Bab 6 Ketaksamaan Linear dalam Dua 40 – 46 Pemboleh Ubah Linear Inequalities in Two Variables Bab 7 Graf Gerakan 47 – 54 Graphs of Motion Bab 8 Sukatan Serakan Data Tak Terkumpul 55 – 63 Measure of Dispersion for Ungrouped Data Bab 9 Kebarangkalian Peristiwa Bergabung 64 – 75 Probability of Combined Events Bab 10 Matematik Pengguna: 76 – 84 Pengurusan Kewangan Consumer Mathematics: Financial Management Jawapan J1 – J12 Bab 1 Fungsi dan Persamaan Kuadratik dalam Satu Pemboleh Ubah 1 – 8 Quadratic Functions and Equations in One Variable Bab 2 Asas Nombor 9 – 13 Number Bases Bab 3 Penaakulan Logik 14 – 23 Logical Reasoning Bab 4 Operasi Set 24 – 30 Operations on Sets Bab 5 Rangkaian dalam Teori Graf 31 – 39 Network in Graph fi eory Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) No. 18, Jalan PJS 5/26, Taman Desaria, 46150 Petaling Jaya, Selangor Darul Ehsan. Tel: 03-7783 3233, 7783 5233 Faks: 03-7783 7233 E-mel: [email protected] Laman web: www.penerbitilmubakti.com © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2024 Pertama kali diterbitkan 2024 ISBN 978- Cetakan: 9 8 7 6 5 4 3 2 1 Semua hak cipta terpelihara. Sebarang bahagian dalam buku ini tidak boleh diterbitkan semula, disimpan dalam cara yang boleh digunakan lagi, ataupun dipindahkan dalam sebarang bentuk atau cara, baik secara elektronik, mekanik, gambar, rakaman dan sebagainya, tanpa kebenaran terlebih dahulu daripada Penerbit Ilmu Bakti Sdn. Bhd. (732516-M). Penyunting: Aznani Hasnor Ahmad Pereka letak: Sarifuddin b. Yusof Pereka kulit buku: Jessica Choo Teks diset dalam Utopia Std 10/12 poin Dicetak di Malaysia oleh xxxxxxxxxxxxxxxxxxxx Kandungan Imbas dan Muat Turun • Peperiksaan Akhir Tahun • Jawapan Tk4 PT24 M kandungan.indd 1 1/12/2024 10:00:29 AM Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) No. 18, Jalan PJS 5/26, Taman Desaria, 46150 Petaling Jaya, Selangor Darul Ehsan. Tel: 03-7783 3233, 7783 5233 Faks: 03-7783 7233 Emel: [email protected] Laman web: www.penerbitilmubakti.com © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2024 Pertama kali diterbitkan 2024 ISBN 978-629-473-322-0 Cetakan: 9 8 7 6 5 4 3 2 1 Semua hak cipta terpelihara. Sebarang bahagian dalam buku ini tidak boleh diterbitkan semula, disimpan dalam cara yang boleh digunakan lagi, ataupun dipindahkan dalam sebarang bentuk atau cara, baik secara elektronik, mekanik, gambar, rakaman dan sebagainya, tanpa kebenaran terlebih dahulu daripada Penerbit Ilmu Bakti Sdn. Bhd. (732516-M). Penyunting: Aznani Hasnor Ahmad Pereka kulit buku: Sarifuddin Yusof Pereka letak: Jessica Choo Teks diset dalam Utopia Std 10/12 poin Dicetak di Malaysia oleh Herald Printers Sdn. Bhd. Imbas dan Muat Turun • Peperiksaan Akhir Tahun dan Jawapan
1 1.1 Fungsi dan Persamaan Kuadratik Quadratic Functions and Equations 1 Antara berikut, yang manakah mewakili nilai a, b dan c bagi fungsi f(x) = (x – 2)(x + 3)? Which of the following represents the values of a, b and c for the function f(x) = (x – 2)(x + 3)? a b c A 1 –2 3 B 1 2 –3 C 1 –1 –6 D 1 1 –6 TP 3 BT ms.4 – 5 2 Rajah 1 menunjukkan punca-punca bagi suatu fungsi kuadratik. Diagram 1 shows the roots of a quadratic function. x y –16 O 6 Rajah 1/Diagram 1 Nyatakan paksi simetri fungsi itu. State the axis of symmetry of the function. A x = –16 C x = –5 B y = –11 D y = 6 TP 3 BT ms.8 – 10 3 Rajah 2 menunjukkan graf bagi suatu fungsi kuadratik. Diagram 2 shows the graph of a quadratic function. x f(x) O Rajah 2/Diagram 2 Persamaan manakah yang paling sesuai mewakili graf itu? Which equation best fi ts the graph? A f(x) = 2x 2 + 5x + 1 C f(x) = 2x 2 – 5x – 1 B f(x) = –2x 2 + 5x + 1 D f(x) = –2x 2 – 5x – 1 TP 3 BT ms.11 – 14 4 Antara berikut, yang manakah ialah puncapunca persamaan kuadratik 2x2 – 19x + 45 = 0? Which of the following are the roots of the quadratic equation 2x2 – 19x + 45 = 0? A x = 0 dan/ and x = 5 B x = 3 dan/ and x = 16 C x = 5 dan/ and x = 9 2 D x = 6 dan/ and x = 2 9 TP 3 BT ms.16 – 19 5 Rajah 3 menunjukkan graf bagi suatu fungsi kuadratik. Diagram 3 shows a graph of a quadratic function. x –2 O 8 (3, 25) 16 f(x) Rajah 3/Diagram 3 Antara pernyataan berikut, yang manakah tidak betul? Which of the following statements is not correct? A Pintasan-y ialah 16 Th e y-intercept is 16 B Paksi simetri graf ialah x = 3 Th e axis of symmetry of the graph is x = 3 C Titik minimum graf ialah (3, 25) Th e minimum point of the graph is (3, 25) D Punca-punca bagi persamaan ialah x = –2 dan x = 8 Th e roots of the equation are x = –2 and x = 8 TP 3 BT ms.7 – 19 Bab 1 Fungsi dan Persamaan Kuadratik dalam Satu Pemboleh Ubah Quadratic Functions and Equations in One Variable Bidang Pembelajaran: Perkaitan dan Algebra Tk4 PT24 M1(1-8).indd 1 1/24/2024 10:35:33 AM
2 6 Diberi bahawa suatu persamaan kuadratik ialah x2 – 2x – k = 0 dan x = 6 merupakan salah satu penyelesaian. Cari nilai k. Given that a quadratic equation is x 2 – 2x – k = 0 and x = 6 is one of the solutions. Find the value of k. A 24 C –12 B 12 D –24 TP 3 BT ms.21 7 Rajah 4 menunjukkan graf f(x) = x 2 + 6x. Diagram 4 shows the graph of f(x) = x2 + 6x. x f(x) O Rajah 4/Diagram 4 Antara berikut, yang manakah graf bagi f(x) = x 2 – 6x? KBAT Mengaplikasi Which of the following is the graph of f(x) = x2 – 6x? A C B D O x f(x) x f(x) x f(x) x f(x) O O O TP 3 BT ms.11 – 14 8 Rajah 5 menunjukkan sebuah segi empat tepat. Diagram 5 shows a rectangle. (x + 4) cm (x + 2) cm Rajah 5/Diagram 5 Diberi bahawa luas segi empat tepat itu ialah 35 cm2 . Berapakah ukuran bagi sisi yang terpanjang? It is given that the area of the rectangle is 35 cm2 . What is the size of the longest side? A 4 cm C 7 cm B 5 cm D 9 cm TP 4 BT ms.26 – 27 9 Diberi laju bagi suatu zarah ialah (5x + 2) m s–1 untuk tempoh (x + 3) s. Cari jarak, dalam m, yang dilalui oleh zarah itu. Given the speed of a particle is (5x + 2) m s–1 for a period of (x + 3) s. Find the distance, in m, travelled by the particle. A 5x 2 + 13x + 6 C 5x 2 – 17x + 5 B 5x 2 + 17x + 6 D 5x 2 – 17x – 5 TP 5 BT ms.26 – 27 10 Rajah 6 menunjukkan graf y = x 2 – x – 12. Diagram 6 shows the graph of y = x 2 – x – 12. x y m O n r Rajah 6/Diagram 6 Tentukan nilai bagi m, n dan r. Determine the values of m, n and r. m n r A 3 –4 –12 B 4 –3 12 C –3 4 –12 D 3 –4 12 TP 3 BT ms.17 – 18 11 Rajah 7 menunjukkan graf y = –x 2 + 8. Diagram 7 shows the graph of y = –x 2 + 8. x y O (p, 4) Rajah 7/Diagram 7 Cari nilai p. Find the value of p. A –1 C –3 B –2 D –4 TP 2 BT ms.13 – 14 Tk4 PT24 M1(1-8).indd 2 1/24/2024 10:35:33 AM
3 12 Graf manakah yang mewakili sebahagian daripada graf y = (x + 2)(x – 3)? Which graph represents part of the graph y = (x + 2)(x – 3)? A C B D y x –2 3 –6 O y x –3 2 –6 O y x –3 2 6 O y x –2 3 6 O TP 3 BT ms.17 – 18 13 Apabila satu sisi segi empat sama digandakan dan sisi satu lagi dikurangkan sebanyak 2 unit, luas segi empat tepat yang terbentuk ialah 126 unit2 . Cari luas, dalam unit2 , untuk segi empat sama tersebut. KBAT Menilai When one side of a square is doubled and the other side is reduced by 2 units, the area of the resulting rectangle is 126 unit2 . Find the area, in unit2 , of the square. A 36 C 64 B 49 D 81 TP 5 BT ms.26 – 27 14 Diberi bahawa hasil darab dua nombor positif yang berturutan ialah 24. Tentukan satu persamaan kuadratik untuk mencari penyelesaian bagi kedua-dua nombor positif tersebut. KBAT Mengaplikasi It is given that the product of two consecutive positive numbers is 24. Determine a quadratic equation to find the solutions of these two positive numbers. A n2 + n + 24 = 0 C n 2 + n – 24 = 0 B n2 + 24 = 0 D n 2 – 24 = 0 TP 5 BT ms.15 15 Graf y = 9 – x 2 dan y = –x + 7 bersilang di titik (p, q) dengan keadaan p dan q ialah integer positif. Cari nilai q. KBAT Menilai The graphs y = 9 – x 2 and y = –x + 7 intersect at point (p, q) where p and q are positive integers. Find the value of q. A 2 C 8 B 3 D 5 TP 4 BT ms.16 – 17 1.1 Fungsi dan Persamaan Kuadratik / Quadratic Functions and Equations 1 Jika x = 5 ialah suatu penyelesaian bagi persamaan kuadratik 5 + mx – 2x 2 = 0, cari nilai m dan nilai x yang satu lagi. TP 4 BT ms.21 If x = 5 is a solution for the quadratic equation 5 + mx – 2x 2 = 0, find the value of m and another value of x. [4 markah/marks] Jawapan/Answer: 2 Lakarkan graf fungsi kuadratik bagi y = x 2 – 6x. Seterusnya, tentukan titik minimum atau maksimum bagi lengkung tersebut. TP 3 BT ms.7 – 9 Sketch the graph of the quadratic function for y = x 2 – 6x. Hence, determine the minimum or maximum points of the curve. [4 markah/marks] Bahagian A / Section A Tk4 PT24 M1(1-8).indd 3 1/24/2024 10:35:34 AM
4 Jawapan/Answer: 3 Selesaikan persamaan kuadratik yang berikut: TP 4 BT ms.21 Solve the following quadratic equation: (3x – 4)(x + 5) = (3x – 4)2 [4 markah/marks] Jawapan/Answer: 4 Jason telah membeli x batang pen dengan harga RM48. Dia memberikan 3 batang pen kepada adiknya sebagai hadiah hari lahir. Dia bercadang untuk menjual baki pen kepada kawan-kawannya dan mendapat keuntungan RM3 daripada setiap batang pen itu. Jason telah mendapat jumlah keuntungan sebanyak RM15 daripada jualan itu. TP 5 BT ms.26 – 27 Jason bought x pens with a total amount of RM48. He gave 3 pens to his brother as his birthday present. He plans to make a profit of RM3 from each of the remaining pens by selling them to his friends. Jason managed to make a total profit of RM15 from the sales. (a) Cari harga kos setiap batang pen dalam sebutan x. Find the cost price of each pen in terms of x. (b) Cari harga jual setiap batang pen dalam sebutan x. Find the selling price of each pen in terms of x. (c) Berdasarkan maklumat di atas, cari nilai x. KBAT Menilai Based on the above information, find the value of x. [5 markah/marks] Jawapan/Answer: 5 Sebuah tangki air berukuran (w + 5) cm panjang, w cm lebar dan 40 cm tinggi. Muatan tangki air itu ialah 12 000 cm3 . Hitung nilai w jika tangki air tersebut diisi penuh dengan air. TP 4 BT ms.26–27 A water tank is (w + 5) cm long, w cm wide and 40 cm tall. The capacity of the water tank is 12 000 cm3 . Calculate the value of w if the water tank is completely filled up with water. [4 markah/marks] Tk4 PT24 M1(1-8).indd 4 1/24/2024 10:35:34 AM
5 Jawapan/Answer: 6 Diberi bahawa sebuah poligon dengan sisi n mempunyai n 2 (n – 3) pepenjuru. TP 5 BT ms.26 – 27 It is given that a polygon with n sides has n 2 (n – 3) diagonals. (a) Berapakah bilangan sisi bagi sebuah poligon dengan 90 pepenjuru? How many sides does a polygon with 90 diagonals have? (b) Adakah terdapat poligon dengan 100 pepenjuru? Justifikasikan jawapan anda. KBAT Menganalisis Is there a polygon with 100 diagonals? Justify your answer. [5 markah/marks] Jawapan/Answer: 7 Rajah 1 menunjukkan sekeping kadbod segi empat tepat PQRS. Diagram 1 shows a rectangular cardboard PQRS. x cm (3x + 4) cm 2x cm 14 cm S U T V P W R Q Rajah 1/Diagram 1 Dua buah segi tiga sama kaki WQV dan STU telah dipotong daripada kadbod tersebut. Luas yang tinggal diwakili oleh kawasan yang berlorek. TP 5 BT ms.26 – 27 Two isosceles triangles WQV and STU are removed from the cardboard. The remaining area is represented by the shaded region. (a) Bentukkan satu ungkapan kuadratik untuk mewakili luas yang tinggal. Form a quadratic expression to represent the remaining area. (b) Diberi bahawa luas kawasan yang berlorek ialah 184 cm2 dan x ialah nombor bulat. Cari nilai x. It is given that the area of the shaded region is 184 cm2 and x is a whole number. Find the value of x. [5 markah/marks] Tk4 PT24 M1(1-8).indd 5 1/24/2024 10:35:34 AM
6 Jawapan/Answer: Bahagian B / Section B 8 (a) Lengkapkan jadual nilai bagi fungsi kuadratik f (x) = x 2 + 8x + 13. TP 2 BT ms.5 – 6 Complete the table of values for the quadratic function f (x) = x 2 + 8x + 13. [3 markah/marks] (b) Plotkan graf fungsi kuadratik dan nyatakan Plot the graph of the quadratic function and state the (i) bentuk graf, TP 2 BT ms.7 shape of the graph, (ii) titik minimum atau maksimum, TP 3 BT ms.7 – 8 minimum or maximum point, (iii) persamaan paksi simetri bagi graf. TP 3 BT ms.8 – 9 equation of the axis of symmetry of the graph. [6 markah/marks] Jawapan/Answer: (a) f (x) = x 2 + 8x + 13 x –8 –5 –4 –1 0 1 2 3 y 13 22 (b) 50 — 40 — 30 — 20 — 10 — –10 — –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 f (x) x Tk4 PT24 M1(1-8).indd 6 1/24/2024 10:35:34 AM
7 9 Semasa hari persaraan Encik Wong, ahli-ahli daripada Persatuan Matematik ingin menghadiahkannya sebuah beg bimbit yang bernilai RM360. Kos beg itu akan dikongsikan antara ahli secara sama rata. Sekiranya terdapat 10 orang lagi ahli daripada Persatuan Sains ingin berkongsi kos hadiah itu, setiap ahli Persatuan Matematik boleh jimat sebanyak RM6. TP 5 BT ms.26 – 27 On Mr Wong’s retirement day, the members of the Mathematics Society wish to buy a briefcase worth RM360 as a present for him. The cost will be shared equally among the members. If there are 10 more members from the Science Society who would like to share the cost of the present, each of the Mathematics Society member can save RM6. Hitung/Calculate (a) bilangan ahli dalam Persatuan Matematik, the number of members in the Mathematics Society, [5 markah/marks] (b) amaun yang disumbangkan oleh setiap ahli yang berkongsi wang untuk membeli hadiah itu, the amount contributed by each member who shared the cost to buy the present, [2 markah/marks] (c) amaun yang perlu dibayar oleh setiap ahli Persatuan Matematik pada asalnya. the amount that each of the Mathematics Society member needs to pay initially. [2 markah/marks] Jawapan/Answer: Bahagian C / Section C 10 Sebiji batu dilontarkan secara menegak ke atas dari puncak sebuah bangunan berketinggian 95 meter. Jarak, s meter, batu dari tanah selepas t saat diberikan oleh s = 95 + 81t – 16t2 . A stone is thrown vertically upwards from the top of a 95-metre building. The distance, s metres, of the stone from the ground after t seconds is given by s = 95 + 81t – 16t2 . (a) Lengkapkan jadual nilai berikut untuk s = 95 + 81t – 16t2 . TP 3 BT ms.5 – 6 Complete the following table of values for s = 95 + 81t – 16t2 . [3 markah/marks] t 0 1 2.5 3 5 6 s Jadual 1/Table 1 (b) Berdasarkan Jadual 1, lukiskan graf s = 95 + 81t – 16t2 dalam ruang yang disediakan. KBAT Menilai TP 3 BT ms.7 – 9 Based on Table 1, draw the graph of s = 95 + 81t – 16t2 in the space provided. [3 markah/marks] (c) Berdasarkan graf yang dilukis di 10(b), jawab soalan-soalan berikut. TP 5 BT ms.26 – 27 Based on the graph drawn in 10(b), answer the following questions. Tk4 PT24 M1(1-8).indd 7 1/24/2024 10:35:34 AM
8 (i) Anggarkan masa yang diambil, dalam saat, batu itu mencecah tanah. Estimate the time taken, in seconds, for the stone to hit the ground. (ii) Berapa saatkah yang diambil oleh batu itu untuk melepasi bahagian atas bangunan semasa jatuh ke bawah? How many seconds are taken by the stone to pass the top of the building on its way down? (iii) Cari ketinggian maksimum yang boleh dicapai oleh batu itu dan nyatakan masa yang sepadan dengannya. Find the maximum height that can be reached by the stone and state its corresponding time. [4 markah/marks] (d) Sebuah roket mainan dilancarkan ke udara dari kaki bangunan pada waktu yang sama dengan batu yang dilontarkan dari puncak bangunan itu. Pergerakan roket itu diberikan oleh s = 20t. TP 4 BT ms.26 – 27 A toy rocket is launched into the air from the foot of the building at the same time as the stone thrown from the top of the building. The movement of the rocket is given by s = 20t. (i) Pada graf di 10(b), lukis graf s = 20t. On the graph in 10(b), draw the graph of s = 20t. (iii) Pada saat ke berapakah batu dan roket itu berada pada ketinggian yang sama dari permukaan tanah? At what time, in seconds, the stone is at the same height with the rocket from the ground? [5 markah/marks] Jawapan/Answer: 0 1 2 3 4 5 6 7 200 — 150 — 100 — 50— Tk4 PT24 M1(1-8).indd 8 1/24/2024 10:35:34 AM
9 2.1 Asas Nombor / Number Bases 1 Rajah 1 menunjukkan satu set nombor yang mempunyai asas yang sama. Diagram 1 shows a set of numbers of the same base. 23, 451, 83, 33, 100 Rajah 1/Diagram 1 Apakah asas bagi nombor-nombor itu? What is the base of the numbers? A 6 C 8 B 7 D 9 TP 2 BT ms.34 – 35 2 Antara pasangan nombor berikut, yang manakah tidak betul? Which of the following pair of numbers is not correct? A 1012 , 1000102 C 667 , 1007 B 2344 , 1204 D 21009 , 34289 TP 2 BT ms.34 – 35 3 Apakah nilai tempat bagi digit 2 dalam 2103 ? What is the place value of the digit 2 in 2103 ? A 9 C 27 B 25 D 45 TP 2 BT ms.35 – 36 4 Nyatakan nilai digit yang bergaris dalam nombor berikut. State the value of the underlined digit in the following number. 41537 A 4 C 343 B 28 D 1 372 TP 2 BT ms.36 – 37 5 (5 × 63 ) + (7 × 62 ) + (3 × 61 ) + (4 × 60 ) = (3 × 73 ) + (6 × 72 ) + (4 × 71 ) + (p × 70 ) Cari nilai p. KBAT Menilai Find the value of p. A 3 C 5 B 4 D 6 TP 4 BT ms.37 – 39 6 Apakah tiga nombor yang berturutan dalam jujukan di bawah? What are the three consecutive numbers in the following sequence? 10102 , 10112 , 11002 , 11012 , , , , A 10002 , 11112 , 101002 B 11102 , 10112 , 101102 C 10002 , 10112 , 111002 D 11102 , 11112 , 100002 TP 3 BT ms.37 – 39 7 Nyatakan nilai bagi digit yang bergaris dalam nombor berikut. State the value of the underlined digit in the following number. 455016 A 5 C 180 B 14 D 1060 TP 3 BT ms.37 – 39 8 Hitung hasil tambah nilai digit 7 dalam 88719 dan 173118 . Calculate the sum of the values of digit 7 in 88719 and 173118 . A 956 C 3 647 B 1 365 D 7 070 TP 3 BT ms.41 – 46 9 Diberi bahawa 1011112 + 3114 = 10p013 , cari nilai p. Given that 1011112 + 3114 = 10p013 , fi nd the value of p. A 0 C 2 B 1 D 3 TP 3 BT ms.45 – 46 10 789 – 1113 = y4 Apakah nilai y? What is the value of y? A 322 C 32 B 58 D 9 TP 3 BT ms.41 – 44, 47 – 48 Bab 2 Asas Nombor Number Bases Bidang Pembelajaran: Nombor dan Operasi Tk4 PT24 M2(9-13).indd 9 1/24/2024 10:34:21 AM
10 11 Diberi bahawa 9(6 × 94 + 3 × 92 + 1) = x 9 , maka x = KBAT Mengaplikasi Given that 9(6 × 94 + 3 × 92 + 1) = x9 , then x = A 60301 C 603100 B 603010 D 630101 TP 4 BT ms.37 – 39 12 Diberi bahawa P8 = 10101101012 , cari nilai P. Given that P8 = 10101101012, find the value of P. A 1265 C 5231 B 1625 D 5321 TP 3 BT ms.43 – 44 13 31436 – 2456 = A 24456 C 25446 B 24546 D 25456 TP 3 BT ms.47 – 48 14 Tukarkan 3 × 74 + 2 × 72 + 6 kepada suatu nombor dalam asas tujuh. Convert 3 × 74 + 2 × 72 + 6 into a number in base seven. A 300267 C 302607 B 302067 D 320067 TP 3 BT ms.37 – 39 15 Diberi bahawa Q5 = 2568 , cari nilai Q. Given that Q5 = 2568, find the value of Q. A 1144 C 2011 B 1414 D 2101 TP 3 BT ms.42 – 43 16 Jadual 1 menunjukkan nilai setara bagi nombor dalam asas 4 dan asas 2. Table 1 shows the equivalent value of numbers in base 4 and base 2. Asas 4/ Base 4 Asas 2/ Base 2 0 00 1 01 2 10 3 11 Jadual 1/Table 1 Cari nilai setara 130214 dalam asas 2. Find the equivalent value of 130214 in base 2. A 1100010012 C 1110010012 B 1101010012 D 10110010012 TP 3 BT ms.42 – 43 17 111002 – 10012 = 1p0q12 A p = 0, q = 0 C p = 1, q = 0 B p = 0, q = 1 D p = 1, q = 1 TP 3 BT ms.47 – 48 18 Nyatakan nilai digit 2 bagi nombor 52317 dalam asas sepuluh. State the value of digit 2 in the number 52317 in base ten. A 14 C 49 B 28 D 98 TP 2 BT ms.36 – 37 19 Diberi 100100000112 = 2x + 4 + 27 + 3. Cari nilai x. Given 100100000112 = 2x + 4 + 27 + 3. Find the value of x. A 5 C 7 B 6 D 8 TP 4 BT ms.37 – 39 20 Ungkapkan 64 + 12 sebagai suatu nombor dalam asas enam. Express 64 + 12 as a number in base six. A 10026 C 100026 B 10206 D 100206 TP 4 BT ms.41 – 42 21 12415 + 3115 = A 16525 C 23025 B 21025 D 22125 TP 3 BT ms.45 – 46 22 Apakah nilai digit 3, dalam asas sepuluh, bagi nombor 13204 ? What is the value of digit 3, in base ten, in the number 13204 ? A 12 C 48 B 16 D 192 TP 2 BT ms.36 – 37 23 120013 – 11103 = A 101213 B 110213 C 111203 D 111213 TP 3 BT ms.47 – 48 24 Diberi bahawa m10 = 2179 , dengan keadaan m ialah suatu integer. Cari nilai m. Given that m10 = 2179, where m is an integer. Find the value of m. A 52 C 178 B 162 D 1 602 TP 3 BT ms.41 – 42 25 32348 + 4568 = A 36908 C 36128 B 36028 D 37128 TP 3 BT ms.45 – 46 Tk4 PT24 M2(9-13).indd 10 1/24/2024 10:34:21 AM
11 1 Isi tempat kosong dengan langkah pengiraan yang betul. TP 3 BT ms.37 – 39 Fill in the blanks with the correct calculation steps. Nilai tempat / Place value 72 71 70 Digit / Digit 1 3 2 1327 = (1 × 72 ) + (3 × 71 ) + (2 × 70 ) = = [3 markah/marks] 2 Tukarkan 1102003 kepada asas 9. TP 3 BT ms.42 – 43 Convert 1102003 to base 9. [3 markah/marks] Jawapan/Answer: 3 Diberi bahawa 14416 + 1356 = p6 , cari nilai p. TP 3 BT ms.45 – 46 Given that 14416 + 1356 = p6, find the value of p. [3 markah/marks] Jawapan/Answer: 4 Bahrin tidak dapat mengingati harga kos bagi semua barangan yang dijual di kedai runcitnya. Untuk mengelakkan harga kos sebenar diketahui oleh pelanggan, dia menulis harga kos bagi setiap barangan dalam asas empat. Jadual 1 menunjukkan harga kos beberapa barang dan keuntungan yang Bahrin ingini. TP 5 BT ms.49 Bahrin cannot remember the cost price of all the items sold in his grocery store. To avoid customers from knowing the cost prices, he writes the cost price of each item in base four. Table 1 shows the labelled cost of some items and the profit Bahrin intends to make. 26 31457 – 2467 = A 25567 B 25667 C 26567 D 26657 TP 3 BT ms.47 – 48 27 Diberi bahawa 2 × 94 + 4 × 92 + 9p = 204309 , cari nilai p. Given that 2 × 94 + 4 × 92 + 9p = 204309, find the value of p. A 0 C 3 B 2 D 4 TP 4 BT ms.37 – 39 2.1 Asas Nombor / Number Bases Bahagian A / Section A Tk4 PT24 M2(9-13).indd 11 1/24/2024 10:34:21 AM
12 Barangan Goods Harga kos yang dilabelkan Labelled cost price Keuntungan Profit 1 paket malt coklat 3 kg 1 packet of 3 kg chocolate malt RM1324 RM5.00 1 kampit beras wangi 10 kg 1 bag of 10 kg fragrant rice RM2314 RM7.00 Jadual 1/Table 1 Berapakah harga jual seunit bagi setiap barang itu? KBAT Menilai What is the selling price per unit for each item? [4 markah/marks] Jawapan/Answer: 5 Rajah 1 menunjukkan bacaan meter elektrik pada awal dan akhir bulan bagi rumah Esha. TP 5 BT ms.49 Diagram 1 shows the readings of an electric meter at the beginning and end of the month for Esha's house. 0 0 5 7 kWh 0 2 3 4 kWh Rajah 1/Diagram 1 Bacaan-bacaan itu adalah dalam asas 8. Jika satu unit elektrik dalam asas 10 ialah RM0.65, berapakah bil elektrik rumah Esha? KBAT Menilai The readings are in base 8. If an electric unit in base 10 is RM0.65, what is Esha’s home electricity bill? [5 markah/marks] Jawapan/Answer: Tk4 PT24 M2(9-13).indd 12 1/24/2024 10:34:21 AM
13 Bahagian B / Section B 6 (a) Chew Teng menulis nombor dua digit yang paling besar dalam asas empat pada sekeping kad. Chew Hui diminta menukarnya kepada nombor binari. Apakah nombor yang Chew Hui dapat? TP 5 BT ms.49 Chew Teng wrote the largest two-digit number in base four on a card. Chew Hui is asked to convert it to a binary number. What is the number that Chew Hui gets? [4 markah/marks] Jawapan/Answer: (b) Kad-kad berikut ditulis dengan nombor dalam asas tertentu. TP 4 BT ms.49 The following cards are written with numbers in a certain base. 11112 235 324 1223 Chew Teng menyusun kad-kad nombor itu mengikut tertib menaik. Nyatakan susunannya. Chew Teng arranges the number cards in ascending order. State the arrangement. [5 markah/marks] Jawapan/Answer: Zon KBAT 1 Harga 5 kg durian ialah RM121203 . Jerrine mempunyai RM300 dan membeli 8 kg durian. Berapakah baki wangnya? TP 5 BT ms.49 KBAT Menilai The price of 5 kg of durians is RM121203 . Jerrine has RM300 and bought 8 kg of durians. How much is her change? Jawapan/Answer: Tk4 PT24 M2(9-13).indd 13 1/24/2024 10:34:21 AM
Praktis Topikal Praktis Topikal Praktis Topikal SPM Matematik (Dwibahasa) Tingkatan 4 KSSM – Jawapan J1 Fungsi dan Persamaan Kuadratik dalam Satu Pemboleh Ubah Bab 1 Kertas 1 1 D 2 C 3 B 4 C 5 C 6 A 7 D 8 C 9 B 10 C 11 B 12 A 13 D 14 C 15 D Kertas 2 Bahagian A 1 5 + m(5) – 2(5)2 = 0 5m – 45 = 0 m = 9 5 + 9x – 2x 2 = 0 2x 2 – 9x – 5 = 0 (2x + 1)(x – 5) = 0 x = – 1 2 ; x = 5 [ Penyelesaian yang satu lagi ialah/ Another solution is x = – 1 2 . 2 a = 1 > 0, graf berbentuk/ the graph is y = x 2 – 6x = x(x – 6) Apabila/ When y = 0 ; x(x – 6) = 0 x = 0 ; x = 6 Graf itu melalui titik-titik (0, 0) dan (6, 0). Th e graph passes through points (0, 0) and (6, 0). Garis x = 3 merupakan paksi simetri bagi graf. Line x = 3 is the axis of symmetry of the graph. Apabila/ When x = 3, y = (3)2 – 6(3) = –9 0 10 — 5 — — –5 — –10 — × y x (3, –9) x = 3 × 2 4 6 8 Titik minimum/ Minimum point = (3, –9) 3 3x 2 + 11x – 20 = 9x 2 – 24x + 16 6x 2 – 35x + 36 = 0 (3x – 4)(2x – 9) = 0 x = 4 3 , x = 9 2 4 (a) Harga kos sebatang pen/ Cost price for each pen = 48 x (b) Harga jual sebatang pen/ Selling price for each pen = 48 x + 3 (c) (48 x + 3)(x – 3) = 48 + 15 (48 x + 3)(x – 3) = 63 (48 + 3x)(x – 3) = 63x 3x 2 – 24x – 144 = 0 x 2 – 8x – 48 = 0 (x – 12)(x + 4) = 0 x = 12, x = –4 (x > 0) ∴ x = 12 5 w(w + 5)(40) = 12 000 w(w + 5) = 300 w2 + 5w – 300 = 0 (w – 15)(w + 20) = 0 w = 15, w = –20 (w > 0) ∴ w = 15 6 (a) n 2 (n – 3) = 90 n2 – 3n – 180 = 0 (n – 15)(n + 12) = 0 n = 15; n = –12 (n . 0) ∴ n = 15 Poligon mempunyai 15 sisi/ Th e polygon has 15 sides. (b) Tiada/ No n 2 (n – 3) = 90 n2 – 3n –100 = 0 Tiada nombor bulat yang positif untuk n. Th ere is no positive whole number for n. 7 (a) Luas kawasan berlorek = Luas PQRS – Luas WQV – Luas STU Area of shaded region = Area of PQRS – Area of WQV – Area of STU = 14(3x + 4) – 1 2(x)(x) – 1 2(2x)(2x) = 42x + 56 – 1 2x2 – 2x2 = – 5 2x2 + 42x + 56 (b) – 5 2 x2 + 42x + 56 = 184 5x2 – 84x + 256 = 0 (x – 4)(5x – 64) = 0 x = 4, x = 64 5 (x = nombor bulat/ whole number) ∴ x = 4 Bahagian B 8 (a) f(x) = x2 + 8x + 13 x –8 –5 –4 –1 0 1 2 3 y 13 –2 –3 6 13 22 33 46 (b) x y 50 40 30 20 10 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 y = x2 + 8x + 13 (i) Bentuk U/ U-shaped (ii) Titik minimum/ Minimum point: (–4, –3) (iii) Persamaan paksi simetri / Equation of axis symmetry: x = –4 9 (a) Andaikan bilangan ahli Persatuan Matematik sebagai x Assume the number of members in Mathematics Society is x 360 x – 360 x + 10 = 6 360(x + 10) – 360x = 6x(x + 10) 6x 2 + 60x – 600 = 0 x 2 + 10x – 600 = 0 (x – 20)(x + 30) = 0 x = 20, x = –30 (x > 0) Bilangan ahli Persatuan Matematik ialah 20 orang. Th e number of members in Mathematics Society is 20. Jawapan Tk4 PT24 M Jaw(J1-12).indd 1 1/24/2024 10:23:32 AM
Praktis SPM Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2024 J2 (b) RM360 (20 + 10) = RM12 (c) RM360 20 = RM18 atau/or RM12 + RM6 = RM18 Bahagian C 10 (a) t 0 1 2.5 3 5 6 s 95 160 197.5 194 100 5 (b) s t s = 20t s = 95 + 81t – 16t 2 0 1 2 3 4 5 6 7 200— 150— 100— 50— × × × × × × (c) (i) Lebih daripada 6 saat atau 6.05 saat. More than 6 seconds or 6.05 seconds. (ii) 5.1 saat/ seconds (iii) 197.5 m , x = 2.5 s (d) (i) x 0 5 y 0 100 (ii) 5 saat/ seconds Batu dan roket berada pada ketinggian 100 m dari tanah pada saat ke-5. The stone and the rocket were at 100 m above the ground at the 5th seconds. Bab 2 Asas Nombor Kertas 1 1 D 2 B 3 A 4 D 5 A 6 D 7 C 8 C 9 C 10 A 11 B 12 A 13 B 14 B 15 A 16 C 17 B 18 D 19 B 20 D 21 B 22 C 23 A 24 C 25 D 26 B 27 C Kertas 2 Bahagian A 1 49 + 21 + 2 = 7210 2 3 1 3 1 3 1 1 1 0 2 0 0 4 2 0 1102003 = 4209 3 [ p = 2020 1 1 1 1441 6 + 135 6 2020 6 4 42 41 40 RM1324 = 1 × 42 + 3 × 41 + 2 × 40 = 16 + 12 + 2 = 3010 Harga jual bagi 1 malt coklat Selling price for 1 packet of chocolate malt = RM30 + RM5 = RM3510 RM2314 = 2 × 42 + 3 × 41 + 1 × 40 = 32 + 12 + 1 = 4510 Harga jual bagi 1 kampit beras wangi Selling price for 1 bag of fragrant rice = RM45 + RM7 = RM5210 5 2348 – 578 = 1558 82 81 80 1 5 58 = 1 × 82 + 5 × 81 + 5 × 80 = 64 + 40 + 5 = 10910 Bil elektrik/ Electricity bill = 109 × RM0.65 = RM70.85 Bahagian B 6 (a) 334 = 21 20 21 20 1 1 1 12 334 = 11112 (b) 23 22 21 20 1 1 1 12 = 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20 = 8 + 4 + 2 + 1 = 1510 52 50 2 35 = 2 × 51 + 3 × 50 = 10 + 3 = 1310 41 40 3 24 = 3 × 41 + 2 × 40 = 12 + 2 = 1410 32 31 30 1 2 23 = 1 × 32 + 2 × 31 + 2 × 30 = 9 + 6 + 2 = 1710 Maka, susunannya Therefore, the arrangement is: 235 , 324 , 11112 , 1223 Zon KBAT 1 34 33 32 31 30 1 2 1 2 03 = 1 × 34 + 2 × 33 + 1 × 32 + 2 × 31 + 0 × 30 = 81 + 54 + 9 + 6 +0 = 15010 Harga 1 kg durian/ Price of 1 kg of durian = RM15010 ÷ 5 = RM3010 Harga 8 kg durian/ Price of 8 kg of durian = RM3010 × 8 = RM24010 Baki wang/ Money change = RM300 – RM40 = RM6010 Bab 3 Penaakulan Logik Kertas 1 1 A 2 B 3 A 4 D 5 A 6 B 7 A 8 C 9 B 10 C 11 D Kertas 2 Bahagian A 1 (a) (i) Palsu/ False (ii) Benar/ True (b) (i) 71 bukan nombor perdana/ 71 is not a prime number (ii) Hasil tambah 2 nombor ganjil bukan nombor genap The sum of 2 odd numbers is not an even number Tk4 PT24 M Jaw(J1-12).indd 2 1/24/2024 10:23:32 AM