JURNAL MATEMATIK TAMBAHAN

PEMENANG PERTANDINGAN MENGHASILKAN JURNAL MATEMATIK TAMBAHAN
ANJURAN PANITIA MATEMATIK TAMBAHAN

(BULAN SAINS DAN MATEMATIK SMK GAJAH BERANG 2021)

TINGKATAN 4 : LEE PEI XIN 4A2
KWEK CHEE LING 4A1
TEMPAT PERTAMA : DAVID CHAN KENG HOE 4A1
TEMPAT KEDUA :
TEMPAT KETIGA :

TINGKATAN 5 : TAN TZE HUEY 5A1
LI YI TING 5A3
TEMPAT PERTAMA : -
TEMPAT KEDUA :
TEMPAT KETIGA :

PERTANDINGAN MENGHASILKAN JURNAL MATEMATIK TAMBAHAN
PANITIA MATEMATIK TAMBAHAN SMK GAJAH BERANG
(MINGGU SAINS DAN MATEMATIK 2021)

Objektif Program

✔ Untuk meningkatkan minat pelajar terhadap mata pelajaran Matematik Tambahan.
✔ Untuk melahirkan para pelajar yang kreatif dan inovatif.

Sasaran :
✔ Pelajar Tingkatan 4 & 5 yang mengambil subjek Matematik Tambahan

Tarikh Pelaksanaan

✔ Tarikh : 1 hingga 23 Julai 2021

Pilihan Tajuk :

1. Circular Measure (Form 5)
✔ Are you more inclined to measure an angle of a circle in degrees or radians? Give
justification and rationale for your answers.
✔ Visit the website to obtain the radius, in m, for the following six Ferris wheels
(a) Eye on Malaysia
(b) Wiener Riesenrad, Vienna
(c) The London Eye
(d) Tianjin Eye, China
(e) High Roller, Las Vegas
(f) The Singapore Flyer
If the coordinates of the centre of each Ferris wheel is (0, 0), determine
(i) the circumference of each Ferris wheel, in m,
(ii) the area, in m2 , covered by each Ferris wheel in one complete
oscillation,
(iii) the equation for each Ferris wheel

2. System of Equations (Form 4)
✔ Think of a problem in your surroundings which can be solved using systems of
linear and non-linear equations.
✔ Formulate the problem in the form of a system of linear equations with proper
definitions for the variables used.
✔ State the relation between the variables.
✔ Solve the system of equations which has been constructed.

Format penghasilan jurnal :

1. Nama, Kelas, Mata pelajaran, Guru Mata Pelajaran, Tajuk, Isi kandungan, Dapatan
penyelesaian masalah

2. Bentuk pelaporan : PDF (perlu dihantar kepada guru mata pelajaran)

Ganjaran kepada pemenang bagi setiap tingkatan :
1. Tempat pertama : RM50 baucer koperasi GBS dan sijil

2. Tempat kedua : RM30 baucer koperasi GBS dan sijil
3. Tempat ketiga : RM20 baucer koperasi GBS dan sijil

TINGKATAN 4

SYSTEM OF EQUATIONS

NAME: LEE PEI XIN
CLASS: 4A2

TEACHER: PN. WAN MALINA

NO. CONTENT PAGE
1. Introduction of System of Equations 1
2. Problems that can be solved by system 2

of equations 3
3. Formulation of Problem 1 4
4. Formulation of Problem 2 5-6
5. Relations of Variables 7
6. Calculation for Problem 1 8-9
7. Calculation for Problem 2

INTRODUCTION
A system of equations is a collection of two or more

equations with a same set of unknowns.
In solving a system of equations, we try to find values

for each of the unknowns that will satisfy every
equation in the system.
SIGNIFICANCES

• Engineers use system of equations to solve
problems involving voltages, currents and
resistances.

• Biomedical, chemical, electrical, mechanical and
nuclear engineers use system of equations to
derive the measurements of solids and liquids.

In daily life, we might face a lot of problems involving
unknowns. Hence, system of equations is useful to
solve our problems. For example:
PROBLEM SOLVED BY SYSTEM OF LINEAR EQUATION:
1. Determine the cost of each item in a package
• Sunny Café has offered some exclusive set meals

for the customers. System of linear equations can
help us to determine the price of each item in the
set meals and decide which is the most suitable
one.

PROBLEM SOLVED BY SYSTEM OF NON-LINEAR EQUATION:

1. Determine the number of items can be packed into
a container

• A chocolate packaging factory wants to design a
container in the shape of a right prism with a square
base. System of non-linear equations can help it to
figure out how many pieces of mini chocolates can be
packed into the container.

Formulate the problem in the form of a system of linear
equations with proper definitions for the variables used.

Problem 1: Determine the cost of each item in a package
Sunny Café has offered some exclusive set meals for the
customers. Determine the price of a piece of cake, a glass of
iced lemon tea and a plate of pasta.

SUNNY CAFÉ SET MEALS 2 = 13
SET A : RM 13 2 = 25
• Iced lemon tea x 1 = 35
• Chocolate lava cake x 1
• Tomato pasta x 1

SET A : RM 25
• Iced lemon tea x 2
• Chocolate lava cake x 1
• Tomato pasta x 2

SET A : RM 35
• Iced lemon tea x 2
• Chocolate lava cake x 2
• Tomato pasta x 3

Formulate the problem in the form of a system of linear
equations with proper definitions for the variables used.
PROBLEM 2: Determine the number of items can be
packed into a container
A chocolate packaging factory wants to design a
container in the shape of a right prism with a square
base. Given its total length is 133cm and its hypotenuse
of the triangle is 25cm. The volume of a piece of mini
chocolate is 14cm³. Determine how many pieces of mini
chocolate can be packed into the container.

-------
5
5 = 83 -----------②

State the relation between the variables.

Problem 1: Determine the cost of each item in a package
Sunny Café has offered some exclusive set meals for the
customers. Determine the price of a piece of cake, a glass of
iced lemon tea and a plate of pasta.

= 13 ------①
2 = 25 -------②
2 = 35 -------③

x = price of a glass of iced lemon tea
y = price of a piece of cake
z = price of a plate of tomato pasta

State the relation between the variables.
PROBLEM 2: Determine the number of items can be
packed into a container
A chocolate packaging factory wants to design a
container in the shape of a right prism with a square
base. Given its total length is 133cm and its hypotenuse
of the triangle is 25cm. The volume of a piece of mini
chocolate is 14cm³. Determine how many pieces of mini
chocolate can be packed into the container.

-------
5
5 = 83 -----------②

x = length of side of base
y = length of height

Solve the system of equations which has been constructed.
Problem 1: Determine the cost of each item in a package
Sunny Café has offered some exclusive set meals for the
customers. Determine the price of a piece of cake, a glass of
iced lemon tea and a plate of pasta.

= 13 ------①
2 = 25 -------②
2 = 35 -------③
③-②: y + z = 10

z = 10 – y -------④
① x 2 : 2x + 2y + 2z = 26 ------⑤
⑤-②: y = 1
sub y = 1 into ④,
z = 10 – 1
z=9
sub y = 1 and z = 9 into ①,
x + 1 + 9 = 13
x=3

Price of a piece of cake = RM 1
Price of a glass of iced lemon tea = RM 3
Price of a plate of pasta = RM 9

Solve the system of equations which has been constructed.
PROBLEM 2: Determine the number of items can be packed
into a container
A chocolate packaging factory wants to design a container in
the shape of a right prism with a square base. Given its total
length is 133cm and its hypotenuse of the triangle is 25cm. The
volume of a piece of mini chocolate is 14cm³. Determine how
many pieces of mini chocolate can be packed into the
container.

-------
5
5 = 83 -----------②

From ②: −−−−−−−−−−

Sub ,

= 25²

² = 625

4x² + 25x² - 830x + 6889 – 2500 = 0
29x² - 830x + 4389 = 0
(29x -627)(x-7) = 0

x = or x=7

sub x = into ③, sub x=7 into ③,

y = 24

y = (rejected)

Solve the system of equations which has been constructed.
PROBLEM 2: Determine the number of items can be packed
into a container
A chocolate packaging factory wants to design a container in
the shape of a right prism with a square base. Given its total
length is 133cm and its hypotenuse of the triangle is 25cm. The
volume of a piece of mini chocolate is 14cm³ Determine how
many pieces of mini chocolate can be packed into the
container.

Volume of container = x 7 x 7 x 24



Volume of container = 588
Total of mini chocolates that can be packed into the
container:
588 4cm³ = 42

588

THE END

PERTANDINGAN MENGHASILKAN JURNAL
MATEMATIK TAMBAHAN

PANITIA MATEMATIK TAMBAHAN
SMK GAJAH BERANG

(MINGGU SAINS DAN MATEMATIK 2021)

NAMA : KWEK CHEE LING
KELAS : 4A1
MATA PELAJARAN : MATEMATIK TAMBAHAN
GURU MATA PELAJARAN : PN. LOW MEI HUA
TAJUK : SYSTEM OF EQUATIONS

ISI
KANDUNGAN

01 Solving Problems Involving
Linear Equation and Non-Linear Equation

02 Understanding the problem

03 Planning a Strategy

04 Implementing the Strategy

05 Making a conclusion

Solving Problems Involving
Linear Equation and Non-Linear Equation

The diagram below show a plan of a rectangular garden
PQRS. The garden consists of a playground in a shape of

right-angled triangle TQU and a grassy area PSRUT.

P TQ

U

SR

It is given that TU = 5m and U is at the center of QR. The
length of RS is 6 times of the length of QT. The perimeter

of the rectangular garden PQRS is 60m. Find the the
length and width of the rectangular garden PQRS.

Understanding the problem

The garden is in a shape of rectangular while the
playground is in a shape of right-angled triangle.

TU=5m. U is the midpoint of QR, thus QU=UR.

The length of RS is 6 times of the length of QT

The perimeter of the rectangular garden PQRS is
60m.

To determine the length and width of the
rectangular garden PQRS

Planning a Strategy

Let the length of the side of triangle TQU,QT be
y m. Let the length of the side of triangle TQU,
QU be x m.
Thus, the length of the rectangular garden
PQRS, RS is (6y) m. The width of the rectangular
garden PQRS, QR is (2x) m.

Form a linear equation for the perimeter of the
rectangular garden PQRS.

Form an non-linear equation for the length of the
sides of the right-angled triangle TQU.

Implementing the Strategy

x² + y² = 5² ......

2(2x) + 2(6y) =60

4x + 12y = 60

x + 3y = 15 ......
x= 15-3y ......

Sub. into

(15-3y)² + y² = 5²
225-90y+9y² + y² = 25

10y²-90y+200= 0
y²-9y+20= 0
(y-4)(y-5)=0
y=4m y=5 (ignored as y<5))

Sub. y=4 into
x=15-3(4)
x=3m

Making a conclusion

Thus,
the length of the rectangular garden PQRS,
RS is 6y= 6(4) m

=24m

the width of the rectangular garden PQRS,
QR is 2x= 2(3) m

=6m

TThhee

EEnndd!!

SMK GAJAH BERANG
BULAN SAINS & MATEMATIK, 2021

'WRITE YOUR JURNAL' COMPETITION

FORM 4 : SYSTEMS OF EQUATIONS

Organised by Panel of Additional Mathematics

ROBBERY CASE

Name: David Chan Keng Hoe
Class: 4A1
Subject: Additional Mathematics
Subject Teacher: Pn. Low Mei Hua

Index

1. Problem Statement …………………………. pg 3
2. Formulate Equation …………………………. pg 4
3. Solving Problem ………………………………. pg 5

2

Problem Statement

Yesterday night, a thief stole a pack of jewelry from a rich merchant's house.
According to the words of the security which was chasing the thief yesterday, both of
them ran in a linear line. The thief jumped through the window and broke the wooden
fence to get into the garden. He was able to jump over the circular pond with a 2m
radius and escaped.

The coordinate of the centre of the pond is (1, 10). The coordinate of the window
and the broken fence are (15, 0) and (12, 2) respectively. As the thief didn't leave any
trace in the building, the police need to find his footprint which is the most important
clue at the surface of the wet soil around the pond. It took time because the soil are
covered by the grasses. How can they find out the coordinate of the footprints more
easily?

3

Formulate Equation

Equation of the linear line tracing by the thief while escaping:

Let W(15, 0) be the coordinate of the window and B(12, 2) be the coordinate of the
broken fence.

Gradient, mWB = (y₂ - y₁) / (x₂ - x₁)
= (2 - 0) / (12 - 15)
= -2/3

y = mx + c
0 = -2/3 (15) + c

= -10 + c
c = 10

y = -2/3x + 10

Equation of the circle of the circular water pond’s edge:

Let P(x, y) is a moving point around the circular water pond’s edge and C(1, 10) is the
coordinate of the centre point of the pond.

PC = 2
√ (x - 1)² + (y - 10)² = 2
(x – 1)² + (y – 10)² = 4
x² - 2x + 1 + y² - 20y + 100 = 4
x² + y² - 2x - 20y + 97 = 0

Simultaneous Equation

y = -2/3x + 10 ……………………………...①

x² + y² - 2x - 20y + 97 = 0……………...②

1. The wet soil covered around the edge of the water pond and form a circle.

2. The first equation shows the route covered by the thief at yesterday night while

the second equation shows the circle formed by the wet soil.

3. (x, y) represent the coordinate of their intersection points. The last footprint

before the thief jumped over the pond and the first footprint after he jumped can

be found around the intersection points. 4

Solving Problem

y = -2/3x + 10 ……………………………...①
x² + y² - 2x - 20y + 97 = 0……………...②
-----------------------------------------------------------------------------------------------------------------
Sub ① into ②.
x² + (-2/3x + 10)² - 2x - 20(-2/3x + 10) + 97 = 0
x² + 4/9x² - 40/3x + 100 - 2x + 40/3x - 200 + 97 = 0
13/9x² - 2x - 3 = 0

x = ( -b ± √b² - 4ac ) / 2a

= [ -(-2) ± √(-2)² - 4(13/9)(-3) ] / 2(13/9)

= ( 2 ± √64/3 ) / 26/9

x = ( 2 + √64/3 ) / 26/9 or x = ( 2 - √64/3 ) / 26/9

x = 2.29 x = -0.91

Sub x = ( 2 ± √64/3 ) / 26/9 into ①.

y = -2/3 [( 2 ± √64/3 ) / 26/9] + 10

y = 2/3 [( 2 +√64/3 ) / 26/9] + 10 or y = 2/3 [( 2 -√64/3 ) / 26/9] + 10

y = 8.47 y = 10.60

⚫ The footprints of the thief can be found around the coordinates (2.29, 8.47) and

(-0.91, 10.60).

5

PERTANDINGAN
MENGHASILKAN JURNAL
MATEMATIK TAMBAHAN

PANITIA MATEMATIK
TAMBAHAN SMK GAJAH

BERANG

System of Equations
(Form 4)

NAME : BRANDON LOH WEI MIN

CLASS : 4A1

SUBJECT : ADDITIONAL MATHEMATICS

SUBJECT TEACHER : PN. LOW MEI HUA

cegrge 01 PROBLEM

- Page 3

02 UNDERSTANDING THE
PROBLEM

- Page 4

03 PLANNING A STRATEGY

- Page 4

04 IMPLEMENTING THE
STRATEGY

- Page 5

05 MAKING A CONCLUSION

- Page 6



Understanding the problem

 Total amount of money for selling horror books, fantasy books
and comic books in January, February and March respectively
are RM 234, RM256 and RM 318.

 In January, two horror books, three fantasy books and six
comic books are sold.

 In February, four horror books, two fantasy books and six
comic books are sold.

 In March, three horror books, four fantasy books and eight
comic books are sold.

Planning a strategy

 Form three equations which involve the total amount of
money for selling the books in January, February and March
respectively.

 Let the price of one book of horror books be x, price of one
book of fantasy books be y and price of one book of comic
books be z.

Implementing the strategy

+ + = ------- 1
+ + = -------- 2
+ + = -------- 3
3 3 −4 1 : x = 18
1 − 2 : −2x + y =− 22 ------- 4

Subtitute x = 18 into 4 :

−2(18) + y =− 22
−36 + y =− 22
y =− 22 + 36
y = 14

Subtitute y = 14 and x = 18 into (1) :

2(18) + 3(14) + 6z = 234
36 + 42 + 6z = 234
78 + 6z = 234
234 − 78
z = 6

z = 26
The price of one book of horror books is RM18, price of one
book of fantasy books is RM 14 and price of one book of comic
books is RM 26.

Making a conclusion

Total amount of money for selling books in January
= 2(18) + 3(14) + 6(26)
= 36 + 42 + 156
= RM 234

Total amount of money for selling books in February
= 4(18) + 2(14) + 6(26)
= 72 + 28 + 156
= RM 256

Total amount of money for selling books in January
= 3(18) + 4(14) + 8(26)
= 54 + 56 + 208
= RM 318



TINGKATAN 5

Ferris Wheels in Malaysia

Name: Tan Tze Huey
Class: 5A1
Subject: Additional Mathematics
Teacher: Pn. Wan Manila
Title: Circular Measure (Form 5)

Contents Page
Number
No. Contents 1
1. Degrees or Radians
2. The radius of Ferris Wheels 1
3. The circumference of Ferris Wheels
1-2
4. The area of Ferris Wheels
5. The equation of Ferris Wheels 2-3

3-4

Circular Measure (Form 5)

1. Are you more inclined to measure an angle of a circle in degrees or radians? Give
justification and rationale for your answers.
I am more inclined to measure an angle of a circle in radians. This is because a lot of
formulas that are related to radian angles. The formula is much easier to be written
and to be understood in radians compared to degrees.

2. The radius, in m, for the following six Ferris wheels
a) Eye on Malaysia
radius = 26 m
b) Wiener Riesenrad, Vienna
radius = 30.48 m
c) The London Eye
radius = 60 m
d) Tianjin Eye, China
radius = 55 m
e) High Roller, Las Vegas
radius = 79.25 m
f) The Singapore Flyer
radius = 75 m

3. The circumference of each Ferris Wheel, in m

3.142
360° = 360° × 180°
360° = 6.284 rad

a) Eye on Malaysia

circumference
=
= (26)(6.284)
= 163.384

b) Wiener Riesenrad, Vienna

circumference
=
= (30.48)(6.284)
= 191.5363

c) The London Eye

circumference
=
= (60)(6.284)
= 377.04

1

d) Tianjin Eye, China
circumference
=
= (55)(6.284)
= 345.62

e) High Roller, Las Vegas
circumference
=
= (79.25)(6.284)
= 498.007

f) The Singapore Flyer
circumference
=
= (75)(6.284)
= 471.3

4. The area, in 2, covered by each Ferris Wheel in one complete oscillation

a) Eye on Malaysia

Area

= 1 2
2
1
= 2 (26)2(6.284)

= 2123.992 2

b) Wiener Riesenrad, Vienna

Area

= 1 2
2
1
= 2 (30.48)2(6.284)

= 2919.0135 2

c) The London Eye

Area

= 1 2
2
1
= 2 (60)2(6.284)

= 11311.2 2

2

d) Tianjin Eye, China

Area

= 1 2
2
1
= 2 (55)2(6.284)

= 9504.55 2

e) High Roller, Las Vegas

Area

= 1 2
2
1
= 2 (79.25)2(6.284)

= 19733.5274 2

f) The Singapore Flyer

Area

= 1 2
2
1
= 2 (75)2(6.284)

= 17673.75 2

5. The equation for each Ferris Wheel
a) Eye on Malaysia

Using point (0,0),
√( − 0)2 + ( − 0)2 = 26
2 + 2 = 262
2 + 2 = 676

b) Wiener Riesenrad, Vienna
Using point (0,0),
√( − 0)2 + ( − 0)2 = 30.48
2 + 2 = 30.482
2 + 2 = 929.0304

c) The London Eye
Using point (0,0),
√( − 0)2 + ( − 0)2 = 60
2 + 2 = 602
2 + 2 = 3600

3

d) Tianjin Eye, China
Using point (0,0),
√( − 0)2 + ( − 0)2 = 55
2 + 2 = 552
2 + 2 = 3025

e) High Roller, Las Vegas
Using point (0,0),
√( − 0)2 + ( − 0)2 = 79.25
2 + 2 = 79.252
2 + 2 = 6280.5625

f) The Singapore Flyer
Using point (0,0),
√( − 0)2 + ( − 0)2 = 75
2 + 2 = 752
2 + 2 = 5625

4

Addmath C

Name: Li Yi Ting
Class: 5A3
Subject: Addmath
Subject Teacher’s Name: Cikgu
Title: Circular Measure (Form 5)

ompetition

Wan Malina
)

Content Slide
3
No Title 4-5
1. Quetion 1 6-13
2. Question 2
3. Question 3



Question 1

Are you more inclined to measu
degrees or radians? Give justific
answers.

Answer: I’m more inclined to measu
This is because Radians give a very n
whereas the idea of 360 degrees ma
arbitrary.

ure an angle of a circle in
cation and rationale for your

ure an angle of a circle in radians.
natural description of an angle
aking a full rotation is very

Question 2

✔Visit the website to obtain the rad
Ferris wheels
(a) Eye on Malaysia
(b) Wiener Riesenrad, Vienna
(c) The London Eye
(d) Tianjin Eye, China
(e) High Roller, Las Vegas
(f) The Singapore Flyer