Matematik T5

Matematik Tingkatan 5 Soalan SPM Kertas 2 Bahagian C 130 (b) Min/ Mean: 1 495 50 = 29.9 Sisihan piawai/ Standard deviation: σ2 = 51 332.5 50 ‒ 29.92 = 1 026.65 – 894.01 = 132.64 σ = ±11.52 Skor Scores Bilangan peserta, f Number of participants, f Titik tengah, x Midpoint, x fx x 2 fx 2 1 – 10 4 5.5 22 30.25 121 11 – 20 6 15.5 93 240.25 1 441.5 21 – 30 13 25.5 331.5 650.25 8 453.25 31 – 40 18 35.5 639 1 260.25 22 684.5 41 – 50 9 45.5 409.5 2 070.25 18 632.25 ∑f = 50 ∑fx = 1 495 ∑fx2 = 51 332.5 ⸫ Min skor peserta tertumpu kepada 29.9 dan dalam julat 18.38 hingga 41.42. Prestasi peserta dalam cabaran matematik boleh dikatakan baik kerana kebanyakan skor peserta pencong ke kiri (skor yang tinggi). The mean of scores of the participants concentrates around 29.9 and within the range from 18.38 to 41.42. The performance of the participants in the mathematics challenge can be said to be good because most of the scores of the participants skewed to the left (high scores). (c) (i) n(ξ) = n(S) + n(M) – n(S ∩ M) (ii) ξ S M 36 24 26 86 = 60 + 50 – n(S ∩ M) n(S ∩ M) = 110 – 86 = 24 3 Jadual 2 menunjukkan bilangan pen yang dihasilkan oleh sebuah mesin di sebuah kilang dan masa yang diambil untuk penghasilan tersebut. Table 2 shows the number of pens produced by a machine at a factory and the time taken for the production. Bilangan pen Number of pens 45 000 60 000 p 360 000 Masa yang diambil (minit) Time taken (minutes) 90 120 300 q Jadual 2/ Table 2 (a) (i) Berdasarkan Jadual 2, nyatakan hubungan antara bilangan pen, n dan masa, t, yang diambil, dan wakilkan hubungan itu dengan satu rumus yang sesuai. Based on Table 2, state the relation between the number of pens, n and the time taken, t, and represent the relation using a suitable formula. (ii) Seterusnya, cari nilai p. Hence, find the value of p. [3 markah/ marks] (b) Diberi bahawa maksimum masa operasi mesin tersebut sebelum terlampau panas ialah 6 jam. Given that the maximum operating hour of the machine before overheating is 6 hours. (i) Cari nilai q. Find the value of q. (ii) Kilang pen tersebut beroperasi selama 8 jam sehari. Andaikan kilang tersebut hanya mempunyai sebuah mesin, terangkan sama ada kilang tersebut dapat menghasilkan 360 000 batang pen dalam sehari dengan satu hujah deduktif yang sah. The pen factory operates for 8 hours a day. Assuming the factory only has one machine, explain whether the factory is able to produce 360 000 pens in a day using a valid deductive argument. [3 markah/ marks] CONTOH

Matematik Tingkatan 5 Soalan SPM Kertas 2 Bahagian C 131 (c) (i) Kilang itu menghasilkan x batang pen merah dan y batang pen biru dalam sehari. Bilangan pen biru yang terhasil tiga kali lebih daripada bilangan pen merah dan beza bilangan pen biru dan pen merah tidak boleh melebihi 50 000 batang. Tuliskan tiga ketaksamaan linear selain daripada x M 0 dan y M 0 bagi penghasilan pen merah dan pen biru oleh kilang tersebut dalam sehari. The factory produces x red pens and y blue pens in a day. The number of blue pens produced is three times more than the number of red pens and the difference between the number of blue pens and red pens must not exceed 50 000. Write three linear inequalities other than x M 0 and y M 0 for the production of red pens and blue pens by the factory in a day. (ii) Lukis satu graf pada Rajah 3 di ruang jawapan untuk mewakili kekangan penghasilan pen dan lorekkan rantau yang memuaskan penghasilan pen kilang tersebut. Draw a graph on Diagram 3 in the answer space to represent the constraints of the production of pens and shade the region which satisfies the factory’s production of pens. (iii) Dengan menggunakan dua persamaan linear yang sesuai, hitung bilangan maksimum pen merah dan pen biru yang boleh dihasilkan oleh kilang tersebut. By using two suitable linear equations, calculate the maximum number of red pens and blue pens which can be produced by the factory. [8 markah/ marks] Jawapan/ Answers: (a) (i) Bilangan pen berubah secara langsung dengan masa yang diambil. The number of pens varies directly as the time taken. n ∝ t n = kt 45 000 = 90k k = 500 ⸫ n = 500t (ii) Apabila/ When t = 300, p = 500(300) = 150 000 ⸫ p = 150 000 batang pen/ pens (b) (i) Apabila/ When n = 360 000, 360 000 = 500q q = 720 minit/ minutes = 12 jam/ hours (ii) Premis 1: Jika mesin itu menghasilkan 360 000 batang pen, maka mesin itu perlu beroperasi selama 12 jam. Premise 1: If the machine produced 360 000 pens, then the machine has to operate for 12 hours. Premis 2: Mesin itu hanya boleh beroperasi selebih-lebihnya 6 jam sehari. Premise 2: The machine can only operate at most for 6 hours a day. Kesimpulan: Mesin itu tidak boleh menghasilkan 360 000 batang pen dalam sehari. Conclusion: The machine cannot produce 360 000 pens in a day. (c) (i) nmax = 500(6 × 60) = 180 000 I : x + y N 180 000 II : y > 3x III : y – x N 50 000 CONTOH

Matematik Tingkatan 5 Soalan SPM Kertas 2 Bahagian C 132 (ii) 0 50 000 180 000 150 000 180 000 150 000 200 000 50 000 100 000 100 000 y x y = 3x y – x = 50 000 y + x = 180 000 Rajah 3/ Diagram 3 (iii) y = 3x …… ➀ x + y = 180 000 ……➁ ➀ ↷ ➁, x + 3x = 180 000 4x = 180 000 x = 45 000 Apabila/ When x = 45 000, y = 3(45 000) = 135 000 ⸫ Bilangan maksimum bagi pen merah ialah 45 000 batang dan bagi pen biru ialah 135 000. The maximum number of red pens is 45 000 and the maximum number of blue pens is 135 000. 4 Cik Zuraini bekerja sebagai seorang setiausaha di sebuah syarikat antarabangsa dan mendapat gaji tahunan sebanyak RM90 000. Dia juga mengutip sewa bulanan sebanyak RM1 800 sebagai pendapatan pasif. Pada tahun 2021, Cik Zuraini mempunyai pelepasan cukai sebanyak RM16 750 dan dia juga membayar zakat berjumlah RM1 800. Jadual 3 menunjukkan sebahagian kadar cukai pendapatan individu untuk tahun taksiran 2021. Ms Zuraini works as a secretary in an international company and receives an annual salary of RM90 000. She also collects a monthly rental of RM1 800 as passive income. In the year 2021, Ms Zuraini has a tax exemption of RM16 750 and she also paid zakat amounting to RM1 800. Table 3 shows a part of the individual income tax rates for assessment year of 2021. Pendapatan bercukai Chargeable income (RM) Pengiraan Calculations (RM) Kadar Rate (%) Cukai Tax (RM) 70 001 – 100 000 Pada 70 000 pertama/ On the first 70 000 4 400 Pada 30 000 berikut/ On the next 30 000 21 6 300 100 001 – 250 000 Pada 100 000 pertama/ On the first 100 000 10 700 Pada 150 000 berikut/ On the next 150 000 24 36 000 Jadual 3/ Table 3 (a) Diberi bahawa gaji Cik Zuraini dipotong sebanyak RM900 setiap bulan untuk potongan cukai bulanan (PCB). Tentukan sama ada Cik Zuraini perlu membuat bayaran baki cukai pendapatan atau mempunyai lebihan PCB yang akan dipulangkan oleh LHDN. Given that Ms Zuraini’s salary was deducted monthly by RM900 for the monthly tax deduction (PCB). Determine whether Ms Zuraini needs to pay the balance of income tax or has excess PCB which will be returned by IRB. [5 markah/ marks] CONTOH

Matematik Tingkatan 5 Soalan SPM Kertas 2 Bahagian C 133 (b) Berdasarkan gaya hidup Cik Zuraini pada tahun 2021, Jadual 4 menunjukkan perbelanjaan bulanan anggarannya. Based on Ms Zuraini’s lifestyle in year 2021, Table 4 shows her estimated monthly expenses. Perbelanjaan bulanan Monthly expense Amaun anggaran (RM) Estimated amount (RM) Ansuran pinjaman rumah (1)/ Housing loan instalment (1) 1 200 Ansuran pinjaman rumah (2)/ Housing loan instalment (2) 1 400 Bil utiliti/ Utility bills 1 250 Perbelanjaan dapur/ Groceries 220 Makanan/ Food 820 Elaun ibu bapa/ Parents’ allowance 1 000 Premium insurans/ Insurance premium 420 Bil telefon/ Telephone bill 100 Belanjaan lain/ Other expenses 350 Jadual 4/ Table 4 Bina satu pelan kewangan bulanan bagi Cik Zuraini sekiranya dia menetapkan 15% daripada jumlah pendapatannya selepas menolak PCB sebagai simpanan untuk kegunaan kecemasan. Construct a monthly financial plan for Ms Zuraini if she fixed 15% of her total income after the deduction of PCB as a saving for emergency. [4 markah/ marks] (c) Rajah 4 menunjukkan dua buah kereta dengan harga jualan masing-masing. Diagram 4 shows two cars with their respective selling price. Proton X50 RM189 000 RM93 200 Toyota Camry Rajah 4/ Diagram 4 Cik Zuraini bercadang membeli salah satu buah kereta yang ditunjukkan di Rajah 4 dalam masa dua tahun. Menurut penjual kereta, Cik Zuraini perlu membayar 8% wang pendahuluan untuk pembelian kereta pilihannya. Ms Zuraini plans to buy one of the cars shown in Diagram 4 within two years. According to the car dealer, Ms Zuraini needs to pay 8% down payment to purchase the car of her choice. (i) Kenal pasti dan definisikan masalah Cik Zuraini. Seterusnya, berikan dua andaian yang perlu dibuat bagi masalah ini. Identify and define Ms Zuraini’s problem. Hence, given two assumptions that need to be made for this problem. (ii) Kereta yang manakah patut dibeli oleh Cik Zuraini? Terangkan jawapan anda. Which car should be bought by Ms Zuraini? Explain your answer. [6 markah/ marks] Jawapan/ Answers: (a) Jumlah pendapatan/ Total income: Pendapatan bercukai/ Chargeable income: RM90 000 + 12(RM1 800) RM111 600 – RM16 750 – RM1 800 = RM93 050 = RM90 000 + RM21 600 = RM111 600 Jumlah cukai pendapatan yang perlu dibayar/ Total payable income tax: RM4 400 + [(RM93 050 – RM70 000) × 21%] = RM4 400 + RM4 840.50 = RM9 240.50 Jumlah PCB yang dipotong/ Total PCB deducted: 12 × RM900 = RM10 800 ⸫ Jumlah PCB yang dipotong melebihi cukai pendapatan yang perlu dibayar (RM10 800 > RM9 240.50), oleh itu LHDN akan memulangkan lebihan PCB, iaitu RM1 559.50 (RM10 800 – RM9 240.50) kepada Cik Zuraini melalui akaun banknya. The total PCB deducted exceeds the payable income tax (RM10 800 > RM9 240.50), therefore IRB will return the balance PCB, which is RM1 559.50 (RM10 800 – RM9 240.50) to Ms Zuraini through her bank account. CONTOH

Matematik Tingkatan 5 Soalan SPM Kertas 2 Bahagian C 134 (b) Gaji bulanan Cik Zuraini selepas dipotong PCB: Ms Zuraini’s monthly salary after deducting PCB: (RM90 000 ÷ 12) – RM900 = RM7 500 – RM900 = RM6 600 Pendapatan dan perbelanjaan Income and expenditure (RM) Pendapatan aktif (selepas menolak PCB)/ Active income (after deducting PCB) Pendapatan pasif/ Passive income 6 600 1 800 Jumlah pendapatan/ Total income 8 400 (‒) Simpanan tetap bulanan/ Fixed monthly savings 1 260 Baki pendapatan/ Income balance 7 140 (‒) Perbelanjaan tetap bulanan/ Monthly fixed expenses Ansuran pinjaman rumah (1)/ Housing loan instalment (1) Ansuran pinjaman rumah (2)/ Housing loan instalment (2) Premium insurans/ Insurance premium 1 200 1 400 420 Jumlah perbelanjaan tetap bulanan/ Total monthly fixed expenses 3 020 (‒) Perbelanjaan tidak tetap bulanan/ Monthly variable expenses Bil utiliti/ Utility bills Perbelanjaan dapur/ Groceries Makanan/ Food Elaun ibu bapa/ Parents’ allowance Bil telefon/ Telephone bill Belanjaan lain/ Other expenses 1 250 220 820 1 000 100 350 Jumlah perbelanjaan tidak tetap bulanan/ Total monthly variable expenses 3 740 Pendapatan lebihan/ Surplus of income 380 (c) (i) Masalah Cik Zuraini ialah dia perlu menyimpan cukup wang dalam masa dua tahun untuk membayar wang pendahuluan bagi pembelian keretanya. Cik Zuraini perlu menentukan berapa wang dia perlu menyimpan daripada pendapatan lebihannya setiap bulan selama dua tahun. Ms Zuraini’s problem is she needs to save enough money within two years in order to pay the down payment for the purchase of her car. Ms Zuraini needs to determine how much she needs to save from her surplus of income every month for two years. Andaian/ Assumptions: • Harga kereta tidak berubah dalam masa dua tahun itu. The price of the car does not change within the two years. • Pendapatan lebihan Cik Zuraini kekal sama selama dua tahun. Ms Zuraini’s surplus of income remains the same for two years. • Peratusan wang pendahuluan yang perlu dibayar tidak berubah dalam masa dua tahun itu. The percentage of down payment which needs to be paid does not change within the two years. (ii) Simpanan bulanan untuk wang pendahuluan bagi pembelian Toyota Camry: Monthly saving for the down payment to purchase Toyota Camry: (RM189 000 × 8%) ÷ 24 bulan/ months = RM15 120 ÷ 24 = RM630 Simpanan bulanan untuk wang pendahuluan bagi pembelian Proton X50: Monthly saving for the down payment to purchase Proton X50: (RM93 200 × 8%) ÷ 24 bulan/ months = RM7 456 ÷ 24 = RM310.67 CONTOH

Matematik Tingkatan 5 Soalan SPM Kertas 2 Bahagian C 135 ⸫ Cik Zuraini patut membeli Proton X50, kerana dia dapat menyimpan RM310.67 daripada pendapatan lebihannya pada setiap bulan dan masih mempunyai baki pendapatan. Manakala, simpanan bulanan untuk wang pendahuluan pembelian Toyota Camry telah melebihi pendapatan lebihannya jikalau Cik Zuraini bercadang membeli kereta dalam masa dua tahun. Ms Zuraini should buy Proton X50, because she is able to save RM310.67 from her surplus of income every month and still have some balance. However, the monthly saving for the down payment of the purchase of Toyota Camry has exceeded her surplus of income if Ms Zuraini plans to buy the car within two years. 5 Jadual 5 menunjukkan jarak antara beberapa pasang lokasi di Selangor dan Kuala Lumpur. Table 5 shows the distances between a few pairs of locations in Selangor and Kuala Lumpur. Pasangan lokasi Location pair Jarak (km) Distance (km) (Kuala Lumpur, Klang) 33.5 (Kuala Lumpur, Petaling Jaya) 14.7 (Klang, Kuala Lumpur) 35.8 (Klang, Petaling Jaya) 25.0 (Petaling Jaya, Kuala Lumpur) 18.3 (Petaling Jaya, Subang Jaya) 12.6 (Subang Jaya, Klang) 21.6 (Subang Jaya, Kuala Lumpur) 22.1 (Subang Jaya, Petaling Jaya) 11.7 Jadual 5/ Table 5 (a) Berdasarkan Jadual 5, lukis satu graf terarah dan berpemberat untuk menunjukkan rangkaian laluan di Selangor dan Kuala Lumpur. Based on Table 5, draw a directed and weighted graph to show the network of routes in Selangor and Kuala Lumpur. [3 markah/ marks] (b) Cik Eliza ialah seorang pereka hias dalaman yang bekerja bebas. Dia sendiri perlu menghantarkan rekaan penting kepada kliennya. Jadual 6 menunjukkan masa janji temu, tempat kediaman klien Cik Eliza dan tempoh masa perundingan pada suatu hari tertentu. Ms Eliza is a freelance interior designer. She needs to send important designs to her clients personally. Table 6 shows the time of appointment, the residence locations of Ms Eliza’ clients and her consultation duration on a certain day. Nama klien Client’s name Tempat kediaman Residence location Masa janji temu Time of appointment Tempoh masa perundingan Consultation duration Edward Kuala Lumpur 1100 2.5 jam/ hours Zuraini Petaling Jaya 1430 1.2 jam/ hours Suk Kuan Klang 1600 1.4 jam/ hours Subramaniam Petaling Jaya 1730 0.9 jam/ hour Jadual 6/ Table 6 Diberi bahawa Cik Eliza meninggal di Subang Jaya, nyatakan perjalanan yang dilalui oleh Cik Eliza sehingga dia balik rumah pada hari itu. Seterusnya, hitung perbelanjaan petrol, dalam RM, bagi seluruh perjalanannya jika harga petrol ialah RM2.80 per liter dan penggunaan petrol kereta Cik Eliza ialah 0.94 liter setiap 10 km. Given that Ms Eliza stays in Subang Jaya, state the journey travelled by Ms Eliza until she got home on that day. Hence, calculate the expense of petrol, in RM, of her whole trip if the price of petrol is RM2.80 per litre and the petrol consumption of Ms Eliza’s car is 0.94 litre for each 10 km. [3 markah/ marks] (c) (i) Diberi Cik Eliza bertolak dari rumahnya pada pukul 10.30 pagi dan menyampai rumahnya pada pukul 6.15 petang. Hitung jumlah masa, dalam jam, yang digunakan oleh Cik Eliza dalam perjalanan. Given Ms Eliza departed from her house at 10.30 a.m. and reached her house at 6.15 p.m. Calculate the total time, in hours, Ms Eliza spent on the road. [2 markah/ marks] (ii) Cari laju purata, dalam km/j, bagi perjalanan Cik Eliza. Find the average speed, in km/h, of Ms Eliza’s trip. [2 markah/ marks] CONTOH

Matematik Tingkatan 5 Soalan SPM Kertas 2 Bahagian C 136 (d) (i) Pendapatan, P, Cik Eliza bergantung kepada masa perundingan, tp , dan masa perjalanan, td . Ungkapkan P dalam sebutan tp dan td . Ms Eliza’s income, P, depends on the consultation duration, tp and the travelling time, td . Express P in terms of tp and td . [2 markah/ marks] (ii) Diberi bahawa Cik Eliza mengenakan bayaran perundingan sebanyak RM200 setiap jam. Hitung jumlah pendapatan bersih, dalam RM, yang diperoleh Cik Eliza pada hari itu. Given that Ms Eliza charges a consultation fee of RM200 per hour. Calculate the total net income, in RM, obtained by Ms Eliza on that day. [3 markah/ marks] Jawapan/ Answers: (a) Kuala Lumpur 18.3 km 14.7 km 25 km 35.8 km 21.6 km 12.6 km 11.7 km 22.1 km 33.5 km Petaling Jaya Klang Subang Jaya (b) Subang Jaya → Kuala Lumpur → Petaling Jaya → Subang Jaya → Klang → Petaling Jaya → Subang Jaya Jumlah jarak/ Total distance: 22.1 km + 14.7 km + 12.6 km + 21.6 km + 25 km + 12.6 km = 108.6 km Perbelanjaan petrol/ Expense of petrol: (108.7 km ÷ 10 km) × 0.94 /km × RM2.80/ = RM28.61 (c) (i) 18 jam/ hours 15 minit/ minutes – 10 jam/ hours 30 minit/ minutes = 7 jam/ hours 45 minit/ minutes Jumlah masa lawatan/ Total visiting time: 2.5 + 1.2 + 1.4 + 0.9 = 6 jam/ hours Masa dalam perjalanan/ Time on the road: 7.75 – 6 = 1.75 jam/ hours (ii) Laju purata/ Average speed: 108.6 km 1.75 jam/ hours = 62.06 km/j/ km/h (d) (i) P ∝ tp td P = ktp td (ii) Pendapatan yang diperoleh pada hari itu/ Income obtained on that day: P = 200(6) 1.75 = RM685.71 Pendapatan bersih/ Net income: RM685.71 – RM28.61 = RM657.10 CONTOH

Matematik Tingkatan 5 Soalan SPM Kertas 2 Bahagian C 137 6 Michael menjual 5 kotak pelitup muka dan 2 botol pensanitasi tangan pada harga RM265. Pada masa yang sama, seorang pelanggan telah membayar RM369 untuk membeli 3 kotak pelitup muka dan 4 botol pensanitasi tangan. Michael sells 5 boxes of face masks and 2 bottles of hand sanitisers at the price of RM265. At the same time, a customer paid RM369 to buy 3 boxes of face masks and 4 bottles of hand sanitisers. (a) Dengan menggunakan kaedah matriks, hitung harga masing-masing bagi sekotak pelitup muka dan sebotol pensanitasi tangan. By using matrix method, calculate the respective price for a box of face masks and a bottle of hand sanitiser. [3 markah/ marks] (b) Jadual 7 menunjukkan jualan pelitup muka dan pensanitasi tangan dalam enam bulan. Table 7 shows the sales of face masks and hand sanitisers in six months. Bulan Month Jualan pelitup muka (kotak) Sales of face masks (box) Jualan pensanitasi tangan (botol) Sales of hand sanitisers (bottle) Januari/ January 64 90 Februari/ February 72 72 Mac/ March 56 60 April/ April 68 52 Mei/ May 60 64 Jun/ June 76 82 Jadual 7/ Table 7 (i) Pada graf yang sama, plot satu histogram bagi jualan pelitup muka dan satu kekerapan poligon bagi jualan pensanitasi tangan. Seterusnya, huraikan bentuk taburan jualan bagi kedua-dua produk. On the same graph, plot a histogram for the sales of face masks and a frequency polygon for the sales of hand sanitisers. Hence, describe the shapes of the sales distribution for both products. [5 markah/ marks] (ii) Hitung min jualan bagi kedua-dua produk dalam enam bulan itu. Calculate the mean of sales for both products in the six months. [2 markah/ marks] (c) (i) Diberi keuntungan bagi sekotak pelitup muka dan sebotol pensanitasi tangan masing-masing ialah RM12 dan RM24. Berdasarkan min jualan di 6(b)(ii), hitung jumlah anggaran pendapatan yang diperoleh Michael daripada jualan kedua-dua produk untuk setahun. Given the profits for a box of face masks and a bottle of hand sanitiser are RM12 and RM24 respectively. Based on the mean sales in 6(b)(ii), calculate the estimated income obtained by Michael from the sales of both products for one year. [2 markah/ marks] (ii) Jadual 8 menunjukkan pelepasan cukai yang boleh dituntut oleh Michael pada tahun taksiran itu. Table 8 shows the tax relief which can be claimed by Michael for that assessment year. Pelepasan cukai Tax relief Amaun/ Amount (RM) Individu/ Individual 9 000 Insurans hayat dan KWSP (had RM7 000)/ Life insurance and EPF (limited to RM7 000) 2 256 Yuran pengajian sendiri (had RM7 000)/ Self-education fees (limited to RM7 000) 7 200 Gaya hidup (had RM2 500)/ Lifestyle (limited to RM2 500) 3 000 Jadual 8/ Table 8 Dengan menggunakan anggaran di 6(c)(i) sebagai pendapatan tahunan, hitung pendapatan bercukai bagi Michael. Seterusnya, nyatakan sama ada Michael layak mendapat rebat cukai sebanyak RM400. Using the estimation in 6(c)(i) as the annual income, calculate the chargeable income for Michael. Hence, state whether Michael is eligible for the tax rebate of RM400. [3 markah/ marks] Jawapan/ Answer: (a) Biar m sebagai harga sekotak pelitup muka dan t sebagai harga sebotol pensanitasi tangan. Let m be the price of a box of face masks and t be the price of a bottle of hand sanitiser. 5m + 2t = 265 3m + 4t = 369 CONTOH

Matematik Tingkatan 5 Soalan SPM Kertas 2 Bahagian C 138 5 3 2 4 m t = 265 369 m t = 1 5(4) ‒ (2)(3) 4 –3 –2 5 265 369 = 1 14 4(265) ‒ 2(369) ‒3(265) + 5(369) = 1 14 322 1 050 = 23 75 ⸫ Harga bagi sekotak pelitup muka ialah RM23 dan harga bagi sebotol pensanitasi tangan ialah RM75. The price of a box of face masks is RM23 and the price of a bottle of hand sanitiser is RM75. (b) (i) 0 60 80 100 Kekerapan/ Frequency 40 20 Bulan Month Jan Feb Mac Apr Mei May Jun Jualan pelitup muka adalah seragam manakala jualan pensanitasi tangan adalah bimod dalam enam bulan ini. The sales of face masks are uniform whereas the sales of hand sanitisers are bimodal in these six months. (ii) Min jualan bagi pelitup muka/ Mean sales for face masks: 64 + 72 + 56 + 68 + 60 + 76 6 = 369 6 = 66 kotak/ boxes Min jualan bagi pensanitasi tangan/ Mean sales for hand sanitiser: 90 + 72 + 60 + 52 + 64 + 82 6 = 420 6 = 70 kotak/ boxes (c) (i) Jumlah pendapatan/ Total income: [66(RM12) + 70(RM24)] × 12 = (RM792 + RM1 680) × 12 = RM2 472 × 12 = RM29 664 (ii) RM29 664 – (RM9 000 + RM2 256 + RM7 000 + RM2 500) = RM29 664 – RM20 756 = RM8 908 ⸫ Pendapatan bercukai Michael adalah kurang daripada RM35 000, maka dia layak mendapat rebat cukai. Michael’s chargeable income is less than RM35 000, therefore he is eligible for the tax rebate. CONTOH

SIJIL PELAJARAN MALAYSIA MATEMATIK KERTAS 1 1449/1 1 1 2 jam Satu jam tiga puluh minit NO. KAD PENGENALAN : ANGKA GILIRAN : JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU ARAHAN: 1 Tulis nombor kad pengenalan dan angka giliran anda pada ruang yang disediakan. 2 Kertas soalan ini adalah dalam dwibahasa. 3 Soalan dalam bahasa Melayu mendahului soalan yang sepadan dalam bahasa Inggeris. 4 Kertas soalan ini mengandungi 40 soalan. Jawab semua soalan. 5 Calon dikehendaki membaca maklumat di halaman belakang kertas peperiksaan ini. CONTOH

CONTOH

141 Kertas soalan ini mengandungi 40 soalan. Jawab semua soalan. Setiap soalan diikuti dengan empat pilihan jawapan, A, B, C dan D. Bagi setiap soalan, pilih satu jawapan sahaja. Rajah yang diberikan tidak dilukis mengikut skala kecuali dinyatakan. Anda boleh menggunakan kalkulator saintifik yang tidak boleh diprogramkan. This question paper consists of 40 questions. Answer all questions. Each question is followed by four choices of answers A, B, C and D. For each question, choose one answer only. The diagrams in the questions provided are not drawn to scale unless stated. You may use a non-programmable scientific calculator. 1 Nyatakan 84 500 321 dalam bentuk piawai. State 84 500 321 in standard form. A 84.5 × 106 B 8.45 × 107 C 8.50 × 107 D 9.00 × 107 2 Bundarkan 1 300.094562 betul kepada lima angka bererti. Round off 1 300.094562 correct to five significant figures. A 0.094562 C 1 300.0 B 13.946 D 1 300.1 3 Senaraikan semua integer bagi nilai x yang memuaskan ketaksamaan 12 – 2x < x + 24 N 42 – 5x. List all the integers for the value of x which satisfy the inequality 12 – 2x < x + 24 N 42 – 5x. A ‒4, ‒3, ‒2, ‒1, 0, 1, 2, 3 B ‒4, ‒3, ‒2, ‒1, 1, 2, 3 C ‒3, ‒2, ‒1, 0, 1, 2, 3 D ‒3, ‒2, ‒1, 0, 1, 2 4 Permudahkan/ Simplify: 4m ‒ 4n m2 ‒ n2 – m ‒ n 6m + 6n A 4 ‒ m ‒ n m + n C 24 ‒ m ‒ n 6(m + n) B 4 ‒ m + n m + n D 24 ‒ m + n 6(m + n) 5 Rajah 1 menunjukkan sebuah bulatan. Diagram 1 shows a circle. y 30° 70° Rajah 1/ Diagram 1 Cari nilai y. Find the value of y. A 40° B 35° C 30° D 15° 6 Rajah 2 menunjukkan dua buah segi tiga, PQR dan PRS. QRS ialah satu garis lurus. Diagram 2 shows two triangles, PQR and PRS. QRS is a straight line. 12 cm 13 cm 21 cm Q R S P Rajah 2/ Diagram 2 Cari luas, dalam cm2 , bagi segi tiga PRS. Find the area, in cm2 , of triangle PRS. A 126 B 104 C 96 D 88 7 Tentukan nisbah bilangan pepenjuru sebuah nonagon kepada bilangan pepenjuru sebuah heksagon. Determine the ratio of the number of diagonals of a nonagon to the number of diagonals of a hexagon. A 27 : 3 B 3 : 27 C 3 : 1 D 1 : 3 8 Diameter bagi sebuah rim roda motosikal ialah 45.72 cm. Hitung jarak minimum, dalam m, yang dilalui oleh roda itu selepas roda itu membuat 80 putaran yang lengkap. The diameter of the rim of a motorcycle wheel is 45.72 cm. Calculate the minimum distance, in m, travelled by the wheel after the wheel has made 80 complete rotations. [Guna/ Use π = 22 7 ] A 57.48 B 114.95 C 5 747.66 D 11 495.2 Kertas Model SPM CONTOH

Matematik Tingkatan 5 Kertas Model SPM 142 9 Rajah 3 menunjukkan dua garis lurus yang selari, AB dan CD. Diagram 3 shows two parallel straight lines, AB and CD. A B C D x 125° 140° Rajah 3/ Diagram 3 Cari nilai x. Find the value of x. A 15° C 175° B 95° D 265° 10 David telah membeli saham Syarikat M pada tahun 2022 dengan RM9 500. Dia menerima dividen sebanyak RM120 pada tahun itu. Kemudian, dia menjualkan semua sahamnya dengan harga RM10 187.50. Hitung nilai pulangan pelaburan bagi David. David had bought the shares of Company M in the year 2022 with RM9500. He received a dividend of RM120 in that year. Then, he sold all his shares for RM10187.50. Calculate the return on investment for David. A 8.5% B 12.5% C 18.5% D 22.5% 11 Rajah 4 menunjukkan dua garis lurus yang selari, PQ dan RS dilukis pada suatu satah Cartes. Diagram 4 shows two parallel straight lines, PQ and RS are drawn on a Cartesian plane. R P QO S x y (1, 5) Rajah 4/ Diagram 4 Diberi pintasan-x dan pintasan-y bagi garis lurus PQ ialah 1. Tentukan persamaan bagi garis lurus RS. Given the x-intercept and the y-intercept of straight line PQ is 1. Determine the equation for straight line RS. A y = x + 6 B y = x – 6 C y = ‒x + 6 D y = ‒x – 6 12 Selesaikan/ Solve: x + 1 2 = x + 7 x – 1 A x = ‒3 atau/ or ‒5 B x = ‒3 atau/ or 5 C x = ‒5 atau/ or 3 D x = 3 atau/ or 5 13 Diberi P05P6 = Q25Q8 = X10. Jika P = 3Q, cari nilai bagi X. Given P05P6 = Q25Q8 = X10. If P = 3Q, find the value of X. A 1 C 247 B 3 D 681 14 Diberi ξ = {x : 1 N x N 50, x ialah suatu integer}, P = {x : x ialah satu nombor perdana} dan Q = {x : x ialah satu nombor berdigit dua yang mengandungi nombor 3}. Senaraikan unsur-unsur bagi P ∩ Q′. Given ξ = {x : 1 N x N 50, x is an integer}, P = {x : x is a prime number} and Q = {x : x is a two-digit number which contains number 3}. List the elements in P ∩ Qꞌ. A 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 41, 43, 47 B 2, 5, 7, 11, 13, 17, 19, 23, 29, 41, 47 C 2, 3, 5, 7, 11, 17, 19, 29, 41, 47 D 2, 5, 7, 11, 17, 19, 29, 41, 47 15 Jadual 1 menunjukkan polisi insurans yang dibeli oleh Halim. Table 1 shows the insurance policy which was bought by Halim. Jenis polisi Type of policy Perubatan dan kesihatan Medical and health Jumlah perlindungan Total coverage RM10 000 Deduktibel Deductible RM850 Tempoh perlindungan Coverage period Mac 2022 – Feb 2023 Jadual 1/ Table 1 Pada 27 September 2022, Halim bercadang menjalani pembedahan katarak tanpa pisau. Jumlah kos pembedahan yang dikenakan ialah RM13 000. Hitung jumlah yang perlu ditanggung oleh Halim. On 27 September 2022, Halim decided to undergo no-blade cataract surgery. The total chargeable cost of surgery is RM13000. Calculate the amount which needs to be borne by Halim. A RM3 000 C RM9 150 B RM3 850 D RM10 000 16 Jadual 2 menunjukkan nilai X, Y dan Z. Diberi bahawa X berubah secara langsung dengan kuasa dua Z dan berubah secara songsang dengan punca kuasa dua Y. Table 2 shows the values of X, Y and Z. Given that X varies directly with square of Z and varies inversely with square root of Y. X Y Z 24 25 2 45 36 p Jadual 2/ Table 2 Hitung nilai p. Calculate the value of p. A 1 C 3 B 2 D 4 CONTOH

Matematik Tingkatan 5 Kertas Model SPM 143 17 Rajah 5 menunjukkan ikatan kimia antara atom-atom di dalam suatu sebatian kimia. Diagram 5 shows the chemical bonds between atoms in a chemical compound. C2 C1 C3 C4 C5 C6 H2 H3 H1 H4 H6 H5 Rajah 5/ Diagram 5 Diberi C dan H masing-masing mewakili atom karbon dan atom hidrogen. Tentukan jumlah bilangan ikatan kimia yang terbentuk oleh semua atom karbon dalam sebatian kimia tersebut. Given C and H represent carbon atom and hydrogen atom respectively. Determine the total number of chemical bonds which are formed by all carbon atoms in the chemical compound. A 15 B 12 C 6 D 4 18 Sebiji kek yang berbentuk silinder mempunyai tinggi 12 cm. Permukaan kek itu selain daripada dasar kek perlu disapukan selapis krim. Jika isi padu kek itu ialah 4 158 cm3 , hitung jumlah luas permukaan, dalam cm2 , yang akan disapu dengan krim. A cylindrical cake has a height of 12 cm. The surface of the cake besides the cake base needs to be spread with a layer of cream. If the volume of the cake is 4 158 cm3 , calculate the total surface area, in cm2 , which will be spread with cream. [Guna/ Use π = 22 7 ] A 1 485 B 1 138.5 C 792 D 346.5 19 Antara berikut, yang manakah bukan suatu transformasi isometri? Which of the following is not an isometric transformation? A Pembesaran C Pantulan Enlargement Reflection B Translasi D Putaran Translation Rotation 20 Diberi satu set data mempunyai Σx = 180 dan Σx2 = 6 850. Tentukan bilangan data dalam set data tersebut jika varians bagi set data itu ialah 74. Given a set of data has Σx = 180 dan Σx2 = 6 850. Determine the number of data in the set of data if the variance of the set of data is 74. A 87 21 37 C 10 B 25 D 5 21 Diberi/ Given: 4 6 2 4 H = 1 0 0 1 Apakah matriks H? What is matrix H? A –0.5 1 1 – 3 2 C 0.5 4 4 3 4 B 0.5 1 1 3 2 D –0.5 4 4 – 3 4 22 Fatimah memiliki sebuah rumah kediaman di Petaling Jaya. Diberi anggaran sewa bulanan rumahnya ialah RM1 200 pada tahun itu. Bil cukai pintu Fatimah dikenakan kadar cukai sebanyak 4%. Hitung cukai pintu yang perlu dibayar oleh Fatimah pada tahun itu. Fatimah owns a residential house in Petaling Jaya. Given the estimated monthly rental is RM1 200 in that year. Fatimah’s property assessment bill is charged with a tax rate of 4%. Calculate the property assessment tax payable by Fatimah in that year. A RM576 B RM567 C RM288 D RM283.50 23 Rajah 6 menunjukkan sebuah graf laju-masa bagi sebuah motosikal. Diagram 6 shows a speed-time graph of a motorcycle. Laju/ Speed Masa 0 Time Rajah 6/ Diagram 6 Antara berikut, yang manakah betul tentang pecutan motosikal itu? Which of the following is correct about the acceleration of the motorcycle? A Tiada perubahan Remains unchanged B Berkurang secara malar Decreases constantly C Meningkat secara malar Increases constantly D Berkurang secara eksponen Decreases exponentially 24 Cari nilai x bagi 32(x + 1) = 97 – x . Find the value of x for 32(x + 1) = 97 – x. A 1 B 2 C 3 D 4 CONTOH

Matematik Tingkatan 5 Kertas Model SPM 144 25 Kenny diberi RM50 untuk membeli x batang pen dan y buah buku nota. Diberi harga sebatang pen dan sebuah buku nota masing-masing ialah RM1.20 dan RM2.80. Bilangan buku nota yang dibeli harus melebihi 4 kali bilangan pen. Antara berikut, persamaan ketaksamaan yang manakah memuaskan situasi ini? Kenny is given RM50 to buy x pens and y notebooks. Given the prices of a pen and a notebook are RM1.20 and RM2.80 respectively. The number of notebooks bought must be 4 times more than the number of pens. Which of the following linear inequalities satisfy this situation? A y M 4x dan/ and 3x – 7y N 125 B y M 4x dan/ and 3x + 7y N 125 C y < 4x dan/ and 3x + 7y M 125 D y < 4x dan/ and 3x – 7y M 125 26 Kelas 5 Petunia mempunyai 40 orang murid. Antara 27 orang murid suka matematik dan 21 orang murid suka sains. Cari bilangan murid yang suka matematik dan sains. Class 5 Petunia has 40 students. Among 27 students like mathematics and 21 students like science. Find the number of students who like mathematics and science. A 19 B 13 C 8 D 6 27 Sebatian glukosa mengandungi 25% atom karbon, 50% atom hidrogen dan 25% atom oksigen. Diberi satu molekul glukosa mempunyai 12 atom hidrogen, cari bilangan atom karbon dalam tiga molekul glukosa. Glucose compound contains 25% carbon atoms, 50% hydrogen atoms and 25% oxygen atoms. Given a glucose molecule has 12 hydrogen atoms, find the number of carbon atoms in three glucose molecules. A 6 B 12 C 18 D 24 28 Rajah 7 menunjukkan sebuah sisi empat selari PQRS. PTR dan QTS ialah dua pepenjuru. Diagram 7 shows a parallelogram PQRS. PTR and QTS are two diagonals. P T S Q R Rajah 7/ Diagram 7 Antara berikut, yang manakah betul? Which of the following is correct? A PS = PQ B PT = TS C ÐPTS = ÐSTR D ÐSQR = ÐPSQ 29 Sebuah beg mengandungi 5 biji bola hitam, 1 biji bola emas dan 9 biji bola putih. Dua biji bola dikeluarkan satu demi satu. Apakah kebarangkalian bahawa warna kedua-dua biji bola itu adalah sama? A bag contains 5 black balls, 1 gold ball and 9 white balls. Two balls are taken out one by one. What is the probability that the colour of both balls is the same? A 1 5 C 3 5 B 2 5 D 93 210 30 Rajah 8 menunjukkan satu corak teselasi. Diagram 8 shows a tessellation pattern. X Y Rajah 8/ Diagram 8 Diberi bentuk Y ialah imej bagi bentuk X. Apakah transformasi yang terlibat untuk menghasilkan bentuk Y daripada bentuk X? Given shape Y is an image of shape X. What is the transformation involved in producing shape Y from shape X? A Putaran sahaja Rotation only B Pantulan sahaja Reflection only C Putaran dan pantulan Rotation and reflection D Pantulan dan translasi Reflection and translation 31 Satu set koleh mempunyai enam saiz yang berkadaran. Jadual 3 menunjukkan isi padu cawan bagi tiga saiz. A set of mugs has six proportional sizes. Table 3 shows the volumes of the mugs of the three sizes. Saiz Size 2 3 4 Isi padu koleh (cm3) Volume of mug (cm3 ) 402 2 7 785 5 7 1 357 5 7 Jadual 3/ Table 3 Jika isi padu koleh mengikut persamaan 44n3 7 , tentukan isi padu, dalam cm3 , koleh dengan saiz yang paling besar. If the volume of mug follows equation 44n3 7 , determine the volume, in cm3 , of the mug with the biggest size. A 169 5 7 B 2 156 C 3 218 2 7 D 4 582 2 7 CONTOH

Matematik Tingkatan 5 Kertas Model SPM 145 32 Encik Hamid mempunyai pendapatan bulanan aktif sebanyak RM3 500. Dia mengutip sewa rumah sebanyak RM800 setiap bulan. Perbelanjaan bulanan tetap dan perbelanjaan bulanan tidak tetap Encik Hamid ialah RM2 200 dan RM1 385 masing-masing. Jika Encik Hamid bercadang membeli sebuah telefon pintar yang berharga RM3 246, cari berapakah bulan diperlukan oleh Encik Hamid menyimpan dan jumlah simpanan bulanan minimumnya untuk pembelian telefon pintar itu. Mr Hamid has an active monthly income of RM3500. He collects house rental of RM800 every month. Mr Hamid’s fixed monthly expenses and variable monthly expenses are RM2200 and RM1385 respectively. If Mr Hamid plans to buy a smartphone which costs RM3246, find how many months does he need to save and his minimum monthly saving for purchasing the smartphone. Tempoh simpanan Period of saving Simpanan bulanan minimum Minimum monthly saving A 4 RM649.20 B 4 RM715 C 5 RM649.20 D 5 RM715 33 Rajah 9 menunjukkan dua buah sisi empat, ABCD dan PQRS dilukis pada suatu satah Cartes. PQRS ialah imej bagi ABCD di bawah suatu pembesaran. Diagram 9 shows two quadrilaterals, ABCD and PQRS are drawn on a Cartesian plane. PQRS is an image of ABCD under an enlargement. x –6 –4 –2 O 2 4 6 8 y 4 6 8 –2 2 R S C D A B P Q Rajah 9/ Diagram 9 Tentukan pusat pembesaran dan faktor skala. Determine the centre of enlargement and the scale factor. Pusat pembesaran Centre of enlargement Faktor skala Scale factor A (‒1, 2) ‒2 B (‒1, 2) 2 C (‒2, 1) 2 D (‒2, 1) ‒2 34 Diberi/ Given: 3x – 6y = ‒18 5x – 3y = 5 Cari nilai x dan nilai y. Find the values of x and y. A x = 4, y = ‒5 C x = 5, y = 4 B x = ‒4, y = 5 D x = 4, y = 5 35 Pakcik Nazir mempunyai sebuah (6x + 4) m × 3x m dusun durian. Dia hendak memagar perimeter dusunnya yang seluas 336 m2 . Diberi harga segulung pagar dengan panjang 12.2 m ialah RM158, hitung jumlah yang perlu dibayar oleh Pakcik Nazir untuk membeli gulungan pagar yang cukup memagar dusunnya. Uncle Nazir has a (6x + 4) m × 3x m durian orchard. He wants to fence up the perimeter of his orchard with an area of 336 m2 . Given the price of a roll of fence with a length of 12.2 m is RM158, calculate the amount which needs to be paid by Uncle Nazir to buy enough rolls of fence to fence up his orchard. A RM1 106 C RM632 B RM1 036.48 D RM518.03 36 Antara garis berikut, yang manakah adalah selari dengan garis 2y = 5 – 6x? Which of the following lines is parallel with line 2y = 5 – 6x? A y = 3x + 7 C 9y = 3x + 4 B 9x = 8 – 3y D 5y = 2 – 3x 37 Rajah 10 menunjukkan jisim, dalam kg, bagi 14 ekor arnab. Diagram 10 shows the mass, in kg, of 14 rabbits. 1.4 1.6 1.8 1.6 1.8 1.75 2.1 1.24 2.2 1.65 1.32 1.36 1.22 2.0 Rajah 10/ Diagram 10 Cari beza antara julat dan julat antara kuartil. Find the difference between the range and the interquartile range. A 0.44 kg C 0.98 kg B 0.54 kg D 1.71 kg 38 Rajah 11 menunjukkan empat titik yang segaris, J, K, L dan M. Titik K ialah titik tengah bagi JM dan titik L ialah titik tengah bagi KM. Diagram 11 shows four collinear points, J, K, L and M. Point K is the midpoint of JM and point L is the midpoint of KM. J(2, 8) K M L(8, 5) Rajah 11/ Diagram 11 Hitung jarak, dalam unit, bagi JM. Calculate the distance, in units, of JM. A 5 C 80 B 20 D 288 CONTOH

Matematik Tingkatan 5 Kertas Model SPM 146 39 Susan hendak pergi ke Cheras dari Petaling Jaya untuk menghantar barang dalam masa yang paling singkat. Antara berikut, yang manakah bukan pemboleh ubah pemodelan matematik untuk menentukan perjalanan Susan? Susan wants to go to Cheras from Petaling Jaya to send a parcel in the shortest time. Which of the following is not a variable of the mathematical modelling for determining Susan’s route? A Masa Time B Jarak Distance C Keadaan cuaca Weather condition D Laju pemanduan Driving speed 40 Rajah 12 menunjukkan sebuah jam randik yang digunakan untuk merekod masa larian 500 m Siva. Diagram 12 shows the stopwatch used to record the time of Siva’s 500 m run. Rajah 12/ Diagram 12 Rekod Siva ialah 1 minit 29 saat. Hitung sin θ jika θ ialah sudut yang dicangkum oleh jam saat pada jam randik itu. Siva’s record is 1 minute 29 seconds. Calculate sin θ if θ is the angle moved by the second hand on the stopwatch. A 0.1405 C ‒0.1045 B 0.1045 D ‒0.1405 CONTOH

Matematik Tingkatan 5 Kertas Model SPM SIJIL PELAJARAN MALAYSIA MATEMATIK KERTAS 2 1449/2 2 1 2 jam Dua jam tiga puluh minit NO. KAD PENGENALAN : ANGKA GILIRAN : JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU ARAHAN: 1 Tulis nombor kad pengenalan dan angka giliran anda pada ruang yang disediakan. 2 Kertas soalan ini adalah dalam dwibahasa. 3 Soalan dalam bahasa Melayu mendahului soalan yang sepadan dalam bahasa Inggeris. 4 Kertas soalan ini mengandungi tiga bahagian: Bahagian A, Bahagian B, dan Bahagian C. 5 Calon perlu menjawab semua soalan dalam Bahagian A dan Bahagian B, dan mana-mana satu soalan daripada Bahagian C. 6 Calon dikehendaki membaca maklumat di halaman belakang kertas peperiksaan ini. Untuk Kegunaan Pemeriksa Nama Pemeriksa: Bahagian Soalan Markah Penuh Markah Diperoleh A 1 4 2 3 3 4 4 4 5 5 6 4 7 4 8 5 9 3 10 4 B 11 9 12 9 13 9 14 9 15 9 C 16 15 17 15 Jumlah 100 CONTOH

CONTOH

Matematik Tingkatan 5 Kertas Model SPM 149 Kertas soalan ini mengandungi tiga bahagian: Bahagian A, Bahagian B dan Bahagian C. Jawab semua soalan dalam Bahagian A dan Bahagian B dan mana-mana satu soalan daripada Bahagian C. Anda boleh menggunakan kalkulator saintifik yang tidak boleh diprogramkan. This question paper consists of three sections: Section A, Section B and Section C. Answer all questions in Section A and Section B and, any one question from Section C. You may use a non-programmable scientific calculator. Bahagian A/ Section A [40 markah/ marks] Jawab semua soalan dalam bahagian ini. Answer all questions in this section. 1 Gambar rajah Venn di ruang jawapan menunjukkan set J, set K dan set L dengan keadaan set semesta, ξ = J ∩ K ∩ L. The Venn diagrams in the answer space show set J, set K and set L such that the universal set, ξ = J ∩ K ∩ L. Lorek/ Shade: (a) (J′ ∩ K) ∪ L′ (b) J′ ∩ (K ∪ L′) [4 markah/ marks] Jawapan/ Answer: (a) J K L (b) J K L 2 Rajah 1 menunjukkan sebuah gabungan pepejal yang terdiri daripada sebuah kubus, PQRSTUVW dan sebuah kuboid, ABCDEFWV. Diagram 1 shows a composite solid which made up of a cube, PQRSTUVW and a cuboid, ABCDEFWV. Z U T A Q B W V E F C D S R P Rajah 1/ Diagram 1 Diberi nisbah tinggi kuboid kepada tinggi kubus ialah 1 : 2 dan nisbah panjang kuboid kepada panjang kubus ialah 2 : 3. Dengan menggunakan unit sebagai unit panjang, lakarkan unjuran ortogon bagi gabungan pepejal pada satah mencancang sebagaimana dilihat dari arah Z. Given the ratio of the height of the cuboid to the height of the cube is 1 : 2 and the ratio of the length of the cuboid to the length of cube is 2 : 3. By using unit as the unit of length, sketch the orthogonal projection of the composite solid on a vertical plane as viewed from the direction of Z . [3 markah/ marks] Jawapan/ Answer: Q/P D/A C/B R/S 1 unit/ unit 1 unit/ unit 2 unit/ units E/V/U F/W/T CONTOH

Matematik Tingkatan 5 Kertas Model SPM 150 3 Cari nilai x dan nilai y yang memuaskan persamaan yang berikut. Find the values of x and y which satisfy the following equations. 2x – 5y 4 = 4x – 3 , 4y – 2x = – 3y 7 [4 markah/ marks] Jawapan/ Answer: 2x – 5y 4 = 4x – 3 4y – 2x = – 3y 7 2x – 5y = 16x – 12 28y – 14x = ‒3y 14x = 12 – 5y……➀ 14x = 31y……➁ ➀ = ➁, 12 – 5y = 31y 12 = 36y y = 1 3 y = 1 3 ↷ ➁, 14x = 31( 1 3 ) x = 31 42 ⸫ x = 31 42 , y = 1 3 4 Rajah 2 menunjukkan sebuah pelan lantai bagi sebuah kolam yang berbentuk segi tiga bersudut tegak. Diagram 2 shows the floor plan of a right-angled triangular pond. (x + 1) m 2x m (x + 3) m Rajah 2/ Diagram 2 Tulis satu persamaan untuk menunjukkan hubungan antara sisi-sisi segi tiga. Seterusnya, hitung perimeter, dalam m, kolam itu. Write an equation to show the relationship between the sides of the triangle. Hence, calculate the perimeter, in m, of the pond. [4 markah/ marks] Jawapan/ Answer: (2x)2 + (x + 1)2 = (x + 3)2 (2x)2 + (x + 1)2 = (x + 3)2 4x2 + (x2 + 2x + 1) = x2 + 6x + 9 4x2 ‒ 4x – 8 = 0 x2 – x – 2 = 0 (x + 1)(x – 2) = 0 x = ‒1 atau/ or 2 ⸫ x > 0, maka/ hence x = 2 Perimeter: 2(2) + (2 + 1) + (2 + 3) = 4 + 3 + 5 = 12 m CONTOH

Matematik Tingkatan 5 Kertas Model SPM 151 5 (a) Nyatakan sama ada pernyataan berikut adalah benar atau palsu. State whether the following statement is true or false. Semua poligon mempunyai pepenjuru. All polygons have diagonals. [1 markah/ mark] (b) Tulis songsang bagi implikasi berikut: Write the inverse for the following implication: Jika n2 N 4, maka n = 0, 1 dan 2. If n2 N 4, then n = 0, 1 and 2. [1 markah/ mark] (c) Tulis Premis 1 untuk melengkapkan hujah berikut: Write Premise 1 to complete the following argument: Premis 1: .................................................................................................................................................................................................................. Premise 1: ................................................................................................................................................................................................................... Premis 2: Nombor 29 ialah satu nombor perdana. Premise 2: 29 is a prime number. Kesimpulan: Nombor 29 hanya boleh dibahagikan dengan diri sendiri dan nombor 1. Conclusion: 29 can only be divided by itself and 1. [1 markah/ mark] (d) Berdasarkan maklumat di bawah, buat satu kesimpulan secara deduksi bagi jumlah luas permukaan kubus dengan sisi 12 cm. Based on the information below, make one conclusion by deduction for the total surface area of a cube with sides of 12 cm. Jumlah luas permukaan bagi suatu kubus dengan sisi n cm ialah 6n2 . The total surface area of a cube with sides n cm is 6n2 . [2 markah/ marks] Jawapan/ Answer: (a) Palsu/ False (b) Jika n2 > 4, maka n ≠ 0, 1 dan 2. If n2 > 4, then n ≠ 0, 1 and 2. (c) Semua nombor perdana hanya boleh dibahagikan dengan diri sendiri dan nombor 1. All prime numbers can only be divided by itself and 1. (d) 6(122 ) = 864 cm2 Jumlah luas permukaan kubus ialah 864 cm2 . The total surface area of the cube is 864 cm2 . CONTOH

Matematik Tingkatan 5 Kertas Model SPM 152 6 Rajah 3 menunjukkan sebuah graf laju-masa bagi pergerakan sebuah kereta dalam masa 18 minit. Diagram 3 shows a speed-time graph of the movement of a car in 18 minutes. 0 4 10 14 18 50 60 Laju (km/j) Speed (km/h) Masa (minit) Time (minute) Rajah 3/ Diagram 3 (a) Hitung jumlah jarak, dalam m, yang dilalui oleh kereta itu. Calculate the total distance, in m, travelled by the car. [2 markah/ marks] (b) Cari pecutan, dalam km/j2 , bagi kereta itu pada 4 minit pertama. Find the acceleration, in km/h2 , of the car in the first 4 minutes. [2 markah/ marks] Jawapan/ Answer: (a) Jumlah jarak/ Total distance: [ 1 2 × 50 × 4 60 ] + [ 1 2 × (50 + 60) × 10 – 4 60 ] + [ 1 2 × 60 × (14 ‒ 10) + (18 ‒ 10) 60 ] = 1 2 3 + 5 1 2 + 6 = 13 1 6 km (b) Pecutan/ Acceleration: (50 ‒ 0) km/j/ km/h (4 ÷ 60) j/ h = 750 km/j2 / km/h2 7 Elaine telah meletakkan 5 biji bola kuning, 4 biji bola merah, 8 biji bola biru dan 3 biji bola hitam ke dalam sebuah beg. Tiga biji bola akan dipilih secara rawak daripada kotak itu tanpa pemulangan. Cari kebarangkalian bahawa Elaine had put 5 yellow balls, 4 red balls, 8 blue balls and 3 black balls into a bag. Three balls will be picked randomly from the box without returning. Find the probability that (a) ketiga-tiga biji bola itu ialah bola merah, all the balls are red balls, [1 markah/ mark] (b) dua biji bola mestilah bola hitam, two of the balls must be black balls, [2 markah/ marks] (c) ketiga-tiga biji bola bukan bola biru. all the three balls are not blue balls. [1 markah/ mark] Jawapan/ Answer: (a) P(M1 , M2 , M3 ) = 4 20 × 3 19 × 2 18 = 1 285 CONTOH

Matematik Tingkatan 5 Kertas Model SPM 153 (b) 3[P(H1 , H2 , K3 ) + P(H1 , H2 , M3 ) + P(H1 , H2 , B3 ) + P(H1 , H2 , H3 )] = 3[( 3 20 × 2 19 × 5 18 ) + ( 3 20 × 2 19 × 4 18 ) + ( 3 20 × 2 19 × 8 18 ) + ( 3 20 × 2 19 × 1 18 )] = 3[ 30 6 840 + 24 6 840 + 48 6 840 + 6 6 840] = 3( 108 6 840) = 9 190 (c) 1 – P(B1 , B2 , B3 ) = 1 – ( 8 20 × 7 19 × 6 18 ) = 1 ‒ 14 285 = 271 285 8 Jadual 1 menunjukkan maklumat pembelian pen oleh Aishah. Table 1 shows the information of pens purchased by Aishah. Jenis pen Type of pen Bilangan pen Number of pens Harga unit (RM) Unit price (RM) Pen biru Blue pen x 1.60 Pen merah Red pen y 1.20 Jadual 1/ Table 1 Aishah telah membeli 14 batang pen dengan RM20. Aishah had bought 14 pens with RM20. (a) Tulis dua persamaan linear dalam sebutan x dan y untuk mewakili maklumat di atas. Write two linear equations in terms of x and y to represent the above information. [2 markah/ marks] (b) Seterusnya, dengan menggunakan kaedah matriks, cari nilai x dan nilai y. Hence, by using the matrix method, find the values of x and y. [3 markah/ marks] Jawapan/ Answer: (a) x + y = 14, 1.6x + 1.2y = 20 16x + 12y = 200 4x + 3y = 50 (b) 1 4 1 3 x y = 14 50 x y = 1 3 – 4 3 –4 –1 1 14 50 = – 3(14) ‒ 50 ‒4(14) + 50 = – –8 –6 = 8 6 ⸫ x = 8 dan/ and y = 6 CONTOH

Matematik Tingkatan 5 Kertas Model SPM 154 9 Seketul batu dilontarkan ke udara. Tinggi, h dalam meter, kedudukan batu diberi oleh h = ‒2t2 + 3t + 2 dengan keadaan t ialah masa, dalam saat, selepas dilontarkan. Cari masa, dalam saat, batu itu mencecah tanah. A stone is thrown into the air. The height, h in metre, of the position of the stone is given by h = –2t2 + 3t + 2 where t is the time, in seconds, after thrown. Find the time, in seconds, when the stone hit the ground. [3 markah/ marks] Jawapan/ Answer: Apabila/ When h = 0, ‒2t2 + 3t + 2 = 0 2t2 – 3t – 2 = 0 (2t + 1)(t – 2) = 0 t = ‒ 1 2 atau/ or 2 ⸫ t > 0, maka t = 2 saat t > 0, therefore t = 2 seconds 10 Penyelesaian menggunakan kaedah matriks tidak dibenarkan untuk soalan ini. Solving using matrix method is not allowed in this question. Rajah 4 menunjukkan kotak pensel berbentuk segi empat tepat dengan perimeter 60 cm. Diagram 4 shows a rectangular pencil case with a perimeter of 60 cm. (4y + 1) cm 3x cm Rajah 4/ Diagram 4 Diberi nisbah panjang kepada lebar ialah 7 : 3. Hitung panjang, dalam cm, kotak pensel itu. Given the ratio of the length to the width is 7 : 3. Calculate the length, in cm, of the pencil case. [4 markah/ marks] Jawapan/ Answer: 4y + 1 3x = 7 3 12y + 3 = 21x 7x – 4y = 1………➀ 2[3x + (4y + 1)] = 60 3x + 4y + 1 = 30 3x + 4y = 29………➁ ➀ + ➁, 7x + 3x = 30 10x = 30 x = 3 x = 3 ↷ ➁, 3(3) + 4y = 29 9 + 4y = 29 4y = 20 y = 5 Panjang/ Length: 4(5) + 1 = 21 cm CONTOH

Matematik Tingkatan 5 Kertas Model SPM 155 Bahagian B/ Section B [45 markah/ marks] Jawab semua soalan dalam bahagian ini. Answer all questions in this section. 11 Diberi P(‒2, 3), Q(2, q) dan R(p, ‒3). PQ dan PR ialah dua garis lurus yang berserenjang. Given P(–2, 3), Q(2, q) and R(p, –3). PQ and PR are two perpendicular straight lines. (a) Tulis satu persamaan bagi garis PQ dalam sebutan q dan satu persamaan bagi garis PR dalam sebutan p. Write an equation for line PQ in terms of q and an equation for line PR in terms of p. [4 markah/ marks] (b) Jarak antara pintasan-y garis PQ dan pintasan-y garis PR ialah 5 unit. Cari nilai p dan nilai q yang mungkin. The distance between the y-intercept of line PQ and the y-intercept of line PR is 5 units. Find the possible values of p and q. [5 markah/ marks] Jawapan/ Answer: (a) Persamaan garis PQ/ Equation of line PQ: Persamaan garis PR/ Equation of line PR: q – 3 2 – (–2) = y – 3 x – (–2) –3 – 3 p – (–2) = y – 3 x – (–2) 4(y – 3) = (x + 2)(q – 3) (y – 3)(p + 2) = ‒6(x + 2) 4y – 12 = qx – 3x + 2q – 6 py + 2y – 3p – 6 = ‒6x – 12 4y = (q – 3)x + 2q + 6 (p + 2)y = ‒6x + 3p – 6 y = q – 3 4 x + 2(q + 3) 4 y = ‒ 6 p + 2x + 3(p – 2) p + 2 y = q – 3 4 x + q + 3 2 (b) (q – 3 4 )(‒ 6 p + 2) = –1 ( q + 3 2 ) ‒ [ 3(p – 2) p + 2 ] = 5 6q – 18 4p + 8 = 1 (p + 2)(q + 3) ‒ 6(p ‒ 2) 2p + 4 = 5 4p + 8 = 6q – 18 pq + 3p + 2q + 6 – 6p + 12 = 10p + 20 4p = 6q – 26 13p – pq – 2q = ‒2………➁ p = 3 2 q ‒ 13 2 ………➀ ➀ ↷ ➁, 13( 3 2 q ‒ 13 2 ) – ( 3 2 q ‒ 13 2 )q – 2q = ‒2 39 2 q ‒ 169 2 ‒ 3 2 q2 + 13 2 q – 2q = ‒2 39q – 169 – 3q2 + 13q – 4q = ‒4 3q2 – 48q + 165 = 0 q2 – 16q + 55 = 0 (q – 5)(q – 11) = 0 q = 5 atau/ or 11 Apabila/ When q = 5, p = 3 2 (5) ‒ 13 2 Apabila/ When q = 11, p = 3 2 (11) ‒ 13 2 = 1 = 10 ⸫ p = 1, q = 5 atau/ or p = 10, q = 11 CONTOH

Matematik Tingkatan 5 Kertas Model SPM 156 12 Jadual 2 menunjukkan pengkadaran premium di bawah Tarif Motor bagi polisi motor yang dikeluarkan di Semenanjung Malaysia dan Jadual 3 menunjukkan kadar NCD yang diberi berdasarkan tempoh insurans. Table 2 shows the premium rates under the Motor Tariff for motor policies issued in Peninsular Malaysia and Table 3 shows the NCD rate given based on the insurance period. Kapasiti enjin (cc) Engine capacity (cc) Polisi komprehensif (RM) Comprehensive policy (RM) Polisi pihak ketiga (RM) Third party policy (RM) N 1 400 273.80 120.60 N 1 650 305.50 135.00 N 2 200 339.10 151.20 N 3 050 372.60 167.40 Jadual 2/ Table 2 Tempoh insurans Insurance period Kereta persendirian Private car Kereta komersial Commercial car Tahun kedua Second year 25% 15% Tahun ketiga Third year 30% 20% Tahun keempat Fourth year 38.33% 25% Tahun kelima Fifth year 45% 25% Jadual 3/ Table 3 (a) Encik Hamid telah membeli sebuah kereta dengan enjin 2 200 cc untuk kegunaan sendiri pada dua tahun yang lalu. Jumlah yang diinsuranskan ialah RM75 000. Hitung premium kasar yang perlu dibayar oleh Encik Hamid untuk polisi komprehensif bagi tahun ini jika tiada tuntutan dibuat. Mr Hamid had bought a car with an engine of 2 200 cc for private use two years ago. The total sum insured is RM75 000. Calculate the gross premium payable by Mr Hamid under the comprehensive policy for this year if there is no claim. [4 markah/ marks] (b) Jadual 4 menunjukkan sebahagian kadar cukai jalan kereta persendirian di Semenanjung Malaysia. Table 4 shows a part of the road tax rates for private car in Peninsular Malaysia. Kapasiti enjin (cc) Engine capacity (cc) Kadar cukai jalan Road tax rate Kadar asas Basic rate Kadar progresif Progressive rate 1 401 – 1 600 RM90 – 1 601 – 1 800 RM200 RM0.40 1 801 – 2 000 RM280 RM0.50 2 001 – 2 500 RM380 RM1.00 2 501 – 3 000 RM880 RM2.50 Jadual 4/ Table 4 Hitung cukai jalan yang perlu dibayar oleh Encik Hamid. Calculate the road tax payable by Mr Hamid. [2 markah/ marks] (c) Pendapatan bulanan Encik Hamid ialah RM4 500 dan dia mendapat sebanyak RM1 200 daripada kerja sambilan. Encik Hamid mempunyai perbelanjaan bulanan tetap sebanyak RM1 850 dan perbelanjaan tidak tetap sebanyak RM2 145. Tentukan sama ada Encik Hamid dapat membaharui cukai jalannya pada bulan itu. Mr Hamid’s monthly income is RM4 500 and he gets RM1200 from his part-time job. Mr Hamid has fixed monthly expenses of RM1 850 and variable monthly expenses of RM2145. Determine whether Mr Hamid is able to renew his road tax. [3 markah/ marks] CONTOH

Matematik Tingkatan 5 Kertas Model SPM 157 Jawapan/ Answer: (a) Premium asas/ Basic premium: RM339.10 + ( 75 000 ‒ 1 000 1 000 × RM26) = RM339.10 + 74(RM26) = RM2 263.10 Premium kasar/ Gross premium: RM2 263.10 – (RM2 263.10 × 30%) = RM2 263.10 – RM678.93 = RM1 584.17 (b) Cukai jalan/ Road tax: RM380 + [(2 200 – 2 000) × RM1.00] = RM380 + RM200 = RM580 (c) Aliran tunai/ Cash flow: RM4 500 + RM1 200 – RM1 850 – RM2 145 = RM1 705 Jumlah yang perlu untuk membaharui cukai jalan: Amount needed to renew road tax: RM1 584.17 + RM580 = RM2 164.17 ⸫ Tidak, kerana RM2 164.17 > RM1 705. Encik Hamid masih perlu RM459.17 tambahan untuk membaharui cukai jalannya. No, because RM2164.17 > RM1705. Mr Hamid still needs an extra of RM459.17 to renew his road tax. 13 Rajah 5 menunjukkan sebuah sektor OPQ dengan pusat O dan berjejari 7 cm. Diagram 5 shows a sector OPQ with centre O and a radius of 7 cm. O T P S R Q 60° Rajah 5/ Diagram 5 (a) Buktikan bahawa OT = RQ. Prove that OT = RQ. [3 markah/ marks] (b) Hitung panjang, dalam cm, SR. Calculate the length, in cm, of SR. [2 markah/ marks] (c) Dengan menggunakan π = 22 7 , hitung luas, dalam cm2 , kawasan yang berlorek. By using � = 22 7 , calculate the area, in cm2 , of the shaded region. [4 markah/ marks] Jawapan/ Answer: (a) kos/ cos 60° = OT 7 OT = 7(0.5) = 3.5 cm Segi tiga OPR dan segi tiga OQT adalah kongruen kerana sifat RHS, maka OTP = ORQ. Triangles OPR and OQT are congruent due to the property of RHS, therefore OTP = ORQ. OT + TP = 7 cm 3.5 cm + TP = 7 cm TP = 3.5 cm = RQ ⸫ OT = RQ = 3.5 cm (Dibuktikan/ Proven) CONTOH

Matematik Tingkatan 5 Kertas Model SPM 158 (b) ÐTQR = 90° ‒ 60° = 30° tan 30° = SR 3.5 SR = 3.5 tan 30° = 2.021 cm (c) sin 60° = PR 7 PR = 7 sin 60° = 6.062 cm Luas kawasan berlorek/ Area of shaded region: Sektor/ Sector OPQ – Segi tiga/ Triangle OPR – Segi tiga/ Triangle QRS = [ 22 7 × 72 × 60° 360°] – ( 1 2 × 3.5 × 6.062) – ( 1 2 × 3.5 × 2.021) = 25.6667 ‒ 10.6085 – 3.5368 = 11.52 cm2 14 Rajah 6 menunjukkan tiga buah sisi empat yang kongruen dilukis pada satah Cartes. Diagram 6 shows three congruent quadrilaterals drawn on a Cartesian plane. x –2 O 2 4 6 8 y 2 4 –4 –2 H′ H′′ H Rajah 6/ Diagram 6 (a) (i) H″ ialah imej bagi H di bawah gabungan transformasi LK. Huraikan dengan penuh transformasi K dan transformasi L. H′′ is an image of H under a combined transformation LK. Describe completely transformation K and transformation L. [4 markah/ marks] (ii) Lukiskan imej H‴ di bawah gabungan transformasi KL di Rajah 7 pada ruang jawapan. Draw the image Hꞌꞌꞌ under the combined transformation KL on Diagram 7 in the answer space. [1 markah/ mark] (iii) Adakah gabungan transformasi LK mematuhi sifat kalis tukar tertib? Terangkan jawapan anda. Does the combined transformation LK satisfy the commutative law? Explain your answer. [2 markah/ marks] (b) Nyatakan dua ciri-ciri bagi bentuk kongruen. State two characteristics of congruent shapes. [2 markah/ marks] Jawapan/ Answer: (a) (i) Transformasi K: Pantulan pada garis x = 2. Transformation K: Reflection on line x = 2. Transformasi L: Translasi ( 2 –5) Transformation L: Translation ( 2 –5) CONTOH

Matematik Tingkatan 5 Kertas Model SPM 159 (ii) x –8 –4–6 –2 O 2 4 6 8 y 2 4 –4 –2 H′′′ H Rajah 7/ Diagram 7 (iii) Tidak, kerana imej di bawah gabungan transformasi LK dan imej di bawah gabungan transformasi KL adalah tidak sama. No, because the image under combined transformation LK and the image under combined transformation KL are not the same. (b) Bentuk kongruen mempunyai bentuk dan saiz yang sama. Congruent shapes have the same shape and size. 15 Rajah 8 menunjukkan graf bagi dua fungsi trigonometri. Diagram 8 shows the graph of two trigonometric functions. 0 90° 180° 270° 360° y 1 2 –2 –1 θ G H Rajah 8/ Diagram 8 (a) Nyatakan amplitud dan tempoh bagi fungsi G dan fungsi H. State the amplitude and period of function G and of function H. [4 markah/ marks] (b) Tulis persamaan bagi fungsi G dan fungsi H. Write the equation for function G and function H. [2 markah/ marks] (c) Tentukan nilai θ apabila fungsi G berada pada titik maksimum. Determine the values of θ when function G is at maximum point. [3 markah/ marks] Jawapan/ Answer: (a) Fungsi/ Function Amplitud/ Amplitude Tempoh/ Period G 1 180° H 2 360° (b) Fungsi G/ Function G: y = sin 2θ Fungsi H/ Function H: y = 2 kos/ cos θ CONTOH

Matematik Tingkatan 5 Kertas Model SPM 160 (c) sin 2θ = 1 sin 90° = 1 Jika 0° < θ < 360°, maka 0° < 2θ < 720°. If 0° < θ < 360°, then 0° < 2θ < 720°. sin 90° berada di sukuan I dan sukuan II. sin 90° is in quadrant I and quadrant II. 2θ = 90°, (360° + 90°) θ = 45°, 225° Bahagian C/ Section C [15 markah/ marks] Jawab satu soalan dalam bahagian ini. Answer one question in this section. 16 Rajah 9 menunjukkan jarak dan kedudukan beberapa perhentian bas. Diagram 9 shows the distances and locations of a few bus stops. P T S Q R 90 km 45 km 78 km 40km 60 km 83 km 45 km 52 km Rajah 9/ Diagram 9 Diberi P ialah sebuah terminal bas di mana semua bas bertolak dan kembali ke terminal tersebut. Given P is a bus terminal where all the buses depart and return to the terminal. (a) Semua bas beroperasi dengan bertolak ke semua perhentian bas tanpa menggunakan semula perjalanan yang telah dilaluinya. Senaraikan semua perjalanan bas yang mungkin. All the buses operate by departing to all the bus stops without reusing the travelled routes. [5 markah/ marks] (b) Tambang bas adalah berkadaran dengan jarak yang dilaluinya. Hitung beza antara tambang bas yang paling mahal dan tambang bas yang paling murah jika kadar tambang bas ialah RM1.20 setiap 20 km. The bus fares are proportional to the distance travelled. Calculate the difference between the most expensive bus fare and the cheapest bus fare if the bus fare rate is RM1.20 per 20 km. [5 markah/ marks] (c) Jadual 5 menunjukkan sebahagian kadar cukai jalan kereta di Semenanjung Malaysia. Table 5 shows a part of the road tax rates for car in Peninsular Malaysia. Enjin (cc) Engine (cc) Kadar Asas (RM) Base Rate (RM) Kadar Progresif (setiap cc) Progressive Rate (per cc) (RM) 1 000 dan ke bawah 1 000 and below 20 – 1 601 – 1 800 200 0.40 1 801 – 2 000 280 0.50 2 001 – 2 500 380 1.00 Jadual 5/ Table 5 CONTOH

Matematik Tingkatan 5 Kertas Model SPM 161 Irsyad mengambil keputusan untuk memandu keretanya berbanding menaiki bas. Dia memiliki sebuah kereta 1 822 cc. Berdasarkan Jadual 5, hitung jumlah cukai jalan yang perlu dibayar oleh Irsyad untuk keretanya. Irsyad decides to drive his own car compared that to ride the bus. He owns a 1 822 cc car. Based on Table 5, calculate the amount of road tax that Irsyad needs to pay for his car. [5 markah/ marks] Jawapan/ Answer: (a) P → Q → R → S → T → P, P → Q → S → R → T → P, P → R → Q → S → T → P, P → R → T → S → Q → P, P → T → S → R → Q → P, P → T → S → Q → R → P, P → T → R → S → Q → P (b) Jarak yang paling panjang/ The longest distance: P → R → Q → S → T → P atau/ or P → T → S → Q → R → P Jumlah jarak/ Total distance: 83 + 52 + 60 + 45 + 90 = 330 km Jarak yang paling pendek/ The shortest distance: P → Q → R → S → T → P atau/ or P → T → S → R → Q → P Jumlah jarak/ Total distance: 45 + 52 + 40 + 45 + 90 = 272 km Beza tambang bas/ Difference in bus fare: 330 ‒ 272 20 × RM1.20 = RM3.48 (c) Kadar asas/ Base rate: RM280 Jumlah kadar progresif/ Total progressive rate = (1 822 cc − 1 800 cc) × RM0.50 = 22 × RM0.50 = RM11.00 Jumlah cukai jalan/ Total road tax = Jumlah kadar progresif / Total progressive rate + Kadar asas/ Base rate = RM11.00 + RM280.00 = RM291 17 (a) Rafiq merupakan seorang pemilik kedai kopi, Secangkir Kopi. Beliau memiliki dua buah cawangan kedai kopi masing-masing di Senaling dan Seri Menanti. Beliau telah merekodkan jumlah hasil jualan kopi selama 13 hari di kedua-kedua buah kedainya. Plot batang-dan-daun dalam Rajah 10 menunjukkan keputusan di cawangan Senaling. Rafiq is the owner of the coffee shop, Secangkir Kopi. He owns two franchises of coffee shops each in Senaling and Seri Menanti. He had recorded the total sales of coffee sold in 13 days in both franchises. The stem-and-leaf plot in Diagram 10 shows the results at the Senaling branch. Batang Stem Daun Leaf 45 3 3 5 50 1 3 5 55 1 2 2 60 3 6 65 5 5 Kekunci/ Keys: 45|3 bermaksud/ means RM453 Rajah 10/ Diagram 10 (i) Hitung sisihan piawai bagi hasil jualan kopi di cawangan Senaling. Calculate the standard deviation of the sales of coffee in Senaling branch. [4 markah/ marks] (ii) Jika sisihan piawai bagi jualan kopi di cawangan Seri Menanti ialah 65.529, cawangan manakah yang mempunyai hasil jualan kopi yang lebih baik? Berikan justifikasi anda. If the standard deviation for the sales of coffee in Seri Menanti branch is 65.529, which branch has better coffee sales revenue? Give your justification. [2 markah/ marks] CONTOH

Matematik Tingkatan 5 Kertas Model SPM 162 (b) (i) Rafiq akan menyertai pesta makanan di Kemaman yang terletak kira-kira 330 km dari Seri Menanti. Dia memandu dengan laju 90 km/j dan singgah di empat perhentian rehat dan rawat sepanjang perjalanan. Jika dia singgah di setiap perhentian selama 5 minit, hitung masa, dalam jam, yang diambil oleh Rafiq untuk melengkapkan satu perjalanan. Rafiq will participate in a food festival at Kemaman which is located about 330 km from Seri Menanti. He drives at a speed of 90 km/h and stops at four rest and treat stops along the journey. If he stops at each stop for 5 minutes, calculate the time, in hour, taken by Rafiq to complete one route. [3 markah/ marks] (ii) Cari laju purata, dalam km/j, bagi satu perjalanan yang lengkap. Find the average speed, in km/h, for one whole journey. [2 markah/ marks] (c) Rafiq bercadang untuk memberi satu kek cawan percuma kepada setiap pelanggan yang datang membeli air kopi vanila dan air kopi karamel di gerainya nanti. Berikut ialah empat jenis kek cawan yang disediakan. Rafiq plans to give a free muffin to each customer who comes to buy vanilla coffee and caramel coffee at his stall soon. Below are the four types of muffins available. Oren/ Orange (O) Pandan/ Pandan (P) Coklat/ Chocolate (C) Strawberi/ Strawberry (S) Dengan menyenaraikan semua kesudahan yang mungkin bagi peristiwa itu, cari kebarangkalian bahawa By listing all the possible outcomes of the event, find the probability that (i) air kopi vanila dan kek cawan buah-buahan dipilih, vanilla coffee and fruits muffin are chosen, (ii) air kopi karamel atau bukan kek cawan coklat dipilih. caramel coffee or not a chocolate muffin are chosen. [4 markah/ marks] Jawapan/ Answer: (a) (i) Min/ mean: 453 + 453 + 455 + 501 + 503 + 505 + 551 + 552 + 552 + 603 + 606 + 655 + 655 13 = 7 044 13 = 541.846 Sisihan piawai/ Standard deviation: (453)2 + (453)2 + (455)2 + (501)2 + (503)2 + (505)2 + (551)2 + (552)2 + (552)2 + (603)2 + (606)2 + (655)2 + (655)2 13 − (541.846)2 = 3 878 382 13 − 293 597.0877 = 4 739.989 = 68.85 (ii) Cawangan di Seri Menanti mempunyai hasil jualan yang lebih baik kerana mempunyai nilai sisihan piawai lebih kecil. Branch in Seri Menanti has better total sales because its standard deviation is smaller. (b) (i) Masa yang diambil dalam perjalanan/ Time taken on the road: 330 km ÷ 90 km/j/ km/h = 3 2 3 jam/ hours Masa yang diambil di semua perhentian bas/ Time taken on all the bus stops: 5 minit/ minutes × 4 = 20 minit/ minutes = 1 3 jam/ hour Jumlah masa yang diambil/ Total time taken: 3 2 3 + 1 3 = 4 jam/ hours (ii) Laju purata/ Average speed: 330 km 4 j/ h = 82.5 km/j/ km/h (c) (i) (V, O), (V, S) 2 8 = 1 4 (ii) (V, O), (V, P), (V, S), (K, O), (K, P), (K, C), (K, S) = 7 8 CONTOH

CONTOH

J1 Bab 1 1.1 Ubahan Langsung Direct Variation 1 (a) (i) Jumlah gaji Amin adalah tiga kali ganda. Amin’s total salary is tripled. (ii) Jumlah gaji Amin bertambah sebanyak 30%. Amin’s total salary increases by 30%. (iii) Bilangan jam bekerja adalah separuh daripada bilangan jam bekerja asal. The number of working hours is halved. (b) (i) Amaun gula adalah separuh daripada amaun asalnya. The amount of sugar is halved. (ii) Amaun air adalah separuh daripada amaun asalnya. The amount of water is halved. (iii) Amaun gula berkurang sebanyak 40%. The amount of sugar decreases by 40%. (c) (i) Nilai rintangan bertambah sebanyak 20%. The value of resistance increases by 20%. (ii) Nilai rintangan berkurang sebanyak 10%. The value of resistance decreases by 10%. (iii) Suhu bertambah dua kali ganda. The temperature is doubled. (d) (i) Bilangan botol jem bertambah sebanyak 40%. The number of bottles of jam increases by 40%. (ii) Bilangan botol jem adalah separuh daripada bilangan asalnya. The number of bottles of jam is halved. (iii) Bilangan botol jem bertambah dua kali ganda. The number of bottles of jam is doubled. 2 (a) x y 1 10 20 30 40 32 4 O Maka, y berubah secara langsung dengan x. Hence, y varies directly as x. (b) x y 5 0.1 0.2 0.3 0.4 10 15 20 O Maka, y berubah secara langsung dengan x. Hence, y varies directly as x. Jawapan Langkah Penyelesaian Lengkap Kertas Model SPM CONTOH

Matematik Tingkatan 5 Jawapan J2 3 (a) x 3 6 9 12 y 1 2 3 6 y x 1 3 1 3 1 3 1 2 Tidak kerana y x bukan pemalar. No because y x is not a constant. (b) √x 0.1 0.2 0.3 0.4 y 0.3 1.2 2.7 4.8 x 0.01 0.04 0.09 0.16 y x 30 30 30 30 Ya kerana y x adalah pemalar. Yes because y x is a constant. ∴ y ∝ x (c) x3 1 8 27 64 y 1 8 27 64 x 1 2 3 4 y x 1 4 9 16 Tidak kerana y x bukan pemalar. No because y x is not a constant. (d) √x √2 2 √6 √8 y3 1 8 27 64 x 2 4 6 8 y 1 2 3 4 y x 1 2 1 2 1 2 1 2 Ya kerana y x adalah pemalar. Yes because y x is a constant. ∴ y ∝ x 4 (a) x3 x 1 2 3 4 y 100 200 300 400 x3 1 8 27 64 y x3 100 25 100 9 25 4 Tidak kerana y x3 bukan pemalar. No because y x3 is not a constant. (b) √x 3 x 1 8 27 64 √y 1 2 3 4 √x 3 1 2 3 4 y 1 4 9 16 y √x 3 1 2 3 4 Tidak kerana y √x 3 bukan pemalar. No because y √x 3 is not a constant. 5 (a) y = kx 4 = k(6) k = 4 6 = 2 3 ∴ y = 2 3 x (b) y = kx2 14 = k(2)2 k = 14 4 = 7 2 ∴ y = 7 2 x2 (c) y = kx3 1 = k(0.1)3 k = 1 (0.1)3 = 1 000 ∴ y = 1 000x3 (d) y = k√x 10 = k√25 k = 10 √25 = 2 ∴ y = 2√x (e) y = k √x 3 6 = k √27 3 k = 6 √27 3 k = 2 ∴ y = 2√x 3 6 (a) y1 xn 1 = y2 xn 2 4 12 = p 22 p = 4 1 × 4 = 16 CONTOH

Matematik Tingkatan 5 Jawapan J3 (b) y1 xn1 = y2 xn2 16p3 = 5433 p3 = 16 × 27 54 = 8 p = √8 3 = 2 (c) y ∝ √x → y ∝ x 12 28 16 12 = 35p12 p 12 = 35 × 4 28 = 5 p = (5)2 = 25 (d) y ∝ √x 3 → y ∝ x13 4 27 13 = 6p13 p 13 = 6 × 34 = 92 p = ( 92)3 = 7298 7 (a) a = kbc2 63 = k × 7 × 32 k = 63 63 = 1 ∴ a = bc2 (b) a = k√bc2 100 = k × √100 × 52 k = 100 250 = 25 ∴ a = 25 √bc2 8 (a) y ∝ x2z2 → y = kx2z2 83 = k(22 )(22) k = 16 y = 16 x2z2 q = 16 (42 )(32) = 24 (b) y ∝ x3z → y = kx3z 2 = k(23 )(0.5) k = 12 y = 12 x3z 32 = 12 (43 )(q) q = 32 32 = 1 9 (a) (i) A ∝ r2 (ii) AO r2 (b) (i) RM USD 4.1 4.1 4.1 4.1 Ya kerana RM USD adalah pemalar. Yes because RM USD is a constant. (ii) RM USD = 4.1 = y 5.5 y = 4.1 × 5.5 = 22.55 Ia boleh ditukar ke RM22.55. It can be exchanged for RM22.55. (c) (i) E ∝ mv2 (ii) E = kmv2 125 = k × 10 × 52 k = 125 10(25) = 125 250 = 12 ∴ E = 12 mv2 (d) (i) r ∝ s√t → r = ks√t 80 = k × 5 × √16 k = 80 5(4) = 4 r = 4s√t √t = r4s t = r2 16s2 (ii) s = 2 dan/ and r = 10, t = 102 16(2)2 = 100 64 = 25 16 / 1 9 16 CONTOH

Matematik Tingkatan 5 Jawapan J4 1.2 Ubahan Songsang Inverse Variation 1 (a) (i) t v 1 21 2 10.5 ×2 ÷2 Laju purata menjadi separuh daripada laju purata asalnya. The average speed is halved. (ii) t v 3 7 1.5 14 ÷2 ×2 Laju purata menjadi dua kali ganda daripada laju asalnya. The average speed is doubled. (b) (i) Bilangan hadiah akan bertambah sebanyak 0.5 kali ganda. The number of gifts will increase by 0.5 times. (ii) Bilangan hadiah akan berkurang sebanyak dua kali ganda. The number of gifts will decrease by two times. 2 x 1 2 4 5 8 d 20 10 5 4 2.5 1 x 1 0.5 0.25 0.2 0.125 x2 1 4 16 25 64 Graf 1 Graf d melawan 1 x adalah satu garis lurus yang bermula dari asalan. The graph d against 1 x is a straight line starts from the origin. ∴ d ∝ 1 x d O 10 5 15 20 0.2 0.4 0.6 0.8 1.0 1 1 x Graf 2 Graf d melawan x2 ialah hiperbola. Graph d against x2 is a hyperbola. ∴ d ∝ 1 x2 x2 d O 30 40 50 60 10 5 15 20 10 20 2 3 x 1.25 2 2.5 4 5 8 10 y 8 5 4 2.5 2 1.25 1 1 x 0.8 0.5 0.4 0.25 0.2 0.125 0.1 y 0.1 1 5 3 7 2 6 4 8 0.2 0.3 0.4 0.5 0.6 0.7 0.8 1 x O Graf y melawan 1 x menunjukkan satu garis lurus yang bermula dari asalan. Maka, y berubah secara songsang dengan x. The graph of y against 1 x shows a straight line that starts from the origin. Hence, y varies inversely as x. CONTOH

Matematik Tingkatan 5 Jawapan J5 4 (a) x 3 6 12 24 y 8 4 2 1 xy 24 24 24 24 Ya kerana xy adalah satu pemalar. Yes because xy is a constant. (b) x 2 5 10 50 y 50 10 5 2 xy 100 50 50 100 Tidak kerana xy bukan satu pemalar. No because xy is not a constant. (c) √x 1 2 4 8 y 64 32 16 8 x 1 4 16 64 xy 64 128 256 512 Tidak kerana xy bukan satu pemalar. No because xy is not a constant. (d) x 24 30 40 60 y3 125 64 27 8 y 5 4 3 2 xy 120 120 120 120 Ya kerana xy adalah satu pemalar. Yes because xy is a constant. (e) √x 2 3 4 5 y3 8 27 64 125 x 4 9 16 25 y 2 3 4 5 xy 8 27 64 125 Tidak kerana xy bukan satu pemalar. No because xy is not a constant. 5 (a) x3 x 1 2 3 4 y 4 3 2 1 x3 1 8 27 64 x3y 4 24 54 64 Tidak kerana x3 y bukan satu pemalar. No because x3 y is not a constant. (b) √x x 1 64 4 096 √y 3 8 4 2 √x 1 8 64 y 512 64 8 √xy 512 512 512 Ya kerana √xy adalah satu pemalar. Yes because √xy is a constant. 6 (a) xy = k 2 × 9 = k k = 18 ∴ y = 18 x (b) x3 y = k (2)3 (3) = k k = 24 ∴ y = 24 x3 (c) √xy = k √0.01 × 1 = k k = 0.1 ∴ y = 1 10√x (d) x2 y = k ( 1 2 ) 2 × 8 = k k = 2 ∴ y = 2 x2 (e) √xy 3 = k √0.001 3 × 6 = k k = 3 5 ∴ y = 3 5√x 3 7 (a) xn y 1 1 = xn y 2 2 (0.1)(36) = (p)(18) p = 0.2 (b) y ∝ 1 √x → y ∝ 1 x 1 2 (11 2)(4) = (p 1 2)(9) p 1 2 = 4 9 p = (4 9) 2 = 16 81 (c) (32 )(p) = (62 )(54) p = 216 (d) (13 )(32) = (p3 )(4) p3 = 8 p = √8 3 = 2 (e) √27 3 × 4 = √125 3 (p) 12 = 5p p = 12 5 CONTOH

Matematik Tingkatan 5 Jawapan J6 8 (a) (i) f ∝ 1 λ (ii) 1 λ f O (b) I ∝ 1 R IR = k (5)(20) = k k = 100 ⇒ I = 100 R Apabila/ When R = 25 Ω, I = 100 25 = 4 A (c) t ∝ 1 √s t(√s ) = k ( 30 60 )(√4) = k k = 1 ⇒ t = 1 √s Apabila s = 16 km j−1 / When s = 16 km h−1, t = 1 √16 = 1 4 jam/ hour = 15 minit/ minutes (d) (i) p ∝ 1 √q 3 → p = k √q 3 12 = k √64 3 k = 48 ⇒ p = 48 √q 3 (ii) x = 48 √8 3 = 24 1.3 Ubahan Bergabung Combined Variation 1 (a) v ∝ h k2 (b) a ∝ bc √d 2 (a) x 1 2 4 8 y 3 6 9 12 z 1 2 16 3 16 x2 1 4 16 64 yz x2 3 3 3 3 Ya kerana yz x2 adalah satu pemalar. Yes because yz x2 is a constant. (b) x 2 3 4 y 846 128 27 z 72 32 18 x3 8 27 64 yx3 z 94 108 96 Tidak kerana yx3 z bukan satu pemalar. No because yx3 z is not a constant. (c) w 10 15 30 40 x 1 4 9 16 y 3 6 9 12 z 10 15 30 40 √x 1 2 3 4 yw z√x 3 3 3 3 Ya kerana yw z√x adalah satu pemalar. Yes because yw z√x is a constant. 3 (a) yx z2 = k 5 × 2 102 = k k = 1 10 ∴ y = z2 10x (b) yx3 √z = k 100 × 53 √25 = k k = 2 500 ∴ y = 2 500√z x3 CONTOH

Matematik Tingkatan 5 Jawapan J7 4 (a) a = kb2 c3 2 = k(2)2 13 4k = 2 k = 1 2 (b) a = kb √c 6 = k(36) √9 36k = 18 k = 1 2 5 (a) (i) w ∝ 1 lh (ii) 1 lh w O (b) (i) V ∝ T P (ii) V = kT P 50 = k × 290 100 k = 500 29 V = 500T 29P = 500(348) 29(60) = 100 ∴ V = 100 m3 (c) r ∝ s2 t√u r = ks2 t√u 1 = k(32 ) 2 × √4 k = 4 9 (d) x = kyz w 1 = k(3)(3) 6 k = 2 3 ∴ x = 2yz 3w a = b2 2c3 4 = (8)2 2p3 2p3 = 16 p = √8 3 = 2 a = b 2√c 2 = 8 2√p √p = 2 p = 22 = 4 r = 4s2 9t√u ∴ t = 4s2 9r√u Praktis Kendiri Kertas 1 1 C x 1 2 3 4 y 2 8 18 32 x2 1 4 9 16 x3 1 8 27 64 y x 2 4 6 8 x2y 2 32 162 512 y x2 2 2 2 2 y x3 2 1 2 3 1 2 y ∝ x2 kerana y x2 adalah satu pemalar. y ∝ x2 because y x2 is a constant. 2 B 3 D 4 A 5 A Kerana y x adalah satu pemalar dalam A sahaja. Because y x is a constant only in A. 6 A Pemalar ubahan ialah 5 × 4 = 20. Hanya rs dalam A tidak sama dengan 20. Constant of variation is 5 × 4 = 20. Only rs in A is not equal to 20. 7 B Kerana a c b adalah satu pemalar hanya dalam B. Because a c b is a constant only in B. 8 C Graf y ∝ x3 adalah satu garisan lurus melalui asalan dalam graf y melawan x3 . The graph of y ∝ x3 is a straight line passing through origin in the graph of y against x3 . 9 C Oleh sebab d = vt, maka d ∝ t. As d = vt, then d ∝ t. d1 t n 1 = d2 t n 2 50 2 = 75 t2 t2 = 75 × 2 50 = 3 jam/ hours CONTOH

Matematik Tingkatan 5 Jawapan J8 10 C Katakan t ialah masa yang diambil dan p ialah bilangan halaman. Let t is the time taken and p is the number of pages. t ∝ p t = k p k = 30 20 t = 30 20 p = 30 20( 80) = 60 minit/ minutes / 1 jam/ hour Kertas 2 Bahagian A/ Section A 1 (a) Tidak kerana graf tersebut bukan satu garis lurus melalui asalan. No because the graph is not a straight line passing through the origin. (b) Graf y melawan z adalah satu garis lurus melalui asalan ketika c = −5 seperti yang ditunjukkan di bawah. Oleh itu, c = −5. The graph of y against z is a straight line passing through the origin when c = −5 as shown below. Hence, c = −5. y z O Bahagian B/ Section B 2 (a) a ∝ b2√c 3 d (b) 2a = 10 a = 5 a = kb2 √c 3 d 5 = k(10)2 √27 3 12 k = 60 100 × 3 = 1 5 (c) a1 d1 b1 √c1 23 = a2 d2 b2 √c2 23 a1 d1 b1 √c1 23 = a2 (8d1 ) (2b1 )2√8c1 3 = a2 d1 b1 √c1 23 ∴ a1 = a2 Oleh itu, a tidak berubah. Hence, a remains the same. Bab 2 2.1 Matriks Matrices 1 (a) 75 83 97 69 79 88 atau/ or 83 79 75 69 97 88 (b) 57 61 102 91 atau/ or 57 102 61 91 2 Bilangan baris Number of rows (a) 1 (b) 3 (c) 2 (d) 2 (e) 3 (f) 3 Bilangan lajur Number of columns 2 1 2 3 2 3 Peringkat matriks Order of matrix 1 × 2 3 × 1 2 × 2 2 × 3 3 × 2 3 × 3 3 (a) a12 = 2 a21 = 6 a13 = –3 a23 = 7 (b) a13 = 8 a23 = 7 a21 = –3 a32 = 9 (c) a12 = 11 a22 = 5 a21 = 3 a32 = –1 4 (a) x = 5 y = 7 (b) 2x + 1 = 15 2y – 3 = –9 2x = 14 2y = –6 x = 7 y = –3 5 (a) x + 5 = 10 8 = y – 9 x = 5 y = 17 (b) x – 5 = 1 x = 6 y = 0 (c) y = 4 2x + 2 = y 2x + 2 = 4 2x = 2 x = 1 (d) 14 – x = 2x – 1 2x + x = 14 + 1 3x = 15 x = 5 16 – 2y = 4 + y 2y + y = 16 – 4 3y = 12 y = 4 CONTOH

Matematik Tingkatan 5 Jawapan J9 2.2 Operasi Asas Matriks Basic Operations on Matrices 1 (a) [2 6 8] + [–3 5 3] = [2 + (–3) 6 + 5 8 + 3] = [–1 11 11] (b) –7 3 + 1 2 = –7 + 1 3 + 2 = –6 5 (c) 5 –16 –12 4 + 2 10 7 3 = 5 + 2 –16 + 10 –12 + 7 4 + 3 = 7 –6 –5 7 2 (a) [4 1] – [3 8] = [4 – 3 1 – 8] = [1 –7] (b) –11 8 16 –2 – –3 1 4 –3 = –11 – (–3) 8 – 1 16 – 4 –2 – (–3) = –8 7 12 1 3 (a) 4 2 8 –13 + –5 6 7 7 – –8 5 10 –7 = 4 + (–5) – (–8) 2 + 6 – 5 8 + 7 – 10 –13 + 7 – (–7) = 7 3 5 1 (b) 6 15 –5 8 – 1 –13 11 6 + 9 7 –3 –4 = 6 – 1 + 9 15 – (–13) + 7 –5 – 11 + (–3) 8 – 6 + (–4) = 14 35 –19 –2 4 (a) 6 – 3 = y 3x – y – 4 = 8 3x – y = 12 3x – 3 = 12 3x = 15 x = 5 y = 3 (b) x + 2 = 5 x = 3 2x + y = –8 2(3) + y = –8 y = –14 (c) –x – x = –2 –2x = –2 x = 1 y – 2 – (–2) = 3 y = 3 5 (a) A = [–7 3 –6] – [9 0 –6] = [–7 – 9 3 – 0 –6 – (–6)] = [–16 3 0] (b) A = 7 –2 4 –3 – 3 0 6 9 = 4 –2 –2 –12 (c) A = 5 4 0 1 – 4 8 9 2 = 1 –4 –9 –1 6 (a) 3 5 2 1 (b) –8 12 2 4 (c) 1.8 0.3 0.9 –2.4 –1.2 0.6 7 (a) 1 2 8 –4 – 1 4 –4 16 = 1 2 (8) – 1 4 (–4) 1 2 (–4) – 1 4 (16) = 5 –6 (b) 0.1 10 0 0 10 + 0.2 0 10 10 0 = 0.1(10) + 0.2(0) 0.1(0) + 0.2(10) 0.1(0) + 0.2(10) 0.1(10) + 0.2(0) = 1 2 2 1 (c) 3 1 3 –2 –1 2 3 – 2 3 –4 3 = 3 1 3 – 2 3(–2) – 3 3(–1) – (–4) 3 2 3 – 3 = –1 –9 1 –1 8 (a) M = 9 –3 4 1 – 3 –6 7 –8 = 9 – 3 –3 – (–6) 4 – 7 1 – (–8) = 6 3 –3 9 (b) 3M = 3 12 –15 6 – –6 6 0 9 = 3 – (–6) 12 – 6 –15 – 0 6 – 9 = 9 6 –15 –3 M = 3 2 –5 –1 CONTOH

Matematik Tingkatan 5 Jawapan J10 9 (a) [6 2] 1 –1 5 –8 –7 4 = [6(1) + 2(–1) 6(5) + 2(–8) 6(–7) + 2(4)] = [4 14 –34] (b) –2 –1 [3 10] = –2(3) –1(3) –2(10) –1(10) = –6 –3 –20 –10 (c) [4 6] 3 0 1 9 = [4(3) + 6(0) 4(1) + 6(9)] = [12 58] (d) –3 –1 4 6 7 2 = (–3)(7) + 4(2) (–1)(7) + 6(2) = –13 5 (e) 1 6 –1 3 5 1 4 8 = 1(5) + (–1)(1) 6(5) + 3(1) 1(4) + (–1)(8) 6(4) + 3(8) = 4 33 –4 48 10 (a) [6(k) + h(0) 6(3) + h(–2)] = [–6 10] [6k 18 – 2h] = [–6 10] 6k = –6 k = –1 18 – 2h = 10 2h = 8 h = 4 (b) a(–2) 3(–2) ab 3b = –4 –6 10 15 –2a –6 ab 3b = –4 –6 10 15 –2a = –4 a = 2 3b = 15 b = 5 (c) 5(m) + m(3) (–2)(m) + n(3) = 16 8 8m –2m + 3n = 16 8 8m = 16 m = 2 –2m + 3n = 8 –2(2) + 3n = 8 3n = 12 n = 4 (d) x(–3) + 0(–5) 2(–3) + w(–5) = 24 w –3x –6 – 5w = 24 w –3x = 24 x = –8 –6 – 5w = w 6w = –6 w = –1 (e) e(4) + 3(–1) 2(4) + f(–1) = 13 10 4e – 3 8 – f = 13 10 4e – 3 = 13 4e = 16 e = 4 8 – f = 10 f = –2 11 (a) 7 –8 2 –3 1 0 0 1 = 7 –8 2 –3 (b) 1 0 0 1 1 6 –3 5 + 7 –8 2 –3 1 0 0 1 = 1 6 –3 5 + 7 –8 2 –3 = 8 –2 –1 2 12 (a) B–1 = 1 0(9) – 4(–1) 9 –4 1 0 = 1 4 9 –4 1 0 = –1 9 4 0 1 4 (b) C–1 = 1 2 1 2 – 4 1 2 1 2 – 1 2 –4 2 = 1 –1 1 2 – 1 2 –4 2 = – 1 2 1 2 4 –2 (c) D–1 = 1 1(4) – 1(3) 4 –3 –1 1 = 1 1 4 –3 –1 1 = 4 –3 –1 1 13 (a) 4 2 –3 –1 m n = 5 2 (b) 2 –4 –1 3 x y = 5 –11 CONTOH

Matematik Tingkatan 5 Jawapan J11 14 (a) 2 –3 –1 4 x y = –7 13 x y = 1 2(4) – (–1)(–3) 4 3 1 2 –7 13 = 1 5 4(–7) + 1(13) 3(–7) + 2(13) = 1 5 –28 + 13 –21 + 26 = 1 5 –15 5 = –3 1 ∴ x = –3, y = 1 (b) 1 3 2 –4 x y = 17 1 x y = 1 1(–4) – 2(3) –4 –3 –2 1 17 1 = – 1 10 (–4)(17) + (–2)(1) (–3)(17) + 1(1) = – 1 10 –68 – 2 –51 + 1 = – 1 10 –70 –50 = 7 5 ∴ x = 7, y = 5 (c) 3 4 –2 5 x y = 24 –14 x y = 1 3(5) – (–2)(4) 5 –4 2 3 24 –14 = 1 23 5(24) + 2(–14) (–4)(24) + 3(–14) = 1 23 120 – 28 –96 – 42 = 1 23 92 –138 = 4 –6 ∴ x = 4, y = –6 15 (a) Katakan x sebagai harga sebatang pen dan y sebagai harga sebiji pemadam. Let x be the price of a pen and y be the price of an eraser. 10x + 5y = 26 y – x = 0.7 10 –1 5 1 x y = 26 0.7 x y = 1 10(1) – 5(–1) 1 1 –5 10 26 0.7 = 1 15 1(26) + (–5)(0.7) 1(26) + 10(0.7) = 1 15 22.5 33 = 1.5 2.2 ∴ Harga bagi sebatang pen ialah RM1.50 dan harga bagi sebiji pemadam ialah RM2.20. The price of a pen is RM1.50 and the price of an eraser is RM2.20. (b) Katakan x sebagai bilangan basikal dan y sebagai bilangan basikal roda tiga. Let x be the number of bicycles and y be the number of tricycles. 2x + 2y = 70 2x + 3y = 85 2 2 2 3 x y = 70 85 x y = 1 2(3) – 2(2) 3 –2 –2 2 70 85 = 1 2 3(70) + (–2)(85) (–2)(70) + 2(85) = 1 2 40 30 = 20 15 ∴ Bilangan basikal ialah 20 buah dan bilangan basikal roda tiga ialah 15 buah. The number of bicycles is 20 and the number of tricycles is 15. (c) Katakan p sebagai harga sebuku roti dan q sebagai harga sepotong kek. Let p be the price of a loaf of bread and q be the price of a slice of cake. 6p + 4q = 28 2p – q = 3.5 6 2 4 –1 p q = 28 3.5 p q = 1 6(–1) – 4(2) –1 –2 –4 6 28 3.5 = 1 –14 (–1)(28) + (–4)(3.5) (–2)(28) + 6(3.5) = 1 –14 –42 –35 = 3 2.5 CONTOH

Matematik Tingkatan 5 Jawapan J12 ∴ Harga bagi sebuku roti ialah RM3 dan harga bagi sepotong kek ialah RM2.50. The price of a loaf of bread is RM3 and the price of a slice of cake is RM2.50. Praktis Kendiri Kertas 1 1 A B dan D ialah matriks lajur manakala C ialah matriks identiti. B and D are column matrices while C is an identity matrix. 2 B A: [ 3 0 6 1 ][ 1 0 0 1 ] = [ 1 0 0 1 ] [ 3 0 6 1 ] = [ 3 0 6 1 ] B: [ 0 1 ] + [ 1 0 ] = [ 0 + 1 1 + 0] = [ 1 1 ] C: [ 2 –1 3 5 ][ 1 0 0 1 ] = [ 1 0 0 1 ][ 2 –1 3 5 ] = [ 2 –1 3 5 ] D: 3[ 7 4 –8 5 ] = [21 12 –24 15 ] 3 C Penambahan dan penolakan matriks tidak mematuhi Hukum Kalis Sekutuan. Addition and subtraction of matrices do not obey the Associative Law. 4 D [ 3 4 –1 –7 5 0 2 –6 1 ] – [ 2 –4 1 1 –5 4 0 6 3 ] = [ 3 − 2 4 − (−4) −1 − 1 −7 − 1 5 − (−5) 0 − 4 2 − 0 −6 − 6 1 − 3 ] = [ 1 8 –2 –8 10 –4 2 –12 –2 ] 5 C [ 1 x + 3] = [ y 2x – y] 1 = y x – y = 4 – 1 = 3 6 C [ 2a –b c ] = [ 4 –1 3 ] 2a = 4 –b = –1 a = 2 b = 1 c = 3 2a – 2b c = 2(2) − 2(1) 3 = 2 3 x + 3 = 2x – y x + 3 = 2x – 1 x = 4 7 C A2 = AA = [a – 1 3 b –1][a – 1 3 b –1] = [a2 − 2a + 1 + 3b 3a − 6 ab – 2b 3b + 1 ] 8 B Suatu matrik tidak mempunyai matriks songsang jika penentu = 0. A matrix does not have inverse matrix if its determinant = 0. 1(2) – (–5)x = 0 2 + 5x = 0 5x = –2 x = – 2 5 9 A Suatu matrik tidak mempunyai matriks songsang jika penentu = 0. A matrix does not have inverse matrix if its determinant = 0. A: 4(1) – 2(2) = 0 B: 3(1) – (–2)(1) = 5 C: 1(1) – 2(0.6) = –0.2 D: (–7)(–1) – (–2)(–3) = 1 10 B [2a c –b cd ] –1 = 1 2a(cd) – (–b)(c) [ cd –c b 2a] = 1 2acd + bc [ cd –c b 2a] Kertas 2 Bahagian A/ Section A 1 (a) Katakan x ialah biskut mentega dan y ialah biskut coklat. Let x be the butter cookies and y be the chocolate cookies. 3x + 5y = 138 4x + 4y = 136 (b) 3 4 5 4 x y = 138 136 x y = 1 3(4) – 5(4) 4 –4 –5 3 138 136 = 1 –8 4(138) + (–5)(136) (–4)(138) + 3(136) = 1 –8 –128 –144 = 16 18 \ Harga bagi biskut mentega ialah RM16 dan harga bagi biskut coklat ialah RM18. The price of butter cookies is RM16 and the price of chocolate cookies is RM18. CONTOH

Matematik Tingkatan 5 Jawapan J13 2 (a) Katakan x ialah markah bagi jawapan yang betul dan y ialah markah bagi jawapan yang salah. Let x be the mark for a correct answer and y be the mark for a wrong answer. 8x + 2y = 70 7x + 3y = 55 (b) 8 7 2 3 x y = 70 55 x y = 1 8(3) – 2(7) 3 –7 –2 8 70 55 = 1 10 3(70) + (–2)(55) (–7)(70) + 8(55) = 1 10 100 –50 = 10 –5 \ Mereka mendapat 10 markah bagi jawapan yang betul dan kehilangan 5 markah bagi jawapan yang salah. They got 10 marks for a correct answer and lost 5 marks for a wrong answer. Bahagian B/ Section B 3 (a) 2 4 3 5 k –4 –10 8 = 0 4 4 0 2k + 3(–4) 4k + 5(–4) 2(–10) + 3(8) 4(–10) + 5(8) = 0 4 4 0 2k – 12 4k – 20 4 0 = 0 4 4 0 2k – 12 = 0 2k = 12 k = 6 (b) G = 2 4 3 5 , G–1 = 1 2(5) – 3(4) 5 –4 –3 2 = – 1 2 5 –4 –3 2 = 2 5 2 –1 3 2 – (c) 2 4 3 5 x y = 4 10 x y = 2 4 3 5 –1 = 2 5 2 –1 3 2 – 4 10 = – 5 2 (4) + 3 2 (10) 2(4) + (–1)(10) = 5 –2 \ x = 5, y = –2 Bab 3 3.1 Risiko dan Perlindungan Insurans Risk and Insurance Coverage 1 2 (a) Aina Care for You Sdn. Bhd Kematian, penyakit kritikal dan hilang upaya (keilatan) Death, critical illness and loss of ability (b) Raju Syarikat SJ SJ Company Kematian, kehilangan bagasi, pasport dan wang, perbelanjaan perubatan dan lain-lain Death, loss of luggage, passport and money, medical expenses and others 3 (a) Kebarangkalian warga emas dimasukkan ke hospital adalah lebih tinggi The probability of senior citizen to be hospitalised is higher (b) Peratusan yang ditanggung oleh Amy ialah 10% Percentage borne by Amy is 10% Jumlah yang ditanggung oleh Amy: Amount borne by Amy: RM10 000 × 10 100 = RM1 000 (c) Peratusan yang ditanggung oleh agensi ialah 75% Percentage borne by agency is 75% Jumlah yang ditanggung oleh agensi: Amount borne by agency: RM100 000 × 75 100 = RM75 000 4 (a) Kos bagi/ Cost per RM1 000 = RM26.02 RM300 000 RM1 000 × RM26.02 = RM7 806 (b) Kos bagi/ Cost per RM1 000 = RM2.12 Premium tahunan/ Annual premium: RM150 000 RM1 000 × RM2.12 = RM318 Premium bulanan/ Monthly premium: RM318 ÷ 12 = RM26.50 CONTOH

Matematik Tingkatan 5 Jawapan J14 (c) Kos bagi/ Cost per RM1 000 = RM2.12 Premium tahunan/ Annual premium: RM200 000 RM1 000 × RM2.12 = RM424 (d) Kos bagi/ Cost per RM1 000 = RM2.27 Premium tahunan/ Annual premium: x RM1 000 × RM2.27 = RM1 135 x = RM500 000 \ Nilai muka insurans adalah RM500 000. The face value of the insurance is RM500 000. (e) Premium tahunan/ Annual premium: RM700 000 RM1 000 × x = RM7 028 x = RM10.04 Memandangkan kos RM1 000 insurans ialah RM10.04, maka Puan Ling berusia 25 tahun. Since the cost per RM1 000 of the insurance is RM10.04, then she is 25 years old. 5 (a) Bagi polisi komprehensif: For comprehensive policy: Kadar untuk RM1 000 pertama The rate for the first RM1 000 RM266.50 Nilai baki jumlah yang diinsuranskan The value of the remaining of sum insured RM20.30 × ( RM80 000 – RM1 000 RM1 000 ) = RM1 603.70 Premium asas Basic premium RM266.50 + RM1 603.70 = RM1 870.20 NCD 35% RM1 870.20 × 0.35 = RM654.57 Premium kasar Gross premium RM1 870.20 – RM654.57 = RM1 215.63 Bagi polisi pihak ketiga, kebakaran dan kecurian/ For third party, fire and theft policy: Premium asas Basic premium RM1 870.20 × 0.75 = RM1 402.65 NCD 35% RM1 402.65 × 0.35 = RM490.93 Premium kasar Gross premium RM1 402.65 – RM490.93 = RM911.72 Bagi polisi pihak ketiga/ For third party policy: Premium asas Basic premium RM93.60 NCD 35% RM93.60 × 0.35 = RM32.76 Premium kasar Gross premium RM93.60 – RM32.76 = RM60.84 (b) Bagi polisi komprehensif: For comprehensive policy: Kadar untuk RM1 000 pertama The rate for the first RM1 000 RM273.80 Nilai baki jumlah yang diinsuranskan The value of the remaining of sum insured RM26 × ( RM40 000 – RM1 000 RM1 000 ) = RM1 014 Premium asas Basic premium RM273.80 + RM1 014 = RM1 287.80 NCD 15% RM1 287.80 × 0.15 = RM193.17 Premium kasar Gross premium RM1 287.80 – RM193.17 = RM1 094.63 Bagi polisi pihak ketiga, kebakaran dan kecurian/ For third party, fire and theft policy: Premium asas Basic premium RM1 287.80 × 0.75 = RM965.85 NCD 15% RM965.85 × 0.15 = RM144.88 Premium kasar Gross premium RM965.85 – RM144.88 = RM820.97 Bagi polisi pihak ketiga/ For third party policy: Premium asas Basic premium RM120.60 NCD 15% RM120.60 × 0.15 = RM18.09 Premium kasar Gross premium RM120.60 – RM18.09 = RM102.51 6 (a) Bulan Month Kerugian Loss (RM) Boleh buat tuntutan? Can make a claim? Bayaran pampasan Amount of compensation (RM) April April 80 Tidak No Tiada None Jun June 290 Boleh Yes 290 – 250 = 40 September September 500 Boleh Yes 500 – 250 = 250 (b) Kereta Car Kerugian Loss (RM) Boleh dituntut Claimable Bayaran pampasan Amount of compensation (RM) A 1 500 Boleh Yes 1 500 – 400 = 1 100 B 250 Tidak No Tiada None C 800 Boleh Yes 800 – 400 = 400 CONTOH

Matematik Tingkatan 5 Jawapan J15 (c) Tahun Year Kos perubatan (RM) Medical cost (RM) Boleh dituntut Claimable Bayaran pampasan (RM) Amount of compensation (RM) Jumlah yang ditanggung (RM) Amount borne (RM) Pertama First 9 000 Tidak No Tiada None 9 000 Kedua Second 45 000 Boleh Yes 45 000 – 25 000 = 20 000 25 000 \ Hasrul perlu menanggung RM9 000 + RM25 000 = RM34 000. Hasrul needs to bear RM9 000 + RM25 000 = RM34 000. 7 (a) Jumlah insurans yang harus dibeli: Amount of required insurance: 50 100 × RM100 000 = RM50 000 Bayaran pampasan Amount of compensation = Jumlah kerugian – Deduktibel = Amount of loss – Deductible = RM50 000 – RM0 = RM50 000 (b) Nisbah ko-insurans/ Co-insurance ratio = Jumlah yang ditanggung oleh syarikat Amount borne by company : Jumlah yang ditanggung oleh pemegang polisi Amount borne by policyholder = 60 000 : 15 000 = 60 000 15 000 : 15 000 15 000 = 4 : 1 \ Peratusan ko-insurans ialah 80/20. The co-insurance percentage is 80/20. (c) Peratusan yang ditanggung oleh Encik Adlishah: Percentage borne by Mr Adlishah: RM3 750 RM25 000 × 100% = 15% \ Peratusan ko-insurans ialah 85/15. Percentage of co-insurance is 85/15. (d) Kos perubatan selepas deduktibel: Medical cost after deductible: RM17 500 – RM500 = RM17 000 Kos yang ditanggung oleh Emma: The cost borne by Emma: 20 100 × RM17 000 + RM500 = RM3 900 Praktis Kendiri Kertas 1 1 C 2 D 3 A 4 D 5 A 6 C Premium tahunan = RM191 Annual premium x RM1 000 × RM1.91 = RM191 x = RM191 RM1.91 × RM1 000 = RM100 000 7 C Jumlah pampasan yang diterima oleh Ainul: The amount of compensation received by Ainul: (RM20 000 – RM10 000) + (RM25 000 – RM10 000) = RM25 000 8 B Jumlah deduktibel/ Amount of deductible: RM650 – RM500 = RM150 9 C Jumlah pampasan/ Amount of compensation: (RM60 000 – RM1 000) × 75 100 = RM44 250 10 A Jumlah insurans yang diperlukan: Amount of required insurance: 80 100 × RM100 000 = RM80 000 RM30 000 < RM80 000, maka jumlah pampasan: RM30 000 < RM80 000, hence the amount of compensation: RM30 000 RM80 000 × RM10 000 = RM3 750 Kertas 2 Bahagian A/ Section A 1 Farhan Zaqwan Kos bagi RM1 000 Cost per RM1 000 = RM2.80 Kos bagi RM1 000 Cost per RM1 000 = RM2.64 Premium tahunan: Annual premium: RM100 000 RM1 000 × RM2.80 = RM280 Premium tahunan: Annual premium: RM80 000 RM1 000 × RM2.64 = RM211.20 \ Perbezaan premium tahunan mereka: The difference of their annual premium: RM280 – RM211.20 = RM68.80 CONTOH

Matematik Tingkatan 5 Jawapan J16 2 Syarikat Company Jumlah tuntutan (RM) Amount of claim (RM) A (10 000 – 800) × 80 100 = 7 360 B (10 000 – 500) × 75 100 = 7 125 \ Polisi yang ditawarkan oleh syarikat A boleh memberikan pampasan yang lebih banyak kepada pemegang polisi. Policy offered by company A can provide more compensation to the policyholder. 3 Bot Boat Kerugian (RM) Loss (RM) Boleh buat tuntutan? Can make a claim? Bayaran pampasan (RM) Amount of compensation (RM) A 5 500 Boleh Yes 5 500 – 5 000 = 500 B 6 000 Boleh Yes 6 000 – 5 000 = 1 000 Jumlah tuntutan/ Amount of claim: RM500 + RM1 000 = RM1 500 Bahagian B/ Section B 4 (a) Sebuah kereta dengan kelengkapan keselamatan yang lebih tinggi mempunyai risiko yang lebih rendah untuk mengalami kemalangan. A car with higher safety fitting has lower risk of getting into an accident. (b) RM75 000 – RM60 000 = RM15 000 \ Kos yang perlu ditanggung ialah RM15 000. The cost that need to be borne is RM15 000. (c) (i) Kehilangan keupayaan kekal Permanent disability (ii) Oktober/ October 2021 – Mei/ May 2022 8 bulan/ months RM210 000 � 8 = RM26 250 \ Pampasan maksimum sebulan = RM26 250 Maximum compensation per month = RM26 250 (ii) RM10 500 × 8 = RM84 000 Bab 4 4.1 Percukaian Taxation 1 (a) ✓ (d) ✓ (e) ✓ (f) ✓ 2 (a) pendapatan/ income (b) Lembaga Hasil Dalam Negeri (LHDN) Inland Revenue Board (IRB) (c) Akta Cukai Pendapatan 1967 (Akta 53) Seksyen 112(1) Income Tax Act 1967 (Act 53) Section 112(1) (d) Akta Cukai Pendapatan 1967 (Akta 53) Seksyen 114(1) Income Tax Act 1967 (Act 53) Section 114(1) (e) cukai jalan/ road tax (f) menggunakan kenderaan continually using a vehicle (g) pihak berkuasa tempatan/ local authority (h) Waran tahanan/ Detention warrant (i) pemilik tanah pertanian/ agricultural land (j) wilayah bebas cukai/ tax-free territories (k) Nilai ambang/ Threshold (l) barangan bercukai/ taxable goods 3 (a) Jumlah pendapatan tahunan: Total annual income: RM7 200 × 12 = RM86 400 Pendapatan bercukai/ Chargeable income: RM86 400 – RM9 250 = RM77 150 Cukai pendapatan/ Income tax: RM4 400 + (RM77 150 – RM70 000) × 21% = RM4 400 + RM1 501.50 = RM5 901.50 (b) Pendapatan bercukai/ Chargeable income = RM120 284 Cukai pendapatan/ Income tax: RM10 700 + (RM120 284 – RM100 000) × 24% = RM10 700 + RM4 868.16 = RM15 568.16 (c) Pendapatan bercukai/ Chargeable income: RM81 100 −RM9 000 −RM3 000 −RM2 000 = RM67 100 Cukai pendapatan/ Income tax: RM1 800 + (RM67 100 − RM50 000) × 13% = RM1 800 + RM2 223 = RM4 023 (d) Cukai pendapatan/ Income tax: RM1 800 + (RM53 000 − RM50 000) × 13% − RM280 = RM1 800 + RM390 – RM280 = RM1 910 CONTOH