Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Contents
CHAPTER Indices 1 HOTS Challenge 57
MRSM Cloned Question, TIMSS/PISA Cloned Question,
1
Online Quick Quiz (QR Code)
57
Indeks
1.1 Index Notation 1
6
1.2 Law of Indices 2 CHAPTER Angles and Tangents of Circles 58
PT3 Standard Practice 1 6 Sudut dan Tangent bagi Bulatan
HOTS Challenge 8
MRSM Cloned Question, TIMSS/PISA Cloned Question, 6.1 Angles at the Circumfenence and Central Angle
Online Quick Quiz (QR code) 8 Subtended by an Arc 58
6.2 Cyclic Quadrilaterals 60
CHAPTER Standard Form 9 6.3 Tangents to Circles 64
61
Angles and Tangents of Circles
63
6.4
2
PT3 Standard Practice 6
Bentuk Piawai
HOTS Challenge 68
2.1 Significant Figures 9 MRSM Cloned Question, TIMSS/PISA Cloned Question,
2.2 Standard Form 10 Online Quick Quiz (QR Code) 68
PT3 Standard Practice 2 13
HOTS Challenge 15
7
Online Quick Quiz (QR Code) 15 CHAPTER Plans and Elevations 69
Pelan dan Dongakan
CHAPTER Consumer Mathematics: Savings and 16 7.1 Orthogonal Projections 69
Investments, Credit and Debits
3
Matematik Pengguna: Simpanan dan Pelaburan,
Kredit dan Hutang 7.2 Plan and Elevations 73
PT3 Standard Practice 7
87
HOTS Challenge 92
3.1 Savings and Investments 16 Online Quick Quiz (QR Code) 92
3.2 Credit and Debt Management 22
PT3 Standard Practice 3 29
HOTS Challenge 32 CHAPTER
Online Quick Quiz (QR Code) 32 8 Loci in Two Dimensions 93
Lokus dalam Dua dimensi
CHAPTER Scale Drawings 33 8.1 Loci 94
93
4
Loci in Two Dimensions
8.2
Lukisan Berskala
101
PT3 Standard Practice 8 105
HOTS Challenge
4.1 Scale Drawings 33 MRSM Cloned Question, Online Quick Quiz (QR Code) 105
PT3 Standard Practice 4 43
HOTS Challenge 47
MRSM Cloned Question, TIMSS/PISA Cloned Question, CHAPTER Straight Lines 106
Online Quick Quiz (QR Code) 47 9 Garis Lurus
CHAPTER Trigonometric Ratios 48 9.1 Straight Lines 106
5
119
PT3 Standard Practice 9
Nisbah Trigonometri HOTS Challenge 123
TIMSS/PISA Cloned Question,
5.1 Sine, Cosine and Tangent for Acute Angles in Online Quick Quiz (QR Code) 123
Right-angled Triangles 48
PT3 Standard Practice 5 54 PT3 Model Paper 124
ii
00 Con TOP MATH F3.indd 2 02/01/2020 6:27 PM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
CHAPTER Standard Form
2
Bentuk Piawai
Textbook
pg. 30 – 49
2.1 Significant Figures
Angka Bererti
NOTES
1. Match each of the following numbers with the correct number of significant figures. PL 1
Padankan setiap nombor yang berikut dengan bilangan angka bererti yang betul.
Numbers Significant figures
Nombor Angka bererti
Example 3 341 4
(a) 3.02 1
(b) 90 000 3
(c) 3.0002 4
(d) 4.348 5
(e) 0.0045 2
2. Round off each of the following integers to the specified significant figures. PL 2
Bundarkan setiap integer yang berikut kepada angka bererti yang dikehendaki.
Example
(a) 948 (b) 590
98 [2 s.f. / a.b.] [1 s.f. / a.b.]
[1 s.f. / a.b.]
100
950 600
(c) 3 863 (d) 622 (e) 8 817
[3 s.f. / a.b.] [2 s.f. / a.b.] [3 s.f. / a.b.]
3 860 620 8 820
3. Round off each of the following decimals to the specified significant figures. PL 2
Bundarkan setiap nombor perpuluhan yang berikut kepada angka bererti yang dikehendaki.
Example
(a) 3.501 (b) 5.562
4.91 [3 s.f. / a.b.] [3 s.f. / a.b.]
[2 s.f. / a.b.]
4.9
3.50 5.56
(c) 81.979 (d) 9.74 (e) 6.98
[4 s.f. / a.b.] [2 s.f. / a.b.] [2 s.f. / a.b.]
81.98 9.7 7.0
9 © Penerbitan Pelangi Sdn. Bhd.
02 TOP MATH F3.indd 9 03/01/2020 8:00 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 2 Standard Form
4. Round off each of the following decimals to the specified significant figures. PL 2
Bundarkan setiap nombor perpuluhan yang berikut kepada angka bererti yang dikehendaki.
Example
(a) 0.491 (b) 0.07028
0.0891 [2 s.f. / a.b.] [3 s.f. / a.b.]
[2 s.f. / a.b.]
0.089 0.49 0.0703
(c) 0.21976 (d) 0.458 (e) 0.0299
[4 s.f. / a.b.] [2 s.f. / a.b.] [1 s.f. / a.b.]
0.2198 0.46 0.03
5. Calculate each of the following and round off your answer to the specified number of significant figures. PL 3
Hitung setiap yang berikut dan bundarkan jawapan anda mengikut bilangan angka bererti yang dikehendaki.
Example
(a) 5.23 × 1 025 ÷ 50
2.12 × 323 + 60 – 13.2 × 10 [1 s.f. / a.b.]
[2 s.f. / a.b.]
3
= 684.76 + 60 – 132 = 5.36075 × 10 ÷ 50
= 744.76 – 132 = 107.215
= 612.76 = 100
= 610
(b) 42.5 × 8.12 ÷ 400 (c) 980 – 6.13 ÷ 0.02
[3 s.f. / a.b.] [2 s.f. / a.b.]
= 345.1 ÷ 400 = 980 – 306.5
= 0.86275 = 673.5
= 0.863 = 670
55.6 – 34.5 (e) (31.2 + 0.9) × (4.5 – 0.87)
(d)
109 + 58 [1 s.f. / a.b.] [3 s.f. / a.b.]
= 21.1 = 32.1 × 3.63
167 = 116.523
= 0.12635 = 117
= 0.1
2.2 Standard Form
Bentuk Piawai
NOTES
6. State each of the following number in standard form. PL 1 PL 2
Nyatakan setiap nombor yang berikut dalam bentuk piawai.
Example
(a) 545 = (b) 651 00 =
3 540 =
5.45 × 10 2 6.51 × 10 4
3.54 × 10 3
(c) 26 = (d) 4 300 000 = (e) 1 037 =
2.6 × 10 4.3 × 10 6 1.037 × 10 3
© Penerbitan Pelangi Sdn. Bhd. 10
02 TOP MATH F3.indd 10 03/01/2020 8:00 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 2 Standard Form
7. State each of the following numbers in standard form. PL 1 PL 2
Nyatakan setiap nombor yang berikut dalam bentuk piawai.
Example
(a) 0.0045 = (b) 0.00025 =
0.033 =
4.5 × 10 –3 2.5 × 10 –4
3.3 × 10 –2
(c) 0.000076 = (d) 0.00000056 = (e) 0.9093 =
7.6 × 10 –5 5.6 × 10 –7 9.093 × 10 –1
8. Convert each of the number in standard form to single number. PL 1 PL 2
Tukar setiap nombor dalam bentuk piawai kepada nombor tunggal.
Example
(a) 5 × 10 = (b) 5.5 × 10 =
3
2 × 10 =
4
5 000 55
20 000
5
3
(c) 3.46 × 10 = (d) 8 × 10 = (e) 3.69 × 10 =
3 460 800 000 36.9
9. Convert each of the number in standard form to single number. PL 1 PL 2
Tukar setiap nombor dalam bentuk piawai kepada nombor tunggal.
Example
(a) 1 × 10 = (b) 6.5 × 10 =
–3
–4
–2
2 × 10 =
0.0001 0.0065
0.02
–4
(c) 8.6 × 10 = (d) 9 × 10 = (e) 3.68 × 10 =
–6
–6
0.00086 0.000009 0.00000368
10. Calculate each of the following and state the answer in standard form. PL 3
Hitung setiap yang berikut dan nyatakan jawapan dalam bentuk piawai.
Example
(a) 4.44 × 10 + 0.21 × 10 3
3
2.3 × 10 + 4.3 × 10 3 = (4.44 + 0.21) × 10 3
2
3
3
= (0.23 × 10 ) + (4.3 × 10 ) Common factor = 4.65 × 10 3
= (0.23 + 4.3) × 10 3 Faktor sepunya
= 4.53 × 10 3
(b) 4.5 × 10 – 3.2 × 10 4 (c) 640 × 110 ÷ 1 800
5
5
= 4.5 × 10 – 0.32 × 10 5 = 70 400 ÷ 1 800
= (4.5 – 0.32) × 10 5 = 7.04 × 10 ÷ 1.8 × 10 3
4
= 4.18 × 10 5 = 3.91 × 10 4 – 3
= 3.91 × 10
11 © Penerbitan Pelangi Sdn. Bhd.
02 TOP MATH F3.indd 11 03/01/2020 8:00 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 2 Standard Form
(d) 2.63 × 10 × 3.2 × 10 –2 (e) 1.5 × 10 × 5.1 × 10 ÷ (4 × 10 )
4
–2
3
2
= 2.63 × 3.2 × 10 4 + (–2) = (1.5 × 5.1 ÷ 4) × 10 –2 + 3 – 2
= 8.42 × 10 2 = 1.91 × 10 –1
11. Solve each of the following problems. PL 5 PL 6
Selesaikan setiap masalah yang berikut.
Example
Malaysia Singapore
Singapura
Population 32 million 4.48 million
Jumlah Penduduk 32 juta 4.48 juta
Land area (km ) 330 803 700
2
Luas tanah (km )
2
Based on the table above,
Berdasarkan jadual di atas,
(i) rewrite the population and land area of Malaysia and Singapore in standard forms,
tulis semula jumlah penduduk dan luas tanah Malaysia dan Singapura dalam bentuk piawai,
2
(ii) express the ratio of the number of people per km of Malaysia and Singapore in the form 1: n.
ungkapkan nisbah bilangan penduduk per km Malaysia dan Singapura, dalam bentuk 1 : n.
2
2
(i) Malaysia Singapore (ii) Number of people per km in Malaysia
7
Singapura = 3.2 × 10 ÷ 3.31 × 10 5
= 0.97 × 10 2
Population 3.2 × 10 7 4.48 × 10 6
Jumlah Penduduk = 9.7 × 10
2
Land area (km ) 3.31 × 10 5 7 × 10 2 Number of people per km in Singapore
2
Luas tanah (km ) = 4.48 × 10 ÷ 7 × 10 2
2
6
= 0.64 × 10 4
= 6.4 × 10 3
Ratio of number of people per km 2
(9.7 × 10 : 6.4 × 10 ) ÷ 9.7 × 10
3
Therefore, 1: n = 1: 66
(a) China Africa
Cina Afrika
Population 1.32 billion 832 million
Jumlah Penduduk 1.32 bilion 832 juta
Land area (km ) 9.60 million 26.6 million
2
Luas tanah (km ) 9.60 juta 26.6 juta
2
Based on the table above,
Berdasarkan jadual di atas,
(i) rewrite the population and land area of China and Africa in standard forms,
tulis semula jumlah penduduk dan luas tanah Cina dan Afrika dalam bentuk piawai,
(ii) find the number of people per km of China and Africa.
2
2
cari bilangan penduduk per km Cina dan Afrika.
© Penerbitan Pelangi Sdn. Bhd. 12
02 TOP MATH F3.indd 12 03/01/2020 8:00 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 2 Standard Form
(i) China Africa
Cina Afrika
Population 1.32 × 10 9 8.32 × 10 8
Jumlah Penduduk
Land area (km ) 9.60 × 10 6 2.66 × 10 7
2
Luas tanah (km )
2
(ii) Number of people per km in China
2
= 1.32 × 10 ÷ 9.60 × 10 6
9
= 0.1375 × 10 3
= 1.375 × 10 2
2
Number of people per km in Africa
8
= 8.32 × 10 ÷ 2.66 × 10 7
= 3.13 × 10
PT3 Standard Practice 2
Section A / Bahagian A 5. Which of the following numbers is rounded off to
3 significant figures to obtain 3.08? 2.1.2
1. Express 6.47 × 10 as single number. 2.2.1 Antara nombor-nombor yang berikut, nombor yang manakah
–4
Ungkapkan 6.47 × 10 sebagai nombor tunggal. dibundarkan kepada 3 angka bererti untuk memperoleh
–4
A 0.000647 C 6 470 3.08?
B 0.0000647 D 647 000 A 3.0862 C 3.0713
B 3.0785 D 3.8045
3.43 × 10 –3
2. = 2.2.2 Section B / Bahagian B
700 000
A 4.9 × 10 C 4.9 × 10 1. (a) Round off each number correct to the number
–3
–9
B 4.9 × 10 D 4.9 × 10 of significant figures stated. 2.1.2
–12
–6
Bundarkan setiap nombor betul kepada bilangan angka
bererti yang dinyatakan.
3. Find the value of 2k – 3s in standard form. Given
3
k = 6.14 × 10 and s = 1.09 × 10 . 2.2.2 [3 marks / 3 markah]
4
Cari nilai bagi 2k – 3s dalam bentuk piawai. Diberi Answer / Jawapan:
k = 6.14 × 10 dan s = 1.09 × 10 . Number of
4
3
A 2.042 × 10 C –2.042 × 10 4 significant
–4
B 2.042 × 10 D –2.042 × 10 –4 Number Answer
4
Nombor figures Jawapan
Bilangan angka
4. Which of the following numbers are written in bererti
standard form? 2.2.1
Antara yang berikut, nombor yang manakah diungkapkan (i) 0.0183 2 0.018
dalam bentuk piawai?
A 365 000 C 3.65 (ii) 299 530 3 300 000
B 36.5 × 10 D 3.65 × 10 (iii) 43 4 43.00
4
5
13 © Penerbitan Pelangi Sdn. Bhd.
02 TOP MATH F3.indd 13 03/01/2020 8:00 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 2 Standard Form
(b) Express 0.000045 in standard form. 2.2.1 Answer / Jawapan:
Ungkapkan 0.000045 dalam bentuk piawai. 5.43 5.43
[1 marks / 1 markah] (i) =
30.1 – 0.54 29.56
Answer / Jawapan: = 1.84 × 10 –1
–5
4.5 × 10
2
(ii) (6.09 × 10 ) ÷ (2.31 × 10 )
–3
= 6.09 ÷ 2.31 × 10 2 – (–3)
2. (a) Round off 299 530 correct to three significant = 2.64 × 10 5
figures. 2.1.2
Bundarkan 299 530 betul kepada tiga angka bererti.
[1 marks / 1 markah] (c) (i) A water fall has a width of 2 356 m, write
Answer / Jawapan: the width in standard form. 2.2.1
+1 Suatu air terjun mempunyai lebar 2 356 m, tulis
299 530 300 000 lebar itu dalam bentuk piawai.
[1 mark / 1 markah]
Answer / Jawapan:
3
(b) Fill in the blanks. 2.2.2 2 356 = 2.36 × 10 m
Isi tempat kosong.
[3 marks / 3 markah]
Answer / Jawapan: (ii) It is estimated that by the year 2060, the
world population will exceed k billion. If
–9
0.000000085 ÷ 1.7 × 10 the world population in the year 2060 is
11
1.05 × 10 , find the value of k. 2.2.3
= 8.5 × 10 –8 ÷ 1.7 × 10 –9 Jumlah bilangan penduduk dunia dijangka akan
mencapai k billion pada tahun 2060. Jika jumlah
= 8.5 ÷ 1.7 × 10 –8 + 9 penduduk dunia pada tahun 2060 ialah 1.05 ×
10 , cari nilai k.
11
= 50 [2 marks / 2 markah]
Answer / Jawapan:
1.05 × 10 = k × 10
9
11
Section C/ Bahagian C k = 1.05 × 10 billion
2
k
1. (a) Given 69 350 000 = 6.935 × 10 and 0.00000123 –1
l
= 1.3 × 10 , find the value of k × l. 2.2.2 2. (a) The speed of a car is 75 kmh . At this speed,
Diberi 69 350 000 = 6.935 × 10 dan 0.00000123 = 1.3 × 10, find the distance, in m, travelled by the car in 9
k
l
cari nilai bagi k × l. 000 seconds. Express your answer in standard
[3 marks / 3 markah] form. 2.2.3
Kelajuan sebuah kereta ialah 75 kmh . Pada kelajuan
–1
Answer / Jawapan: ini, cari jarak, dalam m, yang dilalui oleh kereta itu
k = 7, l = –6 dalam masa 9 000 saat. Nyatakan jawapan anda dalam
bentuk piawai.
∴ k × l = 7 × (–6) = –42
[3 marks / 3 markah]
Answer / Jawapan:
9 000
(b) Evaluate and give your answers in standard 9 000 seconds = 3 600 = 2.5 h
form. 2.2.2
Nilaikan dan beri jawapan anda dalam bentuk piawai. Distance = Speed × Time
–1
= 75 kmh × 2.5
5.43
(i) [2 marks / 2 markah] = 187.5 km
30.1 – 0.54
= 187 500 m
(ii) (6.09 × 10 ) ÷ (2.31 × 10 ) = 1.875 × 10 m
2
–3
5
[2 marks / 2 markah]
© Penerbitan Pelangi Sdn. Bhd. 14
02 TOP MATH F3.indd 14 03/01/2020 8:00 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 2 Standard Form
(b) The diagram below shows a wooden block in Answer / Jawapan:
the shape of cylinder. 2.2.3 Density = Mass
Rajah di bawah menunjukkan sebuah blok kayu Volume
berbentuk silinder. Mass
0.9 kgm =
–3
14 cm 7.7 × 10 –3
–3
Mass = (0.9 × 7.7 × 10 ) kg
–3
= 6.93 × 10 kg
= 0.00693 kg
50 cm
(c) The table below shows the price of some fruits.
Jadual di bawah menunjukkan harga bagi beberapa
buah-buahan. 2.2.3
3
(i) Calculate the volume, in m , of the wooden
block. State your answer in standard form. Fruits Price
Hitung isipadu, dalam m , blok kayu itu. Nyatakan Buah-buahan Harga
3
jawapan anda dalam bentuk piawai. Mango / Mangga RM5.99 / kg
Use / Guna: π = 22 [2 marks / 2 markah] Mangosteen / Manggis RM9.90 / kg
7
Answer / Jawapan: Puan Rosmah bough 3.5 kg of mango and 1 kg
Volume of cylinder = πr h of mangosteen. Calculate the total price that
2
22 she needs to pay. State your answer in three
= × (0.07) × 0.5
2
7 significant figures.
= 7.7 × 10 m 3 Puan Rosmah membeli 3.5 kg mangga dan 1 kg
–3
manggis. Hitung jumlah harga yang perlu dibayarnya.
Nyatakan jawapan anda dalam 3 angka bererti.
(ii) Given the density of the wooden block [3 marks / 3 markah]
–3
is 0.9 kgm . Find the mass, in kg, of the Answer / Jawapan:
block in single number.
Diberi ketumpatan blok kayu tersebut ialah Total price
–3
0.9 kgm . Cari jisim, dalam kg, blok kayu tersebut = RM5.99(3.5) + RM9.90
dalam nombor tunggal. = RM30.865
[2 marks / 2 markah] = RM30.90
HOTS Challenge
Write 3.45 × 10 seconds as nanoseconds.
–7
Tulis 3.45 × 10 saat sebagai nanosaat.
–7
3.45 × 10 –7
= 3.45 × 10 × 10 9
–7
= 3.45 × 10 2
= 345 nanoseconds
HOTS
Extra
15 © Penerbitan Pelangi Sdn. Bhd.
02 TOP MATH F3.indd 15 03/01/2020 8:00 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
CHAPTER Indices Mathematics Form 3 Chapter 1 Indices 9. (a) 8 = 8 = 2 (b) 2 × 8 × 32 –2n 5 –2n
Mathematics Form 3 Chapter 1 Indices
2n
4n
1
1
3
3
4n
= 2 × (2 ) × (2 )
3 2n
–10n
6n
4n
Indeks
1
= 2
4n + 6n – 10n
4
4
(b) 16 = 16 = 2 = 2 × 2 × 2
= 2 0
= 1
3
3
4
(c) 16 = (16) = 8
4
1. (a) False 5 2 10 7 3 1 1 1
(b) True 5. (a) 3 ÷ 3 3 3 3 2 (c) 2 × 5 × 10 2
2
2
2
(c) False 5 3 2 15 8 (d) 216 = (216) = 36 1 1 1
3
3
2
2
(d) True (b) 7 ÷ 7 7 7 7 = 2 × 5 × (2 × 5) 2
1
+
(e) False 1 2 2 1 2 2 1
+
(c) 2 ÷ 2 3 2 11 2 5 8 24 (e) 3 = 1 = 1 = 2 × 5
8
–3
2. (a) 2 4 27 3 3 27 = 2 × 5
(d) m ÷ m 8 m 17 m 2 m = 10
9
1
1
–4
(b) 2 729 (e) h ÷ h 4 h 16 h 8 h 3 (f) 2 = 2 1 4 = 16
12
8
– 1 3
64 3 × 9 2
(c) 3 3 1 0 (d) 4 –3
64 6. (a) ✓ (g) (–9) = 1 1 3
(b) ✗ (4 ) – 3 × (3 )
2 2
3
(d) 9 16 (c) ✗ 0 = 4 –3
3
1
(d) ✓ (h) = 1 1 3
4
(e) 5 625 10 3 × – – (–3) 2 × 2
3
(e) ✗ = 4 × 3
= 4 × 3 3
2
3. (a) 125 = 5 × 5 × 5 = 5 3 × 3 8
3
–4
7. (a) (5 × 2) 2 10. (a) = 16 × 27
2
(b) 36 = 6 × 6 = 6 3 –3
= 5 × 2 2 –4 + 8 – (–3) = 432
2
(c) 1 000 000 = 10 × 10 × 10 × 10 × 10 × 10 (b) (6 × 2) 3 = 3
= 10 = 6 × 2 = 3 7 2
6
3
3
3
2 5
(d) 64 = 8 × 8 = 8 2 (c) (8 × 9 ) 2 × 5 12p q (e) 27 – 2 3 × 64 3
3 5
× 9
3 × 5
= 8
1
1
1
2 –4 –6
(e) 2 = 1 × = 2 = 8 × 9 10 (b) 2 p q 3 – –4 2 2
15
6
6
36
6
3
3 3
4
(d) (6 × 5 ) = 12 × p 3 – (–4 )q 5 – (–6) = (3 ) 3 × (4 )
5 4
3
4. (a) 2 × 2 = 6 4 × 4 × 5 5 × 4 4 3 –4
2
7 11
2
= 2 2 + 3 = 6 × 5 20 = 3p q = 3 3 × – – (–4) × 4 3 × 2 3
16
3
= 2 5 (e) (3 × 4 × 5) 3 = 3 × 4 2
2
3
2
(b) 6 × 6 = 3 3 × 3 × 4 2 × 3 × 5 1 × 3 (c) m n × m –3 = 9 × 16
9 –3
2
6
9
5 –2
= 6 1 + 2 = 3 × 4 × 5 3 m n = 144
n
= 6 3 = m 9 – 3 – 5 –3 – (–2)
4 2
5
8. (a) 3 × (3 ) ÷ 3 9 = mn –1
(c) 4 × 4 5 + 4 × 2 – 9
7
3
1
= 4 7 + 3 = 3 4 = m (f) 1 × (8 ) ÷ 8
3 2
= 4 10 = 3 n 8
(b) 8 × 8 ÷ (8 ) 1 1
3 2
3
7
4
(d) 5 × 5 7 + 3 – 3 × 2 3 – 1 – 3 = 8 – + 3 × – 1
8
2
2
2
2
= 5 8 + 4 = 8 4 11. (a) 3 × 5 × 15 2 – + – 1
1
3
2
2
= 5 12 = 8 3 2 – 1 2 – 3 2 = 8
5 2
(c) (m ) × (m ) ÷ (m ) = 3 × 5 × (3 × 5) = 8 0
3 4
2 8
7
6
(e) w × w 3 × 4 + 5 × 2 – 2 × 8 3 – 3 – – 3 = 1
1
= w 7 + 6 = m 12 + 10 – 16 = 3 2 2 × 5 2 2
0
= w 13 = m = 3 × 5 –2
= m 6 1
= 1 ×
5 2
= 25
1
1 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 2
A2
BOOKLET ANS MATH F3.indd 2 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
CHAPTER Indices Mathematics Form 3 Chapter 1 Indices 9. (a) 8 = 8 = 2 (b) 2 × 8 × 32 –2n 5 –2n
Mathematics Form 3 Chapter 1 Indices
2n
4n
1
1
3
3
4n
= 2 × (2 ) × (2 )
3 2n
–10n
6n
4n
Indeks
1
= 2
4n + 6n – 10n
4
4
(b) 16 = 16 = 2 = 2 × 2 × 2
= 2 0
= 1
3
3
4
(c) 16 = (16) = 8
4
1. (a) False 5 2 10 7 3 1 1 1
(b) True 5. (a) 3 ÷ 3 3 3 3 2 (c) 2 × 5 × 10 2
2
2
2
(c) False 5 3 2 15 8 (d) 216 = (216) = 36 1 1 1
3
3
2
2
(d) True (b) 7 ÷ 7 7 7 7 = 2 × 5 × (2 × 5) 2
1
+
(e) False 1 2 2 1 2 2 1
+
(c) 2 ÷ 2 3 2 11 2 5 8 24 (e) 3 = 1 = 1 = 2 × 5
8
–3
2. (a) 2 4 27 3 3 27 = 2 × 5
(d) m ÷ m 8 m 17 m 2 m = 10
9
1
1
–4
(b) 2 729 (e) h ÷ h 4 h 16 h 8 h 3 (f) 2 = 2 1 4 = 16
12
8
– 1 3
64 3 × 9 2
(c) 3 3 1 0 (d) 4 –3
64 6. (a) ✓ (g) (–9) = 1 1 3
(b) ✗ (4 ) – 3 × (3 )
2 2
3
(d) 9 16 (c) ✗ 0 = 4 –3
3
1
(d) ✓ (h) = 1 1 3
4
(e) 5 625 10 3 × – – (–3) 2 × 2
3
(e) ✗ = 4 × 3
= 4 × 3 3
2
3. (a) 125 = 5 × 5 × 5 = 5 3 × 3 8
3
–4
7. (a) (5 × 2) 2 10. (a) = 16 × 27
2
(b) 36 = 6 × 6 = 6 3 –3
= 5 × 2 2 –4 + 8 – (–3) = 432
2
(c) 1 000 000 = 10 × 10 × 10 × 10 × 10 × 10 (b) (6 × 2) 3 = 3
= 10 = 6 × 2 = 3 7 2
6
3
3
3
2 5
(d) 64 = 8 × 8 = 8 2 (c) (8 × 9 ) 2 × 5 12p q (e) 27 – 2 3 × 64 3
3 5
× 9
3 × 5
= 8
1
1
1
2 –4 –6
(e) 2 = 1 × = 2 = 8 × 9 10 (b) 2 p q 3 – –4 2 2
15
6
6
36
6
3
3 3
4
(d) (6 × 5 ) = 12 × p 3 – (–4 )q 5 – (–6) = (3 ) 3 × (4 )
5 4
3
4. (a) 2 × 2 = 6 4 × 4 × 5 5 × 4 4 3 –4
2
7 11
2
= 2 2 + 3 = 6 × 5 20 = 3p q = 3 3 × – – (–4) × 4 3 × 2 3
16
3
= 2 5 (e) (3 × 4 × 5) 3 = 3 × 4 2
2
3
2
(b) 6 × 6 = 3 3 × 3 × 4 2 × 3 × 5 1 × 3 (c) m n × m –3 = 9 × 16
9 –3
2
6
9
5 –2
= 6 1 + 2 = 3 × 4 × 5 3 m n = 144
n
= 6 3 = m 9 – 3 – 5 –3 – (–2)
4 2
5
8. (a) 3 × (3 ) ÷ 3 9 = mn –1
(c) 4 × 4 5 + 4 × 2 – 9
7
3
1
= 4 7 + 3 = 3 4 = m (f) 1 × (8 ) ÷ 8
3 2
= 4 10 = 3 n 8
(b) 8 × 8 ÷ (8 ) 1 1
3 2
3
7
4
(d) 5 × 5 7 + 3 – 3 × 2 3 – 1 – 3 = 8 – + 3 × – 1
8
2
2
2
2
= 5 8 + 4 = 8 4 11. (a) 3 × 5 × 15 2 – + – 1
1
3
2
2
= 5 12 = 8 3 2 – 1 2 – 3 2 = 8
5 2
(c) (m ) × (m ) ÷ (m ) = 3 × 5 × (3 × 5) = 8 0
3 4
2 8
7
6
(e) w × w 3 × 4 + 5 × 2 – 2 × 8 3 – 3 – – 3 = 1
1
= w 7 + 6 = m 12 + 10 – 16 = 3 2 2 × 5 2 2
0
= w 13 = m = 3 × 5 –2
= m 6 1
= 1 ×
5 2
= 25
1
1 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 2
A2
BOOKLET ANS MATH F3.indd 2 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 1 Indices Mathematics Form 3 Chapter 1 Indices
1 – 2 n 5 3
–1 2 2
3 –3
2
(g) (12 × 27) ÷ 8 3 (b) 5 × 5 = 5 × 5 PT3 Standard Practice 1 6. A (3a b ) × 7a b
1 – 2 5 n + 1 = 5 5 + 3 = 9a b × 7a b
–2 4
3 –3
3 2
= (2 × 3 × 3 ) ÷ (2 ) 3 n + 1 = 8 = (9 × 7)a –2 + 3 4 – 3
3
2
b
1 2 (1 + 3) × 1 Section A
2 × – 3 × –
= 2 2 3 × 3 2 n = 7 = 63ab
1 4
= 2 × 3 2 1. m = m 5 0 3 2 3 6
3
4
5
= 8 × 9 (c) 3 = 9 n 1 B 7c × 7 ÷ 49c × 343c
2
3
1
= 72 3 = 3 2n = m 5 × 4 = 7c × 1 ÷ 7 c × 7c 2
2 2
2
2
3
2 = 2n = m 4 5 = 7 1 – 2 + 1 2 1 – + 2
c
2
12. (a) 2x y × 5xy 2 n = 1 = c
3 2
= 2 × 5x 3 + 1 2 + 2 Answer: C 5 1
y
1
4 4
= 10x y (d) (b ) = b 9 C 1 d 2 – 3 = d 3
n 3
5
5
2
–1 2
–1 2
3n
b = b 9 2. 7 k × 7k 1 3 = 7 k × 7 k 3 1 3
3 2
d
3
(b) 81m n ÷ 27m n 3n = 9 3 2 (7 ) k 3 2 =
4 2
3 2
5
2 2
4
= 3 m n ÷ 3 m n n = 3 (49k) –1 + 3 – 3 2 + – 3 1
3 2
3
4 2
3
n
= 3 4 − 3 m 3 − 4 2 − 2 = 7 k 2 2 ≠ 5d
3
3 = 7 k
–1 2
= (e) 27(5 ) = 125(3 ) D e × e × e × e × e × e = e 6
n
n
m = k 2
1
n = 125 7 Answer: C
5
3
27
a
4 –1
(c) (3a b ) × 3 Answer: A Section B
3 –1 2
n
5
5
b
b
= 9a 3 × 2 – 4 –1 × 2+1 3 = 3 3. 3q × 5q = (3 × 5)q 4 + 5
4
5
= 9a b n = 3 = 15q 9 1. a × a 3 a 6
2 –1
5
a 2
= 9 Answer: A
b
(a ) a 8
3 2
–3k
(hk ) × h k 4. 2 k + 1 × 2 = 32
2 2
3 –3
(d) k + 1 –3k 5
h k 2 = 2 2
4 –4
k
= h 1 × 2 + 3 – 4 2 × 2 – 3 – (–4) 1 – 2k = 5 a –3 a 3
= h 2 +3 – 4 4 – 3 + 4 2k = –4
k
1
= hk 5 k = –2 3 2 —
a
Answer: D a 3
5
2 –3 3
2 4
(2r s ) × r s 3 4
(e) y 6 3r s
1 5. A = y 6 – 5 4 3
8r s y 5 2. × (16r s )
2 2
2 –6 2
= y 5r s
–3 7
2 3 2 × 3 + – 1
5
7
1
4
= × r 2 2 s –3 × 3 + 4 – 2 B y × y = y 7 + 4 — × 4
2 3 = y 11 3 r 3 × 4 s 2 3 3 3
4 4
5
= r 6 + – 1 2 s –9 + 4 – 2 C y + y 8 = 5r s × 4 2 × 2 × r 2 × 2 × s –6 × 2
–3 7
2
3
8 –7
= r s 3 2
D (y ) = 1 = 81r s 3 r s
11 0
–3 7 × 4
3 –9
5r s
12
13. (a) 2 ÷ 8 = 2 n Answer: B 5 184
2 ÷ 2 = 2 n = 5 r 3 + 3 – (–3) × S 2 – 9 – 7
12
3
2 12 – 3 = 2 n
2 = 2 n 5 184 9 –14
9
n = 9 = 5 r s
3 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 4
A3
BOOKLET ANS MATH F3.indd 3 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 1 Indices Mathematics Form 3 Chapter 1 Indices
1 – 2 n 5 3
–1 2 2
3 –3
2
(g) (12 × 27) ÷ 8 3 (b) 5 × 5 = 5 × 5 PT3 Standard Practice 1 6. A (3a b ) × 7a b
1 – 2 5 n + 1 = 5 5 + 3 = 9a b × 7a b
–2 4
3 –3
3 2
= (2 × 3 × 3 ) ÷ (2 ) 3 n + 1 = 8 = (9 × 7)a –2 + 3 4 – 3
3
2
b
1 2 (1 + 3) × 1 Section A
2 × – 3 × –
= 2 2 3 × 3 2 n = 7 = 63ab
1 4
= 2 × 3 2 1. m = m 5 0 3 2 3 6
3
4
5
= 8 × 9 (c) 3 = 9 n 1 B 7c × 7 ÷ 49c × 343c
2
3
1
= 72 3 = 3 2n = m 5 × 4 = 7c × 1 ÷ 7 c × 7c 2
2 2
2
2
3
2 = 2n = m 4 5 = 7 1 – 2 + 1 2 1 – + 2
c
2
12. (a) 2x y × 5xy 2 n = 1 = c
3 2
= 2 × 5x 3 + 1 2 + 2 Answer: C 5 1
y
1
4 4
= 10x y (d) (b ) = b 9 C 1 d 2 – 3 = d 3
n 3
5
5
2
–1 2
–1 2
3n
b = b 9 2. 7 k × 7k 1 3 = 7 k × 7 k 3 1 3
3 2
d
3
(b) 81m n ÷ 27m n 3n = 9 3 2 (7 ) k 3 2 =
4 2
3 2
5
2 2
4
= 3 m n ÷ 3 m n n = 3 (49k) –1 + 3 – 3 2 + – 3 1
3 2
3
4 2
3
n
= 3 4 − 3 m 3 − 4 2 − 2 = 7 k 2 2 ≠ 5d
3
3 = 7 k
–1 2
= (e) 27(5 ) = 125(3 ) D e × e × e × e × e × e = e 6
n
n
m = k 2
1
n = 125 7 Answer: C
5
3
27
a
4 –1
(c) (3a b ) × 3 Answer: A Section B
3 –1 2
n
5
5
b
b
= 9a 3 × 2 – 4 –1 × 2+1 3 = 3 3. 3q × 5q = (3 × 5)q 4 + 5
4
5
= 9a b n = 3 = 15q 9 1. a × a 3 a 6
2 –1
5
a 2
= 9 Answer: A
b
(a ) a 8
3 2
–3k
(hk ) × h k 4. 2 k + 1 × 2 = 32
2 2
3 –3
(d) k + 1 –3k 5
h k 2 = 2 2
4 –4
k
= h 1 × 2 + 3 – 4 2 × 2 – 3 – (–4) 1 – 2k = 5 a –3 a 3
= h 2 +3 – 4 4 – 3 + 4 2k = –4
k
1
= hk 5 k = –2 3 2 —
a
Answer: D a 3
5
2 –3 3
2 4
(2r s ) × r s 3 4
(e) y 6 3r s
1 5. A = y 6 – 5 4 3
8r s y 5 2. × (16r s )
2 2
2 –6 2
= y 5r s
–3 7
2 3 2 × 3 + – 1
5
7
1
4
= × r 2 2 s –3 × 3 + 4 – 2 B y × y = y 7 + 4 — × 4
2 3 = y 11 3 r 3 × 4 s 2 3 3 3
4 4
5
= r 6 + – 1 2 s –9 + 4 – 2 C y + y 8 = 5r s × 4 2 × 2 × r 2 × 2 × s –6 × 2
–3 7
2
3
8 –7
= r s 3 2
D (y ) = 1 = 81r s 3 r s
11 0
–3 7 × 4
3 –9
5r s
12
13. (a) 2 ÷ 8 = 2 n Answer: B 5 184
2 ÷ 2 = 2 n = 5 r 3 + 3 – (–3) × S 2 – 9 – 7
12
3
2 12 – 3 = 2 n
2 = 2 n 5 184 9 –14
9
n = 9 = 5 r s
3 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 4
A3
BOOKLET ANS MATH F3.indd 3 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 1 Indices
Section C (ii) a = 2, r = 2, n = 10 CHAPTER
a(r – 1) 2 Standard Form
n
1 s =
1. (a) (i) 368 × 3 + 2 020 = 1 105 + 2 020 n r – 1 Bentuk Piawai
3 = 3 125 2(2 – 1)
10
=5 5 s = 2 – 1
10
(ii) (6 – 3.808) ÷ 274 = 2.194 ÷ 274 = 2 046
= 0.008
1 55.6 – 34.5
= 1. Numbers Significant (d)
125 figures 109 + 58
= 5 –3 HOTS Challenge Nombor Angka bererti 21.1
=
167
(b) [(–3k) 4m + 6 –3 (a) 3.02 1 = 0.12635
] = 1
(–3k) –12m – 18 = (–3k) 0 Rounds 0 1 2 3 4 5 6 (b) 90 000 3 = 0.1
–12m – 18 = 0 Pusingan
–12m = 18 Players 64 32 16 8 4 2 1 (c) 3.0002 4 (e) (31.2 + 0.9) × (4.5 – 0.87)
18 Pemain (d) 4.348 5 = 32.1 × 3.63
m = – = 116.523
12 Match played
m = –1.5 Perlawanan dimain 32 16 8 4 2 1 – (e) 0.0045 2 = 117
(c) (i) k × k 4 Total match played: 2. (a) 950 6. (a) 545 = 5.45 × 10 2
–1
= k –1 + 4 32 + 16 + 8 + 4 + 2 + 1 = 63 (b) 600 (b) 65 100 = 6.51 × 10 4
= k 3 (c) 3860 (c) 26 = 2.6 × 10
6
(d) 620 (d) 4 300 000 = 4.3 × 10
(e) 8820 (e) 1 037 = 1.037 × 10 3
3. (a) 3.50 7. (a) 0.0045 = 4.5 × 10 –3
(b) 5.56 (b) 0.00025 = 2.5 × 10
–4
(c) 81.98 (c) 0.000076 = 7.6 × 10 –5
(d) 9.7 (d) 0.00000056 = 5.6 × 10 –7
(e) 7.0 (e) 0.9093 = 9.093 × 10
–1
4. (a) 0.49 8. (a) 5 × 10 = 5 000
3
(b) 0.0703 (b) 5.5 × 10 = 55
3
(c) 0.2198 (c) 3.46 × 10 = 3 460
5
(d) 0.46 (d) 8 × 10 = 800 000
(e) 0.03 (e) 3.69 × 10 = 36.9
5. (a) 5.23 × 1 025 ÷ 50 9. (a) 1 × 10 = 0.0001
–4
3
–3
= 5.36075 × 10 ÷ 50 (b) 6.5 × 10 = 0.0065
= 107.215 (c) 8.6 × 10 = 0.00086
–4
= 100 (d) 9 × 10 = 0.000009
–6
–6
(b) 42.5 × 8.12 ÷ 400 (e) 3.68 × 10 = 0.00000368
= 345.1 ÷ 400 10. (a) 4.44 × 10 + 0.21 × 10 3
3
= 0.86275 = (4.44 + 0.21) × 10 3
= 0.863 = 4.65 × 10 3
(c) 980 – 6.13 ÷ 0.02
5
= 980 – 306.5 (b) 4.5 × 10 – 3.2 × 10 4
5
= 673.5 = 4.5 × 10 – 0.32 × 10 5
= 670 = (4.5 – 0.32) × 10 5
= 4.18 × 10 5
5 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 6
A4
BOOKLET ANS MATH F3.indd 4 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 1 Indices
Section C (ii) a = 2, r = 2, n = 10 CHAPTER
a(r – 1) 2 Standard Form
n
1 s =
1. (a) (i) 368 × 3 + 2 020 = 1 105 + 2 020 n r – 1 Bentuk Piawai
3 = 3 125 2(2 – 1)
10
=5 5 s = 2 – 1
10
(ii) (6 – 3.808) ÷ 274 = 2.194 ÷ 274 = 2 046
= 0.008
1 55.6 – 34.5
= 1. Numbers Significant (d)
125 figures 109 + 58
= 5 –3 HOTS Challenge Nombor Angka bererti 21.1
=
167
(b) [(–3k) 4m + 6 –3 (a) 3.02 1 = 0.12635
] = 1
(–3k) –12m – 18 = (–3k) 0 Rounds 0 1 2 3 4 5 6 (b) 90 000 3 = 0.1
–12m – 18 = 0 Pusingan
–12m = 18 Players 64 32 16 8 4 2 1 (c) 3.0002 4 (e) (31.2 + 0.9) × (4.5 – 0.87)
18 Pemain (d) 4.348 5 = 32.1 × 3.63
m = – = 116.523
12 Match played
m = –1.5 Perlawanan dimain 32 16 8 4 2 1 – (e) 0.0045 2 = 117
(c) (i) k × k 4 Total match played: 2. (a) 950 6. (a) 545 = 5.45 × 10 2
–1
= k –1 + 4 32 + 16 + 8 + 4 + 2 + 1 = 63 (b) 600 (b) 65 100 = 6.51 × 10 4
= k 3 (c) 3860 (c) 26 = 2.6 × 10
6
(d) 620 (d) 4 300 000 = 4.3 × 10
(e) 8820 (e) 1 037 = 1.037 × 10 3
3. (a) 3.50 7. (a) 0.0045 = 4.5 × 10 –3
(b) 5.56 (b) 0.00025 = 2.5 × 10
–4
(c) 81.98 (c) 0.000076 = 7.6 × 10 –5
(d) 9.7 (d) 0.00000056 = 5.6 × 10 –7
(e) 7.0 (e) 0.9093 = 9.093 × 10
–1
4. (a) 0.49 8. (a) 5 × 10 = 5 000
3
(b) 0.0703 (b) 5.5 × 10 = 55
3
(c) 0.2198 (c) 3.46 × 10 = 3 460
5
(d) 0.46 (d) 8 × 10 = 800 000
(e) 0.03 (e) 3.69 × 10 = 36.9
5. (a) 5.23 × 1 025 ÷ 50 9. (a) 1 × 10 = 0.0001
–4
3
–3
= 5.36075 × 10 ÷ 50 (b) 6.5 × 10 = 0.0065
= 107.215 (c) 8.6 × 10 = 0.00086
–4
= 100 (d) 9 × 10 = 0.000009
–6
–6
(b) 42.5 × 8.12 ÷ 400 (e) 3.68 × 10 = 0.00000368
= 345.1 ÷ 400 10. (a) 4.44 × 10 + 0.21 × 10 3
3
= 0.86275 = (4.44 + 0.21) × 10 3
= 0.863 = 4.65 × 10 3
(c) 980 – 6.13 ÷ 0.02
5
= 980 – 306.5 (b) 4.5 × 10 – 3.2 × 10 4
5
= 673.5 = 4.5 × 10 – 0.32 × 10 5
= 670 = (4.5 – 0.32) × 10 5
= 4.18 × 10 5
5 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 6
A4
BOOKLET ANS MATH F3.indd 4 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 2 Standard Form Mathematics Form 3 Chapter 2 Standard Form
(c) 640 × 110 ÷ 1 800 PT3 Standard Practice 2 Section C (b) (i) Volume of cylinder = πr h
2
= 70 400 ÷ 1 800 = 22 × (0.07) × 0.5
2
= 7.04 × 10 ÷ (1.8 × 10 ) Section A 1. (a) k = 7, l = – 6 7
3
4
–3
= 3.91 × 10 4 – 3 1. 6.47 × 10 = 0.000647 ∴ k × l = 7 × (–6) = –42 = 7.7 × 10 m 3
–4
= 3.91 × 10 5.43 5.43 Mass
Answer: A (b) (i) = (ii) Density = Volume
29.56
(d) 2.63 × 10 × 3.2 × 10 –2 30.1 – 0.54 = 1.84 × 10 –1 –3 Mass
4
= 2.63 × 3.2 × 10 4 + (–2) 2. 3.43 × 10 –3 = 3.43 × 10 –3 0.9 kgm = 7.7 × 10 –3
–3
2
= 8.42 × 10 2 700 000 7 × 10 5 (ii) (6.09 × 10 ) ÷ (2.31 × 10 ) mass = (0.9 × 7.7 × 10 ) kg
–3
= 0.49 × 10 –3 – 5 = 6.09 ÷ 2.31 × 10 2 – (–3) –3
(e) 1.5 × 10 × 5.1 × 10 ÷ (4 × 10 ) = 4.9 × 10 –9 = 2.64 × 10 5 = 6.93 × 10 kg
2
–2
3
= (1.5 × 5.1 ÷ 4) × 10 –2 + 3 – 2 Answer: C 3 = 0.00693 kg
= 1.91 × 10 (c) (i) 2 356 = 2.36 × 10 m (c) Total price
–1
9
(ii) 1.05 × 10 = k × 10 = RM5.99(3.5) + RM9.90
11
3. 2k – 3s = RM30.865
2
11. (a) (i) k = 1.05 × 10 billion
= 2(6.14 × 10 ) – 3(1.09 × 10 ) = RM30.90
3
4
China Africa = 1.228 × 10 – 3.27 × 10 4
4
Cina Afrika = –2.042 × 10 4 2. (a) 9 000 seconds = 9 000 = 2.5 h
3 600 HOTS Challenge
Population 9 8 Answer: C Distance = Speed × Time
Jumlah Penduduk 1.32 × 10 8.32 × 10 = 75 kmh × 2.5
–1
4. Answer: D 3.45 × 10 –7
Land area (km ) 6 7 = 187.5 km = 3.45 × 10 × 10 9
2
–7
Luas tanah (km ) 9.60 × 10 2.66 × 10 = 187 500 m
2
5. A 3.0862 = 3.09 (3 s.f.) = 1.875 × 10 m = 3.45 × 10 2
5
B 3.0785 = 3.08 (3 s.f.) = 345 nanoseconds
(ii) Number of people per km in China
2
= 1.32 × 10 ÷ 9.60 × 10 6 C 3.0713 = 3.07 (3 s.f.)
9
= 0.1375 × 10 3 D 3.8045 = 3.80 (3 s.f.)
= 1.375 × 10 2 Answer: B
2
Number of people per km in Africa
= 8.32 × 10 ÷ 2.66 × 10 7 Section B
8
= 3.13 × 10
1. (a) Number of
significant
Number Answer
Nombor figures Jawapan
Bilangan
angka bererti
(i) 0.0183 2 0.018
(ii) 299 530 3 300 000
(iii) 43 4 43.00
(b) 0.000045 = 4.5 × 10 –5
+1
2. (a) 299 530 300 000
(b) 0.000000085 ÷ 1.7 × 10
–9
= 8.5 × 10 –8 ÷ 1.7 × 10 –9
= 8.5 ÷ 1.7 × 10 –8 + 9
= 50
7 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 8
A5
BOOKLET ANS MATH F3.indd 5 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 2 Standard Form Mathematics Form 3 Chapter 2 Standard Form
(c) 640 × 110 ÷ 1 800 PT3 Standard Practice 2 Section C (b) (i) Volume of cylinder = πr h
2
= 70 400 ÷ 1 800 = 22 × (0.07) × 0.5
2
= 7.04 × 10 ÷ (1.8 × 10 ) Section A 1. (a) k = 7, l = – 6 7
3
4
–3
= 3.91 × 10 4 – 3 1. 6.47 × 10 = 0.000647 ∴ k × l = 7 × (–6) = –42 = 7.7 × 10 m 3
–4
= 3.91 × 10 5.43 5.43 Mass
Answer: A (b) (i) = (ii) Density = Volume
29.56
(d) 2.63 × 10 × 3.2 × 10 –2 30.1 – 0.54 = 1.84 × 10 –1 –3 Mass
4
= 2.63 × 3.2 × 10 4 + (–2) 2. 3.43 × 10 –3 = 3.43 × 10 –3 0.9 kgm = 7.7 × 10 –3
–3
2
= 8.42 × 10 2 700 000 7 × 10 5 (ii) (6.09 × 10 ) ÷ (2.31 × 10 ) mass = (0.9 × 7.7 × 10 ) kg
–3
= 0.49 × 10 –3 – 5 = 6.09 ÷ 2.31 × 10 2 – (–3) –3
(e) 1.5 × 10 × 5.1 × 10 ÷ (4 × 10 ) = 4.9 × 10 –9 = 2.64 × 10 5 = 6.93 × 10 kg
2
–2
3
= (1.5 × 5.1 ÷ 4) × 10 –2 + 3 – 2 Answer: C 3 = 0.00693 kg
= 1.91 × 10 (c) (i) 2 356 = 2.36 × 10 m (c) Total price
–1
9
(ii) 1.05 × 10 = k × 10 = RM5.99(3.5) + RM9.90
11
3. 2k – 3s = RM30.865
2
11. (a) (i) k = 1.05 × 10 billion
= 2(6.14 × 10 ) – 3(1.09 × 10 ) = RM30.90
3
4
China Africa = 1.228 × 10 – 3.27 × 10 4
4
Cina Afrika = –2.042 × 10 4 2. (a) 9 000 seconds = 9 000 = 2.5 h
3 600 HOTS Challenge
Population 9 8 Answer: C Distance = Speed × Time
Jumlah Penduduk 1.32 × 10 8.32 × 10 = 75 kmh × 2.5
–1
4. Answer: D 3.45 × 10 –7
Land area (km ) 6 7 = 187.5 km = 3.45 × 10 × 10 9
2
–7
Luas tanah (km ) 9.60 × 10 2.66 × 10 = 187 500 m
2
5. A 3.0862 = 3.09 (3 s.f.) = 1.875 × 10 m = 3.45 × 10 2
5
B 3.0785 = 3.08 (3 s.f.) = 345 nanoseconds
(ii) Number of people per km in China
2
= 1.32 × 10 ÷ 9.60 × 10 6 C 3.0713 = 3.07 (3 s.f.)
9
= 0.1375 × 10 3 D 3.8045 = 3.80 (3 s.f.)
= 1.375 × 10 2 Answer: B
2
Number of people per km in Africa
= 8.32 × 10 ÷ 2.66 × 10 7 Section B
8
= 3.13 × 10
1. (a) Number of
significant
Number Answer
Nombor figures Jawapan
Bilangan
angka bererti
(i) 0.0183 2 0.018
(ii) 299 530 3 300 000
(iii) 43 4 43.00
(b) 0.000045 = 4.5 × 10 –5
+1
2. (a) 299 530 300 000
(b) 0.000000085 ÷ 1.7 × 10
–9
= 8.5 × 10 –8 ÷ 1.7 × 10 –9
= 8.5 ÷ 1.7 × 10 –8 + 9
= 50
7 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 8
A5
BOOKLET ANS MATH F3.indd 5 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
CHAPTER Consumer Mathematics: Saving and Investments, Credits and 6. (a) Return / Pulangan (b) (i) Average cost per share
Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts
3
5 000(1.68) + 4 000(1.56) +
(b) Uncertainty / Ketidakpastian
Debts
3 000(1.52)
Matematik Pengguna: Simpanan dan Pelaburan, Kredit dan Hutang (c) Cash / Wang tunai = 5 000 + 4 000 + 3 000
= RM1.60
7. Saving Real (ii) Cost averaging strategy is used.
account Shares
1. Akaun estate Saham This strategy can reduce the
Savings Investment Hartanah investment risk by bringing down
Simpanan Pelaburan simpanan average purchasing cost.
Liquidity High Low Moderate
Kecairan Tinggi Rendah Sederhana (c) (i) Average cost per share
2 000(0.55) + 2 000(0.53) +
Fixed deposits Shares Current account Real estates Unit trust RM2 100 2 000(0.56)
Simpanan tetap Saham Simpanan semasa Hartanah Unit amanah saham 8. (a) 0.75 = total unit = 2 000 + 2 000 + 2 000
acquired = RM0.55
RM2 100 (ii) Jeremy’s strategy is more profitable
2. (a) Simple interest, I 0.032 4(3) Total unit acquired = 0.75
1.6 3 3. (a) Matured value = 5 000 1 + 4 because the average cost per share
= RM2 000 × × = 2 800 units is lower compared to Firdaus’s
100 12 = RM5 501.69
= RM8 purchase.
(b) Matured value
Total saving = RM2 000 + RM8
= RM2008 = 15 000 1 + 0.043 1(10)
1
1.8 0.0325 4(5)
(b) Simple interest, I = 3 000 × × 2 = RM22 852.53 9. (a) 10 000 = x 1 +
100 4
= RM108 (c) Matured value 10 000 = x(1.008125)
20
Total saving = RM3 000 + RM108 = 8 000 1 + 0.035 12(7) x = 10 000
= RM3 108 12 1.008125 20
2.2 9 = RM10 217.33 x = RM8 505.74
(c) Simple interest, I = 5 000 × ×
100 12 4. (a) ✓
= RM82.50 amount paid
(b) ✓ (b) (i) 1.35 =
Total saving = RM5 000 + RM82.50 (c) ✗ 2 000
= RM5 082.50 Amount paid = RM2 700
2 000x – 2 700
(ii) 8% = × 100%
2 700
5. (a) Total return = RM7 000 + RM100 – RM6 500 0.08(2 700) + 2 700 = 2 000x
= RM600 x = RM1.46 per share
600
ROI = × 100%
6 500
= 9.23% 0.032 1(2)
(c) (i) MV = 12 000 1 + 1
(b) 12.5% = (13 200 + 2x) – 12 000 × 100% = RM12 780.29
12 000
0.125(12 000) + 12 000 = 13 200 + 2x ROI = 12 780.29 – 12 000 × 100%
2x = 300 12 000
x = RM150 = 6.50%
[8 000(1.27) + 180] – x
RM1 500 – RM2 000 (ii) 2(6.50%) = × 100%
(c) (i) ROI = × 100% x
RM2 000 0.13x = 10 340 – x
= –25%
1.13x = 10 340
x = RM9 150.44
(ii) Change of government policy could cause loss in investment.
9 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 10
A6
BOOKLET ANS MATH F3.indd 6 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
CHAPTER Consumer Mathematics: Saving and Investments, Credits and 6. (a) Return / Pulangan (b) (i) Average cost per share
Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts
3
5 000(1.68) + 4 000(1.56) +
(b) Uncertainty / Ketidakpastian
Debts
3 000(1.52)
Matematik Pengguna: Simpanan dan Pelaburan, Kredit dan Hutang (c) Cash / Wang tunai = 5 000 + 4 000 + 3 000
= RM1.60
7. Saving Real (ii) Cost averaging strategy is used.
account Shares
1. Akaun estate Saham This strategy can reduce the
Savings Investment Hartanah investment risk by bringing down
Simpanan Pelaburan simpanan average purchasing cost.
Liquidity High Low Moderate
Kecairan Tinggi Rendah Sederhana (c) (i) Average cost per share
2 000(0.55) + 2 000(0.53) +
Fixed deposits Shares Current account Real estates Unit trust RM2 100 2 000(0.56)
Simpanan tetap Saham Simpanan semasa Hartanah Unit amanah saham 8. (a) 0.75 = total unit = 2 000 + 2 000 + 2 000
acquired = RM0.55
RM2 100 (ii) Jeremy’s strategy is more profitable
2. (a) Simple interest, I 0.032 4(3) Total unit acquired = 0.75
1.6 3 3. (a) Matured value = 5 000 1 + 4 because the average cost per share
= RM2 000 × × = 2 800 units is lower compared to Firdaus’s
100 12 = RM5 501.69
= RM8 purchase.
(b) Matured value
Total saving = RM2 000 + RM8
= RM2008 = 15 000 1 + 0.043 1(10)
1
1.8 0.0325 4(5)
(b) Simple interest, I = 3 000 × × 2 = RM22 852.53 9. (a) 10 000 = x 1 +
100 4
= RM108 (c) Matured value 10 000 = x(1.008125)
20
Total saving = RM3 000 + RM108 = 8 000 1 + 0.035 12(7) x = 10 000
= RM3 108 12 1.008125 20
2.2 9 = RM10 217.33 x = RM8 505.74
(c) Simple interest, I = 5 000 × ×
100 12 4. (a) ✓
= RM82.50 amount paid
(b) ✓ (b) (i) 1.35 =
Total saving = RM5 000 + RM82.50 (c) ✗ 2 000
= RM5 082.50 Amount paid = RM2 700
2 000x – 2 700
(ii) 8% = × 100%
2 700
5. (a) Total return = RM7 000 + RM100 – RM6 500 0.08(2 700) + 2 700 = 2 000x
= RM600 x = RM1.46 per share
600
ROI = × 100%
6 500
= 9.23% 0.032 1(2)
(c) (i) MV = 12 000 1 + 1
(b) 12.5% = (13 200 + 2x) – 12 000 × 100% = RM12 780.29
12 000
0.125(12 000) + 12 000 = 13 200 + 2x ROI = 12 780.29 – 12 000 × 100%
2x = 300 12 000
x = RM150 = 6.50%
[8 000(1.27) + 180] – x
RM1 500 – RM2 000 (ii) 2(6.50%) = × 100%
(c) (i) ROI = × 100% x
RM2 000 0.13x = 10 340 – x
= –25%
1.13x = 10 340
x = RM9 150.44
(ii) Change of government policy could cause loss in investment.
9 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 10
A6
BOOKLET ANS MATH F3.indd 6 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts
10. (a) FALSE (b) TRUE (ii) Outstanding balance (b) (i) Total repayment
(c) TRUE (d) TRUE = RM308.82 – RM16 = RM40 000 + (RM40 000 × 0.05 × 4)
= RM292.82 = RM48 000
11. Finance charge RM48 000
Make personal budget. Always late in making repayment. Start saving. Instalment =
Membuat belanjawan peribadi. Selalu membuat pembayaran balik dengan lewat. Memulakan simpanan. 18 18 4 × 12
= RM308.82 × × = RM1 000
100 365 (ii) Total repayment
18 20
+ (RM292.82 × × = RM40 000 + (RM40 000 × 0.05 × 3)
100 365
Smart management of credits and debts. = RM2.74 + RM2.89 = RM46 000
Pengurusan kredit dan hutang yang bijak. RM46 000
= RM5.63 Instalment =
3 × 12
Late payment charge = RM0 = RM1 277.78
Current amount Amount James needs to top up
Use auto billing to pay bill. Manage credit card wisely. Excessive loan. = RM292.82 + RM5.63+ RM0 = RM1 277.78 – RM1 000
Membayar bil secara automatik. Menguruskan kad kredit dengan bijak. Meminjam dengan banyak. = RM298.45 = RM277.78
15. (a) Total repayment Total
= RM110 000 + (RM110 000 × 0.05 × 9) 1 repayment
12. Late payment charge = RM0 = RM159 500 (c) (i) 4 200 × = 7 × 12
4
Total
A Current amount in April statement Instalment = RM159 500 repayment
= RM1 710 + RM30.60 + 0 9 × 12 1 050 =
= RM1 740.60 = RM1 476.85 84
Weakness (b) Total repayment Total repayment = RM88 200
of credit (b) (i) Finance charge 46 = RM80 000 + (RM80 000 × 0.07 × 5) Let r = annual interest rate
1.5
card usage = RM2 500 × 100 × 30 = RM108 000 RM88 200 = RM64 000 +
Kekurangan (RM64 000 × r × 7)
penggunaan = RM57.50 Instalment = RM108 000 r = 0.054
kad kredit D Late payment charge 5 × 12 r = 5.4%
C = (RM2 500 + RM57.50) × 1% = RM1 800
= RM25.58 (c) Total repayment (ii) Total repayment
= RM50 000 + (RM50 000 × 0.06 × 4) = RM64 000 + (RM64 000 × 0.035
Current amount in May statement = RM62 000 × 5)
= RM2 500 + RM57.50 + RM25.58 RM62 000 = RM75 200
13. (a) Finance charge = RM2 583.08 Instalment = 4 × 12 RM75 200
Caj kewangan Instalment =
(ii) Additional repayment amount = RM1 291.67 5 × 12
(b) Late payment charge = RM2 583.08 – RM2 500 = RM1 253.33
Caj bayaran lewat 16. (a) (i) Total repayment
= RM83.08 = RM81 000 + (RM81 000 × 0.08 × 5) Dahniah cannot shorten the loan
(c) Total amount of outstanding balance period because the instalment for 5
Jumlah baki tertunggak (c) (i) USD1.00 = RM3.92 = RM113 400 years is more than of her salary.
1
14. (a) Outstanding balance Toy price in Ringgit Malaysia Instalment = RM113 400 4
5 × 12
= RM1 800 – RM90 = USD78 × RM3.92 = RM1 890
= RM1 710 USD1.00
= RM305.76 (ii) Let t = new loan period
Finance charge Current amount on July statement = RM81 000 + (RM81 000
1.5
= RM1 800 × 100 × 15 = RM305.76 + (RM305.76 × 1%) × 0.08 × t)
= RM81 000 + 6 480t
30
= RM308.82
1.5
+ RM1 710 × 100 × 20 1 290 = RM81 000 + 6 480t
t × 12
30
= RM13.50 + RM17.10 15 480t = 81 000 + 6 480t
= RM30.60 t = 9 years
11 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 12
A7
BOOKLET ANS MATH F3.indd 7 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts
10. (a) FALSE (b) TRUE (ii) Outstanding balance (b) (i) Total repayment
(c) TRUE (d) TRUE = RM308.82 – RM16 = RM40 000 + (RM40 000 × 0.05 × 4)
= RM292.82 = RM48 000
11. Finance charge RM48 000
Make personal budget. Always late in making repayment. Start saving. Instalment =
Membuat belanjawan peribadi. Selalu membuat pembayaran balik dengan lewat. Memulakan simpanan. 18 18 4 × 12
= RM308.82 × × = RM1 000
100 365 (ii) Total repayment
18 20
+ (RM292.82 × × = RM40 000 + (RM40 000 × 0.05 × 3)
100 365
Smart management of credits and debts. = RM2.74 + RM2.89 = RM46 000
Pengurusan kredit dan hutang yang bijak. RM46 000
= RM5.63 Instalment =
3 × 12
Late payment charge = RM0 = RM1 277.78
Current amount Amount James needs to top up
Use auto billing to pay bill. Manage credit card wisely. Excessive loan. = RM292.82 + RM5.63+ RM0 = RM1 277.78 – RM1 000
Membayar bil secara automatik. Menguruskan kad kredit dengan bijak. Meminjam dengan banyak. = RM298.45 = RM277.78
15. (a) Total repayment Total
= RM110 000 + (RM110 000 × 0.05 × 9) 1 repayment
12. Late payment charge = RM0 = RM159 500 (c) (i) 4 200 × = 7 × 12
4
Total
A Current amount in April statement Instalment = RM159 500 repayment
= RM1 710 + RM30.60 + 0 9 × 12 1 050 =
= RM1 740.60 = RM1 476.85 84
Weakness (b) Total repayment Total repayment = RM88 200
of credit (b) (i) Finance charge 46 = RM80 000 + (RM80 000 × 0.07 × 5) Let r = annual interest rate
1.5
card usage = RM2 500 × 100 × 30 = RM108 000 RM88 200 = RM64 000 +
Kekurangan (RM64 000 × r × 7)
penggunaan = RM57.50 Instalment = RM108 000 r = 0.054
kad kredit D Late payment charge 5 × 12 r = 5.4%
C = (RM2 500 + RM57.50) × 1% = RM1 800
= RM25.58 (c) Total repayment (ii) Total repayment
= RM50 000 + (RM50 000 × 0.06 × 4) = RM64 000 + (RM64 000 × 0.035
Current amount in May statement = RM62 000 × 5)
= RM2 500 + RM57.50 + RM25.58 RM62 000 = RM75 200
13. (a) Finance charge = RM2 583.08 Instalment = 4 × 12 RM75 200
Caj kewangan Instalment =
(ii) Additional repayment amount = RM1 291.67 5 × 12
(b) Late payment charge = RM2 583.08 – RM2 500 = RM1 253.33
Caj bayaran lewat 16. (a) (i) Total repayment
= RM83.08 = RM81 000 + (RM81 000 × 0.08 × 5) Dahniah cannot shorten the loan
(c) Total amount of outstanding balance period because the instalment for 5
Jumlah baki tertunggak (c) (i) USD1.00 = RM3.92 = RM113 400 years is more than of her salary.
1
14. (a) Outstanding balance Toy price in Ringgit Malaysia Instalment = RM113 400 4
5 × 12
= RM1 800 – RM90 = USD78 × RM3.92 = RM1 890
= RM1 710 USD1.00
= RM305.76 (ii) Let t = new loan period
Finance charge Current amount on July statement = RM81 000 + (RM81 000
1.5
= RM1 800 × 100 × 15 = RM305.76 + (RM305.76 × 1%) × 0.08 × t)
= RM81 000 + 6 480t
30
= RM308.82
1.5
+ RM1 710 × 100 × 20 1 290 = RM81 000 + 6 480t
t × 12
30
= RM13.50 + RM17.10 15 480t = 81 000 + 6 480t
= RM30.60 t = 9 years
11 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 12
A7
BOOKLET ANS MATH F3.indd 7 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts
PT3 Standard Practice 3 3. Answer: D Section C HOTS Challenge
4. Answer: D 1. (a) Finance charge
Section A
5. P = RM230, r = 5.5% = 0.055 = RM900 × 1.5 × 48
1. Answer: C 100 30 (a) Total repayment
Interest = RM177.10 = RM90 000 + (RM90 000 × 0.034 × 8)
2. Initial capital Interest = Prt = RM21.60 = RM114 480
= (RM2.05 × 6 000) + (RM1.00 × 2 700) 177.10 = 230(0.055)t Late payment charge RM114 480
= RM15 000 Instalment =
177.10 = 12.65t = (RM900 + RM21.60) × 1% 8 × 12
Dividend t = 14 years = RM9.22 = RM1 192.50
= RM135 × 3 Answer: B Late payment charge = RM10.00 0.034 1(8)
= RM405 (b) Matured value = RM10 000 1 + 1
6. Finance charge Current amount in November statement
Capital gains 1.5 50 = RM900 + RM21.60 + RM10 = RM13 066.65
= RM16 500 – RM15 000 = RM1 200 × 100 × 30 = RM931.60 Car loan interest for RM10 000
= RM1 500 = RM30 = RM10 000 + (10 000 × 0.034 × 8)
5
Total returns Late payment charge (b) P = RM20 000, r = 100 per annum, = RM12 720
= RM1 500 + RM405 = (RM1 200 + RM30) × 1% t = 3 years Shamsiah should not use the money in fixed
= RM1 905 = RM12.30 (i) Half yearly, n = 2 deposit account because the interest gained
1 905
ROI = × 100 Total outstanding in August 0.05 2(3) in fixed deposit account is higher than the
15 000 = RM20 000 1 + 2 interest in car loan.
= 12.7% = RM1 200 + RM30 + RM12.30
= RM1 242.30 = RM23 193.87
Answer: A
Answer: C (ii) Quarterly, n = 4
= RM20 000 1 + 0.05 4(3)
4
Section B = RM23 215.09
1. Simple interest Simple interest (c) Down payment
Prinsipal Total savings
Prinsipal rate per annum Duration rate Jumlah simpanan = RM3 699 × 10%
Tempoh
(RM) Kadar faedah mudah Faedah mudah (RM) = RM369.90
setahun diperoleh
Amount financed
(a) 365 000 6.8% 1 year I = 24 820 389 820 = RM3 699 – RM369.90
(b) 2 500 14.3% 4months I = 119.17 2 619.17 = RM3 329.10
4 Bulan Finance charges
= 8% × RM3 329.10
= RM266.33
2. Savings How easily a savings or an asset can be converted into cash.
Simpanan Keupayaan simpanan atau aset diubah menjadi wang tunai.
An action of gaining profit in the future from income and capital
Investment gain.
Pelaburan Tindakan mendapat pulangan pada masa hadapan melalui pendapatan dan
keuntungan modal.
Liquidity Money that is kept in the bank or other than bank like piggy
Kecairan bank.
Wang yang disimpan di bank mahupun bukan bank seperti dalam tabung.
Interest Charge on the money borrowed or invested.
Faedah Caj ke atas wang yang dipinjam atau dilaburkan.
13 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 14
A8
BOOKLET ANS MATH F3.indd 8 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts Mathematics Form 3 Chapter 3 Consumer Mathematics: Saving and Invesments, Credits and Debts
PT3 Standard Practice 3 3. Answer: D Section C HOTS Challenge
4. Answer: D 1. (a) Finance charge
Section A
5. P = RM230, r = 5.5% = 0.055 = RM900 × 1.5 × 48
1. Answer: C 100 30 (a) Total repayment
Interest = RM177.10 = RM90 000 + (RM90 000 × 0.034 × 8)
2. Initial capital Interest = Prt = RM21.60 = RM114 480
= (RM2.05 × 6 000) + (RM1.00 × 2 700) 177.10 = 230(0.055)t Late payment charge RM114 480
= RM15 000 Instalment =
177.10 = 12.65t = (RM900 + RM21.60) × 1% 8 × 12
Dividend t = 14 years = RM9.22 = RM1 192.50
= RM135 × 3 Answer: B Late payment charge = RM10.00 0.034 1(8)
= RM405 (b) Matured value = RM10 000 1 + 1
6. Finance charge Current amount in November statement
Capital gains 1.5 50 = RM900 + RM21.60 + RM10 = RM13 066.65
= RM16 500 – RM15 000 = RM1 200 × 100 × 30 = RM931.60 Car loan interest for RM10 000
= RM1 500 = RM30 = RM10 000 + (10 000 × 0.034 × 8)
5
Total returns Late payment charge (b) P = RM20 000, r = 100 per annum, = RM12 720
= RM1 500 + RM405 = (RM1 200 + RM30) × 1% t = 3 years Shamsiah should not use the money in fixed
= RM1 905 = RM12.30 (i) Half yearly, n = 2 deposit account because the interest gained
1 905
ROI = × 100 Total outstanding in August 0.05 2(3) in fixed deposit account is higher than the
15 000 = RM20 000 1 + 2 interest in car loan.
= 12.7% = RM1 200 + RM30 + RM12.30
= RM1 242.30 = RM23 193.87
Answer: A
Answer: C (ii) Quarterly, n = 4
= RM20 000 1 + 0.05 4(3)
4
Section B = RM23 215.09
1. Simple interest Simple interest (c) Down payment
Prinsipal Total savings
Prinsipal rate per annum Duration rate Jumlah simpanan = RM3 699 × 10%
Tempoh
(RM) Kadar faedah mudah Faedah mudah (RM) = RM369.90
setahun diperoleh
Amount financed
(a) 365 000 6.8% 1 year I = 24 820 389 820 = RM3 699 – RM369.90
(b) 2 500 14.3% 4months I = 119.17 2 619.17 = RM3 329.10
4 Bulan Finance charges
= 8% × RM3 329.10
= RM266.33
2. Savings How easily a savings or an asset can be converted into cash.
Simpanan Keupayaan simpanan atau aset diubah menjadi wang tunai.
An action of gaining profit in the future from income and capital
Investment gain.
Pelaburan Tindakan mendapat pulangan pada masa hadapan melalui pendapatan dan
keuntungan modal.
Liquidity Money that is kept in the bank or other than bank like piggy
Kecairan bank.
Wang yang disimpan di bank mahupun bukan bank seperti dalam tabung.
Interest Charge on the money borrowed or invested.
Faedah Caj ke atas wang yang dipinjam atau dilaburkan.
13 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 14
A8
BOOKLET ANS MATH F3.indd 8 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
CHAPTER Scale Drawings 6. (a) length of drawing = 200 000 8. (a) length of drawing 1 1
Mathematics Form 3 Chapter 4 Scale Drawings
1
4
Scale = 1 :
actual length
2
1
length of drawing
Lukisan Berskala
=
200 000
2
12 000 000 = 60 cm length of object = 1
8 = 1
length of object 1
(b) Length of square field = 900 2
1. (i) 1 (b) Scale = length of drawing = 30 m
(ii) drawing / lukisan actual length Length of object = 8 × 1
(iii) 10 = 40 length of drawing = 1 = 4 cm 2
(iv) actual object / objek sebenar 4 actual length 300
(v) cm 10 length of drawing 1
= =
1 3 000 300 1.5 cm
1 3 000
= Length of drawing =
2. Diagrams Scale drawings Scale 1 300 1.0 cm 1.0 cm
Rajah Lukisan berskala Skala 10 = 10 cm 0.5 cm
1 1.0 cm
I Yes 1 : 2 = 1 : 10 3 cm
(c) Diameter of coin = 5 cm 1.5 cm
II No –
height of drawing length of drawing 1
(c) Scale = =
III No – actual height actual length 1 4 cm
4 2
IV No – = 1.4 × 100 length of drawing 1
1 5 = 1 9. (a) Length of square A = 49
1 =
V Yes 1 : 35 2 = 7 cm
2
= 1 : 35 Diameter on the drawing = 5 × 2 Length of square B = 196
= 10 cm = 14 cm
3. (a) 10 cm : 1 km 3 1 length of drawing 14
= 10 cm : 1 × 100 000 cm 5. (a) actual length = 10 000 length of drawing 1 actual length = 7
= 1: 10 000 7. (a) = 1
actual length = 3 × 10 000 actual length 100 =
= 30 000 cm base of drawing = 12 × 100 × 1 7
(b) 2 cm : 1m = 0.3 km 100 14
= 2 cm : 1 × 100 cm = 12 cm 1
= 1: 50 5 1 height of drawing = 6 × 100 × 1 = 1
(b) = 100 2
actual length 2 000 = 6 cm
(c) 5 cm : 2 km actual length = 5 × 2 000 1
= 5 cm : 2 × 100 000 = 10 000 cm Therefore, the scale used is 1 : 2
= 1 : 40 000
= 100 m
height of drawing 9 1 (b)
4. (a) Scale = (c) = 4 1
actual height actual length 1 =
2 actual distance 3 000 000
= 2 actual distance = 4 × 3 000 000 cm
600 1
1 actual length = 9 × 2 = 12 000 000 ÷ 100 000 km
= = 120 km
300 = 4.5 cm
= 1 : 300
15 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 16
A9
BOOKLET ANS MATH F3.indd 9 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
CHAPTER Scale Drawings 6. (a) length of drawing = 200 000 8. (a) length of drawing 1 1
Mathematics Form 3 Chapter 4 Scale Drawings
1
4
Scale = 1 :
actual length
2
1
length of drawing
Lukisan Berskala
=
200 000
2
12 000 000 = 60 cm length of object = 1
8 = 1
length of object 1
(b) Length of square field = 900 2
1. (i) 1 (b) Scale = length of drawing = 30 m
(ii) drawing / lukisan actual length Length of object = 8 × 1
(iii) 10 = 40 length of drawing = 1 = 4 cm 2
(iv) actual object / objek sebenar 4 actual length 300
(v) cm 10 length of drawing 1
= =
1 3 000 300 1.5 cm
1 3 000
= Length of drawing =
2. Diagrams Scale drawings Scale 1 300 1.0 cm 1.0 cm
Rajah Lukisan berskala Skala 10 = 10 cm 0.5 cm
1 1.0 cm
I Yes 1 : 2 = 1 : 10 3 cm
(c) Diameter of coin = 5 cm 1.5 cm
II No –
height of drawing length of drawing 1
(c) Scale = =
III No – actual height actual length 1 4 cm
4 2
IV No – = 1.4 × 100 length of drawing 1
1 5 = 1 9. (a) Length of square A = 49
1 =
V Yes 1 : 35 2 = 7 cm
2
= 1 : 35 Diameter on the drawing = 5 × 2 Length of square B = 196
= 10 cm = 14 cm
3. (a) 10 cm : 1 km 3 1 length of drawing 14
= 10 cm : 1 × 100 000 cm 5. (a) actual length = 10 000 length of drawing 1 actual length = 7
= 1: 10 000 7. (a) = 1
actual length = 3 × 10 000 actual length 100 =
= 30 000 cm base of drawing = 12 × 100 × 1 7
(b) 2 cm : 1m = 0.3 km 100 14
= 2 cm : 1 × 100 cm = 12 cm 1
= 1: 50 5 1 height of drawing = 6 × 100 × 1 = 1
(b) = 100 2
actual length 2 000 = 6 cm
(c) 5 cm : 2 km actual length = 5 × 2 000 1
= 5 cm : 2 × 100 000 = 10 000 cm Therefore, the scale used is 1 : 2
= 1 : 40 000
= 100 m
height of drawing 9 1 (b)
4. (a) Scale = (c) = 4 1
actual height actual length 1 =
2 actual distance 3 000 000
= 2 actual distance = 4 × 3 000 000 cm
600 1
1 actual length = 9 × 2 = 12 000 000 ÷ 100 000 km
= = 120 km
300 = 4.5 cm
= 1 : 300
15 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 16
A9
BOOKLET ANS MATH F3.indd 9 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 4 Scale Drawings Mathematics Form 3 Chapter 4 Scale Drawings
30 1 height of painting 1 Section C
(c) = PT3 Standard Practice 4 6. =
height of building 600 (1.1 × 100) 20 2 2
height of building = 30 × 600 height of painting = (1.1 100) ÷ 20 1. (a) (i) Length of LM = 15 – 12
= 18 000 cm Section A = 5.5 cm = 9 cm
= 18 000 ÷ 100 m 1. 2 cm : 1 m = (2 cm : 100 cm) ÷ 2 Answer: B Length of drawing = 1
= 180 m = 1 : 50 9 1
3
Section B Length of drawing = 9 × 3
Answer: A
10. (a) (i) Dimension: = 27 cm
Actual length = 2.5 + 1.5 + 4 = 8 m 1. Object Scale drawing Scale
Actual width = 6 m 8 1 Objek Lukisan berskala Skala 1
2. = (ii) Length of KL in drawing =
Actual height 7 500 (a) 12 1
(ii) Area of the living room = 4 × 3 Actual height = (8 × 7 500) cm 3
= 12 m 2 = 60 000 cm 18 cm 1 : 3 Length of KL in drawing = 12 × 3
= 600 m 6 cm = 36 cm
(iii) Area of bedroom 1 Answer: B (b) Area of KLM in drawing
= (3 × 4) – (1 × 0.5) 1
= 11.5 m 5 cm = × 27 × 36
2
2
3. Answer: B 10 cm 1 = 486 cm 2
Total cost of carpet 1 : 2
= 11.5 × RM60
= RM690 4. Radius of the swimming pool (b) (i) 1 cm
= (5 – 0.75 – 0.75) ÷ 2 (c) 1 cm
(iv) Height of bathroom 1 = 1.75 cm
= 2 ÷ (1 × 0.5) 1.75 1 7 m 2 7 m 2 1 : 1
= 4 m Actual radius = 400
Actual radius = 1.75 × 400 (d)
= 700 cm
= 7 m 3 m 15 cm 1 : 20 (ii)
Area of the swimming pool = pr 2 2 cm
22
= × (7) 2 2 cm
7
= 154 m 2 2. (a)
The scale drawing is bigger
Answer: D
than the object.
Lukisan berskala adalah lebih
besar daripada objek.
5. Length of KL in drawing
1 1 : 3
135 = × 15 × KL The scale drawing is the
2
135 × 2 same as the object.
KL = Lukisan berskala adalah sama 0.6825 1
15 1 saiz dengan objek. (c) Actual height = 600 000
= 18 cm 1 : 2
Actual height = 0.6825 × 600 000
Actual length of KL The scale drawing is = 409 500 cm
= (18 × 10 000) cm smaller than the object. = 4.095 km
= 180 000 cm Lukisan berskala adalah lebih
= 1.8 km kecil daripada objek.
Answer: C
(b) 5 cm : 1 m = (5 cm : 100 cm) ÷ 5
= 1 : 20
17 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 18
A10
BOOKLET ANS MATH F3.indd 10 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 4 Scale Drawings Mathematics Form 3 Chapter 4 Scale Drawings
30 1 height of painting 1 Section C
(c) = PT3 Standard Practice 4 6. =
height of building 600 (1.1 × 100) 20 2 2
height of building = 30 × 600 height of painting = (1.1 100) ÷ 20 1. (a) (i) Length of LM = 15 – 12
= 18 000 cm Section A = 5.5 cm = 9 cm
= 18 000 ÷ 100 m 1. 2 cm : 1 m = (2 cm : 100 cm) ÷ 2 Answer: B Length of drawing = 1
= 180 m = 1 : 50 9 1
3
Section B Length of drawing = 9 × 3
Answer: A
10. (a) (i) Dimension: = 27 cm
Actual length = 2.5 + 1.5 + 4 = 8 m 1. Object Scale drawing Scale
Actual width = 6 m 8 1 Objek Lukisan berskala Skala 1
2. = (ii) Length of KL in drawing =
Actual height 7 500 (a) 12 1
(ii) Area of the living room = 4 × 3 Actual height = (8 × 7 500) cm 3
= 12 m 2 = 60 000 cm 18 cm 1 : 3 Length of KL in drawing = 12 × 3
= 600 m 6 cm = 36 cm
(iii) Area of bedroom 1 Answer: B (b) Area of KLM in drawing
= (3 × 4) – (1 × 0.5) 1
= 11.5 m 5 cm = × 27 × 36
2
2
3. Answer: B 10 cm 1 = 486 cm 2
Total cost of carpet 1 : 2
= 11.5 × RM60
= RM690 4. Radius of the swimming pool (b) (i) 1 cm
= (5 – 0.75 – 0.75) ÷ 2 (c) 1 cm
(iv) Height of bathroom 1 = 1.75 cm
= 2 ÷ (1 × 0.5) 1.75 1 7 m 2 7 m 2 1 : 1
= 4 m Actual radius = 400
Actual radius = 1.75 × 400 (d)
= 700 cm
= 7 m 3 m 15 cm 1 : 20 (ii)
Area of the swimming pool = pr 2 2 cm
22
= × (7) 2 2 cm
7
= 154 m 2 2. (a)
The scale drawing is bigger
Answer: D
than the object.
Lukisan berskala adalah lebih
besar daripada objek.
5. Length of KL in drawing
1 1 : 3
135 = × 15 × KL The scale drawing is the
2
135 × 2 same as the object.
KL = Lukisan berskala adalah sama 0.6825 1
15 1 saiz dengan objek. (c) Actual height = 600 000
= 18 cm 1 : 2
Actual height = 0.6825 × 600 000
Actual length of KL The scale drawing is = 409 500 cm
= (18 × 10 000) cm smaller than the object. = 4.095 km
= 180 000 cm Lukisan berskala adalah lebih
= 1.8 km kecil daripada objek.
Answer: C
(b) 5 cm : 1 m = (5 cm : 100 cm) ÷ 5
= 1 : 20
17 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 18
A10
BOOKLET ANS MATH F3.indd 10 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 4 Scale Drawings Mathematics Form 3 Chapter 5 Trigonometric Ratio
4. (a) (i) Scale CHAPTER
= area of drawing : Actual area HOTS Challenge 5 Trigonometric Ratios
= 400 : 25 Nisbah Trigonometri
= 20 : 5 Length of bedroom 1 in the plan
= 1 : 1 = (16 – 5 – 2(2.5))
4 = 6 cm
1. (a) (i) adjacent side / sisi bersebelahan 6. (a) True
(ii) New scale Width of bedroom 1 in the plan (b) False
= Length of drawing : actual length = 3 + 3 (ii) opposite side/ sisi bertentangan
= 2 : 4 = 6 cm (iii) hypotenuse / hipotenus (c) False
(d) False
= 1 : 2
6 1 (b) (i) opposite = BC (e) True
= adjacent AB
(b) length of square = 324 Actual length 50
= 18 cm Actual length = 6 × 50 cm (ii) opposite = BC 7. (a) sin 30° = PQ
= 300 cm hypotenuse AC PR
length of drawing 1
= = 3 m adjacent AB PQ
actual length 4 (iii) hypotenuse = AC PR = 0.50
18 1 Actual width = 3 m
= PQ
actual length 4 10 = 0.50
actual length = 4 × 18 Area of bedroom 1 = 3 × 3 2. (a) 5 PQ = 0.50 × 10
= 9 m
2
= 72 cm tan θ 13 = 5 cm
Height of drawing 40 5
(c) (i) = sin θ AB
actual length 2.8 × 1 000 12 8. (a) cos x =
1 BC
=
70 cos θ 12 = 12
Scale = 1 : 70 13 13
AB : BC = 12 : 13
= 24 : BC
42 1 6 3
(ii) = 3. (a) sin θ = O = = BC = 13 × 2
length of white board 70 H 10 5
length of white board = 42 × 70 A 8 4 = 26 cm
2
= 2 940 mm (b) cos θ = H = 10 = 5 AC = 26 – 24 2
= 294 cm O 6 3 = 10 cm
(a) tan θ = = =
Length of white board : width of A 8 4 3
white board (b) sin x = 4
= 3 : 2 9 3
= 294 : 196 4. (a) cos 29° = 0.8746 AC = 4
(b) tan 80° = 5.671
Width of white board = 196 cm (c) sin 50° = 0.7660 = 9 × 4
3
Total length of red ribbon needed = 12 cm
= (294 × 2) + (196 × 2)
–1
= 980 cm 5. (a) θ = cos 0.55
= 56.63° (c) Let MB be a line perpendicular to BC
2
2
–1
(b) θ = tan 0.15 AM = 10 – 8 = 6 cm
= 8.53° MB
tan z =
(c) θ = sin 0.87 AM
–1
= 60.46° = 8
6
4
=
3
19 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 20
A11
BOOKLET ANS MATH F3.indd 11 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 4 Scale Drawings Mathematics Form 3 Chapter 5 Trigonometric Ratio
4. (a) (i) Scale CHAPTER
= area of drawing : Actual area HOTS Challenge 5 Trigonometric Ratios
= 400 : 25 Nisbah Trigonometri
= 20 : 5 Length of bedroom 1 in the plan
= 1 : 1 = (16 – 5 – 2(2.5))
4 = 6 cm
1. (a) (i) adjacent side / sisi bersebelahan 6. (a) True
(ii) New scale Width of bedroom 1 in the plan (b) False
= Length of drawing : actual length = 3 + 3 (ii) opposite side/ sisi bertentangan
= 2 : 4 = 6 cm (iii) hypotenuse / hipotenus (c) False
(d) False
= 1 : 2
6 1 (b) (i) opposite = BC (e) True
= adjacent AB
(b) length of square = 324 Actual length 50
= 18 cm Actual length = 6 × 50 cm (ii) opposite = BC 7. (a) sin 30° = PQ
= 300 cm hypotenuse AC PR
length of drawing 1
= = 3 m adjacent AB PQ
actual length 4 (iii) hypotenuse = AC PR = 0.50
18 1 Actual width = 3 m
= PQ
actual length 4 10 = 0.50
actual length = 4 × 18 Area of bedroom 1 = 3 × 3 2. (a) 5 PQ = 0.50 × 10
= 9 m
2
= 72 cm tan θ 13 = 5 cm
Height of drawing 40 5
(c) (i) = sin θ AB
actual length 2.8 × 1 000 12 8. (a) cos x =
1 BC
=
70 cos θ 12 = 12
Scale = 1 : 70 13 13
AB : BC = 12 : 13
= 24 : BC
42 1 6 3
(ii) = 3. (a) sin θ = O = = BC = 13 × 2
length of white board 70 H 10 5
length of white board = 42 × 70 A 8 4 = 26 cm
2
= 2 940 mm (b) cos θ = H = 10 = 5 AC = 26 – 24 2
= 294 cm O 6 3 = 10 cm
(a) tan θ = = =
Length of white board : width of A 8 4 3
white board (b) sin x = 4
= 3 : 2 9 3
= 294 : 196 4. (a) cos 29° = 0.8746 AC = 4
(b) tan 80° = 5.671
Width of white board = 196 cm (c) sin 50° = 0.7660 = 9 × 4
3
Total length of red ribbon needed = 12 cm
= (294 × 2) + (196 × 2)
–1
= 980 cm 5. (a) θ = cos 0.55
= 56.63° (c) Let MB be a line perpendicular to BC
2
2
–1
(b) θ = tan 0.15 AM = 10 – 8 = 6 cm
= 8.53° MB
tan z =
(c) θ = sin 0.87 AM
–1
= 60.46° = 8
6
4
=
3
19 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 20
A11
BOOKLET ANS MATH F3.indd 11 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 5 Trigonometric Ratio Mathematics Form 3 Chapter 5 Trigonometric Ratio
4 3
9. (a) cos ∠BAC = PT3 Standard Practice 5 4. sin x = 2. (a) AB
5 5 sin x = AC ✓
AB = 4 cm, AC = 5 cm Section A / Bahagian A 3 = 3
2
AD = AC + CD 3 AC 5 cos x/ kos x = BC ✓
2
1. cos x = AC = 5 cm AC
= 5 + 12 2 5
2
= 13 cm NP 3 tan y = 1.6 AB
= tan x = ✓
12 10 5 CD BC
cos ∠ADC = 3 = 1.6
13 NP = × 10 5
5
(b) (i) tan x = tan ∠DAE NP = 6 cm CD = 8 cm (b) tan x = 0.4 –1
CE = 4 Answer: B x = tan 0.4
BC 3 NK = 2 × 6 cm = 21.8°
4 = DE + 7 = 12 cm
3 12 5. Answer: A
2
2
DE + 7 = 16 NL = 10 – 6 Section C / Bahagian C
DE = 9 cm = 8 cm
NL 6. Angle between the line BK and the Plane 1. (a) (i) ∠PMQ
4 tan y =
(ii) x = tan NK ADK = ∠BKA (ii) ∠PLQ
–1
3 8
= 53.13° = AK = AD + DK
2
2
12 8
2 = 24 + 7 (b) (i) cos x =
2
2
3 = 17
(c) cos x = 3 = 25 cm NL 8
5 =
AB = 3 Answer: C 10 76.5 17
30 5 tan ∠BKA = 25 NL = 36 cm
3 2. Assume the length of sides AB and BC is
AB = × 30 = 18 cm 10
5 2 cm and AD is the height of the equilateral ∠BKA = tan –1 (ii) tan y = 1
25
2
BD = 30 – 18 triangle. ∠BKA = 21.8° NL = 1
2
= 24 cm LM
2
DC AD = 2 – 1 2 Answer: D 36
tan y = = 1
BD = LM
3
10 LM = 36 cm
= AD
24 sin 60° = Section B 36
10 AB tan z =
y = tan –1 36
3
24 1
= 1. (a) tan 45° z = tan 1
–1
= 22.62° 2 z = 45°
2
Answer: B
(b) sin 60° 1
3. SW = UW = 8 cm 2
UW 8 cos 60°
sin x = =
3
VW 17 (c) kos 60° 2
8
x = sin –1
17
(d) sin 45° 1
x = 28.07°
Answer: D
21 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 22
A12
BOOKLET ANS MATH F3.indd 12 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 5 Trigonometric Ratio Mathematics Form 3 Chapter 5 Trigonometric Ratio
4 3
9. (a) cos ∠BAC = PT3 Standard Practice 5 4. sin x = 2. (a) AB
5 5 sin x = AC ✓
AB = 4 cm, AC = 5 cm Section A / Bahagian A 3 = 3
2
AD = AC + CD 3 AC 5 cos x/ kos x = BC ✓
2
1. cos x = AC = 5 cm AC
= 5 + 12 2 5
2
= 13 cm NP 3 tan y = 1.6 AB
= tan x = ✓
12 10 5 CD BC
cos ∠ADC = 3 = 1.6
13 NP = × 10 5
5
(b) (i) tan x = tan ∠DAE NP = 6 cm CD = 8 cm (b) tan x = 0.4 –1
CE = 4 Answer: B x = tan 0.4
BC 3 NK = 2 × 6 cm = 21.8°
4 = DE + 7 = 12 cm
3 12 5. Answer: A
2
2
DE + 7 = 16 NL = 10 – 6 Section C / Bahagian C
DE = 9 cm = 8 cm
NL 6. Angle between the line BK and the Plane 1. (a) (i) ∠PMQ
4 tan y =
(ii) x = tan NK ADK = ∠BKA (ii) ∠PLQ
–1
3 8
= 53.13° = AK = AD + DK
2
2
12 8
2 = 24 + 7 (b) (i) cos x =
2
2
3 = 17
(c) cos x = 3 = 25 cm NL 8
5 =
AB = 3 Answer: C 10 76.5 17
30 5 tan ∠BKA = 25 NL = 36 cm
3 2. Assume the length of sides AB and BC is
AB = × 30 = 18 cm 10
5 2 cm and AD is the height of the equilateral ∠BKA = tan –1 (ii) tan y = 1
25
2
BD = 30 – 18 triangle. ∠BKA = 21.8° NL = 1
2
= 24 cm LM
2
DC AD = 2 – 1 2 Answer: D 36
tan y = = 1
BD = LM
3
10 LM = 36 cm
= AD
24 sin 60° = Section B 36
10 AB tan z =
y = tan –1 36
3
24 1
= 1. (a) tan 45° z = tan 1
–1
= 22.62° 2 z = 45°
2
Answer: B
(b) sin 60° 1
3. SW = UW = 8 cm 2
UW 8 cos 60°
sin x = =
3
VW 17 (c) kos 60° 2
8
x = sin –1
17
(d) sin 45° 1
x = 28.07°
Answer: D
21 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 22
A12
BOOKLET ANS MATH F3.indd 12 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 5 Trigonometric Ratio
(c) (i) cos 40° = AC CHAPTER
22 HOTS Challenge 6 Angles and Tangents of Circles
AC = 22 cos 40°
AC = 16.85 m Let the length of ladder be x Sudut dan Tangent bagi Bulatan
1.6
cos 65° =
(ii) sin 40° = BC x
22 1.6
BC = 22 sin 40° 0.4226 = x 1. 2.
BC = 14.14 m x = 3.79 m Diagrams Values
nilai
Rajah Relationship
°
DC x( )
tan 52° = between x
14.14 (a) Circles /Bulatan and y
DC = 14.14 tan 52° x Hubungan
DC = 18.10 m antara x dan y
O
40
80°
x
y
x + y = 90°
O
(b)
28
O
x° x y x = y
O
(c)
x
y
28°
42 O
x x = y
2 cm
(d)
x y
x
O
126° 90
x = 2y
6 cm
3. (a) x = 60° ÷ 2 = 30° 4. (a) ∠DCB = 150° ÷ 2 = 75°
(b) Major ∠AOC x = 180° – 75° = 105°
= 360° – 100°
= 260°
x = 260° ÷ 2 = 130° Major ∠DOB = 360° – 150°
= 210°
(c) Minor ∠POQ = 360° – 280° = 80° x = 210° ÷ 2 = 105°
∠ PRQ = 80° ÷ 2 = 40°
x = 180° – 40° =140°
23 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 24
A13
BOOKLET ANS MATH F3.indd 13 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 5 Trigonometric Ratio
(c) (i) cos 40° = AC CHAPTER
22 HOTS Challenge 6 Angles and Tangents of Circles
AC = 22 cos 40°
AC = 16.85 m Let the length of ladder be x Sudut dan Tangent bagi Bulatan
1.6
cos 65° =
(ii) sin 40° = BC x
22 1.6
BC = 22 sin 40° 0.4226 = x 1. 2.
BC = 14.14 m x = 3.79 m Diagrams Values
nilai
Rajah Relationship
°
DC x( )
tan 52° = between x
14.14 (a) Circles /Bulatan and y
DC = 14.14 tan 52° x Hubungan
DC = 18.10 m antara x dan y
O
40
80°
x
y
x + y = 90°
O
(b)
28
O
x° x y x = y
O
(c)
x
y
28°
42 O
x x = y
2 cm
(d)
x y
x
O
126° 90
x = 2y
6 cm
3. (a) x = 60° ÷ 2 = 30° 4. (a) ∠DCB = 150° ÷ 2 = 75°
(b) Major ∠AOC x = 180° – 75° = 105°
= 360° – 100°
= 260°
x = 260° ÷ 2 = 130° Major ∠DOB = 360° – 150°
= 210°
(c) Minor ∠POQ = 360° – 280° = 80° x = 210° ÷ 2 = 105°
∠ PRQ = 80° ÷ 2 = 40°
x = 180° – 40° =140°
23 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 24
A13
BOOKLET ANS MATH F3.indd 13 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 6 Angles and Tangents of Circles Mathematics Form 3 Chapter 6 Angles and Tangents of Circles
(b) ∠ACD = ∠ABD = 50° (c) x = 40° 5. x = ∠FEC = 50°
∠BCD = 50° + 60° = 110° y = 75° PT3 Standard Practice 6 ∠FOB = 360° – 78° – 90° – 50°
x = 180° – 110° = 142°
= 70° 9. (a) ∠CGF = 60° Section A 142°
(c) ∠ABC = 180° – 85° = 95° ∠FGE = 70° y = 2 = 71°
x = 180° – 40° – 95° x = 180° – 60° – 70° 1. ∠POQ = 180° – 35° – 35° = 110° ∴ x + y = 50° + 71° = 121°
= 45° = 50° ∠QOR = 180° – 63° – 63° = 54° Answer: D
x = 360° – 110° – 54°
5. (a) ∠DAE = ∠AEB = 30° (b) ∠SOT = 20° × 2 = 40° = 196° Section B
∠ADE = 180° – 30° – 70° ∠OTS = (180° – 40°) ÷ 2 = 70° Answer: B
∠QST = 110°
= 80° 1.
x = ∠ADE = 80° ∠QTS = 180° – 110° – 20° = 50° Statement
x = 70° – 50° = 20° 2. OQ = PQ – OP Pernyataan
= 8 – 5
(b) 2x + x = 180° (c) ∠EDF = 40° = 3 cm
3x = 180° ∠EDB = 60° + 40° = 100° (a) ∠OFE ≠ 90°
x = 60° QT = 5 – 3 2
2
∠ EFB = 180° – 100° = 80° (b) OB || PC ✓
y = 2 × 60° = 120° ∠DFB = 80° – 38° = 42° = 4 cm
∠DEB = ∠DFB = 42° = 40 mm (c) ∠k = ∠m ✓
(c) ∠PQR = 180° – 70° – 45° = 65° x = ∠DEB = 42° ∴ ST = 2 × QT = 80 mm
x = 180° – 65° = 115°
Answer: C (d) ∠m + ∠n = 180° ✓
6. (a) Yes
(b) No 3. ∠ODC = ∠OCD = y
(c) No = 180° – 100° 2. (a)
(d) Yes 2
(e) No y = 40° 30° x = 30°
x
∠ABC + ∠ADC = 180°
7. (a) x = 90° – 43° = 47° 120° + x + 40° = 180°
y = 43° x = 180° – 120° – 40°
x = 20° (b)
x + y = 90°
∴ x + y = 20° + 40° = 60° x O 30° x = 60°
(b) x = 90° Answer: C
y = 40°
4. ∠MNL = 68° – 35° = 33° (c)
(c) ∠ROQ = ∠POQ = 65°
x = 360° – 2(65°) = 230° x = 90° – 33° = 57° x O x = 100°
y = 90° – 65° = 25°
∠NOK = 68° × 2 100°
= 136°
8. (a) x = 60° 180° – 136°
y = 180° – 60° – 35° ∠NKO = 2 = 22° (d)
= 85° 50°
∠NKL = ∠MNL O x = 100°
22° + y = 33° x
(b) x = 70° y = 11°
y = 80° – 60°
= 20° ∴ y + x = 57° + 11° = 68°
Answer: A
25 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 26
A14
BOOKLET ANS MATH F3.indd 14 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 6 Angles and Tangents of Circles Mathematics Form 3 Chapter 6 Angles and Tangents of Circles
(b) ∠ACD = ∠ABD = 50° (c) x = 40° 5. x = ∠FEC = 50°
∠BCD = 50° + 60° = 110° y = 75° PT3 Standard Practice 6 ∠FOB = 360° – 78° – 90° – 50°
x = 180° – 110° = 142°
= 70° 9. (a) ∠CGF = 60° Section A 142°
(c) ∠ABC = 180° – 85° = 95° ∠FGE = 70° y = 2 = 71°
x = 180° – 40° – 95° x = 180° – 60° – 70° 1. ∠POQ = 180° – 35° – 35° = 110° ∴ x + y = 50° + 71° = 121°
= 45° = 50° ∠QOR = 180° – 63° – 63° = 54° Answer: D
x = 360° – 110° – 54°
5. (a) ∠DAE = ∠AEB = 30° (b) ∠SOT = 20° × 2 = 40° = 196° Section B
∠ADE = 180° – 30° – 70° ∠OTS = (180° – 40°) ÷ 2 = 70° Answer: B
∠QST = 110°
= 80° 1.
x = ∠ADE = 80° ∠QTS = 180° – 110° – 20° = 50° Statement
x = 70° – 50° = 20° 2. OQ = PQ – OP Pernyataan
= 8 – 5
(b) 2x + x = 180° (c) ∠EDF = 40° = 3 cm
3x = 180° ∠EDB = 60° + 40° = 100° (a) ∠OFE ≠ 90°
x = 60° QT = 5 – 3 2
2
∠ EFB = 180° – 100° = 80° (b) OB || PC ✓
y = 2 × 60° = 120° ∠DFB = 80° – 38° = 42° = 4 cm
∠DEB = ∠DFB = 42° = 40 mm (c) ∠k = ∠m ✓
(c) ∠PQR = 180° – 70° – 45° = 65° x = ∠DEB = 42° ∴ ST = 2 × QT = 80 mm
x = 180° – 65° = 115°
Answer: C (d) ∠m + ∠n = 180° ✓
6. (a) Yes
(b) No 3. ∠ODC = ∠OCD = y
(c) No = 180° – 100° 2. (a)
(d) Yes 2
(e) No y = 40° 30° x = 30°
x
∠ABC + ∠ADC = 180°
7. (a) x = 90° – 43° = 47° 120° + x + 40° = 180°
y = 43° x = 180° – 120° – 40°
x = 20° (b)
x + y = 90°
∴ x + y = 20° + 40° = 60° x O 30° x = 60°
(b) x = 90° Answer: C
y = 40°
4. ∠MNL = 68° – 35° = 33° (c)
(c) ∠ROQ = ∠POQ = 65°
x = 360° – 2(65°) = 230° x = 90° – 33° = 57° x O x = 100°
y = 90° – 65° = 25°
∠NOK = 68° × 2 100°
= 136°
8. (a) x = 60° 180° – 136°
y = 180° – 60° – 35° ∠NKO = 2 = 22° (d)
= 85° 50°
∠NKL = ∠MNL O x = 100°
22° + y = 33° x
(b) x = 70° y = 11°
y = 80° – 60°
= 20° ∴ y + x = 57° + 11° = 68°
Answer: A
25 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 26
A14
BOOKLET ANS MATH F3.indd 14 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 6 Angles and Tangents of Circles Mathematics Form 3 Chapter 7 Plan and Elevations
Section C 2. (a) ∠MOL = 38° × 2 CHAPTER
= 76° 7 Plans and Elevations
1. (a) (i) ON = (5 – 1) cm x = 180° – 90° – 76° Pelan dan Dongakan
= 4 cm x = 14°
y = 180° – 76°
2
NB = 5 – 4 2
= 3 cm y = 104°
1. View from Orthogonal
2
2
BD = 3 + 9 (b) x = ∠LKM × 2 Pandangan projections (b) Plan / Pelan
= 90 = 30° × 2 dari Unjuran ortogon
= 60° Plan / Pelan
3 ∠KML = 90° 2 cm
(ii) tan ∠NDB = Plan/Pelan 2 cm
9 y = 180° – 90° – 30° 4 cm
∠NDB = tan –1 3 y = 60° Y 4 cm X
9
= 18.43° (b) (i) x = 180° – 38° X
= 142°
∠ADB = 2 × ∠NDB X Y
= 36.9°
(ii) ∠MGP = x (minor arc)
= 142°
(b) (i) x = 90° – 50° ∠MGP = 360° – 142° (major arc) Y 2 cm 2 cm
= 40° = 218°
y = 218° ÷ 2 4 cm
(ii) ∠LOA = 65° × 2 = 109° 2. (a) ✓ ✗ ✓ (b) ✗ ✓ ✓ 2 cm
= 130° 3. (a) (b)
180° – 130° (iii) x + y = 142° + 109°
∠OAL = = 251°
2
= 25° (c)
(c) x = 180° – 90° – 70° 4. Plan / Pelan
∠BAL = ∠BLK
y + 25° = 50° = 20° (a) Plan / Pelan Plan / Pelan Plan / Pelan 1.25 cm 1.25 cm
y = 25°
2 cm 1.5 cm 1.5 cm
(iii) x + y = 40° + 25° HOTS Challenge
= 65° 4 cm 2 cm X 4 cm
S = 2πr = 60 cm 2 cm 4 cm
(c) ∠CAB = 40° 22 A
2 × × OE = 60 cm X 2.5 cm
7 Y
∠ ABC = (180° – 40°) ÷ 2 OE = 60 × 7 ÷ 44 = 9.55cm Y 2 cm
= 70° l 30 cm 2.5 cm
AE = 30 + 9.55 = 31.48cm X Y
2
2
∠ CBE = 180° – 70° X Y
= 110° B / E r O
x = 180° – 110° – 40° 0.8 cm 1.5 cm 1.5 cm
= 30° 4 cm 4 cm
2 cm
2 cm
4 cm
2.5 cm
2 cm 2 cm
27 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 28
A15
BOOKLET ANS MATH F3.indd 15 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 6 Angles and Tangents of Circles Mathematics Form 3 Chapter 7 Plan and Elevations
Section C 2. (a) ∠MOL = 38° × 2 CHAPTER
= 76° 7 Plans and Elevations
1. (a) (i) ON = (5 – 1) cm x = 180° – 90° – 76° Pelan dan Dongakan
= 4 cm x = 14°
y = 180° – 76°
2
NB = 5 – 4 2
= 3 cm y = 104°
1. View from Orthogonal
2
2
BD = 3 + 9 (b) x = ∠LKM × 2 Pandangan projections (b) Plan / Pelan
= 90 = 30° × 2 dari Unjuran ortogon
= 60° Plan / Pelan
3 ∠KML = 90° 2 cm
(ii) tan ∠NDB = Plan/Pelan 2 cm
9 y = 180° – 90° – 30° 4 cm
∠NDB = tan –1 3 y = 60° Y 4 cm X
9
= 18.43° (b) (i) x = 180° – 38° X
= 142°
∠ADB = 2 × ∠NDB X Y
= 36.9°
(ii) ∠MGP = x (minor arc)
= 142°
(b) (i) x = 90° – 50° ∠MGP = 360° – 142° (major arc) Y 2 cm 2 cm
= 40° = 218°
y = 218° ÷ 2 4 cm
(ii) ∠LOA = 65° × 2 = 109° 2. (a) ✓ ✗ ✓ (b) ✗ ✓ ✓ 2 cm
= 130° 3. (a) (b)
180° – 130° (iii) x + y = 142° + 109°
∠OAL = = 251°
2
= 25° (c)
(c) x = 180° – 90° – 70° 4. Plan / Pelan
∠BAL = ∠BLK
y + 25° = 50° = 20° (a) Plan / Pelan Plan / Pelan Plan / Pelan 1.25 cm 1.25 cm
y = 25°
2 cm 1.5 cm 1.5 cm
(iii) x + y = 40° + 25° HOTS Challenge
= 65° 4 cm 2 cm X 4 cm
S = 2πr = 60 cm 2 cm 4 cm
(c) ∠CAB = 40° 22 A
2 × × OE = 60 cm X 2.5 cm
7 Y
∠ ABC = (180° – 40°) ÷ 2 OE = 60 × 7 ÷ 44 = 9.55cm Y 2 cm
= 70° l 30 cm 2.5 cm
AE = 30 + 9.55 = 31.48cm X Y
2
2
∠ CBE = 180° – 70° X Y
= 110° B / E r O
x = 180° – 110° – 40° 0.8 cm 1.5 cm 1.5 cm
= 30° 4 cm 4 cm
2 cm
2 cm
4 cm
2.5 cm
2 cm 2 cm
27 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 28
A15
BOOKLET ANS MATH F3.indd 15 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 7 Plan and Elevations Mathematics Form 3 Chapter 7 Plan and Elevations
5. (a) (b) Quadrant II Quadrant I
Sukuan II Sukuan I
Quadrant II Quadrant I Side elevation Front elevation
Sukuan II Sukuan I Dongakan sisi Dongakan depan
Front elevation Side elevation
Dongakan depan Dongakan sisi A 5 cm E A / E
A 4 cm B 4 cm
A / B
2.17 cm
C / B D / F B / F 2.5 cm C / D
45°
6 cm 6 cm
F E D
C D C / D 5 cm
45°
B 1.25 cm A 1.25 cm C
Plan / Pelan
Quadrant III Quadrant IV
A / C B / D Sukuan III Sukuan IV
Plan / Pelan Quadrant IV
Quadrant III Sukuan IV
Sukuan III
29 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 30
A16
BOOKLET ANS MATH F3.indd 16 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 7 Plan and Elevations Mathematics Form 3 Chapter 7 Plan and Elevations
5. (a) (b) Quadrant II Quadrant I
Sukuan II Sukuan I
Quadrant II Quadrant I Side elevation Front elevation
Sukuan II Sukuan I Dongakan sisi Dongakan depan
Front elevation Side elevation
Dongakan depan Dongakan sisi A 5 cm E A / E
A 4 cm B 4 cm
A / B
2.17 cm
C / B D / F B / F 2.5 cm C / D
45°
6 cm 6 cm
F E D
C D C / D 5 cm
45°
B 1.25 cm A 1.25 cm C
Plan / Pelan
Quadrant III Quadrant IV
A / C B / D Sukuan III Sukuan IV
Plan / Pelan Quadrant IV
Quadrant III Sukuan IV
Sukuan III
29 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 30
A16
BOOKLET ANS MATH F3.indd 16 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 7 Plan and Elevations Mathematics Form 3 Chapter 7 Plan and Elevations
(c) Quadrant II Quadrant I 6. (a)
Sukuan II Sukuan I Quadrant II Quadrant I
Side elevation Front elevation Sukuan II Sukuan I
Dongakan sisi Dongakan depan Side elevation Front elevation
Dongakan sisi Dongakan depan
B 5 cm Q B / Q
A A
R / P A / P C / R
C / A
2 cm
B E
E / B
D / E
S / T E / T 2 cm D / S
45° 4 cm
P T Q S R
D / C C 6 cm D
6 cm
5 cm 45°
A E B D C
Quadrant III Plan / Pelan 6 cm
Sukuan III Quadrant IV
Sukuan IV C B A E D
3 cm
Plan / Pelan
Quadrant III
Sukuan III Quadrant IV
Sukuan IV
31 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 32
A17
BOOKLET ANS MATH F3.indd 17 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 7 Plan and Elevations Mathematics Form 3 Chapter 7 Plan and Elevations
(c) Quadrant II Quadrant I 6. (a)
Sukuan II Sukuan I Quadrant II Quadrant I
Side elevation Front elevation Sukuan II Sukuan I
Dongakan sisi Dongakan depan Side elevation Front elevation
Dongakan sisi Dongakan depan
B 5 cm Q B / Q
A A
R / P A / P C / R
C / A
2 cm
B E
E / B
D / E
S / T E / T 2 cm D / S
45° 4 cm
P T Q S R
D / C C 6 cm D
6 cm
5 cm 45°
A E B D C
Quadrant III Plan / Pelan 6 cm
Sukuan III Quadrant IV
Sukuan IV C B A E D
3 cm
Plan / Pelan
Quadrant III
Sukuan III Quadrant IV
Sukuan IV
31 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 32
A17
BOOKLET ANS MATH F3.indd 17 03/01/2020 10:20 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 7 Plan and Elevations Mathematics Form 3 Chapter 7 Plan and Elevations
(b) Quadrant II Quadrant I
Quadrant II Quadrant I Sukuan II Sukuan I
Sukuan II Sukuan I Front elevation Side elevation
Side elevation Front elevation Dongakan depan Dongakan sisi
Dongakan sisi Dongakan depan
P A P / A P P
3 cm 3 cm
4 cm 4 cm
C / D B / A D / A C / B
Q / T B / E T / E Q / B
3 cm 3 cm
4 cm 4 cm
F / E 6 cm G / H E / H 6 cm F / G
45°
E D A / H
R / S 6 cm C / D S / D 4 cm R / C
45°
E / D A B / C
6 cm
P
6 cm
F C B / G
2 cm 4 cm
Plan / Pelan Quadrant IV
Quadrant III Sukuan IV
Sukuan III
T / S 2 cm P 2 cm Q / R
Quadrant III Plan / Pelan
Sukuan III Quadrant IV
Sukuan IV
(c)
33 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 34
A18
BOOKLET ANS MATH F3.indd 18 03/01/2020 10:21 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 7 Plan and Elevations Mathematics Form 3 Chapter 7 Plan and Elevations
(b) Quadrant II Quadrant I
Quadrant II Quadrant I Sukuan II Sukuan I
Sukuan II Sukuan I Front elevation Side elevation
Side elevation Front elevation Dongakan depan Dongakan sisi
Dongakan sisi Dongakan depan
P A P / A P P
3 cm 3 cm
4 cm 4 cm
C / D B / A D / A C / B
Q / T B / E T / E Q / B
3 cm 3 cm
4 cm 4 cm
F / E 6 cm G / H E / H 6 cm F / G
45°
E D A / H
R / S 6 cm C / D S / D 4 cm R / C
45°
E / D A B / C
6 cm
P
6 cm
F C B / G
2 cm 4 cm
Plan / Pelan Quadrant IV
Quadrant III Sukuan IV
Sukuan III
T / S 2 cm P 2 cm Q / R
Quadrant III Plan / Pelan
Sukuan III Quadrant IV
Sukuan IV
(c)
33 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 34
A18
BOOKLET ANS MATH F3.indd 18 03/01/2020 10:21 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 7 Plan and Elevations Mathematics Form 3 Chapter 7 Plan and Elevations
7. (a) (b)
Quadrant II Quadrant I Quadrant II Quadrant I
Sukuan II Sukuan I Sukuan II Sukuan I
Side elevation Front elevation Front elevation Side elevation
Dongakan sisi Dongakan depan Dongakan depan Dongakan sisi
F / E G / H E / H F / G 6 cm 6 cm
4 cm 4 cm
7 cm
4 cm 4 cm
45°
B / A 6 cm C / D A / D 6 cm B / C
45°
4 cm
D C
6 cm
H G
2 cm
Plan / Pelan Quadrant IV
Quadrant III Sukuan IV
Sukuan III
E F
4 cm
A 6 cm B
Quadrant III Plan / Pelan
Sukuan III Quadrant IV
Sukuan IV
35 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 36
A19
BOOKLET ANS MATH F3.indd 19 03/01/2020 10:21 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 7 Plan and Elevations Mathematics Form 3 Chapter 7 Plan and Elevations
7. (a) (b)
Quadrant II Quadrant I Quadrant II Quadrant I
Sukuan II Sukuan I Sukuan II Sukuan I
Side elevation Front elevation Front elevation Side elevation
Dongakan sisi Dongakan depan Dongakan depan Dongakan sisi
F / E G / H E / H F / G 6 cm 6 cm
4 cm 4 cm
7 cm
4 cm 4 cm
45°
B / A 6 cm C / D A / D 6 cm B / C
45°
4 cm
D C
6 cm
H G
2 cm
Plan / Pelan Quadrant IV
Quadrant III Sukuan IV
Sukuan III
E F
4 cm
A 6 cm B
Quadrant III Plan / Pelan
Sukuan III Quadrant IV
Sukuan IV
35 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 36
A19
BOOKLET ANS MATH F3.indd 19 03/01/2020 10:21 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 7 Plan and Elevations Mathematics Form 3 Chapter 7 Plan and Elevations
(c) 8. (a) (i) Length/Panjang = 5 × 5 = 25 cm PT3 Standard Practice 7
Quadrant II Quadrant I Width/Lebar = 2 × 5 = 10 cm
Sukuan II Sukuan I
Front elevation Side elevation Height/Tinggi = 5 × 5 = 25 cm Section A
Dongakan depan Dongakan sisi
D / A 5 cm C / B A / B 5 cm D / C 1. Answer: B
(ii) Volume/Isi padu
1 2. Answer: B
= (10 × 25 × 5) + (10 × 25 × 20)
3
2 3. Answer: C
3
= 5416 cm
P P 3 4. Answer: B
5. Answer: D
6 cm 6 cm
25 cm
6. Answer: A
20 cm
25 cm Section B
10 cm
1. Othogonal
H / E G / F E / F H / G Diagram projection
Rajah Unjuran
ortogon
45°
(a)
A / E B / F
✓
Y
5 cm (b)
P
Y
D / H 5 cm C / G (c)
Plan / Pelan Quadrant IV
Quadrant III Sukuan IV
Sukuan III
✓
Y
(d)
Y
37 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 38
A20
BOOKLET ANS MATH F3.indd 20 03/01/2020 10:21 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 7 Plan and Elevations Mathematics Form 3 Chapter 7 Plan and Elevations
(c) 8. (a) (i) Length/Panjang = 5 × 5 = 25 cm PT3 Standard Practice 7
Quadrant II Quadrant I Width/Lebar = 2 × 5 = 10 cm
Sukuan II Sukuan I
Front elevation Side elevation Height/Tinggi = 5 × 5 = 25 cm Section A
Dongakan depan Dongakan sisi
D / A 5 cm C / B A / B 5 cm D / C 1. Answer: B
(ii) Volume/Isi padu
1 2. Answer: B
= (10 × 25 × 5) + (10 × 25 × 20)
3
2 3. Answer: C
3
= 5416 cm
P P 3 4. Answer: B
5. Answer: D
6 cm 6 cm
25 cm
6. Answer: A
20 cm
25 cm Section B
10 cm
1. Othogonal
H / E G / F E / F H / G Diagram projection
Rajah Unjuran
ortogon
45°
(a)
A / E B / F
✓
Y
5 cm (b)
P
Y
D / H 5 cm C / G (c)
Plan / Pelan Quadrant IV
Quadrant III Sukuan IV
Sukuan III
✓
Y
(d)
Y
37 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 38
A20
BOOKLET ANS MATH F3.indd 20 03/01/2020 10:21 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 7 Plan and Elevations Mathematics Form 3 Chapter 7 Plan and Elevations
2. (b) The actual plan Object B:
(a) The length of the edge KP on HOTS Challenge Plan/Pelan:
the pyramid is the same as its 1 cm
orthogonal projection. ✗ 1 cm
Panjang sisi KP pada piramid Object A:
itu adalah sama dengan unjuran L/D 2 cm K H 4 cm G/C Plan/Pelan:
ortogonnya.
(b) The size of ∠PNM on the
pyramid is the same as its Front Elevation/Dongakan depan:
orthogonal projection. ✓
Saiz ∠PNM pada piramid itu adalah
sama dengan unjuran ortogonnya.
(c) The size of ∠KLP on the pyramid I/A J E F/B Front Elevation/Dongakan Depan:
is the same as its orthogonal
projection. ✗
Saiz ∠KLP pada piramid itu adalah Plan of solid in scale 1 : 2
sama dengan unjuran ortogonnya.
1 cm Side Elevation/Dongakan sisi:
(d) The length of the edge PL on 1 cm
the pyramid is the same as its
orthogonal projection. ✗
Panjang sisi PL pada piramid L/D K H G/C
tersebut adalah sama dengan unjuran
ortogonnya. Side Elevation/Dongakan sisi:
Section C
I/A J E F/B
3. (a)
E/H F/G (c)
M/E 3 cm Z/W G/H
I/L J/K
5 cm
N/O Y/X
A/D B/C
B/A 6 cm C/D
39 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 40
A21
BOOKLET ANS MATH F3.indd 21 03/01/2020 10:21 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
Mathematics Form 3 Chapter 7 Plan and Elevations Mathematics Form 3 Chapter 7 Plan and Elevations
2. (b) The actual plan Object B:
(a) The length of the edge KP on HOTS Challenge Plan/Pelan:
the pyramid is the same as its 1 cm
orthogonal projection. ✗ 1 cm
Panjang sisi KP pada piramid Object A:
itu adalah sama dengan unjuran L/D 2 cm K H 4 cm G/C Plan/Pelan:
ortogonnya.
(b) The size of ∠PNM on the
pyramid is the same as its Front Elevation/Dongakan depan:
orthogonal projection. ✓
Saiz ∠PNM pada piramid itu adalah
sama dengan unjuran ortogonnya.
(c) The size of ∠KLP on the pyramid I/A J E F/B Front Elevation/Dongakan Depan:
is the same as its orthogonal
projection. ✗
Saiz ∠KLP pada piramid itu adalah Plan of solid in scale 1 : 2
sama dengan unjuran ortogonnya.
1 cm Side Elevation/Dongakan sisi:
(d) The length of the edge PL on 1 cm
the pyramid is the same as its
orthogonal projection. ✗
Panjang sisi PL pada piramid L/D K H G/C
tersebut adalah sama dengan unjuran
ortogonnya. Side Elevation/Dongakan sisi:
Section C
I/A J E F/B
3. (a)
E/H F/G (c)
M/E 3 cm Z/W G/H
I/L J/K
5 cm
N/O Y/X
A/D B/C
B/A 6 cm C/D
39 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 40
A21
BOOKLET ANS MATH F3.indd 21 03/01/2020 10:21 AM
Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd
CHAPTER Loci in Two Dimensions Mathematics Form 3 Chapter 8 Loci in Two Dimensions (c) Q
(c)
8
P
Lokus dalam Dua Dimensi
S Q
1. (a) The path traced by a moving wheel (c)
of a bicycle on a horizontal plane is a
straight line. R
(b) The path traced by a moving swing is
an arc of a circle.
P
(c) The locus of a spinning ball is a sphere. 4. (a)
P
(d) The locus of a rolling rod is a cylinder. O
5. (a)
2. (a) Q
S
3. (a) Q R
O
Q P
R
(b)
(b)
P S
(b) R
(b)
Q
P
Q P
P R
S
O
Q
S
41 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 42
A22
BOOKLET ANS MATH F3.indd 22 03/01/2020 10:21 AM
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Discover the best professional documents and content resources in AnyFlip Document Base.
Search