Praktis Topikal Matematik T5

49 Jawapan/Answer: Sempadan atas Upper boundary Kekerapan Frequency Kekerapan longgokan Cumulative frequency 2 0 0 2 Jadual 1 menunjukkan taburan umur bagi sekumpulan 45 orang pekerja di sebuah kilang. Table 1 shows the distribution of ages for a group of 45 workers in a factory. Lengkapkan Jadual 1. TP 3 BT ms.199–201 Complete the Table 1. [5 markah/marks] Jawapan/Answer: Umur (tahun) Age (years) Kekerapan Frequency Titik tengah Midpoint Sempadan atas Upper boundary Kekerapan longgokan Cumulative frequency 17 – 21 0 19 21.5 22 – 26 3 24 27 – 31 7 32 – 36 10 37 – 41 11 42 – 46 8 47 – 51 6 Jadual 1/Table 1 Tk 5 23 PT Math 7(47-55).indd 49 23/2/2023 5:59:36 PM PENERBIT ILMU BAKTI SDN. BHD.

50 3 Rajah 2 menunjukkan dua histogram. BT ms.204 Diagram 2 shows two histograms. Kekerapan Frequency 6 4 2 0 64.5 69.5 74.5 79.5 84.5 89.5 94.5 99.5 8 10 12 14 Markah Marks Kekerapan Frequency 6 4 2 0 64.5 69.5 74.5 79.5 84.5 89.5 94.5 99.5 8 10 12 14 Markah Marks Ujian 1/Test 1 Ujian 2/Test 2 Rajah 2/Diagram 2 (a) Nyatakan bentuk taburan setiap histogram itu. TP 2 State the shape of distribution for each histogram. (b) Bandingkan serakan dua histogram itu. TP 4 Compare the dispersions of two histograms. [4 markah/marks] Jawapan/Answer: 4 Rajah 3 menunjukkan gabungan dua set data terkumpul berdasarkan poligon kekerapan. BT ms.204–205 Diagram 3 shows a combination of two sets of grouped data based on frequency polygon. 3 2 1 0 5 10 15 20 25 30 35 4 5 6 7 Jisim (kg) Mass (kg) Jisim murid kelas A Masses of pupils in class A Petunjuk/Key: Jisim murid kelas B Masses of pupils in class B Kekerapan Frequency Jisim murid di kelas A dan kelas B Masses of pupils in class A and class B Rajah 3/Diagram 3 Tk 5 23 PT Math 7(47-55).indd 50 23/2/2023 5:59:37 PM PENERBIT ILMU BAKTI SDN. BHD.

51 (a) Nyatakan bentuk taburan setiap poligon kekerapan itu. TP 2 State the shape of distribution for each frequency polygon. (b) Bandingkan serakan dua poligon kekerapan itu. TP 4 Compare the dispersions of two frequency polygons. (c) Kelas manakah yang mungkin mewakili murid prasekolah dan murid sekolah rendah? TP 4 Which class represents a preschool pupils and which class represents an elementary school pupils? [5 markah/marks] Jawapan/Answer: 7.2 Sukatan Serakan / Measures of Dispersion 5 Rajah 4 ialah suatu ogif yang menunjukkan markah yang diperoleh sekumpulan murid dalam suatu ujian Matematik. TP 5 BT ms.209–210 Diagram 4 is an ogive showing the marks obtained by a group of students in a Mathematics test. Berdasarkan ogif, cari persentil ke-10, persentil ke-25 dan persentil ke-70. Based on the ogive,  nd the 10th percentile, the 25th percentile and the 70th percentile. [4 markah/marks] Jawapan/Answer: 15 10 5 0 P10 P25 P70 39.5 44.5 49.5 54.5 59.5 64.5 20 25 30 35 40 Markah Marks × 40 = 28 70 100 × 40 = 10 25 100 × 40 = 4 10 100 Kekerapan longgokan Cumulative frequency Rajah 4/Diagram 4 Tk 5 23 PT Math 7(47-55).indd 51 23/2/2023 5:59:37 PM PENERBIT ILMU BAKTI SDN. BHD.

52 Bahagian B / Section B 6 Banding dan tafsirkan sukatan serakan bagi dua set data terkumpul di bawah. TP 4 BT ms.204–205 Compare and interpret the measures of dispersion of two sets of grouped data below. Poligon kekerapan R/Frequency polygon R Poligon kekerapan S/Frequency polygon S Kekerapan Frequency 4 3 2 1 0 37 42 47 52 57 62 67 72 77 5 6 7 Jisim (kg) Mass (kg) 4 3 2 1 0 37 42 47 52 57 62 67 72 77 5 6 7 Kekerapan Frequency Jisim (kg) Mass (kg) [9 markah/marks] Jawapan/Answer: Bahagian C / Section C 7 Jadual 2.1 menunjukkan taburan kekerapan bagi markah yang dikutip oleh 80 orang murid. BT ms.217–218 Table 2.1 shows the frequency distribution of the marks obtained by 80 students. Markah Mark 50 – 54 55 – 59 60 – 64 65 – 69 70 – 74 75 – 79 80 – 84 Kekerapan Frequency 4 9 12 20 23 9 3 Jadual 2.1/Table 2.1 Tk 5 23 PT Math 7(47-55).indd 52 23/2/2023 5:59:37 PM PENERBIT ILMU BAKTI SDN. BHD.

53 (a) (i) Nyatakan kelas mod. TP 2 State the modal class. (ii) Hitung anggaran min markah yang dikutip oleh kumpulan murid itu. TP 3 Calculate the estimated mean of the marks obtained by the group of students. [5 markah/marks] (b) Berdasarkan Jadual 2.1, lengkapkan Jadual 2.2 untuk menunjukkan taburan kekerapan longgokan bagi markah. TP 3 Based on Table 2.1, complete Table 2.2 to show the cumulative frequency distribution of marks. [3 markah/marks] (c) Menggunakan skala 2 cm kepada 5 markah pada paksi mengufuk dan 2 cm kepada 10 murid pada paksi mencancang, lukis ogif bagi data itu. TP 4 Using the scale of 2 cm to 5 marks on the horizontal axis and 2 cm to 10 students on the vertical axis, draw an ogive for the data. Sempadan atas Upper boundary Kekerapan longgokan Cumulative frequency 49.5 0 54.5 Jadual 2.2/Table 2.2 [4 markah/marks] (d) 25% daripada semua murid dalam kumpulan itu memperoleh markah kurang daripada m. Muridmurid ini akan menghadiri kelas pemulihan. Menggunakan ogif yang anda lukis di 7(c), tentukan nilai m. TP 5 25% of all students in the group have marks less than m.  ese students will attend remedial classes. Using the ogive you had drawn in 7(c), determine the value of m. [3 markah/marks] Jawapan/Answer: Tk 5 23 PT Math 7(47-55).indd 53 23/2/2023 5:59:38 PM PENERBIT ILMU BAKTI SDN. BHD.

54 8 Data dalam Rajah 5 menunjukkan markah yang diperoleh 40 orang murid dalam ujian objektif Biologi. BT ms.198–219  e data in Diagram 5 shows the marks obtained by 40 students in a Biology objective test. 39 31 20 23 31 35 26 41 18 22 40 33 37 25 36 26 27 30 34 49 39 46 32 41 18 32 23 36 21 35 27 40 38 43 18 28 42 47 24 32 Rajah 5/Diagram 5 (a) Berdasarkan data dalam Rajah 5, lengkapkan Jadual 3 di bawah. TP 3 Based on the data in Diagram 5, complete Table 3 below. [3 markah/marks] Selang kelas Class interval Titik tengah Midpoint Kekerapan Frequency 15 – 19 17 Jadual 3/Table 3 Tk 5 23 PT Math 7(47-55).indd 54 23/2/2023 5:59:38 PM PENERBIT ILMU BAKTI SDN. BHD.

55 (b) Berdasarkan Jadual 3, TP 3 Based on Table 3, (i) hitung anggaran min markah yang diperoleh seorang murid, calculate the estimated mean mark obtained by a student, (ii) cari varians,  nd the varians, (iii) cari nilai sisihan piawai.  nd the value of standard deviation. [6 markah/marks] (c) Untuk ceraian soalan ini, gunakan kertas graf yang disediakan. Menggunakan skala 2 cm kepada 5 markah pada paksi mengufuk dan 2 cm kepada 1 murid pada paksi mencancang, lukis poligon kekerapan bagi data itu. TP 4 For this part of the question, use the graph paper provided. Using a scale of 2 cm to 5 marks on the horizontal axis and 2 cm to 1 student on the vertical axis, draw a frequency polygon for the data. [4 markah/marks] (d) Berdasarkan poligon kekerapan di 8(c), nyatakan bilangan murid yang memperoleh lebih daripada 34 markah. TP 3 Based on the frequency polygon in 8(c), state the number of students who obtained more than 34 marks. [2 markah/marks] Jawapan/Answer: Tk 5 23 PT Math 7(47-55).indd 55 23/2/2023 5:59:38 PM PENERBIT ILMU BAKTI SDN. BHD.

56 8.1 Pemodelan Matematik Mathematical Modeling 1 Dalam pemodelan matematik, masalah akan diterjemahkan menjadi masalah . In mathematical modelling, a problem will be translated into a problem. A dunia nyata / simulasi real-world / simulation B dunia sebenar / matematik real-world / mathematical C matematik / simulasi mathematical / simulation D matematik / linear mathematical / linear TP 1 BT ms.228 2 Antara komponen berikut, yang manakah terlibat dalam pemodelan matematik? Which of the following components are involved in mathematical modeling? I Memurnikan model matematik Re ning the mathematical modeling II Membuat anggaran dan mengenal pasti hipotesis Making estimation and identifying hypothesis III Mengenal pasti dan mendefi nisikan masalah Identifying and de ning the problem IV Melaporkan dapatan Report the  ndings A I, II dan/and III C I , III dan/and IV B I, II dan/and IV D II, III dan/and IV TP 2 BT ms.229 3 Kelab iCycle mengadakan kempen kitar semula di kawasan kejiranan. Mereka mendirikan pusat sehenti pengumpulan bahan kitar semula di kawasan sekitar dan ingin mengkaji kadar penggunaannya. Antara berikut, yang manakah merupakan andaian yang betul tentang bilangan penduduk yang akan menggunakan sebaiknya pusat sehenti itu?  e iCycle Club is organising a recycling campaign in a neighbourhood.  ey setup one-stop recycling collection centre around the neighbourhood and would like to study their usage rate. Which of the following is a valid assumption about how many people would make use of the one-stop centre? A Diandaikan bahawa 100% orang yang berada berdekatan pusat sehenti akan menggunakannya It is assumed that 100% of the people near the one-stop center would use it B Diandaikan bahawa tiada orang yang berada berdekatan pusat sehenti yang akan menggunakannya It is assumed that none of the people near the one stop center would use it C Diandaikan bahawa 20% orang yang berada berdekatan pusat sehenti akan menggunakannya berdasarkan kajian penyelidikan sebelumnya It is assumed that 20% of the people near the one-stop center would use it based on a previous research study D Diandaikan bahawa 90% orang yang berada berdekatan pusat sehenti akan menggunakannya kerana rumah mereka berada dalam jarak 0.5 km dari pusat sehenti It is assumed that 90% of the people near the one-stop center would use it as their house is within 0.5 km from the one-stop centers TP 3 BT ms.230–233 4 Nathan menyimpan 30 buah  lem dalam cakera kerasnya. Bilangan ini bertambah sebanyak dua buah  lem setiap minggu. Nathan ingin mengetahui tempoh masa yang diperlukan untuk memiliki sejumlah 100 buah  lem. Apakah pemboleh ubah yang terlibat dalam masalah ini? Nathan stores 30 movies on his hard drive.  is number grows by two movies every week. He wants to know how long it will take him to have a collection of 100 movies. What are the variables involved in this problem? I masa dalam minggu, t/ time in weeks, t II bilangan fi lem, n number of movies, n III bilangan cakera keras, x the number of hard drives, x IV jenis fi lem, y/ the type of movies, y Kertas 1 / Paper 1 Bidang Pembelajaran: Perkaitan dan Algebra Pemodelan Matematik Mathematical Modeling BAB 8 Tk 5 23 PT Math 8(56-58).indd 56 23/2/2023 6:00:12 PM PENERBIT ILMU BAKTI SDN. BHD.

57 A I dan/and II B I dan/and III C II dan/and III D III dan/and IV 5 Diberi bahawa suhu di bandar A pada bulan Julai ialah –5 °C. Suhunya menurun sebanyak –2 °C setiap bulan bagi tempoh enam bulan berturutan. Berapakah suhu di bandar A pada bulan Oktober? It is given that the temperature of town A is –5 °C in July.  e temperature decreases for –2 °C per month for six consecutive months. What is the temperature of town A in October? A –7 °C B –9 °C C –11 °C D –13 °C TP 4 BT ms.234 1 Danish mempunyai simpanan sebanyak RM2 000 di Bank Bersatu dengan kadar faedah 1.8% setahun. Dia mahu membeli sebuah komputer riba berharga RM3 800 dengan wang simpanannya. Berapakah tempoh masa yang diambil oleh Danish untuk menabung bagi membeli komputer riba itu? TP 4 BT ms.230–233 Danish saves RM2 000 at Bank Bersatu with a simple interest rate of 1.8% per annum. He wants to buy a laptop worth RM3 800 with his savings. How long does Danish need to save in order to buy the laptop? (a) Kenal pasti dan tentukan masalahnya. Identify and de ne the problem. (b) Tentukan andaian yang perlu dibuat dan kenal pasti pemboleh ubah dalam menyelesaikan masalah. Determine the assumptions that need to be made and identify the variables in solving the problem. [5 markah/marks] Jawapan/Answer: 2 Kesesakan lalu lintas sering menjadi masalah di Bandar A. Majlis bandaran telah membuat beberapa inisiatif untuk mengurangkan kesesakan lalu lintas di kawasan tersebut. Bolehkah pemodelan matematik digunakan untuk menentukan pelan yang paling berkesan untuk mengatasi masalah tra k ini? TP 3 BT ms.220–230 Tra c congestion has always been a pain in the neck for City A.  e city council has developed several plans to reduce tra c congestion in the area. Can a mathematical model be used to determine the most e ective plan to solve the tra c problems? [3 markah/marks] Jawapan/Answer: 8.1 Pemodelan Matematik/ Mathematical Modeling Kertas 2/Paper 2 Bahagian A / Section A Tk 5 23 PT Math 8(56-58).indd 57 23/2/2023 6:00:12 PM PENERBIT ILMU BAKTI SDN. BHD.

58 Bahagian B / Section B 3 Taman Naruli telah dibangunkan pada tahun 2016. Pada awalnya, bilangan penduduk di situ ialah 450 orang. Menjelang tahun 2021, populasi taman itu telah meningkat menjadi 880 orang. Andaikan penduduk bertambah secara linear, TP 5 BT ms.230–240 Taman Naruli was established in 2016.  e initial population was 450. By 2021, the population grows to 880. Assume the population grows linearly, (a) ramalkan populasi pada tahun 2026. predict the population in 2026. (b) kenal pasti pada tahun ke berapa populasi akan mencecah 1 400 orang. identify the year in which the population will reach 1 400. [8 markah/marks] Jawapan/Answer: Bahagian C / Section C 4 Super Cool Sdn. Bhd. mengeluarkan alat penyaman udara. Pengilang itu mempunyai kos tetap bulanan sebanyak RM360 000 dan kos pengeluaran sebanyak RM730 untuk setiap unit yang dihasilkan. Produk ini dijual pada harga RM1 400 seunit. BT ms.230–240 Super Cool Sdn. Bhd. manufactures air-conditioners.  e manufacturer has a monthly  xed cost of RM360 000 and a production cost of RM730 for each unit produced.  e product is sold for RM1 400 per unit. (a) Bina pemodelan matematik untuk meramalkan keuntungan atau kerugian syarikat. TP 4 Build a mathematical modeling to predict the pro t or loss for the company. [7 markah/marks] (b) Gunakan pemodelan matematik di 4(a) untuk mencari TP 5 Use the mathematical modeling in 4(a) to  nd (i) untung atau rugi sepadan dengan tahap pengeluaran 500 unit. the pro t or loss corresponding to production levels of 500 units. (ii) untung atau rugi sepadan dengan tahap pengeluaran 600 unit. the pro t or loss corresponding to production levels of 600 units. (iii) bilangan unit yang perlu dijual untuk pulang modal. the number of units need to be sold to break even. [8 markah/marks] Jawapan/Answer: Tk 5 23 PT Math 8(56-58).indd 58 23/2/2023 6:00:12 PM PENERBIT ILMU BAKTI SDN. BHD.

59 Arahan: Kertas peperiksaan ini mengandungi 40 soalan. Jawab semua soalan. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. Anda dibenarkan menggunakan kalkulator sainti k. Instructions:  is question paper consists of 40 questions. Answer all questions.  e diagrams provided in the questions are not drawn to scale unless stated. You may use a scienti c calculator. 1 Bundarkan 44 550 kepada 3 angka bererti. Round o 44 550 to 3 signi cant  gures. A 44 500 C 44 600 B 44 550 D 44 700 2 Ungkapkan 30y2 – 3y – 6 dalam bentuk termudah. Express 30y2 – 3y – 6 in its simplest form. A (6y – 3)(5y + 2) C 3(2y + 1)(5y – 2) B (15y + 6)(2y – 1) D 3(2y – 1)(5y + 2) 3 Rajah 1 menunjukkan sebuah tangki air berjejari 20 cm. Diagram 1 shows a water tank with a radius of 20 cm. 700 cm 20 cm Rajah 1/Diagram 1 Jika 40% daripada tangki itu diisi dengan air, hitung isi padu, dalam cm3 , air di dalam tangki tersebut. If 40% of the tank is  lled with water, calculate the volume, in cm3 , of water in the tank. [Guna/ Use π = 22 7 ] A 2.36 × 105 C 3.52 × 105 B 3.48 × 105 D 3.68 × 106 4 Nilai bagi digit 6 dalam nombor 61758 ialah 6 × 8m. Cari nilai m.  e value of digit 6 in the number 61758 is 6 × 8m. Find the value of m. A 3 C 5 B 4 D 6 5 101102 + 1101012 = A 10010112 C 10110112 B 10011112 D 11010112 6 Antara berikut, yang manakah faktor yang boleh mempengaruhi pelan kewangan jangka panjang? Which of the following are the factors that can in uence a long-term  nancial plan? I Kadar in asi/ In ation rates II Bil elektrik/ Electrical bill III Hobi/ Hobby IV Polisi kerajaan/ Government policy A I dan/ and II C II dan/ and III B I dan/ and IV D III dan/ and IV 7 Rajah 2 menunjukkan dua buah segi empat sama, P dan Q, yang dilukis pada grid segi empat sama. Diagram 2 shows two squares, P and Q, drawn on a grid of equal squares. P Q A B D C Rajah 2/Diagram 2 P ialah imej bagi Q di bawah satu pembesaran. Antara titik A, B, C dan D, yang manakah pusat pembesaran itu? P is the image of Q under an enlargement. Which of the points, A, B, C or D, is the centre of enlargement? 8 Dalam Rajah 3, RST ialah tangen kepada bulatan di titik S. In Diagram 3, RST is a tangent to the circle at point S. Kertas 2 / Paper 1 [40 markah/marks] Masa: 1 jam 30 minit KERTAS MODEL SPM Tk 5 23 PT Math KM(59-80).indd 59 24/2/2023 8:21:03 AM PENERBIT ILMU BAKTI SDN. BHD.

60 40° 52° w° R V U T S Rajah 3/Diagram 3 Cari nilai w. Find the value of w. A 40 C 88 B 52 D 92 9 Antara hubungan berikut, yang manakah ialah suatu fungsi? Which of the following relations is a function? A C x 1 3 5 1 2 y x 1 1 2 5 4 y B D x 8 1 2 3 y x 2 1 2 6 4 y 10 Rajah 4 menunjukkan suatu plot-titik bagi suatu data. Diagram 4 shows a dot-plot of a data. 2 3 4 5 6 7 8 9 Rajah 4/Diagram 4 Cari kuartil ketiga bagi data itu. Find the third quartile of the data. A 5 C 7 B 6 D 8 11 Dalam Rajah 5, FGH ialah garis lurus. In Diagram 5, FGH is a straight line. 12 cm 5 cm θ E F G H Rajah 5/Diagram 5 Apakah nilai bagi kos θ? What is the value of cos θ? A – 5 13 C 10 13 B 5 13 D 12 13 12 Antara berikut, yang manakah mewakili graf y = sin x bagi 90° < x < 270°? Which of the following represents the graph y = sin x for 90° < x < 270°? A C B D 1 − 0 90° 180° 270° 1 − y x 1 − 0 90° 180° 270° 1 − y x 1 − 0 90° 180° 270° 1 − y x 1 − 0 90° 180° 270° 1 − y x 13 Dalam Rajah 6, titik M dan titik N terletak di atas lengkok suatu bulatan unit berpusat O. In Diagram 6, point M and point N lie on the arc of a unit circle with centre O. M(0.53, 0.85) N(0.88, 0.47) y O x q° p° Rajah 6/Diagram 6 Cari nilai bagi (kos p° + sin q°). Find the value of (cos p° + sin q°). A 1.015 C 1.410 B 1.321 D 1.730 14 Antara graf berikut, yang manakah menunjukkan objek itu bergerak dengan laju seragam? Which of the following graphs shows the object is moving with uniform speed? A Jarak (m) Distance (m) Masa (saat) Time (seconds) O Tk 5 23 PT Math KM(59-80).indd 60 24/2/2023 8:21:03 AM PENERBIT ILMU BAKTI SDN. BHD.

61 B Laju (m s–1) Speed (m s–1) Masa (saat) Time (seconds) O C Laju (m s–1) Speed (m s–1) Masa (saat) Time (seconds) O D Laju (m s–1) Speed (m s–1) Masa (saat) Time (seconds) O 15 Matriks songsang bagi   3 2 5 4 ialah 1 a   4 –2 –5 3 . Cari nilai bagi 2a.  e inverse matrix of   3 2 5 4 is 1 a   4 –2 –5 3 . Find the value of 2a. A 2 C 4 B –2 D –4 16 Diberi bahawa/ It is given that   3 2 –4 8 – 1 2 Q =   –2 6 7 1 . Cari matriks Q. Find the value of matrix Q. A   5 –4 –11 7 C   10 8 22 14 B   5 4 11 7 D   10 –8 –22 14 17 Cari nilai x bagi [4x 2]  –3 1 = [–22]. Find the value of x for[4x 2]  –3 1 = [–22]. A 2 C 8 B 4 D 10 18 Antara berikut, yang manakah bukan pernyataan? Which of the following is not a statement? A 3 + 6 = 36 B 42 < 24 C 3 –64 + 4 = 0 D 3x – 2 = 7 19 Diberi implikasi: “Jika x = 4, maka 2x – 3 = 5”. Nyatakan songsangan bagi implikasi itu. Given the implication: “If x = 4, then 2x – 3 = 5”. State the inverse of the implication. A Jika x ≠ 4, maka 2x – 3 ≠ 5 If x ≠ 4, then 2x – 3 ≠ 5 B Jika 2x – 3 = 5, maka x ≠ 4 If 2x – 3 = 5, then x ≠ 4 C Jika 2x – 3 = 5, maka x = 4 If 2x – 3 = 5, then x = 4 D Jika 2x – 3 ≠ 5, maka x ≠ 4 If 2x – 3 ≠ 5, then x ≠ 4 20 Rajah 7 menunjukkan suatu graf H. Diagram 7 shows a graph H. p q r t s Rajah 7/Diagram 7 Antara graf berikut, yang manakah bukan subgraf bagi H? Which of the following graphs is not a subgraph of H? A p C s q r q s r p B D p q t s p t q s 21 Rajah 8 menunjukkan suatu graf dengan bucu p, q, r, s dan t. Diagram 8 shows a graph with vertices p, q, r, s and t. p t q s r Rajah 8/Diagram 8 Cari bucu yang mempunyai darjah bersamaan dengan 4. Find the vertex with degree equal to 4. A p B q C r D s H Tk 5 23 PT Math KM(59-80).indd 61 24/2/2023 8:21:03 AM PENERBIT ILMU BAKTI SDN. BHD.

62 22 Rajah 9 menunjukkan suatu graf dengan berbilang tepi. Diagram 9 shows a graph with multiple edges. j k l n m 4 44 3 6 2 2 3 Rajah 9/Diagram 9 Cari kos yang paling optimum untuk laluan dari j ke l. KBAT Menganalisis Find the most optimal cost to travel from j to l. A 6 B 7 C 8 D 10 23 Rajah 10 ialah sebuah ogif yang menunjukkan taburan umur bagi 40 orang peserta dalam suatu pertandingan memasak. Diagram 10 is an ogive showing the age distribution of 40 participants in a cooking competition. 15 10 5 0 9.5 14.5 19.5 24.5 29.5 34.5 39.5 20 25 30 35 40 Umur (Tahun) Age (Years) Kekerapan longgokan Cumulative frequency Rajah 10/Diagram 10 Anggarkan nilai julat antara kuartil. Estimate the interquartile range. A 10 B 11 C 12 D 13 24 Jadual 1 menunjukkan masa menunggu di sebuah restoran sebelum pesanan diambil. Table 1 shows the waiting time at a restaurant before ordering. Masa (minit) Time (minutes) Kekerapan, f Frequency, f Kekerapan longgokan Cumulative frequency 6 – 10 2 2 11 – 15 5 7 16 – 20 9 16 21 – 25 8 24 26 – 30 5 29 31 – 35 1 30 Jadual 1/Table 1 Hitung peratusan masa menunggu melebihi 25 minit. Calculate the percentage of waiting time over 25 minutes. A 3.3% C 20% B 10% D 47% 25 Jadual 2 menunjukkan empat jenis minuman kegemaran bagi sekumpulan murid. Table 2 shows four types of favourite drinks of a group of students. Minuman kegemaran Favourite drinks Kekerapan Frequency Teh/ Tea 25 Kopi/ Co ee m Coklat/ Chocolate 30 Barli/ Barley 19 Jadual 2/Table 2 Diberi bahawa mod minuman kegemaran ialah coklat. Cari nilai maksimum bagi m. Given that the mode of the favourite drinks is chocolate. Find the maximum value of m. A 26 C 30 B 29 D 31 26 Apakah bentuk am bagi pemodelan fungsi kuadratik? What is the general form of quadratic function modeling? A y = mx + c B y = ax2 + bx + c C y = Cax D y = a(1 + r)t Tk 5 23 PT Math KM(59-80).indd 62 24/2/2023 8:21:04 AM PENERBIT ILMU BAKTI SDN. BHD.

63 27 Terdapat 170 ekor rusa dalam sebuah ladang ternakan. Populasi bagi rusa ini meningkat pada kadar 30% setahun.  ere are 170 deer in a farm.  e population of these deer increases at a rate of 30% per annum. Apakah fungsi eksponen yang menunjukkan populasi P(t) pada ladang ternakan itu t tahun dari sekarang? What is the function of the exponent showing the population of P(t) at the farm t years from now? A P(t) = 170 (0.3)t B P(t) = 170 (1.0)t C P(t) = 170 (1.3)t D P(t) = 170 (3.0)t 28 Dalam Rajah 11, gambar rajah Venn menunjukkan bilangan unsur bagi set J, K dan L. In Diagram 11, the Venn diagram shows the number of elements of sets J, K and L. 6 3x 2 8 J K x L Rajah 11/Diagram 11 Diberi ξ = J K L dan n(ξ) = 36, cari nilai bagi x. Given ξ = J K L and n(ξ) = 36,  nd the value of x. A 3 C 5 B 4 D 7 29 Diberi set semesta ξ = P Q R, antara gambar rajah Venn berikut, yang manakah mewakili set (P  Q) R? Given the universal set ξ = P Q R, which of the following Venn diagrams represents the set (P  Q) R? A C B D Q R P Q R P Q R P Q R P 30 Rajah 12 menunjukkan suatu rantau berlorek yang dipenuhi oleh tiga ketaksamaan. Diagram 12 shows a shaded region satis ed by three inequalities. y x O x = 4 y = 5 Rajah 12/Diagram 12 Antara berikut, yang manakah ialah salah satu daripada ketaksamaan itu? Which of the following is one of the inequalities? A 4x + 5y > 20 C 4x + 5y > 20 B 5x + 4y > 20 D 5x + 4y > 20 31 Rajah 13 menunjukkan ketaksamaan pada satah Cartes. Diagram 13 shows the inequalities on the Cartesian plane. y x O y = 2x x + 2y = 4 B D C A Rajah 13/Diagram 13 Antara rantau A, B, C dan D, yang manakah memuaskan ketaksamaan y < 2x dan x + 2y > 4? Which region, A, B, C or D is satis ed by inequalities y < 2x and x + 2y > 4? 32 Rajah 14 menunjukkan graf jarak-masa bagi perjalanan pergi dan balik sebuah kereta dari Bandar P ke Bandar Q. Diagram 14 shows a distance-time graph of a car travels from Town P to Town Q and return journey. 20 O 3015 50 Masa (minit) Time (minutes) Jarak (km) Distance (km) Rajah 14/Diagram 14 Tk 5 23 PT Math KM(59-80).indd 63 24/2/2023 8:21:04 AM PENERBIT ILMU BAKTI SDN. BHD.

64 Hitung laju purata, dalam km j–1, kereta itu pada keseluruhan perjalanan. Calculate the average speed, dalam km h–1, of the car for the whole journey. A 24 C 48 B 34.3 D 68.6 33 Jadual 3 menunjukkan bilangan kad yang berlabel jenis fungsi di dalam sebuah kotak. Table 3 shows the number of cards labeled with di erent types of functions in a box. Jenis kad Type of cards Bilangan kad Number of cards Linear/ Linear 2 Kuadratik / Quadratic 7 Kubik / Cubic 2 Salingan / Reciprocal 9 Jadual 3/Table 3 Ali, Zaki dan Peter bergilir-gilir mencabut sekeping kad secara rawak daripada kotak itu. Ali membuat cabutan pertama dan Peter membuat cabutan terakhir. Hasil cabutan Ali ialah kad berlabel fungsi kuadratik dan hasil cabutan Zaki ialah kad berlabel fungsi linear. Hitung kebarangkalian bahawa hasil cabutan Peter ialah kad berlabel fungsi kuadratik. Ali, Zaki and Peter take turns to draw a card at random from the box. Ali makes the  rst draw and Peter makes the last draw.  e result of Ali’s draw is a card labeled with a quadratic function and the result of Zaki’s draw is a card labeled with a linear function. Calculate the probability that the result of Peter’s draw is a card labeled with a quadratic function. A 1 3 C 7 18 B 3 10 D 7 20 34 Shamila mahu membeli sebuah kereta yang berharga RM60 000 dengan pendahuluan 10%. Pendapatan bulanannya ialah RM3 500, manakala belanja tetap dan belanja berubahnya masing-masing ialah RM600 dan RM400. Jika Shamila membuat simpanan setiap bulan dalam tempoh 2 tahun bagi membayar wang pendahuluan itu, berapakah lebihan gaji bulanannya? KBAT Menilai Shamila wants to buy a car priced at RM60 000 with a down payment of 10%. Her monthly income is RM3 500 while her  xed and variable expenses are RM600 and RM400 respectively. If Shamila makes savings every month for 2 years in order to pay the down payment, how much is her monthly surplus income? A Tiada baki C RM2 500 No balance B RM3 450 D RM2 250 35 Jadual 4 menunjukkan sekumpulan 100 orang murid yang lulus atau gagal dalam suatu ujian tertentu. Table 4 a group of 100 students who passed or failed a certain test. Murid Student Lulus Pass Gagal Fail Lelaki/ Boy 30 15 Perempuan / Girl 45 10 Jadual 4/Table 4 Jika seorang murid dipilih secara rawak daripada murid perempuan, cari kebarangkalian murid itu gagal dalam ujian. If a student is picked at random from the girls,  nd the probability that the student is failed the test. A 2 7 C 2 11 B 2 9 D 1 10 36 Diberi bahawa a berubah secara langsung dengan kuasa tiga b dan secara songsang dengan punca kuasa dua c. Hubungan antara a, b, dan c ialah a = kbx c y , dengan keadaan k ialah pemalar. Cari nilai x dan nilai y. Given that a is varies directly as the cube of b and inversely as the square root of c.  e relationship between a, b and c is a = kbx cy , where k is a constant. Find the values of x and y. A x = 3, y = 1 2 C x = –3, y = 1 2 B x = 3, y = – 1 2 D x = –3, y = – 1 2 37 Jadual 5 menunjukkan pemboleh ubah x, y dan z yang memuaskan hubungan x ∝ z y . Table 5 shows the variables of x, y and z that satisfy the relation of x ∝ z y . x 80 20 y 40 8 z 25 k Jadual 5/Table 5 Hitung nilai k. Find the value of k. Tk 5 23 PT Math KM(59-80).indd 64 24/2/2023 8:21:04 AM PENERBIT ILMU BAKTI SDN. BHD.

65 A k = 4 C k = 8 B k = 1 4 D k = 16 38 Pei Li memiliki sebuah kereta dengan kapasiti enjin 1 799 cc. Pei Li owns a car of 1 799 cc engine capacity. Enjin Engine (cc) Kadar asas Base rate (RM) Kadar progresif (setiap cc) Progressive rates (RM) 1 000 dan ke bawah 1 000 and below 20 – 1 001 – 1 200 55 – 1 201 – 1 400 70 – 1 401 – 1 600 90 – 1 601 – 1 800 200 0.40 1 801 – 2 000 280 0.50 2 001 – 2 500 380 1.00 Jadual 6/Table 6 Berdasarkan maklumat yang diberikan dalam Jadual 6, hitung jumlah cukai jalan yang Pei Li perlu dibayar untuk keretanya. Based on the information given in Table 6, calculate the amount of road tax Pei Li has to pay for her car. A RM279.60 C RM79.60 B RM279.20 D RM79.20 39 Encik Yee memiliki sebuah rumah teres dua tingkat berkeluasan 2 800 kaki persegi di Kuala Lumpur. Anggaran sewa rumah itu ialah RM1 600 sebulan. Hitung jumlah cukai pintu yang Encik Yee perlu bayar untuk tahun 2023 dengan mengambil kira kadar cukai pintu sebanyak 6%. Mr. Yee owns a 2 800-square-foot double-storey terrace house in Kuala Lumpur.  e estimated monthly rent of the house is RM1 600. Calculate the total property assessment tax that Mr. Yee has to pay for the year 2023 assuming an assessment rate of 6%. A RM1 152 C RM400 B RM576 D RM96 40 Encik Kim ialah seorang perokok yang berusia 52 tahun. Dia ingin membeli insurans dengan nilai muka sebanyak RM250 000. Anggarkan kadar premium untuk Encik Kim menggunakan kadar insurans yang ditunjukkan dalam Jadual 7. Mr. Kim is a 52-year-old smoker. He wishes to buy an insurance with a face value of RM250 000. Estimate the premium rate for Mr. Kim using the insurance rate shown in Table 7. Jadual kadar premium tahunan bagi setiap RM1 000 nilai muka Annual premium rate schedule per RM1 000 face value Julat umur Range of ages (Tahun Years) Bukan perokok Non-smoker (RM) Perokok Smoker (RM) Lelaki/ Male Perempuan/ Female Lelaki/ Male Perempuan/ Female < 30 1.035 1.176 1.16 1.282 30 – 34 1.326 1.302 1.405 1.325 35 – 39 1.423 1.395 1.530 1.502 40 – 44 1.563 1.498 1.620 1.599 45 – 49 1.832 1.751 1.952 1.795 50 – 54 2.542 2.465 3.268 3.065 55 – 59 3.952 3.217 4.256 3.954 60 – 64 4.876 4.236 5.953 5.024 Jadual 7/Table 7 A RM100 C RM250 B RM265 D RM817 KERTAS PEPERIKSAAN TAMAT END OF QUESTION PAPER Tk 5 23 PT Math KM(59-80).indd 65 24/2/2023 8:21:04 AM PENERBIT ILMU BAKTI SDN. BHD.

66 Paper 1 Bahagian A [40 markah/marks] Jawab semua soalan. Answer all the questions. 1 Rajah 1 menunjukkan sebuah kuboid. Sebuah kon tegak membulat setinggi 24 cm dan tapak berdiameter MN dikeluarkan daripada kuboid itu. M dan N masing-masing ialah titik tengah bagi tepi EF dan HG. Diagram 1 shows a cuboid. A right circular cone of height 24 cm and a base of diameter MN is removed from the cuboid. M and N are the midpoints of the edges EF and HG respectively. 24 cm 16 cm A B C G N E H F M D V Rajah 1/Diagram 1 Diberi bahawa VN = VM = 25 cm. Dengan menggunakan π = 22 7 , hitung isi padu, dalam cm3 , pepejal yang tinggal. It is given that VN = VM = 25 cm. By using π = 22 7 , calculate the volume, in cm3, of remaining solid. [4 markah/marks] Jawapan/Answer: 2 Dalam Rajah 2, ABCD ialah sebuah segi empat selari dan O ialah asalan. Persamaan garis lurus AB ialah y = x 2 – 1. In Diagram 2, ABCD is a parallelogram and O is the origin.  e equation of the straight line AB is y = x 2 – 1. y x O p B C D A(−4, −3) 5 Rajah 2/Diagram 2 Kertas 2 / Paper 2 [100 markah/marks] Masa: 2 jam 30 minit Tk 5 23 PT Math KM(59-80).indd 66 24/2/2023 8:21:04 AM PENERBIT ILMU BAKTI SDN. BHD.

67 Cari Find (a) nilai p, the value of p, (a) persamaan garis lurus BC dan seterusnya, pintasan-y bagi garis lurus itu. the equation of the straight line BC and hence, the y-intercept of the straight line. [5 markah/marks] Jawapan/Answer: 3 (a) Nyatakan akas bagi pernyataan berikut dan seterusnya, tentukan nilai kebenaran akas itu. State the converse of the following statement and hence, determine the truth value of the converse. Jika x , –12, maka x , –5 If x , –12, then x , –5 [2 markah/marks] (b) Buat satu kesimpulan umum secara induktif bagi jujukan nombor 4, 14, 36, 76, … yang mengikut pola berikut. Make a general conclusion by inductive for the sequence of numbers 4, 14, 36, 76, … which follows the following pattern. 4 = 13 + 3 14 = 23 + 6 36 = 33 + 9 76 = 43 + 12 . . . [2 markah/marks] Jawapan/Answer: Tk 5 23 PT Math KM(59-80).indd 67 24/2/2023 8:21:05 AM PENERBIT ILMU BAKTI SDN. BHD.

68 4 Rajah 3 menunjukkan graf laju-masa bagi pergerakan dua buah kereta, M dan N. Graf OAB mewakili pergerakan kereta M, manakala graf OC mewakili pergerakan kereta N. Diagram 3 shows the speed-time graph for the motion of two cars, M and N.  e graph OAB represents the motion of car M whereas the graph OC represents the motion of car N. Laju (km j–1) Speed (km h–1) Masa (minit) O Time (minutes) 30 50 70 t 100 A B C Rajah 3/Diagram 3 (a) Hitung kadar perubahan laju, dalam km j–2, kereta M dalam 30 minit yang pertama. Calculate the rate of change of speed, in km h–2, of car M in the  rst 30 minutes. [2 markah/marks] (b) Cari nilai t, diberi jarak yang dilalui oleh kedua-dua kereta adalah sama. Find the value of t, given that the distance travelled by both cars are equal. [2 markah/marks] Jawapan/Answer: 5 (a) Rajah 4 ialah gambar rajah Venn yang menunjukkan set P dan Q dengan keadaan set semesta ξ = P Q. Diagram 4 is a Venn diagram shows sets P and Q such that the universal set ξ = P Q. Q P Rajah 4/Diagram 4 Nyatakan hubungan antara set P dengan Q. State the relationship between sets P and Q. [1 markah/mark] Tk 5 23 PT Math KM(59-80).indd 68 24/2/2023 8:21:05 AM PENERBIT ILMU BAKTI SDN. BHD.

69 (b) Rajah di ruang jawapan ialah gambar rajah Venn yang menunjukkan set P, Q dan R dengan keadaan set semesta ξ = P Q R. Pada rajah, lorekkan set P R Q’.  e diagram in the answer space is a Venn diagram shows sets P, Q and R such that the universal set ξ = P Q R. On the diagram, shade the set P R  Q’. [2 markah/marks] Jawapan/Answer: (a) (b) Q P R 6 Ali menjual 4 batang pen yang berharga RM(x + 5) sebatang. Jumlah jualannya ialah RM(x + 6)2 . Hitung jumlah jualan, dalam RM, yang diperoleh Ali. Ali sold 4 pens priced at RM(x + 5) each. His total sales is RM(x + 6)2 . Calculate the total sales, in RM, obtained by Ali. [4 markah/marks] Jawapan/Answer: Tk 5 23 PT Math KM(59-80).indd 69 24/2/2023 8:21:05 AM PENERBIT ILMU BAKTI SDN. BHD.

70 7 Arus elektrik, I berubah secara langsung dengan kuasa, P dan secara songsang dengan voltan, V suatu alat elektrik. Diberi sebuah televisyen dengan kuasa 600 W dan voltan 240 V menggunakan arus elektrik 2.5 A. Cari arus elektrik yang mengalir melalui sebuah ketuhar gelombang mikro dengan kuasa 1 000 W dan voltan 240 V.  e electric current, I varies directly as the power, P and inversely as the voltage, V of an electrical appliance. Given that a television with a power of 600 W and a voltage of 240 V using an electric current of 2.5 A. Find the electric current  owing through a microwave oven with a power of 1 000 W and a voltage of 240 V. [4 markah/marks] Jawapan/Answer: 8 Rajah 5 menunjukkan sebahagian daripada peta Malaysia. Diagram 5 shows part of a Malaysia map. Perak Kelantan Pahang Terengganu Rajah 5/Diagram 5 Wakilkan keempat-empat negeri dalam peta itu sebagai satu graf. Setiap negeri boleh diwakilkan sebagai bucu, manakala negeri yang berkongsi sempadan boleh dihubungkan dengan tepi pada graf itu. Represent the four states in the map as a graph. Each state can be represented as a vertex, while the states that share a common border are connected by edges on the graph. [3 markah/marks] Jawapan/Answer: Tk 5 23 PT Math KM(59-80).indd 70 24/2/2023 8:21:05 AM PENERBIT ILMU BAKTI SDN. BHD.

71 9 Aminah mempunyai polisi insurans perubatan dengan peruntukan deduktibel sebanyak RM2 000 dan peratusan ko-insurans 80/20 dalam polisinya. Lengkapkan jadual di bawah jika kos perubatan Aminah ialah RM6 000. Aminah has a medical insurance policy with a deductible provision of RM2 000 and an 80/20 co-insurance percentage in her policy. Complete the table below if Aminah’s medical cost is RM6 000. [4 markah/marks] Jawapan/Answer: Pelan Plans Deduktibel Deductible Dengan ko-insurans With co-insurance Perjanjian (Nilai) Agreement (Value) RM2 000 80/20 (%) Kos yang dibayar oleh pemegang polisi  e cost paid by the policyholder Kos yang dibayar oleh syarikat insurans  e cost paid by the insurer 10 Jadual 1 menunjukkan jarak yang dilalui oleh staf ke tempat kerja. Table 1 shows the distance travelled by sta to the workplace. Jarak (km) Distance (km) 1 – 5 6 – 10 11 – 15 16 – 20 21 – 25 26 – 30 Kekerapan Frequency 3 7 8 10 7 4 Jadual 1/Table 1 Hitung varians. Caclulate the variance. [5 markah/marks] Jawapan/Answer: 80 20 80 20 Tk 5 23 PT Math KM(59-80).indd 71 24/2/2023 8:21:05 AM PENERBIT ILMU BAKTI SDN. BHD.

72 Bahagian B [45 markah/marks] Jawab semua soalan. Answer all the questions. 11 Min bagi satu set data 3, n, 4, 2n, 6, 10 dan 11 ialah m. Jika setiap data didarab dengan 2 dan hasil darabnya ditambah dengan 1, maka min dan varians bagi set data itu masing-masing ialah 15 dan p 7 . Cari nilai m, n dan p.  e mean of a set of data 3, n, 4, 2n, 6, 10 and 11 is m. If each data is multiplied by 2 and the product is added with 1, then the mean and the variance of the set of data are 15 and p 7 respectively. Find the values of m, n and p. [8 markah/marks] Jawapan/Answer: 12 Sebuah kafeteria menjual dua jenis perisa kopi, S dan T. Harga secawan kopi perisa S dan secawan kopi perisa T masing-masing ialah RM8 dan RM4. Dalam suatu hari tertentu, bilangan cawan kopi perisa S yang dijual ialah selebih-lebihnya dua kali ganda bilangan cawan kopi perisa T. Jumlah bilangan cawan kopi yang dijual tidak melebihi 80 cawan. Jumlah jualan pada hari itu adalah sekurang-kurangnya RM320. Kafeteria itu menjual x cawan kopi perisa S dan y cawan kopi perisa T. A cafeteria sells two types of co ee  avours, which are  avours S and T. A cup of  avoured S co ee and a cup of  avoured T co ee cost RM8 and RM4 respectively. On a particular day, the number of cups of  avoured S co ee sold is at most twice the number of cups of  avoured T co ee.  e total number of co ee sold does not exceed 80 cups.  e sales volume on that day is at least RM320.  e cafeteria sells x cups of  avoured S co ee and y cups of  avoured T co ee. (a) Nyatakan tiga ketaksamaan, selain x > 0 dan y > 0, yang mewakili situasi di atas. State three inequalities, other than x > 0 and y > 0, which represent the above situation. [3 markah/marks] (b) Untuk ceraian soalan ini, gunakan kertas graf yang disediakan di halaman 73. For this part of question, use the graph paper provided on page 73. Menggunakan skala 2 cm kepada 10 cawan kopi pada kedua-dua paksi, bina dan lorek rantau R yang memuaskan sistem ketaksamaan linear yang dinyatakan di 12(a). Using a scale of 2 cm to 10 cups of co ee on both axes, construct and shade the region R that satis es the system of linear inequalities in 12(a). [3 markah/marks] (c) Dari graf dalam 12(b), From the graph in 12(b), (i) cari bilangan maksimum dan minimum cawan kopi perisa T jika bilangan cawan kopi perisa S yang dijual pada hari itu ialah 40.  nd the maximum and minimum number of cups of  avoured T co ee if the number of cups of  avoured S co ee sold on that day is 40. [2 markah/marks] Tk 5 23 PT Math KM(59-80).indd 72 24/2/2023 8:21:05 AM PENERBIT ILMU BAKTI SDN. BHD.

73 (ii) tentukan sama ada syarat-syarat bagi situasi yang diberi dipatuhi jika pengusaha kafeteria itu menjangkakan sebanyak 60 cawan kopi perisa S yang dijual pada hari itu. determine whether the conditions of the given situation are adhered to if the owner of the cafeteria estimates that there are 60 cups of  avoured S co ee sold on that day. [1 markah/mark] (iii) adakah graf itu memenuhi sistem ketaksamaan linear yang dibina jika bilangan cawan kopi perisa T yang dijual ialah 60 cawan? does the graph satisfy the system of linear inequalities constructed if the number of cups for  avoured T co ee sold is 60 cups? [1 markah/mark] Jawapan/Answer: Tk 5 23 PT Math KM(59-80).indd 73 24/2/2023 8:21:05 AM PENERBIT ILMU BAKTI SDN. BHD.

74 13 Jadual 2 menunjukkan perbandingan pelan pinjaman bagi dua jenis kereta yang ditawarkan oleh sebuah bank. Table 2 shows the comparison of loan plans for two types of cars o ered by a bank. Jenis kereta Type of car Jenis A Type A Jenis B Type B Harga Price RM68 000 RM105 000 Pinjaman kereta Car loan 90% 95% Faedah mudah Simple interest 3.2% 2.8% Tempoh Tenure 9 tahun/years 9 tahun/years Jadual 2/Table 2 Gaji bulanan Encik Bong ialah RM3 500. Dia membayar RM300 untuk bil utiliti dan RM700 untuk barangan runcit setiap bulan. Dia berhasrat untuk membeli sebuah kereta baharu dengan mengambil pinjaman bank. Kereta idamannya ialah kereta jenis B. Pada pendapat anda, adakah dia mampu membeli kereta jenis B? Justi kasikan jawapan anda. KBAT Menganalisis Mr. Bong’s monthly salary is RM3 500. He pays RM300 for utilities bills and RM700 on groceries every month. He intends to buy a new car by taking a bank loan. His dream car is the type B car. In your opinion, can he a ord to buy the type B car? Justify your answer. [9 markah/marks] Jawapan/Answer: Tk 5 23 PT Math KM(59-80).indd 74 24/2/2023 8:21:05 AM PENERBIT ILMU BAKTI SDN. BHD.

75 14 Rajah 6 menunjukkan sebuah padang bola sepak mini bagi kelas 5 Berlian yang berbentuk segi empat tepat. Perimeter padang itu ialah 96 m. Diagram 6 shows a mini football  eld in the shape of rectangle for class 5 Berlian.  e perimeter of the  eld is 96 m. (q + 6) m p m Rajah 6/Diagram 6 Diberi bahawa panjang padang itu ialah tiga kali lebarnya. Dengan menggunakan kaedah matriks, hitung panjang, dalam cm, padang itu. Given that the length of the  eld is three times its width. Using the matrix method, calculate the length, in cm, of the  eld. [9 markah/marks] Jawapan/Answer: 15 Rajah 7 menunjukkan segi tiga E, F, G, M dan N dilukis pada satu satah Cartes. Diagram 7 shows triangles E, F, G, M and N drawn on a Cartesian plane. 12 – 10 – 8 – 6 – 4 – 2 – –2 2 4 6 8 10 x y F G N M E O Rajah 7/Diagram 7 Tk 5 23 PT Math KM(59-80).indd 75 24/2/2023 8:21:06 AM PENERBIT ILMU BAKTI SDN. BHD.

76 (a) Segi tiga F ialah imej bagi segi tiga E di bawah satu transformasi P. Segi tiga G ialah imej bagi segi tiga F di bawah satu transformasi Q. Triangle F is the image of triangle E under a transformation P. Triangle G is the image of triangle F under a transformation Q. Huraikan selengkapnya, Describe in full, (i) transformasi P, transformation P, (ii) suatu transformasi tunggal yang setara dengan transformasi QP. a single transformation which is equivalent to the transformation QP. (b) Segi tiga N ialah imej bagi segi tiga M di bawah gabungan transformasi WV. Triangle N is the image of triangle M under the combined transformation WV. Huraikan selengkapnya transformasi V dan transformasi W. Describe in full the transformation V and the transformation W. [9 markah/marks] Jawapan/Answer: Tk 5 23 PT Math KM(59-80).indd 76 24/2/2023 8:21:06 AM PENERBIT ILMU BAKTI SDN. BHD.

77 Bahagian C [15 markah/marks] Jawab satu soalan dalam bahagian ini. Answer one question from this section. 16 Rajah 8 menunjukkan sebuah kuboid. Diagram 8 shows a cuboid. (3 + x) cm (7 – x) cm 4 cm Rajah 8/Diagram 8 (a) Bentukkan persamaan kuadratik bagi isi padu kuboid itu, y cm3 , dalam sebutan x. Form a quadratic expression for the volume of the cuboid, y cm3 , in terms of x. [2 markah/marks] (b) Lakarkan graf bagi fungsi kuadratik y yang dinyatakan di 16(a) bagi –4 < x < 8. Sketch the graph of a quadratic function y that stated in 16(a) for –4 < x < 8. [6 markah/marks] (c) Diberi bahawa isi padu kuboid itu ialah 84 cm3 . Cari nilai x, selain daripada x = 0. It is given that the volume of the cuboid is 84 cm3 . Find the value of x, other than x = 0. [3 markah/marks] (d) Tentukan isi padu maksimum, dalam cm3 , kuboid itu. Determine the maximum volume, in cm3 , of the cuboid. [4 markah/marks] Jawapan/Answer: Tk 5 23 PT Math KM(59-80).indd 77 24/2/2023 8:21:06 AM PENERBIT ILMU BAKTI SDN. BHD.

78 17 Rujuk Jadual 3.1 dan Jadual 3.2 untuk menjawab soalan-soalan berikut. Refer Table 3.1 and Table 3.2 to answer the following questions. Kadar Cukai Pendapatan Income Tax Rates Pendapatan yang dikenakan cukai Chargeable income Pengiraan Calculations (RM) Kadar Rate (%) Cukai Tax (RM) 0 – 5 000 Pada 5 000 pertama On the fi rst 5 000 0 0 5 001 – 20 000 Pada 5 000 pertama On the fi rst 5 000 15 000 seterusnya Next 15 000 1 0 150 20 001 – 35 000 Pada 20 000 pertama On the fi rst 20 000 15 000 seterusnya Next 15 000 3 150 450 35 001 – 50 000 Pada 35 000 pertama On the fi rst 35 000 15 000 seterusnya Next 15 000 8 600 1 200 50 001 – 70 000 Pada 50 000 pertama On the fi rst 50 000 20 000 seterusnya Next 20 000 14 1 800 2 800 70 001 – 100 000 Pada 70 000 pertama On the fi rst 70 000 30 000 seterusnya Next 30 000 21 4 600 6 300 100 001 – 250 000 Pada 100 000 pertama On the fi rst 100 000 150 000 seterusnya Next 150 000 24 10 900 36 000 250 001 – 400 000 Pada 250 000 pertama On the fi rst 250 000 150 000 seterusnya Next 150 000 24.5 46 900 36 750 400 001 – 600 000 Pada 400 000 pertama On the fi rst 400 000 200 000 seterusnya Next 200 000 25 83 650 50 000 Jadual 3.1/Table 3.1 Tk 5 23 PT Math KM(59-80).indd 78 24/2/2023 8:21:06 AM PENERBIT ILMU BAKTI SDN. BHD.

79 80 Kadar Cukai Pendapatan Income Tax Rates Perkara Item RM Diri Self 9 000 Individu kurang upaya (tambahan) Disabled self (additional) 6 000 Perbelanjaan perubatan untuk ibu bapa Medical expenses expended on parents (Terhad) (Limit) 5 000 Perbelanjaan perubatan untuk diri sendiri, pasangan atau anak dengan penyakit serius, termasuk sehingga RM500 untuk pemeriksaan perubatan Medical expenses expended on self, spouse or child with serious disease, including up to RM500 for medical examination (Terhad) (Limit) 6 000 Peralatan dan keperluan khas untuk diri sendiri, pasangan, anak atau ibu bapa Basic supporting equipment for disabled self, spouse, child or parent (Terhad) (Limit) 6 000 Yuran pengajian untuk kemahiran atau kelayakan Study course fees for skills or qualifi cations (Terhad) (Limit) 7 000 Elaun gaya hidup Lifestyle allowance (Terhad) (Limit) 2 500 Anak – kadar asas Child – basic rate (Setiap satu) (Each) 2 000 Yuran penjagaan anak di pusat jagaan kanak-kanak / tadika berdaftar (bawah enam tahun) Childcare fee to registered child care centre / kindergarten (below six years old) (Terhad) (Limit) 1 000 KWSP dan insurans hayat EPF and life insurance (Terhad) (Limit) 6 000 Sumbangan skim persaraan swasta (PRS) Private retirement scheme contributions (PRS) (Terhad) (Limit) 3 000 Insurans perubatan dan / atau pendidikan Medical and/or education insurance (Terhad) (Limit) 3 000 Deposit untuk anak ke dalam Skim Simpanan Pendidikan Kebangsaan (SSPN) Deposit for a child into the National Education Savings Scheme (SSPN) (Terhad) (Limit) 6 000 Sumbangan kepada PERKESO Contribution to SOCSO (Terhad) (Limit) 250 Jadual 3.2/Table 3.2 Peter dan Amy ialah pasangan suami isteri. Mereka mempunyai seorang anak perempuan bernama Jessica yang berusia 5 tahun. Peter bekerja sebagai jurutera dan pendapatan penggajiannya dan amaun yang dibelanjakan untuk tahun taksiran 2020 adalah seperti berikut. Peter and Amy are a married couple.  ey have a 5-year-old daughter, Jessica. Peter works as an engineer and his employment income and the amount expended for the year of assessment 2020 are as follows. Tk 5 23 PT Math KM(59-80).indd 79 24/2/2023 8:21:06 AM PENERBIT ILMU BAKTI SDN. BHD.

80 • Gaji tahunannya ialah RM80 000. His annual salary is RM80 000 • Bonus tahunan sebanyak RM3 200. An annual bonus of RM3 200. • Sumbangan keselamatan sosial (PERKESO) sebanyak RM4 000. Social security contribution (SOCSO) of RM4 000. • Sumbangan tunai RM1 000 yang dibuat kepada pertubuhan amal yang diluluskan. A cash donation of RM1 000 made to approved charity organizations. • Perbelanjaan perubatan sebanyak RM2 000 kepada ibu bapa.  e medical expenses of RM2 000 on parents. Amy ialah seorang pengurus bank dan dia menerima pendapatan tahunan sebanyak RM120 000. Pembayaran yang dibuat oleh Amy pada tahun 2020 adalah seperti berikut: Amy is a bank manager and she receives an annual income of RM120 000.  e payment made by Amy in 2020 are as follows: Perkara Item RM Pembelian komputer riba Purchase of a laptop 3 200 Pembelian buku dan majalah Purchase of books and magazines 800 Pembayaran kepada skim persaraan swasta yang diluluskan Payments to an approved private retirement scheme 4 500 Pembayaran yuran sekolah Jessica Payments of Jessica’s school fee 4 800 (Pembelian komputer riba, buku dan majalah adalah di bawah elaun gaya hidup) ( e purchase of laptop, books and magazines is under the lifestyle allowance) Peter dan Amy bersetuju bahawa Amy akan menuntut pelepasan anak. Jika pasangan tersebut memilih penilaian berasingan, hitung cukai pendapatan yang perlu dibayar untuk tahun taksiran 2020 oleh: Peter and Amy have agreed that Amy would claim the child relief. If the couple elect for separate assessment, calculate the income tax payable for the year assessment of 2020 by: (a) Peter, (b) Amy. [15 markah/marks] Jawapan/Answer: KERTAS PEPERIKSAAN TAMAT END OF QUESTION PAPER Tk 5 23 PT Math KM(59-80).indd 80 24/2/2023 8:21:06 AM PENERBIT ILMU BAKTI SDN. BHD.

J1 Praktis Topikal SPM Matematik Tingkatan 5 - Jawapan BAB 1 Ubahan KERTAS 1 1 C 2 B 3 A 4 B 5 D 6 B 7 D 8 D 9 A 10 C 11 B 12 B 13 C KERTAS 2 BAHAGIAN A 1 E ∝ v2 E = kv2 200 = k(52 ) k = 8 E = 8v2 650 = 8v2 v2 = 81.25 v = 81.25 = 9.014 m s–1 2 (a) A∝ θr 2 A = k θr 2 33 = k(11 6 )(62 ) 33 = 66k k = 0.5 A = 0.5 θr 2 (b) A = 0.5(1.8)(72 ) = 44.1 cm2 3 Andaikan a = pecutan dan m = jisim Assume that a = acceleration and m = mass a∝ 1 m a = k m 1.5 = k 4 k = 6 a = 6 m a = 6 10 = 0.6 4 I ∝ 1 L I = k L 1.5 = k 60 k = 90 I = 90 L 4 = 90 m m = 22.5 cm 5 F ∝ v 2 r F = kv 2 r F = k( 1 3 v) 2 2r = k 1 9 v 2 2r = kv 2 18r = 1 18 F BAHAGIAN B 6 f ∝ 1 λ f = k λ 100 = k 200 (a) f = 20 000 λ = 20 000 500 = 40 MHz (b) f = 20 000 λ 80 = 20 000 λ = 20 000 80 = 250 m 7 R ∝ 1 d2 , l p = 2l Q dan/and dp dQ = 1 3 R = kl d2 dP = 1 3 dQ RP = klp dp RQ = klQ dQ 2 RP = klP dQ 2 = k(2l Q k( 1 3 dQ) 2 = 18klQ dQ 2 = 18 RQ RP RQ = 18 1  RP : RQ = 18 : 1 8 T = Masa yang diambil/Time taken A = Luas padang/Area of the  eld m = Bilangan pekerja/Number of workers T ∝ A m T = k A m 3 = k 4 × 104 5 k = 15 4 × 104 = 3.75 × 10–4 T = 3.75 × 10–4 A m (a) T = 3.75 × 10–4 × 2× 104 3 = 2.5 jam/hours (b) 5 = 3.75 × 10–4 × 8× 104 m m = 6 9 g ∝ 1 r 2 g = k r 2 10 = k 6 4002 = 4.096 × 108 g = 4.096 × 108 12 8002 = 2.5 N kg–1 BAHAGIAN C 10 (a) (i) F ∝ v 2 r (b) (ii) 12.8 = 0.2v 2 4 v  = 12.8 × 4 0.2 = 256 v = 16 m s–1 F = kv 2 r 18 = k(3)2 0.1 1.8 = 9k k = 0.2 F = 0.2v 2 r (b) T = tempoh ayunan/oscillation period l = panjang bandul/length of pendulum g = pecutan graviti/gravitational acceleration Diberi/Given that T ∝ l g T = l g k dengan k ialah pemalar/where k is a constant JAWAPAN Tk 5 23 PT Math J(J1-14).indd 1 23/2/2023 6:04:36 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J2 Apabila/When l = 0.4 dan/and g = 10, T = 1.26 1.26 = 0.4 10 k 1.26 = 0.2k k = 1.26 0.2 = 6.3 Maka/ erefore T = l g 6.3 Apabila/When l = 0.9 dan/and T = 4.73 4.73 = 0.9 g 6.3 4.73 g = 6.3 0.9 g = 0.9 4.73 6.3 = 1.264 g = 1.2642 = 1.598  Pecutan graviti di permukaan bulan ialah 1.598 m s–2  e gravitational acceleration on the surface of the moon is 1.598 m s–2 ZON KBAT 1 h ∝ 1 x2 h = k x2 9 = k 62 k = 324 h = 324 x2 4 = 324 x2 x2 = 81 x = 9 cm BAB 2 Matriks KERTAS 1 1 A 2 A 3 A 4 B 5 A 6 C 7 B 8 B 9 C 10 D 11 D 12 B 13 C 14 C 15 A 16 A 17 B 18 C 19 C 20 C 21 A 22 C 23 A 24 B 25 C 26 B KERTAS 2 BAHAGIAN A 1 Tingkatan/ Form Lelaki/ Boys Perempuan/ Girls 5 Amanah 26 20 5 Bestari 17 26 5 Cekal 20 18 Matriks yang terbentuk ialah/ e matrix formed is  26 20 17 26 20 18 . 2 5x + 6y = 28 7x + 9y = 41   5 6 7 9   x y =   28 41   x y = 1 5(9) – 6(7)   9 –6 –7 5   28 41 = 1 3  9 × 28 + (–6) × 41 –7 × 28 + 5 × 41  = 1 3   6 9 =   2 3  x = 2, y = 3 3 (a) x + y = 56 .......... 1 y – 2 = 3(x – 2) y – 2 = 3x – 6 –3x + y = –4 atau/or y – 3x = –4 .......... 2 (b) x + y = 56 –3x + y = –4  1 –3 1 1  x y  =  56 –4   x y  = 1 1 – [–3]  1 3 –1 1  56 –4  = 1 4  60 164 =  15 41  x = 15 dan/and y = 41 4 (a) x + y = 150 .......................... 1 1 4 x = 1 6 y 3x = 2y 3x – 2y = 0 .............................. 2 (b) x + y = 150 3x – 2y = 0  1 3 1 –2   x y  =  150 0   x y  = 1 –2 – 3  –2 –3 –1 1  150 0  = 1 –5  –300 –450  =  60 90   x = 60 dan/and y = 90 5 (a) x + y = 11 3.5x + 21y = 126 (b) x + y = 11 3.5x + 21y = 126  1 3.5 1 21 x y  =  11 126   x y  = 1 21 – 3.5  21 –3.5 –1 1   11 126  = 1 17.5  105 87.5  =  6 5   x = 6 dan/and y = 5 6 (a) Panjang/Length = x, lebar/width = y x + y = 420 x – y = 30 (b)   1 1 1 –1  x y  =  420 30   x y  = 1 –1 – 1   –1 –1 –1 1  420 30  = – 1 2  –450 –390  =  225 195  7 (a) Harga sebatang pen/Price of a pen = RMx Harga sebatang pensel/Price of a pencil = RMy 2x + 7y = 12 6x + 5y = 20 Tk 5 23 PT Math J(J1-14).indd 2 23/2/2023 6:04:37 PM PENERBIT ILMU BAKTI SDN. BHD.

J3 Praktis Topikal SPM Matematik Tingkatan 5 - Jawapan (b)   2 7 6 5  x y  =  12 20   x y  = 1 2(5) – 7(6)   5 –7 –6 2  12 20  = 1 –32  5(12) + (–7)(20) –6(12) + 2(20)  = – 1 32  –80 –32  =  2.5 1  x = 2.5, y = 1 BAHAGIAN B 8 (a)   5 6 3 4 –1 = 1 (5 × 4) – (6 × 3)   4 –6 –3 5 = 1 2   4 –6 –3 5 = 2 –3 –3 2 5 2 (b) x = harga sebiji oren/price of an orange y = harga sebiji epal/price of an apple 5x + 6y = 19 3x + 4y = 12   5 6 3 4   x y =   19 12 =   x y = 1 2   4 –6 –3 5   19 12 = 1 2   4 × 19 + (–6) × 12 –3 × 19 + 5 × 12 = 1 2   4 3 = 2 3 2  x = 2, y = 3 2 9 (a) Q = P–1 = 1 3(–5) – (–2)(4)   –5 2 –4 3 = – 1 7   –5 2 –4 3 Bandingkan dengan/Compare with Q = m  –5 n –4 3  m = – 1 7 , n = 2 (b)   3 –2 4 –5   x y =   12 23   x y = – 1 7   –5 2 –4 3   12 23 = – 1 7   –5 × 12 + 2 × 23 –4 × 12 + 3 × 23 = – 1 7   –14 21 =   2 –3  x = 2, y = –3 10 (a) (p, q)   0 –1 –1 0 (3, 2)   –1 0 0 1 (r, s).   0 –1 –1 0   p q =   3 2   –1 0 0 1   3 2 =   r s   0 + (–q) –p + 0 =   3 2   –3 2 =   r s –q = 3 , –p = 2 q = –3 , p = –2 r = –3 , s = 2 (b) Dari/From   0 –1 –1 0   p q =   3 2 Darabkan sebelumnya kedua-dua belah dengan   –1 0 0 1 Multiplied both sides of matrices with   –1 0 0 1   –1 0 0 1   0 –1 –1 0   p q =   –1 0 0 1   3 2   0 1 –1 0   p q =   r s Maka, transformasi matriks tunggal yang memetakan (p, q) kepada (r, s) ialah   0 1 –1 0 .  erefore, the transformation of a single matrix that mapped (p, q) onto (r, s) is   0 1 –1 0 . 11 (a) M =   7 –4 3 –2 –1 = 1 –14 – (–12)   –2 4 –3 7 = – 1 2   –2 4 –3 7 =   1 –2 3 2 – 7 2 (b)   7 –4 3 –2  x y  =  5 2   x y  = – 1 2   –2 4 –3 7  5 2  = – 1 2  –2 –1 =   1 1 2  x = 1, y = 1 2 12 (a)   –3 4 –5 8 –1 = 1 –24 – (–20)   8 –4 5 –3 = – 1 4   8 –4 5 –3 =   –2 1 – 5 4 3 4 (b)  x y  = – 1 4   8 –4 5 –3  7 12  x y  = – 1 4   56 – 48 35 – 36 = – 1 4  8 –1 =   –2 1 4  x = –2, y = 1 4 13 (a) M = 4 × 8 – 3 × 9 = 32 – 27 = 5 k = –9 Tk 5 23 PT Math J(J1-14).indd 3 23/2/2023 6:04:38 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J4 (b)   4 3 9 8  x y  =  6 11  x y  = 1 5   8 –3 –9 4  6 11 = 1 5  48 – 33 –54 + 44  = 1 5  15 –10 =  3 –2  x = 3, y = –2 14 Murid/ Student Kertas 1/ Paper 1 Kertas 2/ Paper 2 Adam 80 70 Bella 60 85 Christ 85 90 Pemberat/Weightage: Kertas/Paper 1 0.4 Kertas/Paper 2 0.6 80 70 60 85 85 90 0.4 0.6 = 80 × 0.4 + 70 × 0.6 60 × 0.4 + 85 × 0.6 85 × 0.4 + 90 × 0.6 = 74 75 88  Markah Adam ialah 74%, markah Bella ialah 75% dan markah Christ ialah 88%. Adam’s marks is 74%, Bella’s marks is 75% and Christ’s marks is 88%. BAB 3 Matematik Pengguna: Insurans KERTAS 1 1 A 2 B 3 C 4 D 5 B 6 B KERTAS 2 BAHAGIAN A 1 (a) Syarikat insurans ialah TAKAFUL Insurance dan pemegang polisi ialah Manisha.  e insurance company is TAKAFUL Insurance and the policyholder is Manisha. (b) Had perlindungan ialah RM300 000.  e coverage limit is RM300 000. (c) Premium bulanan ialah RM200.  e monthly premium is RM200. (d) Risiko yang diinsuranskan ialah penyakit kritikal.  e risk to be insured is critical illnesses. 2 (a) RM1 000 yang pertama  e  rst RM1 000 RM220.00 (b) RM26 × 57 (setiap RM 1000 baki / each RM1 000 balance) RM1 482.00 (c) Premium asas / Basic premium = (a) + (b) RM1 702.00 (d) NCD 50% RM851.00 (e) Premium kasar / Gross premium = (c) – (d) RM851.00 Rujuk jadual Refer to the table 58 000 – 1 000 1 000 = 57 0.5 × 1 702.00 = 851 3 Anggaran premium tahunan/Estimated annual premium = RM250 000 RM100 × 1.321 = RM3 302.50 BAHAGIAN B 4 (a) Jumlah insurans yang harus dibeli Jumlah insurans yang harus dibeli = 70 100 × RM1 000 000 = RM700 000 (b) (i) Jumlah pampasan/ Amount of compensation RM50 000 – RM20 000 = RM30 000 (ii) RM600 000 < RM700 000 Jumlah yang diinsuranskan adalah kurang daripada jumlah insurans yang harus dibeli.  e sum insured is less than the amount of required insurance. Jumlah pampasan/ Amount of compensation = RM600 000 RM700 000 × RM50 000 – RM20 000 = RM42 857.14 – RM20 000 = RM22 857.14 Penalti ko-insurans / Co-insurance penalty = RM50 000 – RM42 857.14 = RM7 142.86 BAHAGIAN C 5 (a) Panduan untuk memilih insurans terbaik untuk Puan Reena: Guidelines to choose the best insurance for Puan Reena: (i) Ketahui jumlah perlindungan yang diperlukan Find out the total coverage needed (ii) Memahami skop perlindungan, terma dan syarat polisi Understanding the scope, the terms and conditions of the policy (iii) Bandingkan kadar premium dan manfaat insurans Compare the premium rates and insurance bene ts (iv) Elakkan perlindungan yang tidak perlu Avoid unnecessary coverage (b) (i) Dari jadual, kadar premium ialah RM1.21 From the table, the premium rate is RM 1.21 Premium tahunan Puan Reena Puan Reena’s annual premium = RM200 000 RM1 000 × RM1.21 = RM242 (ii) Jumlah perlindungan untuk penyakit kritikal  e total coverage for critical illness = 25 100 × RM200 000 = RM50 000 Premium tambahan tahunan untuk penyakit kritikal/ Annual additional premium for critical illness = RM50 000 RM10 000 × RM1.50 = RM75.00 Jumlah premium tahunan yang perlu dibayar Total annual premium payable = RM242.00 + RM75.00 = RM317 (c) Dari jadual, kadar premium untuk Encik Jaya ialah RM2.72 From the table, the premium rate for Encik Jaya is RM2.72 Premium tahunan Encik Jaya/ Encik Jaya’s annual premium = RM200 000 RM1 000 × RM2.72 = RM544.00 (d) Berdasarkan jumlah perlindungan insurans yang sama, Encik Jaya harus membayar harga yang jauh lebih tinggi daripada isterinya (RM544 berbanding RM242). Ini disebabkan oleh usia Encik Jaya adalah 4 tahun lebih tua daripada Puan Reena dan dia seorang perokok. Kebarangkalian seorang perokok terdedah kepada penyakit lebih tinggi daripada seorang bukan perokok. Based on the same amount of insurance coverage, Encik Jaya has to pay a price much higher than his wife (RM544 as compared to RM242).  is is due to the condition that Encik Jaya is 4 years older than Puan Reena and he is a smoker.  e probability of a smoker being exposed to illness is higher than that of a nonsmoker. 6 Untuk polisi komprehensif/For comprehensive policy: (a) RM1 000 yang pertama  e  rst RM1 000 RM273.80 ← Rujuk pada jadual Refer to the table Tk 5 23 PT Math J(J1-14).indd 4 23/2/2023 6:04:38 PM PENERBIT ILMU BAKTI SDN. BHD.

J5 Praktis Topikal SPM Matematik Tingkatan 5 - Jawapan (b) RM26 × 83 (setiap RM1 000 baki) RM26 × 83 (each RM1 000 balance) RM2 158 ← 84 000 – 1 000 1 000 = 83 Bagi Semenanjung Malaysia, kadar bagi RM1 000 yang pertama + RM26 bagi setiap RM1 000 For Peninsular Malaysia, the rate for the  rst RM1 000 + RM26 for each RM1 000 (c) Premium asas Basic premium = (a) + (b) RM2 431.80 (d) NCD 30% RM729.54 ← 0.30 × 2 431.80 (e) Premium kasar Gross premium = (c) – (d) RM1 702.26 Untuk polisi pihak ketiga, kebakaran dan kecurian/For third party,  re and theft policy: (a) Premium asas Basic premium RM1 823.85 (b) NCD 30% RM547.16 (c) Premium kasar Gross premium = (a) – (b) RM1 276.70 ← 0.75 × 2 431.80 ← 0.30 × 1 823.85 Untuk polisi pihak ketiga/For third party policy: (a) Premium asas Basic premium RM120.60 (b) NCD 30% RM36.18 (c) Premium kasar Gross premium = (a) – (b) RM84.42 ← Rujuk jadual Refer to the table ← 0.30 × 120.60 ZON KBAT 1 Hutang semasa dan keperluan kewangan Current debt and  nancial needs = Pinjaman rumah + Pinjaman kereta/Home loan + Car loan = RM350 000 + RM45 000 = RM395 000 Keperluan kewangan masa depan/Future  nancial needs: = Dana persaraan/Retirement fund = RM2 500 000 Aset/Assets = Pelaburan/Investments = RM80 000 Oleh itu, anggaran jumlah perlindungan insurans hayat bagi Encik Wong  erefore, the estimated amount of life insurance coverage for Mr Wong = RM395 000 + RM2 500 000 – RM80 000 = RM2 815 000 BAB 4 Matematik Pengguna: Percukaian KERTAS 1 1 C 2 C 3 B 4 A 5 C 6 A KERTAS 2 BAHAGIAN A 1 Pelepasan cukai ialah insentif bagi mengurangkan amaun cukai dengan membuat potongan daripada jumlah pendapatan tahunan. Sebaliknya, rebat cukai mengurangkan jumlah cukai daripada jumlah cukai pendapatan untuk mendapatkan cukai kena dibayar. Tax relief is an incentive to reduce the amount of taxes to be paid by making a deduction from the total annual income. Meanwhile, the tax rebate reduces the amount of taxes from the total chargeable income to get the payable income tax. 2 Encik Suvindran dan Puan Suvindran harus memilih taksiran cukai bersama kerana Encik Suvindran telah menanggung kerugian dalam perniagaan. Maka, taksiran cukai bersama di bawah nama Puan Suvindran lebih sesuai dipilih untuk memaksimumkan potongan cukai akibat kerugian perniagaan dan pelepasan cukai. Mr. Suvindran and Mrs. Suvindran should opt for joint income tax assessment as Mr. Suvindran has incurred losses in the business.  us, a joint tax assessment under Mrs. Suvindran’s name is more appropriately chosen to maximise the amount of tax deductions (due to business losses) and tax relief. 3 Encik Tan dan Puan Tan harus memilih taksiran cukai secara berasingan. Kedua-duanya berhak mendapat pelepasan peribadi dan pelepasan yang lain. Contohnya, mereka boleh membuat tuntutan untuk pelepasan cukai anak secara berasingan (seperti yuran penjagaan anak dan tadika). Selain itu, mereka juga boleh mendapatkan pelepasan perbelanjaan perubatan ke atas ibu bapa kandung masing-masing dengan had RM5 000. Jika mereka membuat taksiran cukai bersama, pasangan itu hanya dapat menuntut sehingga had bagi seorang individu sahaja. Mr Tan and Mrs Tan should choose for separate income tax assessment. Both of them are entitled for personal relief and other relief. For example, they can claim for child relief separately (e.g., childcare and kindergarten fees). Besides, they also entitle for medical treatment expenses on their biological parents' relief with a limit of RM5 000. If they compile for joint tax assessment, the couple can only claim up to the limit for one individual only. 4 Kapasiti enjin kereta terletak antara 1 801 cc dan 2 000 cc. Jadi, kadar asas ialah RM280.  e engine capacity of the car is between 1 801 cc and 2 000 cc. So, the base rate is RM280. Jumlah kadar progresif/Total progressive rate = (1 822 cc – 1 800 cc) × RM0.50 = 22 × RM0.50 = RM11.00 Jumlah jumlah cukai jalan/Total amount of road tax = Jumlah kadar progresif + Kadar asas Total progressive rate + Base rate = RM11.00 + RM280.00 = RM291 BAHAGIAN B 5 (a) Jumlah cukai taksiran harta Total property assessment tax = (RM2 000 × 12) × 0.05 = RM1 200 (b) Cukai pintu yang perlu dibayar pada awal tahun Amount of property assessment tax payable in the beginning of the year = RM1 200 ÷ 2 = RM600 (c) Cukai tanah/Quit rent = 4 000 × RM0.03 = RM120.00 (d) Tanahnya akan dirampas oleh pihak berkuasa negeri Her land can be seized by the state land authority 6 Bagi Jonathan/For Jonathan: Pendapatan bercukai/Chargeable income = RM40 000 – RM11 800 = RM28 200 Bagi pendapatan bercukai sebanyak RM28 200, kadar cukai ialah 3%. For chargeable income of RM28 200, the tax rate is 3%. Jumlah yang perlu dibayar bagi RM20 000 yang pertama Tax on the  rst RM20 000 = RM150 RM28 200 – RM20 000 = RM8 200 Jumlah yang perlu dibayar bagi RM8 200 seterusnya Tax on the next balance of RM8 200 = RM8 200 × 0.03 = RM246  Jumlah cukai perlu dibayar/Payable income tax = RM150 + RM246 – RM400 Rebat diri RM400 Self rebate RM400 = –RM4 = RM0 (Tidak perlu bayar cukai)/ (No need to pay taxes) Bagi Samuel/For Samuel: Pendapatan bercukai/Chargeable income = RM40 000 Tk 5 23 PT Math J(J1-14).indd 5 23/2/2023 6:04:38 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J6 Bagi pendapatan bercukai sebanyak RM40 000, kadar cukai ialah 8%. For chargeable income of RM40 000, the tax rate is 8% Jumlah yang perlu dibayar pada RM35 000 pertama = RM600 Tax on the  rst RM35 000 = RM600 RM40 000 – RM35 000 = RM5 000 Jumlah yang perlu dibayar pada RM5 000 seterusnya Tax on the next balance of RM5 000 = RM5 000 × 0.08 = RM400  Jumlah cukai perlu dibayar Payable income tax = RM600 + RM400 = RM1 000 BAHAGIAN C 7 (a) Percukaian penting kepada kerajaan untuk membiayai perkhidmatan awam seperti infrastruktur pengairan dan jalan raya, pendidikan, rawatan perubatan, keselamatan awam, dan pertahanan negara. Taxation is important to the government to fund public services such as irrigation and road infrastructures, education, medical treatment, public safety and national defence. (b) Pendapatan yang dikenakan cukai/Chargeable income = RM73 000 – RM9 000 – RM1 100 – RM2 040 – RM6 000 = RM54 860 (c) Bagi pendapatan yang dikenakan cukai sebanyak RM54 860, kadar cukai ialah 14%. For chargeable income of RM54 860, the tax rate is 14%. Jumlah yang perlu dibayar pada RM50 000 yang pertama/Tax on the  rst RM50 000 = RM1 800 RM54 860 – RM50 000 = RM4 860 Jumlah yang perlu dibayar pada RM4 860 seterusnya Tax on the next balance of RM4 860 = RM4 860 × 0.14 = RM680.40 Jadi, jumlah cukai yang perlu dibayar Payable income tax = RM1 800 + RM680.40 = RM2 480.40 (d) Jumlah cukai yang kena dibayar Payable income tax = Cukai yang kena dibayar – (Zakat dan  trah) Income tax – (Zakat and  trah) = RM2 480.40 – RM1 500 = RM980.40 Maka, En Daud perlu membayar cukai pendapatan sebanyak RM980.40.  erefore, the income tax payable by En Daud is RM980.40. (e) Ya, En. Daud and Puan Shima harus memilih penilaian cukai pendapatan bersama. Ini kerana penilaian bersama membolehkan En. Daud menuntut pelepasan isteri sebanyak RM3 000 kerana isterinya tidak bekerja. Ini akan mengurangkan jumlah cukai yang perlu dibayarnya. Yes, En. Daud and Puan Shima should choose a joint income tax assessment.  is is because the joint assessment allows En. Daud to claim for spouse relief of up to RM3 000 because his wife is not working.  is will reduce the amount of taxes he has to pay. 8 (a) Jadual berikut menunjukkan cukai pendapatan untuk Encik Gan dan isterinya menggunakan taksiran cukai bersama dan taksiran cukai yang berasingan.  e following table shows the income tax for Mr Gan and his wife using a joint tax assessment and a separate tax assessment. Perkara Item Taksiran cukai bersama Joint tax assessment Taksiran cukai berasingan Separate tax assessment Suami dan isteri Husband and wife Suami Husband Isteri Wife Jumlah pendapatan Total income RM60 000 + RM72 000 = RM132 000 RM60 000 RM72 000 Jumlah pengecualian (Sumbangan) Total exemption (Contributions) –(RM1 000 + RM1 400) = – RM2 400 – RM1 000 – RM1 400 Pelepasan cukai/ Tax relief Individu/ Individual – RM9 000 – RM9 000 – RM9 000 Gaya hidup (terhad kepada RM2 500) Lifestyle (limited to RM2 500) – RM2 500 – RM2 300 – RM2 500 KWSP & Insurans Hayat (Terhad kepada RM7 000) EPF & Life insurance (limited to RM7 000) – RM7 000 – RM6 000 – RM6 500 Pendapatan bercukai/ Chargeable income RM111 100 RM41 700 RM52 600 Cukai dasar/ Base tax RM10 900 RM600 RM1 800 Cukai pada baki seterusnya Tax on the next balance Baki/Balance = RM111 100 – RM100 000 = RM11 100 RM11 100 × 24% = RM2 664 Baki/Balance = RM41 700 – RM35 000 = RM6 700 RM6 700 × 8% = RM536 Baki/Balance = RM52 600 – RM50 000 = RM2 600 RM2 600 × 14% = RM364 Cukai pendapatan yang perlu dibayar Income tax payable RM10 900 + RM2 664 = RM13 564 RM600 + RM536 = RM1 136 RM1 800 + RM364 = RM2 164 Jumlah/Total: RM1 136 + RM2 164 = RM3 300 (b) Taksiran cukai berasingan lebih sesuai untuk Encik Gan dan isterinya kerana jumlah cukai pendapatan jauh lebih rendah (RM3 300) berbanding dengan taksiran cukai bersama (RM13 564). A separate tax assessment is more suitable for Mr Gan and his wife as their total income tax is much lower (RM3 300) as compared to the joint tax assessment (RM13 564). Tk 5 23 PT Math J(J1-14).indd 6 23/2/2023 6:04:39 PM PENERBIT ILMU BAKTI SDN. BHD.

J7 Praktis Topikal SPM Matematik Tingkatan 5 - Jawapan BAB 5 Kekongruenan, Pembesaran dan Gabungan Transformasi KERTAS 1 1 C 2 B 3 C 4 D 5 C 6 D 7 B 8 D KERTAS 2 BAHAGIAN A 1 (a) Pantulan pada garis lurus yang melalui O dan selari dengan BF. A re ection on a straight line passing through O and parallel to BF. (b) Pembesaran dengan faktor skala 3 pada titik D. An enlargement with a scale factor of 3 at point D 2 (a) BCDE V BAFE V ialah pantulan pada garis BE V is a re ection on line BE (b) BAFE W FAKJ W ialah putaran 120° ikut arah jam pada titik A W is a clockwise rotation of 120° at point A 3 (a) K(4, 1) T   –2 3 K9(2, 4) T   –2 3 K99(0, 7) (b) K(4, 1) R K9(4, 5) T   –2 3 K99(2, 8) BAHAGIAN B 4 (a) (i) X: Pantulan pada garis x = –2 X: A re ection on line x = –2 (ii) Y: Pembesaran dengan faktor skala 2 pada titik A(–2, 5) Y: An enlargement with a scale factor of 2 at point A(–2, 5) (b) Luas/Area of AMNPQ = 22 (luas/area of AJKLE) = 4 × 50 m2 = 200 m2  Luas kawasan berlorek/Area of the shaded region = (200 – 50) m2 = 150 m2 5 (b) (i) (a) W: Putaran 90° ikut arah jam pada (4, 0). W: A clockwise rotation of 90° at point (4, 0). (b) X: Pembesaran dengan faktor skala 3 pada titik T atau (7, 2). X: An enlargement with a scale factor of 3 at point T or (7, 2). (ii) Luas/Area of PQRST = k2 (luas/area of ABCDE) 108 = 32 (luas/area of ABCDE) Luas/Area of ABCDE = 108 9 = 12 m2 BAHAGIAN C 6 (a) (i) A(5, 4) R (–2, 3) (ii) A(5, 4) P (1, 4) R (–2, –1) Transformasi tunggal titik A: Translasi   –7 –5 . Single translation of point A: Translation   –7 –5 . (b) (i) U: Putaran 90° lawan arah jam pada (4, 2) U: An anticlockwise rotation of 90° at point (4, 2) V: Pembesaran dengan faktor skala 2 pada (4, 0) V: An enlargement with a scale factor of 2 at (4, 0) (c) (i) k = 1 2 (ii) Luas/Area of KLMN = ( 1 2) 2 × 154 m2 = 38.5 m2 BAB 6 Nisbah dan Graf Fungsi Trigonometri KERTAS 1 1 B 2 C 3 B 4 D 5 A 6 B 7 B 8 B 9 A 10 B 11 B 12 C 13 D KERTAS 2 BAHAGIAN A 1 tan θ = –2.5 tan θ = –tan ∠PTQ Maka/ erefore, tan ∠PTQ = 2.5 20 PT = 2.5  PT = 20 2.5 = 8 cm PT : TS = 2 : 1 PT TS = 2 1 8 TS = 2 TS = 4 cm 2 y x 1 1 –1 –1 0 θ X(0.71, –0.71) sin θ = koordinat-y = –0.71/sin θ = y-coordinate = –0.71 kos θ = koordinat-x = 0.71/cos θ = x-coordinate = 0.71 tan θ = koordinat-y koordinat-x tan θ = y-coordinate x-coordinate = –0.71 0.71 = –1.0 3 (a) y = kos x/ cos x (b) y = –kos x/ –cos x (c) y = sin x (d) y = –sin x BAHAGIAN B 4 (a) sin θ = koordinat-y/y-coordinate = 0.26 kos θ = koordinat-x/x-coordinate = –0.97 tan θ = koordinat-y koordinat-x tan θ = y-coordinate x-coordinate = 0.26 –0.97 = –0.27 Tk 5 23 PT Math J(J1-14).indd 7 23/2/2023 6:04:39 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J8 (b) sin θ = koordinat-y /y-coordinate = –0.52 kos θ = koordinat-x /x-coordinate = –0.86 tan θ = koordinat-y koordinat-x tan θ = y-coordinate x-coordinate = –0.52 –0.86 = 0.60 (c) sin θ = koordinat-y /y-coordinate = –0.37 kos θ = koordinat-x /x-coordinate = 0.93 tan θ = koordinat-y koordinat-x tan θ = y-coordinate x-coordinate = –0.37 0.93 = –0.40 5 (a) sin θ = –0.6428 = –sin 40° y x S A T C 320° 220° 40° 40° Sukuan/Quadrant III 180° < θ < 270° Sukuan/Quadrant IV 270° < θ < 360° θ = 180° + 40° atau/or θ = 360° – 40° = 220° = 320° θ = 220°, 320° (b) kos θ/cos θ = 0.6820 = kos 47°/cos 47° y x S A T C 313° 47° 47° Sukuan/Quadrant I 0° < θ < 90° Sukuan/Quadrant IV 270° < θ < 360° Maka/ erefore, θ = 47° atau/or 360° – 47° = 313° (c) tan θ = –1.732 = –tan 60° y x S A T C 120° 300° 60° 60° Sukuan/Quadrant II 90° < θ < 180° Sukuan/Quadrant IV 270° < θ < 360° Maka/ erefore, θ = 180° – 60° atau/or θ = 360° – 60° = 120° = 300° θ = 120°, 300° 6 (a) tan x = 1.428, 90° < x < 270° = tan 55° Maka/ erefore, x = 180° + 55° = 235° y x S A T C 235° 55° Sukuan/Quadrant III 180° < x < 270° (b) kos y/cos y = –0.9511, 180° < y < 360° = –kos 18° Maka/ erefore, y = 180° + 18° = 198° y x S A T C 198° 18° Sukuan/Quadrant III 180° < y < 270° 7 (a) Menggunakan Teorem Pythagoras, Using Pythagoras’ theorem, PR = AP2 – AR2 =  102 – 82 = 6 m Sisi bertentangan Opposite side Sisi bersebelahan Adjacent side 8 m 10 m x A R P Maka/ erefore, tan x = Sisi bertentangan Opposite side Sisi bersebelahan Adjacent side = 6 8 = 3 4 (b) Q ialah titik tengah PR/Q is the midpoint of PR Maka/ erefore, PQ = QR = 3 m Menggunakan teorem Pythagoras, Using Pythagoras’ theorem, BQ =  32 + 42 = 5 m y ialah sudut cakah yang terletak pada sukuan II dengan keadaan nilai sin sahaja yang positif. y is an obtuse angle that lies in quadrant II where only the values of the sine are positive. sin x = sin ∠BQR = 4 5 P Q R B y 3 m 5 m 4 m BAB 7 Sukatan Serakan Data Terkumpul KERTAS 1 1 A 2 D 3 B 4 A 5 C 6 C Tk 5 23 PT Math J(J1-14).indd 8 23/2/2023 6:04:39 PM PENERBIT ILMU BAKTI SDN. BHD.

J9 Praktis Topikal SPM Matematik Tingkatan 5 - Jawapan KERTAS 2 BAHAGIAN A 1 Sempadan atas Upper boundary Kekerapan Frequency Kekerapan longgokan Cumulative frequency 2 0 0 5 4 4 8 6 10 11 14 24 14 12 36 17 6 42 20 4 46 2 Umur tahun) Age (years) Kekerapan Frequency Titik tengah Midpoint Sempadan atas Upper boundary Kekerapan longgokan Cumulative frequency 17 – 21 0 19 21.5 0 22 – 26 3 24 26.5 3 27 - 31 7 29 31.5 10 32 – 36 10 34 36.5 20 37 – 41 11 39 41.5 31 42 – 46 8 44 46.5 39 47 – 51 6 49 51.5 45 3 (a) Bentuk taburan histogram bagi Ujian 1 ialah bentuk loceng, manakala bagi Ujian 2 ialah bentuk-U.  e distribution shape of the histogram for Test 1 is bell-shaped whereas for Test 2 is U-shaped distribution. (b) Ujian 1 mempunyai serakan yang lebih luas berbanding dengan Ujian 2 kerana julat markahnya lebih besar (97 – 67 = 30). Test 1 has a wider dispersion than Test 2 because the range of scores is greater (97 – 67 = 30). 4 (a) Bentuk taburan jisim murid kelas A ialah pencong ke kanan, manakala bagi kelas B ialah pencong ke kiri.  e mass distribution shape of students in class A is skewed to the right whereas in class B is skewed to the left. (b) Jisim murid kelas A mempunyai serakan yang sama dengan jisim murid kelas B walaupun mempunyai bentuk taburan yang berbeza.  e dispersions of the mass of pupils in class A and class B are the same even though their distribution shapes are di erent. (c) Kelas A ialah murid prasekolah kerana kebanyakan jisim mereka lebih ringan, manakala kelas B ialah murid sekolah rendah kerana kebanyakan jisim mereka lebih berat. Class A is preschool pupils because most of their mass is lighter whereas Class B is primary school pupils because most of their mass is heavier. 5 15 10 5 0 P10 P25 P70 39.5 44.5 49.5 54.5 59.5 64.5 20 25 30 35 40 Markah/Marks × 40 = 28 70 100 × 40 = 10 70 100 × 40 = 4 10 100 Kekerapan longgokan Cumulative frequency P10 = 43.5, P25 = 47.5, P70 = 54.0 BAHAGIAN B 6 Poligon kekerapan R/Frequency polygon R Varians/Variance, s2 = ∑fx2 ∑f – ( ∑fx ∑f ) 2 = 2 × 372 + 6 × 422 + 7 × 472 + 5 × 522 + 5 × 572 + 4 × 622 + 5 × 672 + 2 × 722 36 – ( 2 × 37 + 6 × 42 + 7 × 47 + 5 × 52 + 5 × 57 + 4 × 62 + 5 × 67 + 2 × 72 36 ) 2 = 99.75 Sisihan piawai/Standard deviation, s =  ∑fx2 ∑f –  ∑fx ∑f  2 =  99.75 = 9.987 Poligon kekerapan S/Frequency polygon S Varians/Variance, s2 = ∑fx2 ∑f – ( ∑fx ∑f ) 2 = 2 × 372 + 3 × 422 + 5 × 472 + 6 × 522 + 7 × 572 + 6 × 622 + 4 × 672 + 3 × 722 36 – ( 2 × 37 + 3 × 42 + 5 × 47 + 6 × 52 + 7 × 57 + 6 × 62 + 4 × 67 + 3 × 72 36 ) 2 = 91.13 Sisihan piawai/Standard deviation, s =  91.13 = 9.546 Sisihan piawai bagi data dalam poligon kekerapan S , Sisihan piawai bagi data dalam poligon kekerapan R Maka, data dalam poligon kekerapan S kurang diserakkan berbanding dengan data dalam poligon kekerapan R. Standard deviation of the data for frequency polygon S < Standard deviation of the data for frequency polygon S  erefore, the data in frequency polygon S is less dispersed than the data in frequency polygon R. BAHAGIAN C 7 (a) (i) Kelas mod/Modal class = 70 – 74 (ii) Min anggaran/Estimated mean = 5 400 80 = 67.5 (b) Sempadan atas Upper boundary Kekerapan longgokan Cumulative frequency 49.5 0 54.5 4 59.5 13 64.5 25 69.5 45 74.5 68 79.5 77 84.5 80 Tk 5 23 PT Math J(J1-14).indd 9 23/2/2023 6:04:40 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J10 (c) Kekerapan longgokan/ Cumulative frequency 80 — 70 — 60 — 50 — 40 — 30 — 20 — 10 — 0 Markah Marks 49.5 54.5 59.5 64.5 69.5 74.5 79.5 84.5 x x x x x x x 62.5 (d) 25% × 80 = 1 4 × 80 = 20 Pada graf, kekerapan longgokan = 20, markah, m = 62.5 On the graph, cumulative frequency = 20, marks, m = 62.5 8 (a) Selang kelas Class interval Titik tengah Midpoint Kekerapan Frequency 15 – 19 17 3 20 – 24 22 6 25 – 29 27 6 30 – 34 32 8 35 – 39 37 8 40 – 44 42 6 45 – 49 47 3 (b) (i) Min anggaran markah/Estimated mean mark = 1 290 40 = 32.25 (ii) Varians/Variance = 44 500 40 – ( 1 290 40 )2 = 72.4375 (iii) Sisihan piawai/Standard deviation = varians/variance = 72.4375 = 8.511 (c) Kekerapan/Frequency 4 3 2 1 0 12 17 22 27 32 37 42 47 52 5 6 7 8 Markah/Marks (d) Bilangan murid yang memperoleh lebih daripada 34 markah = 17 Number of students who obtained more than 34 marks = 17 BAB 8 Pemodelan Matematik KERTAS 1 1 B 2 C 3 C 4 A 5 C KERTAS 2 BAHAGIAN A 1 Dalam masalah ini, prinsipal dan kadar faedah diberi. Faedah simpanan ialah nilai yang diperlukan oleh Danish selain daripada simpanannya sebanyak RM2 000 untuk membeli komputer riba. Kita perlu menentukan tempoh yang diperlukan oleh Danish untuk menyimpan di bank bagi membolehkannya membeli komputer riba itu. In this problem, the principal and interest rates are given.  e savings interest is the amount that Danish need besides his savings of RM2 000 to buy a laptop. We need to determine how long Danish needs to save his money in the bank so that he can buy the laptop. (a) Andaian/Assumptions: • Andaikan bahawa kadar faedah tidak berubah dalam jangka masa pengiraan faedah. Assume that the interest rate does not change during the interest calculation period. • Andaikan bahawa harga komputer riba tidak berubah ketika Danish telah berjaya mengumpulkan wang yang diperlukan. Assume that the price of the laptop does not change when Danish has managed to save the required amount of money. Pemboleh ubah/Variables: • Pemboleh ubah bebas: Masa dalam tahun, t. Independent variable:  e time in years, t. • Parameter tetap: (i) Prinsipal, P; (ii) Kadar faedah, r. Fixed parameters: (i)  e principal, P; (ii)  e interest rate, r. • Pemboleh ubah bersandar: Faedah I, I = Prt. Dependent variables: Interest I, I = Prt. 2 Ya, pemodelan matematik boleh digunakan untuk menentukan pelan yang paling berkesan untuk mengatasi masalah tra k. Model matematik dibina untuk mengenal pasti jumlah bilangan kenderaan yang memasuki Bandar A dan jalan yang terlibat. Model ini juga dapat digunakan untuk menganalisis keadaan dan mencadangkan penyelesaian untuk mengurangkan kesesakan lalu lintas. Yes, mathematical modeling can be used to determine the most e ective plan to solve the tra c problems. A mathematical model was constructed to identify the total number of vehicles entering City A and the a ected roads.  e model can also be used to analyse the situation and propose the solutions to reduce the tra c congestion. BAHAGIAN B 3 1. Tentukan masalah: Cari bilangan penduduk di Taman Naruli. Determine the problem: Find the number of residents (population) in Taman Naruli. 2. Buat andaian: Penduduk bertambah secara linear. Make an assumption:  e population increases linearly. 3. Tentukan pemboleh ubah: Terdapat dua kuantiti yang berubah iaitu saiz penduduk dan masa dalam tahun. Determine the variables:  ere are two quantities that change i.e. population size and the time in years. (a) Pemboleh ubah bebas: t, tahun sejak 2016 Independent variable: t, time in years since 2016 (b) Pemboleh ubah bersandar: P(t), populasi di Taman Naruli Dependent variable: P (t), the population in Taman Naruli 4. Bina penyelesaian/Solve the problem: (a) Andaikan bahawa tahun 2016 → t = 0. Pertumbuhan penduduk dapat ditunjukkan oleh persamaan linear P(t) = mt + b. Tk 5 23 PT Math J(J1-14).indd 10 23/2/2023 6:04:40 PM PENERBIT ILMU BAKTI SDN. BHD.

J11 Praktis Topikal SPM Matematik Tingkatan 5 - Jawapan Diberi pasangan input-output (0, 450) dan (5, 880). Jadi, kita boleh menggunakan nilai-nilai ini untuk mengira kecerunan, m, Assume that the year 2016 → t = 0.  e population grows is given by a linear equation P(t) = mt + b. Given that the pair of input-output (0, 450) and (5,880).  erefore, we can use these values to calculate the gradient, m, Assume that the year 2016 → t = 0. m = 880 – 450 5 – 0 = 86 orang setahun/residents per year Maka, persamaan ialah P(t) = 86t + 450  erefore, the equation is P(t) = 86t + 450 Bilangan penduduk pada tahun 2026,  e population in 2026, P(10) = 86(10) + 450 = 1 310 (b) 1 400 = 86t + 450 86t = 950 t = 11.04 2016 + 11 = 2027 Jadi, populasi akan mencecah 1 400 orang pada tahun 2027.  erefore, the population will reach 1 400 by 2027. BAHAGIAN C 4 (a) 1 Mengenal pasti masalah: Cari untung / rugi dan titik balik modal pengeluaran. Identifying the problem: Find the pro t/loss and break even point of production capital. 2 Membuat andaian: Fungsi kos dan fungsi pendapatan ialah fungsi linear. Making assumptions:  e cost function and the revenue function are linear functions. 3 Mengenal pasti pemboleh ubah: Pemboleh ubah yang berkaitan dengan fungsi kos ialah kos tetap dan kos membuat unit tambahan produk. Sebaliknya, pemboleh ubah yang berkaitan dengan fungsi pendapatan adalah harga jualan seunit. Identifying the variables:  e variables related to the cost function are the  xed cost and the cost of making an additional unit of the product. Meanwhile, the variable related to the revenue function is the selling price per unit. 4 Menyelesaikan masalah/Solving the problem: (a) Fungsi kos keseluruhan ialah C(x) = cx + F dengan keadaan c mewakili kos per unit tambahan produk, dan F mewakili kos tetap yang tidak bergantung pada jumlah unit yang dihasilkan.  e total cost function is C(x) = cx + F where c represents the cost of making an additional unit of the product, and F represents the  xed cost that is not depend on the number of units produced. C(x) = 730x + 360 000 (b) Fungsi pendapatan → R(x) = sx dengan keadaan s ialah harga jualan per unit. R(x) = 1 400x  e revenue function is → R(x) = sx where s is the selling price per unit. R(x) = 1 400x (c) Fungsi keuntungan ialah P(x) = R(x) – C(x)  e pro t function is P(x) = R(x) – C(x) (b) (i) Untuk 500 unit/For 500 units: C(500) = 730(500) + 360 000 = 725 000 R(500) = 1 400(500) = 700 000 P(500) = 700 000 – 725 000 = –25 000 Oleh itu, terdapat kerugian RM25 000 yang setara dengan tahap pengeluaran 500 unit.  erefore, there is a loss of RM25 000 which is corresponding to the production level of 500 units. (ii) Untuk 600 unit/For 600 units: C(600) = 730(600) + 360 000 = 798 000 R(600) = 1 400(600) = 840 000 P(600) = 798 000 – 840 000 = 42 000 Jadi, terdapat keuntungan RM42 000 dengan tahap pengeluaran sebanyak 600 unit.  erefore, there is a pro t of RM42 000 which is corresponding to the production level of 600 units. (iii) Balik modal/Break-even: R(x) = C(x) 1400x = 730x + 360 000 670x = 360 000 x = 537.3 ≈ 538 unit/units KERTAS MODEL SPM KERTAS 1 1 C 2 D 3 C 4 A 5 A 6 B 7 B 8 B 9 A 10 C 11 A 12 C 13 A 14 B 15 C 16 D 17 A 18 D 19 A 20 B 21 D 22 A 23 A 24 C 25 B 26 B 27 C 28 C 29 C 30 D 31 B 32 C 33 A 34 D 35 C 36 B 37 D 38 A 39 A 40 D KERTAS 2 BAHAGIAN A 1 Menggunakan Teorem Pythagoras, Using Pythagoras’ theorem, 252 – 242 = 49 = 7 MN = (2 × 7) cm = 14 cm Isi padu kuboid/Volume of cuboid 14 × 24 × 16 = 5 376 cm3 Isi padu kon/Volume of cone = 1 3 × 22 7 × 72 × 24 = 1 232 cm3 Baki isi padu/Remaining volume = 5 376 – 1 232 = 4 144 cm3 2 (a) y =  2 x – 1 2y = x – 2, (y = 0) x = 2  p = 2 (b) mAD =  – (–) 0 – (–4) =  4 = 2 mAD = mBC  mBC = 2 B(2, 0) y – 0 = 2(x – 2) y = 2x – 4 y = 2x – 4, pintasan-y/y-intercept = –4 3 (a) Jika/If x , –5, maka/then x , –12 (Palsu/False) (b) 4 = 13 + 3 = 13 + 3 × 1 14 = 23 + 6 = 23 + 3 × 2 36 = 33 + 9 = 33 + 3 × 3 76 = 43 + 12 = 43 + 3 × 4  n3 + 3n, n = 1, 2, 3, 4, … Tk 5 23 PT Math J(J1-14).indd 11 23/2/2023 6:04:41 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J12 4 (a) Laju seragam kereta M = 50 km j–1 Uniform speed of car M = 50 km h–1 Kadar perubahan laju kereta M Rate of change of speed of car M = ( 30 60 ) 50 = 100 km j–2/100 km h–2 (b) 1 2 (t – 30 + t)50 = 1 2 × 100 × 70 t – 15 = 70 t = 85 5 (a) P’ ∩ Q (b) Q P R 6 (x + 6)2 = 4(x + 5) x2 + 12x + 36 = 4x + 20 x2 + 8x + 16 = 0 (x + 4)(x + 4) = 0 x + 4 = 0 x = –4 Jumlah jualan/Total sales: (x + 6)2 = [(–4) + 6]2 = (2)2 = 4 Jumlah jualan/Total sales: RM4 7 I ∝ P V I = kP V Apabila/When: I = 2.5 P = 600 dan/and V = 240 2.5 = k(600) 240 k = (2.5)(240) 600 = 1 Oleh itu/ erefore, I = P V Apabila/When: P = 1 000 dan/and V = 240 I = 1 000 240 = 4.17 A 8 Pahang Terengganu Perak Kelantan 9 Pelan Plan Deduktibel Deductible Dengan ko-insurans With co-insurance Perjanjian (Nilai) Agreement (Value) RM2 000 80/20 (%) Kos yang dibayar oleh pemegang polisi  e cost paid by the policyholder RM6 000 – RM2 000 = RM4 000 0.2 × RM4 000 = RM800 RM800 + RM2 000 = RM2 800 Kos yang dibayar oleh syarikat insurans  e cost paid by the insurer 0.8 × RM4 000 = RM3 200 10 Jarak (km) Distance (km) Kekerapan, f Frequency, f Titik tengah x, Midpoint, x fx x 2 fx 2 1 – 5 3 3 9 9 27 6 – 10 7 8 56 64 448 11 – 15 8 13 104 169 1 352 16 – 20 10 18 180 324 3 240 21 – 25 7 23 161 529 3 703 26 – 30 4 28 112 784 3 136 Σf = 39 Σfx = 622 Σf 2 = 11 906 Min/Mean, – x = Σfx Σf – x = 622 39 = 15.95 Varians/Variance, s2 = Σfx 2 Σf – – x = 11 906 39 –15.952 = 50.88 BAHAGIAN B 11 2m + 1 = 15 2m = 14 m = 7 3 + n + 4 + 2n + 6 + 10 + 11 7 = min/mean 3n + 34 7 = m 3n + 34 = 7(7) 2m + 1 = 15 2m = 14 m = 7 3n + 34 = 7m 3n + 34 = 7(7) 3n = 15 n = 5 Set data/Set of data: 3, 5, 4, 10, 6, 10, 11 s2 = Σx2 N – (x – )2 = 32 + 52 + 42 + 102 + 62 + 102 + 112 7 – 72 = 64 7 64 7 × 22 = p 7 p = 64 × 4 = 256 12 (a) 8x + 4y > 320 x + y < 80 x < 2y Tk 5 23 PT Math J(J1-14).indd 12 23/2/2023 6:04:41 PM PENERBIT ILMU BAKTI SDN. BHD.

J13 Praktis Topikal SPM Matematik Tingkatan 5 - Jawapan (b) 30 20 10 0 10 20 30 40 50 60 70 80 40 50 60 70 80 x + y = 80 8x + 4y = 320 2y = x x R y (perasa T) y (flavour T) x (perasa S) x (flavour S) (c) (i) Bilangan minimum = 20 cawan Minimum number = 20 cups Bilangan maksimum = 40 cawan Maximum number = 40 cups (ii) Tidak, kerana nilai x = 60 berada di luar kawasan berlorek No, because the value of x = 60 lies outside the shaded region (iii) Ya, kerana nilai y = 60 berada dalam kawasan berlorek jika nilai x terletak di antara 10 hingga 20 cawan. Yes, because the value of y = 60 lies inside the shaded region if the values of x are between 10 to 20 cups. 13 Lebihan gaji/Surplus of income = RM3 500 – RM300 – RM700 = RM2 500 Kereta jenis A/Typed A car Pinjaman bank/Bank loan = 90% × RM68 000 = RM61 200 Faedah pinjaman/Loan interest = RM61 200 × 0.032 × 9 = RM17 625.60 Ansuran bulanan/Monthly instalment = RM61 200 + RM17 625.60 9 × 12 = RM729.87 Kereta jenis B/Typed B car Pinjaman bank/Bank loan = 95% × RM105 000 = RM99 750 Faedah pinjaman/ Loan interest = RM99 750 × 0.028 × 9 I = p × r × t = RM25 137 Ansuran bulanan/Monthly instalment = RM99 750 + RM25 137 9 × 12 = RM1 156.36 Dengan lebihan gaji sebanyak RM2 500, Encik Bong mampu membeli kereta idamannya. Tetapi, kereta jenis A lebih murah dan kurang membebankan berbanding dengan kereta jenis B. With an income surplus of RM2 500, Encik Bong can a ord to buy his dream car. But, typed A car is cheaper and less burdensome compared to type B car. 14 2p + 2(q + 6) = 96 2p + 2q = 96 – 12 2p + 2q = 84 2(p + q) = 84 p + q = 42 q + 6 = 3p 3p – q = 6  1 3 1 –1p q  =  42 6   p q  = 1 (1)(–1) – (1)(3)  –1 –3 –1 1  42 6  = 1 –4  –1 –3 –1 1  42 6  = 1 –4  –48 –120  p q  =  12 30 Oleh itu, panjang padang (q + 6)  erefore, the length of the  eld (q + 6) = 30 + 6 = 36 m = 3 600 cm 15 (a) (i) P = Translasi/Translation ( –4 0 ) (ii) Putaran 90° ikut arah jam pada pusat (3, 11) A clockwise rotation of 90° at point (3, 11) (b) V = Translasi/Translation ( –3 0 ) W = Pembesaran dengan faktor skala 3 pada pusat (7, 6) W = An enlargement with a scale factor of 3 at centre (7, 6) BAHAGIAN C 16 (a) y = 4(7 – x)(3 + x) y = –4x2 + 16x + 84 (b) x = –4, y = 4[7 – (–4)][3 + (–4)] = –44 x = 8, y = 4(7 – 8)(3 + 8) = –44 Pintasan-x/x-intercept: y = 0, 4(7 – x)(3 + x) = 0 x = 7 atau/or x = –3 Pintasan-y/y-intercept: x = 0 , y = 84 y x O 84 −44 −4 −3 7 8 (c) Apabila/When y = 84, 84 = –4x2 + 16x + 84 4x2 – 16x = 0 4x(x – 4) = 0 x = 0 atau/or x = 4  x = 4 (d) y adalah maksimum apabila/y is maximum when x = (–3) + 7 2 = 2 paksi simetri/axis of symmetry x = 2, y = 4(7 – 2)(3 + 2) = 100  Isi padu maksimum kuboid = 100 cm3 Maximum volume of cuboid = 100 cm3 Tk 5 23 PT Math J(J1-14).indd 13 23/2/2023 6:04:41 PM PENERBIT ILMU BAKTI SDN. BHD.

 Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J14 17 (a) Jumlah pendapatan tahunan Peter Total annual income for Peter = RM80 000 + RM3 200 = RM83 200 Jumlah pelepasan cukai/Total tax relief = Sumbangan PERKESO + Perbelanjaan perubatan ibu bapa SOCSO + Medical expenses for parents = RM250 + RM2 000 = RM2 250 Jumlah pendapatan bercukai/Total chargeable income = Jumlah pendapatan tahunan – Potongan cukai (pendermaan) – Pelepasan cukai Total annual income – Tax exemption (donation) – Tax relief = RM83 200 – RM1 000 – RM2 250 = RM79 950 Bagi pendapatan bercukai sebanyak RM79 950, kadar cukai ialah 21%. For the chargeable income of RM79 950, the tax rate is 21%. Jumlah yang perlu dibayar pada RM70 000 pertama Base tax on the  rst RM70 000 = RM4 600 RM79 950 – RM70 000 = RM9 950 Jumlah yang perlu dibayar pada RM9 950 seterusnya Tax on the next balance of RM9 950 = RM9 950 × 0.21 = RM2 089.50 Jadi, cukai kena dibayar  erefore, the income tax payable = RM4 600 + RM2 089.50 = RM6 689.50 (b) Jumlah pendapatan tahunan Amy Total annual income for Amy = RM120 000 Jumlah pelepasan cukai = elaun gaya hidup (pembelian komputer riba + buku & majalah) + bayaran kepada PRS + Yuran penjagaan anak ke tadika + Pelepasan anak Total tax relief = Lifestyle allowance (for the purchase of laptop, books and magazines) + payment to PRS + childcare fee to kindergarten + Child relief = RM2 500 + RM3 000 + RM1 000 + RM2 000 = RM8 500 Jumlah pendapatan bercukai/Total chargeable income = Jumlah pendapatan tahunan/Total annual income – Jumlah pelepasan cukai/Tax relief = RM120 000 – RM8 500 = RM111 500 Bagi pendapatan bercukai sebanyak RM111 500, kadar cukai ialah 24%. For the chargeable income of RM111 500, the tax rate is 24%. Jumlah yang perlu dibayar pada RM100 000 yang pertama = RM10 900 Base tax on the  rst RM100 000 = RM10 900 RM111 500 – RM100 000 = RM11 500 Jumlah yang perlu dibayar pada RM11 500 seterusnya Tax on the next balance of RM11 500 = RM11 500 × 0.24 = RM2 760 Jadi, cukai kena dibayar  erefore, the income tax payable = RM10 900 + RM2 760.00 = RM13 660 Tk 5 23 PT Math J(J1-14).indd 14 23/2/2023 6:04:41 PM PENERBIT ILMU BAKTI SDN. BHD.

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