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O Level E Maths Tuition Singapore/Tuition O Level E Maths/Tutor

Division of Fractions

1. Convert all mixed numbers to improper fractions first.

2. To divide two fractions, turn the second one upside down and then multiply

Example

3 1/3 divide 5/6
= 10/3 * 6/5
= 4

For Exam based questions with full worked solution. Please contact Mr Ong @98639633 or Angie @96790479

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A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

Chemical Bonding

Modification to the general shapes predicted by VSEPR

Bond angles depend on repulsion between electron pairs. The closer the electron pairs are to the nucleus of the central atom, the greater is their repulsion power. Bond angles may thus deviate from the ideal as electron pairs present may have different repulsion power.

Factors affecting the bond angles are:
(i) Number of lone pairs present

lone pair – lone pair > lone pair – bond pair > bond pair – bond pair
repulsion repulsion repulsion

A lone pair exerts greater repulsion than a bond pair as it is closer to the
nucleus as it is attracted by only one positive nucleus compared to bond–pair
electrons which are attracted by two nuclei.

(ii) Electronegativity of central/ side atom

For species with the same central atom and the same number of lone pairs of
electrons on the central atom, the bond angles may differ depending on whether the atoms attached to the central atom are more or less electronegative than the
central atom.

Atoms which are more electronegative than the central atom tend to draw electron
density of the bond pair towards itself. The bond pair electrons are thus further from the nucleus and exert less repulsion.

On the other hand, atoms which are less electronegative than the central atom has less attraction for the bond pair of electrons so that electron density is drawn towards the central atom. The bond pair electrons are thus closer to the nucleus and exert greater repulsion resulting in bond angle being bigger than that mentioned above.

(iii) Size of atoms attached

For species with the same central atom and the same number of lone pairs of
electrons on the central atom, the bond angles may differ depending on the size of the atoms attached to the central atom.

For side atoms which are very much bigger than the central atom, the bond angle is bigger than the bond angle in the species with smaller side atoms. That is because the electron clouds on the side atoms may exert a repulsion which prevents the bonds from coming too close together.

Please contact Angie @ 96790479 or Mr Ong @ 98639633 if you need help in Chemistry

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A Level GP/General Paper Tuition Singapore

PAPER 1

Answer one question from this Paper.
Answers should be between 500 and 800 words in length.

1. Do you agree that the world would be a better place without religion?

2. Consider the view that historical buildings and monuments have no place in your society today.

3. How far should governments have the right to monitor what is being said online?

4. ‘A multicultural society gives rise to more problems than benefits.’ Comment.

5. ‘The family should be solely responsible for taking care of the elderly.’ Do you agree?

6. How important is funding in determining sporting excellence?

7. ‘The key to a successful business is advertising.’ Discuss.

8. Can the use of animals in scientific research ever be justified?

9. ‘The youth of today are forced to grow up too quickly.’ Do you agree?

10. How effective are harsh punishments in dealing with crime?

11. Is obedience always a virtue?

12. Consider the view that gender equality could and should be achieved in today’s world.

For model essay Please contact Angie 96790479 or Mr Ong 98639633

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A-Level Biology Tuition Singapore/H2 Biology Tuition/JC Biology Tutor

TOPIC 3: GENETICS OF VIRUSES AND BACTERIA – Part 3

LEARNING OUTCOME

(c)Describe the reproductive cycles of the following virus types:
i. bacteriophages that reproduce via a lytic cycle, e.g. T4 phage;

ESSAY ANSWER
T4 bacteriophage:

Adsorption – attachment of T4 phages onto specific receptor sites on surface
of host
 – specific receptor sites: cell wall lipopolysaccharides or proteins
 – attachment: via tail fibres

Penetration – Injection of phage DNA
– baseplate is firmly attached on the host cell surface
- tail sheath contracts and tail tube penetrates the outer membrane
–  phage enzyme (lysozyme) “drills” a hole in the bacterial wall
–  viral DNA is injected through the bacterial cell wall
- phage DNA enters bacteria cytosol

Replication – Replication of phage DNA & Synthesis of phage proteins
– Phage genome is expressed, using the RNA polymerase and ribosomes from its host cells.
- Phage proteins, including enzymes, encoded by the phage genome,
are produced.
- Enzyme degrade bacterium host DNA, shutting down thebacterium’s gene expression and its macromolecular synthesis (protein, RNA and DNA).
- Replication of phage DNA occurs using the bacterium’s metabolic
machinery, e.g DNA polymerase
- Transcription of phage DNA.
- Synthesis of phage proteins, enzymes and structural components
using bacterium’s metabolic machinery, e.g. ribosomes

Assembly – spontaneous assembly of viral particles
– Spontaneous assembly

Release – causing lysis of bacterium host cell
- Phage-encoded lysozyme breaks down the bacterial peptidoglycan
- osmotic lysis and release of the intact new bacteriophages,
- host cell lysis

For exam based questions and solutions, please contact Hp @9863 9633

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A-Level Physics Tuition Singapore/H2 Physics Tuition/JC Physics Tutor

Topic 2: Kinematics Part 1

1. Define displacement, speed, velocity and acceleration.

Displacement – The linear / shortest distance in a given direction from a
reference point.

Speed – The rate of change of distance.

Velocity – The rate of change of displacement.

Acceleration – The rate of change of velocity.

Please contact Angie @ 96790479 or Mr Ong @ 98639633 if you need help in Physics

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A-Level Economics Tuition Singapore/H2/H1 Economics Tuition

Surprise Test: Demand and Supply

Notes: Factors affecting Demand, Supply and Price Mechanism

Section 1: Simple Recall/Definitions Questions

a) State the four types of factors essential for every production process

b) State other interchangeable words for factors of production

c) Define demand
Demand refers to the quantities of a product that buyers are __________ and _________ to buy at various prices per period of time, _________ paribus

d) State the law of demand
The law of demand states that ceteris paribus, the ________ the price of a good, the lower the quantity demanded of a good. This implies an ___________ relationship between the price and quantity demanded for a good.

e) Define supply
Supply refers to the quantities of a product that sellers are _____________ and
_____________ to produce at various prices per period of time, ___________ paribus.

f) State the law of supply
The law of supply states that ceteris paribus, the higher the price of a good, the greater the quantity supplied of the good. This implies a _____________ relationship between price and quantity supplied of a good.

g) Define complements and state 2 examples

h) Define complementary demand

i) Define substitutes and state 2 examples

j) Define competitive demand

k) State 3 determinants of demand

l) State 3 determinants of supply

m) State how equilibrium prices are determined in a free market

Please contact Angie Hp 96790479 or Mr Ong 98639633 if you need help in Economics and complete Notes and Test questions.

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A-Level Mathematics Tuition Singapore/JC Maths/H2 Math Tuition and Tutor

Permutations and Combinations – Skill Set

1. Box Method
Consider consecutive operations one after another and use multiplication
principle

2. Cases Method
Consider mutually exclusive cases and use addition principle

3. Restrictions Method
Take note of restrictions or constraints and arrange the others

4. Complementation Method
Apply method of ‘Complementation’

5. Grouping Method
Apply method of ‘Grouping’

6. Slotting Method
Apply method of ‘Slotting’

7. Polygons Method
Arrangements around polygons

8. Identical Group Size Method
Groups with identical size

Please contact Angie @96790479 or Mr Ong @98639633 if you need help in Mathematics

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O-Level Singapore/O-Level Combine Chemistry and Physics Tuition/Physics Tutor

Elements, Compounds and Mixtures – Concise Notes

Elements

· Elements can be classified as metal, metalloid or non-metal.
· Elements can exist as atoms (as the noble gases) or as diatomic/polyatomic molecules

Compounds

· Compounds are formed when elements react together, e.g. water (a compound) is formed when the elements hydrogen and oxygen react together.
· Compounds can exist as molecules (e.g. H2O molecule) or as ions.
· Some compounds decompose to give their elements, e.g. mercury(II) oxide
decomposes to give mercury and oxygen

Mixtures

· A mixture can be made up of elements (e.g. hydrogen gas and oxygen gas).
· A mixture can be made up of compounds (e.g. hydrogen chloride and sodium chloride).
· A mixture can be made up of elements and compounds (e.g. oxygen gas and iron(II) sulfide).

If you need help in the O level Chemistry, please contact Angie @96790479 or Mr Ong 98639633

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O-Level Singapore/O-Level Combine Chemistry and Physics Tuition/Physics Tutor

Chapter 2 Kinematics Part 3

Exam Tip 7
An important point for a free falling body:
A free falling body may also move up-wards. When a free falling body is rising instead of falling, its direction of motion is upwards but the acceleration is -still 10 m 5 2 downwards.
This upward motion will slow down even-tually because the acceleration due to gravity is opposing its direction of motion.
In fact, this free falling body is deceler-ating. When its velocity becomes zero, it will stop at that position momentarily and then start falling again. The free falling body is now accelerating.

If you need help in the above topics, please contact Angie @96790479 or Mr Ong @98639633

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O-Level Singapore/O-Level/Pure Physics Tuition/Physics Tutor

Chapter 2 Kinematics Part 3

Exam Tip 7
An important point for a free falling body:
A free falling body may also move up-wards. When a free falling body is rising instead of falling, its direction of motion is upwards but the acceleration is -still 10 m 5 2 downwards.
This upward motion will slow down even-tually because the acceleration due to gravity is opposing its direction of motion.
In fact, this free falling body is deceler-ating. When its velocity becomes zero, it will stop at that position momentarily and then start falling again. The free falling body is now accelerating.

If you need help in the above topics, please contact Angie @96790479 or Mr Ong @98639633

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O Level Chemistry Tuition Singapore/Chemistry O Level Tuition/Tutor

Elements, Compounds and Mixtures – Concise Notes

Elements

· Elements can be classified as metal, metalloid or non-metal.
· Elements can exist as atoms (as the noble gases) or as diatomic/polyatomic molecules

Compounds

· Compounds are formed when elements react together, e.g. water (a compound) is formed when the elements hydrogen and oxygen react together.
· Compounds can exist as molecules (e.g. H2O molecule) or as ions.
· Some compounds decompose to give their elements, e.g. mercury(II) oxide
decomposes to give mercury and oxygen

Mixtures

· A mixture can be made up of elements (e.g. hydrogen gas and oxygen gas).
· A mixture can be made up of compounds (e.g. hydrogen chloride and sodium chloride).
· A mixture can be made up of elements and compounds (e.g. oxygen gas and iron(II) sulfide).

If you need help in the O level Chemistry, please contact Angie @96790479 or Mr Ong 98639633

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O-Level Additional Mathematics Tuition Singapore

Question 1

A rhombus PQRS has coordinates P(-3, 12) and Q(9,11). The diagonals of the
rhombus intersect at the point M(−1, 7).
(i) Find the coordinates of R.
(ii) Find the equation of the diagonal QS.

Ans : (i) R (1, 2) (ii) 5y = 2x + 37

Question 2

(i) Expand (1 – 2x)^4 in ascending powers of x, up to and including the term
in x2.

(ii) Find the value of a such that the coefficient of x is zero in the expansion of (2 + ax)^2 * (1 – 2x)^4

Ans : (i) 2 – 8x + 24x^2…….. (ii) a = 8

For exam based question with full worked solution, please contact Mr Ong
@98639633 0r Angie @96790479

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O Level E Maths Tuition Singapore/Tuition O Level E Maths/Tutor

Multiplication of Fractions

1. Convert all mixed numbers to improper fractions first.

2. Simplify the fractions first by ‘cancelling’, to see if anything will divide into any of the top numbers and also the bottom numbers.

3. Multiply the numerators, then the denominators together.

Example

2 2/3 * 1 1/5
= 8/3 * 6/5
= 48/15
= 3 3/15

For Exam based questions with full worked solution. Please contact Mr Ong @98639633 or Angie @96790479

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A-Level Chemistry Tuition Singapore/H2 Chemistry Tuition/JC Chemistry Tutor

Chemical Bonding

Shapes of Molecules – Valence Shell Electron Pair Repulsion Theory (VSEPR)

Covalent bonds are directional. Hence covalent molecules or ions with covalent bonds have definite shapes.

VSEPR is used to predict the molecular geometry of a species. The key essential principle underlying the theory is that regions of electron density arrange themselves as far apart as possible so as to minimise electronic repulsion.

Steps to deduce the molecular shape of a given species:

1. Draw the dot–and–cross diagram of the required species. Remember that lone pairs must be included.

2. Count the total number of regions of electron density around the central atom to determine the “electron pair geometry”

(i) Each lone pair OR lone electron = ONE region of electron density

(ii) Each single bond OR coordinate bond = ONE region of electron density

(iii) Each double bond OR triple bond = ONE region of electron density

3. Count the total number of lone pairs around the central atom.

4. Deduce the molecular geometry / shape of the species using Table 1

Please contact Angie @ 96790479 or Mr Ong @ 98639633 if you need help in Chemistry

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A Level GP/General Paper Tuition Singapore

Question 1: Are women more disadvantaged than men in your society? (Singapore)

Topic: male & female issues in your society
Issue: whether women are more disadvantaged
Context: your society
Keywords: more disadvantaged

Definition of disadvantaged (OED): An unfavourable circumstance or condition that reduces the chances of success.

Yes – Women are more disadvantaged

1. Women are seen as a liability to their work which thus affects their opportunities at a career

a. Discriminatory practices against pregnant employees still occur in this day and age
i. In 2011, MOM received 112 pregnancy and maternity-related complaints compared to 84 in 2010. On average, there have been 97 complaints each year between 2006 and 2010, with the highest number recorded in 2009
http://www.hrmasia.com/forum/pregnant-youre-fired/128433/

2. Women are perceived to have a tougher time dealing with the societal expectations of beauty

a. Notions of beauty in the Singapore context: Slim, tall, thigh gap, fair…
b. Because of these pressures of conforming to the perception of beauty, women are more likely to fall prey to beauty related illnesses e.g. anorexia, bulimia, depression…
i. Figures from the Singapore General Hospital (SGH) show that 95 youngsters aged 13 to 19 suffered from anorexia or bulimia last year, up from 65 in 2011 and 75 the year before.
http://www.straitstimes.com/breaking-news/singapore/story/more-youngsters-diagnosed-eating-disorders-20131013
ii. Angeline Yap Siling (Ms World Singapore 2014 finalist) – suffered from anorexia from the tender age of 15 because of society’s perceived notion of beauty. Had to battle with her illness for seven years before making a comeback
http://yourhealth.asiaone.com/content/miss-world-singapore-finalist-was-anorexic

No – women are not more disadvantaged (does not necessarily mean they have an advantage either)

1. Meritocracy is an ideal that is practiced in many aspects of society
a. In school, career etc meritocracy is practiced. This means that women have equal opportunities as men. For example, in the public service sector, employees are put into different grades that can be roughly mapped to experience. Those who work harder and take up more responsibilities are promoted faster and thus move up the grades quicker. Using this type of meritocratic system, both men and women have equal opportunities, thus women are not more disadvantaged than men.
This notion is also applicable in the education context, where students, regardless of gender are able to receive monetary and other forms of rewards based on their results, which can be seen as a tangible measure of how hard a student works.

2. Women can be seen to be in a position of greater advantage as they are better protected by the law.

a. Women’s Charter is a legislative act that was passed in 1961 to protect and advance the rights of women and girls in Singapore. It is a wide-ranking charter that brings together the regulation of the relationship between husband and wife and the relationship between parents and their children, termination of marriages and division of matrimonial assets. It covers areas such as child custody, divorce and matrimonial rights.

3. Women are recognised for their outstanding contributions to society

a. Her World magazine organises the Women of the Year award and gives them to women who have a positive impact on the community through their selfless dedication and unflagging spirit.
b. E.g. Rachel Eng 2014 winner – notable contribution as a mother of three and as a dedicated legal professional.
c. Comparatively, there is no such award for men. This could suggest that women are at a greater advantage since the recognition could open the route to more opportunities in career or social aspects.
d. In addition to the womens’ awards, there are also other awards that have seen women awardees. E.g. The prestigious cultural medallion was presented to Jennifer Tham in 2012 to recognise her achievements in artistic excellence in the music scene in Singapore

For complete articles Please contact Angie 96790479 or Mr Ong 98639633

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