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

Solve the pair of simultaneous equations
6x + 5y = 6
2 y – 3x = 15

Worked Solution

6x + 5y = 6 …… (1)
2 y -3x = 15 ⇒ 4 y – 6x = 30 …… (2)
(1) + (2): 9 y = 36
y = 4
Then (1): 6x + 5(4)= 6
6x = -14
x = -7/3

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O-Level Additional Mathematics Tuition Singapore

Question

Find the range of values of k for which the line x + 3y = k intersects the curve
y^2 = 2x + 3 at 2 distinct points.

(ii) State the value of k if the line x + 3y = k is tangent to the curve
y^2 = 2x + 3.

Answer

i) k > -6
ii) k = -6

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

Chromatography

So how does chromatography separate a mixture?
The components in a mixture have
(1) different adsorption by the chromatography paper, and
(2) different solubility in the moving solvent.

– Solutes which are more soluble in the solvent will travel further while the less soluble solutes are found closer to the base of the chromatography paper.

Note. To identify and locate colourless spots, the chromatogram is sprayed a locating agent which can reacts with the substances on the paper to produce a coloured product.

Chromatography can be used
(a) To separate very complex mixtures which contain many components.
(b) When only very small amount of sample.
(c) To identify unknown substances ***

Please contact Angie @ 96790470 or Mr Ong @ 98639633 if you need Pure or Combine Chemistry Tuition

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

Essay

The price of organic fertilizers for organic crops has been rising. Rising income and healthy living campaigns are encouraging households to switch to consuming organic food.

With reference to the above events, assess the relevance of different concepts of elasticities of demand in explaining the effects on expenditure by consumers on organic and non-organic food. [25]

Suggested formats

Introduction
Focus: Discuss the relevance of PED, YED & XED in affecting CE in each market given a rise in the price of organic fertilizers, a rise in income and healthy living campaigns.

Key Definitions:

-Consumer expenditure (CE): Total amount of money that consumers spend on a product.[CE = P x Q]
-PED: Measures the degree of responsiveness of quantity demanded to a change in the price of the good itself, ceteris paribus.
-YED: Measures the degree of responsiveness of demand of a good given a change in income, ceteris paribus.
-XED: Measures the degree of responsiveness of demand of a good to a change in the price of another good B (substitutes / complements), ceteris paribus.

Please contact Mr Ong @ 98639633 if you need the complete model essay

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

Light – Reflection

1. The diagram below shows a ray of light being reflected from a plane surface

2. The following terms are commonly used in the reflection of light

Normal – Imaginary line perpendicular to the surface of reflection
Angle of incidence, i – Angle between the incident ray and the normal
Angle of reflection, r – Angle between the reflected ray and the normal

3. Laws of reflection:
(a) Angle i = Angle r
(b) The incident ray, reflected ray and the normal at the point of incidence all lie on the same plane.

4. Characteristics of an image formed in a plane mirror:
(a) Upright
(b) Virtual (Cannot be captured on a screen)
(c) Laterally inverted
(d) Same size as the object
(e) Image distance from the other side of the surface of reflection is the same as the object’s distance from the surface of reflection.

Please contact Angie @ 96790479 or Mr Ong if you need O level Pure or Combined Physics Tuition

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O Level Physics Tuition Singapore/Tuition O Level Physics/Tutor

Light – Reflection

1. The diagram below shows a ray of light being reflected from a plane surface

2. The following terms are commonly used in the reflection of light

Normal – Imaginary line perpendicular to the surface of reflection
Angle of incidence, i – Angle between the incident ray and the normal
Angle of reflection, r – Angle between the reflected ray and the normal

3. Laws of reflection:
(a) Angle i = Angle r
(b) The incident ray, reflected ray and the normal at the point of incidence all lie on the same plane.

4. Characteristics of an image formed in a plane mirror:
(a) Upright
(b) Virtual (Cannot be captured on a screen)
(c) Laterally inverted
(d) Same size as the object
(e) Image distance from the other side of the surface of reflection is the same as the object’s distance from the surface of reflection.

Please contact Angie @ 96790479 or Mr Ong if you need O level Pure or Combined Physics Tuition

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A-Level Tuition Singapore/GP Tuition/General Paper Tutor

A picture is always more powerful than mere words.’ What is your view?

A picture is worth a thousand words’ is an oft-bandied maxim which depicts the power images hold to convey messages, ideas and opinions to people. Since the dawn of mankind, pictures have been used to communicate, a classic example being
the Palaeolithic cave paintings our prehistoric ancestors used to create. With
the advent of modern technology, pictures today take a multitude of different
forms. The creation of new methods of capturing a moment in time on canvas
has become even more complex and intricate and the invention of cameras has
paved the way for an entirely new form of pictures in the form of photographs. With such technological advances in the ability of pictures to purposefully deliver content and convey thoughts, it does seem like a picture can be more powerful than the spoken or written word. However, this is not always the case as words are also undeniably an important form of communication that is biquitous throughout the entire world.

For complete essay please contact Mr Ong @ 98639633

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

DNA and Genomics Mutation Cancer

Important concepts

1. DNA and Genomics
· DNA replication process (leading and lagging strand synthesis), end replication problem, Meselson-Stahl experiment (proves semi-conservative replication)
· Gene expression
· Genetic code
· Transcription, amino acid activation, translation
· Roles of various enzymes, ribosomes, mRNA, tRNA, rRNA in these processes
· How structure is related to function – ribosomes, mRNA, tRNA, rRNA, DNA
· Comparisons between processes
· Replication vs transcription
· Transcription vs translation
· Replication vs translation
· Replication vs reverse transcription

2. Mutation
· Types of mutation
· Effect on DNA -* mRNA -* protein STRUCTURE -* protein PROPERTY -* protein FUNCTION -* effect on PHENOTYPE

3. Cancer
· Molecular level – oncogenes , tumour suppressor gene
· Cellular level – loss of control of cell cycle checkpoint
· Possible modes of treatment

Please contact Angie @ 96790479 or Mr Ong @ 98639633 if you need H1 H2 Biology Tuition

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

Math Tuition Center Singapore

Arithmetic Progression & Geometric Progress

The sum of the first 100 terms of an arithmetic progression is 10,000. The first, second and fifth terms of this progression are three consecutive terms of a geometric progression. Find the first term and the non-zero common difference of the arithmetic progression.

Answer

First Term = 1
Common difference = 2

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

Physics Tuition Center Singapore

Water is pumped through a hosepipe at a rate of 90 kg per minute. It emerges from the hosepipe horizontally with a speed of 20 m s–1.

Which force is required from a person holding the hosepipe to prevent it moving backwards?

A 30 N
B 270 N
C 1800 N
D 10800 N

Answer

A
90 kg min-1 = 1.5 kg s-1
F = d(mv)/t
= v dm/t
= 20*1.5
= 30 N

Please contact Angie @ 96790479 or Mr Ong @ 98639633 if you need A level/JC Physics Tuition

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

Chemistry Tuition Center Singapore

Question

A 2.85 g sample of haematite iron ore, Fe O , was dissolved in hydrochloric acid and the solution diluted to 250 cm3 in a standard flask. A 25.0 cm3 of this solution was completely reduced with excess tin (II) chloride to form a solution of iron(II) ions.

After the remaining tin(II) ions were removed with a suitable reagent, the solution of iron(II) ions was titrated against an acidified solution of 0.020 mol dm-3 potassium dichromate(VI) and required 26.40 cm3 for complete oxidation back to iron(III) ions.

(i) Give the balanced equation for the reaction between iron(II) and dichromate(VI) ions.

(ii) Calculate the percentage of iron(III) oxide, Fe2O3, in the ore.

Answer

(i) Cr2O72–(aq) + 14H+(aq) + 6Fe2+(aq) —> 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)

(ii) 88.7% – Contact Mr Ong @ 98639633 if you need the full work solution

Please contact Angie @ 96790479 or Mr Ong @ 98639633 if you need A level/JC Chemistry Tuition

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

Question
(a) Explain how named factor(s) may increase the chance of cancerous growth [5 marks]
(b) Therapeutic genes can be introduced into stem cells. Discuss why the genes used are more likely to be obtained from a cDNA library, than a genomic DNA library.[5 marks]

Ans
(A)
1. Source: Chemical carcinogens like tar from cigarettes/ionising radiation from UV rays or X-ray/viruses like the human papillomavirus/Epstein barr virus (reject “diet” or “smoking”)
2. Mutations in DNA take place in the form of change in sequence of bases/base substitution/deletion/insertion/inversion.
3. Mutations can also convert proto-oncogenes such as the Ras to oncogenes
4. When proto-oncogenes are converted to oncogenes, it is a gain of function mutation,
5. Resulting in (i) increased protein activity / (ii) increased protein quantity / (iii) active promoter / upregulation of gene expression
6. Carcinogens can also cause the mutation of the tumour suppressor genes such as the p53 gene.
7. This leads to the loss of function mutation.
8. Resulting in inability to control cell cycle / DNA repair / apoptosis / OWTTE
9. With the accumulations of mutations, this leads to excessive cell division/proliferation/lack of programmed cell death or apoptosis.

(b)
1. Genes obtained from the cDNA library lack introns, whereas genes obtained from genomic DNA library still possess introns.
2. Therefore, genes that are obtained from cDNA library can bypass the need for post-transcriptional modifications.
3. Genes obtained from the cDNA library can be genetically-engineered to be placed directly after a constitutive promoter to ensure its continual expression in target cells.
4. Genes obtained from the genomic DNA library may contain an existing promoter that is either weak or not active in the target cells.
5. Sometimes, alternative splicing of genes may derive gene product which is different from the desired gene product, using genes from the cDNA library will ensure the correct gene product is utilised.
6. Genes from cDNA library will be shorter in length as compared to genes from the genomic DNA library due to the absence of introns.
7. A shorter DNA length will ensure facilitate ease of packaging into vectors such as viruses or liposomes.
8. function and sequence of cDNA are known
9. in cDNA specific genes are obtained
10. easier to find the gene sequence with cDNA
REJECT: any view point from Genomic Library perspective.

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

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A-Level Tuition Singapore/GP Tuition/General Paper Tutor

Most of us would know that driving a vehicle, switching on the lights would have a carbon footprint. However, carbon footprint also exists for electronics usage. It’s just that we probably don’t perceive the sending of an e-mail or forwarding of a text message to also leave a carbon footprint. Find out more about how much carbon footprint we leave by reading this article. Perhaps our online activities are not as environmentally friendly as what it appears to be.

For more information on GP Tuition Classes / Secondary English Tuition Classes conducted by Mr Aaron, click here to view his profile and call 96790479 or 98639633 to enquire.

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

Measurement

Question 1

Which pair of units both measure the same quantity?
A km/h and kg/m3
C V and J/C
B N/m3 and Pa
D W and J

Ans C
Potential Difference V = Work done per unit charge = J/C

Question 2

Which instrument is used to measure directly the circumference of a golf
ball?
A calipers
B micrometer
C rule
D tape

Ans : D
A cloth tape can be used to directly measure the length along a curved surface, such as the circumference of a golf ball.

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

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O Level Physics Tuition Singapore/Tuition O Level Physics/Tutor

Measurement

Question 1

Which pair of units both measure the same quantity?
A km/h and kg/m3
C V and J/C
B N/m3 and Pa
D W and J

Ans C
Potential Difference V = Work done per unit charge = J/C

Question 2

Which instrument is used to measure directly the circumference of a golf
ball?
A calipers
B micrometer
C rule
D tape

Ans : D
A cloth tape can be used to directly measure the length along a curved surface, such as the circumference of a golf ball.

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

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