Matematik (Book 2) Tingkatan 3

MODULPENTAKSIRAN UASA

Proaktif KSSM

Matematik
Dwibahasa
2Buku
Ujian Akhir
Sesi Akademik

Bonus untuk
Guru
(Bab 2, 4, 6, 8)

Praktis DSKP

Tahap Penguasaan (TP)

Standard Pembelajaran (SP) e-RPH

Soalan STEM 3Tingkatan

Ujian Akhir Sesi Akademik

(UASA)

0 Soalan KBAT
90
Sisipan PAK-21

Yong Sze Wei・Ong Jia Sheng

Pentaksiran Mata Pelajaran Matematik Tingkatan 3 KSSM iii - vi

Bab 1 - 13
1
2 Bentuk Piawai 13
Standard Form
Praktis DSKP
Praktis Formatif 1

CONTOHBab 14 - 29
14
4 Lukisan Berskala 28
Scale Drawings
Praktis DSKP 30 - 45
Praktis Formatif 2 30
44
Bab 36

6 Sudut dan Tangen bagi Bulatan 46 - 62
Angles and Tangents of Circles 46
Praktis DSKP 59
Praktis Formatif 3

Bab

8 Lokus dalam Dua Dimensi
Loci in Two Dimension
Praktis DSKP
Praktis Formatif 4

Ujian Akhir Sesi Akademik (UASA) 63 - 78

Jawapan J1 - J12

PENTAKSIRAN BAHARU Ujian Akhir Jawapan
(Set Pentaksiran ini akan dimuat naik Sesi Akademik
ke dalam QR sebaik-baik sahaja KPM
mengeluarkan format baharu)

ii

Nama: Tarikh:

Bab Bentuk Piawai

2 Standard Form

Praktis D S K P

PBD 1 Angka Bererti / Significant Figures Hal. Buku Teks: 32-36

A Tentukan bilangan angka bererti bagi nombor-nombor tersebut.
State the number of significant figures for the following numbers. SP 2.1.1 Menguasai TP 1
CONTOH
Contoh / Example: 1. 18.6 2. 397
1 058

Jawapan / Answer: 4 Jawapan / Answer: 3 Jawapan / Answer: 3

3. 4.00 4. 25 000 5. 0.00080

Jawapan / Answer: 3 Jawapan / Answer: 2 Jawapan / Answer: 2

PAK-21

B Bundarkan setiap nombor yang berikut kepada bilangan angka bererti yang
diberikan. / Round off each of the following numbers to the given significant figures.

SP 2.1.2 Menguasai TP 2

Contoh / Example: 1. 361 (1 a.b. / 1 s.f.) 2. 6 184 (2 a.b. / 2 s.f.)
295 (2 a.b. / 2 s.f.)

295 = 300 361 = 400 6 184 = 6 200

3. 762 118 (3 a.b. / 3 s.f.) 4. 8 143 (2 a.b. / 2 s.f.) 5. 57 309 (2 a.b. / 2 s.f.)
762 118 = 762 000 8 143 = 8 100 57 309 = 57 000

6. 294 (2 a.b. / 2 s.f.) 7. 4 102 (3 a.b. / 3 s.f.) 8. 571 548 (3 a.b. / 3 s.f.)
294 = 290 4 102 = 4 100 571 548 = 572 000

TP1 Mempamerkan pengetahuan asas tentang angka bererti dan bentuk piawai. 1

C Bundarkan setiap nombor yang berikut kepada bilangan angka bererti yang diberikan.
Round off each of the following numbers to the given significant figures.

SP 2.1.2 Menguasai TP 2

Contoh / Example: 1. 7.99 (1 a.b. / 1 s.f.) 2. 3.502 (3 a.b. / 3 s.f.)
6.59 (2 a.b. / 2 s.f.) 7.99 = 8 3.502 = 3.50

6.59 = 6.6

3. 12.103 (1 a.b. / 1 s.f.) 4. 11.588 (3 a.b. / 3 s.f.) 5. 11.000 (3 a.b. / 3 s.f.)
12.103 = 10 11.588 = 11.6 11.000 = 11.0

CONTOH
D Bundarkan setiap nombor yang berikut kepada bilangan angka bererti yang diberikan.
Round off each of the following numbers to the given significant figures.

SP 2.1.2 Menguasai TP 2

Contoh / Example: 1. 0.03334 (2 a.b. / 2 s.f.) 2. 0.2298 (3 a.b. / 3 s.f.)
0.00263 (1 a.b. / 1 s.f.)

0.00263 = 0.003 0.03334 = 0.033 0.2298 = 0.230

3. 0.02152 (3 a.b. / 3 s.f.) 4. 0.15391 (1 a.b. / 1 s.f.) 5. 0.3319 (4 a.b. / 4 s.f.)
0.02152 = 0.0215 0.15391 = 0.2 0.3319 = 0.3319

E Hitung setiap yang berikut dan bundarkan jawapan betul kepada 1 angka bererti.
Calculate each of the following and round off the answer correct to 1 significant figure.

SP 2.1.2 Menguasai TP 3

Contoh / Example: 1. 123 – 456 + 719 2. 4.25 ÷ 2.3 + 0.72
65 – 3.25 × 6
= –333 + 719 = 1.848 + 0.72
= 65 – 19.5 = 386 = 2.568
= 45.5 = 400 =3
= 50

2 TP2 Mempamerkan kefahaman tentang angka bererti dan bentuk piawai.

3. 45 – 1.143 4. 3.2 – 1.2 5. 2.2 × 2.9
67 2.23 + 1.2 7.6

= 0.6716 – 1.143 =2 = 2.2 × 0.3816
= –0.4714 3.43 = 0.8395
= –0.5 = 0.8
= 0.5831
= 0.6

F Hitung setiap yang berikut dan bundarkan jawapan anda betul kepada 2 angka bererti.
Calculate each of the following and round off the answer correct to 2 significant figures.

SP 2.1.2 Menguasai TP 3

Contoh / Example:CONTOH 1. 4.56 × 5.2 ÷ 103 2. (2.56 + 7.54) ÷ 56
3 366 – 210 + 487
= 23.712 ÷ 103 = 10.1 ÷ 56
= 3 156 + 487 = 0.2302 = 0.1804
= 3 643 = 0.23 = 0.18
= 3 600

3. 8.59 ÷ 2.1 – 2.25 4. 15.23 – 3.36 × 1.3 5. 10.25 ÷ 0.75 – 4.2 × 0.5

= 4.0905 – 2.25 = 15.23 – 4.368 = 13.6667 – 2.1
= 1.8405 = 10.862 = 11.5667
= 1.8 = 11 = 12

6. 36 × 24 + 15 ÷ 0.31 7. 7.21 – 2.3 × 1.4 + 6.62 ( )8.  × 6
0.521 + 4.3
= 864 + 48.39 = 7.21 – 3.22 + 6.62 1.34
= 912.39 = 10.61
= 910 = 11 = 4.821 × 6
1.34

= 3.5978 × 6
= 21.5868
= 22

TP3 Mengaplikasikan kefahaman tentang angka bererti dan bentuk piawai untuk melaksanakan tugasan mudah. 3

PBD 2 Bentuk Piawai / Standard Form Hal. Buku Teks: 37-44

A Nyatakan setiap nombor yang berikut dalam bentuk piawai.
State each of the following numbers in standard form. SP 2.2.1 Menguasai TP 1

Contoh / Example: 1. 7 560 2. 32
1 234

= 1.234 × 103 = 7.56 × 103 = 3.2 × 10

3. 15.3 4. 536.21 5. 14 500 000
= 1.53 × 10 = 5.3621 × 102 = 1.45 × 107
CONTOH
B Nyatakan setiap nombor yang berikut dalam nombor tunggal.
State each of the following numbers in single number. SP 2.2.1 Menguasai TP 1

Contoh / Example: 1. 3.5 × 10 2. 4.15 × 105
2.1 × 103

= 2 100 = 35 = 415 000

3. 0.00322 × 107 4. 0.032 × 105 5. 0.6654 × 103
= 32 200 = 3 200 = 665.4

C Nyatakan setiap nombor yang berikut dalam bentuk piawai.
State each of the following numbers in standard form. SP 2.2.1 Menguasai TP 1

Contoh / Example: 1. 0.00062 2. 0.00000315

0.0012

= 1.2 × 10–3 = 6.2 × 10–4 = 3.15 × 10–6

4 TP1 Mempamerkan pengetahuan asas tentang angka bererti dan bentuk piawai.

3. 0.00000002697 4. 0.00048 5. 0.000213
= 2.697 × 10–8 = 4.8 × 10–4 = 2.13 × 10–4

D Nyatakan setiap nombor yang berikut dalam nombor tunggal.
State each of the following numbers in single number. SP 2.2.1 Menguasai TP 1

Contoh / Example: 1. 1.25 × 10–2 2. 9.98 × 10–4
5.4 × 10–5 = 0.0125 = 0.000998

= 0.000054
CONTOH
3. 5.78 × 10–7 4. 7.3 × 10–6 5. 3.54 × 10–4
= 0.000000578 = 0.0000073 = 0.000354

E Hitung setiap yang berikut dan berikan jawapan anda dalam bentuk piawai.

Calculate each of the following and give your answer in standard form.

SP 2.2.2 Menguasai TP 3

Contoh / Example: 1. 7.2 × 103 + 6.3 × 103
2.3 × 106 + 5.3 × 106

= (2.3 + 5.3) × 106 = (7.2 + 6.3) × 103
= 7.6 × 106 = 13.5 × 103
= 1.35 × 104

2. 8.9 × 105 + 4.2 × 105 3. 6.2 × 10–5 + 7.9 × 10–5

= (8.9 + 4.2) × 105 = (6.2 + 7.9) × 10–5
= 13.1 × 105 = 14.1 × 10–5
= 1.31 × 106 = 1.41 × 10–4

TP1 Mempamerkan pengetahuan asas tentang angka bererti dan bentuk piawai. 5
TP3 Mengaplikasikan kefahaman tentang angka bererti dan bentuk piawai untuk melaksanakan tugasan mudah.

F Hitung setiap yang berikut dan berikan jawapan anda dalam bentuk piawai.

Calculate each of the following and give your answer in standard form.

SP 2.2.2 Menguasai TP 3

Contoh / Example: 1. 8.4 × 106 – 4.6 × 106
3.5 × 107 – 2.1 × 107

= (3.5 – 2.1) × 107 = (8.4 – 4.6) × 106
= 1.4 × 107 = 3.8 × 106

2. 7.2 × 104 – 5.8 × 104 3. 9.1 × 102 – 1.5 × 102

= (7.2 – 5.8) × 104 = (9.1 – 1.5) × 102
= 1.4 × 104 = 7.6 × 102
CONTOH
G Hitung setiap yang berikut dan berikan jawapan anda dalam bentuk piawai.

Calculate each of the following and give your answer in standard form.

SP 2.2.2 Menguasai TP 3

Contoh / Example: 1. 5 × 9 × 105
6 × 2 × 102

= 12 × 102 = 45 × 105
= 1.2 × 103 = 4.5 × 106

2. 7 × 8 × 106 3. 12 × 11 × 107

= 56 × 106 = 132 × 107
= 5.6 × 107 = 1.32 × 109

4. 5.4 × 15 × 10–2 5. 3.1 × 24 × 104

= 81 × 10–2 = 74.4 × 104
= 8.1 × 10–1 = 7.44 × 105

6 TP3 Mengaplikasikan kefahaman tentang angka bererti dan bentuk piawai untuk melaksanakan tugasan mudah.

H Hitung setiap yang berikut dan berikan jawapan anda dalam bentuk piawai.

Calculate each of the following and give your answer in standard form.

SP 2.2.2 Menguasai TP 3

Contoh / Example: 1. 4.2 × 108
6
2.1 × 106
7

= 0.3 × 106 = 0.7 × 108
= 3 × 105 = 7 × 107

2. 3.15 × 109 3. 5.12 × 103
9 8

= 0.35 × 109 = 0.64 × 103
= 3.5 × 108 = 6.4 × 102
CONTOH
I Hitung setiap yang berikut dan berikan jawapan anda dalam bentuk piawai.

Calculate each of the following and give your answer in standard form.

SP 2.2.2 Menguasai TP 3

Contoh / Example: 1. 8.1 × 104 + 9.2 × 105
6.3 × 103 + 1.2 × 104

= 0.63 × 104 + 1.2 × 104 = 0.81 × 105 + 9.2 × 105
= (0.63 + 1.2) × 104 = (0.81 + 9.2) × 105
= 1.83 × 104 = 10.01 × 105
= 1.001 × 106

2. 7.5 × 107 + 8.5 × 106 3. 7.4 × 105 + 6.6 × 106

= 7.5 × 107 + 0.85 × 107 = 0.74 × 106 + 6.6 × 106
= (7.5 + 0.85) × 107 = (0.74 + 6.6) × 106
= 8.35 × 107 = 7.34 × 106

TP3 Mengaplikasikan kefahaman tentang angka bererti dan bentuk piawai untuk melaksanakan tugasan mudah. 7

J Hitung setiap yang berikut dan berikan jawapan anda dalam bentuk piawai.

Calculate each of the following and give your answer in standard form.

SP 2.2.2 Menguasai TP 3

Contoh / Example: 1. 6.7 × 105 – 2.8 × 104
5.5 × 103 – 2.3 × 102

= 5.5 × 103 – 0.23 × 103 = 6.7 × 105 – 0.28 × 105
= (5.5 – 0.23) × 103 = (6.7 – 0.28) × 105
= 5.27 × 103 = 6.42 × 105

CONTOH2. 9.3 × 107 – 2.6 × 106 3. 8.5 × 106 – 8.5 × 105

= 9.3 × 107 – 0.26 × 107 = 8.5 × 106 – 0.85 × 106
= (9.3 – 0.26) × 107 = (8.5 – 0.85) × 106
= 9.04 × 107 = 7.65 × 106

K Hitung setiap yang berikut dan berikan jawapan anda dalam bentuk piawai.

Calculate each of the following and give your answer in standard form.

SP 2.2.2 Menguasai TP 3

Contoh / Example: 1. 8.1 × 103 × 1.3 × 105
3.8 × 106 × 2.2 × 104
= 8.1 × 1.3 × 103 + 5
= 3.8 × 2.2 × 106 + 4 = 10.53 × 108
= 8.36 × 1010 = 1.053 × 109

2. 1.5 × 108 × 7.4 × 102 3. 2.7 × 104 × 3.26 × 10–2

= 1.5 × 7.4 × 108 + 2 = 2.7 × 3.26 × 104 – 2
= 11.1 × 1010 = 8.802 × 102
= 1.11 × 1011

8 TP3 Mengaplikasikan kefahaman tentang angka bererti dan bentuk piawai untuk melaksanakan tugasan mudah.

4. 4.1 × 10–3 × 0.0077 5. 3.3 × 106 × 0.00094

= 4.1 × 10–3 × 7.7 × 10–3 = 3.3 × 106 × 9.4 × 10–4
= 4.1 × 7.7 × 10–3 – 3 = 3.3 × 9.4 × 106 – 4
= 31.57 × 10–6 = 31.02 × 102
= 3.157 × 10–5 = 3.102 × 103

L Hitung setiap yang berikut dan berikan jawapan anda dalam bentuk piawai.

Calculate each of the following and give your answer in standard form.

SP 2.2.2 Menguasai TP 3
CONTOH
Contoh / Example: 1. 4.5 × 106 ÷ (3 × 102)
2.4 × 103 ÷ (8 × 107)

= 2.4 ÷ 8 × 103 – 7 = 4.5 ÷ 3 × 106 – 2
= 0.3 × 10–4 = 1.5 × 104
= 3 × 10–5

2. 9 × 10–3 3. 8.4 × 102 ÷ (6 × 10–4)
450 000
9 × 10–3 = 8.4 ÷ 6 × 102 – (–4)
= 4.5 × 105 = 1.4 × 106
= 2 × 10–3 – 5
= 2 × 10–8 5. 2.4 × 1011 ÷ (4 × 10–3)
5 × 104
4. 27 000 ÷ 0.03 ÷ 0.00003
= 2.4 ÷ 4 × 1011 – (–3) – 4
= 2.7 × 104 ÷ (3 × 10–2) ÷ (3 × 10–5) 5
= 2.7 ÷ 3 ÷ 3 × 104 – (–2) – (–5)
= 0.3 × 1011 = 0.12 × 1010
= 3 × 1010 = 1.2 × 109

TP3 Mengaplikasikan kefahaman tentang angka bererti dan bentuk piawai untuk melaksanakan tugasan mudah. 9

M Selesaikan setiap masalah yang berikut dan beri jawapan anda dalam bentuk piawai.

Solve each of the following problems and give your answers in standard form.

SP 2.2.3 Menguasai TP 4 TP 5

Contoh / Example:

Fatimah telah membeli 4.5 kg gula, 5.9 kg tepung dan 2 300 g minyak masak. Hitung jumlah

jisim barang yang dibeli olehnya dalam gram.

Fatimah bought 4.5 kg of sugar, 5.9 kg of flour and 2 300 g of cooking oil. Evaluate the total mass
of things she bought in grams.

Jumlah jisim / Total mass
= 4.5 kg + 5.9 kg + 2 300 g
= 4 500 g + 5 900 g + 2 300 g
= 12 700 g
= 1.27 × 104 g
CONTOH
1. Tinggi Ami ialah 1.69 m dan tinggi abangnya ialah 3.6 cm lebih tinggi daripadanya.
Hitung jumlah tinggi Ami dan abangnya dalam cm.
The height of Ami is 1.69 m and her brother’s height is 3.6 cm more than her. Evaluate the total
height of Ami and her brother in cm.

Jumlah tinggi / Total height
= 169 cm + 169 cm + 3.6 cm

= 341.6 cm
= 3.416 × 102 cm

2. Sebuah segi empat tepat masing-masing dengan panjang dan lebar sebanyak 14 cm
dan 8 cm. Cari luas segi empat tepat tersebut dalam cm2.
A rectangle with length and width 14 cm and 8 cm respectively. Find the area of the rectangle in
cm2.

Luas segi empat tepat / Area of rectangle
= 14 cm × 8 cm

= 112 cm2

= 1.12 × 102 cm2

10

3. Di dalam sebuah kilang, 5 000 kg garam telah sedia diisi ke dalam peket. Setiap peket
telah diisi oleh garam berjisim 50 g. Berapakah peket garam yang boleh diisi?
In a factory, 5 000 kg of salt is ready to be filled into packets. Each packet is filled with 50 g of
salt. How many packets of salt can be filled? KBAT Mengaplikasi

Bilangan peket garam / Number of packet of salt

= 5 000 000 g
50 g

= 100 000 peket / packets
= 1 × 105 peket / packets

CONTOHN Selesaikan masalah berikut.
Solve the following problems. SP 2.2.3 Menguasai TP 6

Contoh / Example:

Rajah di bawah menunjukkan sebuah kuboid. Hitung / Calculate
The diagram below shows a cuboid. (a) jumlah luas permukaan dalam cm2.

2 cm total surface area in cm2.
(b) isi padu dalam cm3.
4 cm
volume in cm3.
Berikan jawapan anda dalam bentuk piawai.

Give your answers in standard form.

12 cm

(a) Jumlah luas permukaan / Total surface area (b) Isi padu / Volume
= [(12 × 4) + (4 × 2) + (12 × 2)] × 2 = 12 × 4 × 2
= 160 cm2 = 96 cm3
= 1.6 × 102 cm2 = 9.6 × 10 cm3

1. Sebuah jag mengandungi 2 300 m jus oren. Jus oren ini telah dibahagi kepada 6 orang.
Kira isi padu jus oren bagi setiap orang dalam mℓ. Berikan jawapan anda dalam 4 angka
bererti.
A jug contains of 2 300 m of orange juice. The orange juice is given to 6 persons. Calculate the
volume of orange juice of each person in mℓ. Give your answer in 4 significant figures.

Isi padu jus oren bagi setiap orang / Volume of orange juice of each person
= 2 300 ÷ 6

= 383.33 mℓ

= 383.3 mℓ

TP4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang bentuk piawai dalam konteks 11
penyelesaian masalah rutin yang mudah.

TP5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang bentuk piawai dalam konteks
penyelesaian masalah rutin yang kompleks.

2. Sebuah kereta bergerak dengan kelajuan 34 ms–1 selama 1 jam 2 minit. Cari jarak,
STEM dalam m, kereta itu bergerak. Berikan jawapan anda dalam bentuk piawai.

A car moved with speed 34 ms–1 for 1 hour 2 minutes. Find the distances, in m, of the car moved.
Give your answer in standard form.

1 jam 2 minit / 1 hour 2 minutes
= 1 × 60 × 60 + 2 × 60

= 3 600 + 120

= 3 720 saat / seconds
∴ Jarak kereta itu bergerak / Distance of the car moved

= 34 × 3 720

= 126 480 m

= 1.2648 × 105 m
CONTOH
3. Sebuah silinder masing-masing dengan jejari dan tinggi sebanyak 2.3 × 103 cm dan

4.5 × 102 cm. Cari

A cylinder with radius and height 2.3 × 103 cm and 4.5 × 102 cm respectively. Find

(a) jumlah luas permukaan dalam cm2.

total surface area in cm2.

(b) isi padu dalam m3.

volume in m3.

Berikan jawapan anda dalam bentuk piawai. ( )Guna π = 22
Give your answers in standard form. / Use 7

(a) Jumlah luas permukaan / Total surface area

= 2 × 22 × (2.3 × 103)2 + 2 × 22 × 2.3 × 103 × 4.5 × 102
77

= 2 × 22 × 2.32 × 106 + 2 × 22 × 2.3 × 4.5 × 105
7 7

= 33.25 × 106 + 65.06 × 105
= 33.25 × 106 + 6.506 × 106
= (33.25 + 6.506) × 106
= 39.756 × 106
= 3.9756 × 107 cm2

(b) Isi padu / Volume = 22 × (2.3 × 103)2 × 4.5 × 102
7

= 22 × 2.32 × 4.5 × 106 + 2
7

= 74.82 × 108
= 7.482 × 109 cm3
= 7.482 × 103 m3

12 TP6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang bentuk piawai dalam konteks
penyelesaian masalah bukan rutin secara kreatif.

Praktis F o r m a t i f 1

Jawab semua soalan.
Answer all questions.

Bahagian A

1. Pilih nombor yang mempunyai bilangan 2. Apakah nilai 8 899 jika ia dibundarkan

angka bererti yang terkecil. kepada 2 angka bererti?

Choose the number with the smallest number What is the value of 8 899 if it is rounded off

of significant figures. to 2 significant figures?

A 0.0120 B 1.200 A 9 000 B 8 900

CONTOHC 120.0 D 1 200 C 8 800 D 8 090

Bahagian C

3. (a) (i) Hitung 1.5 × 104 – 3.6 × 102 dan berikan jawapan anda dalam bentuk piawai.
Calculate 1.5 × 104 – 3.6 × 102 and give your answer in standard form.
[3 markah / marks]

(ii) Hitung 84.9 – 2.3 × 1.5 dan bundarkan jawapan anda betul kepada 1 angka bererti.

Calculate 84.9 – 2.3 × 1.5 and round off your answer correct to 1 significant figure.
[3 markah / marks]

Jawapan / Answer:

(i) 1.5 × 104 – 3.6 × 102 (ii) 84.9 – 2.3 × 1.5 = 84.9 – 3.45
= 1.5 × 104 – 0.036 × 104 = 81.45
= (1.5 – 0.036) × 104 = 80
= 1.464 × 104

(b) Terdapat sebuah bekas yang berbentuk kon dengan jejari 5 cm dan tinggi 8 cm. Diberi

bahawa 65% daripada bekas tersebut diisi dengan air. Hitung isi padu, dalam cm3, air

di dalam bekas itu. Bundarkan jawapan anda betul kepada 3 angka bererti.

There is a cone-shaped container with radius 5 cm and height 8 cm. Given that 65% of the

container is filled with water. Calculate the volume, in cm3, of water in the container. Round
[4 markah / marks]
off your answer correct to 3 significant figures.

( )Guna 22
/ Use π = 7

Jawapan / Answer:

Isi padu air / Volume of water

= 65 × 1 × 22 × 52 × 8 PErkaskttrias
100 3 7

= 136.2 cm3
= 136 cm3

13

Jawapan Jawapan
untuk QR
Bab 2 3. 4.0905 – 2.25
J1
Praktis D S K P = 1.8405

PBD 1 = 1.8

A 1. 3 4. 15.23 – 4.368
2. 3
3. 3 = 10.862
4. 2
5. 2 = 11

B 1. 400 5. 13.6667 – 2.1
2. 6 200
3. 762 000 = 11.5667
4. 8 100
5. 57 000 = 12
6. 290
7. 4 100 6. 864 + 48.39
8. 572 000
= 912.39
C 1. 8CONTOH
2. 3.50 = 910
3. 10
4. 11.6 7. 7.21 – 3.22 + 6.62
5. 11.0
= 10.61
D 1. 0.033
2. 0.230 = 11
3. 0.0215
4. 0.2 8. 4.821 × 6
5. 0.3319 1.34

E 1. –333 + 719 = 3.5978 × 6
= 386
= 400 = 21.5868

2. 1.848 + 0.72 = 22
= 2.568
=3 PBD 2

3. 0.6716 – 1.143 A 1. 7.56 × 103
= –0.4714 2. 3.2 × 10
= –0.5 3. 1.53 × 10
4. 5.3621 × 102
4. 2 5. 1.45 × 107
3.43
= 0.5831 B 1. 35
= 0.6 2. 415 000
3. 32 200
5. 2.2 × 0.3816 4. 3 200
= 0.8395 = 0.8 5. 665.4

F 1. 23.712 ÷ 103 C 1. 6.2 × 10–4
= 0.2302 2. 3.15 × 10–6
= 0.23 3. 2.697 × 10–8
4. 4.8 × 10–4
2. 10.1 ÷ 56 5. 2.13 × 10–4
= 0.1804
= 0.18 D 1. 0.0125
2. 0.000998
3. 0.000000578
4. 0.0000073
5. 0.000354

E 1. (7.2 + 6.3) × 103
= 13.5 × 103
= 1.35 × 104

2. (8.9 + 4.2) × 105
= 13.1 × 105
= 1.31 × 106

3. (6.2 + 7.9) × 10–5 4. 4.1 × 10–3 × 7.7 × 10–3
= 14.1 × 10–5 = 4.1 × 7.7 × 10–3 – 3
= 1.41 × 10–4 = 31.57 × 10–6
= 3.157 × 10–5
F 1. (8.4 – 4.6) × 106
= 3.8 × 106 5. 3.3 × 106 × 9.4 × 10–4
= 3.3 × 9.4 × 106 – 4
2. (7.2 – 5.8) × 104 = 31.02 × 102
= 1.4 × 104 = 3.102 × 103

3. (9.1 – 1.5) × 102 L 1. 4.5 ÷ 3 × 106 – 2 = 1.5 × 104
= 7.6 × 102
2. 9 × 10–3 = 2 × 10–3 – 5
G 1. 45 × 105 4.5 × 105
= 4.5 × 106 = 2 × 10–8

2. 56 × 106 3. 8.4 ÷ 6 × 102 – (–4) = 1.4 × 106
= 5.6 × 107 4. 2.7 × 104 ÷ (3 × 10–2) ÷ (3 × 10–5)

3. 132 × 107 = 2.7 ÷ 3 ÷ 3 × 104 – (–2) – (–5)
= 1.32 × 109 = 0.3 × 1011
= 3 × 1010
4. 81 × 10–2CONTOH
= 8.1 × 10–1 5. 2.4 ÷ 4 × 1011 – (–3) – 4 = 0.12 × 1010
5 = 1.2 × 109
5. 74.4 × 104
= 7.44 × 105 M 1. Jumlah tinggi / Total height
= 169 cm + 169 cm + 3.6 cm
H 1. 0.7 × 108
= 7 × 107 = 341.6 cm

2. 0.35 × 109 2. Luas segi empat tepat / Area of rectangle
= 3.5 × 108 = 14 cm × 8 cm

3. 0.64 × 103 = 112 cm2
= 6.4 × 102
3. Bilangan peket garam
I 1. 0.81 × 105 + 9.2 × 105
= (0.81 + 9.2) × 105 Number of packet of salt
= 10.01 × 105 = 5 000 000 g
= 1.001 × 106
50 g
2. 7.5 × 107 + 0.85 × 107
= (7.5 + 0.85) × 107 = 100 000 peket / packets
= 8.35 × 107 = 1 × 105 peket / packets

3. 0.74 × 106 + 6.6 × 106 N 1. Isi padu jus oren bagi setiap orang
= (0.74 + 6.6) × 106
= 7.34 × 106 Volume of orange juice of each person

J 1. 6.7 × 105 – 0.28 × 105 = 2 300 ÷ 6
= (6.7 – 0.28) × 105
= 6.42 × 105 = 383.33 mℓ

2. 9.3 × 107 – 0.26 × 107 = 383.3 mℓ
= (9.3 – 0.26) × 107
= 9.04 × 107 2. 1 jam 2 minit / 1 hour 2 minutes
= 1 × 60 × 60 + 2 × 60
3. 8.5 × 106 – 0.85 × 106
= (8.5 – 0.85) × 106 = 3 720 saat / seconds
= 7.65 × 106 ∴ Jarak kereta itu bergerak

K 1. 8.1 × 1.3 × 103 + 5 Distance of the car moved
= 10.53 × 108 = 34 × 3 720
= 1.053 × 109
= 126 480 m
2. 1.5 × 7.4 × 108 + 2
= 11.1 × 1010 = 1.2648 × 105 m
= 1.11 × 1011
3. (a) Jumlah luas permukaan
3. 2.7 × 3.26 × 104 – 2
= 8.802 × 102 Total surface area
22
J2 = 2 × 7 × (2.3 × 103)2 +

2 × 22 × 2.3 × 103 × 4.5 × 102
7

= 2 × 22 × 2.32 × 106 +
7

2 × 22 × 2.3 × 4.5 × 105 3. 1.1 = 1
7 2.2 2

= 33.25 × 106 + 65.06 × 105 Skala yang digunakan ialah 1 : 2
= 33.25 × 106 + 6.506 × 106 The scale used is 1 : 2
= (33.25 + 6.506) × 106
= 39.756 × 106 D 1. 5 = 1
= 3.9756 × 107 cm2 400 80

(b) Isi padu / Volume Skala yang digunakan ialah 1 : 80
22 The scale used is 1 : 80
= 7 × (2.3 × 103)2 × 4.5 × 102

= 22 × 2.32 × 4.5 × 106 + 2 2. 100 = 8
7 12.5

= 74.82 × 108 Skala yang digunakan ialah 1 : 1
= 7.482 × 109 cm3 8
= 7.482 × 103 m3 1
The scale used is 1 : 8
CONTOH
Praktis Formatif 1 3. 3 = 1
600 200
Bahagian A
1. D Skala yang digunakan ialah 1 : 200
2. B The scale used is 1 : 200

Bahagian C E 1
1. Skala / Scale = 500
3. (a) (i) 1.5 × 104 – 3.6 × 102
2
= 1.5 × 104 – 0.036 × 104 =

= (1.5 – 0.036) × 104 Tinggi sebenar / Actual height

= 1.464 × 104 Tinggi sebenar / Actual height
= 2 × 500 ÷ 100
(ii) 84.9 – 2.3 × 1.5 = 84.9 – 3.45 = 10 m

= 81.45

= 80 1
2. Skala / Scale = 250
(b) Isi padu air / Volume of water

= 65 × 1 × 22 × 52 × 8 200
100 3 7 =

= 136.2 cm3 Tinggi sebenar / Actual height

Bab 4 Tinggi sebenar / Actual height
= 200 × 250 ÷ 100
Praktis D S K P = 500 m

PBD 1 1
3. Skala / Scale = 10 000
A 1. I
2. III

B 1. IV 60
2. IV =

Jarak sebenar / Actual distance

C 1. 8 =4 Jarak sebenar / Actual distance
2 = 60 × 10 000 ÷ 100 000
= 6 km
1
Skala yang digunakan ialah 1 : 4 F 1. Skala / Scale

The scale used is 1 : 1 = Jarak pada peta / Distance on the map
4 4 000
75
2. 100 = 3 = 1
4 100

Skala yang digunakan ialah 1 : 1 1 Jarak di atas peta / Distance on the map
3 = 4 000 ÷ 100
1
The scale used is 1 : 1 3 = 40 m

J3

2. Skala = Tinggi pada lukisan berskala (c)
80 000

Scale = Height on the scale drawing
80 000

= 1
2 000
H 1. Luas sebenar / Actual area
Tinggi di atas lukisan berskala = (20 × 1 000 ÷ 100) × (20 × 1 000 ÷ 100)
Height on the scale drawing
= 80 000 ÷ 2 000 = 40 000 m2

= 40 cm 2. Lilitan bulatan / Circumference

Panjang di atas peta = 2 × 22 × (7 × 8)
3. Skala = 7

1 000 000 = 352 cm

Scale = Length on the map 3. Luas segi tiga / Area of triangle
1 000 000 1
CONTOH 1 = 2 × (2 × 5) × (3 × 5)
100 000
=

Panjang di atas peta / Length on the map = 75 cm2

= 1 000 000 ÷ 100 000 4. Panjang sisi / Length of square = √ 36 × 3
= 18 cm
= 10 cm

G 1. (a) I 1. (a) Skala lukisan pelan / Scale of the plan
= 3 cm + 3 cm + 2.5 cm : 34 m

= 8.5 : 3 400

= 1 : 400

(b) (b) Luas sebenar rumah

Actual area of the house
= (8.5 × 400) × (5 × 400) ÷ 10 000
= 3 400 × 2 000 ÷ 10 000
= 680 m2

(c) (c) Kos memasang jubin
2. (a)
Cost of installing tiles
(b) = (2 × 400) × (3 × 400) ÷ 10 000 × 15
= 800 × 1 200 ÷ 10 000 × RM15
= 96 × RM15

= RM1 440

2. (a) Skala / Scale = 12.5 : 2 500
= 1 : 200

(b) Luas sebenar rumah

Actual area of the house
= (12.5 × 200) × (5 × 200) ÷ 10 000
= 250 m2

(c) Luas Bilik 1 dan 2 / Area of Room 1 and 2
= (2 × 200) × (3 × 200) ÷ 10 000 × 2
= 48 m2

∴ Kos sewa / Rental cost = 48 × RM20
= RM960

Praktis Formatif 2

Bahagian C
1. (a) Luas rumah / Area of the house

= [(10 × 200)2 + (6 × 200)2] ÷ 10 000
= 544 m2

J4

∴ Kos memasang jubin 3. AO = BO, ∠ABO = ∠BAO
Cost of installing tiles x = 90° – 40°
= 544 × RM30
= RM16 320 = 50°
4. OP = OQ, ∠PQO = ∠QPO
(b) Skala / Scale = 5
1 000 OQ = OR, ∠QRO = ∠RQO

1 ∠QPO = 180° – 108° = 36°
200 2
=

= 1 : 200 x = 90° – 36° = 54°

(c) Skala / Scale 5. ∠ADB = ∠BDC

= Jarak pada peta / Distance on the map x = 90° – (34° × 2) = 22°
500 000
D 1. (a) ∠QPR = 36°
= 1 (b) ∠PQR = 90°
100 000 OP = OQ , ∠OPQ = ∠OQP = 36°
∠OQR = 90° – 36° = 54°
Jarak pada peta / Distance on the map
= 500 000 ÷ 100 000 2. (a) ∠ABD = 37°
(b) ∠CAD = ∠CBD
= 5 cm
= 90° – 37°

= 53°
2. (a) Luas sebenar / Actual areaCONTOH
= 22 × (4 × 14)2 PBD 2
7
A 1. PRST dan / and PQST
= 9 856 cm2 2. ABCD

(b)

B 1. ∠SUV dan / and ∠STV,
∠TSU dan / and ∠TVU

2. ∠JKM dan / and ∠JNM,
∠KJN dan / and ∠KMN

Bab 6 C 1. ∠QRS 2. ∠BCD

Praktis D S K P D 1. x + 87° = 180° y + 106° = 180°
x = 93° y = 74°
PBD 1
2. x + 92° = 180° 2y + y = 180°
A 1. ∠ADB, ∠AEB x = 88° 3y = 180°
2. ∠BAC y = 60°
3. ∠ABC 3. x + 82° = 180°
x = 98° y + 88° = 180°
B 1. θ = 74° y = 92°
2

= 37° 4. x = 32° y + 76° + 32° = 180°
y = 72°
2. θ = 107° × 2

= 214° 5. x = 240° y + 120° = 180°
2 y = 60°
3. ∠AOB = 48° × 2
x = 120°
= 96°

AO = BO, ∠ABO = ∠BAO E 1. ∠ABO = (180° – 96°) ÷ 2 = 42°
x = 180° – 99° – 42° = 39°
θ = 180° – 96° = 42° ∠ABO = ∠BAO = 42°
2 y = 23° + 42° = 65°

C 1. 5x = 90° 2. x = 2 × 26° = 52°
x = 90° y + 26° + 38° + 52° = 180°
5 y = 64°
= 18°
z = x = 52°
2. 4x + 2x = 90°
6x = 90° 3. ∠AOB = 360° – 275° = 85°
∠OAB = (180° – 85°) ÷ 2 = 47.5°
x = 90°
6

= 15°

J5

∠OAB = ∠ABO = 47.5° D 1. x = 180° – 134°
x + 23° + 47.5° = 180° = 46°

x = 109.5° y = 15 cm
y + 80° + 47.5° = 180°
2. x = √ 32 + 42
y = 52.5° = 5 cm

4. OC = OB, ∠OBC = ∠OCB = 60° tan y° = 3
∠BOC = 180° – (60° × 2) = 60° 2 4
x = 60° ÷ 2 = 30°
∠ADC = 90° – 34° = 56° y = 2 tan–1 3
y + 60° + 56° = 180° 4
y = 64°
= 73.74°

PBD 3 3. x = 90° – 35°
= 55°
A 1. Tangen / Tangent
2. Bukan tangen / Not tangent y = 17 cm
3. Bukan tangen / Not tangent
CONTOH E 1. ∠ABC = 180° – 87°
B 1. Tangen = 93°
Tangent
x = 180° – 53° – 93°
O = 34°
P
2. ∠BAQ = x
2. x + 65° + x + 41° = 180°
2x = 74°
O x = 37°

Tangen 3. ∠CAQ = x
P Tangent ∠BAC = 90°
x + 90° + x + 2x = 180°
4x = 90°
x = 22.5°

F 1.

3.

Tangen
O P Tangent

C 1. ∠OBA = 90° 2.
∠AOB = 25° × 2 3.
= 50° 4.

x = 90° – 50° 5.
= 40°

2. ∠CBE = 90°
∠BEC = 90° – 68°
= 22°

x = 22° × 2
= 44°

3. ∠OBC = 90°
∠BOC = 90° – 43°
= 47°

x = (180° – 47°) ÷ 2
= 66.5°

J6

G 1. (a) tan ∠ABQ = 7 Bab 8
6–4
Praktis D S K P
∠ABQ = tan–1 7 = 74.05°
2 PBD 1

(b) ∠PAS = 360° – 90° – 90° – 74.05° 1. Suatu bentuk lengkung / A curve shape
= 105.95° 2. Satu garis lurus mencancang

∠PTS = 105.95° ÷ 2 = 52.98° A vertical straight line
(c) ∠PRS = 180° – 52.98° = 127.02° 3. Satu garis lurus mengufuk

PBD 4 A horizontal straight line
4. Satu garis lurus mencancang
1. (a) ∠SUT = 90°
∠UST = 180° – 90° – 60° = 30° A vertical straight line
∠UTC = ∠UST = 30° 5. Suatu bentuk lengkung / A curve shape
SU
(b) sin ∠UTS = ST PBD 2

A 1.

CONTOH ST = 7 = 14 3 cm
sin 60° √ 3
X
PT = 14 × 3 = 12.12 cm
√ 3 2 2.

2. (a) Jadikan / Let ∠DFB = x Lokus B
∠DBF = 34° Locus of B
Dari / From ΔEFB,
35° + 34° + 34° + x = 180° X
x = 77°
∠ABE = ∠DFB Lokus B
= 77° Locus of B

(b) ∠BFG = 180° – 34° – 77° = 69° 3.
∠FGB = ∠ABE = 77°
∠GBF = 180° – 69° – 77° = 34° X

Praktis Formatif 3

Bahagian A

1. C 2. D 3. B 4. C

Bahagian C

5. (a) x = 65° B 1. P Lokus B
∠ABD = ∠DCE = 43° Locus of B
y = 180° – 30° – 65° – 43° = 42° Y
X
(b) x = 180° – 120° = 60° 2.
∠FBQ = 90° Lokus B
∠EQB = 180° – 90° = 90° X Locus of B
y = ∠EQP Q
= 90° – 60° = 30°
Lokus B
(c) Jadikan panjang OS = y Locus of B
Let the length OS = y
Y
RS = 2OS

∠POS = 90° – 40° = 50°
PS
tan ∠POS = OS

PS = y tan 50°

tan x = PS = y tan 50°
RS 2y

tan 50°
2
( )∴ x = tan–1 = 30.79°

J7

3. 2. P Q
Lokus Y
X 3. R Locus of Y
Z Lokus B
P Lokus Y S
Locus of B Locus of Y
Y R
Q
C 1. S

B

CONTOH A 4. Lokus Y
Lokus Z Locus of Y
Locus of Z P
R
2.
SQ
A
B Lokus Z 5. Lokus Y
3. Locus of Z Locus of Y
R P
A
S
Q

B E 1. A B
Lokus Z D
D 1. Locus of Z Lokus S
Locus of S
P Lokus Y
Locus of Y C
R
2. A
C

QS

Lokus S
Locus of S

BD

J8

3. A 4. P Q

Lokus S
Locus of S

Lokus X

Locus of X Lokus Z

Locus of Z

UR

B C B TS
Lokus Y
F 1. A Locus of Y
Lokus X
Locus of X

CONTOH Lokus Y 5. Q

Locus of Y P
CD

B Lokus Z
Locus of Z

2. Lokus Y
Locus of Y
Lokus X
Locus of X

Lokus X
Locus of X

Lokus Y SR
Locus of Y
A
C
3. D

B A Praktis Formatif 4

Lokus X Bahagian C
Locus of X 1. (a) (i) Sebuah bulatan berpusat O dengan

E panjang jejari sebanyak 2 cm.
A circle of centre O with radius 2 cm.
Lokus Y Lokus Z (ii) Suatu garis pembahagi dua sama
Locus of Y Locus of Z serenjang bagi garis AB.
A perpendicular bisector of line AB.
C D (iii) Suatu pembahagi dua sama sudut
bagi sudut ∠BAC.
An angle bisector of the angle ∠BAC.
(iv) Sepasang garis selari dengan garis
AB dan berjarak 2 cm dari AB.
A pair of parallel line with 2 cm from AB.
(b) (i) Lokus X / Locus of X:
Garis MN / Line MN
(ii) Lokus Y / Locus of Y:
Lengkok APQ / Arc APQ
(iii) Lokus Z / Locus of Z:
Lengkok COD / Arc COD

J9

2. (a) (i) Lokus X ialah garis lurus BD. (b) (i) 2
Locus of X is straight line BD. (ii) 4

(ii) dan / and (iii) 24. (a)

B

A C
LLookkussYY
LLooccuuss ooffYY

D (b) (i) y = 4

(ii) x = –6
8
Lokus Z 25. (a) 15, 17
Locus of Z
E F

(b) (i) Lokus X ialah garis yang selari (b) ✓
dengan AB dan melalui titik C. ✗
Locus of X is a line parallel to AB and
passes through point C.

(ii) dan / and (iii)
CONTOH Bahagian C

26. (a) Kos purata seunit saham

Average cost per share unit

AB = 12 × RM250
4 000

= RM3 000
4 000

LLookuss ZZ = RM0.75
LLooccus ooffZZ
(b) Jumlah pinjaman / Total loan
LLookuss YY
LLooccus ooffYY C ( )= RM60 000 + RM60 000 × 4 × 11
D 100
E
= RM60 000 + RM26 400

= RM86 400

Ujian Akhir Sesi Akademik (UASA) Jumlah baki selepas bayaran dibuat

Bahagian A Total balance after payment is made
= RM500 × (12 × 11 – 2)
1. D 2. C 3. A 4. B 5. B = RM500 × 130
8. A 9. A 10. D = RM65 000
6. A 7. B 13. B 14. A 15. D ∴ Jumlah pembayaran dibuat
18. D 19. A 20. A
11. B 12. C Total payment made
= RM86 400 – RM65 000
16. B 17. C
= RM21 400
Bahagian B
(c) Caj bayaran lewat / Late payment charge
21. (a) (i) 210 cm = 850 × 1%

= 8.5 (<10)

= RM10

(ii) 2.5 cm Faedah yang dikenakan / Interest charged

(b) (i) 1 : 2 ( )= RM850 × 18 × 15
100 365
(ii) 1: 1
4 = RM6.29
∴ Bayaran bulan hadapan
22. (a) (i) PALSU / FALSE
Payment for next month
(ii) BENAR / TRUE = RM850 + RM10 + RM6.29

(b) (i) 9 = RM866.29

(ii) –5

23. (a) (i) Bererti / Significant 27. (a) x = 32°
y = 180° – 32° – 61°
(ii) Tidak bererti / Not significant = 87°

J10

(b) (i) ∠PSR = ∠POQ = 50° 29. (a) Kecerunan / Graident

kos / cos 50° = OP = 3–6
20 4 – 10

OP = kos / cos 50° × 20 = 1
2
= 12.86 cm

(ii) PQ = √ 202 – 12.862 Persamaan baharu / New equation:
= 15.32 cm
y= 1 x + c
∴ Luas / Area of PORQ 2

( )= 2 1 × 15.32 × 12.86 Daripada titik A / From point A
2 1
3= 2 (4) + c
= 197.02 cm2
c=1
(c) (i) ∠FOD = 360° – 220° ∴ Persamaan garis lurus:
= 140°
Equation of straight:
∠FED = 140° ÷ 2
= 70° y= 1 x + 1
2
∴ x = 70° – 40°
= 30°

(ii) ∠OFE, ∠ABF
CONTOH (b) kx – 2y = 10
2y = kx – 10
k
y = 2x – 5

28. (a) (i) k
2
6 cm ∴ = –4

9 cm k = –8

(c) (i) Kecerunan / Gradient of PQ
3 – (–3)
(ii) 5 cm = 2 – (–6)

6 cm = 6
8
3 cm

4 cm = 3
4
(b) (i) D/F H/L C/K
(ii) x = –4, y = 6
y = mx + c
3
6= 4 (–4) + c

c=6+3

=9

y = mx + c
3
A/E G/I B/J y= 4 x + 9
(ii) E/I/J F/L/K
(iii) y = 0
G H 3
4 x + 9 = 0

3 x = –9
4

x = –9 × 3
4

x = –12

A/B C/D 30. (a) (i) AC = √ 92 + 132
= √ 250
= 15.81 cm

J11

(ii) tan ∠CAH = 4 31. (a) (i), (ii) dan / and (iii)
15.81

∠CAH = tan–1 4 P Q
15.81
Lokus Y
= 14° 12ʹ Locus of Y

(b) tan y = MN
24

5 = MN
12 24
UR
5
MN = 12 × 24

= 10 cm

OM = √ 242 + 102 Lokus X
= 26 cm Locus of X

∴ PM = 26 – 12
= 14 cm
CONTOH T S
(b) (i), (ii) dan / and (iii)
(c) tan θ = DA =2
AB
DA
8 = 2 Lokus X A
Locus of X
DA = 16 cm
BD = √ 82 + 162

= 17.89 cm

CD = 17.89 − 9
= 8.89 cm

Lokus Y

B Locus of Y
C

J12

MODULPENTAKSIRAN UASA 3

Proaktif KSSM ModulPentaksiran PROAKTIF KSSM Matematik (Buku 2)

Dihasilkan dalam dua Buku, iaitu Buku 1 dan Buku 2
Memudahkan murid membuat latih tubi sebelum menghadapi
Pentaksiran Bilik Darjah
Menepati format, standard dan keperluan Pentaksiran Bilik Darjah
(PBD)
Merujuk Buku Teks dan Dokumen Standard Kurikulum dan
Pentaksiran (DSKP)
Persembahannya diolah secara integrasi untuk memudahkan murid
memahami kesemua elemen yang diterapkan
Jawapan lengkap disertakan

Judul-judul dalam siri ini:

Subjek Tingkatan 1 2 3

Matematik (Buku 1&2) Dwibahasa

Sains Dwibahasa

Pendidikan Islam

Reka Bentuk & Teknologi

Pendidikan Jasmani &
Pendidikan Kesihatan

1 & 2, JalaLontB1S &7/21,AJ, aTalamnaBnS B7u/1kAit,STearmdaanngB, uSkeitksSyeernda7n, g, Seksyen 7, SeSme.mM.’sMia’:sRiaM: R11SM.9e80m.9(.s0eMt)’sia: RM8.90
00 Seri Ke4m3b3a0n0gSaenri,KSeemlabnagnogr aDna,ruSleElahnsagno.r Darul Ehsan. SaSbaabha/ hS/’wSa’kw:aRkM:S1Ra2Mb.9a90h.9/(s0Se’tw) ak: RM9.9
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g. Editoria•l:B0h3g.–E6d1it5o0ria3l0: 0093 •– 6Fa1k5s0: 03300–96•15F0ak1s0: 0073 – 6150 1007 ISBN 978-967-0048-53-6
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