Module & More Math Tg 4

Matematik  Tingkatan 4  Bab 6 Ketaksamaan Linear dalam Dua Pemboleh Ubah 

(a) y > – 1 x + 2, y . x, x < 5 (b) y , –2x + 4, y < 2x + 2, y > 1 x (c) y< 1 x + 2, x > 2, x < 4, y > 0
2 2 2

y y=x y y x=2 x=4

y = 2x + 2

4 y = 1 x
2

x=5 1 2
2
2 2 y = x + 2

x x O24 x
O4 2 y = –2x + 4
–1 O

y = – 1 x+2
2

12. Lorekkan rantau yang memuaskan sistem ketaksamaan linear berikut. TP 4

Shade the region that satisfies the following system of linear inequalities.

CONTOH (a) 2y – 2x < 3, 2y – x + 3 . 0, x + y > –5

2y < 3x + 4, x + 2y – 2 . 0, x < 2 y 2y – 2x = 3

Penyelesaian: Ungkapkan y sebagai perkara

y rumus.
2y = 3x + 4 Express y as the subject of the
formula.
x=2  2y < 3x + 4
( bya <wa32h
Ox x+2 the line) O x
x + 2y – 2 = 0 garis/below 2y – x + 3 = 0

 x + 2y – 2 . 0 1 BAB 6
2
y . – x + 1 x + y = –5

(atas garis/above the line)
 x < 2
(sebelah kiri garis/left of the
line)

(b) 2x – 3y > –6, x + y – 10 , 0, 2x + 3y + 2 > 0 (c) x > 0, y > 0, x + y < 60, x + 4y < 120

y y
2x – 3y = –6
x + y = 60
x + y – 10 = 0 x + 4y = 120
Ox
Ox
2x + 3y + 2 = 0

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  Matematik  Tingkatan 4  Bab 6 Ketaksamaan Linear dalam Dua Pemboleh Ubah

13. Lakar dan lorek rantau yang memuaskan setiap sistem ketaksamaan linear berikut. TP 4

Sketch and shade the region that satisfies each of the following system of linear inequalities.

(a) y , 2x + 3, x > 0, y > 0 (b) y . − 1 x +2, y < x + 2, x , 4 (c) y – x < 8, y > 2x + 3, y , 4
2

x=0 y

8 y x =4 10
6 y=x 8
4 y = 2x +3 6 + 2
2 y=0 4 y–x=8 6
4 y = –4
–2 O 246 2 2 x
–2
–4 –2–O2 246 x –8 –6 – 4 –2–O2 24
y = 2x + 3
y = – 1 x + 2
2

14. Nyatakan sistem ketaksamaan linear yang memuaskan rantau berlorek berikut. TP 4

State the system of linear inequalities which satisfies the following shaded regions.

CONTOH (a) y
y=2
y 2x – 3y – 3 = 0

y=2x+5
3

y = –2x + 7 O x
x
BAB 6 x = –2

O Kesalahan Lazim

2y + x + 4 = 0

y < –2x + 7 ; y. 2 x + 5; y < 2
3
Penyelesaian: 2y + x + 4 < 0 x > –2

y < –2x + 7 • Tidak dapat mentakrifkan 2x – 3y – 3 > 0
rantau berlorek dengan
y, 2 x + 5 ketaksamaan yang betul.
3
Unable to define the shaded
2y + x + 4 > 0 region with correct inequalities.

(b) y = –3x – 15 y y = 4x – 12 (c) y

2y – x = 15 2y = 5x + 3

Ox

y + 1 x = –6 Ox
2 2y + 3x + 5 = 0

y . –3x – 15 2y > 5x + 3
1 2y – x , 15
y+ 2 x > –6 2y + 3x + 5 > 0

y > 4x – 12

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Matematik  Tingkatan 4  Bab 6 Ketaksamaan Linear dalam Dua Pemboleh Ubah 

15. Selesaikan setiap yang berikut.

Solve each of the following.

(a) Rajah di bawah menunjukkan rantau (b) Rajah di bawah menunjukkan rantau
penyelesaian bagi bilangan dua jenis kraftangan,
penyelesaian bagi bilangan roti tuna, x dan roti iaitu bakul, x dan sarung tisu, y yang boleh dibuat
oleh Andak Hanum dalam sehari. TP 5
sardin, y yang boleh dijual di sebuah kedai bakeri
The diagram below shows the solutions region of the
dalam tempoh seminggu. TP 5 number of two types of handicrafts, basket, x and tissue

The diagram below shows the solutions region of the bag, y that can be made by Andak Hanum in a day.
number of tuna breads, x and sardine breads, y that can
be sold in a bakery in a week. y

y

y = –2x + 160 8 y = – 1 x + 8
120 3

100 (80, 100) 6
80

60 y = 40 4 y=4
40

20 y = –x + 120 2

O x
20 40 60 80 100120

O 2 4 6 8 10 12 14 16 18 20 22 24 x

Berdasarkan graf di atas: Berdasarkan graf di atas:

Based on the above graph: Based on the above graph:

(i) Tulis semua ketaksamaan linear dalam sistem (i) Tulis semua ketaksamaan linear dalam sistem

yang mewakili rantau berlorek itu. yang mewakili rantau berlorek itu.

Write all linear inequalities in the system that Write all linear inequalities in the system that
represent the shaded region.
represent the shaded region.
(ii) Jika bilangan roti sardin yang boleh dijual
(ii) Nyatakan bilangan maksimum sarung tisu BAB 6
ialah 80 biji, nyatakan bilangan maksimum
yang boleh dibuat oleh Andak Hanum dalam
roti tuna yang boleh dijual.
sehari.
If the number of sardine breads that can be sold
is 80 pieces, state the maximum number of tuna State the maximum number of tissue bags can be
breads that can be sold.
made by Andak Hanum in a day.
(iii) Adakah kedai bakeri itu boleh menjual 80
(iii) Harga bagi sebuah bakul dan satu sarung tisu
biji roti tuna dan 100 biji roti sardin dalam
masing-masing ialah RM15 dan RM6. Antara
masa seminggu? Berikan sebab anda.
titik (10, 3) dan (12, 4), yang manakah
Can the bakery sell 80 pieces of tuna breads and
100 pieces of sardine breads in a week? State your memberikan hasil jualan maksimum?
reason.
The prices of a basket and a tissue bag are RM15
and RM6 respectively. Which of the point (10, 3)
and (12, 4) will give the maximum sales?

(i) y < –2x + 160 (i) y< – 1 x + 8, y < 4, x > 0, y > 0
y < –x + 120 3
y > 40
x>0 (ii) Bilangan maksimum sarung tisu yang boleh

(ii) Bilangan maksimum roti tuna yang boleh dibuat oleh Andak Hanum ialah 4.
dijual ialah 40 biji.
The maximum number of tissues bags that can be
The maximum number of tuna breads that can be made by Andak Hanum is 4.
sold is 40 pieces.
(iii) Bagi titik (10, 3), hasil jualan maksimum
(iii) Tidak, kerana titik (80, 100) berada luar
daripada kawasan berlorek, maka ia bukan For point (10, 3), the maximum sales
titik penyelesaian bagi sistem tersebut.
= 15(10) + 6(3) = RM168
No, because point (80, 100) is out of the shaded
region, thus it is not the solution point of the system. Bagi titik (12, 4), hasil jualan maksimum

For point (12, 4), the maximum sales

= 15(12) + 6(4) = RM204

Maka, titik (12, 4) memberikan hasil jualan

yang maksimum.

Thus, point (12, 4) gives the maximum sales.

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  Matematik  Tingkatan 4  Bab 6 Ketaksamaan Linear dalam Dua Pemboleh Ubah

(c) Xin Yi mempunyai 50 unit pencelup warna merah dan Bilangan unit diperlukan
42 unit pencelup warna kuning. Dia ingin menghasilkan
dua jenis pencelup warna jingga, iaitu jingga pekat dan Number of units needed

jingga cair. Jadual di sebelah menunjukkan bilangan unit Merah Kuning
pencelup warna merah dan pencelup warna kuning yang
Red Yellow

diperlukan untuk menghasilkan setiap liter pencelup Jingga pekat 5 2
1 2
warna jingga pekat dan pencelup warna jingga cair. Dark orange

Xin Yi has 50 units of red dye and 42 units of yellow dye. She Jingga cair
wants to produce two types of orange dye, which are dark orange
and light orange. The table shows the number of units of red dye Light orange

and yellow dye needed to produce each litre of dark orange and

light orange. TP 6

(i) Diberi x ialah isi padu pencelup warna jingga pekat yang terhasil dan y ialah isi padu pencelup

warna jingga cair yang terhasil. Tulis satu sistem yang terdiri daripada dua ketaksamaan linear bagi

mewakili situasi tersebut.

Given x is the volume of dark orange dye produced and y is the volume of light orange dye produced. Write a
system of two linear inequalities representing the situation.

(ii) Dengan menggunakan skala 2 cm kepada 10 liter pada kedua-dua paksi, bina dan lorek rantau yang

memuaskan sistem ketaksamaan linear tersebut.

By using a scale of 2 cm to 10 litres on both axes, construct and shade the region that satisfies the system of linear
inequalities.

(iii) Berdasarkan graf: / Based on the graph:

(a) Adakah Xin Yi boleh menghasilkan 15 liter pencelup warna jingga pekat dan 20 liter pencelup

warna jingga cair? Berikan justifikasi anda.

Does Xin Yi able to produce 15 litres of dark orange dye and 20 litres of light orange dye? Give your justification.

(b) Berapakah isi padu maksimum pencelup warna jingga pekat yang boleh dihasilkan jika Xin Yi

menghasilkan 18 liter pencelup warna jingga cair? Kemudian, nyatakan baki unit pencelup

BAB 6 warna merah dan kuning yang tinggal.
What is the maximum volume of dark orange dye can be produced if Xin Yi produces 18 litres of light orange

dye? Then, state the remaining units of red dye and yellow dye. 2 cm

(i) Sistem ketaksamaan linear / System of linear inequalities : y
60
5x + y < 50  →  y < –5x + 50
2x + 2y < 42 →  y < –x + 21

(ii) Jadual nilai bagi / Table of values for y = –5x + 50 50
x 0 10 y = –5x + 50
y 50 0
40
Jadual nilai bagi / Table of values for y = –x + 21

x 0 21 30
y 21 0

(iii) (a) Tidak kerana (15, 20) bukan titik dalam rantau penyelesaian. 20 (15, 20)

No because (15, 20) is not in the solution region.

(b) 3 liter pencelup warna jingga pekat boleh dihasilkan jika 18 10

liter pencelup warna jingga cair dihasilkan. O3 10 y = –x + 21 x
20 30
3 litres of dark orange dye can be produced if 18 litres of light orange
dye is produced.

Bilangan unit pencelup warna merah yang digunakan

Number of units of red dye used

= 5(3) + 1(18) = 33

Bilangan unit pencelup warna kuning yang digunakan / Number of units of yellow dye use

= 2(3) + 2(18) = 42

Maka, baki pencelup warna merah ialah 17 (50 – 33) dan pencelup warna kuning tiada baki.

Thus, the remaining units of red dye is 17 (50 – 33) and the yellow dye has no remaining.

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Matematik  Tingkatan 4  Bab 6 Ketaksamaan Linear dalam Dua Pemboleh Ubah 

PRAKTIS SPM 6

Kertas 1 4. Titik yang manakah tidak memuaskan sistem
ketaksamaan x + y < 6 dan x . –2?
1. Antara graf yang berikut, yang manakah mewakili
Which point is not satisfying the system of inequalities
4y + 3x , 24? x + y < 6 and x . –2?
Which of the following graphs represents 4y + 3x , 24?
A (–1, 4) C (0, 5)
A y C y B (4, 2) D (–2, 3)

66 Kertas 2

1. Lorek rantau yang memuaskan ketiga-tiga

O x O x 2015 ketaksamaan y < x + 4, y + x , 3 dan 4y + x > –4
8 8 pada graf yang diberikan.

Shade the region which satisfies all three inequalities
y < x + 4, y + x , 3 and 4y + x > –4 on the graph
B y D y provided.

6 6 y
y=x+4

O x O x
8 8

2. Titik yang manakah bukan dalam rantau y+x=3 x BAB 6
penyelesaian bagi 6x + y > 10? O

Which point is not in the solution region for 6x + y > 10? 4y + x = –4

A (2, –2) C (1, 2) 2. Lorek rantau yang memuaskan ketiga-tiga
B (3, 1) D (2, –1)
2016 ketaksamaan y > –2x + 6, y . x – 1 dan y < 5 pada
3. Laila ingin membeli x buah buku tulis dan y buah graf yang diberikan.
buku lukisan. Sebuah buku tulis dan sebuah buku
lukisan masing-masing berharga RM3 dan RM7. Shade the region which satisfies all three inequalities
Dia ingin berbelanja tidak melebihi wangnya yang y > –2x + 6, y . x – 1 and y < 5 on the graph provided.
berjumlah RM30 dan mempunyai baki wang
selebih-lebihnya RM4. Dia juga memerlukan y
sekurang-kurangnya dua buah buku lukisan.
Nyatakan sistem ketaksamaan linear bagi situasi 6
tersebut. y=5

Laila wants to buy x notebooks and y drawing books. 4
The prices of a notebook and a drawing book are RM3 y=x–1
and RM7 respectively. She wants to spend of not more
that her amount of money, which is RM30 and has the 2
balance of at most RM4. She also needs at least two
drawing books. State the system of linear inequalities for y = –2x + 6 Praktis
the situation. SPM
O x Ekstra
A 3x + 7y < 30 C 3x + 7y , 30 246
3x + 7y > 26 3x + 7y . 26
y > 2 y > 2

B 3x + 7y < 30 D 3x + 7y < 30
3x + 7y > 4 3x + 7y < 26
y < 2 y > 2

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  Matematik  Tingkatan 4  Bab 6 Ketaksamaan Linear dalam Dua Pemboleh Ubah

Sudut KBAT KBAT

Ekstra

BAB 6 Dalam suatu eksperimen, Ahn Foo ingin menentukan ketumpatan suatu objek dalam sejenis cecair.
Objek akan tenggelam jika ketumpatannya lebih daripada ketumpatan cecair manakala objek akan timbul
jika ketumpatannya kurang daripada ketumpatan cecair itu. Diberi bahawa ketumpatan cecair ialah 1.26 g/cm3
dan rumus bagi jisim objek, m = dV dengan keadaan d ialah ketumpatan, dalam g/cm3, dan V ialah isi padu,
dalam cm3, objek itu.

In an experiment, Ahn Foo wants to determine the density of an object in a liquid. The object will submerge if the density
is more that the density of the liquid whereas the object will float if the density is less than the density of the liquid. Given
that the density of the liquid is 1.26 g/cm3 and the formula for the mass of object, m = dV where d is the density, in g/cm3,
and V is the volume, in cm3, of the object.

(a) Tulis satu sistem ketaksamaan linear yang mengaitkan jisim dan isi padu objek itu apabila ia tenggelam
dan timbul dalam cecair itu.

Write a system of linear inequalities that relates the mass and volume of the object when it is submerged and floated
in the liquid.

(b) Sebuah bongkah kayu dengan jisim 21.6 g dan panjang, lebar dan tinggi masing-masing ialah 4 cm, 2 cm
dan 3 cm dimasukkan ke dalam cecair itu. Adakah bongkah kayu itu akan tenggelam atau timbul? Jelaskan
jawapan anda.

A wooden block with a mass of 21.6 g and the length, width and height are 4 cm, 2 cm and 3 cm respectively are
put into the liquid. Will the wooden block submerge or float? Explain your answer.

Jawapan / Answer:

(a) m , 1.26v (objek timbul / object float)
m . 1.26v (objek tenggelam / object submerge)

(b) Isi padu bongkah kayu Ketumpatan bongkah kayu, d

Volume of the wooden block Density of the wooden block, d
m
= 4 × 2 × 3 = V
= 24 cm3 21.6
= 24

= 0.9 g/cm3

0.9 g/cm3 , 1.26 g/cm3.

Maka, bongkah kayu itu timbul kerana ketumpatannya lebih kecil daripada ketumpatan cecair itu.

Thus, the wooden block will float because its density is less than the density of the liquid.

Kuiz 6

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BAB Graf Gerakan

7 Graphs of Motion

7.1 Graf Jarak-Masa

Distance-Time Graphs

NOTA IMBASAN

Graf jarak-masa: (a) Kecerunan OP mewakili kadar perubahan jarak, iaitu laju, suatu objek dari
D istance-time graph: 0 hingga t1. Objek itu bergerak dengan laju seragam dari 0 hingga t1.

Jarak The gradient of OP represents the rate of change of distance,which is speed,of an object
Distance from 0 to t1.The object moves with a uniform speed from 0 to t1.

P (b) Q (b) Kecerunan PQ ialah 0. Maka, laju objek ialah sifar. Objek itu berada dalam
keadaan pegun dari t1 hingga t2.
(a) (c)
The gradient of PQ is 0. Then, the speed of the object is zero. The object is at stationary
O t1 R from t1 to t2.
t2 t3
(c) Kecerunan QR bernilai negatif, ini bermaksud objek itu bergerak dalam
Masa arah yang bertentangan.
Time The gradient of QR is a negative value,this means the object moves in opposite direction.

1. Lukis graf jarak-masa berdasarkan situasi berikut. TP 3

Draw distance-time graph based on the following situations.

CONTOH (a) Encik Murugan memandu lori dari Seremban

Adam sedang bermain bola di dalam rumahnya. ke Nilai. Jarak dari Seremban ke Nilai ialah 25

Jadual di bawah menunjukkan pergerakan bola km. Jadual di bawah menunjukkan pergerakan

yang ditolak ke dinding. lori tersebut.

Adam is playing a ball in his house. The table below shows Mr. Murugan drives a lorry from Seremban to Nilai.
the movement of the ball pushed onto a wall. The distance between Serenban and Nilai is 25 km.
The table below shows the movement of the lorry.
Masa (s)
0 23 7 Masa (min) 0 20 30 60
Time (s) 0 60 90 0 0 15 15 25
Time (min)
Jarak (cm)
Jarak (km)
Distance (cm)
Distance (km)
Penyelesaian:

Jarak (cm) Jarak (km)
Distance (cm) Distance (km)

100 25

80 20

60 15

40 10

20 5

O 1 23 45 67 Masa (s) O 10 20 30 40 50 60 Masa (min)
Time (s) Time (min)

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  Matematik  Tingkatan 4  Bab 7 Graf Gerakan

(b) Jadual di bawah menunjukkan masa yang diambil oleh Amira Jarak/ Distance(m)

untuk berjalan dari rumahnya ke stesen bas. Jadual di bawah 200
150
menunjukkan pergerakan dari rumahnya. 100

The table below shows the time taken by Amira to walk from her house 50
to the bus station every day. The table below shows the movement 0
from her house. 246

Masa (min) 0 2 4 6 8 8 Masa/Time(min)

Time (min)

Jarak (m) 0 50 100 150 200

Distance (m)

CONTOH (c) Sebuah roket dilancarkan secara menegak dari
sebuah pentas. Jarak roket itu dari tanah, d m,
Sebuah kereta bergerak dari bandar X ke bandar diberi oleh d = 1.6t + 8, dengan t ialah masa
Y. Jarak kereta itu dari bandar X, d km, diberi oleh dalam saat selepas roket itu dilancarkan dan
d = 60t, dengan t ialah masa dalam jam selepas kereta
itu memulakan perjalanan dan 0 < t < 4. 0 < t < 20.

A car moves from city X to city Y. The distance of the car A rocket is launched upward from a stage. The
from city X, d km, is given by d = 60t, where t is the time distance of the rocket from the ground, d m, is given
in hour after the car starts the journey and 0 < t < 4. by d = 1.6t + 8, where t is the time in second after the

Penyelesaian: rocket is launched and 0 < t < 20.

t (j) / (h) 0 2 t (s) 0 10

d (km) 0 120 d (m) 8 24

Jarak (m)
Distance (m)

Jarak (km) 40
Distance (km) 32

240

BAB 7 180 24

120 16

60 8

O 1 2 34 Masa ( j) O 5 10 15 20 Masa (s)
Time (h) Time (s)

(d) Suatu zarah bergerak dari X ke Y. Pergerakan ini diwakili oleh Jarak/ Distance(cm) Masa/ Time(s)
persamaan d = 3t + 4, dengan keadaan d ialah jarak dalam
cm dan t ialah masa dalam saat, bagi tempoh masa 0 < t < 5. 20
15
A particle moves from X to Y. The motion is represented by d = 3t + 10
4, where d is the distance in cm and t is the time in seconds, for the
period of 0 < t < 5. 5
0
t (s) 0 5
12345
d (cm) 4 19

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Matematik  Tingkatan 4  Bab 7 Graf Gerakan 

2. Pergerakan sebuah objek diwakili oleh graf berikut. Nyatakan laju objek tersebut. TP 1 Tip

The movement of an object is represented by the following graph. State the speed of the object.

(a) (b) (c) Kecerunan bagi graf
jarak-masa ialah laju.
Jarak/ Distance(m) Jarak/ Distance(m) Jarak/ Distance(km) The gradient of a distance-
10 20 time graph is speed.
15
10

5 Masa/ Time(min) 3 Masa/ Time(s) 0 Masa(j)
3 5 1 6 Time(h)
01 03 Laju = 
5 m min–1 3.5 m s–1 0 km j–1 / km h–1
Laju =  Laju =  Speed =

Speed = Speed =

3. Rajah di bawah menunjukkan graf gerakan Ben berlari di sebuah padang. Padankan graf tersebut dengan

pergerakan Ben yang berikut. TP 2

The diagram shows the graph of motion of Ben running on a field. Match the graph with the following Ben’s movements.

Jarak (m) A B Pergerakan Graf
Distance (m)
350 Movement Graph

C (a) Ben berhenti berehat selepas berlari. AB
Masa (min) BC
O 4.25 Time (min) Ben stops to rest after running. OA

1.5 2.5 (b) Ben berlari ke arah titik permulaan.

Ben runs toward the starting point.

(c) Ben mula berlari dari titik permulaan.

Ben starts to run from a starting point.

4. Jawab soalan berdasarkan situasi berikut. TP 4 Tip

Answer the question based on the following situations. Huraian gerakan perlu melibatkan jarak, masa dan laju.
Description of motion needs to involve distance, time and speed.
CONTOH
BAB 7
Rajah di sebelah menunjukkan graf jarak-masa bagi Linggam Jarak (m) Q R
Distance (m)
berjalan dari rumahnya. 150 P

The diagram shows the distance-time graph of Linggam walking from his 110
house.
O 75 100 150 S
(a) Apakah jarak maksimum Linggam berjalan dari rumahnya? Masa (s)

What is the maximum distance of Linggam walking from his house? 250 Time (s)

(b) Hitung laju Linggam berjalan dari rumahnya.

Calculate the speed of Linggam walking from his house.

(c) Huraikan gerakan perjalanan pulang Linggam ke rumahnya.

Describe the motion of Linggam going back to his house?

Penyelesaian: (c) Perjalanan pulang Linggam ke rumahnya ditunjukkan oleh graf PQRS. Graf
(a) 150 m
The journey of Linggam going back to his house is shown by the graph PQRS. Graph
(b) Laju / Speed
PQ – Linggam berjalan sejauh 40 m dalam masa 25 saat dengan kelajuan 1.6 m/s
= 150 (40 ÷ 25).
75
= 2 m/s Linggam walks 40 m in 25 seconds with the speed of 1.6 m/s (40 ÷ 25).

QR – Dia berhenti selama 50 saat. / He stops for 50 seconds.
RS – Menyambung perjalanannya sejauh 110 m dalam masa 1 minit 40 saat dengan

kelajuan 1.1 m/s (110 ÷ 100).

Continue his journey for 110 m in 1 minute 40 seconds with the speed of 1.1 m/s
(110 ÷ 100).

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  Matematik  Tingkatan 4  Bab 7 Graf Gerakan

(a) Rajah di bawah menunjukkan graf jarak-masa (b) Rajah di bawah menunjukkan graf jarak-masa

bagi Faiza melawat kawannya di hospital dan bagi Encik Tan dari pekan J ke pekan M, melalui

pulang ke rumah. pekan K dan pekan L.

The diagram below shows the distance-time graph of The diagram below shows the distance-time graph of
Faiza visiting her friend in a hospital and back to home. Mr. Tan from town J to town M, passing through town
K and town L.
Jarak (km)
Distance (km) Q Jarak (m) Pekan M
132 P R Distance (m) Town M

92 120
Pekan K

80 Town K

S 60 Pekan L
Town L

O Masa (j) Pekan J
1.2 2.5 3.3 4.0 Time (h) Town J O
Masa (min)
(i) Berapakah jauh hospital tersebut dari rumah 8:00 8:06 8:16 8:30 9:00 Time (min)
Faiza? a.m.
a.m. a.m. a.m. a.m.
How far was the hospital from Faiza’s house?
(i) Berapakah jauh pekan L dari pekan K?
(ii) Berapa lamakah Faiza melawat kawannya di
hospital, dalam jam dan minit? How far is town L from town K?

How long did Faiza visit her friend in the hospital, in (ii) Berapakah jauh pekan M dari pekan J?
hour and minute?
How far is town M from town J?
(iii) Hitung masa yang diambil oleh Faiza untuk
pulang ke rumahnya dari hospital. (iii) Berapa lamakah Encik Tan berhenti di
pekan K?
Calculate the time taken by Faiza to return to her
house from the hospital. How long did Mr. Tan stop in town K?

(iv) Huraikan pergerakan Faiza bagi seluruh (iv) Berapa lamakah Encik Tan bergerak ke
perjalanan tersebut. pekan M dari pekan K?

Describe the movement of Faiza for the whole How long did Mr. Tan travel to town M from town
journey. K?

BAB 7 (v) Huraikan pergerakan Encik Tan dari pekan J
ke pekan M.

Describe the movement of Mr. Tan from town J to
town M.

(i) 132 km (i) Jarak antara pekan L dengan pekan K

(ii) 2.5 – 1.2 = 1.3 jam / hours = 1 jam 18 minit Distance between town L and town K

1 hours 18 minutes = 80 – 60

(iii) 4.0 – 2.5 = 1.5 jam / hours = 20 m

(iv) Laju / Speed (ii) 120 m

= JaMraaksa/ Distance (iii) 8:06 a.m. hingga / to 8:16 a.m.

/ Time = 10 minit / minutes

= 132 + 132 (iv) 8:16 a.m. hingga / to 9:00 a.m.
4
= 66 km j–1 = 44 minit / minutes

66 km h–1 (v) Jarak / Distance

Faiza bergerak sejauh 264 km dalam tempoh = (80 – 0) + (80 – 60) + (120 – 60)
4 jam dengan kelajuan 66 km j–1.
= 160 km

Faiza travelled for a distance of 264 km in 4 hours Laju / Speed
160
with the speed of 66 km h–1. = 1

Tip = 160 km j–1
160 km h–1

Jumlah jarak Encik Tan bergerak sejauh 160 km dalam
Jumlah masa
Laju purata = tempoh 1 jam dengan kelajuan 160 km j–1.

Average speed = Total distance Mr. Tan travelled for a distance of 160 km in 1 hour
Total time with the speed of 160 km h–1.

© Penerbitan Pelangi Sdn. Bhd. 102

Matematik  Tingkatan 4  Bab 7 Graf Gerakan 

5. Selesaikan setiap yang berikut. TP 5 (a)

Solve each of the following. 21

CONTOH

18
Jarak (m)
Distance (m) 8 Masa (s)
O5 11 15 Time (s)
Jarak (m)
Distance (m)
Masa (s)
O6 14 20 Time (s) (i) Hitung laju dalam 5 saat yang pertama.

(i) Hitung laju dalam 6 saat yang pertama. Calculate the speed in the first 5 seconds.

Calculate the speed in the first 6 seconds. (ii) Nyatakan tempoh masa apabila laju adalah

(ii) Nyatakan tempoh masa apabila laju adalah sifar.

sifar. State the period when the speed is zero.

State the period when the speed is zero. (iii) Hitung laju purata.

(iii) Hitung laju purata. Calculate the average speed.

Calculate the average speed. (i) Laju / Speed = —85 = 1.6 m s–1
(ii) Tempoh masa / Time period = 11 – 5 = 6 s
Penyelesaian:
(i) Laju / Speed = —168– = 3 m s–1 (iii) Laju purata / Average speed
(ii) Tempoh masa / Time period = 14 – 6 = 8 s
= —2115–
(iii) Laju purata / Average speed = —18–2—+0—18– = 1.4 m s–1

= 1.8 m s–1

(b) (c)

60 120
Jarak (m)
Distance (m)20 80
Jarak (km) 50
Distance (km)
Masa (s) Masa (j)
BAB 7O Time (s) Time (h)
10 20 25 O2 56

(i) Hitung laju dalam 10 saat yang pertama. (i) Hitung laju dalam 2 jam yang pertama.
Calculate the speed in the first 10 seconds. Calculate the speed in the first 2 hours.

(ii) Nyatakan tempoh masa apabila laju adalah (ii) Hitung laju dalam 1 jam yang terakhir.
Calculate the speed in the last 1 hour.
sifar.
(iii) Hitung jumlah jarak yang dilalui.
State the period when the speed is zero.

(iii) Hitung laju dalam 5 saat yang terakhir. Calculate the total distance travelled.

Calculate the speed in the last 5 seconds. (iv) Hitung laju purata.

(iv) Hitung laju purata. Calculate the average speed.

Calculate the average speed. (i) Laju / Speed = —12–0—2––5—0 = 35 km j–1
(ii) Laju / Speed = —12–0—1––8—0 = 40 km j–1
(i) Laju / Speed = —60–1––0—20– = 4 m s–1
(iii) Jumlah jarak / Total distance
(ii) Tempoh masa / Time period = 20 – 10 = 10 s = (120 – 50) + (120 – 80) + (120 – 80)
= 150 km
(iii) Laju / Speed = —205–––—62—00 = –12 m s–1­ (iv) Laju purata / Average speed = —156—0 = 25 km j–1
= 12 m s–1


(iv) Laju purata / Average speed = —40–2+—56—0 = 4 m s–1

103 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 7 Graf Gerakan

6. Selesaikan setiap yang berikut. TP 5

Solve each of the following.

(a) Rajah di bawah menunjukkan graf jarak-masa (b) Graf jarak-masa di bawah menunjukkan
bagi sebuah perjalanan.
perjalanan antara Kuala Lumpur dengan Tanjung
The diagram below shows the distance-time graph of a
journey. Malim. Graf OPQR mewakili perjalanan sebuah

bas dari Kuala Lumpur ke Tanjung Malim,

Jarak/ Distance(km) manakala graf JQK mewakili perjalanan sebuah

80 P kereta dari Tanjung Malim ke Kuala Lumpur.

KQ The distance-time graph below shows a journey between

0 27 120 Masa/ Time(min) Kuala Lumpur and Tanjung Malim. The graph OPQR
represents the journey of a bus from Kuala Lumpur to
Tanjung Malim, while the graph JQK represents the
journey of a car from Tanjung Malim to Kuala Lumpur.

Jarak/ Distance(km)

(i) Diberi laju purata bagi perjalanan dari P ke Q 120 J R
ialah 50 km j–1. Tentukan nilai K.
80 P Q
Given the average speed for the journey from P to
Q is 50 km h–1. Determine the value of K. 0 40 50 K Masa/ Time(min)
t
(ii) Hitung jarak, dalam km, antara P dan Q.
(i) Cari jarak, dalam km, dari Tanjung Malim
Calculate the distance, in km, between P and Q.
apabila dua kenderaan tersebut bertemu.
(iii) Hitung laju purata, dalam km j–1, bagi
Find the distance, in km, from Tanjung Malim
keseluruhan perjalanan. when the two vehicles meet.

Calculate the average speed, in km h–1, of the (ii) Jika kadar perubahan jarak OP dan QR
whole journey.
adalah sama, hitung nilai t.
80 – K = 50
27 If the rate of changes in distance of OP and QR is
 (i) the same, calculate the value of t.
60
(iii) Kereta tersebut sampai ke destinasinya 10
80 – K = 22.5
minit lebih awal berbanding dengan bas.
K = 80 – 22.5
Hitung laju purata, dalam km j–1, kereta itu.
= 57.5 km
The car reached its destination 10 minutes earlier
BAB 7 (ii) 80 km – 57.5 km = 22.5 km than the bus. Calculation the average speed, in
km h–1, of the car.
(iii) Laju purata / Average speed

  80 = 40 km j–1 / km h–1 (i) 120 – 80 = 40 km
120

60 (ii) 80 – 0 = 120 – 80
40 – 0 t – 50

t – 50 = 120 – 80
2

t = 20 + 50

= 70 min

(iii) Laju purata / Average speed

  = 120 km
70 –10 j / h
60

= 120 km j–1

120 km h–1

© Penerbitan Pelangi Sdn. Bhd. 104

Matematik  Tingkatan 4  Bab 7 Graf Gerakan 

7.2 Graf Laju-Masa

Speed-Time Graphs

NOTA IMBASAN

1. Graf laju-masa: (b) Q zero.The object moves with a uniform speed from t1 to t2.
Speed-time graph:
(c) Kecerunan QR bernilai negatif. Objek itu dikatakan
Laju bergerak dengan nyahpecutan. Laju objek itu
Speed berkurang dari t2 hingga t3.

P The gradient of QR is a negative value. The object moves
with deceleration. The speed of the object decreases from
(a) (c) t2 to t3.

O t1 R Masa 2. Bagi graf laju-masa,
t2 t3 Time For speed-time graph,
• kecerunan mewakili pecutan.
the gradient represents acceleration.
• luas di bawah graf mewakili jarak yang dilalui.
the area under the graph represents the distance travelled.

(a) Kecerunan OP mewakili kadar perubahan laju, O t1Laju Masa
iaitu pecutan, suatu objek sepanjang masa dari 0 Speed t2 t3 Time
hingga t1. Laju objek itu meningkat dari 0 hingga t1.
BAB 7
The gradient of OP represents the rate of change of speed, Jumlah jarak yang dilalui = Luas di bawah graf laju-masa
which is acceleration, of an object from 0 to t1.The speed of Total distance travelled = The area under the speed-time graph
the object increases from 0 to t1.

(b) Kecerunan PQ ialah 0. Maka, pecutan objek itu ialah
sifar. Objek itu bergerak dengan laju seragam dari t1
hingga t2.

The gradient of PQ is 0.Then, the acceleration of the object is

7. Lukis graf laju-masa berdasarkan situasi berikut. TP 3

Draw speed-time graph based on the following situations.

CONTOH (a) Jadual di bawah menunjukkan pergerakan

Jadual di bawah menunjukkan pergerakan sebiji sebuah kereta.

bola. The table shows the motion of a car.

The table shows the motion of a ball. Masa (min) 0 48 10
60 110 110 0
Masa (s) 0 2 6 10 Time (min)

Time (s) Laju (km/j)

Laju (m/s) 4 8 8 14 Speed (km/h)

Speed (m/s)

Laju (m/s) Laju (km/j)
Speed (m/s) Speed (km/h)

14 120

12 100

10 80

8 60

6 40

4 20

2 O Masa (min)
2 4 6 8 10 Time (min)
O Masa (s)
2 4 6 8 10 Time (s)

105 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 7 Graf Gerakan

(b) Jadual di bawah menunjukkan pergerakan (c) Jadual di bawah menunjukkan pergerakan suatu

sebuah motosikal. objek.

The table shows the motion of a motorcycle. The table shows the motion of an object.

Masa (min) 0 20 40 50 Masa (min) 0 2 3 5

Time (min) Time (min)

Laju (km/j) 20 10 30 30 Laju (km/j) 40 40 60 0

Speed (km/h) Speed (km/h)

Laju (m/s) Laju (km/j)
Speed (m/s) Speed (km/h)

30 60

25 50

20 40

15 30

10 20

5 10

O 10 20 30 40 50 Masa (s) O 12 3 45 Masa (min)
Time (s) Time (min)

8. Hitung jumlah jarak yang dilalui bagi graf laju-masa berikut. TP 3 Tip

Calculate the total distance travelled of the following speed-time graphs. Jumlah jarak yang dilalui = Luas di bawah graf
Total distance travelled = Area under the graph
CONTOH (a)

Laju (km j–1) Pastikan unit Laju (m/min)
Speed (km h–1) Speed (m/min)
70 adalah sepadan. 24
Ensure the units 10
are equivalent.

km × j = km
j

Masa (j) O Masa (min)
8 Time (h) 7 Time (min)
O3
Jumlah jarak yang dilalui
BAB 7 Jumlah jarak yang dilalui/ Total distance travelled
= Luas segi tiga / Area of triangle Total distance travelled
= —12 × 8 × 70
= 280 km = Luas trapezium / Area of trapezium
= —12 × (10 + 24) × 7
= 119 m

(b) Laju (m/s) (c) Laju (m s–1)
Speed (m/s) Speed (m s–1)
8 16

O Masa (s) O7 Masa (s)
40 Time (s) 24 Time (s)

Jumlah jarak yang dilalui / Total distance travelled Jumlah jarak yang dilalui / Total distance travelled
= Luas segi tiga / Area of trapezium = Luas trapezium / Area of trapezium
= —21 × 40 × 8 = —21 × (24 + 7) × 16
= 160 m = 248 m

© Penerbitan Pelangi Sdn. Bhd. 106

Matematik  Tingkatan 4  Bab 7 Graf Gerakan 

9. Hitung jumlah jarak yang dilalui dan laju purata bagi graf laju-masa berikut. TP 3

Calculate the total distance travelled and average speed of the following speed-time graphs.

CONTOH Jumlah jarak yang dilalui / Total distance travelled
= —12 × (90 + 40) × 1.5 + —12 × (40 + 75) × (3 – 1.5)
Laju (km/j) = 97.5 + 86.25 = 183.75 km
Speed (km/h)
Laju purata / Average speed
90 = 1—8—33.7—5 = 61.25 km j–1 / km h–1
75

40 B
A

O Masa (j)
1.5 3.0 Time (h)

(a) Laju (cm/s) Jumlah jarak yang dilalui / Total distance travelled
Speed (cm/s) = —12 × 3 × 110 + —12 × (110 + 180) × (9 – 3)
= 165 + 870 = 1 035 cm
180
Laju purata / Average speed
110 = 1—09—35– = 115 cm s–1

(b) O3 Masa (s) Jumlah jarak yang dilalui / Total distance travelled
Laju (km j–1) 9 Time (s) = 45 × 0.5 + —12 × (45 + 100) × (1.4 – 0.5)
Speed (km h–1) = 22.5 + 65.25 = 87.75 km
Masa (j)
100 1.4 Time (h) Laju purata / Average speed
= 8—17—..475– = 62.68 km j–1 / km h–1
45 BAB 7
O 0.5

10. Bagi setiap graf laju-masa berikut, hitung pecutan dan kemudian, Tip

tentukan sama ada ia pecutan, nyahpecutan atau laju seragam. TP 3 Pecutan = Kecerunan graf laju-masa
Acceleration = Gradient of speed-time graph
For each of the following speed-time graphs, calculate the acceleration and then,
determine whether it is an acceleration, deceleration or uniform speed.

(a)  Laju (m/s) (b)  Laju (m/s) (c)  Laju (m/s)

Speed (m/s) Speed (m/s) Speed (m/s)

12 30 35

O Masa (s) O Masa (s) O Masa (s)
30 Time (s) 50 Time (s) 25 Time (s)

Pecutan / Acceleration Pecutan / Acceleration Pecutan / Acceleration
12 – 0 30 – 30 0 – 35
= 30 – 0 = 0.4 m s–2 = 50 – 0 = 0 m s–2 = 25 – 0 = –1.4 m s–2

∴ Pecutan / Acceleration ∴ Laju seragam / Constant speed ∴ Nyahpecutan / Deceleration

107 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 7 Graf Gerakan

11. Jawab soalan berdasarkan situasi berikut. TP 3 Tip

Answer the questions based on the following situations. Huraian gerakan graf laju-masa perlu melibatkan jarak, masa,
laju dan pecutan.
CONTOH Description of motion in speed-time graph needs to involve distance,
time, speed and acceleration.
Rajah di bawah menunjukkan graf laju-masa bagi
(a) Rajah di sebelah menunjukkan graf laju-masa
pergerakan sebuah bot.
bagi pergerakan sebuah objek.
The diagram shows speed-time graph for the motion of
a boat. The diagram shows speed-time graph for the motion of
an object.
Laju (m s–1) / Speed (m s–1)
Laju (m s–1) / Speed (m s–1)
20 Q R
15 P
Q R
16

S 10 P

O5 15 20 4S

Masa (s) / Time (s) O2

7 10
Masa (s) / Time (s)
(i) Hitung kadar
(i) Hitung kadar
perubahan laju dalam 5 saat yang pertama.
perubahan laju dalam 2 saat yang pertama.
Calculate the rate of change of speed in the first
5 seconds. Calculate the rate of change of speed in the first
2 seconds.
(ii) Hitung kadar perubahan laju dalam 5 saat
(ii) Nyatakan tempoh masa apabila laju adalah
yang terakhir.
seragam.
Calculate the rate of change of speed in the last
5 seconds. State the period of time when the speed is uniform.

(iii) Huraikan gerakan bot itu pada 5 saat yang (iii) Hitung kadar perubahan laju dalam 3 saat

pertama. yang terakhir.

Describe the motion of the boat for the first 5 Calculate the rate of change of speed in the last
seconds. 3 seconds.

Penyelesaian: (iv) Huraikan gerakan bot itu pada 3 saat yang

BAB 7 (i) Kadar perubahan laju / Rate of change of speed terakhir.
= —250–—–– —105– = 1 m s–2 Describe the motion of the boat in the last 3 seconds.

(ii) Kadar perubahan laju / Rate of change of speed (i) Kadar perubahan laju = —126–—–– —100– = 3 m s–2
= —200––—–2—105– = –4 m s–2
Rate of change of speed
(iii)Jarak yang dilalui / Distance travelled
= —12 × (15 + 20) × 5 = 87.5 m (ii) Tempoh masa = 7 – 2 = 5 s
Kadar perubahan laju/ Rate of change of speed
= 1 m s–2 Time period —410––—–1—76
Bot itu bergerak sejauh 87.5 m pada 5 saat
(iii) Kadar perubahan laju = = –4 m s–2
pertama dengan pecutan 1 m s–2. Rate of
change of speed
The boat travelled for 87.5 m in the first 5 seconds —21
with an acceleration of 1 m s–2. (iv) Jarak yang dilalui = 30 × (4 + 16) × (10 – 7)
Distance travelled = m

Bot itu bergerak sejauh 30 m dalam masa

3 saat dengan kelajuan menurun dan

mengalami nyahpecutan 4 m s–2 sebelum

berhenti.

The boat travelled 30 m in 3 seconds with
decreasing speed and decelerating at 4 m s–2 before
stop.

© Penerbitan Pelangi Sdn. Bhd. 108

Matematik  Tingkatan 4  Bab 7 Graf Gerakan 

12. Selesaikan setiap yang berikut. (a)

Solve each of the following. v

CONTOH

v

7
Laju (m s–1) 4
Speed (m s–1) O 7 15 Masa (s)
25 Time (s)
Laju (m s–1)
Speed (m s–1)
Masa (s)
O4 12 16 Time (s) Rajah di atas menunjukkan graf laju-masa bagi

Rajah di atas menunjukkan graf laju-masa bagi suatu zarah dalam tempoh 25 saat. TP 4

suatu zarah dalam tempoh 16 saat. Diberi zarah The diagram above shows the speed-time graph of a
particle for a period of 25 seconds.
itu bergerak dengan laju seragam, 12 m s–1, selama
(i) Nyatakan nilai v, sekiranya kadar perubahan
8 saat.
laju dalam 7 saat yang pertama ialah 3 ms–2.
The diagram above shows the speed-time graph of a
particle for a period of 16 seconds. Given the particle State the value of v, if the rate of change of speed in
moves with a uniform speed, 12 m s–1, for 8 seconds. the first 7 seconds is 3 m s–2.

(i) Nyatakan nilai v. (ii) Hitung jumlah jarak yang dilalui oleh zarah
State the value of v.
itu.
(ii) Hitung kadar perubahan laju dalam 4 saat
Calculate the total distance travelled by the
yang pertama. particle.

Calculate the rate of change of speed in the first (iii) Hitung laju purata zarah itu bagi seluruh
4 seconds.
perjalanan.
(iii) Hitung jumlah jarak yang dilalui oleh zarah
Calculate the average speed of the particle for the
itu. whole journey.

Calculate the total distance travelled by the particle.

(iv) Hitung laju purata zarah itu bagi seluruh (i) Kadar perubahan laju
The rate of change of speed
perjalanan. —7v –––—00 = 3
—7v = 3
Calculate the average speed of the particle for the v = 21

whole journey. (ii) Jumlah jarak yang dilalui

Penyelesaian: Total distance travelled

(i) v = 12 = —12 × [(15 – 7) + 15] × 21 BAB 7
+ —12 × (21 + 4) × (25 – 15)
(ii) Kadar perubahan laju / Rate of change of speed = —12 × 23 × 21 + —12 × 25 × 10
= —142––—–07– = 241.5 + 125
= 1.25 m s–2 = 366.5 m

(iii) Jumlah jarak yang dilalui (iii) Laju purata
Total distance travelled Average speed
= —12 × (7 + 12) × 4 + —12 × [(12 – 4)
= —362–65—.5–
  + (16 – 4)] × 12 = 14.66 m s–1

= —12 × 19 × 4 + —12 × 20 × 12
= 38 + 120
= 158 m

(iv) Laju purata / Average speed
= —115–6—8

= 9.875 m s–1

109 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 7 Graf Gerakan (c)

(b) 120

v
Laju (m s–1)
Speed (m s–1)
Laju (km j–1)
Speed (km h–1)
8 Masa (s) v
O4 Time (s)
16 20 O Masa (j)
Time (h)
1.2 t2

Rajah di atas menunjukkan graf laju-masa bagi Rajah di atas menunjukkan graf laju-masa bagi

suatu zarah dalam tempoh 20 saat. Diberi jarak suatu zarah dalam tempoh 2 jam. Diberi kadar

yang dilalui oleh zarah itu dalam 4 saat yang perubahan laju zarah itu dalam 1.2 jam yang

pertama ialah 52 m. TP 5 pertama ialah –75 km j–2 dan jarak yang dilalui

The diagram shows the speed-time graph of a particle oleh zarah itu dengan laju seragam ialah 18 km.
for a period of 20 seconds. Given the distance travelled
by the particle in the first 4 seconds is 52 m. The diagram shows the speed-time graph of a
particle for a period of 2 hours. Given the rate of
(i) Nyatakan laju seragam. change of speed of the particle in the first 1.2 hours is
–75 km h–2 and the distance travelled with the uniform
State the uniform speed.

(ii) Hitung nilai v. speed is 18 km. TP 6

Calculate the value of v. (i) Hitung nilai v.

(iii) Hitung jumlah jarak yang dilalui oleh zarah Calculate the value of v.

itu. (ii) Hitung nilai t.
Calculate the total distance travelled by the particle. Calculate the value of t.

(v) Hitung laju purata zarah itu bagi seluruh (iii) Hitung jumlah jarak yang dilalui oleh zarah

perjalanan. itu.

Calculate the average speed of the particle for the Calculate the total distance travelled by the

whole journey. particle.

(i) 8 m s–1 (iv) Hitung laju purata zarah itu bagi tempoh 2

jam.

(ii) —12 × (v + 8) × 4 = 52 Calculate the average speed of the particle for
2 hours.

BAB 7 2(v + 8) = 52 (i) —v1.––2—1–2–00– = –75
v + 8 = 26 v – 120 = –90
v = 18

(iii) Jumlah jarak yang dilalui v = 30

Total distance travelled (ii) (t – 1.2) × 30 = 18
= 52 + (16 – 4) × 8 + —12 × (18 + 8) × (20 – 16)
= 52 + 96 + 52 t – 1.2 = 0.6

t = 1.8

= 200 m (iii) Jumlah jarak yang dilalui

(iv) Laju purata Total distance travelled
Average speed
= —21 × (120 + 30) × 1.2

= —220–00– + —21 × [(1.8 – 1.2) + (2 – 1.2)] × 30

= 10 m s–1 = 90 + 21

= 111 km

(iv) Laju purata
Average speed

= —112–1–
= 55.5 km j–1

= 55.5 km h–1

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Matematik  Tingkatan 4  Bab 7 Graf Gerakan 

(d) Rajah di sebelah menunjukkan graf laju-masa pergerakan sebuah

kenderaan bagi tempoh t saat. Hitung Laju/ Speed (m s-1)

The diagram shows a speed-time graph of the motion of a vehicle for a 28
period of t seconds. Calculate TP 6

(i) kadar perubahan laju terhadap masa, dalam m s–2, untuk 7 16
saat yang pertama.
the rate of change of speed, in m s–2, for the first 7 seconds.
7 10 t Masa/ Time(s)
(ii) laju purata, dalam m s–1, untuk 7 saat yang pertama.
the average speed, in m s–1, for the first 7 seconds.

(iii) nilai t, jika jarak yang dilalui oleh kenderaan untuk tempoh 7 saat yang pertama ialah separuh

daripada jarak yang dilalui pada saat ke-10 sehingga saat t.

the value of t, if the distance travelled by the vehicle for the first 7 seconds is half the distance travelled at 10th to

t seconds.

(i) (—2(87–—–– –10–6)–) = 1—72 = 1.71 m s–2

(ii) Jumlah jarak / Total distance
= —21 × (16 + 28) × 7 = 154 m

Laju purata / Average speed
= —157–4– = 22 m s–1

( iii ) —21 × 28 × (t – 10) = 22
4 + t = 22

t = 18

PRAKTIS SPM 7 BAB 7

Kertas 1 A Jarak C Pecutan

1. Rajah di bawah menunjukkan sebuah graf jarak- Distance Acceleration
masa.
B Laju D Nyahpecutan
The diagram below shows a distance-time graph.
Speed Deceleration
Jarak/ Distance (km)
M 2. Tentukan persamaan yang mewakili graf laju-
masa di bawah, dengan keadaan a ialah laju,
0 Masa(j) dalam m s–1 dan t ialah masa, dalam saat.
Time(h)
Determine the equation that represents the speed-time
Berdasarkan rajah tersebut, apakah yang diwakili graph below, where a is the speed, in m s–1, and t is the
oleh kecerunan OM? time, in seconds.

Based on the diagram, what is represented by the Laju/ Speed(m s-1)
gradient OM?
50
40
30
20
10

0 Masa/Time(s)
2 4 6 8 10

111 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 7 Graf Gerakan

A a = 5t – 40 C a = 40 + 5t (a) Diberi laju perjalanan dari Rembau ke Muar
B a = 40 – 5t D a = 5t × 8
ialah 2.5 km min–1. Cari nilai t.
3. Rajah di bawah menunjukkan graf laju-masa bagi
Given that the speed of the journey from Rembau to
perjalanan sebuah motosikal. Muar is 2.5 km min–1. Find the value of t.

The diagram below shows the speed-time graph of the (b) Berapakah masa yang diambil, dalam minit,
journey of a motorcycle.
untuk sampai ke Muar dari Rembau?
Laju/ Speed (m s-1)
What is the time taken, in minutes, to reach Muar
30 from Rembau?
k
(c) Hitung purata laju, dalam km j–1, bagi
10
keseluruhan perjalanan.

Calculate the average speed, in km h–1, for the whole
journey.

10 18 Masa/ Time(s) (a) 2.5 = —9t0––––4–35—0
t – 45 = —90–2––.5–3—0
Jumlah jarak yang dilalui oleh motosikal itu ialah t = 24 + 45
= 69 minit / minutes
330 km. Hitung nilai k.
The total distance travelled by the motorcycle is 330 km. (b) 69 – 45 = 24 minit / minutes

Calculate the value of k.

A 21 m s–1 C 26 m s–1

B 24 m s–1 D 29 m s–1

4. Rajah di bawah menunjukkan sebuah graf laju- (c) Purata laju / Average speed
masa.
  = 90
The diagram below shows a speed-time graph. 69

Laju/ Speed (m s-1) 60

40 = 78.26 km j–1

78.26 km h–1

18 2. Rajah di bawah menunjukkan graf laju-masa bagi

0 12 17 20 Masa/ Time(s) KBAT sebuah bas dan sebuah kereta yang bergerak dari
BUKAN A ke B.

RUTIN The diagram below shows the speed-time graph for a
bus and a car travel from A to B.
BAB 7
Laju (km j–1)
Speed (km h–1)
Hitung pecutan, dalam m s–2, bagi 12 saat yang

pertama.
Calculate the acceleration, in m s–2, for the first 12
100 P
seconds. vQ R

A 1.5 C 12

B 3 D 18

S Masa (j)

Kertas 2 O 1.6 T Time (h)

Graf garis lurus OP mewakili pergerakan bas itu

1. Rajah di bawah menunjukkan graf jarak-masa dan graf QRS mewakili pergerakan kereta itu.

bagi perjalanan David dari Seremban ke Muar. Kedua-dua kenderaan itu memulakan perjalanan

2019 The diagram below shows a distance-time graph for the mereka pada masa yang sama dan melalui laluan

journey of David from Seremban to Muar. yang sama.

The straight line graph OP represents the movement of

Jarak/ Distance(km) the bus and the graph QRS represents the movement of
90
Muar the car. Both vehicles start at the same time and move

along the same route.

(a) Kereta itu bergerak dengan laju seragam

Rembau 60 km j–1 dalam suatu tempoh tertentu.
30
Nyatakan nilai v.
The car moves with the uniform speed 60 km h–1 for

0 t Masa/Time(min) a certain period. State the value of v.
Seremban 30 45


© Penerbitan Pelangi Sdn. Bhd. 112

Matematik  Tingkatan 4  Bab 7 Graf Gerakan 

(b) Diberi kedua-dua kenderaan itu bertolak dari Bas dan kereta itu melalui jarak yang sama.
A pada jam 0800 dan sampai di B pada masa
yang sama. Nyatakan masa, dalam sistem The bus and the car travelled through the same
24 jam, kenderaan itu sampai di B.
distance.
Given both the vehicles left A at 0800 hours and
reached at B at the same time. State the time, in Maka / Thus, 30(1.6 + T) = 50T
24-hour system, when the vehicles reached at B.
4.8 + 3T = 5T
(c) Hitung jarak, dalam km, antara A dengan B.
T = 2.4
Calculate the distance, in km, between A and B.
= 2 jam 24 minit
(a) v = 60 2 hours 24 minutes
(b) Jumlah jarak yang dilalui oleh bas
Maka, kenderaan itu sampai di B pada jam
Total distance travelled by bus
1024.
= —21 × T × 100 = 50T Thus, the vehicles will arrive at B at 1024.

Jumlah jarak yang dilalui oleh kereta (c) Jarak antara A dengan B

Total distance travelled by car Distance between A and B

= —12 × (1.6 + T) × 60 = 30(1.6 + T) = 50 × 2.4 = 120 km

Praktis
SPM
Ekstra

Sudut KBAT KBAT

Ekstra

Rajah di sebelah menunjukkan graf laju-masa bagi pergerakan kereta Encik Hassim Laju (km j–1)
dalam tempoh 1 jam. Diberi kereta Encik Hassim memerlukan 1 liter petrol untuk Speed (km h–1)

bergerak sejauh 16 km pada laju 80 km j–1. 1.5 liter petrol telah digunakan apabila 100

kereta Encik Hassim bergerak pada laju seragam. Hitung 80

The diagram below shows the speed-time graph for the movement of Encik Hassim’s car for a
period of 1 hour. Given Encik Hassim’car needs 1 litre of patrol to move 16 km at the speed of
80 km h–1. 1.5 litres of patrol has been used when Encik Hassim’s car moved at the uniform

speed. Calculate O 0.2 t 1 BAB 7

(a) nilai t, Masa (j) / Time (h)

the value of t,

(b) kadar perubahan laju, dalam km j–1 per min, zarah itu dalam tempoh 0.2 jam yang pertama.
the rate of change of speed, in km h–1per min, of the particle for the first 0.2 hour.

Jawapan / Answer : (b) Kadar perubahan laju / Rate of change of speed
= —80—k—m0.–2j––1j—//—khm—h—–1
(a) Jarak yang dilalui dengan laju seragam: = 8—(00—.k2m—×–j6––10—/) k—mm–ihn—–1
The distance travelled with uniform speed:
1.5 × 16 = 24 km = 6.67 km j–1 / min

(t – 0.2) × 80 = 24 6.67 km h–1 / min

t – 0.2 = 0.3

t = 0.5

Kuiz 7

113 © Penerbitan Pelangi Sdn. Bhd.

BAB Sukatan Serakan Data Tak Terkumpul

8 Measures of Dispersion for Ungrouped Data

8.1 Serakan

Dispersion

NOTA IMBASAN Tip

1. Serakan bagi suatu set data menunjukkan Sukatan kecenderungan memusat / Measures of central tendencies
bagaimana sebaran nilai-nilai dalam set data
itu di sekitar satu nilai memusat seperti min. • Mod – Nilai yang paling kerap berulang dalam set data.
Mode – The most frequence value in a set of data.
Dispersion of a set of data indicate how the values
in the data are spread around a central value such • Median – Nilai yang berada di tengah-tengah apabila data disusun mengikut
as mean.
tertib menaik atau menurun.
Median – The value in the middle of a data when arranged in ascending or descending order.
Jumlah nilai data / Sum of values of data
• MMeinan, ,x––x= Bilangan data / Number of data = Σfx
Σf

1. Bandingkan dan tafsirkan serakan bagi dua set data berikut. TP 1

Compare and interpret the dispersion of the following two sets of data.

CONTOH

Data di bawah menunjukkan taburan umur pesakit bagi dua buah klinik.

The data below shows the ages of patients in two clinics.

Umur Pesakit di Klinik A Umur Pesakit di Klinik B
Ages of Patients in Clinic A Ages of Patients in Clinic B

10 30 53 50 7 44 19 33 42 26 37 45 38 48

27 16 2 32 47 9 25 41 21 39 33 24 25 36

53 25 38 48 12 21 31     38 45 27 46 49 33 30

Lukis plot batang-dan-daun untuk mempamerkan taburan umur pesakit bagi kedua-dua klinik tersebut.

Berikan komen anda tentang taburan tersebut.

Draw a stem-and-leaf plot to show the distribution of ages of patients in the two clinics. Comment on the distributions.

Penyelesaian:

Umur Pesakit di Klinik A Umur Pesakit di Klinik B
Ages of Patients in Clinic A Ages of Patients in Clinic B

Batang Daun Batang Daun
Stem Leaf Stem Leaf

0 2 79 Kekunci: 1 | 2 bermaksud 12 tahun. 0
1 0 269 Key: 1 | 2 means 12 years old. 1
2 1 557 2 14567
3 0 128 3 0 3 3 3 67 8 8 9
4 4 78 4 1 2 5 5 68 9
5 0 33 5

Serakan bagi umur pesakit di klinik A adalah lebih besar berbanding dengan di klinik B, kerana umur
pesakit di klinik A berada antara 2 – 53 tahun dan klinik B antara 21 – 49 tahun.

The dispersion of ages of patients in clinic A is larger than in clinic B, because the ages of patients in clinic A are between
2 – 53 years old and clinic B are between 21 – 49 years old.

© Penerbitan Pelangi Sdn. Bhd. 114

Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul 

(a) Data di bawah menunjukkan masa yang diambil, dalam minit, oleh sekumpulan pelajar menulis sebuah
karangan dalam kelas tingkatan 4 Amanah dan 4 Bijak.

The data below show the time taken, in minutes, by a group of students writing an essay in class of form 4 Amanah
and 4 Bijak.

Masa yang Diambil bagi kelas 4 Amanah Masa yang Diambil bagi kelas 4 Bijak
Time Taken in class 4 Amanah Time Taken in class 4 Bijak

42 39 55 49 38 43 45 38 48 46 59 27 63 33

45 50 38 42 49 44 52       44 55 24 46 61 33 55

Lukis plot batang-dan-daun untuk mempamerkan taburan masa yang diambil bagi kelas 4 Amanah dan

4 Bijak untuk menulis sebuah karangan. Berikan komen anda tentang data tersebut.

Draw a stem-and-leaf plot to show the distribution of time taken in class 4 Amanah and 4 Bijak to write an essay.
Comment on the distributions.

Masa yang Diambil bagi kelas 4 Amanah Masa yang Diambil bagi kelas 4 Bijak
Time Taken in class 4 Amanah Time Taken in class 4 Bijak

Batang Daun Batang Daun
Stem Leaf Stem Leaf

2 Kekunci: 3 | 8 bermaksud 38 minit. 2 47
3 889 Key: 3 | 8 means 38 minutes. 3 338
4 2 2 3 4 55 9 9 4 4668
5 025 5 559
6 6 13

Serakan bagi kelas 4 Bijak adalah lebih besar daripada kelas 4 Amanah, kerana masa yang diambil oleh
4 Amanah adalah antara 38 – 55 minit dan 4 Bijak antara 24 – 63 minit.

The dispersion of class 4 Bijak is larger than class 4 Amanah, because the time taken for 4 Amanah are between

38 – 55 minutes and 4 Bijak between 24 – 63 minutes.

(b) Data di bawah menunjukkan bilangan emel yang dihantar oleh tiga orang pekerja, P, Q dan R, dalam

sebuah syarikat dalam suatu bulan tertentu.

The data below show the number of emails sent by three workers, P, Q and R, in a company in a certain month.

Bilangan Emel Dihantar oleh Pekerja P Bilangan Emel Dihantar oleh Pekerja Q Bilangan Emel Dihantar oleh Pekerja R BAB 8
Number of Emails Sent by Worker P Number of Emails Sent by Worker Q Number of Emails Sent by Worker R

20 40 60 30 70 50 60 50 40 30 50 50 70 60 50 40 50 60 50 50 40
70 80 50 20 40 80 30 60 50 60 40 40 50 50 50 50 40 50 60 60 50

Bilangan Emel Dihantar oleh Pekerja P Bilangan Emel Dihantar oleh Pekerja Q
Number of Emails Sent by Worker P Number of Emails Sent by Worker Q

20 30 40 50 60 70 80 20 30 40 50 60 70 80

Bilangan emel / Number of emails     Bilangan emel / Number of emails

115 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul
Bilangan Emel Dihantar oleh Pekerja R
Number of Emails Sent by Worker R

20 30 40 50 60 70 80

Bilangan emel / Number of emails
Serakan bilangan emel dihantar oleh pekerja P adalah lebih besar diikuti oleh pekerja Q dan serakan

bagi pekerja R adalah yang terkecil dengan kebanyakan emel dihantar tertumpu pada 50 buah emel.
The distribution of the number of emails sent by employee P is larger, followed by employee Q and the distribution of

employee R is the smallest, that most of the emails sent concentrated on 50 emails.

2. Jawab soalan bagi setiap perwakilan data berikut. TP 3

Answer the questions for each of the following data representations.

(a) Plot batang-dan-daun di bawah menunjukkan jangka hayat, dalam jam, bagi dua jenama mentol, S dan T.

The stem-and-leaf plots show the lifespan, in hours, of two brands of bulbs, S and T.

Jangka Hayat Mentol Jenama S Jangka Hayat Mentol Jenama T
Lifespan of Bulb Brand S Lifespan of Bulb Brand T

60 70 80 90 100 110 60 70 80 90 100 110
Bilangan jam / Number of hours Bilangan jam / Number of hours

(i) Hitung min jangka hayat, dalam jam, bagi kedua-dua jenama mentol.

Calculate the mean of the lifespan, in hours, for both brands of the bulbs.

(ii) Bandingkan serakan bagi jangka hayat dengan min untuk kedua-dua jenama mentol.

Compare the dispersion of the lifespan with the mean for both brands of the bulbs.

(iii) Buat satu kesimpulan tentang jenama mentol yang lebih baik.

Make a conclusion about the better brand of the bulb.

(i) Min jangka hayat bagi mentol jenama S Min jangka hayat bagi mentol jenama T
Mean of the lifespan for bulb brand S Mean of the lifespan for bulb brand T

= 60 + 2(70) + 2(75) + 3(80) + 85 + 90 = 2(95) + 4(100) + 3(105) + 110
10 10
BAB 8
= 76.5 jam / hours = 101.5 jam / hours

(ii) Jangka hayat bagi mentol jenama S tertabur di antara 60 hingga 90 jam manakala bagi jenama T

pula tertabur di antara 95 hingga 110 jam. Serakan nilai-nilai data bagi jenama S adalah tertabur

luas daripada nilai min manakala nilai-nilai data bagi jenama T tertabur hampir dengan nilai min.

Oleh itu, serakan jangka hayat bagi mentol jenama S adalah lebih besar daripada jenama T.

The lifespan for bulbs brand S are distributed between 60 and 90 hours whereas for brand T are distributed

between 95 and 110 hours. The dispersion of the values of data for brand S are widely distributed from the mean
value whereas the values of data for brand T are distributed close to the mean value. Thus, the dispersion of the
lifespan for bulbs brand S are larger than of brand T.

(iii) Mentol jenama T adalah lebih baik daripada jenama S kerana nilai min jangka hayatnya lebih lama

dan serakannya lebih kecil yang menunjukkan ia lebih konsisten.

Bulb of brand T is better than brand S because the mean of the lifespan is longer and the dispersion is smaller,

which shows that it is more consistent.

© Penerbitan Pelangi Sdn. Bhd. 116

Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul 

(b) Plot titik di bawah menunjukkan harga, dalam RM, bagi 12 keping cakera padat yang dijual di Kedai
Muzika dan Kedai Irama.

The dot plots show the prices, in RM, of 12 compact discs sold in Kedai Muzika dan Kedai Irama.

Harga Cakera Padat di Kedai Muzika Harga Cakera Padat di Kedai Irama
Prices of Compact Discs in Kedai Muzika Prices of Compact Discs in Kedai Irama

Batang Daun Batang Daun
Stem Leaf Stem Leaf
258
2 0 358 2
3 3 45
4 0 9 3
5
6 4 259

5 0 1 1 4 88
6 013

Kekunci: 4 | 0 bermaksud RM40.
Key: 4 | 0 means RM40.

(i) Hitung min harga bagi sekeping cakera padat di Kedai Muzika dan Kedai Irama.

Calculate the mean price of a compact disc in Kedai Muzika and Kedai Irama.

(ii) Bandingkan serakan harga daripada min bagi cakera padat yang dijual di Kedai muzika dengan

Kedai Irama.

Compare the dispersion of the prices from the mean for compact discs sold in Kedai Muzika and Kedai Irama.

(iii) Kedai yang manakah menjual cakera padat dengan harga lebih murah? Berikan sebab anda.

Which shop sells compact disk at the cheaper price? State your reason.

(i) Min harga di Kedai Muzik Min harga di Kedai Irama

Mean price in Kedai Muzika Mean price in Kedai Irama

20 + 22 + 25 + 28 + 33 + 33 + 35 + 38 42 + 45 + 49 + 50 + 51 + 51 + 54 + 58

= + 39 + 40 + 44 + 45 = + 58 + 60 + 61 + 63
12 12

= RM33.50 = RM53.5

(ii) Serakan bagi harga cakera padat yang dijual di Kedai Muzika adalah di antara RM20 hingga RM45 BAB 8
manakala di Kedai Irama adalah antara RM42 hingga RM63. Serakan nilai-nilai data bagi Kedai
Muzika dan Kedai Irama adalah tertabur hampir dengan nilai min masing-masing. Oleh itu, serakan
harga cakera padat di kedua-dua kedai adalah hampir sama, iaitu tertabur secara berkelompok.

The dispersion of the prices for compact discs sold in Kedai Muzika are between RM20 and RM45 whereas in
Kedai Irama are between RM42 and RM63. The dispersion of the values of data in Kedai Muzika and Kedai Irama
are distributed close to each mean value. Thus, the dispersion of the prices of compact discs in both shops are

approximately the same, which are distributed in a group.

(iii) Kedai Muzika menjual cakera padat dengan harga yang lebih murah kerana nilai minnya yang lebih
kecil berbanding dengan Kedai Irama.

Kedai Muzika sells compact disc at the cheaper price because the mean value is smaller compared to Kedai Irama.

117 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul

8.2 Sukatan Serakan

Measures of Dispersion

NOTA IMBASAN

1. Sukatan serakan bagi data tak terkumpul: If each value in the set of data is added or subtracted with a constant
Measures of dispersion of an ungrouped data: value, the range, interquartile range, variance and standard deviation

(a) Julat = Nilai terbesar – Nilai terkecil remain unchanged.
Range = Largest value – Smallest value
4. Jika setiap nilai dalam set data itu didarabkan dengan suatu
(b) Julat antara kuartil pemalar k, maka
If each value in the set of data is multiplied by a constant k,
= Kuartil atas (Q3) – Kuartil bawah (Q1) then
Interquartile range (a) julat baharu / new range
= k × julat asal / original range
= Upper quartile (Q3) – Lower quatile (Q1) (b) julat antara kuartil baharu / new interquartile range
= k × julat antara kuartil asal / original interquartile range
(c) Varians / Variance x–)2 (c) varians baharu / new variance
∑x2 ∑(x – = k2 × varians asal / original variance
2 = N – (x–)2 atau/or 2 = N (d) sisihan piawai baharu / new standard deviation
= k × sisihan piawai asal / original standard deviation
(d) Sisihan piawai / Standard deviation
∑(x – x–)2
= ∑x2 – (x–)2 atau/or  = N
N

2. Plot kotak / Box plot 5. Jika setiap nilai dalam set data itu dibahagi dengan suatu
pemalar k, maka
Kuartil qbuaawrtailhe,,QQ11 UKppuearrtqil uaatartsil,e,QQ33 If each value in the set of data is divided by a constant k, then
Lower
julat asal / original range
(a) julat baru / new range = k

Nilai minimum Median Nilai maksimum (b) julat antara kuartil baru / new interquartile range
Minimum value Median Maximum value

= julat antara kuartil asal / original interquartile range
k

(c) sisihan piawai baru / new standard variation

1234567 = sisihan piawai asal / original standard variation
k
3. Jika setiap nilai dalam set data itu ditambah atau varians asal / original
ditolak dengan suatu nilai pemalar, julat, julat antara (d) varians baru / new variance = k2 variance
kuartil, varians dan sisihan piawai tidak berubah.

BAB 8 3. Hitung julat bagi set data berikut. TP 2

Calculate the range of the following sets of data.

CONTOH (a) 28, 37, 27, 40, 26, 35, 31, 23 (b) 49, 51, 43, 57, 52, 47, 35, 53,
44
Nilai terbesar = 28
Largest value Julat / Range = 40 – 23

9, 13, 5, 25, 17, 28 = 17 Julat / Range = 57 – 35

Nilai terkecil = 5 = 22

Penyelesaian: Smallest value (c) 15.8, 8.4, 19.7, 25.5, 31.8 (d) 2.8, 1.1, 2.4, 3.9, 3.1, 4.2
Julat / Range
= 28 – 5 Julat / Range = 31.8 – 8.4 Julat / Range = 4.2 – 1.1

= 23 Tip = 23.4 = 3.1

Julat / Range
= Nilai terbesar – Nilai terkecil
= Largest value – Smallest value

© Penerbitan Pelangi Sdn. Bhd. 118

4. Hitung julat bagi setiap data berikut. TP 2 Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul 

Calculate the range for each of the following data. Penyelesaian:
90 Julat / Range
CONTOH
= 90 – 50
Skor 50 60 70 80 14 = 40

Score

Bilangan pelajar 3 7 12 9

Number of students

(a) Julat / Range
7 8 9 10 11 12 = 12 – 7
Umur
=5
Age
435972
Bilangan Kanak-kanak

Number of children

(b) 01234 Julat / Range
3 9 16 14 8 =4–0
Bilangan adik-beradik =4

Number of siblings

Bilangan pelajar

Number of students

5. Hitung julat antara kuartil bagi set data berikut. TP 3

Calculate the interquartile range of the following sets of data.

CONTOH Susun data mengikut tertib menaik. 2 29, 40, 73, 85, 16, 31, 59, 48
1 30, 42, 25, 71, 58, 37, 49 Arrange data in ascending order.
Penyelesaian: Penyelesaian:
25, 30, 37, 42, 49, 58, 71 16, 29, 31, 40, 48, 59, 73, 85

Q1 Median Q3 Q1 Median Q3 BAB 8

Julat antara kuartil Q1 = 29 + 31 Q3 = 59 + 73
2 2
Interquartile range = 30 = 66

= 58 – 30 Julat antara kuartil
= 28
Interquati le range
Tip
= 66 – 30

= 36

• Kuartil bbaawhaahgi,anQb1awiaalhahdamtaetdeirasnebubta.gi
separuh
Lower qthueatdilaet,a.Q1 is the median of the lower
half of

• Kuartil bataahsa, giaQn3 ialah median bagi Info
separuh atas data tersebut.
Upper tqhueadtialet,a.Q3 is the median of the upper Kuartil bawah juga dikenali sebagai kuartil pertama manakala
half of kuartil atas dikenali sebagai kuartil ketiga.
Lower quartile is also known as first quartile whereas upper quartile is
• Julat antara kuartil / Interquartile range known as third quartile.

= Q3 – Q1

119 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul

(a) 25, 17, 35, 40, 29, 43, 9 (b) 43, 27, 52, 60, 31, 28, 16, 41 (c) 4.7, 3.6, 3.8, 3.9, 2.9, 4.2, 5.8,
3.8, 3.2
9, 17, 25, 29, 35, 40, 43 16, 27, 28, 31, 41, 43, 52, 60
Q 1 Med ian Q3 2.9, 3.2, 3.6, 3.8, 3.8, 3.9, 4.2, 4.7, 5.8
Q1 Med ian Q3
Julat antara kuartil
Q1 Median Q3
Interquartile range 27 + 28
Q1 = 2 = 27.5
= Q3 – Q1
= 40 – 17 Q3 = 43 + 52 = 47.5 Q1 = 3.2 + 3.6 Q3 = 4.2 + 4.7
= 23 2 2 2
= 3.4 = 4.45
Julat antara kuartil
Julat antara kuartil
Interquartile range
Interquartile range
= Q3 – Q1
= 47.5 – 27.5 = Q3 – Q1
= 4.45 – 3.4
= 20
= 1.05

6. Hitung julat antara kuartil bagi set data berikut. TP 3

Calculate the interquartile range of the following sets of data.

CONTOH

Skor / Score 1 234 5 Tip
Kekerapan / Frequency 4 5 11 9 3
Rumus menghitung Q1 dan Q3 daripada data dalam
Penyelesaian: bentuk jadual kekerapan:
Formula to calculate Q1 and Q3 from data in the form of
Skor / Score frequency table:
Kekerapan / Frequency
Kekerapan longgokan 12345  Q1 ke- 1 n  Q3 = Data ke- 3 n
= data 4 4
Cumulative frequency 4 5 11 9 3
+ = += += += 1    nth 3 n
4 data 4 th data
4 9 20 29 32
dengan keadaan / where
BAB 8 QQ11 QQ33 n = Jumlah kekerapan
Total frequency

terletak di sini. terletak di sini.
lies here. lies here.

 Q1 = 1  Q3 = Data ke- 3
Data ke- 4 × 32 4 × 32

    1 3 Kesalahan Lazim
4 × 32 th data 4 × 32 th data
Menggunakan nilai kedudukan kuartil bawah dan
= Data ke-8 = Data ke-24 kuartil atas.
Using the position values of lower quartile and upper quartile.
8th data 24th data Julat antara kuartil / Interquartile range
= 24 – 8
= 2 = 4 = 16

Julat antara kuartil / Interquartile range = 4 – 2 Data skor.
Score data.

=2

© Penerbitan Pelangi Sdn. Bhd. 120

Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul 

(a) Mata 01234 (b) Bilangan anak 0123 4
5 13 17 4 1
Points Number of children

Kekerapan 3 5 9 6 1 Kekerapan
Frequency
Frequency

Mata 01234 Bilangan anak 0 1 2 3 4

Points Number of children

Kekerapan 3 5 9 6 1 Kekerapan 5 13 17 4 1

Frequency Frequency

Kekerapan Kekerapan

longgokan 3 8 17 23 24 longgokan 5 18 35 39 40

Cumulative Cumulative
frequency
frequency

QQ11liteesrlheetraek. di sini. QQ33liteesrlheetraek. di sini. Q1 terletak di sini. Q3 terletak di sini.
Q1 lies here. Q3 lies here.

1 2 1 2 Q1 = Nilai ke-1 1 1 2 1 2 Q1 = Data ke-1 1
4 × 24 / 4 × 24 th value 4 × 40 / 4 × 40 th value

= Nilai ke-6 / 6th value = 1 = Data ke-10 / 10th value = 1

1 2 1 2 Q3 = Nilai ke-3 / 3 1 2 1 2 Q3 = Nilai ke-3 3
4 × 24 4 × 24 th value 4 × 40 / 4 × 40 th value

= Nilai ke-18 / 18th value = 3 = Nilai ke-30 / 30th value = 2

Julat antara kuartil / Interquartile range = 3 – 1 Julat antara kuartil / Interquartile range = 2 – 1

=2 =1

7. Hitung varians dan sisihan piawai bagi set data berikut. TP 3

Calculate the variance and standard deviation of the following sets of data.

CONTOH Tip 2   1.6, 1.8, 2.3, 2.5, 2.8, 3.4
1   9, 12, 14, 20, 35
• Min / Mean, –x = Σx
Penyelesaian: • Varians / Variance N Penyelesaian:

Min / Mean σ2 = Σx2 – (–x)2 atau/or x x2
σ2 = N – x–)2
x– = 90 = 18 Σ(x 1.6 2.56 BAB 8
5 N
1.8 3.24
• Sisihan piawai / Standard deviation

x (x – x–)2 σ= Σx2 – (–x)2 atau/or 2.3 5.29
9 81 σ= N
12 36 Σ(x – x–)2 2.5 6.25
14 16
N 2.8 7.84

dengan keadaan / where 3.4 11.56
N = bilangan data / number of data

20 4 ∑x = 14.4 ∑x2 = 36.74

35 289 Min / Mean
∑x = 90 ∑(x – x–)2 = 426 14.4
Sisihan piawai ialah x– = 6 = 2.4
punca kuasa dua
Varians Sisihan piawai varians. Varians Sisihan piawai
Standard deviation is the
Variance x–)2 Standard deviation square root of variance. Variance Standard deviation
∑x2
σ2 = ∑(x – σ = 85.2 σ2 = N – (x–)2 σ = 0.36
N = = 0.6
426 = 9.23 36.74
= 5 = 85.2 – 2.42
6

= 0.36

121 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul

(a) 142, 144, 148, 149, 153, 156, 157, 163, 165 (b) 6.1, 6.3 6.3, 6.5, 7.2, 7.4, 7.8

Min / Mean, x– = 1 377 = 153 x x2
9
6.1 37.21
x (x – x–)2
6.3 39.69

142 121 6.3 39.69

144 81 6.5 42.25

148 25 7.2 51.84

149 16 7.4 54.76

153 0 7.8 60.84

156 9 ∑x = 47.6 ∑x2 = 326.28

157 16 Min / Mean, x– = 47.6 = 6.8
7
163 100

165 144 Varians / Variance Sisihan piawai
∑x = 1 377 ∑(x – x–)2 = 512
σ2 = ∑x2 – (x–)2 Standard deviation
N
V σ2a r==ia∑n5s19( x2/N–V=ax–r)i52a6 n .c8e9 SσS tias==inhd7a5a.n5r6d4.p8di9aewviaatiion 326.28 σ = 0.37
= 7 – 6.82
= 0.61

= 0.37

8. Jawab soalan berdasarkan dua set data yang diberi. TP 3

Answer the following questions based on the two sets of data given.

BAB 8 CONTOH (a) Data di bawah menunjukkan jisim, dalam kg,
bagi sekumpulan murid.
1 Data di bawah menunjukkan markah yang
The data shows the masses, in kg, of a group of
diperoleh Kavya dalam satu pertandingan. students.

The data shows the marks obtained by Kavya in a 43, 51, 52, 53, 54, 54, 55, 57
competition.
(i) Hitung julat dan julat antara kuartil.
74, 75, 76, 76, 77, 77, 86
Calculate the range and interquartile range.
(i) Hitung julat dan julat antara kuartil.
(ii) Tentukan sukatan serakan yang paling
Calculate the range and interquartile range. tepat, untuk mengukur serakan set data
tersebut.
(ii) Tentukan sukatan serakan yang paling tepat,
untuk mengukur serakan set data tersebut. Determine the most appropriate measure of
dispersion, to measure the distribution of the
Determine the most appropriate measure of dispersion, data set.
to measure the distribution of the data set.
(i) Julat / Range
Penyelesaian: = 57 – 43 = 14

(i) Julat markah Kavya / Range of Kavya’s mark Julat antara kuartil / Interquartile range
= 86 – 74 = 12 54 + 55 51 + 52
Julat antara kuartil bagi markah Kavya = 2 – 2 = 3

Interquartile range for Kavya’s mark (ii) Julat antara kuartil adalah sukatan yang

= 77 – 75 = 2 paling tepat kerana terdapat kewujudan

(ii) Julat antara kuartil adalah sukatan yang paling pencilan , 43.
tepat kerana terdapat kewujudan pencilan , 86.
Interquartile range is the most appropriate
Interquartile range is the most appropriate measure of measure of dispersion because of the existence
dispersion because of the existence of the outlier, 86. of the outlier, 43.

© Penerbitan Pelangi Sdn. Bhd. 122

Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul 

2 Perbelanjaan harian bagi Syafiq ialah RM1.80, (b) Tinggi anak pokok dalam sebuah tapak semaian

RM1.90, RM2.20, RM2.40, RM2.40, RM2.50, ialah11.0cm,10.7cm,11.3cm,10.9cm,10.2cm,
RM2.60 dan RM2.80.
11.9 cm, 10.6 cm, 11.5 cm dan 11.3 cm.
Syafiq’s daily expenses are RM1.80, RM1.90, RM2.20,
RM2.40, RM2.40, RM2.50, RM2.60 and RM2.80. The heights of young plants in a nursery are 11.0 cm,
10.7 cm, 11.3 cm, 10.9 cm, 10.2 cm, 11.9 cm, 10.6 cm,
(i) Hitung julat antara kuartil, varians dan sisihan
piawai bagi perbelanjaan harian Syafiq. 11.5 cm and 11.3 cm.

Calculate the interquartile range, variance and (i) Hitung julat antara kuartil, varians dan
standard deviation of Syafiq’s daily expenses.
sisihan piawai bagi data di atas.
(ii) Antara sisihan piawai dan julat antara kuartil,
yang manakah menunjukkan sukatan yang Calculate the interquartile range, variance and
lebih baik bagi menghuraikan serakan data di standard deviation of the above data.
atas?
(ii) Antara sisihan piawai dan julat antara
Which one, standard deviation or interquartile range,
shows the better measure to describe the dispersion kuartil, yang manakah menunjukkan
of the above data?
sukatan yang lebih baik bagi menghuraikan

serakan data di atas?

Which one, standard deviation or interquartile
range, shows the better measure to describe the
dispersion of the above data?

Penyelesaian: (i) 10.2, 10.6, 10.7, 10.9, 11.0, 11.3, 11.3,

11.5, 11.9

(i) Julat antara kuartil / Interquartile range Julat antara kuartil / Interquartile range

= 2.5 + 2.6 – 1.9 + 2.2 = RM0.50 = 11.3 + 11.5 – 10.6 + 10.7
2 2 2 2

Min / Mean, = 11.4 – 10.65 = 0.75

–x = 18.60 = 2.325 Min / Mean
8 99.4
x– = 9 = 11.04
Varians / Variance

σ2 = ∑x2 – (x–)2 Varians / Variance
N
σ2 = ∑x2 – (x–)2
1.82 + 1.92 + 2.22 + 2.42 + 2.42 N

= + 2.52 + 2.62 + 2.82 – 2.3252 10.22 + 10.62 + 10.72
8
+ 10.92 + 112 + 11.32

= 44.06 – 2.3252 = + 11.32 + 11.52 + 11.92 – 11.042
8 9
BAB 8
= 0.102 = 1 099.94 – 11.042
9
Sisihan piawai / Standard deviation
= 0.334
σ = 0.102
= 0.32 Sisihan piawai / Standard deviation

(ii) Sisihan piawai adalah sukatan yang lebih baik σ = 0.334
= 0.58
berbanding julat antara kuartil kerana setiap
(ii) Sisihan piawai adalah sukatan yang
nilai dalam data diambil kira dalam pengiraan.
lebih baik berbanding julat antara kuartil
Standard deviation is a better measu re compared to
interquartile range because each value in the data are kerana setiap nilai dalam data diambil kira
taking into the calculation.
dalam pengiraan.

Standard deviation is a better measure compared
to interquartile range because every values in
the data are taking into the calculation.

123 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul

9. Hitung varians dan sisihan piawai bagi set data berikut. TP 3

Calculate the variance and standard deviation of the following sets of data.

CONTOH (a) Umur / Age 13 14 15 16 17
Kekerapan 21 18 25 16 20
Mata / Points 01 234
431 Frequency

Kekerapan / Frequency 3 5 fx2 xf fx fx2
0 13 21 273 3 549
Penyelesaian: 5 14 18 252 3 528
16 15 25 375 5 625
xf fx 27 16 16 256 4 096
03 0 16 17 20 340 5 780
15 5 ∑fx2 = 64 ∑fx = 1 496 ∑fx2 = 22 578
24 8 ∑f = 100
33 9
41 4 Min / Mean, x– = 1 496 = 14.96
∑fx = 26 100
f = 16

Varians / Variance Sisihan piawai

Min / Mean σ2 = ∑fx2 – (x–)2 Standard deviation
26 ∑f
x– = 16 = 1.625 σ = 1.98

= 22 578 – 14.962 = 1.41
100

Varians Sisihan piawai = 1.98

Variance Standard deviation

σ2 = ∑fx2 – (x–)2 σ = 1.36
∑f
= 1.17
= 64 – 1.6252 (b) Jisim (kg) 40 50 60 70 80
16 4 15 11 13 7
Mass (kg)
= 1.36
Kekerapan

Frequency

Tip xf fx fx2

Bagi data dalam jadual kekerapan, 40 4 160 6 400
For data in frequency table,
BAB 8 50 15 750 37 500

• Min / Mean, x– = Σfx 60 11 660 39 600
Σf
70 13 910 63 700
• Varians / Variance
Σfx2 Σf(x – x–)2 80 7 560 44 800
σ2 = Σf – (x–)2 atau / or σ2 = Σf
∑f = 50 ∑fx = 3 040 ∑fx2 = 192 000
• Sisihan piawai / Standard deviation
x– = 3 040
σ= Σfx2 – (x–)2 atau/ or Min / Mean, 50 = 60.8
σ= Σf
Σf(x – x–)2 Varians / Variance Sisihan piawai

Σf σ2 = ∑fx2 – (x–)2 Standard deviation
∑f
dengan keadaan / where σ = 143.36

f = kekerapan / frequency = 192 000 – 60.82 = 11.97
50

= 143.36

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Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul 

10. Bina plot kotak bagi set data yang diberi. TP 3 Tip

Construct the box plot for the given sets of data. Langkah-langkah membina plot kotak:
Steps to construct box plot:
CONTOH

70, 55, 60, 75, 60, 85, 40

Penyelesaian:
40, 55, 60, 60, 70, 75, 85  

 Susun data dalam tertib menaik.

Q1 Median Q3 Arrange data in ascending order.

Nilai minimum / Minimum value = 40  Cari tnheilami inmiminuimmuvmalu,eQ, Q1,1,mmeeddiiaann,, QQ33 dan nilai maksimum.
Q1 = 55 Find and maximum value.
Median / Median = 60
Q3 = 75  Lukis skala yang merangkumi nilai minimum dan maksimum.
Nilai maksimum / Maximum value = 85 Draw a scale that includes minimum value and maximum value.

 Lukis sebuah kotak yang menghubungkan nilai Q1 dan Q3.
Draw a box that connects the values of Q1 and Q3.

 Lukis satu garis dalam kotak tersebut yang mewakili median.
Draw a line in the box that represents median.

 Tandakan titik nilai minimum pada skala dan sambungkan dengan

MsisairkkothteakmyinainmgummevawluaekiwliitQh a1.point on the scale and join with the side of

4 the box that represents Q1.
 Tandakan titik nilai maksimum pada skala dan sambungkan
5
6 dengan sisi kotak vyaalunegwmitehwaapkoiilni tQon3. the scale and join with the side of
Mark the maximum

7 the box that represents Q3.

3
40 50 60 70 80 90

(a) 20, 19, 12, 18, 24, 17, 22 (b) 3.3, 3.9, 5.0, 5.7, 4.4, 2.9, 5.2, 1.8, 5.4
1.8, 2.9, 3.3, 3.9, 4.4, 5.0, 5.2, 5.4, 5.7
12, 17, 18, 19, 20, 22, 24
Q1 Median Q3
Q1 Median Q3
Nilai minimum Q1 = 2.9 + 3.3 Q3 = 5.2 + 5.4 BAB 8
2 2
Minimum value = 3.1 = 5.3

= 12 Nilai minimum / Minimum value = 1.8
Q1 = 17 Q1 = 3.1
Median = 19
Q3 = 22 Median = 4.4
Nilai maksimum Q3 = 5.3
Nilai maksimum / Maximum value = 5.7
Maximum value

= 24

12 14 16 18 20 22 24 123456
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125

  Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul 012345
40 50 60 70 80 90 100
11. Bagi setiap plot kotak, nyatakan TP 2

For each of the box plot, state

CONTOH

(i) nilai minimum / the minimum value = 0
(ii) nilai maksimum / the maximum value = 4.5
(iii) julat / the range = 4.5
(iv) kuartil pertama / the first quartile = 1
(v) kuartil ketiga / the third quartile = 2.5
(vi) julat antara kuartil / the interquartile range = 1.5
(vii) nilai median / the median = 1.5

(a) (i) nilai minimum / the minimum value = 48
(ii) nilai maksimum / the maximum value = 94
(iii) julat / the range = 46
(iv) kuartil pertama / the first quartile = 66
(v) kuartil ketiga / the third quartile = 88
(vi) julat antara kuartil / the interquartile range = 22
(vii) nilai median / the median = 80

12. Panjang tali, dalam cm, yang dipotong oleh Misha ialah 2, 3, 5, 7, 9, 11 dan 12. Diberi min panjang tali ialah

7 cm. TP 3
The length of rope, in cm, cut by Misha are 2, 3, 5, 7, 9, 11 and 12. Given the mean of length of a rope is 7 cm.

(a) Hitung julat, julat antara kuartil, varians dan sisihan piawai bagi data di atas.

Calculate the range, interquartile range, variance and standard deviation of the data above.

Julat / Range Varians / Variance
= 12 – 2 22 + 32 + 52 + 72 + 92 + 112 + 122
= 10 = 7 – 72

= 12.86

Julat antara kuartil / Interquartile range Sisihan piawai piawai / Standard deviation
= 11 – 3 = 12.86
=8
= 3.59

BAB 8 (b) Hitung julat, julat antara kuartil, varians dan sisihan piawai yang baharu bagi setiap perubahan berikut.

Kemudian, nyatakan kesan perubahan tersebut ke atas nilai sukatan serakan baharu.

Calculate the new range, interquartile range, variance and standard deviation for each of the following changes. Then,
state the effect of the changes on the new values of measures of dispersion.

(i) Setiap panjang tali ditambah 2 cm. / Each length of rope is added by 2 cm.

Data baharu / New data: 4, 5, 7, 9, 11, 13, 14

Julat / Range Julat antara kuartil Min / Mean
=7+2
= 14 – 4 Interquartile range =9

= 10 = 13 – 5 Sisihan piawai / Standard deviation
= 12.86
=8 = 3.59

Varians / Variance
42 + 52 + 72 + 92 + 112 + 132 + 142
= 7 – 92

= 12.86

Kesan perubahan: Setiap nilai sukatan serakan tidak berubah.

Effect of changes: Each value of measures of dispersion is unchanged.

© Penerbitan Pelangi Sdn. Bhd. 126

Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul 

(ii) Setiap panjang tali didarab 3 cm. / Each length of rope is multiplied by 3 cm.

Data baharu / New data: 6, 9, 15, 21, 27, 33, 36

Julat / Range Julat antara kuartil Min / Mean
=7×3
= 36 – 6 Interquartile range = 21

= 30 = 33 – 9 Sisihan piawai / Standard deviation
= 115.71 = 10.76
Varians / Variance = 24

= 62 + 92 + 152 + 212 + 272 + 332 + 362 – 212 = 115.71
7

Kesan perubahan: Julat asal × 3, Julat antara kuartil asal × 3, Varians asal × 32, Sisihan piawai asal × 3.
Effect of changes: Original range × 3, Original interquartile range × 3, Original variance × 32, Original standard

deviation × 3.

(iii) Nilai data 12 cm diganti dengan nilai ekstrem 19 cm.
The data value 12 cm is substituted with an extreme value 19 cm.

Data baharu / New data: 2, 3, 5, 7, 9, 11, 19

Julat / Range Julat antara kuartil Min / Mean
= 19 – 2 56
= 17 Interquartile range = 7 = 8

Varians / Variance = 11 – 3 = 8 Sisihan piawai / Standard deviation

= 22 + 32 + 52 + 72 + 92 + 112 + 192 – 82 = 28.86 berubah dan = 28.86 = 5.37 tidak berubah.
Kesan perubahan:7Julat, varians dan sisihan piawai julat antara kuartil

Effect of changes: Changes in range, variance and standard deviation and no changes in interquartile range.

(iv) Nilai ekstrem 21 cm ditambah. / Extreme value of 21 cm is added.

Data baharu / New data: 2, 3, 5, 7, 9, 11, 12, 21

Julat / Range Julat antara kuartil Min / Mean
70
= 21 – 2 Interquartile range = 8 = 8.75

= 19 = 11.5 – 4 = 7.5 Sisihan piawai / Standard deviation
Varians / Variance = 32.69
22 + 32 + 52 + 72 + 92 + 112 + 122 + 212
= 8 – 8.752 = 5.72

= 32.69

Kesan perubahan: Setiap sukatan serakan berubah secara tidak tetap. BAB 8

Effect of changes: Each measure of dispersion changes irregularly.

(v) Nilai data 8 cm ditambah. / The value of data 8 cm is added.

Data baharu / New data: 2, 3, 5, 7, 8, 9, 11, 12

Julat / Range Julat antara kuartil Min / Mean
57
= 12 – 2 Interquartile range = 8 = 7.125

= 10 = 10 – 4 = 6

Varians / Variance Sisihan piawai / Standard deviation
= 11.36 = 3.37
= 22 + 32 + 52 + 72 + 82 + 92 + 112 + 122 – 7.1252 = 11.36
8

Kesan perubahan: Julat tidak berubah, julat antara kuartil berubah mengikut perubahan Q1 dan Q3.
Manakala, varians dan sisihan piawai berubah.

Effect of changes: No changes in range, interquartile range changes according to the changes in Q1 and Q3. Whereas,
there are changes in variance and standard deviation.

127 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul

13. Satu tinjauan tentang bilangan anak telah dilakukan terhadap sekumpulan guru di sebuah sekolah. Hasil

dapatan kajian adalah seperti berikut. TP 4

A survey about the number of children has been conducted towards a group of teachers in a school. The results of the survey
are as follows.

1, 2, 2, 3, 3, 3, 4, 4, 5, 6, 9

(a) Bina satu plot kotak bagi data tersebut.

Construct a box plot for the data above.

Nilai minimum / Minimum value = 1 1 23 45 6 7 8 9
Q1 = 2
Median = 3
Q3 = 5
Nilai maksimum / Maximum value = 9

(b) Kenal pasti nilai ekstrem di dalam data tersebut. Bagaimanakah nilai ekstrem tersebut memberi kesan

kepada serakan plot kotak yang dibina?

Identify the extreme value in the data. How does the extreme value effect the dispersion of the constructed box plot?

Nilai ekstrem ialah 9. Nilai ini memberi kesan bahawa data telah terserak luas daripada yang sebenar, iaitu

di antara 1 hingga 6 orang anak.

Extreme value is 9. This value implies that the data is spread out more widely than it actually is, between 1 and 6

children.

(c) Bina semula plot kotak yang dapat menggambarkan serakan yang betul.

Re-construct a box plot that can visualise the correct dispersion.

x

1 23 45 6 7 8 9

14. Jawab soalan-soalan berikut berdasarkan set-set data yang diberi. TP 4

Answer the following questions based on the given sets of data.

CONTOH

BAB 8 Jadual di sebelah menunjukkan markah ujian bagi dua orang murid. Murid Markah

The table shows the test marks of two students. Student Marks

(i) Hitung min dan sisihan piawai bagi murid P dan Q. P 68, 73, 62, 79
Q 72, 65, 59, 86
Calculate the mean and the standard deviation of students P and Q.

(ii) Prestasi murid yang manakah lebih konsisten? Terangkan jawapan anda.

Which student’s performance is more consistent? Explain your answer.

Penyelesaian:

(i) Bagi murid P, / For student P, Bagi murid Q, / For student Q,

Min, x = 68 + 73 + 62 + 79 Min, x = 72 + 65 + 59 + 86

Mean 282 4 Mean 282 4
4 4
= = 70.5 = = 70.5

∑x2 = 682 + 732 + 622 + 792 ∑x2 = 722 + 652 + 592 + 862
= 20 038 = 20 286

Sisihan piawai, s = 20 038 – 70.52 Sisihan piawai, s = 20 286 – 70.52
4 4
Standard deviation Standard deviation
= 6.265 = 10.06

(ii) Prestasi murid P lebih konsisten kerana sisihan piawainya lebih rendah.

Performance of student P is more consistent because the standard deviation is smaller.

© Penerbitan Pelangi Sdn. Bhd. 128

Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul 

(a) Jadual di bawah menunjukkan keputusan permainan ragbi yang diperoleh dua pasukan dalam 10
perlawanan.

The table shows the results of rugby games obtained by two teams in 10 matches.

Pasukan Merah / Red Team 67 66 65 68 60 71 69 61 70 60

Pasukan Biru / Blue Team 64 63 64 65 64 71 65 66 70 65

(i) Hitung min dan sisihan piawai bagi kedua-dua pasukan.

Calculate the mean and standard deviation of both teams.

(ii) Prestasi pasukan yang manakah lebih baik? Terangkan jawapan anda.

Which team’s performance is better? Explain your answer.

(i) Bagi pasukan Merah / For Red team,

Min / Mean, x– = 657 = 65.7
10

∑x2 = 672 + 662 + 652 + 682 + 602 + 712 + 692 + 612 + 702 + 602 = 43 317

Sisihan piawai / Standard deviation, s = 43 317 – 65.72 = 3.9
10

Bagi pasukan Biru / For Blue team,

Min / Mean, x– = 657 = 65.7
10

∑x2 = 642 + 632 + 642 + 652 +642 + 712 + 652 + 662 + 702 + 652 = 43 229

Sisihan piawai / Standard deviation, s = 43 229 – 65.72 = 2.53
10

(ii) Prestasi pasukan Biru lebih baik kerana min bagi kedua-dua pasukan adalah sama, tetapi sisihan

piawai bagi pasukan biru lebih kecil berbanding pasukan Merah. Ini menunjukkan pasukan Biru

lebih konsisten.

The performance of Blue team is better because the mean for both teams is the same, but the standard deviation

of the Blue team is smaller than the Red team. It shows that the Blue team is more consistent.

(b) Plot kotak di sebelah menunjukkan bilangan Bandar Z
pelanggan bagi sebuah restoran francais di tiga lokasi City Z

yang berbeza. Bandar Y
City Y
The box plot shows the number of customers of a franchise
Bandar X
restaurant in three different locations. City X BAB 8

(i) Jika bilangan data adalah sama, lokasi manakah

yang menunjukkan kebanyakan pelanggan 100 150 200 250 300 350 400 450 500
adalah melebihi 275 orang? Jelaskan jawapan

anda.
If the number of data is the same, which location shows the most customers are more than 275? Explain your

answer.

(ii) Restoran francais tersebut terpaksa menutup sebuah cawangan kerana kegawatan ekonomi. Pada

pendapat anda, cawangan di lokasi manakah yang patut dihentikan operasinya? Berikan sebab

anda.

The franchise restaurant had to close a branch due to economic downturn. In your opinion, in which location of
branch needs to stop the operation? Give your reason.

(i) Bandar Y kerana nilai 75% pelanggannya adalah melebihi 275 orang.

City Y because 75% of the customers are more than 275 people.

(ii) Cawangan di Bandar X kerana 50% pelanggannya adalah paling rendah berbanding dengan
pelanggan di Bandar Y dan Bandar Z.

Branch in City X as 50% of the customers are lower compared to the customers in City Y and City Z.

129 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul

15. Selesaikan setiap yang berikut.

Solve each of the following.

(a) Jadual di sebelah menunjukkan hasil tambah dan hasil tambah kuasa ∑x RM20 000

dua bagi x, dengan keadaan x ialah pendapatan bulanan, dalam RM,

bagi Encik Rosli untuk 8 bulan yang pertama pada tahun 2019. TP 4 ∑x2 RM50 385 000

The table shows the sum and the sum of square of x, where x is the monthly
income, in RM, of Encik Rosli for the first eight months in 2019.

(i) Cari varians dan sisihan piawai bagi pendapatan bulanan Encik Rosli.

Find the variance and standard deviation of Encik Rosli’s monthly income.

(ii) Jika anaknya memberi RM600 kepadanya pada setiap bulan untuk tempoh masa tersebut, cari

varians dan sisihan piawai yang baharu.
If his child gives him RM600 each month for the period of time, find the new variance and standard deviation.

 (i) ∑x2 – ∑x 2 (ii) Varians baharu / New variance
σ2 = n n


  = 2 = 48 125
50 385 000 – 20 000
8
8

= 48 125 Sisihan piawaian baharu

New standard deviation

= 219.37
σ = 48 125

= 219.37

(b) Jadual di bawah menunjukkan markah Adrian dan Bryan dalam lima set soalan Matematik. TP 4

The table below shows the marks obtained by Adrian and Bryan in five sets of Mathematics questions.

Set / Set 12345
Adrian 85 96 94 88 87
Bryan 82 93 96 95 84

Encik Phang ingin memilih salah seorang daripada mereka untuk mewakili sekolah dalam pertandingan

Matematik. Siapakah yang paling layak mewakili sekolah dalam pertandingan itu?

Mr. Phang wants to choose one of them to represent school in a Mathematics competition. Who is the most qualified to
represent the school in the competition?

BAB 8 Bagi markah Adrian / For Adrian’s marks, Bagi markah Brian / For Bryan’s marks,

Min / Mean, x– = 450 = 90 Min / Mean, x– = 450 = 90
5 5

∑x2 = 852 + 962 + 942 + 882 + 872 = 40 590 ∑x2 = 822 + 932 + 962 + 952 + 842 = 40 670

Sisihan piawai, s = 40 590 – 902 = 4.24 Sisihan piawai, s = 40 670 – 902 = 5.83
5 5
Standard deviation Standard deviation

Min bagi markah mereka adalah sama, iaitu 90. Adrian adalah yang paling layak mewakili sekolah
dalam pertandingan itu kerana nilai sisihan piawainya yang lebih kecil yang menunjukkan markahnya
lebih konsisten berbanding Bryan.

The mean for their marks is the same, which is 90. Adrian is the most qualified to represent the school in the competition

because the standard deviation is smaller, which shows that his marks are more consistent than Bryan.

© Penerbitan Pelangi Sdn. Bhd. 130

Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul 

(c) Diberi set data 20, 16, 4, 12, 20, 28, a dan b dengan min 20 dan varians 84. Hitung nilai a dan b jika

b , a. TP 5

Given a set of data 20, 16, 4, 12, 20, 28, a and b with a mean of 20 and a variance of 84. Calculate the value of a and
b if b , a.

∑x = 20 Daripada / From ,
n a = 60 – b …… 
20 + 16 + 4 + 12 + 20 + 28 + +
8 a b = 20 Gantikan  ke dalam ,

100 + a + b = 160 Substitute  into ,

a + b = 60 ……  (60 – b)2 + b2 = 1 872
3 600 – 120b + b2 + b2 = 1 872
∑x2 – (–x)2 = 84 2b2 – 120b + 1 728 = 0
n b = 36, b = 24

202 + 162 + 42 + 122 + 202 + 282 + a2 + b2 – 202 = 84 Apabila / When b = 36, a = 60 – 36 = 24
8 Apabila / When b = 24, a = 60 – 24 = 36
2 000 + a2 + b2
8 = 484 b , a, maka / thus b = 24, a = 36.

2 000 + a2 + b2 = 3 872

a2 + b2 = 1 872 …… 

(d) Carta palang di bawah menunjukkan bilangan peserta mengikut umur dalam suatu pertandingan

mewarna sempena Hari Merdeka di dua buah sekolah. TP 6

The bar chart shows the number of participants according to age in a colouring competition during Hari Merdeka in
two schools.

Peserta di Sekolah S Bilangan peserta Peserta di Sekolah T
Participants in School S Number of participants Participants in School T

Bilangan peserta
Number of participants

10 10
88
66
44
22

O 7 8 9 10 11 12 13 O 8 9 10 11 12
Umur (tahun) / Age (years old)
Umur (tahun) / Age (years old)

(i) Cari min umur peserta bagi setiap sekolah. BAB 8

Find the mean of participant’s age in each school.

(ii) Hitung sisihan piawai umur peserta bagi setiap sekolah.

Calculate the standard deviation of participant’s age in each school.

(i) Min umur peserta sekolah S = 10 tahun / Mean of participant’s age in school S = 10 years old
Min umur peserta sekolah T = 10 tahun / Mean of participant’s age in school T = 10 years old

(ii) Bagi umur peserta di sekolah S / For participants’ ages in school S,
∑x2 = 2(72) + 3(82) + 4(92) + 5(102) + 4(112) + 3(122) + 2(132) = 2 368

s= 2 368 – 102 = 1.72
23

Bagi umur peserta di sekolah T / For participants’ ages in school T,

∑x2 = 4(82) + 6(92) + 10(102) + 6(112) + 4(122) = 3 044

s= 3 044 – 102 = 1.21
30

131 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul SPM 8

PRAKTIS

Kertas 1 Kertas 2

1. Antara yang berikut, yang manakah tidak 1. Suatu set nombor 5, 12, x, 2x, 10 dan 18

dipengaruhi nilai ekstrem dalam suatu set data? BUKAN mempunyai min h. Apabila setiap nombor itu
RUTIN 6
Which of the following is not affected by an extreme ditambah dengan 3, min menjadi 5 h.

value in a data set? The set of numbers 5, 12, x, 2x, 10 and 18 has a mean

A Min C Julat 3, 6
5
Mean Range of h. When each number is added by the mean is h.

B Median D Sisihan piawai (a) Cari nilai x.

Median Standard deviation Find the value of x.

2. Plot batang-dan-daun di bawah menunjukkan (b) Cari sisihan piawai asal bagi nombor itu.

suatu set data. Find the original standard deviation of the set of

The stem-and-leaf plot below shows a data set. numbers.

(c) Dua nombor, h + 4 dan h − 4, ditambahkan

Batang Daun ke dalam set nombor itu. Tentukan sama ada
Stem Leaf
sisihan piawai bagi set 8 nombor itu adalah

5 034 lebih besar atau lebih kecil daripada set asal
6 256
7 1257 itu. Beri sebab anda.
8 14 Two numbers, h + 4 and h − 4, are added to the

set of numbers. Determine whether the standard
deviation for the set of 8 numbers is greater or
Kekunci: 6 | 2 bermaksud 62 mm
Key: 6 | 2 means 62 mm smaller than the original set. State your reason.

Apakah julat bagi data di atas? (a) 5 + 12 + x + 2x + 10 + 18 = h
6 45 h
What is the range of the above data?

A 27 C 30 + 3x = …… 1
6
B 31 D 34 45 + 3x + 6(3) 6
6 5
3. Plot kotak di bawah menunjukkan laju, dalam = h
km j–1, kereta di sebuah lebuh raya.
63 + 3x = 6 h …… 2
The box plot shows the speeds, in km h–1, of cars in a highway. 6 5

Gantikan / Substitute 1 ke / into 2:

63 + 3x = 6 1 45 + 3x 2
6 5 6
BAB 8 70 80 90 100 110
5(63 + 3x) = 6(45 + 3x)

Separuh daripada pemandu memandu kurang 315 + 15x = 270 + 18x

daripada x km j–1. Apakah nilai x? 45 = 3x
Half of the drivers drive less than x km h–1. What is the
x = 15
value of x?
(b) 5, 12, 15, 30, 10, 18
A 75 C 85
5 + 12 + 15 + 30 + 10 + 18
B 90 D 100 x = 6 = 15

4. Diberi min bagi set data di bawah ialah 12.9. s = 52 + 122 + 152 + 302 + 102 + 182 – 152
6
Given mean for the data set below is 12.9.

Kekerapan 3 3 2 11 = 7.8316

Frequency (c) Sisihan piawai bagi set 8 nombor adalah lebih
kecil daripada set 6 nombor kerana dua nombor
x 10 15 P 16 14 yang ditambahkan itu dekat dengan min.

Hitung nilai P. C 11 Praktis The standard deviation for the set of 8 numbers is
D 12 SPM smaller than the set of 6 numbers because the two
Calculate the value of P. Ekstra added numbers are close to the mean.

A 6
B 8

© Penerbitan Pelangi Sdn. Bhd. 132

Matematik  Tingkatan 4  Bab 8 Sukatan Serakan Data Tak Terkumpul 

Sudut KBAT KBAT

Ekstra

1. Plot kotak di sebelah menunjukkan jisim, dalam kg, sampel Mesin B
bungkusan gula yang diambil daripada dua buah mesin, A Machine B

dan B, yang dihasilkan di bahagian pembungkusan. Mesin A
Machine A
The box plot shows the masses, in kg, of sugar sample packs taken

from two machines, A and B, produced in packaging section.

(a) Hitung julat antara kuartil jisim bungkusan gula bagi 1.4 1.5 1.6

setiap mesin. Apakah kesimpulan yang dapat dibuat

daripada dua nilai ini?

Calculate the interquatile range of the masses of sugar packs for each machine. What conclusion can be made from

these two values?

(b) Jika sebungkus gula diambil daripada sampel tersebut, berapakah kebarangkalian jisim gula tersebut

adalah kurang daripada 1.48 kg bagi setiap mesin?
If a pack of sugar is taken from the sample, what is the probability of the mass of the sugar is less than from 1.48 kg

for each machine?

(a) Bagi mesin A, julat antara kuartil = 1.53 – 1.48
For machine A, interquatile range = 0.05

Bagi mesin B, julat antara kuartil = 1.55 – 1.42
For machine B, interquatile range = 0.13

Julat antara kuartil bagi mesin A adalah lebih kecil berbanding mesin B. Ini menunjukkan 50% daripada
sampel bagi mesin B terserak lebih besar berbanding dengan mesin A.

The interquartile range of machine A is smaller than machine B. It shows that 50% of the sample for machine B have
a wider dispersion than machine A.

(b) Bagi mesin A, kebarangkalian jisim bungkusan gula tersebut adalah kurang daripada 1.48 kg ialah 0.25
(25%), manakala bagi mesin B ialah 0.5 (50%).

For machine A, the probability of the mass of the sugar less than 1.48 kg is 0.25 (25%), while for machine B is 0.5

(50%).

2. Set A mempunyai enam nombor dengan sisihan piawai 3 dan set B mempunyai empat nombor dengan sisihan BAB 8

piawai 5. Kedua-dua set nombor itu mempunyai min yang sama. Jika dua set nombor itu digabungkan, cari

nilai varians.

Set A of six numbers has a standard deviation of 3 and set B of four numbers has a standard deviation of 5. Both sets of
numbers have an equal mean. If the two sets of numbers are combined, find the variance.

Katakan min bagi set A dan set B ialah m. Katakan σ2 mewakili varians bagi gabungan set

Let the mean for set A and set B is m. Let σ2 represents the variance of the combined set

m = ∑a ∑a2 + ∑b2 ∑a + ∑b 2
6 6+4 6+4
1 2 s2 = –
∑a = 6m

a2 6m + 4m 2
6 6+4
1 2
32 = – m2 = 6m2 + 54 + 4m2 + 100 –
6+4
54 = ∑a2 – 6m2
10m 2
10
1 2
∑a2 = 6m2 + 54 = 10m2 + 154 –
10
m = ∑b
4 10m2 + 154
= 10 – m2
∑b = 4m
Kuiz 8

52 = b2 – m2 = 10m2 + 154 – 10m2
4 10

100 = ∑b2 – 4m2 = 15.4

∑b2 = 4m2 + 100

133 © Penerbitan Pelangi Sdn. Bhd.

BAB Kebarangkalian Peristiwa Bergabung

9 Probability of Combined Events

9.1 Peristiwa Bergabung

Combined Events

NOTA IMBASAN

1. Persilangan dua peristiwa, A dan B ialah semua 3. Persilangan dan kesatuan dua peristiwa ini dikenali
kesudahan yang terdapat di dalam kedua-dua A dan B. sebagai peristiwa bergabung.
Ia ditulis sebagai A  B.
The intersection and union of these two events are known as
The intersection of two events, A and B is all of the outcomes that combined events.
are in both A and B. It is written as A  B.
4. Peristiwa bergabung boleh terhasil daripada satu
2. Kesatuan dua peristiwa, A dan B ialah semua kesudahan
sama ada dalam A atau B. Ia ditulis sebagai A  B. atau lebih eksperimen. Ia juga terdiri daripada dua

The union of two events, A and B is all of the outcomes that are in atau lebih peristiwa.
either A or B. It is written as A  B.
Combined events are resulted from one or

more experiments. It is also consists of two

or more events. NOTA

1. Tentukan ruang sampel bagi setiap yang berikut. TP 3

the sample space of each of the following.

CONTOH (a) Rajah di bawah menunjukkan dua keping kad

Rajah di sebelah bernombor di dalam kotak A dan empat keping

menunjukkan kad berhuruf di dalam kotak B.

sekeping duit syiling X The diagram below shows two pieces of numbered
cards in box A and four pieces of lettered cards in box B.

dan sekeping cakera. YZ 59 CAKE
Duit syiling itu
Kotak A / Box A Kotak B / Box B
dilambung sekali dan

cakera itu diputar Sekeping kad bernombor dipilih secara rawak

sekali. dari kotak A dan sekeping kad berhuruf dipilih

The diagram shows a coin and a disc. The coin is tossed secara rawak dari kotak B.

once and the disc is spun once. A piece of numbered card is chosen randomly from
box A and a piece of lettered card is chosen randomly
Penyelesaian: from box B.
Duit syiling
Cakera Kesudahan Kotak A Kotak B Kesudahan
Coin
Disc Outcome Box A Box B Outcome
A
X (A, X) C (5, C)
G Y (A, Y)
Z (A, Z) 5 A (5, A)
K (5, K)
X (G, X)
Y (G, Y) E (5, E)
Z (G, Z)
C (9, C)

9 A (9, A)
K (9, K)

E (9, E)

© Penerbitan Pelangi Sdn. Bhd. 134

Matematik  Tingkatan 4  Bab 9 Kebarangkalian Peristiwa Bergabung 

(b) Rajah di bawah menunjukkan empat keping (c) Dua batang aiskrim dipilih secara rawak dari
sebuah bekas yang mengandungi 2 batang
kad berhuruf di dalam sebuah kotak. aiskrim berperisa durian, sebatang aiskrim
berperisa jagung dan sebatang aiskrim berperisa
The diagram below shows four lettered cards in a box. anggur.

KUA T Two ice-creams are selected randomly from a
container that contains 2 durian-flavored ice-creams,
Dua keping kad dipilih secara rawak dari kotak a corn-flavored ice-cream and a grape-flavored ice
cream.
tersebut satu demi satu tanpa pemulangan.
Aiskrim Aiskrim kedua Kesudahan
Two cards are chosen randomly one by one from the pertama Second ice-cream Outcome
box without replacement.
First ice-cream
Kad pertama Kad kedua Kesudahan
First card Second card Outcome D2 (D1, D2)
D1 J (D1, J)
U (K, U)
K A (K, A) A (D1, A)

T (K, T) D1 (D2, D1)
D2 J (D2, J)
K (U, K)
U A (U, A) A (D2, A)

T (U, T) D1 (J, D1)
J D2 (J, D2)
K (A, K)
A U (A, U) A (J, A)

T (A, T) D1 (A, D1)
A D2 (A, D2)
K (T, K)
T U (T, U) J (A, J)

A (T, A)

9.2 Peristiwa Bersandar dan Peristiwa Tak Bersandar

Dependent Events and Independent Events

NOTA IMBASAN Kebarangkalian peristiwa bergabung BAB 9
bagi peristiwa A dan B, P(A dan B)
1. Dua peristiwa adalah bersandar jika kesudahan satu peristiwa memberi The probability of combined events for event
kesan terhadap peristiwa yang satu lagi. A and event B, P(A and B)

Two events are dependent if the outcome of one event affects the outcome of P(A ∩ B) = P(A) × P(B)
another event.
NOTA
2. Dua peristiwa adalah tak bersandar jika kesudahan satu peristiwa tidak
memberi kesan terhadap peristiwa yang satu lagi.

Two events are independent if the outcome of one event does not affect the
outcome of another event.

135 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 9 Kebarangkalian Peristiwa Bergabung

2. Tentukan sama ada setiap peristiwa berikut adalah peristiwa bersandar atau peristiwa tak bersandar. TP 1

Determine whether each of the following events is dependent event or independent event.

CONTOH (a) Puan Mariam mempunyai 5 biji telur. Sebiji
Terdapat 5 batang pensel dan 7 batang pen dalam daripada telur tersebut adalah rosak. Puan
sebuah kotak. Dua alat tulis tersebut dipilih secara Mariam telah memasak sebiji telur pada hari
rawak tanpa mengembalikan alat tulis pertama ke semalam dan hari ini. Dia telah memasak telur
dalam kotak. rosak pada hari ini.

There are 5 pencils and 7 pens in a box. Two stationeries Puan Mariam has 5 eggs. One of the eggs is rotten. Puan
are chosen at randon without returning the first stationery Mariam has cooked an egg yesterday and today. She
to the box. has cooked the rotten egg today.

Penyelesaian: Peristiwa bersandar / Dependent events
Peristiwa bersandar / Dependent events

(b) Dua keping kad dikeluarkan daripada sebuah (c) Ramalan kaji cuaca melaporkan bahawa hari ini

laci satu demi satu secara rawak tanpa dan lusa dijangkakan akan hujan.

mengembalikan kad yang pertama. The weather forecast reported that today and the day

Two cards are taken out from a drawer one after the after are expected it will rain.

other randomly without replacing the first card.

Peristiwa bersandar / Dependent events Peristiwa tak bersandar / Independent events

3. Hitung kebarangkalian peristiwa bergabung menggunakan rumus. TP 3

Calculate the probability of combined events using formula.

CONTOH

BAB 9 Dua buah majalah dipilih secara rawak daripada rak yang mengandungi 4 buah majalah dekorasi dan 6 buah

majalah kesihatan. Hitung kebarangkalian mendapat kedua-dua buah majalah ialah majalah kesihatan

dengan keadaan

Two magazines are chosen at random from a rack consists of 4 decoration magazines and 6 health magazines. Calculate
the probability of getting both magazines are health magazines such that

(i) majalah pertama dikembalikan sebelum memilih majalah kedua.

first magazine is replaced before choosing the second magazine.

(ii) majalah pertama tidak dikembalikan sebelum memilih majalah kedua.

first magazine is not replaced before choosing the second magazine.

Penyelesaian:

(i) P(kesihatan  kesihatan) / P(health  health) (ii) P(kesihatan  kesihatan) / P(health  health)

= 6 × 6 = 6 × 5 1 daripada 6 majalah
10 10 10 9 kesihatan telah dipilih.
1 of the 6 magazines was
9 chosen.
= 25 = 1 1 daripada 10 majalah
3

telah dipilih.
1 of the 10 magazines
was chosen.

© Penerbitan Pelangi Sdn. Bhd. 136

Matematik  Tingkatan 4  Bab 9 Kebarangkalian Peristiwa Bergabung 

(a) Terdapat 8 biji guli biru, 7 biji guli merah dan (b) Sebuah kotak mengandungi 5 batang kapur

5 biji guli kuning di dalam sebuah beg. Sebiji putih dan 10 batang kapur merah. Dua batang

guli dipilih secara rawak dan diletakkan kembali kapur dipilih tanpa mengembalikan kapur

ke dalam beg sebelum mengambil guli kedua. pertama. Hitung kebarangkalian mendapat dua

Hitung kebarangkalian sebiji guli biru dan sebiji batang kapur merah.

guli kuning dipilih. A box contains 5 white chalks and 10 red chalks. Two
chalks are chosen without replacing the first chalk.
There are 8 blue marbles, 7 red marbles and 5 yellow Calculate the probability of getting two red chalks.
marbles in a bag. A marble is chosen at random and
returned into the bag before picking the second marble. P(kapur merah  kapur merah)
Calculate the probability of a blue marble and a yellow
marble are chosen. P(red chalk  red chalk)

P(guli biru  guli kuning) = 10 × 9
15 14
P(blue marble ∩ yellow marble)
8 5 = 3
= 20 × 20 7

= 1
10

(c) Rajah di bawah menunjukkan sebuah papan (d) Kelab Matematik tingkatan 4 mempunyai 20 ahli

pemutar yang dibahagikan kepada 5 bahagian. berbangsa Melayu, 18 ahli berbangsa Cina dan

Seorang murid memutar sebanyak dua kali. 14 ahli berbangsa India. 3 daripada mereka akan

Hitung kebarangkalian murid tersebut mendapat dipilih mewakili sebuah pertandingan. Hitung

huruf C bagi putaran pertama dan mendapat kebarangkalian setiap bangsa mewakili dalam

huruf E bagi putaran kedua. pertandingan itu.

The diagram below shows a spinner board that is Mathematics Club of form 4 has 20 Malay members,
divided into 5 parts. A student spins twice. Calculate 18 Chinese members and 14 Indian members. 3 of
the probability of that student obtained letter C for the them will be chosen to represent in a competition.
first spin and obtained letter E for the second spin. Calculate the probability of each race represents in the
competition.

A P(seorang Melayu ∩ seorang Cina ∩ seorang India) BAB 9
EB
P(a Malay ∩ a Chinese ∩ an Indian)

D C = 20 × 18 × 14
52 51 50
42
= 1 105

P(C  E) = 1 × 1
5 5

= 1
25

137 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 9 Kebarangkalian Peristiwa Bergabung

4. Hitung kebarangkalian bagi setiap yang berikut. TP 3

Calculate the probability for each of the following.

CONTOH (a) Rajah di bawah menunjukkan dua keping kad

Rajah di bawah menunjukkan dua keping kad berhuruf di dalam kotak A dan tiga keping kad
bernombor di dalam kotak P dan tiga keping kad
berhuruf di dalam kotak Q. bernombor di dalam kotak B.

The diagram below shows two pieces of numbered cards The diagram below shows two pieces of lettered cards
in box P and three pieces of lettered cards in box Q. in box A and three pieces of numbered cards in box B.

ON 569

36 YES Kotak A Kotak B
Box A Box B

Kotak P Kotak Q Sekeping kad berhuruf dipilih secara rawak
Box P Box Q
daripada kotak A dan sekeping kad bernombor
Sekeping kad bernombor dipilih secara rawak
dipilih secara rawak daripada kotak B. Cari
daripada kotak P dan sekeping kad berhuruf
kebarangkalian mendapat
dipilih secara rawak daripada kotak Q. Hitung
A piece of lettered card is chosen randomly from Box A
kebarangkalian mendapat
and a piece of numbered card is chosen randomly from
A piece of numbered card is chosen randomly from box P Box B. Calculate the probability of getting
and a piece of lettered card is chosen randomly from box
Q. Calculate the probability of getting (i) huruf ‘N’ dan nombor 5.

(i) nombor 3 dan huruf ‘S’. the letter ‘N’ and number 5.

number 3 and the letter ‘S’. (ii) satu huruf vokal dan satu nombor ganjil.

(ii) satu nombor ganjil dan satu huruf vokal. a vowel and an odd number.

an odd number and a vowel letter. (i) P(huruf ‘N’)

Penyelesaian: P(the letter ‘N’)
1
(i) P(nombor 3) / P(number 3) = 1 = 2
2
P(nombor 5)

P(huruf ‘S’) / P(the letter ‘S’) = 1 P(number 5)
3 1
= 3
1 1 1
P(nombor 3 dan huruf ‘S’) = 2 × 3 = 6 P(huruf ‘N’ dan nombor 5)

P(number 3 and the letter ‘S’) P(the letter ‘N’ and number 5)

Kaedah Alternatif = 1 × 1 = 1
2 3 6
Kesudahan / Outcome: {(3, S)}
(ii) P(huruf vokal)
n(S) = 2 × 3 = 6 1
6
P(nombor 3 dan huruf ‘S’) = P(vowel letter)
P(number 3 and the letter ‘S’) 1
= 2

BAB 9 (ii) P(nombor ganjil) / = 1 P(nombor ganjil)
2
P(odd number) P(odd number)

P(huruf vokal) / P(vowel letter) = 1 = 2
3 3

P(nombor ganjil dan huruf vokal) = 1 × 1 = 1 P(huruf vokal dan nombor ganjil)
2 3 6 P(vowel letter and odd number)
P(odd number and vowel letter) 1 2 1
= 2 × 3 = 3

© Penerbitan Pelangi Sdn. Bhd. 138

Matematik  Tingkatan 4  Bab 9 Kebarangkalian Peristiwa Bergabung 

(b) Kotak X mengandungi tiga keping kad berlabel (c) Sebiji dadu adil dilambung sekali dan

‘E, A, T’ dan kotak Y mengandungi lima sekeping duit syiling dilambung sekali. Hitung

kad berlabel ‘L, U, N, C, H’. Sekeping kad kebarangkalian

dipilih secara rawak dari setiap kotak. Hitung A fair dice is rolled once and a coin is tossed once.
Calculate the probability that
kebarangkalian mendapat
(i) dadu itu menunjukkan nombor ganjil dan
Box X contains three cards labelled ‘E, A, T’ and box
Y contains five cards labelled ‘L, U, N, C, H’. A card duit syiling menunjukkan gambar.
is chosen randomly from each box. Calculate the
probability of getting the dice shows an odd number and the coin shows
head.
(i) huruf ‘T’ dari kotak X dan huruf ‘N’ dari
(ii) dadu itu menunjukkan nombor perdana dan
kotak Y.
duit syiling menunjukkan angka.
the letter ‘T’ from box X and the letter ‘N’ from box
Y. the dice shows a prime number and the coin shows
tail.
(ii) dua kad berhuruf vokal.
(i) P(dadu menunjukkan nombor ganjil)
two vowel letter cards.

P(the dice shows an odd number)
3 1
(i) P(huruf ‘T’) = 6 = 2

P(the letter ‘T’)
1
= 3 P(duit syiling menunjukkan gambar)
P(the coin shows head)
1
P(huruf ‘N’) = 2

P(the letter ‘N’)
1
= 5 P(dadu menunjukkan nombor ganjil dan

P(huruf ‘T’ dan huruf ‘N’) duit syiling menunjukkan gambar)

P(the letter ‘T’ and the letter ‘N’) P(the dice shows an odd number and the coin
1 1 1
= 3 × 5 = 15 shows head)
1 1 1
= 2 × 2 = 4

(ii) P(huruf vokal dalam kotak X) (ii) P(dadu menunjukkan nombor perdana)
P(vowel letter in box X)
2 P(the dice shows a prime number)
3 3 1
= = 6 = 2

P(huruf vokal dalam kotak Y) P(duit syiling menunjukkan angka)
P(vowel letter in box Y)
1 P(the coin shows tail)
5 1
= = 2

P(huruf vokal dalam kotak X dan kotak Y) P(dadu menunjukkan nombor perdana dan

P(vowel letter in box X and Y) duit syiling menunjukkan angka)

= 2 × 1 P(the dice shows a prime number and the coin BAB 9
3 5
shows tail)
2 1 1 1
= 15 = 2 × 2 = 4

139 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 9 Kebarangkalian Peristiwa Bergabung

5. Hitung kebarangkalian bagi setiap yang berikut. TP 3

Calculate the probability for each of the following.

CONTOH (a) Dua ekor anak kucing dipilih oleh seorang

Dua orang murid dipilih untuk mengadakan suatu pelanggan di sebuah kedai haiwan untuk dibeli

perbentangan di dalam kelas. Mereka dipilih sebagai haiwan peliharaan. Mereka dipilih

daripada kumpulan yang mempunyai dua orang daripada kumpulan yang mempunyai empat

murid lelaki dan tiga orang murid perempuan ekor anak kucing jantan dan dua ekor anak

tanpa pemulangan. kucing betina tanpa pemulangan.

Two students are chosen to carry out a presentation in the Two kittens are chosen by a customer in a pet shop
class. They are chosen from a group of two male students
and three female students without replacement. to be bought as a pet. They are chosen from a group
of four male kittens and two female kittens without
(i) Lakar satu gambar rajah pokok untuk replacement.

menyenaraikan semua kesudahan yang (i) Lakar satu gambar rajah pokok untuk

mungkin. menyenaraikan semua kesudahan yang

Sketch a tree diagram to list all the possible outcomes. mungkin.

(ii) Hitung kebarangkalian bahawa Sketch a tree diagram to list all the possible

Calculate the probability that outcomes.

(a) seorang murid lelaki dan seorang murid (ii) Hitung kebarangkalian bahawa

perempuan dipilih. Calculate the probability that

a male student and a female student are chosen. (a) seekor anak kucing jantan dan seekor

(b) sekurang-kurangnya seorang murid lelaki anak kucing betina dipilih.

dipilih. a male kitten and a female kitten are chosen.

at least a male student is chosen. (b) sekurang-kurangnya seekor anak

kucing betina dipilih.

at least a female kitten is chosen.

Penyelesaian: (i)

(i) Anak kucing Anak kucing

Murid Murid kedua Kesudahan pertama kedua Kesudahan
pertama First kitten Second kitten
Second student Outcome 3 Outcome
First student
1 (L, L) (J, J)
L (L, P) 5 J (J, B)
2 4 P (P, L) 4 J B (B, J)
5 L L (P, P) 6 J (B, B)
P 2 B
3 2 3 2 4 5
5 4 4 6 5

P2 B1

4 5

(ii) (a) P(L, P) + P(P, L) (ii) (a) P(J, B) + P(B, J)

BAB 9 =  2 × 3  +  3 × 2  =  4 × 2  +  2 × 4 
5 4 5 4 6 5 6 5

= 6 + 6 = 8 + 8
20 20 30 30

= 12 = 3 = 16 = 8
20 5 30 15

(b) P(L, P) + P(P, L) + P(L, L) (b) P(J, B) + P(B, J) + P(B, B)

=  2 × 3  +  3 × 2  +  2 × 1  =  4 × 2  +  2 × 4  +  2 × 1 
5 4 5 4 5 4 6 5 6 5 6 5

= 6 + 6 + 2 = 8 + 8 + 2
20 20 20 30 30 30

= 14 = 7 = 18 = 3
20 10 30 5

© Penerbitan Pelangi Sdn. Bhd. 140

Matematik  Tingkatan 4  Bab 9 Kebarangkalian Peristiwa Bergabung 

(b) Sebuah bekas mempunyai 24 biji gula-gula. (c) Sebuah bekas mempunyai 18 biji guli. Satu per

8 biji daripadanya berperisa epal manakala tiga daripadanya adalah guli hitam manakala

bakinya berperisa mangga. Dua biji gula-gula bakinya adalah guli biru. Dua biji guli dipilih

dipilih secara rawak daripada bekas itu dengan secara rawak daripada bekas itu dengan adanya

adanya pemulangan. pemulangan.

A container contains 24 sweets. 8 of the sweets are A container contains 18 marbles. One third of the

apple flavored while the remainder is mango flavored. marbles are black marbles while the remainder are blue
Two sweets are chosen randomly from the container marbles. Two marbles are chosen randomly from the
with replacement. container with replacement.

(i) Lakar satu gambar rajah pokok untuk (i) Lakar satu gambar rajah pokok untuk

menyenaraikan semua kesudahan yang menyenaraikan semua kesudahan yang

mungkin. mungkin.

Sketch a tree diagram to list all the possible Sketch a tree diagram to list all the possible

outcomes. outcomes.

(ii) Hitung kebarangkalian bahawa (ii) Hitung kebarangkalian bahawa

Calculate the probability that Calculate the probability that

(a) kedua-dua biji gula-gula yang dipilih (a) kedua-dua biji guli yang dipilih adalah

adalah berperisa epal. berwarna biru.

both sweets chosen are apple flavored. both marbles chosen are blue marbles.

(b) gula-gula yang dipilih mempunyai (b) guli yang dipilih mempunyai warna

perisa berlainan. berlainan.

the sweets chosen are of different flavors. the marbles chosen are of different colours.

(i) (i)

Gula-gula Gula-gula Kesudahan Guli pertama Guli kedua Kesudahan
pertama kedua First marble Second marble Outcome
Outcome
First sweet Second sweet 6
(E, E)
8 E 8 (E, M) 6 H 18 H (H, H)
24 24 E (M, E) 18 B (H, B)
(M, M) H (B, H)
8 16 M 12 6 12
24 24 E 18 18

16 M B
24 18 12 B (B, B)
16 M 18
24
(ii) (a) P(B, B)
(ii) (a) P(E, E) 12 12
8 8 = 18 × 18
= 24 × 24
144
= 64 = 324
576
= 4
= 1 9 BAB 9
9
(b) P(H, B) + P(B, H)
(b) P(E, M) + P(M, E)
    =6 12 12 6
    = 8 × 16 + 16 × 8 18 × 18 + 18 × 18
24 24 24 24
= 2 + 2
= 2 + 2 9 9
9 9
= 4
= 4 9
9

141 © Penerbitan Pelangi Sdn. Bhd.

  Matematik  Tingkatan 4  Bab 9 Kebarangkalian Peristiwa Bergabung

6. Selesaikan setiap yang berikut. TP 4

Solve each of the following.

CONTOH

Kebarangkalian Kenji masuk universiti X, Y atau Z masing-masing adalah 0.3, 0.6 dan 0.8. Hitung
kebarangkalian Kenji

The probability Kenji gains entry into university X, Y or Z is 0.3, 0.6 and 0.8 respectively. Calculate the probability that
Kenji is admitted into

(i) masuk universiti Z sahaja, / university Z only,
(ii) tidak masuk semua universiti itu, / none of the universities,
(iii) masuk salah satu universiti itu. / only one of the universities.

Penyelesaian:

(i) P(universiti Z) = 0.8 × 0.7 × 0.4
P(university Z) = 0.224

(ii) P(tidak masuk) = 0.7 × 0.4 × 0.2

P(none) = 0.056

(iii) P(salah satu universiti)

P(one of the university)

= (0.3 × 0.4 × 0.2) + (0.7 × 0.6 × 0.2) + (0.7 × 0.4 × 0.8)
= 0.332

(a) Sebuah beg mengandungi 25 biji epal dan 5 (b) Suatu tinjauan mendapati kebarangkalian

daripadanya didapati rosak. Jika tiga biji epal seorang murid melanjutkan pelajaran selepas

dikeluarkan satu demi satu tanpa dikembalikan SPM adalah 0.7. Jika tiga orang murid lepasan

ke dalam beg, hitung kebarangkalian bahawa SPM dipilih secara rawak, hitung kebarangkalian
A bag contains 25 apples and 5 of the apples are rotten.
bahawa
If three apples are picked one after the other without A survey shows the probability of a student to further
returning into the bag, calculate the probability that his studies after SPM is 0.7. If three SPM leavers are

(i) dua biji epal itu adalah rosak, chosen at random, calculate the probability that

two of the apples are rotten, (i) hanya seorang melanjutkan pelajaran
selepas SPM,
(ii) dua biji epal itu adalah baik,
only one of them furthers his studies after SPM,
two of the apples are good,
(ii) lebih daripada dua orang melanjutkan
(iii) selebih-lebihnya dua biji epal adalah rosak. pelajaran selepas SPM,

at most two apple is rotten.

(i) P(2 biji rosak) = 5 × 4 × 20 ×3 more than two of them further their studies after
25 24 23 SPM,
P(2 rotten apples) 2
23 (iii) tiada orang yang melanjutkan pelajaran
BAB 9 = selepas SPM.

none of them further their studies after SPM.

(ii) P(2 biji baik) = 20 × 19 × 5 ×3 (i) P(hanya seorang) = 0.7 × 0.3 × 0.3 × 3
25 24 23
P(2 good apples) P(only one) = 0.189
19
= 46

(iii) P(< 2 biji rosak) = 1 – P(semua rosak) (ii) P(semua orang) = 0.7 × 0.7 × 0.7

P(< 2 rotten apples) = 1 – P(all rotten) P(everyone) = 0.343

= 1 – 5 × 4 × 3 (iii) P(tiada orang) = 0.3 × 0.3 × 0.3
25 24 23 P(none of them) = 0.027

= 229
230

© Penerbitan Pelangi Sdn. Bhd. 142