Top One (Maths F1) Penerbitan Pelangi Sdn Bhd

Top One (MATHS F1) Penerbitan Pelangi Sdn Bhd

Top One (MATHS F1) Penerbitan Pelangi Sdn Bhd

Contents

CHA CHA CHA CHA1PTER Rational Numbers 1 7.2 Linear Inequalities in One Variable 81
Nombor Nisbah PT3 Standard Practice 7 85
HOTS Challenge 87
1.1 Integers 1 MRSM & TIMSS/PISA Cloned Question, Online Quick Quiz (QR Code) 87
1.2 Basic Arithmetic Operations involving Integers 3
1.3 Positive and Negative Fractions 7 CHA CHA 8PTER Lines and Angles 88
1.4 Positive and Negative Decimals 10 Garis dan Sudut
1.5 Rational Numbers 12
PT3 Standard Practice 1 14 8.1 Lines and Angles 88
HOTS Challenge 16 8.2 Angles related to Intersecting Lines 95
TIMSS/PISA Cloned Question, Online Quick Quiz (QR Code) 17 8.3 Angles related to Parallel Lines and Transversals 96
18 PT3 Standard Practice 8 100
2PTER Factors and Multiples HOTS Challenge 103
Faktor dan Gandaan MRSM & TIMSS/PISA Cloned Question, Online Quick Quiz (QR Code) 104

9PTER Basic Polygons 105
Poligon Asas
2.1 Factors, Prime Factors and Highest Common Factor (HCF) 18
2.2 Multiples, Common Multiples and Lowest Common
Multiple (LCM) 22
PT3 Standard Practice 2 25 9.1 Polygons 105
9.2 Properties of Triangles and the Interior and
HOTS Challenge 27 Exterior Angles of Triangles 106
MRSM & TIMSS/PISA Cloned Question, Online Quick Quiz (QR Code) 27
Squares, Square Roots, Cubes and Cube Roots 28 9.3 Properties of Quadrilaterals and the Interior and
Kuasa Dua, Punca Kuasa Dua, Kuasa Tiga dan Exterior Angles of Quadrilaterals 109
3PTER PT3 Standard Practice 9 113
HOTS Challenge 116
Punca Kuasa Tiga MRSM & TIMSS/PISA Cloned Question, Online Quick Quiz (QR Code) 116

3.1 Squares and Square Roots 28 CHA CHA CHA CHA10PTER Perimeter and Area 117
3.2 Cubes and Cube Roots 33 Perimeter dan Luas
PT3 Standard Practice 3 38
HOTS Challenge 40 10.1 Perimeter 117
MRSM & TIMSS/PISA Cloned Question, Online Quick Quiz (QR Code) 40 10.2 Area of Triangles, Parallelograms, Kites and Trapeziums 119
10.3 Relationship between Perimeter and Area 123
4PTER Ratios, Rates and Proportions 41 PT3 Standard Practice 10 125
Nisbah, Kadar dan Kadaran HOTS Challenge 129
TIMSS/PISA Cloned Question, Online Quick Quiz (QR Code) 129
4.1 Ratios 41 130
4.2 Rates 42 11PTER Introduction to Set
4.3 Proportions 43 Pengenalan Set
4.4 Ratios, Rates and Proportions 44
4.5 Relationship between Ratios, Rates and Proportions 47 11.1 Set 130
49 11.2 Venn Diagrams, Universal Sets, Complement 133
with Percentages, Fractions and Decimals 51 of a Set and Subsets 139
PT3 Standard Practice 4 51 PT3 Standard Practice 11 142
HOTS Challenge HOTS Challenge 142
MRSM & TIMSS/PISA Cloned Question, Online Quick Quiz (QR Code) Online Quick Quiz (QR Code) 143

CHA CHA 5PTER Algebraic Expressions 52
Ungkapan Algebra
12PTER Data Handling
5.1 Variables and Algebraic Expressions 52 Pengendalian Data
5.2 Algebraic Expressions involving Basic Arithmetic Operations 55
PT3 Standard Practice 5 57 12.1 Data Collection, Organisation and Representation
HOTS Challenge 59 Process, and Interpretation of Data Representation 143
MRSM & TIMSS/PISA Cloned Question, Online Quick Quiz (QR Code) 59 PT3 Standard Practice 12 157

6PTER Linear Equations 60 HOTS Challenge 161
Persamaan Linear MRSM & TIMSS/PISA Cloned Question, Online Quick Quiz (QR Code) 161

13PTER The Pythagoras Theorem 162
Teorem Pythagoras
6.1 Linear Equations in One Variable 60
6.2 Linear Equations in Two Variables 64 13.1 The Pythagoras Theorem 162
6.3 Simultaneous Linear Equations in Two Variables 69 13.2 The Converse of Pythagoras Theorem 165
PT3 Standard Practice 6 74 PT3 Standard Practice 13 167
HOTS Challenge 77 HOTS Challenge 171
MRSM & TIMSS/PISA Cloned Question, Online Quick Quiz (QR Code) 77

7PTER Linear Inequalities 78 MRSM & TIMSS/PISA Cloned Question, Online Quick Quiz (QR Code) 172
Ketaksamaan Linear
CHA Year-End Assessment 173

7.1 Inequalities 78

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Top One (MATHS F1) Penerbitan Pelangi Sdn Bhd

CHAPTER

2 Factors and Multiples
Faktor dan Gandaan

2.1 Factors, Prime Factors and Highest Common Factor (HCF) Textbook
Faktor, Faktor Perdana dan Faktor Sepunya Terbesar (FSTB) pg. 30 – 45

1. Determine whether 8 is a factor for each of the following numbers. NOTES

Tentukan sama ada 8 ialah faktor bagi setiap nombor yang berikut.

Example (a) 96

64 96 ÷ 8 = 12 (No remainder)
\ 8 is a factor of 96.
64 ÷ 8 = 8 (No remainder / Tiada baki)
\ 8 is a factor of 64 / 8 ialah faktor bagi 64.

(b) 108 (c) 1 080

108 ÷ 8 = 13 remainder 4 1 080 ÷ 8 = 135 (No remainder)
\ 8 is not a factor of 108. \ 8 is a factor of 1 080.

2. Circle the factors for the following numbers. i-THiNK Circle map

Bulatkan faktor-faktor bagi nombor berikut. (b)
812
Example 812 (a)
  7 16 3 812 7 192 3
64
64 7 42 3 5
5 64
5

3. List all the factors for the following numbers.

Senaraikan semua faktor bagi nombor berikut.

Numbers / Nombor Factors / Faktor

Example   10 1, 2, 5, 10

(a) 45 1, 3, 5, 9, 15, 45

(b) 72 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

(c) 100 1, 2, 4, 5, 10, 20, 25, 50, 100

(d) 168 1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168

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4. Find all the prime factors of the following numbers. Mathematics  Form 1  Chapter 2 Factors and Multiples 

Cari semua faktor perdana bagi setiap nombor yang berikut. (a) 54
2 54
Example 3 27
3  9
30 Repeated divisions with the smallest prime 3  3
number until the quotient becomes 1.
2 30 Pembahagian berulang dengan nombor  1
3 15 perdana terkecil sehingga hasil bahagi 54 = 2 × 3 × 3 × 3
5  5 menjadi 1. The prime factors of 54 are 2
 1
and 3.
30 = 2 × 3 × 5

The prime factors of 30 are 2, 3 and 5.

Faktor perdana bagi 30 ialah 2, 3 dan 5.

(b) 85 (c) 165 (d) 294

 5 85 3 165 2 294
3 147
17 17 5  55 7  49
7   7
 1 11  11   1
85 = 5 × 17   1 294 = 2 × 3 × 7 × 7
The prime factors of 294 are
The prime factors of 85 are 5 165 = 3 × 5 × 11 2, 3 and 7.
The prime factors of 165 are
and 17.
3, 5 and 11.

5. List all the common factors for each of the following.

Senaraikan semua faktor sepunya bagi setiap yang berikut.

Example (a) 10, 15 (b) 14, 28
2, 8
Factors of Factors of
Factors of / Faktor bagi 10 : 1 , 2, 5 , 10 14 : 1 , 2 , 7 , 14
2 : 1, 2 15 : 1 , 3, 5 , 15 28 : 1 , 2 , 4, 7 , 14 , 28
8 : 1 , 2 , 4, 8 Common factors:
Common factors : 1 and 2 Common factors : 1 and 5 1, 2, 7 and 14

Faktor sepunya : 1 dan 2

(c) 6, 9, 18 (d) 8, 21, 42 (e) 13, 26, 39, 65

Factors of Factors of Factors of
6 : 1 , 2, 3 , 6 13 : 1 , 13
9 : 1, 3, 9 8 : 1 , 2, 4, 8 26 : 1 , 2, 3, 13 , 26
18 : 1 , 2, 3 , 6, 9, 18 39 : 1 , 3, 13 , 39
Common factors : 1 and 3 21 : 1 , 3, 7, 21 65 : 1 , 5, 13 , 65
Common factors : 1 and 13
42 : 1 , 2, 3, 6, 7, 14, 21, 42

Common factor : 1



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  Mathematics  Form 1  Chapter 2 Factors and Multiples

6. Find the highest common factor (HCF) for each of the following.

Cari faktor sepunya terbesar (FSTB) bagi setiap yang berikut.

Example (a) 36, 68 (b) 60, 80

  12, 50 2 36 68 2 60 80
2 18 34 2 30 40
2 12 50  9 17 5 15 20
 6 25 HCF : 2 × 2 = 4  3  4
HCF / FSTB : 2 HCF : 2 × 2 × 5 = 20

(c) 16, 40, 72 (d) 60, 75, 150 (e) 56, 84, 112

2 16 40 72 3 60 75 150 2 56 84 112
2  8 20 36 5 20 25  50 2 28 42  56
2  4 10 18  4  5  10 7  4 21  28
 2  5  9 HCF : 3 × 5 = 15  2  3  4
HCF : 2 × 2 × 2 = 8 HCF : 2 × 2 × 7 = 28

7. Solve each of the following problems.

Selesaikan setiap masalah yang berikut.

Example
Given x is a common factor of 50 and 70. State the possible values of x.

Diberi x ialah satu faktor sepunya bagi 50 dan 70. Nyatakan nilai-nilai yang mungkin bagi x.

Factors of / Faktor bagi 50 :  1 , 2 , 5 , 10 , 25, 50

70 :  1 , 2 , 5 , 7, 10 , 14, 35, 70

The possible values of x = 1, 2, 5 and 10.

Nilai-nilai yang mungkin bagi x = 1, 2, 5 dan 10.

(a) Calculate the difference between the highest and the lowest common factors for the following numbers.

Hitung beza antara faktor sepunya terbesar dan faktor sepunya terkecil bagi nombor berikut.

9, 18 and  / dan 27.
Factors of   9 : 1 , 3 , 9

18 : 1 , 2 , 3 , 6 , 9 , 18
27 : 1 , 3 , 9 , 27

Common factor = 1, 3, 9

Largest value – smallest value = 9 – 1

=8

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Mathematics  Form 1  Chapter 2 Factors and Multiples 

(b) Calculate the sum of all the prime factors of 195.

Hitung hasil tambah kesemua faktor perdana bagi 195.

3 195

5  65

13  13

  1

The sum of prime factors = 3 + 5 + 13

= 21

(c) In a training session of a school sports day, the number of male and female students in blue team is 90
and 108 respectively. Mr. Lee asked them to line up in rows according to their gender. If the number
of students in each row is the same, what is the maximum number of students in a row?

Dalam suatu sesi latihan hari sukan sebuah sekolah, bilangan murid lelaki dan murid perempuan pasukan biru masing-
masing ialah 90 dan 108 orang. Encik Lee meminta mereka berbaris dalam beberapa barisan mengikut jantina masing-
masing. Jika bilangan murid dalam setiap baris adalah sama, berapakah bilangan maksimum murid dalam satu barisan?

Applying, Analysing


2 90 108
9 45  54

 5   6
Maximum number of students in a row
=2×9
= 18 students

(d) Encik Rahmat has 72 cabbage seedlings and 81 chilli seedlings. He wants to plant all the seedlings into

several rows according to their types. Given the number of seedlings in every row are same. Calculate

the minimum number of rows that will be formed. Applying, Analysing

Encik Rahmat ada 72 anak benih kubis dan 81 anak benih cili. Dia ingin menanam semua anak benih itu dalam beberapa
barisan mengikut jenis masing-masing. Diberi bilangan anak benih dalam setiap baris adalah sama. Hitung bilangan
minimum baris yang akan dibentuk.

3 72 81
3 24 27
 8  9
Minimum number of rows
=8+9
= 17 rows

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  Mathematics  Form 1  Chapter 2 Factors and Multiples

2.2 Multiples, Common Multiples and Lowest Common Multiple (LCM)
Gandaan, Gandaan Sepunya dan Gandaan Sepunya Terkecil (GSTK)

NOTES

8. Tick (3) if the number in the brackets is a common multiple of the following numbers and (7) if otherwise.

Tandakan (3) jika nombor di dalam kurungan adalah gandaan sepunya bagi nombor berikut dan (7) jika tidak.

Example (126) (a) 28, 56 (112)
3
16, 42 7

126 ÷ 16 = 7 remainder / baki 14 112 ÷ 28 = 4
126 ÷ 42 = 3 112 ÷ 56 = 2

126 is not a multiple of 16 because it cannot be exactly divided by 16.
126 bukan gandaan bagi 16 kerana tidak dapat dibahagi tepat dengan 16.

(b) 9, 18, 27 (324) (c) 17, 28, 68 (544) (d) 6, 18, 21, 63 (126)
3 3
  324 ÷ 9 = 36 7   126 ÷ 6 = 21
324 ÷ 18 = 18 126 ÷ 18 = 7
324 ÷ 27 = 12 544 ÷ 17 = 32 126 ÷ 21 = 6
126 ÷ 63 = 2
544 ÷ 28 = 19 remainder

12

544 ÷ 68 = 8

9. Fill in the blanks with the given multiples. i-THiNK Double bubble map

Isi tempat kosong berikut dengan nombor gandaan yang diberi.

Example (a) 4, 5, 8, 10, 12, 15, 20, 40, 60 5
2, 3, 4, 6, 8, 9, 12, 15, 18

26 4 20
3

8 4 40 5 10

4 2 12 3 9

8 12 60 15
18 15

(b) 6, 8, 12, 16, 18, 24, 32, 48, 72 (c) 14, 21, 28, 42, 56, 63, 84, 105, 126

8 24 6 14 42 21

16 8 48 6 12 28 14 84 21 63

32 72 18 56 126 105


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Mathematics  Form 1  Chapter 2 Factors and Multiples 

10. Find the lowest common multiple (LCM) of the following whole numbers.

Cari gandaan sepunya terkecil (GSTK) bagi nombor bulat yang berikut.

Example (a) 8, 12

6, 9 2 8 12
2 4  6
Divide the numbers by a factor that can divide at least 2 2  3
3 1  3
one of the number. Move down the number that cannot 1  1
LCM = 2 × 2 × 2 × 3
36 9 be divided and continue the division until get 1. = 24
22 3 Bahagi nombor dengan faktor yang boleh bahagi
31 3 sekurang-kurangnya satu daripada dua nombor yang
diberi. Turunkan nombor yang tidak dapat dibahagi dan
11 teruskan pembahagian sehingga mendapat 1.


LCM / GSTK = 3 × 2 × 3 = 18

(b) 15, 60 (c) 2, 4, 9 (d) 15, 21, 30


3 15 60 22 4 9 3 15 21 30
5  5 20 21 2 9 5  5  7 10
4  1  4 31 1 9 2  1  7  2
 1  1 31 1 3 7  1  7  1
LCM = 3 × 5 × 4 111  1  1  1
= 60 LCM = 2 × 2 × 3 × 3 LCM = 3 × 5 × 2 × 7
= 36 = 210

11. Solve each of the following problems.

Selesaikan setiap masalah yang berikut.

Example

  Diagram below shows some multiples of a number which is less than 100. State the other three multiples

which are not listed. Analysing

Rajah di bawah menunjukkan sebahagian gandaan suatu nombor yang kurang daripada 100. Nyatakan tiga gandaan lain
yang tidak disenaraikan.

39 13 65 78
        

Difference between / Beza antara:
13 and / dan 39 = 26
39 and / dan 65 = 26
65 and / dan 78 = 13

Therefore, the list above is the multiples of 13.

Maka, senarai di atas adalah gandaan bagi 13.

The multiples of 13 which are less than 100 are 13, 26, 39, 52, 65, 78 and 91.

Gandaan bagi 13 yang kurang daripada 100 ialah 13, 26, 39, 52, 65, 78 dan 91.

Therefore, the three multiples which are not in the list are 26, 52 and 91.

Oleh itu, tiga gandaan yang tiada dalam senarai di atas ialah 26, 52 dan 91.

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  Mathematics  Form 1  Chapter 2 Factors and Multiples

(a) x is a common multiple of 5 and 6. Given the value of x is less than 67. State the number of possible
values of x.

x ialah satu gandaan sepunya bagi 5 dan 6. Diberi nilai x adalah kurang daripada 67. Nyatakan bilangan nilai yang
mungkin bagi x.

Multiple of 5 : 5, 10, 15, 20, 25, 30 , 35, 40, 45, 50, 55, 60 and 65.
Multiple of 6 : 6, 12, 18, 24, 30 , 36, 42, 48, 54, 60   and 66.

Common multiples of 5 and 6 are 30 and 60.

Number of possible values of x is 2.

(b) On the launch day of Butik Maya, lucky customers got special gifts for free. The 11th, 22nd, 33rd, ...

(multiples of 11) customers received a shirt each, while the 31st, 62nd, 93rd, ... (multiples of 31) customers

received a cap each. Danny was the first customer to get a shirt and a cap. What is Danny’s turn? Why

Danny got both of the free gifts? Give your reason. Applying, Analysing

Pada hari pelancaran Butik Maya, pelanggan bertuahnya mendapat hadiah istimewa secara percuma. Setiap pelanggan
ke-11, ke-22, ke-33, … (gandaan 11) mendapat sehelai baju, manakala setiap pelanggan ke-31, ke-62, ke-93, …
(gandaan 31) mendapat sebuah topi. Danny merupakan pelanggan pertama yang mendapat sehelai baju dan sebuah
topi. Berapakah nombor giliran Danny? Mengapakah Danny mendapat kedua-dua hadiah percuma tersebut? Berikan
alasan anda.

11 11 31
31  1 31
 1  1

LCM of 11 and 31 = 11 × 31

= 341

Danny’s turn is 341. Danny got both free gifts because his number, 341 is the lowest common multiple

of 11 and 31.

(c) Jessie is preparing a gift containing pencils and sweets for each of her classmates. A shop sells pencils
in packages of 15 and sweets in packages of 20. She wants to wrap the same number of pencil and
sweets in each gift. What is the minimum number of packages of sweets Jessie has to buy?

Jessie menyediakan sebuah hadiah yang mengandungi pensel dan gula-gula untuk rakan-rakannya. Sebuah kedai menjual

15 batang pensel sekotak dan 20 biji gula-gula sebungkus. Dia ingin membungkus pensel dan gula-gula dalam kuantiti
yang sama dalam setiap bungkusan hadiah. Berapakah bilangan bungkusan gula-gula yang minimum perlu dibeli oleh
Jessie?

Applying, Evaluating

5 15 20
3  3  4
4  1  4
 1  1

LCM of 15 and 20 = 5 × 3 × 4

= 60

The number of packages of sweets Jessie has to buy = 60 ÷ 20 = 3 packages

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Mathematics  Form 1  Chapter 2 Factors and Multiples 

PT3 Standard Practice 2

Section A / Bahagian A Section B / Bahagian B

1. Which of the following shows how 48 can be 1. Match the prime factors for each of the following
numbers.
expressed as a product of prime factors?
Padankan faktor perdana bagi setiap nombor yang berikut.
Antara yang berikut, yang manakah menunjukkan cara 48
dapat diungkapkan sebagai hasil darab faktor perdana? [4 marks / 4 markah]
Answer / Jawapan:
A 2 × 3
B 2 × 2 × 3 2
C 2 × 2 × 2 × 2 × 3 14
D 2 + 2 + 2 + 2 + 3
3
2. Which of the following are the common factors of 27

10 and 15? 5
39
Antara yang berikut, yang manakah faktor sepunya bagi
10 dan 15? 7

A 1 and / dan 2 2. Complete the steps to find the prime factors of 72
B 1 and / dan 3 using the factor tree method.
C 1 and / dan 5
D 1 and / dan 10 Lengkapkan langkah-langkah untuk mencari faktor-faktor
perdana bagi 72 menggunakan kaedah pokok faktor.
3. The highest common factor (HCF) of 60, 75 and
[4 marks / 4 markah]
150 is 3a. Find the value of a. Answer / Jawapan:
Faktor sepunya terbesar (FSTB) bagi 60, 75 dan 150 ialah
3a. Cari nilai a. 72

A 3 C 10 89

B 5 D 15 2433

4. The lowest common multiple (LCM) of 2, m and 22

9 is 36. Find the value of m.

Gandaan sepunya terkecil (GSTK) bagi 2, m dan 9 ialah
36. Cari nilai m.

A 2 C 5

B 4 D 6

5. Which of the following are the common multiples Section C / Bahagian C

of 8 and 24? 1. (a) The list below are the factors of 128 which are
arranged in descending order.
Antara yang berikut, yang manakah gandaan sepunya bagi
8 dan 24? Senarai di bawah ialah faktor bagi 128 yang disusun
mengikut tertib menurun.
A 2, 4 C 24, 42

B 8, 16 D 48, 120

6. Determine the lowest common multiple (LCM) of 128, n, 32, 16, m, 4, 2, 1

4, 6 and 14. (i) State the values of m and n.
Tentukan gandaan sepunya terkecil (GSTK) bagi 4, 6 dan
14. Nyatakan nilai m dan nilai n.

A 2 C 42 [2 marks / 2 markah]

B 24 D 84 Answer / Jawapan:
m = 8, n = 64

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  Mathematics  Form 1  Chapter 2 Factors and Multiples

(ii) By using the values of m and n in (a)(i), Answer / Jawapan:
find the highest common factor (HCF) of
both values. 2 30 50

Dengan menggunakan nilai m dan nilai n di 3 15 25
(a)(i), cari faktor sepunya terbesar bagi kedua-
dua nilai tersebut. 5 5 25

[1 mark / 1 markah] 515

Answer / Jawapan: 11

2 8 64 LCM = 2 × 3 × 5 × 5 = 150

2 4 32 The minimum number of packet of chocolates

2 2 16 = 150 ÷ 50 = 3

1  8 2. (a)
HCF = 2 × 2 × 2 = 8 The common multiples of 8 and 12 are

(b) Express the following in the form of prime Gandaan sepunya bagi 8 dan 12 ialah
factorisation.
x, 48, 72, y, z, ...
Ungkapkan setiap yang berikut dalam bentuk
pemfaktoran perdana. From the list of common multiples above,
find the values of x, y and z.
(i) 56     (ii) 170
[4 marks / 4 markah] Daripada senarai gandaan sepunya di atas, cari nilai-
nilai x, y dan z.
Answer / Jawapan:
(i) 56 [3 marks / 3 markah]

78 Answer / Jawapan:

24 x: 24   y: 96   z: 120

2 2 (b) The highest common factor (HCF) and the
56 = 2 × 2 × 2 × 7 lowest common multiple (LCM) of P and 60
(ii) 170 are 6 and 180 respectively. Find the smallest
possible value of P.
17 10
Faktor sepunya terbesar (FSTB) dan gandaan sepunya
2 5 terkecil (GSTK) bagi P dan 60 masing-masing ialah 6
170 = 2 × 5 × 17 dan 180. Cari nilai terkecil yang mungkin bagi P.

(c) Susan wants to wrap a gift containing pens [3 marks / 3 markah]
and chocolates for each teacher for Teacher’s
Day. A shop sells 30 pens for each packet and Answer / Jawapan:
50 chocolates for each packet. She wants to HCF = 6 = 2 × 3
wrap the same number of pens and chocolates LCM = 180 = 2 × 2 × 3 × 3 × 5
in each gift. Calculate the minimum number 60 = 2 × 2 × 3 × 5
of packets of chocolates Susan has to buy. P = 2 × 3 × 3 = 18

Applying (c) Three traffic lights along a street turn red at

Susan ingin membungkus sebuah hadiah yang the intervals of 1 minute, 1 minute 10 seconds
mengandungi pen dan coklat untuk setiap guru
sempena Hari Guru. Sebuah kedai menjual 30 batang and 1 minute 18 seconds. Occasionally, all
pen sebungkus dan 50 biji coklat sebungkus. Dia ingin
membungkus pen dan coklat dalam kuantiti yang three traffic lights will turn red simultaneously
sama dalam setiap hadiah. Hitung bilangan bungkus
coklat yang minimum perlu dibeli oleh Susan. at 0930. Find the next time when it occurs

[3 marks / 3 markah] again. Applying

Tiga buah lampu isyarat sepanjang jalan bertukar
merah pada julat 1 minit, 1 minit 10 saat dan 1 minit 18
saat. Pada keadaan tertentu, ketiga-tiga lampu isyarat
tersebut akan bertukar merah secara serentak pada
0930. Cari masa yang seterusnya apabila keadaan ini
berlaku sekali lagi.

[4 marks / 4 markah]

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Answer / Jawapan: Mathematics  Form 1  Chapter 2 Factors and Multiples 
1 minute = 60 seconds
1 minute 10 seconds = 70 seconds LCM = 2 × 2 × 3 × 5 × 7 × 13
1 minute 18 seconds = 78 seconds = 5 460 s
= 1 hour 31 minutes
2 60 70 78
2 30 35 39 0930 + 1 hour 31 minutes = 1101
3 15 35 39
5 5 35 13
7 1 7 13
13 1 1 13

111

HOTS Challenge

A fruit seller has 60 apples, 72 oranges and 108 guavas. He packed all the fruits into bags with the same number of each type

of fruits. Applying

Seorang penjual buah mempunyai 60 biji epal, 72 biji oren dan 108 biji jambu batu. Dia membungkus semua buah itu ke dalam beg dengan
bilangan yang sama bagi setiap jenis buah.

(a) Find the maximum number of bags he packed.
Cari bilangan maksimum beg yang telah dibungkusnya.

2 60 72 108

2 30 36 54

3 15 18 27

569
HCF = 2 × 2 × 3 = 12

(b) How many guavas are there in each bag?
Berapakah bilangan jambu batu yang ada di dalam setiap beg?

Number of guavas in each bag

= 108
12
=9

HOTS Extra

MRSM TIMSS/PISA
Cloned Question Cloned Question

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CHAPTER

1 Rational Numbers
Nombor Nisbah

1. (a) 4. (a) 2 , 0, 0.5, 0.8
3

(b) 1.4, 5 4 , –8, –17.5
7

(c) 4, 5.6, 1.7, 5
2

–25°C (d) 1 000, 0.01, 100.1, 1
(b)
(e) 1 , 44.5, 24.5, 5
2

5. (a) 2 3 (b) –5 3

–2 m 6. (a)
(c)

–15 –6 –3 0 3 6

(b)

–10 –5 0 5 15 25

+60°C (c)

2. (a) –5 3 –35 –21 –14 –7 7 14

(b) –2 3 7. (a) p = −9, q = –4, r = 11
(b) p = −5, q = 0, r = 10
(c) +5 3 (c) p = 0, q = 12, r = 24

3. (a) Integer 8. (a) 21, 5, 0, –7, –24
(b) Not integer (b) –1 010, –111, –110, 10, 1 001
(c) Integer (c) 8 008, 88, –880, –8 808, –80 000
(d) Not integer
(e) Integer 9. (a) 6 + (–4) = 2
(b) –2 + (+5) = 3
(c) 11 – (+4) = 7
(d) –4 – (–5) = 1
(e) –8 – (+5) = –13
(f) –6 + (–5) = –11

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Mathematics Form 1 Answers –27 112 ÷ 8
14
10. 42 3 × (–9)
6×7 9
48 –12 × (–4)
36 ÷ (–4) –9
–48 –98 ÷ 7
–27 ÷ (–3) –14
(b) 2 999 × 14
–8 × 6 = 14 × 2 999
= 14 × (3 000 – 1)
11. (a) (i) –5 + 18 = 14 × 3 000 – 14
(ii) 13 = 42 000 – 14
(iii) (–3) × (–6) = 18 = 41 986

(b) (i) 6 – (–7) – 8 (c) (100 + 27) + 73
(ii) 13 – 8 = 100 + (27 + 73)
(iii) 5 = 100 + 100
(iv) 21 ÷ (–3) = –7 = 200
(v) 6 – (–7) = 13
(d) 2 × 100 + 2 × 45 + 2 × 5
(c) (i) –14 + 10 = 2 × (100 + 45 + 5)
–4 = 2 × (150)
= 300
(ii) –4
–4 (e) 15 × 29 × 40
= 15 × 40 × 29
(iii) 1 = 600 × 29
= 600 × (30 – 1)
(iv) (–2) × (–5) = 10 = 600 × 30 – 600 × 1
28 – 32 = –4 = 18 000 – 600
= 17 400
12. (a) Yes
(b) No 14. (a) (34 + 7) – (–18 + 5)
(c) No = 41 – (–13)
(d) Yes = 54 m
(e) Yes The distance between the aeroplane and
the shark is 54 m.
13. (a) 2 030 × 25
= 25 × 2 030 2
= (2 000 + 30) × 25
= 2 000 × 25 + 30 × 25
= 50 000 + 750
= 50 750

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Mathematics Form 1 Answers

(b) 25 × 8 ÷ 40 (b) 5 ÷ 1– 14 + 9 2
= 200 ÷ 40 7 15 10
=5
Each children will receive 5 sweets. = 5 ÷ 1 –28 + 27 2
7 30
(c) 8 – 20 + 5
= –12 + 5 = 5 ÷ 1– 1 2
= –7°C 7 30
The final temperature of the solution is
–7°C. = 5 × (–30)
7
15. (a)
= – 150
0 7
– —32 – —31 – —61 —61
= –21 3
7

(c) 2 3 × 1 2 – 7 2
4 5 10

= 11 × 1 4–7 2
4 10

(b) = 11 × 1– 3 2
4 10
– —141
0 1—41 11—11 11—51 = – 33
40

(c) (d) 3 + 1– 4 2 × 1– 2 2 – 1 3
5 5 3 5
–1 —81 – —43 – —38
= 3 + 1 4 × 2 2 – 8
5 5 3 5
0
—34
= 3 + 8 – 8
3 5 15 5

= 9 + 8 – 24
15 15 15
16. (a) 11 , 3 , 2 , – 1 , – 7
14 8 7 3 9 = 17 – 24
15 15
2 4 3 2 1
(b) – 3 , – 9 , – 7 , – 5 , – 3 = – 7
15
11 3 5 4 5
(c) 12 , 4 , 8 , – 9 , – 7

(e) – 10 × 1 2 – 121 ÷ 2 1
3 7 15 5

1 217.(a) 1 3 × 2 – 5 = – 10 × 9 – 121 ÷ 11
4 3 12 3 7 15 5

= 7 × 1 8–5 2 = – 90 – 121 × 5
4 12 21 15 11

= 7 × 3 = – 9 – 11
4 12 21 3

= 7 × 1 = –72201
4 4

= 7
16

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Mathematics Form 1 Answers 21. (a) 7.3 – (– 4.9) – 1.8
= 7.3 + 4.9 – 1.8
18. (a) The number of questions not answered = 12.2 – 1.8
= 50 – 35 – 8 = 10.4
= 7 questions
(b) 3.9 × (– 4.8) ÷ 5.2
Set Hui’s total mark = –18.72 ÷ 5.2
= –3.6
1 2=35 × 3 – 8 × 1 1 – 7 × 1
2 5 (c) 12.8 + (–0.52) ÷ [14.6 + (–10.6)]
= 12.8 + (–0.52) ÷ (4)
= 105 – 12 – 7 = 12.8 – 0.13
5 = 12.67

= 91 3 marks (d) –0.6 × (7.15 – 0.7 + 0.07)
5 = –0.6 × (6.45 + 0.07)
= –0.6 × 6.52
(b) Money spent on shoes = 160 × 5 = –3.912
8
(e) (43.25 – 17.5) ÷ (–6.4 + 5.9)
= RM100 = 25.75 ÷ (–0.5)
= –51.5
Remaining money = 160 – 100
= RM60

Money spent on stationery = 60 × 1
5
= RM12

\ Remaining money = 60 – 12
= RM48

(c) The mass of flour in container R 22. (a) Total mass
= (2.15 × 8) + (0.84 × 5) + 0.15
= 4 × 11 – 3 2 × 350 = 17.2 + 4.2 + 0.15
5 7 = 21.55 kg

= 4 × 1 4 × 3502
5 7

= 4 × 200 3 17
5 7 7
23. (a) 2 =
= 160 g

19. (a) P Q R 2 3 is a rational number
2.25 2.45 7

1.05 1.15 1.25 1.4 3 (b) –0.54 = – 54
100
33 R
0.98 1.05 –27
(b) P = 50
3
Q
R
–0.15 –0.14 0.7 0.8 0.75 3.2 –0.54 is a rational number.

33 3 (c) –6 = –6
1
(c) P Q
–6 is a rational number
2.3 –0.5 –0.25 3.0

33 (d) –2.4 = –24
4.8 48

20. (a) –0.101, –0.027, 0.085, 0.402 = –1
(b) 54.645, 54.564, –54.456, –54.546 2
(c) 89.880, 88.809, –88.098, –88.908
(d) –8.41, –6.32, 3.85, 7.24, 9.47 –2.4 is a rational number
4.8

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Mathematics Form 1 Answers

(e) – 4 = –4 PT3 Standard Practice 1
9 9

– 4 is a rational number Section A
9 1. Answer: C

24. (a) 1 + 3.5 × 2 = 1 + 7 2. 3 7 – (–5) – 3.5 = 5.28
3 3 3 3 9

= 8 3 7 – (–5) + 3.5 = 12.28
3 9

= 2 2 3 7 + (–5) + 3.5 = 2.28
3 9

3 7 + (–5) – 3.5 = –4.72
9
7 1 7 5 13
(b) 2 ÷ 4.2 + 3 4 = 2 × 21 + 4 Answer: B

= 5 + 13 3. Answer: C
6 4
( )4.1 9 3
= 20 + 78 6 – × 7 = – 10
24
1 – = – 3 7
= 4112 6 10 × 9

1 – = – 7
6 30

1 2(c) 3 – 0.8 × 1 + 7.2 = 1 + 7
5 4 6 30

= 0.6 – 0.8 × 7.45 = 2
5
= 0.6 – 5.96

= –5.36 Answer: C

25. (a) Housing loan for a month 5. –12.4 + 4.5 ×5
= RM34 560 ÷ 12 –6.6 – 0.4
= RM2 880
= – 7.9 × 5
–7
2
5 × Monthly salary = 2 880 = 79
14
Monthly salary = 2 880 × 5
2 = 5 9
14
= RM7 200
Monthly salary of Nora is RM7 200. Answer: C

(b) 13.6 + (13.6 + 0.3) + (13.6 – 0.4) + 10.9 6. –5 – 9 = –14
= 51.6 s Answer: B
Therefore, they cannot complete the
run within 50 s as they took 51.6 s to Section B
finish.
1. (a) – 3
5
1
(b) 5

(c) –1.75

(d) 2.82

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Mathematics Form 1 Answers (b) 225 × 40 × 39
= (200 + 25) × 40 × (40 – 1)
2. 9 × 96 = 9 × (100 – 4) = (200 × 40 + 25 × 40) × (40 – 1)
= 9 × 100 – 9 × 4 = (8 000 + 1 000) × (40 – 1)
= 900 – 36 = 9 000 × (40 – 1)
= 864 = 9 000 × 40 – 9 000 × 1
= 360 000 – 9 000
Section C
= 351 000
1. (a) x: –0.05
y: –9.25 (c) Science members Mathematics members
z: 1.725

(b) The temperature after 30 minutes 16 girls
= 30 + (–1.5 × 30)
= 30 – 45 2 units → 16 Science members
= –15oC 1 unit → 8 Science members
The liquid has reached the freezing 5 units → 40 Science members
point, –15oC < –10oC. Thus, the liquid
has frozen. 4 units → 40 Science members
1 unit → 10 Science members
(c) (i) Perimeter 5 units → 50 Mathematics members
( ) ( )= 2 7 7 1 The total number of Mathematics
9 + 2 9 – 3 Society members is 50.

= 22 m HOTS Challenge
9

= 244.44 cm

(ii) Number of pieces of cakes received

by each guest

= (4 × 5) ÷ 8
1
= 2 2

2. (a) (i) Positive integer = 1 The remaining sweets
Negative integer = –1
1 – 7 = 5
12 12

–1.75 –1 0 1 The number of sweets given to Selena

(ii) Largest integer = 10 3 × 5 = 1
Smallest integer = –17 5 12 4
10 × –17 = –170
The number of sweets given to Jolin

1 – 7 – 1 = 1
12 4 6

1 → 72 sweets
6

7 → 7 ÷ 1 × 72 = 252 sweets
12 12 6

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Mathematics Form 1 Answers

CHAPTER

2 Factors and Multiples
Faktor dan Gandaan

1. (a) 96 ÷ 8 = 12 (No remainder) (c) 3 165
\ 8 is a factor of 96. 5 55

(b) 108 ÷ 8 = 13 remainder 4 11 11
\ 8 is not a factor of 108. 1

(c) 1 080 ÷ 8 = 135 (No remainder) 165 = 3 × 5 × 11
\ 8 is a factor of 1 080. The prime factors of 165 are 3, 5 and
11.
2. (a)
(d) 2 294
812 3 147
7 42 3 7 49
77
64 1
5 294 = 2 × 3 × 7 × 7
The prime factors of 294 are 2, 3 and 7.
(b)
812 5. (a) Factors of
10 : 1 , 2, 5 , 10
7 192 3 15 : 1 , 3, 5 , 15
64 Common factors : 1 and 5
5
(b) Factors of
3. (a) 1, 3, 5, 9, 15, 45 14 : 1 , 2 , 7 , 14
(b) 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 28 : 1 , 2 , 4, 7 , 14 , 28
(c) 1, 2, 4, 5, 10, 20, 25, 50, 100 Common factors: 1, 2, 7 and 14
(d) 1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42,
56, 84, 168 (c) Factors of
6 : 1 , 2, 3 , 6
4. (a) 2 54 9 : 1, 3, 9
3 27
39 18 : 1 , 2, 3 , 6, 9, 18
33 Common factors : 1 and 3

1 (d) Factors of
8 : 1 , 2, 4, 8
54 = 2 × 3 × 3 × 3
The prime factors of 54 are 2 and 3. 21 : 1 , 3, 7, 21
42 : 1 , 2, 3, 6, 7, 14, 21, 42
(b) 5 85 Common factor : 1
17 17
1

85 = 5 × 17
The prime factors of 85 are 5 and 17.

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Mathematics Form 1 Answers (c) 2 90 108
9 45 54
(e) Factors of 5
13 : 1 , 13 6
26 : 1 , 2, 3, 13 , 26
39 : 1 , 3, 13 , 39 Maximum number of students in a row
65 : 1 , 5, 13 , 65 =2×9
Common factors : 1 and 13 = 18 students

6. (a) 2 36 68 (d) 3 72 81
2 18 34 3 24 27
17 8 9
9
Minimum number of rows
HCF : 2 × 2 = 4 =8+9
= 17 rows
(b) 2 60 80
2 30 40 8. (a) 3
5 15 20 112 ÷ 28 = 4
3 4 112 ÷ 56 = 2

HCF : 2 × 2 × 5 = 20 (b) 3
324 ÷ 9 = 36
(c) 2 16 40 72 324 ÷ 18 = 18
2 8 20 36 324 ÷ 27 = 12
2 4 10 18
259 (c) 7
544 ÷ 17 = 32
HCF : 2 × 2 × 2 = 8 544 ÷ 28 = 19 remainder 12
544 ÷ 68 = 8
(d) 3 60 75 150
5 20 25 50 (d) 3
4 5 10 126 ÷ 6 = 21
126 ÷ 18 = 7
HCF : 3 × 5 = 15 126 ÷ 21 = 6
126 ÷ 63 = 2
(e) 2 56 84 112
2 28 42 56 9. (a)
7 4 21 28 4 20 5
23 4

HCF : 2 × 2 × 7 = 28

7. (a) Factors of 9 : 1 , 3 , 9 8 4 40 5 10
18 : 1 , 2, 3 , 6, 9 , 18
27 : 1 , 3 , 9 , 27 12 60 15
(b)
Common factor = 1, 3, 9
8 24 6
Largest value – smallest value
=9–1=8 16 8 48 6 12

(b) 3 195 32 72 18
5 65

13 13
1

The sum of prime factors
= 3 + 5 + 13
= 21

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(c) Mathematics Form 1 Answers

14 42 21 (b) 11 11 31
31 1 31
28 14 84 21 63 11
LCM of 11 and 31 = 11 × 31
56 126 105 = 341
Danny’s turn is 341. Danny got both
10. (a) 2 8 12 free gifts because his number, 341 is the
24 6 lowest common multiple of 11 and 31.
22 3
31 3 (c) 5 15 20
11 33 4
LCM = 2 × 2 × 2 × 3 41 4
= 24 11

(b) 3 15 60 LCM of 15 and 20 = 5 × 3 × 4 = 60
5 5 20 The number of sweets packages Jessie
41 4 has to buy = 60 ÷ 20 = 3 packages
11
LCM = 3 × 5 × 4 PT3 Standard Practice 2
= 60
Section A 48
(c) 2 2 4 9 1.
21 2 9
31 1 9 4 12
31 1 3
111 2226
LCM = 2 × 2 × 3 × 3
= 36 Answer: C 23

(d) 3 15 21 30 2. Factors of
5 5 7 10 10: 1 , 2, 5 , 10
21 7 2 15: 1 , 3, 5 , 15
71 7 1 Answer: C
111
3. 3 60 75 150
LCM = 3 × 5 × 2 × 7
= 210 5 20 25 50

11. (a) Multiple of 5 : 5, 10, 15, 20, 25, 30 , 4 5 10
35, 40, 45, 50, 55, 60 and 65.
HCF = 3 × 5 = 15
Multiple of 6 : 6, 12, 18, 24, 30 , 36,
42, 48, 54, 60 and 66. 3a = 15
15
Common multiples of 5 and 6 are 30 a = 3 = 5
and 60.
Answer: B
Number of possible values of x is 2.
4. Choose m = 4

22 4 9
21 2 9
31 1 9
31 1 3

111

LCM = 2 × 2 × 3 × 3 = 36
Answer: B

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Mathematics Form 1 Answers

5. 48 ÷ 8 = 6 (ii) 170
48 ÷ 24 = 2
120 ÷ 8 = 15 17 10
120 ÷ 24 = 5
25
Answer: D 170 = 2 × 5 × 17

6. 2 4 6 14 (c) 2 30 50
2 2 37 3 15 25
3 1 37 5 25
7 1 17 5 5
1 11 1 5
1 1

LCM = 2 × 2 × 3 × 7 = 84 LCM = 2 × 3 × 5 × 5 =150
The minimum number of packets of
Answer: D
chocolate = 150 ÷ 50 = 3

Section B 2. (a) x: 24, y: 96, z: 120
1.
14 2 (b) HCF = 6 = 2 × 3
3 LCM = 180 = 2 × 2 × 3 × 3 × 5
27 5 60 = 2 × 2 × 3 × 5
7 P = 2 × 3 × 3 =18
39
(b) 2 60 70 78
2. 72 2 30 35 39
3 15 35 39
8 9 5 5 35 13
7 1 7 13
2 43 3 13 1 1 13
11 1
22
1 minute = 60 seconds
Section C 1 minute 10 seconds = 70 seconds
1 minute 18 seconds = 78 seconds

LCM = 2 × 2 × 3 × 5 × 7 × 13
= 5 460 s
= 1 hour 31 minutes

0930 + 1 hour 31 minutes = 1101

1. (a) (i) m = 8, n = 64 HOTS Challenge

(ii) 2 8 64 (a) 2
2 4 32 2
2 2 16 3
1 8
60 72 108
HCF = 2 × 2 × 2 = 8 30 36 54
15 18 27
(b) (i) 56 5 69

78 HCF = 2 × 2 × 3 = 12

24 (b) Number of guavas in each bag

22 = 108
56 = 2 × 2 × 2 × 7 12

=9

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CHAPTER Mathematics Form 1 Answers

3 Squares, Square Roots, Cubes and Cube Roots

Kuasa Dua, Punca Kuasa Dua, Kuasa Tiga dan Punca Kuasa Tiga

1. (a) 3 × 3 = 9 unit2 (c) 900 = √(–30) × (–30)
(b) 4 × 4 = 16 unit2
(c) 5 × 5 = 25 unit2 = 30

2. (a) Yes 11 4. (a) –122
121 = –(12 × 12)
= –144
11

121 = 11 × 11 (b) (– 0.6)2
= (–0.6) × (–0.6)
(b) Yes = 0.36
196
(c) 1– 25 2
14 14
2 72 7 11

196 = 2 × 7 × 2 × 7 = 1– 1512 × 1– 1512

= 25
121

(c) Not 90 (d) 2.52

9 10 = 52
3 22
32 5
= 25
4
90 = 3 × 3 × 2 × 5

(d) Yes 225 (e) 1 3 22
4

9 25 = 32
3 35 5 42

= 9
16
225 = 3 × 5 × 3 × 5

3. (a) 49 = √7 × 7 5. (a) –3.92
=7 = –(3.9 × 3.9)
= –15.21
(b) √121 = 11 × 11
= 11 (b) (–6.1)2
= (–6.1) × (–6.1) = 37.21

(c) 23.42
= 23.4 × 23.4
= 547.56

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Mathematics Form 1 Answers

(d) 1– 29 2 7. (a) √71 = 8.43

23 (b) √432 = 20.78

= 1– 2932 × 1– 2932 = 81
529
(c) √0.025 = 0.16

(e) 14 7 22 (d) 4 = 0.29
8 49

= 13892 × 13892 (e) 61101 = 2.63

= 1521 8. (a) 900
64

= 234649 (b) 10 000

6. (a) √100 (c) 0.16

(d) 1
10 000
= 10 × 10 = 10

(b) √0.16 9. (a) √65 is between √64 and √81, that is,
= 0.4 × 0.4 = 0.4 √65 is between 8 and 9. Thus, √65 ≈ 8.

(c) √225 (b) √30 is between √25 and √36, that is,
= 15 × 15 √30 is between 5 and 6. Thus, √30 ≈ 5.

= 15 10. (a) √70 × √70
= (√70)2
= 70

(b) √2 × √32
= √2 × 32
= √64 = 8

(d) 4 = 22 (c) 6 × 6
49 72 16 16

22 =1 26 2
=
16
72
= 6
16

= 2 = 3
7 8

18 9 (d) 1 × √45
32 16 1 125
(e) =
1
= 1 125 × 45

32 = 1
= 25

42 = 1
5
3
= 4

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Mathematics Form 1 Answers

11. (a) The length of chess board (b) 144
=8×3 4 36
= 24 cm
22 4 9
Area of chess board
= 242 2 23 3
= 24 × 24
= 576 cm2 144 = 2 × 2 × 2 × 2 × 3 × 3

(b) Length of paper \ 144 is not a perfect cube.
= √702.25
= 26.5 cm (c) 216
54
The number of sides of square 4
produced 22 69
= 26.5 ÷ 5 cm 2 33 3
= 5.3
≈ 5 sides of square 216 = 2 × 3 × 2 × 3 × 2 × 3
\ 216 is a perfect cube.
The number of squares can be cut
=5×5 15. (a) 3√27 = 3√3 × 3 × 3 = 3
= 25 squares
(b) 3√–0.001
(c) The area of coloured paper that may be = 3√(–0.1) × (–0.1) × (–0.1)
used is 169 cm2 or 196 cm2 or 225 cm2. = –0.1
√169 cm2 = 13 cm
√196 cm2 = 14 cm (c) 3 64 =3 4 × 4 × 4 = 4
√225 cm2 = 15 cm 125 5 5 5 5
The length of coloured paper that may
have to be used is 13 cm or 14 cm or (d) 3 216 = 3√8 or 3 216 = 3√216
15 cm. 27 27 3√27

= 3√2 × 2 × 2 = 6
3
12. (a) 6 × 6 × 6 =2

=2

(b) (–0.5) 3

(c) (–11) × (–11) × (–11) 16. (a) 53 = 5 × 5 × 5
= 125
1 1 2 3
4 (b) 0.43 = 0.4 × 0.4 × 0.4
(d) = 0.064

(e) 16.8 × 16.8 × 16.8

13. (a) 3 1 2(c)2 3= 2 × 2 × 2
(b) 3 5 5 5 5
(c) 7
= 8
125

14. (a) 320 17. (a) 323 = 32 768
8 40 (b) 643 = 262 144
(c) (– 4.1)3 = –68.921
24 4 10

2 2 2 22 5

320 = 2 × 2 × 2 × 2 × 2 × 2 × 5 1 2(d)– 7 3 = – 343
11 1 331
\ 320 is not a perfect cube.
1 2(e) 3 3 205641
2 4 =

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Mathematics Form 1 Answers

18. (a) 3 22170 Volume of water after added
= 62.5 + 37.5
=3 64 = 100 m3
27
Height of water in the tank
=3 4 × 4 × 4 = 4 = 100 ÷ 25
3 3 3 3 =4m

(b) 3 32 (c) Estimation of volume of a cube is :
108 1) 6 000 ÷ 81 ≈ 74 cm3
2) 2 000 ÷ 81 ≈ 25 cm3
= 3 8
27 Length of a cube :
1) 3√74 ≈ 4.2 cm
=3 2 × 2 × 2 = 2 2) 3√25 ≈ 2.9 cm
3 3 3 3
The length of a cube is a whole number
19. (a) 3√100 = 4.64 which is more than 2.9 cm and less
(b) 3√0.05 = 0.37 than 4.2 cm, therefore the length that
(c) 3√9.08 = 2.09 might be used is 3 cm or 4 cm.

22. (a) 4 1 + 3√512
144
(d) – 5 = –0.68
3 16 1
12
= 4 × + 8

(e) 3 23 1 = 2.85 = 1 + 8
5 3

20. (a) 2.63 is between 23 and 33, that is, 2.63 = 8 1
is between 8 and 27. Thus, 2.63 ≈ 27. 3

(b) 3√84 is between 3√64 and 3√125, that is, 1 2(b)3 + 2 2 × √64
3√84 is between 4 and 5. Thus, 3√84 ≈ 4. 4

21. (a) Perfect cubes : 1, 8, 27, 64, 125, … = 1 11 22 × 8
The difference in volume of the two 4
dice is 37 cm3.
8–1 =7 = 121 × 8
27 – 8 = 19 16
64 – 8 = 56, 64 – 27 = 37
= 60.5
3√64 = 4 cm and 3√27 = 3 cm
Therefore, the length of side of each die (c) (√0.04 × 3√–8)3
is 3 cm and 4 cm respectively. = (0.2 × (–2))3
= (–0.4)3
(b) Length of the tank = 3√125 = –0.064
=5m
(d) 3 343 × 42
Area of the base = 5 × 5 8
= 25 m2
= 3√343 × 16
Initial volume of water in the tank 3√8
= 125 ÷ 2 7
= 62.5 m3 = 2 × 16

= 56

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Mathematics Form 1 Answers

PT3 Standard Practice 3 2. (a) 3
(b) 0.81
Section A (c) –3
(d) 8
1. √8 = 2.83
√82 = 9.056 Section C
√100 = 10
√512 = 22.63 1. (a) 216; 343; 512
Answer: C
(b) (i) 7 – 4 = 47
4 9 36
( ) ( )(ii)
2. Sides of the square 8 – 1 2= 7 2= 49
= 64 ÷ 4 9 2 18 324
= 16 cm
Area of the square (c) 3√3.375 = 1.5 m = 150 cm
= 16 × 16 Base area = 150 cm × 150 cm
= 256 = 22 500 cm2

Answer: C 2. (a) (i) √2 025 = √5 × 5 × 9 × 9

9 = √45 × 45 = 45
4
3. √8p – 4 × = –1 (ii) p = 3√512 = 8

√8p – 9 = –1 p = 8 = 4
2 2
√8p = 8
8p = 64 (b) (i) 32 = 4 = 2
72 9 3
p=8
8
Answer: D (ii) 125 – 1.2 × 100

4. 1 3√27 = 3 3 = 8 – 120
II 3√48 = 3.63 7 125
III 3√215 = 5.99 7
IV 3√343 = 7 3 = –119.936
= –119.94 (2 d.p)
Answer: B
(c) The volume of the big cube

5. 3√8 = 3√(–2) × (–2) × (–2) = 60 × 60 × 60
Answer: A = 216 000 cm3

The volume of each small cube

6. 3√64 = 4 = 216 000
Answer: C 125

= 1 728 cm3

The length of side of each small cube

Section B = 3√1 728
= 12 cm
1. 3
5 3 5116
2 8 + 4 – HOTS Challenge
81
3 16 (a) √400 = 20 students in a row
(b) 400 = 144 + 256
27
= 8– √144 = 12 students in a row
√256 = 16 students in a row
= 3 – 9
2 4 15 © Penerbitan Pelangi Sdn. Bhd.

= – 3 A
4

CHAPTERMathematics Form 1 Answers

4 Ratios, Rates and Proportions
Nisbah, Kadar dan Kadaran

1. (a) 7 : 10 : 14 Rate = 486 N
(b) 3 : 9 : 10 2 m2
(c) 3 : 7 : 21
6. (a) 10 m/s = 10 m
2. (a) 2 : 5 : 8 ; 8 : 20 : 32 1 s
(b) 3 : 7 : 16 ; 9 : 21 : 48
(c) 6 : 9 : 27 ; 2 : 3 : 9 = 10 m ÷ 1 s

3. (a) 800 g : 0.5 kg : 0.25 kg = 10 km ÷ 60 1 60 h
800 g : 500 g : 250 g 1 000 ×
16 : 10 : 5
Equivalent ratios: = 10 × 60 × 60
(i) 32 : 20 : 10 1 000
(ii) 48 : 30 : 15
(Any possible equivalent ratios.) = 36 km/h

(b) (b) 1 025 kg/m3

= 1 025 kg
1 m3

= (1 025 × 1 000) g
1m×1m×1m

= 100 1 025 × 1 000 cm
cm × 100 cm × 100
34
6 12 = 1.025 g/cm3

Equivalent ratios: i) 6 : 4 (c) 350 km/h = 350 km
ii) 3 : 2 1h

(Any possible equivalent ratios.) = 350 × 1 000 m
1 × 60 × 60 s

4. (a) 42 cm : 5.5 m : 200 mm = 97.2 m/s
= 420 mm : 5 500 mm : 200 mm
= 21 : 275 : 10 7. (a) 500 g = 2 250 g 3
2 9 7
1 1 3 9
(b) 2 : 6 : 4 = 3 : 1 : 2 (b) 4.5 m≠ 9 m
3 5
=6:2:9

(c) 3.2 : 6.4 : 4.8 8. (a) 7 × 4.5 = xg
= 0.4 : 0.8 : 0.6 2 × 4.5 9g
= 0.2 : 0.4 : 0.3
=2:4:3 7 × 4.5 = 31.5

The mass of fat in 7 packets of snacks

is 31.5 g.

5. (a) Quantity : (i) Speed (b) RM0.50 × 16 = x
(ii) Time 1 m3 × 16 16 m3

Rate = 943 m/s 0.50 × 16 = 8
20 s
The water tariff of 16 m3 of water used
(b) Quantity : (i) Force is RM8.
(ii) Area

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Mathematics Form 1 Answers

(c) 2 ×2 = x 12. (a) The number of students who do not
8 ×2 16 bring dictionary
= 30 – 12 = 18 students
2×2=4

The building has 4 levels. Let the percentage of students who do
not bring dictionary is x.
9. (a) 6 : 11 : 15
(b) 6 : 21 : 35 Ratio of number of students who do not
(c) 8 : 20 : 45 bring dictionary to total students

= 18 : 30

10. (a) Let A = The number of women 18 = x
B = The number of men 30 100
C = The number of children
x= 18 × 100 = 60
30

A:B=2:1 A:C=3:1 There are 60% of students who do not
=6:3 =6:2
bring dictionary to school every day.
B:A=3:6
13. (a) (i) Savings in a month
\B:A:C=3:6:2
3
(b) Let A = The price of a comic = 2 000 – 500 – 5 × 2 000
B = The price of a novel
C = The price of a dictionary = 2 000 – 500 – 1 200 = RM300

A:B=2:5 B:C=4:7 Ratio of savings to salary
= 8 : 20 = 20 : 35 = 300 : 2 000
= 3 : 20

\ A : B : C = 8 : 20 : 35 (ii) Let the percentage of money saved

11. (a) Ben Sister in a month is x.
=5 : 4
3 : 20
Difference in part = 5 – 4 3 x
=1 20 = 100

1 part = 3 years x = 3 × 100
Ben’s age = 5 × 3 20

= 15 years old x = 15%
Ben is 15 years old.
15% of Mr. Lai’s monthly salary is saved
in his bank account.

(b) Let the mass of 3 types of cookies is a, (b) (i) Increase rate of year 2010-2014
b and c.
ab c = 18 000 – 15 000
= 2:3:5 15 000

2 + 3 + 5 = 10 = 3 000
10 parts = 2 000 g 15 000

1 part = 2 000 = 200 g = 1
10 5

= 0.2

a = 2 × 200 = 400 g Let the population for year 2014-
b = 3 × 200 = 600 g
c = 5 × 200 = 1 000 g 2018 is x.

The mass of each types of cookies is 400 g, 1 = x
600 g and 1 000 g respectively. 5 18 000

x = 3 600 people

Estimated population in 2018
= 18 000 + 3 600
= 21 600 people

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Mathematics Form 1 Answers Latifah’s age now
= 2 × 6 = 12 years
(ii) Ratio of population, Mother’s age now
2010 : 2014 = 15 000 : 18 000 = 9 × 6 = 54 years
=5:6 In 6 years’ time
2014 : 2018 = 18 000 : 21 600 = (54 + 6) : (12 + 6)
=5:6 = 60 : 18
= 10 : 3
If the rate of increase in population Answer: D
is the same, then the ratios of the
population will be the same. Section B

PT3 Standard Practice 4 1. 20 : 12 21 : 6
5 : 10
Section A 8 : 10 12 : 15
15 : 9
1. p : q = 4 : 5, q : r = 4 : 9 9 : 18
p : q = 4 × 4 : 5 × 4 = 16 : 20
q : r = 4 × 5 : 9 × 5 = 20 : 45 14 : 4
p : q : r = 16 : 20 : 45
p : r = 16 : 45

Answer: C

2. Alvin : Benjamin = 3 : 4 2. (a) (i) 7
The ratio of Benjamin’s mass to the total (ii) 3
mass = 4 : 7
(b) (i) 36
Answer: D (ii) 54

3. F : G = 5 × 5 : 3 × 5 = 25 : 15 Section C
G : H = 5 × 3 : 4 × 3 = 15 : 12
1. (a) 15 units → RM2 700
F : G : H = 25 : 15 : 12 1 unit → RM180
F : H = 25 : 12 (6 – 4) units → 2 × RM180 = RM360

Answer: A mass 44.8
volume 5
4. Length : Width = 7 : 2 (b) (i) Density = =
Perimeter = 2 × (7 + 2) = 18 units
= 8.96 g/cm3

18 units → 108 cm (ii) 5kg = 5 000 g
1 m2 100 × 100 cm2
1 unit → 108 = 6 cm
18 = 50 g per 100 cm2

Length = 7 × 6 = 42 cm (c) Fare for the first 1.5 km = RM3.60
Width = 2 × 6 = 12 cm Remaining distance = 8 km – 1.5 km
Area = 42 × 12 = 504 cm2 = 6.5 km
= 6 500 m
Answer: C Fare for the next 6 500 m
= (6 500 ÷ 100) × RM0.50
5. 1 hour = 3 600 seconds = 65 × RM0.50 = RM32.50
24 : 3 600
1 : 150 RM3.60 + RM32.50 = RM36.10

Answer: B 2. (a) (i) 8  → 106 km
1  → 106 ÷ 8 = 13.25 km
6. Total units = 2 + 9 = 11 units The rate of petrol used is
78 – 6 – 6 = 66 years
11 units → 66 years 13.25 km per litre.
1 unit → 6 years

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(ii) 1  → 13.25 km Mathematics Form 1 Answers
12  → 12 × 13.25 km
= 159 km 3 units → 24 stickers
1 unit → 8 stickers
(iii) 13.25 km → 1  (8 + 2) units → 80 stickers

530 km → 530 = 40  HOTS Challenge
13.25
Before: Male : Female = 6 : 5
(b) Cynthia → 2 units × 4 = 8 units After: Male : Female = (3 : 2) × 2 = 6 : 4
The ratio of Cynthia, Mary and Natalia
(5 – 4) = 1 unit → 8
after Natalia gave 3 units 6 + 5 = 11 units → 88 guests

=8:2:4

CHAPTER

5 Algebraic Expressions
Ungkapan Algebra

1. (a) c 5. (a) (i) The number of apples owned by
(b) y Siti = 26 – d
(c) k
(ii) 26 – d = 26 – 8 = 18 apples

2. (a) Varies. The quantity of sugar depends (b) (i) Let the numbers of
on the customer’s taste.
male students = a,
(b) Fixed. The distance between Ali’s house
and school is always the same. female students = b

(c) Varies. The room temperature of a The number of students who do
kitchen increases when it is under use. 2
not wear glasses = a– 4 + 3 b

(ii) The number of students who do

not wear glasses

3. (a) The difference = 10 – z = 12 – 4 + 2 (24)
(b) Total number of pen = c + d + e 3
(c) Number of sunny days = 7 – x
= 12 – 4 + 16

= 24 students

4. (a) d2e3 + 11d 6. (a) 908t
= (4)2(–1)3 + 11(4)
= 28 (b) 1 x2
8

(b) 7v – (5 + vw) (c) –kh + 0.3
3w
(d) 3s – 1

= 7(–4) – [5 + (–4)(–1)] (e) 3 y + 2.3t
3(–1) 5

= 9 1 –9 (f) 7y + 2y – 4y
3
(g) 1 + 2e – 3f
= 1
3 (h) 0.5a – 3b + 11c

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A1

Mathematics Form 1 Answers

7. (a) 2m 14n 10. (a) 7
is coefficient of (b) 3
as (c) 3
(d) 7
Relating factor –3m (e) 7
–21n

(b) p 1 11. (a) 6y
is coefficient of as 2
Relating factor q (b) –24k3p3
2r pq
(c) r (c) (–3a)
is coefficient of 3a
Relating factor 1 (d) 7p
2 as 2 8
b
12a (e) 4 xy2
b 9

12. (a) a2b × (–2ab) × 3a
= –6a4b2

8. (b) 6m2n ÷ 3n2 × 4mn

21
6m2n × 4mn
= 3n2 = 8m3

(c) 18x2y ÷12x ÷ 4y

9
18x2y
= 2x × 4y

1

= 9x
4

(d) 35r2st ÷ (7rs × 2t)

9. (a) ab – bc + 3ab 5 5r
= ab + 3ab – bc 35r2st 2
= 4ab – bc = 7rs × 2t =

1

(b) 3 m – 1 n + 1 m (e) 105m2n2 × 5m ÷ 6mn
4 4 2
35
105m2n2 × 5m
= 3 m + 1 m – 1 n = 6mn
4 2 4
2

= 5 m – 1 n = 175m2n
4 4 2

(c) 0.5pq – 3qr – 0.7pq PT3 Standard Practice 5
= 0.5pq – 0.7pq – 3qr
= –0.2pq – 3qr Section A

(d) –2xy + 4yz – 2xy – 3yz 1. Answer: C
= –2xy – 2xy + 4yz – 3yz
= –4xy + yz 2. Answer: C

(e) 3p + 4q – 5 + 6q + 1 3. –x × x = – 2x
= 3p + 4q + 6q – 5 + 1 –2 × 2 = –2(2)
= 3p + 10q – 4 –4 = –4

Answer: C

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Mathematics Form 1 Answers

4. (12 – 5p) – (p – 4) (c) (i) 1 500 + 150M
= 12 – 5p – p + 4
= 12 + 4 – 5p – p (ii) RM1 500 + RM150(6) = RM2 400
= 16 – 6p He did not manage to buy that
smart phone as he was short of
Answer: B RM199.

5. (y – 7)2 = (y – 7)(y – 7) 2. (a) x3 – (5xz – 3y) = x3 – 5xz + 3y
Answer: B = x3 + 3y – 5xz
= –7 – 5(–2)
27x3y4 × –3x2y =3
3xy
6. (b) (i) p= RM30 – RM0.50 = RM5.90
5
= 27 × x × x × x × y× y×y× y× –3 × x × x × y
3 ×x× y
(ii) 4.60(3) + 1.80(6) = RM24.60

= –27x4y4 (c) (3 + 2m)3 = (3 + 2m)(3 + 2m)(3 + 2m)
The length of the cube
Answer: A = 3 + 2m
Total surface area of the cube
Section B = (3 + 2m) × (3 + 2m) × 6
= 6(3 + 2m)2 cm2
1. 25x, 3 r, q , –2g
4 2

2. (a) –3b2c HOTS Challenge
(b) ac
(c) –3b (a) Amount of money spent
(d) c = [(35 × a) + (56 × b)] sen
= (35a + 56b) sen
Section C
(b) 35 ÷ 5 = 7 and 56 ÷ 8 = 7
1. (a) (i) (4p – q)3 He packed 7 bags altogether.
The amount sold
(ii) (4p – q)3 = [4(–2) – 1]3 = 7(8a + 9b) sen
= (–9)3 = (56a + 63b) sen
= –729
(c) Amount of money he earned
(b) (4m2 + 8n – 1) – (m2 + 3n – 2) + (7m2 – 2) = (56a + 63b) sen – (35a + 56b) sen
= 4m2 + 8n – 1 – m2 – 3n + 2 + 7m2 – 2 = (56a + 63b – 35a – 56b) sen
= 4m2 – m2 + 7m2 + 8n – 3n – 1 + 2 – 2 = (56a – 35a + 63b – 56b) sen
= 10m2 + 5n – 1 = (21a + 7b) sen
= am2 – bn – c

a = 10, b = –5, c = 1

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A1

CHAP Mathematics Form 1 Answers

TER

6 Linear Equations
Persamaan Linear

1. (a) No (c) 4x – 6 = 10
(b) No 3
(c) Yes
(4x – 6) × 3 = 10 × 3
3

2. (a) Let the number be p. 12x – 18 = 10
4p + 5 = 20
12x – 18 + 18 = 10 + 18
(b) Let the length of the square is x.
2x + 2(x – 7) = 33 12x = 28
12 12
(c) Let m be the number of pages read.
130 – 4m = 45 x = 7
3

3. (a) The product of 7 with a number which 5. (a) Let the mass of Anson is x.
has been added by 4 is 12. Then, mass of Ali = x + 7
and mass of Amir = x – 6
(b) RMq is equally divided to 3 siblings.
Each of them received an amount of Their total mass
RM88. = x + (x + 7) + (x – 6)
= 3x + 1
(c) Perimeter of an equilateral triangle with = 163 kg
length of sides of y cm is 210 cm.
3x + 1 = 163

4. (a) 4x – 5 = 5(x – 3) 3x + 1 – 1 = 163 – 1

4x – 5 = 5x – 15 3x = 162

4x – 5 + 5 = 5x – 15 + 5 3x = 162
3 3
4x = 5x – 10
x = 54 kg
4x – 5x = 5x – 10 – 5x
Mass of Amir
–x = –10 =x–6
= 54 – 6
–x = –10 = 48 kg
–1 –1

x = 10

(b) 4x + 6 = 22 \ Amir’s mass is 48 kg.
5
4x (b) Let the money received by
5 + 6 – 6 = 22 – 6 Maniam = x
Nurul = x – 50
4x = 16 Asyraf = 2x
5
4x Their total money is
5 ×5 = 16 × 5 = 2x + x + (x – 50)
4x = 80 = 4x – 50
= RM890
4x = 80
4 4 4x – 50 = 890
4x – 50 + 50 = 890 + 50
x = 20
4x = 940

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Mathematics Form 1 Answers

4x = 940 (b) Let the price of a pair of trousers is x
4 4 and a T-shirt is y.
x = 2y
x = 940 = RM235
4
(c) Let the score of the first student is x and
Therefore, the money received by ;
Asyraf = 2x the second student is y.
= 2(235) x+y
= RM470 2 = 75
Nurul = x – 50
= 235 – 50 8. (a) When a number is subtracted from
= RM185 another number that has been
multiplied by two, the result is 12.
Maniam = x
= RM235 (b) When RMp is divided between three
boys, each of them received RMq.

The money received by Asyraf, Nurul (c) The perimeter of an isosceles triangle is
and Maniam are RM470, RM185 and 45 cm.
RM235 respectively.

(c) The number of T-shirt is in multiple of 3. 9. (a) 3
Let the number of T-shirt of 3
Aniq = x, 7
Chow = x + 3,
Ravi = x + 6. (b) 7
3
x + x + 3 + x + 6 = 45 7
3x + 9 = 45
(c) 7
3x + 9 – 9 = 45 – 9 7
3x = 36 7

3x = 36
3 3
10. (a) 3a – 2b =15
x = 12
When a = 1,
Number of Ravi’s T-shirt
=x+6 3(1) – 2b = 15
= 12 + 6
= 18 3 – 2b – 3 = 15 – 3

–2b = 12

Total number of T-shirt of Aniq and –2b = 12
Chow –2 –2
=x+x+3
= 12 + 12 + 3 = 27 b = –6

When a = 2,

The number of Ravi’s T-shirt is not the 3(2) – 2b = 15
same as the total number of T-shirt of
Aniq and Chow because the number of 6 – 2b – 6 = 15 – 6
Ravi’s T-shirt is 18 and sum of number
of Aniq’s and Chow’s T-shirt is 27. –2b = 9

–2b = 9
–2 –2

b = – 9
2
6. (a) 7
(b) 3 When a = 3,
(c) 7
3(3) – 2b = 15

9 – 2b – 9 = 15 – 9

–2b = 6

7. (a) Let the two numbers are x and y. –2b = 6
2x – y = 4 –2 –2

b = –3

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Mathematics Form 1 Answers

Then, the possible solutions are a = 1, When x = 1,
9
b = –6; a = 2, b= – 2 and a = 3, b = –3. 30(1) + 15y = 120

1 30 + 15y – 30 = 120 – 30
2
(b) p + q = 6 15y = 90

When p = 2, 15y = 90
15 15
1
2 (2) + q = 6 y=6

1+q=6 When x = 2,

1+q–1=6–1 30(2) + 15y = 120

q=5 60 + 15y = 120

When 1 p = 4, 60 + 15y – 60 = 120 – 60
2
(4) + q = 6 15y = 60

2+q=6 15y = 60
15 15
2+q–2=6–2
y=4
q=4

When p = 6, When x = 3,

1 (6) + q = 6 30(3) + 15y = 120
2
3+q=6 90 + 15y = 120

3+q–3=6–3 90 + 15y – 90 = 120 – 90

q=3 15y = 30

Then, the possible solutions are p = 2, 15y = 30
q = 5; p = 4, q = 4 and p = 6, q = 3. 15 15

y=2

(c) Let the number of male kittens = x, The number of pack of red and black
and female kittens = y. marbles respectively is 1 and 6 or 2 and
4 or 3 and 2.
So, x + y = 4

When x = 1, 11. (a) 2y + x = 11
x –7 –5 –3 –1 1
1+y=4 y9 8 7 6 5

1+y–1=4–1 y

y=3

When x = 2,

2+y=4

2+y–2=4–2 9
8
y=2 7
6
When x = 3, 5
4
3+y=4 3
2
3+y–3=4–3 1

y=1 –7 –6 –5 –4 –3 –2 –1 0 1

The number of male kitten and female
kitten respectively is 1 and 3 or 2 and
2 or 3 and 1.

(d) Let the number of red marbles = x, x
and the number of black marbles = y.

So, 30x + 15y = 120

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(b) y = –x Mathematics Form 1 Answers

x –1 1 3 5 7 4x + 3y = 120
x 0 30
y1 –1 –3 –5 –7 y 40 0

y y
40
1 1 2 34 567 x
30
–1 0
–1 20
–2
–3 10
–4
–5
–6
–7

(c) y = x + 1 0 x
10 20 30 40
x –7 –5 –3 –1
y –6 –4 –2 0 1 13. (a) 2y = 4 – x ..... 1
2

y 2x + 3y = 5 ........... 2

From 1, x = 4 – 2y .. 3

2 Substitute 3 into 2,
1
–7 –6 –5 –4 –3 –2 –1 0 1 x 2(4 – 2y) + 3y = 5
–1
–2 8 – 4y + 3y = 5
–3
–4 –y = 5 – 8
–5
–6 –y = –3

–y = –3
–1 –1

y=3

Substitute y = 3 into 3.
x = 4 – 2(3) = –2

\ x = –2 and y = 3

(b) 2x – 3y = 5................ 1
2x + 3y = –6.............. 2

12. (a) Let the number of adult and children From 1, x = 5 + 3y ...... 3
tickets are x and y respectively. 2

x + y = 35 Substitute 3 into 2,

x 0 35 2( 5 + 3y ) + 3y = –6
y 35 0 2

5 + 3y + 3y = –6

6y = –6 – 5

y = – 11
6

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A1

Mathematics Form 1 Answers

Substitute y = – 11 into 3. 3 + 4 : 13x = 26
6
x = 26
5 + 3(– 161) 13
2
x= =2

= (5 – 121) ÷ 2 Subsitute x = 2 into 1.
2(2) – 3y = 7
= 10 – 11 × 1 = – 1 –3y = 7 – 4
2 2 4 –3y = 3
y = –1
\ x = – 1 and y= – 161.
4 \ x = 2 and y = –1

(c) 2m + n = 19........... 1 (c) 4p + 3q = 6 .................. 1
m + 3n = 7............. 2 3p – 2q = 13 ................ 2

From 2, m = 7 – 3n ... 3

Substitute 3 into 1, 1 × 2 : 8p + 6q = 12 ... 3
2(7 – 3n) + n = 19 2 × 3 : 9p – 6q = 39 ... 4
3 + 4 : 17p = 51
14 – 6n + n = 19
p=3
–5n = 19 – 14
Substitute p = 3 into 1.
n = – 5
5 4(3) + 3q = 6

= –1 12 + 3q = 6

Substitute n = –1 into 3. 12 + 3q – 12 = 6 – 12
m = 7 – 3n
3q = –6
= 7 – 3(–1)
= 10 3q = –6
\ m = 10 and n = –1 3 3

q = –2

\ p = 3 and q = –2

14. (a) p – 2q = 4 .................... 1 15. (a) Let Wawa’s mass = x
3p – 4q = 10 .................. 2 and her sister’s mass = y

1 × 3 : 3p – 6q = 12 ..... 3 x + y = 86 ……… 1
y – x = 12 ……… 2
2 – 3: 2q = –2
1 + 2: 2y = 98
q = –1 y = 49

Substitute q = –1 into 1. Substitute y = 49 into 1.
p – 2(–1) = 4 x + 49 = 86
p=4–2 x = 86 – 49 = 37
=2
\Wawa’s mass is 37 kg and her sister’s
\ p = 2 and q = –1 mass is 49 kg.

(b) 2x – 3y = 7 .................. 1 (b) RM80 = 8 000 cent
3x + 2y = 4 .................. 2 Let the number of 50 cent coins = x

1 × 2: 4x – 6y = 14 .... 3 and the number of 20 cent coins = y
2 × 3: 9x + 6y = 12 .... 4 x + y = 250 ......... 1

50x + 20y = 8 000 ...... 2

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From 1, x = 250 – y .... 3 Mathematics Form 1 Answers

Substitute 3 into 2, 2. p = 2q – 10 ------- ➀
2p + 7q = 2 ------- ➁
50(250 – y) + 20y = 8 000
Substitute ➀ into ➁:
12 500 – 50y + 20y = 8 000 2(2q – 10) + 7q = 2

12 500 – 30y = 8 000 4q – 20 + 7q = 2
11q = 22
–30y = 8 000 – 12 500 q = 2 -------------- ➂

–30y = –4 500 Substitute ➂ into ➀:
p = 2(2) – 10 = –6
y = –4 500 = 150 p + q = –6 + 2 = –4
–30
Answer: B
Substitute y = 150 into 3.
x = 250 – 150 = 100 3. 6(2 + 7m) = –72
72
\ There are 100 and 150 coins of 50 cent 2 + 7m = – 6
and 20 cent respectively.

(c) Let the number of cows = x 2 + 7m = –12
and the number of ducks = y.
7m = –14
So, the number of cow’s legs = 4x,
the number of duck’s legs = 2y, m = –2
the number of cow’s eyes = 2x,
Answer: A
and the number of duck’s eyes = 2y.
4. Let x be the price of chicken and y be the
4x + 2y = 124 ……… 1 price of fish.
2x + 2y = 88 ……… 2 6x + 10y = 80
3x + 5y = 40
1 – 2 : 2x = 124 – 88
2x = 36 Answer: A
x = 18
5. A linear equation is any equation that can
Substitute x = 18 into 2. be written in the form of ax + b = 0.

2(18) + 2y = 88 Answer: C

36 + 2y = 88 6. 8m – 2 = m – 3
4 2
2y = 88 – 36

2y = 52 8m – 2 = 4m – 6

y = 52 = 26 4m = –4
2
m = –1

\ Yes, the number of ducks is more than Answer: B
the number of cows because there are
Section B
18 cows and 26 ducks in Daniel’s farm.
1. (a) (i) Yes
PT3 Standard Practice 6 (ii) No

Section A (b) 2v – 3w = 4, 7h + 2g = –1, 4p + 12 = 2q

1. 2a + 5b = 15 -------- ➀ 2. (a) (i) x + y = 256
–2a + 3b = 9 -------- ➁
➀+➁: (ii) Cynthia → 8 years old
8b = 24 Brother → (8 + x) years old
b=3 8 + (8 + x) = 20
x + 16 = 20
Answer: B

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Mathematics Form 1 Answers (ii) m – n = 7
2m – 2n = 14 ------ ➀
(b) (i) 6(2) – y = 20 2m + 5n = 91 ------ ➁
12 – y = 20 ➁ – ➀ : 7n = 77
y = 12 – 20 n = 11
y = –8 m – 11 = 7
m = 18
(ii) 6x – (–2) = 20 m = 18, n = 11
6x = 18
x=3

Section C (c) Q = 55x + 20y

1. (a) (i) (a) x – 7 = 9 595 = 55(5) + 20y

(b) 2x = 6 20y = 595 – 275

(ii) (a) A number is added to 1 and 20y = 320
equals to 5.
y = 320
(b) The total mass of three 20
packets of sand is 150 kg.
y = 16

(b) 3x – 2 = 40 HOTS Challenge

3x = 42

x = 42 Width = (x + 4) cm
3 Length of rectangle = 3(x + 4)

x = 14 = (3x + 12) cm

(c) First integer = x Perimeter of rectangle = 2(x + 4) + 2(3x + 12)
Second integer = 3x = 2x + 8 + 6x + 24
Third integer = 3x + 1 = (8x + 32) cm
x + 3x + 3x + 1 = 50
7x + 1 = 50 Perimeter of triangle = 2x + 3 + 4x + 4x + 1
7x = 49 = (10x + 4) cm
x=7
Perimeter of rectangle = perimeter of triangle
3x = 3(7) = 21
3x + 1 = 3(7) + 1 = 22 8x + 32 = 10x + 4
The three integers are 7, 21 and 22.
32 – 4 = 10x – 8x

2. (a) p + q = 3 ------ ➀ 28 = 2x
p – q = 7 ------ ➁
➀ + ➁ : 2p = 10 x = 28
p=5 2

5+q=3 x = 14 cm
q=3–5
q = –2 Perimeter of square = perimeter of rectangle
= 8x + 32
(b) (i) 2m + 5n + 89o = 180o = 8(14) + 32
2m + 5n = 91o = 144 cm

Length of a side of square = 144 ÷ 4
= 36 cm

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Mathematics Form 1 Answers

CHAPTER

7 Linear Inequalities
Ketaksamaan Linear

1. (a) 8 , 12 (c)

(b) –11 . –13 –52

(d)

(c) 3 . 1 3.5
8 7
(e)

(d) –20 , 20 – —23

(f)

(e) 0.089 , 0.098 225

(f) – 5 . – 0.9 5. (a) – 1 , 1
6 3 2

2. (a) x is greater than –4. (b) – 3 . – 7
x . –4 5 8

(b) x is less than –0.25. (c) 2 , 2.8
x , –0.25 (d) 1 . –1
(e) 10 , 100
4
(c) x is less than 3 .
x 4
, 3 6. (a) 2 , 2
5 3

(d) x is greater than 105. (b) – 10 , – 1
x . 105 11 15

3. (a) less than or equal to; (c) –1.2 , 2
x < 60
(d) 2 , 6
(b) greater than or equal to;
x > 165 (e) –3 , –1

(c) greater than or equal to; 7. (a) .
x > 15 (b) ,
(c) .
4. (a) –8 (d) ,
(b) 42 (e) .

8. (a) .
(b) ,

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Mathematics Form 1 Answers

9. (a) 2.6 . 2.06 (f) 10 – 6x > 25 – x

1 , 1 10 – 6x – 10 > 25 – x – 10
2.6 2.06
−6x > 15 – x
1 1
\ 2.6 . 2.06 7 −6x + x > 15 – x + x
3
1 1 −5x > 15
5 4
(b) – . – –5x < 15
–5 –5
1 1
, 1 x < –3
1 4
– 5 (g) 2p – q , p + 3
2p – 9 + 9 , p + 3 + 9
–5 , –4 2p , p + 12
2p – p , p + 12 − p
10. (a) x > 850 p , 12
(b) 7x . 18
(c) x < 4 (h) 6 – 5x . 21

6 – 5x − 6 . 21 − 6

11. (a) The number of residents in Kampung −5x . 15
Teluk is more than 750.
–5x , 15
(b) The number of members in 5 group –5 –5
singer is at least 20.
x , −3
(c) The number of doughnuts that Husna
can eat at most is 5. 13. (a) 21 + 2x + x , 144

21 + 3x , 144

12. (a) x + 5 < −2 21 + 3x – 21 , 144 – 21

x + 5 – 5 < −2 – 5 3x , 123

x < −7 3x , 123
3 3
x , 41
(b) p– 1 . 2
3 3 Biscuits bought by mother = 2x.

p– 1 + 1 . 2 + 1 x × 2 , 41 × 2
3 3 3 3 2x , 82

p .1 The maximum number of biscuits
bought by mother is 81.
x
(c) 4 .9 (b) 100 –15.30y . 56

x × 4 . 9 × 4 100 − 100 –15.30y . 56 − 100
4
x . 36 –15.30y . −44

(d) – x < 1 –15.30y , –44
4 2 –15.30 –15.30

x 1 y , 2.87…
4 2
– × (– 4) > × (– 4) The number of slippers Lin Dan can buy
at most is 2.
x > –2

(e) 8y > 64 14. (a) 3(4 – y) , 2(y + 1)

8y > 64 12 – 3y , 2y + 2
8 8
12 – 12 – 3y , 2y + 2 – 12

y >8 –3y , 2y – 10

–3y – 2y , 2y – 2y – 10

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Mathematics Form 1 Answers

–5y , –10 x ≤ 2—21 x ≥ –9

–5y . –10
–5 –5
–9 2—21
y.2

2y – 5 , 7 The solution is –9 < x < 2 1 .
2
2y – 5 + 5 , 7 + 5
PT3 Standard Practice 7
2y , 12

2y , 12 Section A
2 2

y,6 1. 5x > 16 – 3
5x > 13
y<6 y>2 x > 2.6
x=3
26
Answer: B
The solution is 2 , y , 6.

(b) 5 – 3p –2 2. 1 < 3x – 1 3x – 1 < 4
2 2 2
>

5 – 3p × 2 > –2 × 2 2 < 3x – 1 3x – 1 < 8
2
3 < 3x 3x < 9

5 – 3p > –4 x>1 x<3

5 – 3p – 5 > –4 – 5 1<x<3
Answer: C
–3p > –9

–3p < –9 2
–3 –3 3
3. g – 4 > –6
p <3
2
– p .1 3 g > –2
4
p g > –3
– 4 × (–4) , 1 × (–4)
p , –4 Answer: C

p < –4 p≤3 4. x < 23 – 3
x < 20
–4 3
Answer: A
The solution is p , –4.

(c) 3(7 – 2x) > 6 5. 1 < 1
r s
21 – 6x > 6
=s<r
21 – 6x – 21 > 6 – 21
=r>s
–6x > –15

–6x < –15 Answer: B
–6 –6

x < 2 1 6. Answer: D
2
2
4 – 3 x < 10 Section B

4 – 2 x –4 < 10 – 4 1. (a) (i) <
3 < 6 (ii) >
2
– 3 x (b) (i) –15 < –6
(ii) –7 < 7
– 2 x × 1– 3 2 > 6 × 1– 3 2
3 2 2

x > –9

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A1

Mathematics Form 1 Answers (c) 4(50) + 2(20) + 3(10) + n(5) < 290
270 + 5n < 290
2. (a) (i) R 5n < 290 – 270
(ii) P 5n < 20
n<4
(b) (i) 3 n = 1, 2, 3
(ii) 7
HOTS Challenge
Section C

1. (a) (i) p > 4

(ii) – p –5 < –2
3

–p – 5 < –6 9.2 + 8.7 + 7.4 + p + q
5
p > –5 + 6 (a)  9.5

p>1 25.3 + p + q  47.5

(b) 3(2x + 5) < 9x p + q  47.5 – 25.3
6x + 15 < 9x
15 < 9x – 6x p + q  22.2
15 < 3x
x>5 (b) The solution is not reasonable.
x = 5 cm The total score for p and q should be 20.0
because the maximum score for each event
Perimeter = 3(2(5) + 5) = 45 cm is 10.0.
The solution obtained is p + q  22.2.

CHAPTER

8 Lines and Angles
Garis dan Sudut

1. (a) No 5. (a) ∠ABC is less than 90°.
(b) No \ ∠ABC  70°
(c) Yes
(d) Yes (b) ∠ABC is more than 90°.
\ ∠ABC  120°
2. (a) 3
(b) 3 6. (a) ∠ABC = 75°
(c) 7 (b) ∠ABC = 125°
(d) 7
7. (a) C
3. (a) AB = 5 × 3 (b) A
= 15 cm (c) E
(d) D
(b) AB = 3 × 3 (e) F
= 9 cm
32
4. (a) AB = 3.5 cm
(b) AB = 2.8 cm

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8. (a) (i) (ii) Mathematics Form 1 Answers

60° 60° 60° 60° (iii)
60° 60° 60° 60°
60° 60°
60° 60°

9.

y x and y are xy
x supplementary

angles

xy x and y are x
complementary y

angles

xy x and y are xy
conjugate angles

10. (a) 90° − 34° = 56° (c) 2x + 36° + x = 180°
(b) 90° − 53° = 37°
(c) 90° − 76° = 14° 3x + 36° = 180°

11. (a) 180° − 98° = 82° 3x = 180° − 36°
(b) 180° − 128° = 52°
(c) 180° − 155° = 25° 3x = 144°

12. (a) 360° − 77° = 283° x = 144°
(b) 360° − 179° = 181° 3
(c) 360° − 256° = 104°
= 48°
13. (a) x + 190° = 360°
x = 360° − 190° (d) x + 125° + 60° + 90° = 360°
= 170° x + 275° = 360°
x = 360° − 275°
(b) x + 120° + 20° = 180° = 85°
x + 140° = 180°
x = 180° − 140° (e) x + 2x + 100° + 65° = 360°
= 40°
3x + 165° = 360°

3x = 195°

x = 195°
3

= 65°

14. (a) 4.4 cm Y

X

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A1

Mathematics Form 1 Answers K (b)

(b) P

J 3.3 cm

15. (a) B
C
B

A

18. (a)

A 60°
B
(b)

A

A

(b)

A

B 120°
B
(c)

B

C

19. (a)

16. (a) A (b)

A P
B
(b) P
B

A 20. (a)

17. (a) A B 30°

P 34

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Mathematics Form 1 Answers

(b) (b)

75° 24. (a) AB
(b) KL
(c)
25. (a) a
135° (b) d

26. (a) b
(b) c

21. (a) (i) x and y 27. (a) f and h
(ii) x and z,
y and z (b) r and s

(b) (i) x and z 28. (a)
(ii) x and y,
y and z

22. (a) x = 180° − 25° y = 25° + 90° Corresponding
= 155° = 115° angles

x + y = 155° + 115° Q
= 270°
P 45° 45°

(b) ∠CPD = ∠APB = 59° S

x = 180° − 59° y = 180° – 90° – 59° R 45°
Corresponding
= 121° = 31° angle

x + y = 121° + 31° 45° = 45°
= 152° PQ and RS are parallel because the
corresponding angles are the same.
(c) 4x + 40° + 80° = 180°
4x = 180° – 120° (b) Q S
4x = 60°
x = 15°

4y = 80° 72°
80° 100°
y = 80° = 20°
4

x + y = 15° + 20° = 35° Alternate

P angle R

23. (a) 72° ≠ 80°
PQ and RS are not parallel because the

alternate angles are not the same.

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Mathematics Form 1 Answers 4. 2a + 18o + 90o + 24o = 180o
2a + 132o = 180o
(c) P R 2a = 48o
a = 24o
120°
70° 60° Answer: A
110°
5. x = 53o + 35o
QS = 88o

70° + 120° = 190° Answer: C
≠ 180°
6.
PQ and RS are not parallel because the
sum of interior angles is not 180°. 75° 23°
75° 23°
29. (a) b

Danny

a m = 75o
Answer: B
(b) Salmi
Section B
Danny 1. (a) False
(b) True
a (c) True
b (d) False

30. (a) (i) The angle of depression of 2. (a) d, f
(ii) student E from student A is 30°. (b) b, c
AE and BD are parallel, thus (c) e, f
∠BDA and ∠DAE are alternate (d) f, g R
angles.
Section C
∠BEF − ∠AEF = 30° 1. (a) (i)
∠BEF − 30° = 30°
∠BEF = 30° + 30°
= 60°

PT3 Standard Practice 8 P 60°
Q
Section A (ii) A

1. 130o – 90o = 40o BC
∠y = 180o – 40o = 140o
Answer: D

2. ∠POQ = 180o – 113o = 67o
∠QOR = 110o – 67o = 43o
Answer: B

3. ∠y = 360o – 45o – 90o – 10o = 215o
Answer: D

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Mathematics Form 1 Answers

(b) ∠TUV = 360o – 250o (c) Q
= 110o
R
y = 110o
z = 125o – 60o PX
S
= 65o

(c) x = 180o – 62o – 32o

= 86o T

y = 32o + 62o

= 94o

4z = 86o

z= 86° HOTS Challenge
4

= 21.5o

2. (a) 3m = 50o – 2m + 20o

3m + 2m = 50o + 20o P
25°
5m = 70o 45°

m = 70° = 14o
5

n + 3(14o) = 180o Q

n + 42o = 180o 110°
45°
n = 180o – 42o = 138o

(b) q + p + p = 180o

q + 2p = 180o The angle of elevation of point Q from the student
= 180o – 45o – 110o = 25o
2p = 180o – q

p= 180o – q
2

CHAPTER

9 Basic Polygons
Poligon Asas

1. (a) 7 (b) 9 (c) 15
The number of sides 73 3 15 3
14 93 30
(i) The number of 14 3 9 30
vertices 35 27 3 90 3

(ii) The number of
diagonals

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A2

Mathematics Form 1 Answers

2. (a) AG

F

BE

CD

Heptagon ABCDEFG

(b) G
H
A
B F

D
E

C

Octagon ABCDEFGH

3.

Triangle Decagon Pentagon Octagon Quadrilateral

4. The length of all
sides are not
All interior angles equal.
are 60°.

Equilateral triangles

All interior angles Isosceles triangles The length of all
are different. sides are equal.

Two of the interior Scalene triangles The length of two
angles have the of its sides is
same value. equal.

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