform soon and that might be able to placate this issue.

- Testimonial 8
## 8. What is the rate per lesson?

The rate differs across tutors depending on the experience and qualification of the tutors. Typically, the rate for one-to-one tuition ranges from $30 to $55 per hour and the rate for group tuition ranges from $20 to $30 per hour for a group of 3. Each group is limited to 3 students.

- FAQ testing
## FAQ testing

Testing

## A niche education company

specializing in private tuition

Breeze education is neither a tuition agency nor a company operating tuition centre. The limited effectiveness of big class tuition and the ability to attract good teachers/tutors while offering customer a competitive price is a main obstacle affecting most tuition centre.

The founder of Breeze education, Dr Leow, believes that for tuition to be effective it has to be conducted individually if not preferably limited to three students. The rationale is that student seeks tuition because they need additional help and it is of little value replicating the format in school.

Unlike tuition agency, Breeze Education does not operate as a middleman. Tutors are chosen by the founder of Breeze Education, Dr Leow, who ensures that tutors have the necessary attitude and aptitude to coach and mentor students. Tutors are supported by further coaching and mentoring by Dr Leow who himself have 13 years of tutoring experience.

## Focus on exam

preparation techniques

The second differentiator of Breeze Education is our focus on exam preparation techniques. We aim to make studying for students a more relaxing and fulfilling experience. There is generally a limit on how hard a student can work but good exam preparation skills can liberate students and ensure that they get better results by working on the areas that matter. Dr Leow has tremendous success with students by following this mantra and has imparted these techniques to tutors of Breeze Education.

## Producing

Results

One-to-one and small group size tuition is essential for producing results. Together with the correct exam preparation techniques, students have a good chance of producing results within a shorter time as compared to group tuition. We are confident that the combination of these two will be a brilliant formula for success with students.

For enquiry, Breeze Education can be reached by contacting Dr Leow via:

9851 4057