Praktis Hebat! TG2 (MATEMATIK) Penerbitan Pelangi Sdn Bhd

1 Pola dan Jujukan 1
Patterns and Sequences

2 Pemfaktoran dan Pecahan Algebra 8
Factorisation and Algebraic Fractions

3 Rumus Algebra 14
Algebraic Formulae

4 Poligon 20
Polygons

5 Bulatan 26
Circles

6 Bentuk Geometri Tiga Dimensi 34
Three-Dimensional Geometrical Shapes

7 Koordinat 41
Coordinates

8 Graf Fungsi 49
Graphs of Functions

9 Laju dan Pecutan 58
Speed and Acceleration

10 Kecerunan Garis Lurus 65
Gradient of A Straight Line

11 Transformasi Isometri 73
Isometric Transformations

12 Sukatan Kecenderungan Memusat 80
Measures of Central Tendencies

13 Kebarangkalian Mudah 89
Simple Probability `
Penilaian Pra PT3
: 99

Jawapan J1 - J12

Prak tis PT3 Pemfaktoran dan Pecahan Algebra

2 Factorisation and Algebraic Fractions

Bahagian A BUKU TEKS
m.s. 18 – 41
1. Apakah faktor sepunya terbesar (FSTB) bagi 16y2
dan 8y?   TP1 6. Faktorkan g2 + 5g – 14.   TP2
What is the highest common factor (HCF) of 16y2 Factorise g2 + 5g –14.
A (g + 2)(g + 7)
and 8y? B (g – 2)(g – 7)
A 2y C (g – 2)(g + 7)
B 4y D (g + 2)(g – 7)
C 8y
D 8y2 7. Faktorkan yang berikut.   TP2
PT3 Factorise the following.
2. Senaraikan semua faktor sepunya bagi 3x dan
6xy.   TP1 2019
List all the common factors of 3x and 6xy.
A 1, x 4m – 8
B 1, 3, x
C 1, 3, x, 3x A 4(m – 2)
D 1, 2, 3, x, 3x B 2(4m – 8)
C 2(4m – 4)
3. (p – 1)(p + 4) =    TP2 D 4(m – 8)
A p2 – 4
B p2 – 4p 8. Apakah faktor sepunya terbesar (FSTB) bagi 6,
C p2 – 3p – 21 18 dan 54?   TP2
D p2 + 3p – 4 What is the highest common factor (HCF) of 6, 18
and 54?
4. Kembangkan 7q(q – 2).   TP2 A 2
Expand 7q(q – 2). B 3
A –14q C 6
B 7q – 14 D 12
C 7q2 – 14
D 7q2 – 14q 9. Permudahkan (a – 2b)2 + a(4b – a).   TP2
Expand (a – 2b)2 + a(4b – a).
5. Kembangkan (3y – 2)2.   TP2 A 4b2 + 2a
Expand (3y – 2)2. B 4b2
A 3y2 – 12y + 4 C 4b2 – 2a + 3
B 61y – 12y + 4 D 4b2 – a
C 9y2 – 12y + 4
D 9y2 – 12y – 4 10. 9x2 – 25 =    TP2
A 3x – 5
B (3x – 5)(3x + 5)
C (3x – 5)(3x – 5)
D (9x – 25)(9x + 25)

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Matematik  Tingkatan 2  Praktis PT3 2  Pemfaktoran dan Pecahan Algebra 

Bahagian B

1. Lengkapkan peta buih berganda di bawah dengan 3. Isi tempat kosong yang diberi.  Subtopik 2.1

BUKU menyenaraikan faktor dan faktor sepunya bagi BUKU Fill in the blanks given.
TEKS TEKS

m.s. 28 sebutan 6p dan 4pq.   TP2   Subtopik 2.2 m.s. 24

Complete the double bubble map below by listing PT3 (a) Lengkapkan yang berikut.
2019 Complete the following.
the factors and common factors of terms 6p and 4pq.

5mn 10n2
4q m+5 ÷ m2 – 25

31 2q = 5mn ×
m+5

6 6p 2 = 5mn × 10n­2
6p 4pq m+5

p 4q =

4pq 2pq [3 markah / 3 marks]

(b) Faktorkan / Factorise:

6p – 12 =  (p – 2)

[1 markah / 1 mark]

[4 markah / 4 marks] 4. Lengkapkan pemfaktoran menggunakan kaedah

2. Padankan setiap ungkapan algebra yang berikut BUKU pendaraban silang dengan jawapan yang betul.
TEKS

BUKU dengan jawapan yang betul.   TP3   m.s. 30 Complete the factorisation using cross multiplication
TEKS
Subtopik 2.1 method using the correct answers: TP4 Subtopik 2.2

m.s. 23 Match each of the following algebraic expressions

with the correct answers. 2r2 – 5rs – 3s2 = (2r + s)( )

 a 2 a2 2r s
4 4 –3s –6rs

– a2 2r2 –3s2
9
– a   a 
2 × 3 a2
6
– [4 markah / 4 marks]

– a  × – a  a2
2 2 16
5. Bulatkan jawapan yang betul.   TP4 Subtopik 2.3

 a 2 BUKU Circle the correct answers. 
3 TEKS

m.s. 34

a2 (i) 8xy + x2 = x(8y + x) 8x(y + x)
9

[4 markah / 4 marks] (ii) 3(x + y) = 3x + y 3x + 3y

(iii) 2 – 3 = 2y – 3x 6
x y xy xy

(iv) x2 – y2 = (x + y)(x + y) (x – y)(x + y)

[4 markah / 4 marks]

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 Matematik  Tingkatan 2  Praktis PT3 2  Pemfaktoran dan Pecahan Algebra

Bahagian C

1. (a) (i) Kembangkan:  TP2 Subtopik 2.1 (c) Rajah di bawah menunjukkan sebuah

BUKU Expand: BUKU papan tanda yang berbentuk segi tiga.
TEKS TEKS

m.s. 23 m2(3 – n) m.s. 37 Diberi luas papan tanda tersebut ialah

[1 markah / 1 mark] 0.5(15x2 – 2x – 1) cm2. 

Jawapan / Answer : The diagram below shows a triangular
signboard. Given the area of the signboard is
0.5(15x2 – 2x – 1) cm2.

(ii) Faktorkan: TP3 Subtopik 2.2 !

BUKU Factorise: (3x – 1) cm
TEKS
(i) Hitung tinggi, dalam cm, papan tanda
m.s. 28 (a) p2 – 64 tersebut.  TP5 Subtopik 2.3

(b) 16 – 24x Calculate the height, in cm, of the
signboard.
[3 markah / 3 marks]

Jawapan / Answer :

[2 markah / 2 marks]

Jawapan / Answer :

(b) Permudahkan setiap ungkapan algebra (ii) Hitung luas papan tanda tersebut
apabila x = 3.  TP4 Subtopik 2.3
BUKU yang berikut.  TP4 Subtopik 2.3
TEKS Calculate the area of the signboard when
x = 3.
m.s . 37 Simplify each of the following algebraic [2 markah / 2 marks]

expressions. Jawapan / Answer :

(i) q – q
4p 4pq

(ii) 9a2 – 1 × 3a2
9ab 3a – 1

[2 markah / 2 marks]
Jawapan / Answer :

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Matematik  Tingkatan 2  Praktis PT3 2  Pemfaktoran dan Pecahan Algebra 

2. (a) Rajah di bawah menunjukkan beberapa (b) Diberi panjang sisi segi empat sama A dan

BUKU sebutan algebra.  BUKU segi empat sama B masing-masing ialah
TEKS TEKS

m.s . 28 The diagram below shows some algebraic m.s. 25 (6p + 1) cm dan (5p – 7) cm. Subtopik 2.1

terms. Given the length of sides of square A and square
B are (6p + 1) cm and (5p – 7) cm respectively. 

21q2t2 42qt3 – 42qt2 (i) Cari luas dan perimeter kedua-dua
segi empat sama tersebut. TP4
   Find the area and perimeter of both
squares.
(i) Cari Faktor Sepunya Terbesar bagi
sebutan di atas. TP2   Subtopik 2.2 [4 markah / 4 marks]

Find the Highest Common Factor for the Jawapan / Answer :
terms above.

[3 markah / 3 marks]

Jawapan / Answer :

(ii) Kemudian, faktorkan. TP3 Subtopik 2.2 (ii) Hitung luas kedua-dua bentuk apabila
Hence, factorise. p = 7.   TP4

21q2t2 + 42qt3 – 42qt2 Calculate the area of both shapes when
p = 7.
[1 markah / 1 mark]
[2 markah / 2 marks]

Jawapan / Answer :

Jawapan / Answer :

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 Matematik  Tingkatan 2  Praktis PT3 2  Pemfaktoran dan Pecahan Algebra

3. (a) (i) Ungkapkan 3p × (q2 + 2pq) (iii) Faktorkan 15x2 – 2 + x. 
(2p + q)q2
BUKU BUKU Factorise 15x2 – 2 + x. TP3 Subtopik 2.1
TEKS

TEKS sebagai satu pecahan tunggal dalam m.s. 28

m.s. 37 [2 markah / 2 marks]

bentuk termudah. TP3   Subtopik 2.3

Express 3p × (q2 + 2pq) as a Jawapan / Answer :
(2p + q)q2

single fraction in its simplest form.

[3 markah / 3 marks]

Jawapan / Answer :

(b) Diberi luas sebuah padang bola ialah

BUKU (8p2 + 56p) m2. Cari perimeter, dalam m,
TEKS

m.s. 32 padang tersebut. TP4   Subtopik 2.2

(ii) Permudahkan (2a – 2b)2 + a(4b – a). Given the area of a football field is (8p2 + 56p) m2.
Find the perimeter, in m, of the field.

BUKU Simplify (2a – 2b)2 + a(4b – a). Menganalisis
TEKS

m.s. 25 TP3 Subtopik 2.1 [3 markah / 3 marks]

[2 markah / 2 marks] Jawapan / Answer :

Jawapan / Answer :

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Matematik  Tingkatan 2  Praktis PT3 2  Pemfaktoran dan Pecahan Algebra 

4. (a) Permudahkan:  TP4 Subtopik 2.3 (c) Sebuah kilang dapat menghasilkan

BTUEK KSU Simplify: BUKU (72r + 24) buah telefon bimbit dalam masa
TEKS
m.s. 36
8y – 4yz ÷ 12y2z m.s. 34 enam hari. 
y2 – 9 y–3
A factory can produce (72r + 24) mobile phones

[3 markah / 3 marks] in six days.

Jawapan / Answer : (i) Berapakah bilangan telefon bimbit

yang boleh dihasilkan dalam masa

12 jam? TP4 Subtopik 2.3

How many mobile phones can be produced

in 12 hours?   Menganalisis

[3 markah / 3 marks]

Jawapan / Answer :

(b) Kembangkan:  TP2 Subtopik 2.1

BTUEK KSU Expand: [2 markah / 2 marks]
m.s. 21 (i) –5(3r – 5t)

(ii) – j  (3m + 72) (ii) Berapa harikah masa yang diambil jika
9
kilang tersebut telah menghasilkan

(48r + 16) telefon bimbit?

Jawapan / Answer : How many days taken if the factory has

produced (48r + 16) mobile phones?

TP5 Subtopik 2.3 Menganalisis

[2 markah / 2 marks]

Jawapan / Answer :

KBAT (10k) cm

Rajah di sebelah menunjukkan sebuah papan tanda yang berbentuk segi tiga. BERI
Diberi bahawa luas papan tanda tersebut ialah (25k2 + 15k) cm2. Hitung tinggi, LALUAN
dalam cm, papan tanda tersebut.
The diagram beside shows a triangular signboard. Given that the area of the signboard
is (25k2 + 15k) cm2. Find the height, in cm, of the signboard.

Jawapan / Answer :

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Prak 1tis PT3 Pola dan Jujukan C
Patterns and Sequences
1. (a) (i) Menolak 3 dari nombor sebelumnya.
A Subtract 3 from the previous number.

1. C 2. C 3. D 4. D 5. C (ii) Bahagi nombor sebelumnya dengan 2.
6. B 7. D 8. C 9. A 10. A Divide the previous number by 2.

(iii) Darab nombor sebelumnya dengan 4.
Multiply the previous number with 4.

B (b) (i) p = 38, q = 77 (ii) 11

1. 4, 8, 12, 16, … • (c) 96
1, 8, 27, 64, … •
3, 7, 11, 15, … • • 2n + 2; n = 1, 2, 3, … 2. (a) (i)
4, 6, 8, 10, … • • 4n – 1; n = 1, 2, 3, …
• 4n; n = 1, 2, 3, …
• n3; n = 1, 2, 3, …

2. 8 9 16 (ii) Menambah 4 batang pensel kepada bentuk
16
sebelumnya.
4 25 Add 4 pencils to the previous shape.
32 (iii) 33 batang / pencils

1 36 (b) (i) 11:45 a.m. (ii) 1:15 p.m.

64 2 3. (a) (i) 11
(ii) 6th: (7, 29)
3. (a) (i) Jujukan / A sequence 7th: (8, 37)
(ii) Bukan jujukan / Not a sequence (b) (i) P = 11, Q = 20
(b) (ii) 2 + 3n; n = 1, 2, 3, …

1 4. (a) (i) Bukan jujukan  / Not a sequence
(ii) Jujukan / Sequence
11 (iii) Jujukan / Sequence
(iv) Bukan jujukan / Not a sequence
121 (b) (i) 6, 11, 16, …
1 + 5n; n = 1, 2, 3, …
1331 (ii) Ya, kedudukan ke-29.
Yes, 29th position
14641
5. (a) x = 15, y = 32
1 5 10 10 5 1 (b) (i) 4n – 3, n = 1, 2, 3, …
(ii) 42 – 4n, n = 1, 2, 3, …
4. (a) (i)
(ii)

(b) (i) 1, 8, 13 (c) (i) Masa (minit) Isi padu air (ml)
(ii) 14, 23 Time (minutes) Volume of water (ml)

5. 1 50
32 2 100
3 150
16 16 4 200
5 250
8 88 8 6 300
16 (ii) 9 minit  / minutes

KBAT

4 12 12 4 5, 14, 23, 32, 41, 50

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  Matematik  Tingkatan 2  Jawapan

tis PT3 5. (i) x(8y + x) (ii) 3x +3y
(iv) (x − y)(x + y)
Prak2 Pemfaktoran dan Pecahan Algebra (iii) 2y – 3x
PrakFactorisation and Algebraic Fractionsxy

C

A 1. (a) (i) 3m2 – m2n

1. C 2. C 3. D 4. D 5. C (ii) (a) (p + 8)(p – 8)
6. C 7. A 8. C 9. B 10. B
(b) 8(2 – 3x)
1. 2q
B (b) (i) q–1 (ii) 3a32 b+ a
4p

(c) (i) (5x + 1) cm (ii) 64 cm2

4q 2. (a) (i) 21qt 2 (ii) 21qt 2(q + 2t – 2)

31 (b) (i) Segi empat sama A / Square A :
Luas / Area = (36p2 + 12p + 1) cm2
Perimeter = (24p + 4) cm

6 6p 2 4pq pq Segi empat sama B / Square B :
p 4q Luas / Area = (25p2 – 70p + 49) cm2
3p Perimeter = (20p – 28) cm
6p 2p 4p
4pq 2pq (ii) Segi empat sama A / Square A = 1 849 cm2
Segi empat sama B / Square B = 784 cm2

3. (a) (i) 3p (ii) 3a2 + 4b2 – 4ab
q

(iii) (3x – 1)(5x + 2)

(b) (18p + 14) cm

2. 4. (a) 2–z
3yz(y + 3)
jm
 a 2 a2 (b) (i) –15r + 25t (ii) – 3 – 8j
4• •4 (ii) 4 hari / days
(c) (i) 6r + 2
a2
–   a   a  • – 9
2 3
× • a2 KBAT
6
   –  a a • – 5k + 3
2 2
× –   • a2
16
 a 2 • 3tis PT3
3
• a2 Rumus Algebra
9 Algebraic Formulae
•

3. (a) 5mn ÷ 10n2 A
m+5 m2 – 25
2. D
= 5mn × m2 – 25 1. A 7. B 3. C 4. C 5. A
m+5 10n2 6. B 8. A 9. A 10. D
11. D
5mn (m + 5)(m – 5)
= m+5 × 10n2 B

= m2 – 5m 1. (a) p–3
2n 2
q=

(b) 6p – 12 = 6 (p – 2) p = 2q + 3

4. 2r2 – 5rs – 3s2 = (2r + s)( r – 3s ) q = 2(p + 3)

2r s rs p= q –3 q= p–3
r –3s –6rs 2 2
2r2 –3s2 –5rs
(ii) 2
(b) (i) 3

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2. (a) (i) 3 (ii) 7 Matematik  Tingkatan 2  Jawapan 
(ii)

(b) (ii) 8 (iv) 8
3

3. (a) p – q2 + r √x2 + y2 = 2 xy + xyz

y = mx + 5 2pq + 3qp a(2b) ÷ 4ab Bilangan paksi simetri: 6
Number of axis of symmetry: 6

(b) (i) I (ii) y 2. (a) 3 (b) 108° (c) 8 (d) 144°

4. z = √10x – y  3. (i) 3 (ii) 3 (iii) 3 (iv) ✗

10x – y = z2 4. (a) P : Sudut pedalaman / Interior angle
Q: Sudut peluaran / Exterior angle
10x = z2 + y
(b) (i) 6 (ii) 6

x = z2 + y 5. (a) (i) 3 (ii) 7
10 (b)

C

1. (a) a = 4(P – 18)2
f
(b) (i) s = –1 (ii) 8
2t (ii) 25 A
(c) K = RM225p
(d) (i) V = IR

2. (a) x = 4L – y – y2 – 1
2y + 2
p+1
(b) (i) q = 2r (ii) –1

(c) (i) N = m–5 (ii) 6 biji / bulbs C
2
3. (a) (i) W = 2h – 10 (ii) 158 cm 1. (a) x = 108°, y = 72°
(b) (i) C = 17x + 50 (ii) 4 jam / 4 hours (b) x = 120°, y = 132°
(c) (i) xy + 60y = 1 680 (ii) 20 (c) 129°

4. (a) q = pr 2. (a) (i) 36 (ii) 5 sisi / sides, Pentagon
3p – 2r (b) 132°
(b) (i) P = 1.15x + 1.1y (ii) 43 kg (c) 12
(c) (i) H = 12p + 8q (ii) J = 9.6p + 6.4q
3. (a) 108°
KBAT (b) 96°
(c) 12 sisi / sides
C

tis PT3 KBAT

4Prak Poligon 52.5°
PrakPolygons
5tis PT3
A Bulatan
Circles
2. D
1. C 7. C 3. D 4. A 5. B A
6. C 8. D 9. C 10. A
B 2. C
1. (i) 1. C 7. B 3. B 4. B 5. C
6. D 8. A 9. D 10. A

Bilangan paksi simetri: 5 B
Number of axis of symmetry: 5
1. (a) Sektor minor / Minor sector
(b) Jejari / Radius
(c) Perentas / Chord
(d) Lengkok minor / Minor arc

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  Matematik  Tingkatan 2  Jawapan

2. Panjang lengkok PQ  / Length of arc PQ 3. (a) 1, 0
(b) 1, 1
= 60° × 2 × 22 × 42
360° 7 4. (a) silinder / cylinder
(b) kuboid / cuboid
= 44 cm (c) piramid / pyramid
(d) kon / cone
3. (a) Benar / True
(b) Benar / True C
(c) Palsu / False
(d) Palsu / False 1. (a) 144 cm2
(b) 14 cm
4. (a) DOG, EOB (c) 10 cm
(b) (i) 3
(ii) 3 2. (a)

5. Jejari bulatan Perimeter Luas bulatan (b) 550 cm2
Radius of bulatan Area of (c) 4 cm
circle circle
Perimeter of
4.5 cm circle 63.6 cm2
37.8 cm 4 490.6 cm2
28.3 cm

C 237.6 cm

3. (a) (i) Isi padu silinder / Volume of cylinder = π x2y

1. (a) 77 cm Isi padu kon / Volume of cone = 1 π x2y
(b) 56.76 cm2 3

(c) (i) 33 cm (ii) 47 cm (ii) Apabila sebuah silinder dan sebuah kon
(ii) 56 cm
2. (a) 1.32 cm (ii) 63.81 cm mempunyai tapak dan tinggi yang sama, maka

(b) 9 cm isi padu silinder adalah tiga kali isi padu kon

(c) 30 4 cm2 itu.
7 When a cylinder and a cone have the same
bases and heights, then the volume of the
3. (a) (i) 75.46 cm2 cylinder is three times of the volume of the
cone.
(b) 35.5 cm

4. (a) 24 cm (b)
(b) 71.17 cm2

(c) (i) 17.81 cm

KBAT

(a) 87 cm
(b) 173.25 cm2

Prak tis PT3 Bentuk Geometri Tiga Dimensi
Three-Dimensional Geoemtrical Shapes
6

1. A A 3. B 4. A 5. A
6. B 8. B 9. B 10. C
2. D
7. C

B (c) 30 492 cm3

1. (a) kon / cone, hemisfera / hemisphere 4. (a) (i) 4.445 m2
(b) kubus / cube, piramid / pyramid (ii) 88.9 ml

2. (a) 3 (b) 120 kotak / 120 boxes
(b) 3
(c) 3 KBAT
(d) 3
408 1 π cm3
3

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Matematik  Tingkatan 2  Jawapan 

tis PT3 4. (a) (i) 3 (ii) 28 unit2

7Prak Koordinat (b) (10, –8)
PrakCoordinates
(c) (8, 6), (–8, 6), (–8, –6), (8, –6)

A KBAT

2. A B
7. C
1. B 3. C 4. D 5. B tis PT3
6. B 8. B 9. B 10. C
8
B Graf Fungsi
Graph of Functions
1. Pasangan titik
Pair of points Jarak (Unit) 1. C A 3. A 4. C 5. D
A dan /  and B Distance (Units) 6. B 8. C 9. B 10. C
2. B
A dan /  and C 3 7. D
4
C dan /  and D 5 B Hubungan satu
6 kepada satu
B dan /  and E 7 1. One-to-one relation
8
9 2. (a) BENAR / TRUE Hubungan satu
10 (b) PALSU / FALSE kepada banyak
(c) BENAR / TRUE One-to-many relation
2. (a) 3 (b) 3 (c) 3 (d) 3 (d) PALSU / FALSE
Hubungan banyak
3. AB = [6 – ( 0 )]2 – [4 – ( 4 )]2 (–2, 4) 3. (a) 2, 3, 4, 5, 6 kepada satu
(8, 9) (b) 10, 12 Many-to-one relation
= 62 – 0 2
= 6 unit 4. (a) 7 (b) 3 Hubungan banyak
kepada banyak
4. Many-to-many relation
(0, 0)
(2, 1)

(3, 9) (3, –1)

C (c) {10, 12}
(d) {2, 3, 4, 5, 6}
1. (a) Q(10, –4)
(b) (12, 5), (12, –5), (–12, 5) dan / and (–12, –5) (c) 3 (d) ✗

(c) (i) S (ii) 37.5 unit2 C

  2. (a) (i) E 6, 7 (ii) 3 unit / 3 units 1. (a) a = 3, b = 16
2
(b) (i) 5.39 unit / 5.39 units
(b) (i) x –3
(ii) –1 –2 –1 0 12
(c) P(5, 14) 6 2 0 02
3. (a) (i) E
(b) A(2, 9) (ii) (–6, –1) y 12 y
x
(c) (i) Q
(ii) P dan Q / P dan R / Q dan R / S dan Q. (ii)

Kerana titik tengah antara titik-titik tersebut –3

berada di dalam bulatan. Bangku tersebut tidak –2 0
–1 2
boleh berada di dalam kolam. 06
P and Q / P and R / Q and R / S and Q. 1 12
Because the midpoints between these points 2
are located in the circle. The bench cannot be
placed in the pool.
(iii) SR = 12.17 unit  / units
PS = 12 unit  / units Jenis hubungan: Fungsi banyak kepada satu
Type of relation: Many-to-one function

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  Matematik  Tingkatan 2  Jawapan

(c) (i) Satu kepada banyak 4. (a) (i) 56 m
One-to-many (ii) 53 m, 9 saat / seconds
(iii) Tidak, kerana ketinggian maksimum ialah 56 m
(ii) Kerana objek dalam domain mempunyai lebih (53 , 56).
dari satu imej. No, because the maximum height is 56 m
(53 , 56).
Because the objects in the domain have more
than one image.

2. (a) 80 (b) 12 (b) (i) p = 20, q = 12
(ii)
(c) (i) x –2 –1 0 1 2 3 y

y –12 2 10 12 8 –2 50

(ii) y 40

10 30

5 20

–2 –1 O x 10 x
–5 1 23 1 23
–4 –3 –2 –1
–10

–10 –20

3. (a) (i) Banyak kepada satu  / Many-to-one –30

(ii) Fungsi kerana setiap objek hanya mepunyai –40

satu imej. 5. (a) (i) Keuntungan jualan, K
Function because each object only has one The profit, K
(ii) Bilangan burger dijual, x
image. The number of burgers sold, x
(iii) K = 2.2x
(b) (i) 16 (ii) –2, 2

(iii) y

16 (b) (i) RM28
(ii) Kereta syarikat B kerana apabila jarak

1 pemanduan melebihi 30 km, bayaran sewa
kereta syarikat B lebih murah berbanding
syarikat A.
Car from company B because when the
travelled distance is more than 30 km, the rent
payment the car from company B is cheaper
than company A.
x (iii) Syarikat A / Company A, RM55
–2 –1 0 1 2 Syarikat B / Company B, RM40

(c) 10:30, 6 m
(c)
y
KBAT
–2 –1 0 1234567 x
–2 Panjang = 80 cm, lebar = 80 cm
–4 Length = 80 cm, width = 80 cm

–6 Prak 9tis PT3 Laju dan Pecutan
–8 Speed and Acceleration
–10
–12 1. C A 3. A 4. A 5. D
6. A 8. D 9. A 10. A
–14 2. C
–16 7. B



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Matematik  Tingkatan 2  Jawapan 

B 4. (a) (i) 380 saat / 380 seconds

1. (ii) 47 26 m/s
m/s 27
(b) 3
m/minit km/j m2/minit 4
m/minute km/h m2/minute mm/s2 5. (a) (i) 14 9 m/s
cm/s2
km/j2 km/s (ii) 0.05 km/j per saat / 0.05 km/h per second
km/h2 (b) Masa yang diambil / Time taken:
Bas / Bus A = 4.4 jam / hours
2. Bas / Bus B = 3.75 jam / hours
Bas / Bus C = 3.58 jam / hours
• Bas / Bus D = 3.02 jam / hours
Bas D mengambil masa paling singkat untuk tiba di

• • Laju seragam destinasinya.
Uniform speed Bus D takes the shortest time to reach its destination.

• • Laju tak seragam KBAT
Non-uniform speed
36 km/j / 36 km/h

• 10tis PT3

3. (a) 3 Prak Kecerunan Garis Lurus
(b) 7 Gradient of a Straight Line
(c) 7
(d) 3 1. D A 3. C 4. A 5. A
6. D 8. B 9. A 10. D
4. (a) 23 m/s 2. C
(b) 25 m/s 7. A
(c) 10 saat / seconds
(d) 0.2 m/s B

C 1. Jarak Jarak
mencancang mengufuk
1. (a) (i) Jarak yang dilalui dalam masa 1 saat ialah Vertical Horizontal
5 meter. distance distance

The distance travelled in 1 second is 5 metres. Rajah / Diagram (i) 3m 4m
(ii) Jarak dan masa
Distance and time Rajah / Diagram (ii) 2 unit / units 7 unit / units
(b) (i) 33 km/j / 33 km/h
(ii) (a) 17.5 km 2. Positif
(b) 60 km/j / 60 km/h Positive
(c) Nyahpecutan = 25 cm/s per saat
Deceleration = 25 cm/s per second Negatif
Kelajuan objek itu berkurang sebanyak 25 cm/s Negative

setiap saat. Sifar
The speed of the object decreases 25 cm/s for Zero

every second.

2. (a) Laju Julius / Speed of Julius = 10 m/s Tak tertakrif
Laju Hashim / Speed of Hashim = 10.17 m/s Undefined
Laju Chong / Speed of Chong = 10 m/s
Laju Bala / Speed of Bala = 9.6 m/s 3. (a) (i) AB  3
Hashim adalah yang paling pantas. / Hashim is the (ii) EF 3
fastest.

(b) 80 km/j2 / 80 km/h2

(c) 6 8 m/s
77
(b) (i) Tak tertakrif / Undefined
3. (a) –1.6 m/s2 (ii) Sifar / Zero
(b) 40 cm/s
(c) (i) Pecutan = 2 km/j per saat 4. (a) – b (b) – q
Acceleration = 2 km/h per second a p
(ii) Nyahpecutan = 8 km/j per saat
Deceleration = 8 km/h per second (c) –12 (d) 10

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  Matematik  Tingkatan 2  Jawapan

C Prak tis PT3 Transformasi Isometri
Isometric Transformations
1. (a) Garisan PQ mempunyai kecondongan yang lebih 11
tinggi berbanding dengan garisan ST. Garisan PQ
lebih curam berbanding dengan garisan ST. 1. A A 3. A 4. C 5. C
6. A 8. B 9. C 10. B
Kesimpulan: Semakin besar nilai sudut, semakin 11. D 2. B
tinggi nilai kecerunan. 7. B

Line PQ has higher inclination than line ST. Line
PQ is steeper than line ST.

Conclusion: The greater the value of angle, the
higher the value of gradient.

(b) (i) 5 B (ii) 5
(ii) (25, 0)
1. (a) (i) 4
(c) (i) 110 000 m (b)
(ii) 80. Kecerunan graf jarak-masa ialah kadar

perubahan jarak terhadap masa, iaitu kelajuan.
Maka, kecerunan garis OP menunjukkan kereta 3 3
tersebut bergerak pada kelajuan tetap 80 km/j.
80. The gradient of distance-time graph is the 2. Transformasi Saiz Orientasi
rate of change of distance with respect to time. Transformation Size Orientation
Hence, the gradient of line OP shows the car is Pantulan Ya
moving at a constant speed of 80 km/h. Reflection Yes Tidak
Putaran Ya No
2. (a) (i) 4 Rotation Yes Ya
(ii) 2.4 Yes

(b) (i) 5 saat / 5 seconds

(ii) –3 m/s2 1
=2
(c) Kecerunan AB = 1, Kecerunan AC tinggi daripada
Nilai kecerunan garis AB lebih
garis AC. Maka, garis AB lebih curam berbanding
dengan garis AC. 3. (a) P : Putaran 90° mengikut arah jam. 3
1 90° clockwise rotation. 3
Gradient of AB = 1, Gradient of AC = 2
Q :  Pantulan.
The gradient of line AB is higher than line AC. Reflection.
Hence, line AB is steeper than line AC.

3. (a) (0, 2) (b) (i) 7
(b) (i) 1 (ii) 3
(ii) 2.83 unit / units
4. (a), (b)
(c) Pintasan-x = –4 / x-intercept = –4
Pintasan-y = 9 / y-intercept = 9 BC
(d) –15

4. (a) (i) 0 Q
(ii) –1 A OD
(iii) (–10, 0)
(b) Kecerunan / Gradient = 2, P
Jarak MN = 6.708 unit
Distance of MN = 6.708 units
(c) 36

5. (a) pintasan-y / y-intercept = 10 FE

m = 2.5 / 2 1 / 5 C
2 2

(b) (2, 3) 1. (a) (i) Objek P bergerak 3 unit ke kiri dan 2 unit ke

(c) (i) Q(–12, 6) atas.
(ii) Kecerunan SQ = – 1 / Gradient of SQ = – 1 Object P moves 3 units to the left and 2 units
upwards.
44

KBAT   (ii) –23

–8

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Matematik  Tingkatan 2  Jawapan 

(b) (i) y (b) J
S 4

2 H M K
L R
Z x
–2 O 24 JЈ KЈ
G –2
T LЈ
MЈ
(ii) m = –1
(iii) (1, 4) (c) 20°
2. (a) (i) P: Putaran / Rotation
Q: Translasi / Translation KBAT

(ii) Saiz imej dan objek adalah sama. 70°Prak Sukatan Kecenderungan Memusat
Size of image and object are the same. Measures of Central Tendencies
12tis PT3
Orientasi imej dan objek adalah sama.
Orientation of image and object are the same. A

(b) 25° 2. D
7. B
(c) (i) (1, 4)
(ii) (–7, –3)

3. (a)

1. D 3. D 4. A 5. C
6. B 8. D 9. C 10. C

B

1. (a) Min / Mean 44

Peringkat simetri putaran: 3 Mod / Mode 52
Order of rotational symmetry: 3 Median / Median 54
56
(b) y
4P
(b) (i) D

AЈ 2 x 2. (a) (i) 3 (ii) 7
(b) (i) Benar  / True (ii) Palsu  / False
–4 –2 O
2 4 3. (a) Median / Median
(b) Min / Mean
–2 A (c) Mod / Mode
(d) Median / Median
–4

(c) (i) T C
(ii) 180°
1. (a) 53.03 kg
4. (a) P (b) (i) RM38.50
(ii) (a) RM38.50
Q (b) Tidak, kerana harga beg tersebut selepas
potongan adalah lebih kecil daripada nilai
median (RM19 < RM38.50).
No, because the price of the bag after
discount is smaller than the value of the
median (RM19 < RM38.50).
(c) Sukatan kecenderungan memusat yang sesuai
ialah min kerana min mengambil kira markah
keseluruhan bagi keempat-empat kriteria dan boleh
digunakan untuk menentukan pemenang. Mod dan
median tidak memberi maklumat bererti tentang
pemenang pertandingan itu.

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  Matematik  Tingkatan 2  Jawapan 5. (a) 6
The appropriate measure of central tendency is (b) 3
mean because mean takes the marks for all criteria (c) (i) p = 12, q = 68, r = 72, s = 79
and can be used to determine the winner. Mode (ii) Agus yang layak mewakili sekolah kerana julat
and median cannot give any significant meaning markahnya adalah kecil berbanding Sawing.
about the winner of the competition. Ini bermakna, Agus adalah lebih konsisten
berbanding Sawing. Median dan min bagi Roy
2. (a) (i) Bilangan buku Kekerapan Titik tengah adalah yang paling rendah berbanding Agus
Number of books Frequency Midpoint dan Sawing.
Agus is qualified to represent the school
5–9 97 because the range of his mark is smaller
than Sawing. This means that, Agus is more
10 – 14 11 12 consistent than Sawing. Median and mean
15 – 19 x 17 for Roy is the lowest compared to Agus and
Sawing.
20 – 24 2 22
KBAT
(ii) 8
(b) (i) Bagi gerai A, sukatan kecenderungan memusat Pasukan B / Team B

yang sesuai ialah median kerana terdapat nilai Prak13tis PT3 Kebarangkalian Mudah
ekstrem, iaitu 45 dalam data tersebut. Simple Probability
Median = 80
Bagi gerai B, sukatan kecenderungan memusat 1. D A 3. B 4. A 5. C
yang sesuai ialah min kerana taburan markah 6. C 8. C 9. A 10. D
adalah sekata. Min = 79 2. B
For stall A, the appropriate central of tendency 7. B
is median because there is an extreme value,
which is 45 in the data. Median = 80 B
For stall B, the appropriate central of tendency
is min because the distribution of the marks is 1. (a) 6, 7, 8, 9 3
uniform. Mean = 79
(ii) Julat gerai A / Range of stall A = 42 (b) (i) 3 (ii) 3 (iii) 7
Julat gerai B / Range of stall B = 12
Gerai B kerana secara keseluruhan, markah 2.
bagi gerai B adalah lebih tinggi (min B > RM200
min A) dan lebih konsisten (julat B < julat A)
berbanding gerai A.
Stall B because as overall, the mark for stall
B is higher (mean B > mean A) and more
consistent (range B < range A) than stall A.

(c) (i) Median = 75 Pulau Langkawi RM300
(ii) Median tidak berubah. Langkawi Island RM500
Median would not change.

3. (a) Min. Min = 7.6% RM200
Mean. Mean = 7.6%
(b) Mod / Mode = 5 Pulau Tioman
Median = 3 Tioman Island
Min / Mean = 3.5
(c) (i)

Perbelanjaan (RM) Gundalan Bilangan keluarga RM300
Expenses (RM)
Tally Number of families

100 – 149 |||| 4 RM500

150 – 199 |||| || 7 S = {(Pulau Langkawi / Langkawi Island, RM200),
(Pulau Langkawi / Langkawi Island, RM300),
200 – 249 |||| |||| |||| ||| 18 (Pulau Langkawi / Langkawi Island, RM500),
(Pulau Tioman / Tioman Island, RM200),
250 – 299 |||| |||| |||| |||| 19 (Pulau Tioman / Tioman Island, RM300),
(Pulau Tioman / Tioman Island, RM500)}
300 – 349 |||| |||| || 12
3. (a) (i) Tidak mungkin / Not possible
(ii) RM247.83 (ii) Mungkin /  Possible

4. (a) 0 (b) S = {27, 29 , 31 , 37, 41, 43 ,  47, 49}
(b) (i) Mod = Rabu
Mode = Wednesday P(A) = 8
(ii) Min / Mean = RM54 23
(c) (i) Melayu / Malay
(ii) 180 orang / people

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Matematik  Tingkatan 2  Jawapan 

4. 5. (a) 1
Kesudahan putaran ialah (4, 5). 3
The outcome of rotation is (4, 5). (b) (i) 1 (ii) 3
44
Kesudahan putaran ialah (5, 6). Mungkin
The outcome of rotation is (5, 6). Possible (c) 805
Tidak mungkin (d) Kotak G / Box G
Terdapat 9 kesudahan yang Impossible
mungkin bagi putaran itu. 6. (a) {E, A, I }
There are 9 possibility outcomes Pasti
for the rotation. Sure (b) (i) 52
55
Ruang sampel bagi hasil tambah
dua nombor yang diperolehi ialah (ii) RM1 248
{8, 9, 10, 11, 12, 13, 14}.
The sample space of the sum of (c) (i) 0.35
both numbers obtained is (ii) 0.3
{8, 9, 10, 11, 12, 13, 14}. (iii) Apabila bilangan percubaan cukup besar,

C kebarangkalian eksperimen mendapat butang
hitam akan menghampiri kebarangkalian teori.
Pilihan When the number of trials is large enough,
pertama the experimental probability of getting a black
First picked button will be converging to the theoretical
probability.

1. (a) Pilihan kedua KBAT
Second
picked Keudahan (i) 50
C Outcomes (ii) 12

CC MARKAH

C A CA PPeenniillaaiiaann PPrraa--PPTT33

T CT A

C AC 1. B 2. C 3. C 4. D 5. A
A A AA 6. D 7. C 8. C 9. B 10. A
11. D 12. D 13. A 14. B 15. D
T AT 16. C 17. C 18. D 19. B 20. D

C TC B (ii) Perentas / Chord
T A TA (iv) Tembereng / Segment
21. (i) Diameter
T TT (iii) Sektor / Sector

22.

(b) (i) {(C, A), (A, C)} (ii) {(C, C), (A, A), (T, T)}

(c) (i) 1 (ii) 12 biji / marbles
5

2. (a) (i) M = {1, 2} (ii) N = {1, 2, 3, 6}
(iii) P = {2, 3, 5}

(b) (i) 2 (ii) 52
3
(c) 1 batang / pen
23. (x + y)2  ÷  (x + y)3
3. (a) 15010 2xy2 2x2 – 4xy
(b) (i) R =
{2, 3, 5, 7, 11, 13, 17, 19}, P(R) = 2 = (x + y)2 × 2x2 – 4xy
(ii) G = {1, 3, 5, 7, 5 2xy2 (x + y)3

9, 11, 13, 15, 17, 19}, P(G) = 1 (x + y)2 × 2x(x – 2y)
2 2xy2 (x + y)3
(c) 10 =

4. (a) 5 (b) (i) 3 (ii) 170 = x – 2y
26 10 y2(x +y)
1
(c) 6

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  Matematik  Tingkatan 2  Jawapan

24. (a) (i) AB (ii) CD (iii) BC 29. (a) –2
(b) 0 m/min2 (b)

25. (a) melengkung / curved 30°
(b) segi empat tepat; bulatan 3 cm
rectangle; circles
(c) kubus / cube

C

26. (a) (i) m = 2(k + 1) + n (c) 79.2 l
(ii) –10
(b) (i) Pentagon 30. (a) (6, 0)
(ii) 66° (b) 124°
(c) (i) 55 km/j / 55 km/h (c) x = 3, y = 5
(ii) 90 km/j / 90 km/h
31. (a) A: Putaran / Rotation
     27. (a) (i) Translasi7 7 B: Pantulan / Reflection
– 2 / Translation – 2 C: Translasi / Translation

(ii) T

(b) P

M

S M'
Q
(b) 35 9x(5x – 4)
(c) (i) (c) Min  / Mean = 75
(ii) 13.83 cm Kumpulan P kerana kumpulan P lebih konsisten

28. (a) (p – 5) m2 berbanding Kumpulan Q (Julat P , Julat Q).
Manakala min bagi kedua-dua kumpulan adalah
(b) (i) 0123 sama iaitu 75. Mod dan median tidak memberi
x – 4 –3 –2 –1 – 4 –14 –18 –10 maklumat bererti tentang pertandingan itu.
y –24 2 10 6 Group P because Group P is more consistent
y compared to Group Q (Range P , Range Q).
10 Whereas the mean of both groups is the same that
is 75. Mode and median cannot give any significant
meaning about the competition.

(ii)

5 1 2 3x

–4 –3 –2 –1 O
–5

–10

–15

–20

–25

(c) 184 cm2

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