Chapter 1: Map Scales
Scale Factor
a. Transformation resulted in a figure that has the same shape as original, but a different
size.
b. Transformation can be divided into two: enlargement and reduction.
c. Scale factor: The ratio of a pair of corresponding sides or the enlarged or reduced figure to
the original figure.
Scale factor=
Length of a side of the enlarged or reduced figure
Length of the corresponding side of the original figure
d. Scale factor = 1, the transformed figure and its original is the same.
Scale factor > 1, the transformed figure is bigger than its original size.
Scale factor < 1, the transformed figure is smaller than its original size.
Length Scale
The scale of a drawing can be expressed as a ratio of 1: 𝑛
Example: 1:500 means 1 cm on the drawing represent 500 cm on the actual object.
Area Scale
If the map scale is 1 : n, the area scale will be 1 × 1 : n × n.
Example
A football field map was drawn with a scale of 1 : 1500.
Scale = 1cm : 15 m
Area Scale = (11) cm 2 : (15 15 ) m 2
= 1 cm 2 : 225 m 2
33
Chapter 2: Direct and Inverse Proportion
Direct Proportion
2 quantities x and y are said to be directly proportional to each other if x  ky , where k is a
constant.
Example
A is directly proportional to B, given that A  125 when B  5 , find an equation connecting A
and B
A  kB
125  k  5
k  25
A  25B
Inverse Proportion
2 quantities X and Y are said to be inversely proportional to each other if x 
k
, where k is
y
a constant.
Example
(i) A is inversely proportional to B, given that A  125 when B  5 , find an equation
connecting A and B
k
A
B
k
125 
5
k  625
625
A
B
34
Chapter 3: Algebraic Expansion and Factorisation
Algebraic Expansion
Type 1: Expand and Simplify (Rainbow Method)
a)
a  b  c   ab  ac
Example
x 5  y 
 x(5)  x( y)
 5x  xy
Take note that when we remove the bracket, the signs are included in each multiplication.
Type 2: Expand with formula (Algebraic Identity)
a)
(a  b)2  a 2  2ab  b2
b)
(a  b)2  a 2  2ab  b 2
c)
(a  b)(a  b)  a 2  b2
Example 1
( x  2)2  x 2  4 x  4
Example 2
( x  2)2  x 2  4 x  4
Example 3
( x  2)( x  2)  x 2  4
35
Algebraic Factorisation
Type 1: Taking out Highest Common Factor (HCF)
To factorise the algebraic expression given, we need to identify the common factor.
Example 1
5x  xy
 x(5  y)
Example 2
a2b  2ab  ab2
 ab(a  2  b)
Type 2: Group Factorisation
We need to group the terms to find the common factors.
In some situations, we may need to change the operator in order to find the common factor.
Example 1
2 x 2  8x  4  x
 2x( x  4) 1(4  x)
 2x( x  4) 1( x  4)
 (2x 1)( x  4)
Example 2
Factorise 2x  2 y  ax  ay completely.
2x  2 y  ax  ay
 2( x  y)  a( x  y)
 (2  a)( x  y)
36
Type 3: Factorisation by formula (Algebraic Identity)
a 2  2ab  b2  (a  b)2
a 2  2ab  b2  (a  b)2
a 2  b 2  (a  b)(a  b)
Example 1
Factorise x2 18x  81 completely.
x2 18x  81
 x 2  2( x)(9)  (9)2
 ( x  9) 2
Example 2
Factorise x2  8x  16 completely.
x2  8x  16
 x 2  2( x)(4)  (4)2
 ( x  4) 2
Example 3
Factorise x4  81 completely.
x4  81
 ( x 2 )2  (9)2
 ( x 2  9)( x 2  9)
 ( x 2  9)( x 2  32 )
 ( x 2  9)( x  3)( x  3) (This is a complete factorisation.)
37
Type 4: Quadratic Factorisation
Example 1
Factorise a2  5a  6
Observe that
a) the quadratic term, a 2 , is the product of a and a ;
b) the coefficient of x , 5, is the sum of 2 and 3; and
c) the constant term, 6, is the product of 2 and 3.
Using cross-multiplication method to factorise a2  5a  6 , follow the following steps.
Step 1: Factorise the quadratic term, a 2 , into two linear factors.
a2  a  a
Step 2: Factorise the constant term, 6, into product of two numbers.
6  1 6
6  23
Step 3: Place the factors of the quadratic term and the constant term in two columns.
Cross multiply the factors and write the products in the third column. Add up the products in
the third column to see which pair of factors gives +5a as the middle term.
a
-1
+6a
a
+2
+2a
a
+6
−a
a
+3
+3a
a2
−6
+5a
a2
+6
+5a
When we consider 6  1 6 , the sign for 6 will be negative and is different from the algebraic
expression required, therefore (a 1)(a  6) cannot be the factorisation of a2  5a  6 .
∴ a 2  5a  6  (a  2)(a  3)
38
Chapter 4: Algebraic Simplification
Type 1: Algebraic Simplification by Cancellation
Example 1
xy y

x2 x
x x
Type 2: Algebraic Simplification by Cancellation – Factorisation
Example 1
ac  bc  ad  bd c (a  b)  d (a  b)

a 2  b2
(a  b)(a  b)

(a  b)(c  d )
(a  b)(a  b)

(c  d )
( a  b)
Example 2
2 x 2  3x  5 (2 x  5)( x  1)

(2 x  5)2
(2 x  5) 2

( x  1)
(2 x  5)
2x
5
2x
x
1
5x
2x 2
5
3x
39
Chapter 5: Algebraic Fractions and Making Subject
Type 1: Algebraic Fractions
Example 1
3
a
3(a  b)
a



2
2
( a  b) ( a  b)
( a  b) ( a  b) 2
3a  3b  a

( a  b) 2
2a  3b
=
( a  b) 2
Example 2
3
1
3
1



(Change sign to make denominator same)
a b ba a b a b
2

a b
Type 2: Algebraic Fractions with Factorisation
Example 1
5
2
5
2



x  3x x  3 x( x  3) x  3
2

5
2x

x( x  3) x ( x  3)

5  2x
x ( x  3)
Example 2
5
1
5
1


 2
x  3x x  9 x( x  3) ( x  3)( x  3)
2

5( x  3)
x

x( x  3)( x  3) x ( x  3)( x  3)

5 x  15  x
x( x  3)( x  3)

6 x  15
x( x  3)( x  3)

3(2 x  5)
x( x  3)( x  3)
40
Type 3: Making Subject
Example 1
Make x the subject,
2ax  b  3c
2ax  3c  b
x
3c  b
2a
Example 2
Make b the subject,
b 2  4ac
x
2a
2ax  b2  4ac
(2ax)2  b2  4ac
4a2 x2  b2  4ac
4a2 x2  4ac  b2
b2  4a2 x2  4ac
b   4a 2 x2  4ac
41
Chapter 6: Quadratic Equations
Type 1: Basic Concept of Quadratic Equations
Any Number  0  0
0  Any Number  0
If A  B  0, then either A  0 or B  0
x( x  1)  0
(x  2)( x  1)  0
x  0 or x  1  0
x  2  0 or x  1  0
x  0 or x  1
x  2 or x  1
Type 2: Fraction Form
Cross Multiplication Method
a c

b d
ad  bc
Example 1
x7
4

0
6
x 3
x7
4

6
x 3
 x  7  x  3  24
x2  4x  21  24
x2  4 x  3  0
 x  3 x  1  0
x  3  0 or x  1  0
x  3
x  1
42
Chapter 7: Quadratic Graphs
1. The general form for the equation of a quadratic function is y  ax 2  bx  c , where a , b
and c are constants and a is not equal to 0.
2. The graph of a quadratic equation is called a parabola.
3. The graph of y  ax 2  bx  c passes through the point  0,c  where c is the y-intercept.
4. a)
When a is positive  a  0  ,
Line of symmetry


0
The quadratic curve is an
upward ‘U’ shape.
The graph opens upwards
indefinitely and has a
minimum point.
Minimum point
b) When a is positive  a  0  ,
Maximum point
Line of symmetry 

0
The quadratic curve is an
downward ‘U’ shape.
The graph opens
downwards indefinitely
and has a maximum
point.
5. The line of symmetry of the graph is the vertical line that passes through its minimum /
maximum point.
43
Example 1
Match the possible equation to each graph by writing down the letter corresponding to the
equation. Use each letter once only.
a) y   x 2
b) y  3x 2
c) y  x 2  1
d) y   x 2  2
Solution
(c)
(d)
(a)
(b)
44
Chapter 8: Simultaneous Equations
Elimination
Solve the following pairs of Simultaneous Equations by Elimination.
2 x  3 y  1
3x  4 y  7
Solutions:
2x  3 y  1 ------ (1)
3x  4 y  7 ----- (2)
(1) × 3
6x  9 y  3 ----- (3)
(2) × 2
6x  8 y  14 ----- (4)
(3) − (4)
Be Careful
9 y  (8 y)  3 14
17 y  17
y  1
Sub y  1 into (1)
2x  3(1)  1
2 x  3  1
2x  2
x 1
x  1, y  1 (Ans)
45
Substitution
2 x  3 y  1
3x  4 y  7
Solutions:
2x  3 y  1 ------ (1)
3x  4 y  7 ----- (2)
From (1)
2x  1  3 y
x
1  3 y
----- (3)
2
Sub (3) into (2)
 1  3 y 
3
  4y  7
 2 
3(1  3 y)  8 y  14
3  9 y  8 y  14
17 y  17
y  1
Sub y  1 into (3)
x
1  3(1)
2
1  3
2
2

2
1

x  1, y  1 (Ans)
46
Graphical Method
Solve the following pairs of Simultaneous Equations graphically.
2 x  3 y  1
3x  4 y  7
2 x  3 y  1
3x  4 y  7
(1,1)
Ans: x  1, y  1
47
Chapter 9: Financial Mathematics
Simple Interest
For a sum of money (Principal sum), P, deposited in a bank at R% interest per annum for T
years, the simple interest (I) is given by:
I
PRT
100
Compound Interest
Compound interest can also be computed half-yearly, quarterly or monthly. In this situation,
we need to find out the interest rate per half year, per quarter or per month.
R 

Amount  P 1 

 100 
n
where for a sum of money (Principal sum), P, deposited in a bank at R% interest per
annum for n, the number of times compounded according to the time span given in a
question.
Money Exchange
Exchange rate = Home currency ÷ New currency
Example
The exchange rate between the US dollars (US$) and the Singapore dollars (S$) on a certain
day was US $1 = S $1.3542
Convert US $200 to Singapore dollars:
Solution:
US $200 = S $1.3542 × 200 = S $270.84
Convert S $500 to US dollars:
Solution:
S $500 ÷ 1.3542 = US $369.22
48
Income Tax:
Chargeable Income = Assessment Income – Personal Relief
Assessment Income = Annual Income – Donations
Example
Alan’s chargeable income was $75,000. Using the table for Year of Assessment 2017
(below), calculate Alan’s income tax payable.
Solution: (The computation does not include reliefs and donations)
Chargeable Income
Income Tax Rate (%)
Gross Tax Payable ($)
7
550
2,800
First $40,000
Next $40,000
Income tax payable = 550 +
7
× (75,000 – 40,000)
100
= 550 + 2,450
= $3,000 (Ans)
Hire Purchase
We can buy goods by paying for it immediately (cash payment) or by spreading the payment
over a period of time.
To buy on hire-purchase, often a deposit or initial down payment is required. The loan plus
interest is then calculated and paid through regular instalments.
Total Interest = Loan × Flat Rate x Loan Period (in years)
Repayment Amount = Loan + Total Interest
Monthly Repayment (Instalment) = Repayment Amount ÷ Loan Period (in months)
49
Chapter 10: Speed-Time Graph
- Definitions
Initial = at the beginning
Instantaneously at rest
Stationary
Constant velocity
1.
t=0
v=0
v = 0,
a=0
Velocity is the rate of change of displacement or distance in a particular direction with
respect to time. (Can be Negative  Going in Negative Direction)
Formula :
Velocity 
distance moved in a given direction
time taken
Positive / negative velocity = the object is moving forward / backward respectively
2.
Speed is the rate of change of distance. (Cannot be Negative)
Formula :
3.
distance travelled
time taken
Average speed
Formula :
4.
Speed 
Average Speed 
Total distance travelled
Total time taken
Acceleration is the rate of change of velocity with respect to time.
Formula :
Acceleration 
change in velocity
time taken
Acceleration 
final velocity - initial velocity
time taken
50
Distance-time
The diagram shows a distance-time graph of a car travelling from City A to City B.
Distance (m)
(2) At rest and not moving
B
(1)
(3) Returning to City A
A
0
10
30
Time (s)
t
The gradient of a distance‒time graph gives the speed.
Sign: (1) Positive (+) gradient indicates car travelling away from City A to City B.
(2) Zero gradient indicate car is at rest and not moving.
(3) Negative (‒) gradient indicates car is returning towards City A.
Speed-time
The diagram shows a speed-time graph of a car
Speed (m/s)
20
0
10
25
30
Time (s)
The gradient of a speed‒time graph gives the acceleration.
The area under a speed‒time graph gives the distance travelled.
Average Speed 
Total distance travelled
Total time taken
51
Distance‒time graph
Speed‒time graph
Acceleration‒time graph
Stationary
object
Uniform motion
Motion with
constant
acceleration
52
Chapter 11: Congruence And Similarity
Congruence
Two figures are said to be congruent if they are identical in shape and size.
Example:
F
C
Reflected
K
I
Rotated
Translated
J
A
B
D
E
G
a. All of the triangles above are congruent.
H
L
b. Congruency statement: ∆ABC ≡ ∆ DEF ≡ ∆ HGI ≡ ∆ JLK (‘≡’ denotes ‘is congruent
to’).
When we write congruency statement, the letters representing the vertices of the
shape have to be written in marching pairs.
Example:
∆ABC ≡ ∆ JLK
This means that A and J, B and L, C and K are corresponding vertices.
53
Similarity
Two figures are said to be similar if they have the same shape but different sizes.
Example:
F
C
Enlarged
5 cm
4 cm
A
3 cm
10 cm
8 cm
B
D
6 cm
E
a. The two triangles above are similar but not congruent due to the different sizes.
b. Similarity statement: ∆ABC is similar to ∆ DEF. When we write similarity statement,
the letters representing the vertices of the shape have to be written in marching pairs.
For example: ∆ABC is similar to ∆ DEF
c. The corresponding angles of similar figures are equal.
Example: ∆ABC is similar to ∆ DEF
∠ BAC =∠ EDF, ∠ ABC =∠ DEF, ∠ ACB =∠ DFE
d. The ratios of corresponding sides of similar figures are equal.
Example: ∆ABC is similar to ∆ DEF.
Therefore:
AB BC CA
=
=
DE EF FD
3 cm 4 cm 5 cm 1
=
=
=
6 cm 8 cm 10 cm 2
e. The above mentioned rules are applicable to all polygons.
54
Chapter 12: Pythagoras Theorem and Trigonometry
Pythagoras Theorem
Hypotenuse
Trigonometry
opposite
hypotenuse
adjacent
cos θ 
hypotenuse
opposite
tan θ 
adjacent
sin θ 
SOH
CAH
TOA
Example
In the right-angled triangle ΔABC , AB = 12 cm, BC = 16 cm and AC = 20 cm.
Find the values of
(a)
(b)
(c)
(d)
tan θ ,
cos θ ,
sin θ ,
θ .
( a ) tan θ 
A

20
12
16
3
16
20
4

5
(b) cos θ 
4
12
20
3

5
(c) sin θ 
12
C
θ
16
B
1  3 
(d) θ = tan    36.9
4
55
Chapter 13: Measurement
Pyramid
A pyramid is a solid that has a base with a perpendicular vertex and slant lateral faces.
The base can be a triangle, a square or a rectangle.
s
Nets of pyramid
s
Volume =
1
 base area  h
3
Base area = s²
1
 s  l  4 sides
2
 2 sl
Surface area 
Surface area of pyramid = base area + total area of slant ∆ faces
= s2 + 2sl
Cone
A cone is a solid with a circular base and a vertex.
l
vertex
l
θ°
Arc length = circumference of circular base
Volume =
Curved surface area =  rl
1 2
r h
3
Surface area of a cone = Area of circle + Area of curved surface
  r 2   rl
r = radius
l = length of slant height =
r 2  h2
56
Sphere and Hemisphere (half-sphere)
Every point on the surface of a sphere is equidistance from the centre.
A hemisphere is half a sphere.
r
Volume =
r
4 3
r
3
1
2
 volume of sphere =  r 3
2
3
1
Surface area =  spherical surface area + area of circle
2
Volume =
Surface area of a sphere = 4 r 2
 2 r 2   r 2  3 r 2
Prism
Volume = base area  height
Or area of cross-section  length
Surface area of a prism = Area of all the faces
l
b
l
h
h
b
1
Volume =   b  h  l
2
57
Chapter 14: Data Analysis
Dot Diagram
A dot diagram consists of a horizontal number line and dots placed above the corresponding
number line value. The number of dots above each value indicates how many times each
value occurred.
A sample of dot diagram
Scores of students in a common test
Histogram
We use histogram to represent grouped data with class interval.
A histogram consists of rectangles whose area are proportional to the frequency.
Grades on the Math Test
Grade on the test
The data is grouped into different class interval
There is no spacing between the different rectangles.
The area of the rectangles are proportional to the frequencies.
58
Stem and Leaf
Stem-and-leaf diagram is used to organize a large amount of data.
In a stem and leaf diagram, each value is split into two parts, the stem and the leaf.
The leaf of a number is usually the last digit in a number.
The stem is the remaining digit or digits to the left.
Title: Masses of marbles
Stem
5
6
7
Key:
Leaf
0 6
1 3
2 4
8
5
6
7
5 | 8 means 58 g
Advantages and disadvantages of different forms of statistical representation:
The choice of statistical representation depends on the type of data collected and the purpose
of collecting the data.
Dot diagram
An advantage is that it is an easy way to display small sets of data that do not contain many
distinct values.
A disadvantage is that if a set of data is too large, the diagram may appeared packed.
Histogram
An advantage is that the data sets with the highest and the lowest frequencies can be easily
determined.
Stem-and-leaf diagram
An advantage is that individual data values are retained.
The shape of the data distribution can be easily observed if appropriate units are used.
A disadvantage is that it is not useful if there are extreme values in a set of data as many
stems with no leaves will be present.
59
Measures of central tendency
A measure of central tendency refers to a value that indicates the approximate centre of a set
of data. The 3 most common measures of central tendency are the mean, median, mode.
MEAN
1. The mean of a data set is the sum of all the data values divided by the total number of data.
Sum of all data
Mean 
Total number of data
MEDIAN
1. The middle value of a set of ordered numbers (ascending or descending) is the median.
2. For a set of n data arranged in ascending or descending order
If the total number of data is odd, the median is the middle value.
If the total number of data is even, the median is the mean of the 2 middle values.
n 1
The middle position of a set of data 
2
MODE
1. Mode of a data set is the value which occurs the most frequently (i.e. the data that has the
highest frequency) or
2. Modal class is the class interval with the highest frequency.
Example
Consider the data: 1, 2, 3, 4, 5, 5,
1 2  3  4  5  5
6
10
=
3
3 4
Median =
2
= 3.5
Mode = 5
Mean =
60
Chapter 15: Probability
Introduction of Probability
Probability is a measure of chance.
The probability of an event, A is:
P(A)=
k
m
Where k is the number of outcomes of A while m is the total number of possible outcomes.
When the probability is 0, it is impossible to happen.
When the probability is 1, it will certainly happen.
0  Probability  1
Mutually Exclusive Events and Addition Law
Two events are called mutually exclusive events if they cannot occur at the same time.
Eg. In tossing a coin, the event A of getting 1 head and the event of event B of getting 1 tail
are exclusive events. They cannot happen at the same time.
Addition Law
If A and B are exclusive events, then the probability that either A or B occurring is given by
P(A or B) = P(A) + P(B)
Example
A bag contains 4 red balls, 5 green balls and 6 yellow balls. If a ball is drawn at random from
the bag, find the probability that the ball is either green or yellow.
Total of balls in the bags
 456
 15
P(ball is either green or yellow)
= P(ball is green) + P(ball is yellow)
5 6
 
15 15
11

15
61
Independent Events and Multiplication Law
Two events are called independent events if the occurrence of one event does not affect the
probability of occurrence of the other event.
Eg. In tossing a coin and a die, the event A that the coin is a head and the event B that the
number on the die is even are independent events.
Multiplication Law
If A and B are independent events, then the probability that both A and B occurring is given
by
P(A and B) = P(A) × P(B)
Example
A bag contains 4 red balls and 6 green balls. Two balls are drawn from the bag one by one with
replacement. Find the probability that
i)
both balls are red,
ii) one ball is red and one ball is green.
i)
P(both balls are red)
= P(1st ball is red) × P(2nd ball is red)
4 4
 
10 10
16

100
4

25
ii)
P(one ball is red and one ball is green)
= P[(1st ball is red and 2nd ball is green) or (1st ball is green and 2nd ball is red)]
 4 6 6 4
     
 10 10   10 10 
24 24


100 100
48

100
12

25
62