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N1 NOTA EKSPRES Bab 1 Nombor Nisbah 1.1 Integer 1 Integer ialah nombor bulat positif atau negatif atau sifar. Integers are positive or negative whole numbers or zero. Contoh/Example: +17, –23, 92, –56 2 Pecahan dan perpuluhan bukan integer. Fractions and decimals are not integers. 3 Integer boleh diwakili oleh satu garis nombor. Integers can be represented by a number line. 4 Pada garis nombor, integer negatif terletak pada sebelah kiri sifar, manakala integer positif terletak pada sebelah kanan sifar. On a number line, negative integers are on the left of zero, while positive integers are on the right of zero. Negatif/Negative –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 Positif/Positive 1.2 Operasi Asas Aritmetik yang Melibatkan Integer 1 Peraturan bagi menentukan tanda integer yang terlibat dalam operasi boleh diringkaskan seperti berikut: Rules to determine the sign of integers involved in the operation can be simplified as follows: Tanda Sign Operasi Operation Tanda Sign Tanda Sign Contoh Example + × + = + 6 × 6 = 36 ÷ 6 ÷ 6 = 1 – × – = + –6 × –6 = 36 ÷ –6 ÷ –6 = 1 + × – = – 6 × –6 = –36 ÷ 6 ÷ –6 = –1 – × + = – –6 × 6 = –36 ÷ –6 ÷ 6 = –1 2 Operasi asas aritmetik melibatkan integer adalah sama seperti pengiraan bagi nombor bulat. Susunan operasi adalah seperti berikut: Basic operations involving integers are the same as calculations of whole numbers. The order of operations is as follows: Kurungan Brackets Bahagi/Darab (kiri ke kanan) Divide/multiply (left to right) Tambah/Tolak (kiri ke kanan) Add/subtract (left to right) Contoh/Example: 32 ÷ 8 2 – 5 × (–9) = (32 ÷ 4) – [5 × (–9)] = 8 + 45 = 53 3 Hukum operasi aritmetik boleh diringkaskan seperti berikut: Laws of arithmetic operations can be simplified as follows: Hukum Laws Penerangan Explanation Contoh Example Hukum Identiti Identity Law Identiti penambahan ialah 0. Jika 0 ditambah kepada sebarang nombor, nombor itu kekal tidak berubah. Addition identity is 0. If 0 is added to any number, the number remains unchanged. –45 + 0 = –45 0 + 17 = 17 Identiti pendaraban ialah 1. Jika 1 didarab dengan sebarang nombor, nombor itu kekal tidak berubah. Multiplication identity is 1. If 1 is multiplied with any number, the number remains unchanged. 23 × 1 = 23 1 × (–7) = –7 Hukum Kalis Tukar Tertib Commutative Law Bagi sebarang integer a dan b, For any integers a and b, a + b = b + a dan/and a × b = b × a 12 + (−9) = –9 + 12 49 × 25 = 25 × 49 Hukum Kalis Sekutuan Associative Law Bagi sebarang integer a, b dan c, For any integers a, b and c, (a + b) + c = a + (b + c) dan/and (a × b) × c = a × (b × c) (14 + 56) + 71 = 14 + (56 + 71) (9 × 3) × 8 = 9 × (3 × 8) Operasi penolakan dan pembahagian tidak kalis sekutuan. Subtraction and division are not associative. (3 – 9) + 12 ≠ 3 – (9 + 12) 36 ÷ (4 ÷ 3) ≠ (36 ÷ 4) ÷ 3 Strategi A+ Maths Tg1-Nota 3rd.indd 1 30/10/2023 10:44:58 AM PENERBIT ILMU BAKTI SDN. BHD.
N2 Hukum Laws Penerangan Explanation Contoh Example Hukum Kalis Agihan Distributive Law Bagi sebarang integer a, b, dan c, hasil darab a dengan jumlah b dan c adalah sama dengan jumlah hasil darab ab dan ac. For any integers a, b, and c, the product of a and the sum of b and c is equal to the sum of the products of ab and ac. a(b + c) = ab + ac 5 (7 + 11) = (5 × 7) + (5 × 11) Operasi penolakan juga adalah kalis agihan. Subtraction is also distributive. a(b – c) = ab – ac 5 (12 – 8) = (5 × 12) − (5 × 8) 1.3 Pecahan Positif dan Pecahan Negatif 1 Pecahan ialah nombor yang mewakili sebahagian daripada keseluruhan, ditulis sebagai x y dengan keadaan y > 0. Nombor di atas ialah pengangka dan nombor di bawah ialah penyebut. Fractions are numbers representing a part of a whole, written as x y where y > 0. The number on top is the numerator and the number below is the denominator. 2 Pecahan boleh bernilai positif atau negatif. Fractions can have positive or negative values. 3 Pecahan juga boleh diwakili pada garis nombor. Pecahan yang berada di sebelah kiri pada garis nombor mempunyai nilai yang lebih kecil berbanding dengan pecahan yang berada di sebelah kanan. Fractions can also be represented on a number line. Fractions on the left side of a number line have smaller values compared to fractions on the right side. 4 Kaedah penambahan pecahan adalah seperti berikut: Addition method of fractions is as follows: (i) Bagi pecahan dengan penyebut yang sama, tambahkan pengangka pecahan yang terlibat dan kekalkan penyebut pecahan. Permudahkan jawapan jika perlu. For fractions with the same denominator, add the numerators of the fractions involved and retain the denominator. Simplify the answer if needed. Contoh/Example: 5 13 + 6 13 = 5 + 6 13 = 11 13 (ii) Bagi pecahan dengan penyebut yang berbeza, tukarkan pecahan-pecahan kepada pecahan setara yang mempunyai penyebut yang sama terlebih dahulu. To add fractions with different denominators, first convert the fractions to equivalent fractions with the same denominator. Contoh/Example: 2 5 + 4 7 = 2 × 7 5 × 7 + 4 × 5 7 × 5 = 14 35 + 20 35 = 34 35 5 Pengiraan yang melibatkan gabungan operasi bagi pecahan adalah sama seperti pengiraan bagi nombor bulat. Calculations involving combined operations of fractions are the same as calculations of whole numbers. Kurungan Brackets Bahagi/Darab (kiri ke kanan) Divide/multiply (left to right) Tambah/Tolak (kiri ke kanan) Add/subtract (left to right) 1.4 Perpuluhan Positif dan Perpuluhan Negatif 1 Perpuluhan ialah nombor yang mewakili pecahan dengan penyebutnya 10, 100, 1 000 dan seterusnya. Decimals are numbers representing fractions with denominators 10, 100, 1 000 and so on. 2 Perpuluhan juga boleh diwakilkan pada garis nombor seperti nombor bulat dan pecahan. Decimals can also be represented on a number line as for whole numbers and fractions. 3 Pengiraan yang melibatkan gabungan operasi bagi perpuluhan adalah sama seperti nombor bulat dan pecahan. Calculations involving combined operations of decimals are the same as for whole numbers and fractions. Kurungan Brackets Bahagi/Darab (kiri ke kanan) Divide/multiply (left to right) Tambah/Tolak (kiri ke kanan) Add/subtract (left to right) 1.5 Nombor Nisbah 1 Nombor nisbah ialah nombor yang boleh ditulis dalam bentuk p q dengan keadaan p dan q ialah integer, dan q ≠ 0. Rational numbers are numbers that can be written in the form of p q where p and q are integers, and q ≠ 0. 2 Integer, pecahan, perpuluhan dan perpuluhan berulang ialah nombor nisbah. Integers, fractions, decimals and recurring decimals are rational numbers. Contoh/Example: 27, 0, 22 7 , 4.86, 5.1515, ... 3 Pengiraan yang melibatkan gabungan operasi bagi nombor nisbah adalah menyamai pengiraan yang melibatkan integer. Calculations involving combined operations of rational numbers are the same as calculations involving integers. Strategi A+ Maths Tg1-Nota 3rd.indd 2 30/10/2023 10:44:59 AM PENERBIT ILMU BAKTI SDN. BHD.
Nota Ekspres N1 – N20 Aktiviti Pembelajaran Abad ke-21 P1 – P2 Rekod Prestasi Murid R1 – R6 Bab 1 Nombor Nisbah 1 – 7 Bab 2 Faktor dan Gandaan 8 – 12 Bab 3 Kuasa Dua, Punca Kuasa Dua, Kuasa Tiga dan Punca Kuasa Tiga 13 – 20 Bab 4 Nisbah, Kadar dan Kadaran 21 – 27 Bab 5 Ungkapan Algebra 28 – 33 Bab 6 Persamaan Linear 34 – 42 Bab 7 Ketaksamaan Linear 43 – 50 Bab 8 Garis dan Sudut 51 – 62 Bab 9 Poligon Asas 63 – 69 Bab 10 Perimeter dan Luas 70 – 75 Bab 11 Pengenalan Set 76 – 84 Bab 12 Pengendalian Data 85 – 91 Bab 13 Teorem Pythagoras 92 – 95 Ujian Akhir Sesi Akademik 96 – 108 Jawapan 109 – 114 Kandungan Masteri Math Tg1 Kandg 2nd.indd 2 30/10/2023 11:51:06 AM PENERBIT ILMU BAKTI SDN. BHD.
1 Kelaskan integer berikut dalam kotak yang sesuai. SP 1.1.1 TP 1 Classify the following integers into suitable boxes. 1 –21, 78, –4, –136, 90, 258 78 (a) Positif/ Positive 90 258 –4 (b) Negatif/ Negative –21 –136 Integer/ Integers 0 Wakilkan integer berikut pada garis nombor. Kemudian, lengkapkan pernyataan di bawah berdasarkan kedudukan integer pada garis nombor itu. SP 1.1.2 SP 1.1.3 TP 1 Represent the following numbers on a number line. Then, complete the statements below based on the integer positions on the number line. 2 15, –5, –20, 10 –25 –20 –5 0 10 15 20 (a) 10 lebih kecil daripada 15 kerana 10 berada di sebelah kiri 15 pada garis nombor. 10 is smaller than 15 because 10 is on the left of 15 on the number line. (b) –5 lebih besar daripada –20 kerana –5 berada di sebelah kanan –20 pada garis nombor. –5 is greater than –20 because –5 is on the right of –20 on the number line. Susun semula nombor berikut dalam tertib yang betul. SP 1.1.4 TP 1 Rearrange the following numbers in the correct order. Nombor/ Numbers Tertib menaik/ Ascending order 3 0, –3, 5, –7, 9, –8 –8, –7, –3, 0, 5, 9 4 49, –54, 11, –37, –30, 23 –54, –37, –30, 11, 23, 49 Nombor/ Numbers Tertib menurun/ Descending order 5 7, –2, 1, –10, 11, –3, 16 16, 11, 7, 1, –2, –3, –10 6 –30, –27, –32, –81, –56 –27, –30, –32, –56, –81 Bab 1 Nombor Nisbah Rational Numbers Bidang Pembelajaran: Nombor dan Operasi Buku Teks: Halaman 1 – 29 Buku Teks: Halaman 2 – 7 Integer 1.1 Masteri Math Tg1 Bab 1 2nd.indd 1 30/10/2023 10:47:38 AM PENERBIT ILMU BAKTI SDN. BHD.
2 Selesaikan setiap yang berikut menggunakan garis nombor. SP 1.2.1 TP 3 Solve each of the following using a number line. 1 5 + (+2) 2 4 – (+5) 3 7 + (–3) 4 –9 + (–1) 5 –6 – (–4) 6 –1 – (+8) Selesaikan setiap yang berikut. SP 1.2.2 TP 3 Solve each of the following. 7 5 × (–8) = –(5 × 8) = –40 8 –7 × (–6) = +(7 × 6) = 42 9 –9 × 12 = –(9 × 12) = –108 10 27 ÷ (–9) = –(27 ÷ 9) = –3 11 –32 ÷ 4 = –(32 ÷ 4) = –8 12 –44 ÷ (–11) = +(44 ÷ 11) = 4 Buku Teks: Halaman 7 – 10 Operasi Asas Aritmetik yang Melibatkan Integer 1.2 = 5 + 2 = 7 = 7 – 3 = 4 = – 6 + 4 = –2 012345678 = 4 – 5 = –1 = – 9 – 1 = –10 = –1 – 8 = –9 01234567 –11 –10 –9 –8 –7 –6 –5 –4 –2 –1 0 1 2 3 4 5 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 –7 –6 –5 –4 –3 –2 –1 0 1 Masteri Math Tg1 Bab 1 2nd.indd 2 30/10/2023 10:47:39 AM PENERBIT ILMU BAKTI SDN. BHD.
3 Cari nilai bagi setiap yang berikut. SP 1.2.3 TP 3 Find the value of each of the following. 1 6 – 8(–3) = 6 + 24 = 30 2 –5 × (4 − 11) = –5 × (−7) = 35 3 2 + (–20) ÷ 5 – (−9) = 2 + (–4) – (−9) = 2 – 4 + 9 = 7 4 17 – (–19) –4 + (–2) = 17 + 19 –4 – 2 = 36 –6 = –6 Kenal pasti hukum yang terlibat bagi operasi berikut. SP 1.2.4 TP 3 Identify the laws involved in the following operations. 5 –91 + 91 = 0 Hukum Identiti/ Identity Law 6 100 + (18 + 67) = (100 + 18) + 67 Hukum Kalis Sekutuan/ Associative Law 7 17 × 45 = 45 × 17 Hukum Kalis Tukar Tertib/ Commutative Law 8 (107 × 8) × 5 = 107 × (8 × 5) Hukum Kalis Sekutuan/ Associative Law 9 12(25 + 40) = 12(25) + 12(40) Hukum Kalis Agihan/ Distributive Law 10 1 44 × 44 = 1 Hukum Identiti/ Identity Law Selesaikan setiap yang berikut dengan pengiraan efisien. SP 1.2.5 TP 3 Solve each of the following with efficient computations. 11 –73 × 66 – 27 × 66 = (–73 – 27) × 66 = –100 × 66 = –6 600 12 24 + 33 + 67 = 24 + (33 + 67) = 24 + 100 = 124 13 11 × 2 060 = 11 × (2 000 + 60) = 11 × 2 000 + 11 × 60 = 22 000 + 660 = 22 600 14 6 × 81 × 5 = 81 × (6 × 5) = 81 × 30 = 2 430 Buku Teks: Halaman 10 – 12 Operasi Asas Aritmetik yang Melibatkan Integer 1.2 Masteri Math Tg1 Bab 1 2nd.indd 3 30/10/2023 10:47:39 AM PENERBIT ILMU BAKTI SDN. BHD.
4 Wakilkan nombor berikut pada garis nombor. SP 1.3.1 TP 1 Represent the following number on a number line. 1 1, 5 7 , – 3 7 , –12 7 , 14 7 –1 0 –1 2 7 1 –3 7 5 7 1 4 7 2 2 3 , – 1 2 , – 5 6 , 1 1 3 , – 1 6 – 1 2 – 1 6 0 2 3 1 1 3 – 5 6 1 Susun semula pecahan berikut dalam tertib yang dinyatakan. SP 1.3.2 TP 2 Rearrange the following fractions in the orders stated. 3 Tertib menaik/ Ascending order –2 8 9 , 2 2 3 , –15 6 , 1 2 , 1 –2 8 9 , –1 5 6 , 1 2 , 1, 2 2 3 4 Tertib menurun/ Descending order 23 4 , –111 12, – 5 8 , 1 2 , 7 8 2 3 4 , 7 8 , 1 2 , – 5 8 , –1 11 12 Cari nilai bagi setiap yang berikut. SP 1.3.3 TP 3 Find the value of each of the following. 5 6 7 Buku Teks: Halaman 14 – 16 Pecahan Positif dan Pecahan Negatif 1.3 2 1 3 + (– 2 5) × 11 9 = 7 3 + (– 2 5) × 10 9 = 7 3 + (– 4 9) = 21 9 – 4 9 = 17 9 = 18 9 – 1 6 + 13 4 ÷ (– 5 8) = – 1 6 + 7 4 × (– 8 5 ) = – 1 6 + (– 14 5 ) = – 5 30 + (– 84 30 ) = – 89 30 = –2 29 30 8 × (21 4 – 12 3) = 8 × ( 9 4 – 5 3 ) = 8 × ( 27 12 – 20 12 ) = 8 × 7 12 = 14 3 = 4 2 3 2 1 × 3 × 3 2 1 × 5 × 5 × 6 × 6 × 4 × 4 × 3 × 3 2 3 Masteri Math Tg1 Bab 1 2nd.indd 4 30/10/2023 10:47:39 AM PENERBIT ILMU BAKTI SDN. BHD.
5 Wakilkan nombor berikut pada garis nombor. SP 1.4.1 TP 1 Represent the following number on a number line. 1 –1.5, 0.6, –0.3, 1.2, –0.9 –2.1 –1.5 –0.9 –0.3 0 0.6 1.2 1.8 2 1.6, –0.4, –0.8, –2.0, 2.4 –2.0 –1.6 –0.8 –0.4 0 1.2 1.6 2.4 Susun semula perpuluhan berikut dalam tertib yang dinyatakan. SP 1.4.2 TP 2 Rearrange the following decimals in the orders stated. 3 Tertib menaik/ Ascending order 0.702, 0.724, –0.076, 0.781, – 0.735 –0.735, – 0.076, 0.702, 0.724, 0.781 4 Tertib menurun/ Descending order –2.34, –2.16, 2.08, –2.79, 2.52 2.52, 2.08, –2.16, –2.34, –2.79 Cari nilai bagi setiap yang berikut. SP 1.4.3 TP 3 Find the value of each of the following. 5 –0.8 × (–5.02 + 9.42) = –0.8 × 4.4 = –3.52 6 7.51 – (–2.3) × –0.5 = 7.51 – 1.15 = 6.36 7 (9.63 – 2.88) ÷ (–1.25) = 6.75 4 (–1.25) = –5.4 8 –1.34 + 4.56 1.5 + 0.72 = –1.34 + 3.04 + 0.72 = 1.7 + 0.72 = 2.42 Buku Teks: Halaman 19 – 21 Perpuluhan Positif dan Perpuluhan Negatif 1.4 Masteri Math Tg1 Bab 1 2nd.indd 5 30/10/2023 10:47:39 AM PENERBIT ILMU BAKTI SDN. BHD.
6 Buku Teks: Halaman 23 – 24 Nombor Nisbah 1.5 Tandakan (✓) bagi nombor nisbah dan (✗) bagi yang bukan. Beri sebab bagi jawapan yang bukan nombor nisbah. SP 1.5.1 TP 2 Mark ( ✓) for a rational number and mark (✗) for an irrational number. Give a reason for the answer which is an irrational number. 1 –26 ✓ 2 13 5 ✓ 3 2π π tidak dapat diungkapkan sebagai nisbah bagi dua integer. π cannot be expressed as a ratio of two integers. Selesaikan. SP 1.5.2 TP 2 Solve. 4 2 5 + (–2) – 7 3 4 = 2 5 + (–2) – 31 4 = 8 – 40 – 155 20 = – 187 20 = –9 7 20 5 5 8 × 3 3 5 + 0.8 ÷ 2 = 5 8 1 × 18 5 1 + 0.4 = 18 8 + 0.4 = 9 4 + 0.4 = 2.25 + 0.4 = 2.65 6 0.62 × 5 + 2 3 8 ÷ 1 4 = 3.1 + 19 8 × 4 1 = 31 10 + 19 × 5 2 × 5 = 126 10 = 12.6 7 8.82 ÷ (5.12 – 2.67) × 3 = 8.82 ÷ 2.45 × 3 = 3.6 × 3 = 10.8 ✗ 1 2 Untuk tujuan pembelajaran Imbas kod QR atau layari https://www.youtube.com/ watch?v=SQ4cB9yXkHM untuk menonton video tentang cara-cara mengenal nombor nisbah. Video Tutorial 9 10 × 3 + (–2 1 2) ÷ 2 3 = Masukkan nilai./ Key in the values. 9 (–) a b/c 2 a b/c 10 → 3 × 1 3 ↓ + 2 2 ( ) SHIFT ÷ a b/c Tekan = untuk mendapatkan jawapan. Press = to get the answer. Jawapan/Answer: – 21 20 Sudut Kalkulator 3.71 – 2 × (– 4 5) + 0.6 = Masukkan nilai./Key in the values. 3 4 . a b/c 7 5 1 ) – + 2 0 × . ( 6 – Tekan = untuk mendapatkan jawapan. Press = to get the answer. Jawapan/Answer: 5.91 Sudut Kalkulator Video Tutorial Masteri Math Tg1 Bab 1 2nd.indd 6 30/10/2023 10:47:40 AM PENERBIT ILMU BAKTI SDN. BHD.
7 Buku Teks: Halaman 12 – 25 Operasi Asas Aritmetik yang Melibatkan Integer 1.2 Pecahan Positif dan Pecahan Negatif 1.3 Perpuluhan Positif dan Perpuluhan Negatif 1.4 Nombor Nisbah 1.5 Selesaikan setiap masalah berikut. SP 1.2.6 SP 1.3.4 SP 1.4.4 SP 1.5.3 TP 4 TP 5 KBAT Menilai Solve each of the following problems. 1 Sebuah kereta bergerak sejauh 955.9 km dalam masa 84 5 jam. Berapa jauh kereta itu bergerak dalam masa 20.4 jam jika kelajuannya adalah sama? A car travels 955.9 km in 8 4 5 hours. How far does the car travel in 20.4 hours if the speed of the car remains constant? Jarak yang dilalui dalam tempoh sejam/ Distance travelled in one hour: 955.9 ÷ 8 4 5 = 955.9 ÷ 8.8 = 108.625 km Jarak yang dilalui dalam tempoh 20.4 jam/ Distance travelled in 20.4 hours: 108.625 km × 20.4 = 2 215.95 km 2 Adila mempunyai RM170. Meena mempunyai 3 4 daripada wang Adila. Wang simpanan Kim Liang pula ialah tiga kali ganda simpanan Meena. Berapakah jumlah wang yang dimiliki oleh Adila, Meena dan Kim Liang? Adila has RM170. Meena has 3 4 of Adila’s amount of money. Kim Liang’s money is thrice the amount of Meena’s money. What is the total amount of money that Adila, Meena and Kim Liang have? Simpanan Adila/ Adila's money: RM170 Simpanan Meena/ Meena's money: 3 4 × RM170 = RM127.50 Simpanan Kim Liang/ Kim Liang's money: 3 × RM127.50 = RM382.50 Jumlah wang/ Total amount of money: RM170 + RM127.50 + RM382.50 = RM680 3 Seekor kura-kura bergerak ke hadapan sejauh 16 m dalam tempoh sejam. Kura-kura itu berpatah balik ke belakang dan melangkah 103 5 m pada jam berikutnya. Akhirnya, dalam jam yang ketiga, kura-kura itu bergerak ke hadapan semula dan bergerak sejauh 8.5 m lagi. Berapakah jarak, dalam m, dari titik mula kura-kura itu bergerak dalam tempoh 3 jam tersebut? A tortoise moved 16 m forwards in an hour. The tortoise moved 10 3 5 m backwards in the next hour. Finally, in the third hour, the tortoise moved for another 8.5 m forwards. What is the distance travelled by the tortoise, in m, from the starting point in the duration of 3 hours? Jarak ke hadapan dari titik mula/ Distance travelled from the starting point: 16 – 10 3 5 + 8.5 = 16 – 10.6 + 8.5 = 5.4 + 8.5 = 13.9 m Masteri Math Tg1 Bab 1 2nd.indd 7 30/10/2023 10:47:40 AM PENERBIT ILMU BAKTI SDN. BHD.
8 Buku Teks: Halaman 32 – 33 Faktor, Faktor Perdana dan Faktor Sepunya Terbesar (FSTB) 2.1 Gariskan nombor perdana dalam senarai yang berikut. TP 1 Underline the prime numbers in the following lists. 1 13 , 19 , 27 , 29 , 31 , 41 , 43 , 51 2 71 , 81 , 73 , 83 , 87 , 89 , 93 , 97 3 113 , 119 , 127 , 133 , 139 , 161 , 179 , 193 4 257 , 309 , 313 , 459 , 499 , 507 , 541 , 543 Senaraikan faktor bagi nombor berikut. SP 2.1.1 TP 2 List down the factors of the following numbers. 5 36 6 485 7 126 8 990 Tentukan sama ada 8 ialah faktor bagi nombor berikut. SP 2.1.1 TP 2 Determine whether 8 is a factor of the following numbers. 9 54 54 ÷ 8 = 6 baki/ remainder 6 Bukan/ Incorrect 10 140 140 ÷ 8 = 17 baki/ remainder 4 Bukan/ Incorrect 11 240 240 ÷ 8 = 30 Ya/ Correct 12 2 032 2 030 ÷ 8 = 254 Ya/ Correct Bab 2 Faktor dan Gandaan Factors and Multiples Bidang Pembelajaran: Nombor dan Operasi Buku Teks: Halaman 30 – 45 2 36 2 18 3 9 3 3 1 2 126 3 63 3 21 7 7 1 1, 2, 3, 4, 6, 9, 12, 18, 36 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126 5 485 97 97 1 1, 5, 97, 485 2 990 3 495 3 165 5 55 11 11 1 1, 2, 3, 5, 6, 9, 11, 15, 18, 22, 30, 33, 45, 55, 66, 90, 110, 165, 198, 330, 495, 990 Masteri Math Tg1 Bab 2 3rd.indd 8 30/10/2023 10:50:26 AM PENERBIT ILMU BAKTI SDN. BHD.
9 Buku Teks: Halaman 33 – 36 Faktor, Faktor Perdana dan Faktor Sepunya Terbesar (FSTB) 2.1 Senaraikan faktor perdana bagi nombor-nombor berikut. SP 2.1.2 TP 2 List down the prime factors of the following numbers. 1 54 2 90 3 420 4 1 365 Ungkapkan nombor berikut dalam bentuk pemfaktoran perdana. SP 2.1.2 TP 2 Express the following numbers in the form of prime factorisation. Nombor Number Faktor perdana Prime factors Nombor Number Faktor perdana Prime factors 5 105 3 × 5 × 7 6 1 530 2 × 3 × 3 × 5 × 17 7 144 2 × 2 × 2 × 2 × 3 × 3 8 3 430 2 × 5 × 7 × 7 × 7 Cari faktor sepunya bagi nombor-nombor yang diberi. SP 2.1.3 TP 2 Find the common factors of the given numbers. 9 6, 18 dan/ and 24 10 12, 15 dan/ and 30 420 2 210 3 70 7 10 2 5 Faktor perdana/ Prime factors: 2, 3, 5 dan/ and 7 54 2 27 3 9 3 3 Faktor perdana/ Prime factors: 2 dan/ and 3 90 2 45 5 9 3 3 Faktor perdana/ Prime factors: 2, 3 dan/ and 5 1 365 5 273 3 91 Faktor perdana/ Prime factors: 3, 5 dan/ and 91 6 : 1, 2, 3, 6 18 : 1, 2, 3, 6, 9, 18 24 : 1, 2, 3, 4, 6, 8, 12, 24 Faktor sepunya/ Common factors: 1, 2, 3 dan/ and 6 12 : 1, 2, 3, 4, 6, 12 15 : 1, 3, 5, 15 30 : 1, 2, 3, 5, 6, 10, 15, 30 Faktor sepunya/ Common factors: 1 dan/ and 3 Masteri Math Tg1 Bab 2 3rd.indd 9 30/10/2023 10:50:26 AM PENERBIT ILMU BAKTI SDN. BHD.
10 Buku Teks: Halaman 38 – 40 Gandaan, Gandaan Sepunya dan Gandaan Sepunya Terkecil (GSTK) 2.2 Senaraikan gandaan bagi nombor bulat berikut. SP 2.2.1 TP 2 List down the multiples of the following whole numbers. 1 Enam gandaan 5 yang pertama selepas 25 First six multiples of 5 after 25 30, 35, 40, 45, 50, 55 2 Gandaan 11 yang kurang daripada 100 Multiples of 11 that are less than 100 11, 22, 33, 44, 55, 66, 77, 88, 99 Tentukan sama ada 1 932 ialah gandaan bagi setiap nombor yang berikut. SP 2.2.1 TP 2 Determine whether 1 932 is a multiple of each of the following numbers. 3 9 4 12 Tentukan sama ada setiap nombor berikut ialah gandaan bagi 7. SP 2.2.1 TP 2 Determine whether each of the following numbers is a multiple of 7. 5 564 6 1 792 Tulis dua gandaan sepunya yang pertama bagi nombor-nombor berikut. SP 2.2.1 TP 2 Write down the first two common multiples of the following numbers. 7 5 dan/ and 6 8 3, 4 dan/ and 15 5 : 5, 10, 15, 20, 25, 30 , 35, 40, 45, 50, 55, 60 … 6 : 6, 12, 18, 24, 30 , 36, 42, 48, 54, 60 … Gandaan sepunya/ Common multiples: 30 dan/ and 60 3 3, 4, 15 2 1, 4, 5 2 1, 2, 5 5 1, 1, 5 1, 1, 1 Gandaan sepunya/ Common multiple = 3 × 2 × 2 × 5 = 60 Gandaan seterusnya/ The next multiple = 60 × 2 = 120 1 932 9 = 214 baki 6 1 932 bukan gandaan bagi 9 1 932 is not a multiple of 9 564 7 = 80 baki 4 564 bukan gandaan bagi 7 546 is not a multiple of 7 1 932 12 = 161 1 932 ialah gandaan bagi 12 1 932 is a multiple of 12 1 792 7 = 256 1 792 ialah gandaan bagi 7 1 792 is a multiple of 7 Masteri Math Tg1 Bab 2 3rd.indd 10 30/10/2023 10:50:26 AM PENERBIT ILMU BAKTI SDN. BHD.
11 Buku Teks: Halaman 36, 40 – 41 Faktor, Faktor Perdana dan Faktor Sepunya Terbesar (FSTB) 2.1 Gandaan, Gandaan Sepunya dan Gandaan Sepunya Terkecil (GSTK) 2.2 Tentukan FSTB dengan menyenaraikan semua faktor bagi setiap nombor yang diberi. SP 2.1.4 TP 3 Determine the HCF by listing out all the factors of each of the given numbers. 1 14 dan/ and 20 2 18, 24 dan/ and 42 Tentukan FSTB bagi setiap nombor berikut menggunakan kaedah pembahagian berulang. SP 2.1.4 TP 3 Determine the HCF for each of the following numbers using the repeated division method. 3 6 dan/ and 27 4 28 dan/ and 56 5 6, 8 dan/ and 10 6 18, 24 dan/ and 36 Tentukan GSTK bagi setiap nombor berikut menggunakan kaedah pemfaktoran perdana. SP 2.2.2 TP 3 Determine the LCM for each of the following numbers using the prime factorisation method. 7 12 dan/ and 18 8 10, 15 dan/ and 20 Tentukan GSTK bagi setiap nombor berikut menggunakan kaedah pembahagian berulang. SP 2.2.2 TP 3 Determine the LCM for each of the following numbers using the repeated division method. 9 4, 8 dan/ and 12 10 7, 14 dan/ and 21 18 : 1, 2, 3, 6, 9, 18 24 : 1, 2, 3, 4, 6, 8, 12, 24 42 : 1, 2, 3, 6, 7, 14, 21, 42 FSTB/ HCF = 6 2 28, 56 2 14, 28 7 7, 14 1, 2 FSTB/ HCF = 2 × 2 × 7 = 28 2 18, 24, 36 3 9, 12, 18 3, 4, 6 FSTB/ HCF = 2 × 3 = 6 FSTB/ HCF = 2 2 6, 8, 10 3, 4, 5 12 : 2 × 2 × 3 18 : 2 × 3 × 3 GSTK/ LCM = 2 × 2 × 3 × 3 = 36 10 : 2 × 5 15 : 3 × 5 20 : 2 × 2 × 5 GSTK/ LCM = 2 × 2 × 3 × 5 = 60 2 4, 8, 12 2 2, 4, 6 2 1, 2, 3 3 1, 1, 3 1, 1, 1 7 7, 14, 21 2 1, 2, 3 3 1, 1, 3 1, 1, 1 GSTK/ LCM = 2 × 2 × 2 × 3 = 24 GSTK/ LCM = 7 × 2 × 3 = 42 14 : 1, 2, 7, 14 20 : 1, 2, 4, 5, 10, 20 FSTB/ HCF = 2 FSTB/ HCF = 3 3 6, 27 2, 9 Masteri Math Tg1 Bab 2 3rd.indd 11 30/10/2023 10:50:26 AM PENERBIT ILMU BAKTI SDN. BHD.
12 Buku Teks: Halaman 37 – 38, 41 – 42 Faktor, Faktor Perdana dan Faktor Sepunya Terbesar (FSTB) 2.1 Gandaan, Gandaan Sepunya dan Gandaan Sepunya Terkecil (GSTK) 2.2 Selesaikan setiap yang berikut. SP 2.1.5 SP 2.2.3 TP 4 TP 5 KBAT Menilai Solve each of the following. 1 Untuk menggalakkan penggunaan pengangkutan awam, Encik Khairul ingin memberikan beberapa sampul yang mengandungi tiket bas, tiket komuter dan tiket LRT kepada rakan-rakan anaknya. Jika dia mempunyai 16 keping tiket bas, 20 keping tiket komuter dan 12 keping tiket LRT untuk dimasukkan sama banyak ke dalam setiap sampul tanpa berbaki sebarang tiket, berapakah bilangan terbanyak sampul yang Encik Khairul perlu sediakan? In order to encourage the usage of public transports, Encik Khairul wants to give some envelopes containing bus tickets, commuter tickets and LRT tickets to his children’s friends. If he has 16 pieces of bus tickets, 20 pieces of commuter tickets and 12 pieces of LRT tickets to be put equally into each envelope without any tickets remaining, what is the highest number of envelopes that Encik Khairul has to prepare? 2 Kamelia, Alan dan Kanna ingin menyusun buku di pusat sumber. Kamelia boleh menyusun 5 buah buku serentak, Alan boleh menyusun 6 buah buku serentak, manakala Kanna boleh menyusun 8 buah buku serentak. Jika mereka menyusun bilangan buku yang sama banyak, berapakah bilangan buku paling sedikit yang mereka boleh susun? Kamelia, Alan and Kanna want to arrange some books in the library. Kamelia can arrange 5 books simultaneously, Alan can arrange 6 books simultaneously, whereas Kanna can arrange 8 books simultaneously. If they arrange an equal number of books, what is the smallest number of books that they can arrange? Bilangan buku paling sedikit/ The least number of books = 2 × 2 × 2 × 3 × 5 = 120 3 Rosalinda membuat 32 keping biskut coklat, 24 keping biskut kopi dan 48 keping biskut oren. Biskut-biskut itu dibungkus ke dalam beberapa bekas plastik untuk diagihkan kepada guru-gurunya. Dia ingin membahagikan biskut itu ke dalam bekas yang serupa supaya setiap bekas mempunyai bilangan biskut yang sama bagi setiap jenis. Jika setiap bekas itu mempunyai bilangan biskut yang maksimum, berapakah bilangan bekas plastik yang diperlukannya? Rosalinda baked 32 pieces of chocolate biscuits, 24 pieces of coffee biscuits and 48 pieces of orange biscuits. The biscuits were packed into several plastic containers to be given to her teachers. She wanted to divide the biscuits into similar containers so that each container contained an equal number of biscuits of the same type. If each container had the maximum number of biscuits, how many containers were needed? Bilangan bekas plastik/ Number of plastic containers = 2 × 2 × 2 = 8 Untuk tujuan pembelajaran Imbas kod QR atau layari https://www.youtube. com/watch?v=FAuVYRVfios&list=PLJMyIAdDi Qlkh9Csz4uj4g6ntRZmVTVr2&index=3 untuk mempelajari dengan lebih lanjut tentang faktor dan gandaan. Video Tutorial Video Tutorial 2 16, 20, 12 2 8, 10, 6 4, 5, 3 Bilangan sampul/ Number of envelopes = 2 × 2 = 4 2 5, 6, 8 2 5, 3, 4 2 5, 3, 2 3 5, 3, 1 5 5, 1, 1 1, 1, 1 2 32, 24, 48 2 16, 12, 24 2 8, 6, 12 4, 3, 6 Masteri Math Tg1 Bab 2 3rd.indd 12 30/10/2023 10:50:27 AM PENERBIT ILMU BAKTI SDN. BHD.
109 Jawapan Bab 1 Nombor Nisbah Halaman 1 1 (a) 78, 90, 258 (b) –4, –21, –136 2 (a) 10, 15, 10, 15 (b) –5, –20, –5, –20 3 –8, –7, –3, 0, 5, 9 4 –54, –37, –30, 11, 23, 49 5 16, 11, 7, 1, –2, –3, –10 6 –27, –30, –32, –56, –81 Halaman 2 1 7 0 1 2 3 4 5 6 7 2 –1 –2 –1 0 1 2 3 4 5 3 4 0 1 2 3 4 5 6 7 8 4 –10 –11 –10 –9 –8 –7 –6 –5 –4 5 –2 –7 –6 –5 –4 –3 –2 –1 0 1 6 –9 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 7 −40 8 42 9 –108 10 –3 11 –8 12 4 Halaman 3 1 30 2 35 3 7 4 –6 5 Hukum Identiti/ Identity Law 6 Hukum Kalis Sekutuan/ Associative Law 7 Hukum Kalis Tukar Tertib/ Commutative Law 8 Hukum Kalis Sekutuan/ Associative Law 9 Hukum Kalis Agihan/ Distributive Law 10 Hukum Identiti/ Identity Law 11 –6 600 12 124 13 22 600 14 2 430 Halaman 4 1 –1 0 –12 7 1 – 3 7 14 7 5 7 2 – 1 2 – 1 6 0 11 3 – 5 6 2 1 3 3 –2 8 9 , –1 5 6 , 1 2 , 1, 2 2 3 4 2 3 4 , 7 8 , 1 2 , – 5 8 , –111 12 5 1 8 9 6 –229 30 7 4 2 3 Halaman 5 1 –2.1 –1.5 –0.9 –0.3 0 0.6 1.2 1.8 2 –2.0 –1.6 –0.8 –0.4 0 1.2 1.6 2.4 3 –0.735, –0.076, 0.702, 0.724, 0.781 4 2.52, 2.08, –2.16, –2.34, –2.79 5 –3.52 6 6.36 7 –5.4 8 2.42 Halaman 6 1 ✓ 2 ✓ 3 ✓ π tidak dapat diungkapkan sebagai nisbah bagi dua integer. π cannot be expressed as a ratio of two integers. 4 –9 7 20 5 2.65 6 12.6 7 10.8 Halaman 7 1 2 215.95 km 2 RM680 3 13.9 m Bab 2 Faktor dan Gandaan Halaman 8 1 13, 19, 29, 31, 41, 43 2 71, 73, 83, 89, 97 3 113, 127, 139, 179, 193 4 257, 313, 499, 541 5 1, 2, 3, 4, 6, 9, 12, 18, 36 6 1, 5, 97, 485 7 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126 8 1, 2, 3, 5, 6, 9, 11, 15, 18, 22, 30, 33, 45, 55, 66, 90, 110, 165, 198, 330, 495, 990 9 54 ÷ 8 = 6 baki/ remainder 6 Bukan/ Incorrect 10 140 ÷ 8 = 17 baki/ remainder 4 Bukan/ Incorrect 11 240 ÷ 8 = 30 12 2 030 ÷ 8 = 254 Ya/ Correct Ya/ Correct Halaman 9 1 2 dan/ and 3 2 2, 3 dan/ and 5 3 2, 3, 5 dan/ and 7 4 3, 5 dan/ and 91 5 3 × 5 × 7 6 2 × 3 × 3 × 5 × 17 7 2 × 2 × 2 × 2 × 3 × 3 8 2 × 5 × 7 × 7 × 7 9 1, 2, 3 dan/ and 6 10 1 dan/ and 3 Halaman 10 1 30, 35, 40, 45, 50, 55 2 11, 22, 33, 44, 55, 66, 77, 88, 99 3 1 932 9 = 214 baki 6 1 932 bukan gandaan bagi 9 1 932 is not a multiple of 9 4 1 932 12 = 161 1 932 ialah gandaan bagi 12 1 932 is a multiple of 12 5 564 7 = 80 baki 4 564 bukan gandaan bagi 7 546 is not a multiple of 7 6 1 792 7 = 256 1 792 ialah gandaan bagi 7/ 1 792 is a multiple of 7 7 30 dan/ and 60 8 60 dan/ and 120 Halaman 11 1 2 2 6 3 3 4 28 5 2 6 6 7 36 8 60 9 24 10 42 Halaman 12 1 4 2 120 3 8 Bab 3 Kuasa Dua, Punca Kuasa Dua, Kuasa Tiga dan Punca Kuasa Tiga Halaman 13 1 Ya/ Yes 2 Bukan/ Not 3 Ya/ Yes 4 64 = 8 × 8 = 8 5 289 = 17 × 17 = 17 6 441 = 21 2 = 21 7 1 156 = 34 2 = 34 8 Salah/ Wrong 9 Betul/ Correct 10 Salah/ Wrong 11 13 × 13 = 169 12 (−24) × (−24) = 576 13 2 5 × 2 5 = 4 25 14 (–0.9) × (–0.9) = 0.81 15 289 16 8 1 36 17 16.81 18 144 1 225 Halaman 14 1 9 2 4 6 3 2.5 4 3 2 5 19 6 16 25 7 3.7 8 1 1 6 9 9 10 0.16 11 640 000 12 7.862 terletak antara 49 dengan 64. 7.862 is between 49 and 64. 13 0.2342 terletak antara 0.04 dengan 0.09. 0.2342 is between 0.04 and 0.09. 14 39 terletak antara 6 dengan 7. 39 is between 6 and 7. 15 0.068 terletak antara 0.2 dengan 0.3. 0.068 is between 0.2 and 0.3. Halaman 15 1 27 2 20 63 3 0.88 4 (a) 40 (b) 63 5 2.3 6 12 400 terletak antara 100 dengan 200. 12 400 is between 100 and 200. Halaman 16 1 –27 2 8 125 3 1 8 100 –27 64 –9 216 36 –0.125 –8 000 169 4 Benar/ True 5 Salah/ False 6 Salah/ False 7 Benar/ True 8 – 64 343 9 1.331 10 6 859 11 –17.576 12 – 343 1 728 13 4 149 512 Halaman 17 1 3 4 2 –1 2 3 3 –0.1 4 17 5 1 3 7 6 – 4 5 7 27 8 0.008 9 343 000 10 4.6783 terletak antara 64 dengan 125. 4.6783 is between 64 and 125. 11 0.2643 terletak antara 0.008 dengan 0.027. 0.2643 is between 0.008 and 0.027. Halaman 18 1 3 197 terletak antara 5 dengan 6. 3 197 is between 5 and 6. 2 3 5.73 terletak antara 1 dengan 2. 3 5.73 is between 1 and 2. 3 17.25 4 –361 5 1 5 6 17 18 7 11 1 2 2 3 ÷ 3 3 3 8 = 1 3 2 2 3 ÷ 3 27 8 = 27 8 ÷ 3 2 = 27 8 × 2 3 = 9 4 8 12 2 3 2 2 – (43 ) + 3 27 64 = 1 8 3 2 2 – 64 + 3 4 = 64 9 – 8 + 3 4 = – 5 36 Masteri Math Tg1 Jawapan 3rd.indd 109 30/10/2023 11:05:30 AM PENERBIT ILMU BAKTI SDN. BHD.