Firasat SPM - Matematik

PENERBIT ILMU BAKTI SDN. BHD.

MATEMATIK Kertas 1 KM1–1 Kertas peperiksaan ini mengandungi 8 halaman bercetak. JANGAN BUKA KERTAS PEPERIKSAAN INI SEHINGGA DIBERITAHU 1. Kertas soalan ini mengandungi 40 soalan. This question paper consists of 40 questions. 2. Jawab semua soalan. Answer all questions. 3. Setiap soalan diikuti oleh empat pilihan jawapan, iaitu A, B, C dan D. Bagi setiap soalan, pilih satu jawapan sahaja. Each question is followed by four options, A, B, C and D. For each question, choose one answer only. 4. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. The diagrams in the questions are not drawn to scale unless stated. 5. Anda dibenarkan menggunakan kalkulator saintifi k yang tidak boleh diprogram. You may use a non-programmable scientifi c calculator. 1 1 2 jam Satu jam tiga puluh minit 1449/1 SIJIL PELAJARAN MALAYSIA 2023 Kertas Model SPM 2 3 4 5 6 7 8 1 SULIT Firasat SPM Maths_Set1 K1_vim3p(2023).indd 1 19/6/2023 9:06:50 AM PENERBIT ILMU BAKTI SDN. BHD.

KM1–2 SULIT 1449/1 1449/1 Firasat SPM 2023: Matematik SULIT 1 2.422 + 4.26 × 8.5 betul kepada tiga angka bererti ialah 2.422 + 4.26 × 8.5 correct to three significant figures is A 42.0 B 42.1 C 42.06 D 42.07 2 Antara berikut, yang manakah merupakan satu pernyataan? Which of the following is a statement? A 22 + 32 B 32 = –9 C w – 1  3 D 2b – 4 = 6 3 Hitung beza nilai digit 4 dalam nombor 334029 dan 41256 . Calculate the difference between the values of digit 4 in 334029 and 41256 . A 540 B 828 C 900 D 1 200 4 43215 – 12345 = A 39210 B 60610 C 3 03210 D 3 08710 5 Rajah 1 menunjukkan sebuah pentagon sekata ABCDE dan sebuah segi tiga sama kaki BSE. DSC ialah garis lurus. Diagram 1 shows a regular pentagon ABCDE and an isosceles triangle BSE. DSC is a straight line. B A E D C y 47° x S Rajah 1/Diagram 1 Cari nilai x + y. Find the value of x + y. A 108° B 111° C 115° D 133° 6 Diberi bahawa 1 mn = 3–4. Cari nilai m dan n. Given that 1 mn = 3–4. Find the value of m and n. A m = 3, n = 4 B m = 3, n = –4 C m = –3, n = 4 D m = –3, n = –4 7 Diberi bahawa 3g + 1 = 2g – 2 2h . Ungkapkan g dalam sebutan h dalam bentuk termudah. Given that 3g + 1 = 2g – 2 2h . Express g in terms of h in the simplest form. A h + 1 1 – 3h B 2h + 2 2 – 6h C 3 2 – 6h D 3 2(1 – 3h) 8 Rajah 2 menunjukkan sepasang garis selari yang direntas oleh garis lurus PQRS. Diagram 2 shows a pair of parallel line crossed by straight line PQRS. 4x – 30° 2x 2y + 16° S P Q R Rajah 2/Diagram 2 Cari nilai 2y – x. Find the value of 2y – x. A 12° B 23° C 43° D 59° 9 Rajah 3 menunjukkan sebuah bulatan yang berpusat di O. JKL ialah tangen kepada bulatan KNM di titik K. Diberi bahawa MN = MK dan ∠JKN = 52°. Diagram 3 shows a circle centred at O. JKL is a tangent to the circle KNM at point K. Given that MN = MK and ∠ JKN = 52°. • 52° N J M K L x O Rajah 3/Diagram 3 Hitung nilai x. Calculate the value of x. A 14° B 22° C 26° D 38° Firasat SPM Maths_Set1 K1_vim3p(2023).indd 2 19/6/2023 9:06:50 AM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 1449/1 KM1–3 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT 10 Rajah 4 menunjukkan dua buah bulatan yang berpusat di X dan Y. KLMN ialah tangen sepunya kepada bulatan-bulatan tersebut. PLQR ialah tangen kepada bulatan berpusat di Y. Diagram 4 shows two circles with centres X and Y. KLMN is a common tangent to the circles. PLQR is a tangent to the circle at centre Y. P b K L X a M N Q R 145° Y Rajah 4/Diagram 4 Cari nilai a + b. Find the value of a + b. A 105° C 120° B 115° D 125° 11 Rajah 5 menunjukkan sebuah bulatan. Diagram 5 shows a circle. x 22° z y • 52° Rajah 5/Diagram 5 Cari nilai x + y + z. Find the value of x + y + z. A 116° C 172° B 150° D 180° 12 Sebuah kotak mengandungi p keping kad merah, q keping kad biru dan 12 keping kad kuning. Sekeping kad dipilih secara rawak daripada kotak itu. Kebarangkalian memilih sekeping kad merah ialah 1 6 dan kad kuning ialah 2 7 . Cari nilai p dan nilai q. A box contains p red cards, q blue cards and 12 yellow cards. A card is chosen at random from the box. The probability of choosing a red card is 1 6 and a yellow card is 2 7 . Find the values of p and q. p q A 5 12 B 5 23 C 7 12 D 7 23 13 Permudahkan 2m + n 8m – 6n ÷ 4m2 – n2 4mn – 2n2 Simplify 2m + n 8m – 6n ÷ 4m2 – n2 4mn – 2n2 A n2 m – n B 2 4m – 3n C n 4m – 3n D (2m + n) 2 4m – 3n 14 (2p – 2)2 – (p + 1)(p + 4) = A 3p2 – 13p B 3p2 – 9p C 3p2 + p + 8 D 3p2 – 5p – 8 15 Antara graf berikut, yang manakah mewakili y = 7 – x3 ? Which of the following graphs represents y = 7 – x3 ? A C B D 16 Jadual 1 menunjukkan saiz baju yang ditempah oleh Kelab STEM. Table 1 shows the size of the shirts ordered by the STEM Club. Saiz/Size XS S M L XL Bilangan murid Number of students 5 6 3 2 4 Jadual 1/Table 1 Cari median bagi saiz baju tersebut. Find the median of the size of the shirt. A XS B S C L D XL y x 0 –7 y x 0 7 y x 0 –7 y x 0 7 Firasat SPM Maths_Set1 K1_vim3p(2023).indd 3 19/6/2023 9:06:51 AM PENERBIT ILMU BAKTI SDN. BHD.

KM1–4 SULIT 1449/1 1449/1 Firasat SPM 2023: Matematik SULIT 17 Rajah 6 menunjukkan dua buah segi tiga bersudut tegak. Diagram 6 shows two right-angled triangles. Rajah 6 /Diagram 6 Cari nilai bagi tan x. Find the value of tan x. A 5 12 C 12 13 B 5 13 D 13 12 18 Rajah 7 menunjukkan graf laju-masa bagi suatu objek dalam tempoh 18 saat. Diagram 7 shows the speed-time graph of an object for a period of 18 seconds. Laju/Speed (m s–1) Masa/Time (s) 20 5 0 4 k 18 Rajah 7/Diagram 7 Jumlah jarak yang dilalui oleh objek itu ialah 0.25 km. Hitung nilai k. The total distance travelled by the object is 0.25 km. Calculate the value of k. A 9 B 10 C 12 D 13 19 Rajah 8 menunjukkan suatu graf mudah. Diagram 8 shows a simple graph. Rajah 8/Diagram 8 Nyatakan bilangan bucu, tepi dan darjah bagi graf itu. State the number of vertices, edges and degrees of the graph. Bucu Vertices Tepi Edges Darjah Degrees A 4 7 14 B 4 8 16 C 6 7 14 D 6 8 16 20 Rajah 9 menunjukkan gambar rajah Venn dengan set semester, ξ, set G dan set H. Diagram 9 shows a Venn diagram with universal set, ξ, set G and set H. G H ξ Rajah 9/ Diagram 9 Antara set berikut, yang manakah mewakili rantau berlorek dalam gambar rajah Venn di atas? Which of the following sets represents the shaded region in the Venn diagram above? A G' ∩ H B G ∪ H' C G' ∪ H D G' ∪ H' 21 Rajah 10 menunjukkan set X, Y dan Z. Diagram 10 shows sets X, Y and Z. Z Y X Rajah 10/Diagram 10 Diberi set semesta, ξ = X ∪ Y ∪ Z, n(X) = 36, n(Y) = 26, n(ξ) = 53 dan n[Y' ∩ (X ∩ Z')] = 6. Cari n[Y' ∩ Z]. It is given that the universal set, ξ = X ∪ Y ∪ Z, n(X) = 36, n(Y) = 26, n(ξ) = 53 and n[Y' ∩ (X ∩ Z’)] = 6. Find n[Y' ∩ Z]. A 17 C 23 B 21 D 26 22 Rajah 11 menunjukkan graf f(x) = 3 – 2x² dan g(x) = 3 – kx². Diagram 11 shows the graph f(x) = 3 – 2x² and g(x) = 3 – kx². y x 3 f(x) g(x) Rajah 11/Diagram 11 3 cm 4 cm 13 cm x Firasat SPM Maths_Set1 K1_vim3p(2023).indd 4 19/6/2023 9:06:51 AM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 1449/1 KM1–5 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT Antara berikut, yang manakah mewakili nilai k yang mungkin? Which of the following represents the possible value of k? A –4 B –1 C 1 D 4 23 Rajah 12 merupakan akas bagi suatu implikasi. Diagram 12 is the converse of an implication. Jika x = 2, maka 2x + 7 = 11 If x = 2, then 2x + 7 = 11 Rajah 12/Diagram 12 Antara berikut, yang manakah merupakan songsangan bagi implikasi tersebut? Which of the following is the inverse of the implication? A Jika 2x + 7 = 11, maka x = 2 If 2x + 7 = 11, then x = 2 B Jika x ≠ 2, maka 2x + 7 ≠ 11 If x ≠ 2, then 2x + 7 ≠ 11 C Jika 2x + 7 ≠ 11, maka x ≠ 2 If 2x + 7 ≠ 11, then x ≠ 2 D Jika x = 2, maka 2x + 7 ≠ 11 If x = 2, then 2x + 7 ≠ 11 24 Diberi satu jujukan nombor 4, 8, 12, … Given that a sequence of numbers 4, 8, 12, … 4 = 2 + 1(2) 8 = 2 + 2(3) 14 = 2 + 3(4) . . . Cari sebutan ke-n. Find the nth term. A 2 + (n + 1) B 2 + (n – 1) C 2 + n (n + 1) D 2 + (n – 1) + n 25 Diberi bahawa K berubah secara langsung dengan L kuasa dua dan secara songsang dengan punca kuasa dua M. Cari hubungan antara K, L dan M. Given that K varies directly with the square of L and inversely with the square root of M. Find the relationship between K, L and M. A K ∝ M L2 B K ∝ L M2 C K ∝ M2 L D K ∝ L2 M 26 Rajah 13 menunjukkan graf y = kos x. Diagram 13 shows the graph of y = cos x. 3 2 1 0 –1 90° 180° 270° 360° y x y = p kos x + 1 y = p cos x + 1 y = kos x y = cos x Rajah 13/Diagram 13 Cari nilai p. Find the value of p. A – 1 2 C 2 B 1 2 D 3 27 Dalila menyimpan sebanyak RM3 200 di sebuah bank pada awal tahun dengan kadar faedah 3.5% setahun dan faedah dikompaun 6 bulan sekali. Dia merancang untuk menyimpan jumlah wang tersebut di bank yang sama dalam masa 6 tahun. Berapakah jumlah wang simpanan Dalila pada akhir tahun keenam? Dalila saves RM3 200 in a bank at the beginning of the year with an interest rate of 3.5% per annum and interest compounded once every 6 months. She plans to save this amount of money in the same bank for 6 years. What is the amount of money at the end of the sixth year? A RM3 312.00 C RM3 940.61 B RM3 431.33 D RM3 945.36 28 Jadual 2 menunjukkan markah yang diperoleh murid-murid di SMK Seri Damai dalam mata pelajaran Geografi. Table 2 shows the marks obtained by the students at SMK Seri Damai in the Geography subject. Markah Mark 50-59 60-69 70-79 80-89 90-99 90-100 Kekerapan longgokan Cumulative frequency 16 42 77 109 121 130 Jadual 2/Table 2 Diberi bahawa murid yang mendapat markah 70 dan ke atas akan diberikan sijil kecemerlangan. Cari bilangan murid yang akan mendapat sijil tersebut. Given that the students who scored 70 and above will be awarded a certificate of excellence. Find the number of students who will receive the certificate. A 21 B 53 C 77 D 88 Firasat SPM Maths_Set1 K1_vim3p(2023).indd 5 19/6/2023 9:06:52 AM PENERBIT ILMU BAKTI SDN. BHD.

KM1–6 SULIT 1449/1 1449/1 Firasat SPM 2023: Matematik SULIT 29 Rajah 14 menunjukkan heksagon ABCDEF dan heksagon PQRSTU yang dilukis pada grid segi empat sama. Heksagon ABCDEF ialah imej bagi heksagon PQRSTU di bawah suatu pembesaran. Diagram 14 shows hexagon ABCDEF and hexagon PQRSTU drawn on square grids. Hexagon ABCDEF is the image of hexagon PQRSTU under an enlargement. A E F D C B P Q R T S U Rajah 14/Diagram 14 Nyatakan faktor skala pembesaran itu. State the scale factor of the enlargement. A 1 3 B – 1 3 C 3 D –3 30 Rajah 15 menunjukkan suatu gabungan kubus. Diagram 15 shows a composite cube. Y Rajah 15/Diagram 15 Antara berikut, yang manakah menunjukkan unjuran ortogon dari pandangan sisi Y? Which of the following shows the orthogonal projection from side elevation Y? A B C D 31 Jadual 3 menunjukkan bilangan pengunjung ke Pesta Kebudayaan di Putrajaya. Table 3 shows the number of visitors to the Cultural Festival in Putrajaya. Hari Day Jumaat Friday Sabtu Saturday Ahad Sunday Bilangan pengunjung Number of visitors 132546 3 29410 3100314 Jadual 3/Table 3 Hitung purata bilangan pengunjung yang datang sepanjang tiga hari pesta itu diadakan. Calculate the average number of visitors who came throughout the three days of the festival. A 2 085 B 2 489 C 2 823 D 2 895 32 Rajah 16 menunjukkan aktiviti ikatan yang dilakukan oleh seorang ahli pengakap. Dua ikatan tali diikat pada L dan K. Kemudian, dua batang kayu, JK dan LM diikat dalam keadaan selari. Diagram 16 shows knot-tying activity by a scout. Two knots are tied at L and K. Then, two woods, JK and LM are tied in parallel. • M • K L J O x y Rajah 16/Diagram 16 Diberi kecerunan kayu JK ialah 5 3 . Jarak mengufuk J dan tinggi tegak M dari O masing-masing ialah 2.4 m dan 8.5 m. Cari jarak MK, dalam m. Given that the gradient of the wood JK is 5 3 . The horizontal distance of J and the vertical height of M from O are 2.4 m and 8.5 m respectively. Find the distance of MK, in m. A 4.0 B 4.5 C 4.8 D 5.1 Firasat SPM Maths_Set1 K1_vim3p(2023).indd 6 19/6/2023 9:06:52 AM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 1449/1 KM1–7 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT 33 Rajah 17 menunjukkan dua buah segi tiga yang kongruen. Diagram 17 shows two triangles that are congruent. x cm 84 cm 68 cm Rajah 17/Diagram 17 Satu riben dengan panjang 4.5 m digunakan untuk mengelilingi perimeter kedua-dua segi tiga. Cari nilai x. A ribbon of length 4.5 m is used to surround the perimeter of both triangles. Find the value of x. A 28 B 73 C 76 D 82 34 Rajah 18 menunjukkan lima titik pada satah Cartesan. Diagram 18 shows five points on a Cartesian plane. A B C D V 2 O –2 –4 –8 –6 –4 –2 2 4 6 8 10 y x Rajah 18/Diagram 18 Antara titik A, B, C dan D, yang manakah menunjukkan imej V di bawah pantulan pada paksi-x diikuti dengan putaran 90° ikut arah jam pada pusat (–1, 1)? Which of the points, A, B, C and D, is the image of V under a reflection in the x-axis followed by a rotation of 90° clockwise at point (–1, 1)? 35 Puan Yatt mempunyai polisi insurans perubatan dengan deduktibel sebanyak RM350 dan fasal penyertaaan peratusan ko-insurans 80/20 dalam polisinya. Hitung bayaran kos yang ditanggung oleh Puan Yatt jika kos perubatan yang ditanggung oleh syarikat insurans ialah RM12 232. Puan Yatt has a medical insurance policy with a deductible provision of RM350 and an 80/20 coinsurance percentage participation clause. Calculate the cost borne by Puan Yatt if the medical cost covered by the insurance company is RM12 232. A RM2 708 B RM3 058 C RM3 408 D RM3 758 36 Diberi bahawa matriks songsang bagi [ 8 –4 5 2] ialah 1 m[ n 4 –5 8 ]. Cari nilai m dan n. Given that the inverse matrix of [ 8 –4 5 2] is 1 m[ n 4 –5 8 ]. Find the value of m and n. A m = 4, n = 2 B m = –4, n = –2 C m = –4, n = 2 D m = 36, n = 2 37 Jadual 4 menunjukkan kadar premium tahunan bagi setiap RM1 000 nilai muka insurans sementara boleh baharu tahunan yang ditawarkan oleh syarikat insurans X. Table 4 shows the annual premium rate per RM1 000 face value of a yearly renewable term insurance offered by insurance company X. Umur Age Lelaki/Male (RM) Perempuan/Female (RM) Bukan Perokok NonSmoker Perokok Smoker Bukan Perokok NonSmoker Perokok Smoker 37 2.28 2.70 1.81 2.09 38 2.56 2.93 2.14 2.32 Jadual 4/Table 4 Encik Siva berumur 37 tahun dan merupakan seorang perokok. Dia membeli polisi insurans bernilai RM200 000. Setahun kemudian, dia ambil keputusan untuk menambah polisi penyakit kritikal. Syarikat insurans X menawarkan polisi penyakit kritikal dengan memberikan perlindungan sebanyak 30% nilai muka asas dan kadar premium bagi setiap RM1 000 seperti ditunjukkan dalam Jadual 4. Berapakah premium tahunan yang perlu ditambah oleh Encik Siva bagi penambahan polisi penyakit kritikal? Encik Siva is 37 years old and a smoker. He bought an insurance policy worth RM200 000. One year later, he decided to add on a critical illness policy. Insurance company X offers a critical illness policy with a coverage of 30% of basic face value and the premium rate per RM1 000 as shown in Table 4. How much is the addition of annual premium for critical illness policy? A RM81 B RM162 C RM232 D RM464 Firasat SPM Maths_Set1 K1_vim3p(2023).indd 7 19/6/2023 9:06:53 AM PENERBIT ILMU BAKTI SDN. BHD.

KM1–8 SULIT 1449/1 1449/1 Firasat SPM 2023: Matematik SULIT 38 Puan Shela memperoleh pendapatan tahunan sebanyak RM57 630. Pada tahun 2022, dia menderma sebanyak RM400 kepada hospital kerajaan dan membayar zakat berjumlah RM200. Pelepasan cukai yang dituntut olehnya ditunjukkan dalam Jadual 5. Puan Shela has an annual income of RM57 630. In 2022, she donated RM400 to the public hospital and paid zakat for RM200. The tax relief claimed by her is shown in Table 5. Pelepasan cukai Tax relief Amaun Amount (RM) Individu Individual 9 000 Insurans hayat & KWSP Life insurance & EPF (Had/Limited to RM7 000) 2 600 Insurans perubatan Medical insurance (Had/Limited to RM3 000) 3 400 Jadual 5/Table 5 Hitung pendapatan bercukai Puan Shela. Calculate Puan Shela’s chargeable income. A RM42 230 B RM42 430 C RM42 830 D RM43 030 39 Rajah 19 menunjukkan Azim yang berada di atas sebuah bangunan setinggi 12 m sedang memerhati dua ekor burung, P dan Q di sebuah gelanggang tenis. Diberi bahawa jarak di antara burung P dengan burung Q ialah 5 m. Sudut dongakan Azim dari burung P ialah 50°. Diagram 19 shows Azim at the top of a building with a height of 12 m, observing two birds, P and Q in the tennis court. Given that the distance between bird P and bird Q is 5 m. The angle of elevation of Azim from bird P is 50°. R Q 12 m Azim Rajah 19/Diagram 19 Hitung sudut tunduk burung Q dari Azim. Calculate the angle of depression of bird Q from Azim. A 31.87° B 38.53° C 51.47° D 58.13° 40 Rajah 20 menunjukkan suatu plot kotak yang tidak lengkap. Diagram 20 shows an incomplete box plot. 29 43 K2 K 83 1 Rajah 20/Diagram 20 Julat antara kuartil dan julat bagi plot kotak tersebut masing-masing ialah 47 dan 76. Tentukan nilai K1 dan K2 . The interquartile range and range of the box plot are 47 and 76 respectively. Determine the value of K1 and K2 . K1 K2 A 7 18 B 7 76 C 26 18 D 26 76 KERTAS PEPERIKSAAN TAMAT END OF QUESTION PAPER Firasat SPM Maths_Set1 K1_vim3p(2023).indd 8 19/6/2023 9:06:53 AM PENERBIT ILMU BAKTI SDN. BHD.

Kertas peperiksaan ini mengandungi 16 halaman bercetak. 2 1 —2 jam Dua jam tiga puluh minit JANGAN BUKA KERTAS PEPERIKSAAN INI SEHINGGA DIBERITAHU 1. Kertas soalan ini mengandungi tiga bahagian: Bahagian A, Bahagian B dan Bahagian C. This question paper consists of three sections: Section A, Section B and Section C. 2. Jawab semua soalan dalam Bahagian A dan Bahagian B dan hanya satu soalan daripada Bahagian C. Answer all questions in Section A and Section B and only one question from Section C. 3. Tulis jawapan anda pada ruang yang disediakan dalam kertas soalan ini. Write your answers in the space provided in the question paper. 4. Tunjukkan kerja mengira anda. Ini boleh membantu anda untuk mendapatkan markah. Show your working. It may help you to get marks. 5. Anda dibenarkan menggunakan kalkulator saintifi k yang tidak boleh diprogram. You may use a non-programmable scientifi c calculator. Untuk Kegunaan Pemeriksa Kod Pemeriksa: Bahagian Soalan Markah Penuh Markah Diperoleh A 1 4 2 3 3 4 4 4 5 3 6 5 7 5 8 4 9 5 10 3 B 11 8 12 10 13 9 14 9 15 9 C 16 15 17 15 Jumlah 1449/2 SIJIL PELAJARAN MALAYSIA 2023 NO. KAD PENGENALAN ANGKA GILIRAN KM1–9 Kertas Model SPM 2 3 4 5 6 7 8 1 SULIT MATEMATIK Kertas 2 Firasat SPM Maths_Set1 K2_vim3p(2023).indd 9 19/6/2023 9:07:58 AM PENERBIT ILMU BAKTI SDN. BHD.

KM1–10 SULIT 1449/2 1449/2 Firasat SPM 2023: Matematik SULIT Untuk Kegunaan Pemeriksa 1 4 2 3 Bahagian A Section A [40 markah/marks] Jawab semua soalan dalam bahagian ini. Answer all questions in this section. 1 Rajah 1 menunjukkan papan tanda lintasan pejalan kaki. Papan tanda tersebut berbentuk segi tiga sama sisi. Diberi panjang tapak papan tanda ialah 80 cm. Diagram 1 shows pedestrian crossing signage. The sign age is an equilateral triangle. Given that the base length of the signage is 80 cm. Rajah 1/ Diagram 1 Hitung luas, dalam cm2 , bagi papan tanda tersebut. Calculate the area, in cm2 , of the signage. [4 markah/marks] Jawapan/Answer: 2 Rajah 2 menunjukkan harga kereta A dan B. Harga bagi kedua-dua kereta tersebut adalah dalam asas yang berlainan. Diagram 2 shows the price of cars A and B. The prices of the two cars are in different bases. RM2223768 Kereta A / Car A Kereta B / Car B RM2223768 RM43243015 RM2223768 Kereta A / Car A Kereta B / Car B RM2223768 RM43243015 A B Rajah 2/ Diagram 2 Hitung beza harga antara kereta A dengan kereta B dalam asas lima. Calculate the price difference between cars A and B in base five. [3 markah/marks] Jawapan/Answer: Firasat SPM Maths_Set1 K2_vim3p(2023).indd 10 19/6/2023 9:07:58 AM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 1449/2 KM1–11 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT Untuk Kegunaan Pemeriksa 4 4 3 4 3 (a) Luqman bekerja sebagai juruaudit. Pendapatan bulanannya ialah RM6 200. Perbelanjaan tetap dan perbelanjaan tidak tetap bulanannya masing-masing ialah RM2 350 dan RM1 700. Dia merancang untuk membeli sebuah kerusi urut berjenama yang berharga RM12 500 secara tunai dalam tempoh setengah tahun untuk ibunya. Adakah Luqman akan mencapai matlamat kewangannya? Jelaskan. Luqman works as an auditor. His monthly income is RM6 200. His monthly fixed expenses and variable expenses are RM2 350 and RM1 700 respectively. He plans to buy a branded massage chair for his mother that costs RM12 500 in cash within half a year. Will Luqman achieve his financial goal? Explain. (b) Dhania merupakan seorang akauntan muda dan dia menerima pendapatan bulanan sebanyak RM3 800. Setiap hujung minggu, Dhania bekerja sebagai seorang jurugambar. Pada suatu bulan tertentu, perbelanjaan tetap dan perbelanjaan tidak tetap Dhania adalah sebanyak RM2 935. Jika pada bulan tersebut, Dhania mempunyai aliran tunai positif sebanyak RM1 720, hitung pendapatan yang diperoleh dari kerja sambilannya. Dhania is a junior accountant and she receives a monthly income of RM 3 800. Every weekend, Dhania works as a photographer. In a particular month, her monthly fixed and variable expenses is RM2 935. If in that particular month, Dhania has a positive cash flow of RM1 720, calculate her income earned from the part time job. [4 markah/marks] Jawapan/Answer: (a) (b) 4 Rajah 3 menunjukkan suatu gambar rajah Venn dengan keadaan set semester, ξ = X ∪ Y ∪ Z. Diagram 3 shows a Venn diagram such that the universal set, ξ = X ∪ Y ∪ Z. Rajah 3/ Diagram 3 Diberi n(X ∪ Y ∩ Z') = n[X' ∩ (Y ∪ Z)]. Hitung n(Y' ∩ X). Given n(X ∪ Y ∩ Z')= n[X' ∩ (Y ∪ Z)]. Calculate n(Y' ∩ X). [4 markah/marks] Jawapan/Answer: x – 2 X Y Z 6 x 2 2x + 3 3x – 7 3 Firasat SPM Maths_Set1 K2_vim3p(2023).indd 11 19/6/2023 9:07:58 AM PENERBIT ILMU BAKTI SDN. BHD.

KM1–12 SULIT 1449/2 1449/2 Firasat SPM 2023: Matematik SULIT Untuk Kegunaan Pemeriksa 5 3 5 Rajah 4(a) menunjukkan sebuah bekas dengan gabungan kuboid dan semi silinder. Rajah 4(b) menunjukkan sebuah bekas berbentuk silinder dengan diameter 7 m dan tinggi 10 m. Afi ingin mengisi air di dalam bekas Rajah 4(a) dengan menggunakan bekas silinder dalam Rajah 4(b). Diagram 4(a) shows a container with a combination of a cuboid and a semi-cylinder. Diagram 4(b) shows a cylindrical container with a diameter of 7 m and a height of 10 m. Afi wants to fill water in the container in diagram 4(a) by using the cylindrical container in Diagram 4(b). 15 m 14 m 20 m (a) (b) 10 m 7 m Rajah 4/Diagram 4 Dengan menggunakan π = 22 7 , hitung bilangan minimum bekas silinder yang diperlukan untuk mengisi air di dalam bekas dalam Rajah 4(a) sehingga menjadi penuh. By using π = 22 7 , calculate the minimum number of cylindrical containers needed to fill the container in Diagram 4(a) pool with water until it is full. [3 markah/marks] Jawapan/Answer: 6 (a) Tulis dua implikasi bagi pernyataan berikut. Write two implications based on the following statement. sin θ = 1 jika dan hanya jika θ = 90° sin θ = 1 if and only if θ = 90° (b) Lengkapkan Premis 2 di bawah untuk membentuk hujah deduktif yang sah dan munasabah. Complete Premise 2 below to form a valid and sound deductive argument. Premis 1: Jika m ialah gandaan 9, maka m boleh dibahagi tepat dengan 9. Premise 1: If m is multiples of 9, then m can be divided exactly by 9. Premis 2: Premise 2: Kesimpulan: 21 bukan gandaan 9. Conclusion: 21 is not multiples of 9. Firasat SPM Maths_Set1 K2_vim3p(2023).indd 12 19/6/2023 9:07:59 AM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 1449/2 KM1–13 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT Untuk Kegunaan Pemeriksa (c) Buat satu kesimpulan umum secara aruhan bagi urutan nombor yang mengikut pola berikut. Make a general conclusion by induction for the sequence of numbers which follows the following pattern. 4 = 3(1) + 1 14 = 3(4) + 2 30 = 3(9) + 3 52 = 3(16) + 4 . . . [5 markah/marks] Jawapan/Answer: (a) (b) (c) 7 Rajah 5 menunjukkan keratan rentas sebuah kilang yang mempunyai tiga tingkap berbentuk segi empat sama dan satu pintu laluan kecemasan. Diagram 5 shows the cross section of a factory that has three squared windows and an emergency exit door. Rajah 5/Diagram 5 Diberi panjang dinding kilang ialah 4 m melebihi 5 kali tingginya. Luas kawasan berlorek ialah 4 440 m2 . Cari luas, dalam m2 , pintu laluan kecemasan itu. Given that the length of the factory wall is 4 m more than 5 times of its height. The area of the shaded region is 4 440 m2 . Find the area, in m2 , of the emergency exit door. [5 markah/marks] Jawapan/Answer: 6 5 7 5 5x m 2x – 3 m x m x + 2 m Firasat SPM Maths_Set1 K2_vim3p(2023).indd 13 19/6/2023 9:07:59 AM PENERBIT ILMU BAKTI SDN. BHD.

KM1–14 SULIT 1449/2 1449/2 Firasat SPM 2023: Matematik SULIT Untuk Kegunaan Pemeriksa 8 Jadual 1 menunjukkan bilangan murid yang memakai cermin mata dan tidak memakai cermin mata dalam kelas 5 Cekal. Table 1 shows the number of students who wear spectacles and do not wear spectacles in the class of 5 Cekal. Murid Students Memakai cermin mata Wear spectacles Tidak memakai cermin mata Do not wear spectacles Lelaki/Male p 7 Perempuan/Female q 11 Jadual 1/Table 1 Diberi bahawa kebarangkalian murid yang memakai cermin mata ialah 2 5 dan bilangan murid lelaki yang memakai cermin mata ialah dua kali bilangan murid perempuan yang memakai cermin mata. Cari nilai p dan q. Given that the probability of students who wear spectacles is 2 5 and the number of male students who wear spectacles is twice the number of female students who wear spectacles. Find the value of p and q. [4 markah/marks] Jawapan/Answer: 9 (a) Jadual 2 menunjukkan pelepasan cukai yang dituntut oleh Encik Rizal. Table 2 shows the tax reliefs claimed by Encik Rizal. Pelepasan cukai Tax reliefs Amaun (RM) Amount (RM) Individu Individual 9 000 Insurans hayat & KWSP (had RM7 000) Life insurance & EPF (limited to RM7 000) k Insurans perubatan (had RM3 000 ) Medical insurance (limited to RM3 000) 3 000 Jadual 2/Table 2 Pada tahun 2022, Encik Rizal menerima gaji tahunan sebanyak RM63 000. Dia menderma RM500 kepada badan kebajikan negeri pada tahun tersebut. Hitung nilai k sekiranya pendapatan bercukai Encik Rizal pada tahun 2022 ialah RM46 140. In 2022, Encik Rizal received an annual salary of RM63 000. He donated RM500 to the state charitable organisation that year. Calculate the value of k if his chargeable income in 2022 is RM46 140. 8 4 Firasat SPM Maths_Set1 K2_vim3p(2023).indd 14 19/6/2023 9:07:59 AM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 1449/2 KM1–15 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT Untuk Kegunaan Pemeriksa (b) Adli mempunyai polisi insurans perubatan dengan peruntukan deduktibel sebanyak RM1 500 dan fasal penyertaan peratusan ko-insurans 80/20. Dia mengalami kecederaan semasa perlawanan bola sepak dan telah menjalani suatu pembedahan di Hospital Kasih. Jumlah yang perlu ditanggung oleh Adli ialah RM2 940. Hitung jumlah kos rawatannya? Adli has a medical insurance policy with a deductible provision of RM1 500 and an 80/20 co-insurance percentage participation clause. He was injured during a football match and underwent a surgery at Hospital Kasih. The amount borne by Adli is RM2 940. Calculate the total amount of his treatment? [5 markah/marks] Jawapan/Answer: (a) (b) 10 Rajah 6 di ruang jawapan menunjukkan sebuah rombus ABCD dilukis pada grid segi empat sama bersisi 1 unit. Diagram 6 in the answer space shows a rhombus ABCD drawn on a square grid with side of 1 unit. Pada rajah di ruang jawapan, On the diagram in the answer space, (a) lukis lokus P yang sentiasa bergerak 5 unit dari titik A. draw the locus of P which always moves at 5 units from point A. (b) lukis lokus Q yang bergerak dengan keadaan jaraknya sentiasa sama dari titik B dan titik D. draw the locus of Q which is moving equidistantly from point B and point D. (c) tanda ⊗ pada persilangan antara lokus P dan lokus Q. mark ⊗ at the intersection of the locus of P and the locus of Q. [3 markah/marks] Jawapan/Answer: (a)(b)(c) A B C D Rajah 6/Diagram 6 9 5 10 3 Firasat SPM Maths_Set1 K2_vim3p(2023).indd 15 19/6/2023 9:07:59 AM PENERBIT ILMU BAKTI SDN. BHD.

KM1–16 SULIT 1449/2 1449/2 Firasat SPM 2023: Matematik SULIT Untuk Kegunaan Pemeriksa Bahagian B Section B [45 markah/marks] Jawab semua soalan dalam bahagian ini. Answer all questions in this section. 11 Rajah 7 menunjukkan segi empat selari yang dilukis pada suatu satah Cartes yang mewakili kedudukan rumah Aina, padang, pejabat pos dan kedai runcit. Diagram 7 shows a parallelogram drawn on a Cartesian plane representing the position of Aina’s house, field, post office and convenient store. (a) Diberi bahawa jarak di antara rumah Aina dengan kedai runcit adalah sama dengan jarak di antara padang dengan pejabat pos. Tentukan koordinat pejabat pos. Given that the distance between Aina’s house and the convenient store is the same as the distance between the field and the post office. Determine the coordinates of the post office. [1 markah/mark] (b) Hitung jarak, dalam km, di antara rumah Aina dengan padang. Kemudian, cari persamaan garis lurus yang menghubungkan rumah Aina dengan padang. Calculate the distance, in km, between Aina’s house and the field. Then, find the equation of the straight line connecting Aina’s house and the field. [4 markah/marks] (c) Sebuah restoran akan dibina di bahagian utara padang. Jarak di antara restoran dengan padang adalah sama dengan jarak di antara pejabat pos dengan padang. Cari persamaan garis lurus yang menghubungkan rumah Aina dengan restoran. A restaurant will be built in the north direction of the field. The distance between the restaurant and the field is the same as the distance between the post office and the field. Find the equation of the straight line that connects Aina’s house to the restaurant. [3 markah/marks] Jawapan/Answer: (a) (b) (c) 11 8 Rajah 7/Diagram 7 Pejabat pos Rumah Aina Aina’s house (−3, 6) Padang/Field (1, 9) Pejabat pos Post office Kedai runcit Convenient store (−3, 0) x y Firasat SPM Maths_Set1 K2_vim3p(2023).indd 16 19/6/2023 9:07:59 AM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 1449/2 KM1–17 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT Untuk Kegunaan Pemeriksa 12 (a) Rajah 8.1 menunjukkan graf laju-masa bagi pergerakan suatu zarah dalam tempoh 7 4 minit. Diagram 8.1 shows a speed-time graph for the motion of particles in the period of 7 4 minutes. (i) Hitung kadar perubahan laju, dalam m s–2, zarah itu dalam tempoh 1 minit yang terakhir. Calculate the rate of change of speed, in m s–2, of the particles in the last 1 minute. (ii) Hitung nilai t, dalam saat, jika nisbah jarak yang dilalui dalam tempoh 7 12 minit pertama kepada jumlah jarak yang dilalui dalam tempoh 7 4 minit ialah 1 : 2. Calculate the value of t, in seconds, if the ratio of the distance travelled in the first 7 12 minutes to the distance travelled in 7 4 minutes is 1 : 2. (iii) Hitung laju purata, dalam m s-1, zarah dalam tempoh 7 4 minit. Calculate the average speed, in m s-1, of the particles for the period of 7 4 minutes. [8 markah/marks] (b) Rajah 8.2 menunjukkan pergerakan Intan dari bandar A ke bandar D melalui bandar B dan bandar C. Diagram 8.2 shows the movement of Intan from town A to town D through town B and town C. Diberi laju purata dari bandar A ke bandar D ialah 96 km j–1. Hitung jarak, dalam km, dari bandar B ke bandar C. Given that the average speed from town A to town D is 96 km h–1. Calculate the distance, in km, from town B to town C. [2 markah/marks] Jawapan/Answer: (a) (i) (ii) (iii) (b) 12 10 Rajah 8.1/Diagram 8.1 Rajah 8.2/Diagram 8.2 • • • 142 km 97 min • 95 min 26 km 18 min A C D B 30 10 Laju/Speed (m s–1) Masa Time (min) O 1 3 3 4 7 4 t Firasat SPM Maths_Set1 K2_vim3p(2023).indd 17 19/6/2023 9:08:00 AM PENERBIT ILMU BAKTI SDN. BHD.

KM1–18 SULIT 1449/2 1449/2 Firasat SPM 2023: Matematik SULIT Untuk Kegunaan Pemeriksa 13 Rajah 9 menunjukkan beberapa poligon yang dilukis pada suatu satah Cartes. Diagram 9 shows a few polygons drawn on a Cartesian plane. Rajah 9/Diagram 9 (a) Senaraikan dua pasangan poligon yang kongruen. List two pairs of polygons which are congruent. [2 markah/marks] (b) Diberi transformasi X ialah pembesaran. Poligon ABCD ialah imej bagi poligon PUTS di bawah gabungan transformasi XY. Huraikan selengkapnya transformasi XY. Given that transformation X is an enlargement. Polygon ABCD is the image of polygon PUTS under the combined transformation XY. Describe, in full, the transformation XY. [4 markah/marks] (c) Diberi bahawa luas PQRS ialah 35 m². Hitung luas kawasan berlorek, dalam m². Given that the area of PQRS is 35 m². Calculate the area of the shaded region, in m². [3 markah/marks] Jawapan/Answer: (a) (b) (c) 13 9 y x D B C F E A Q R P S U T Firasat SPM Maths_Set1 K2_vim3p(2023).indd 18 19/6/2023 9:08:00 AM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 1449/2 KM1–19 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT Untuk Kegunaan Pemeriksa 14 (a) Lengkapkan Jadual 3 di ruang jawapan bagi persamaan y = –x3 + 3x. Complete Table 3 in the answer space for equation y = –x3 + 3x. [2 markah/marks] (b) Untuk ceraian soalan ini, gunakan kertas graf yang disediakan. Anda boleh menggunakan pembaris fleksibel. Dengan menggunakan skala 2 cm kepada 1 unit pada paksi-x dan 2 cm kepada 5 unit pada paksi-y, lukis graf y = –x3 + 3x bagi −3 < x < 3. For this part of the question, use the graph paper provided. You may use a flexible curve ruler. By using a scale of 2 cm to 1 unit on x-axis and 2 cm to 5 units on y-axis, draw the graph of y = –x3 + 3x for −3 < x < 3. [4 markah/marks] (c) Daripada graf di 14(b), cari nilai x apabila y = −1. From the graph in 14(b), find the value of x when y = –1. [3 markah/marks] Jawapan/Answer: (a) x –3 –2 –1 0 1 2 3 y 18 2 0 2 –18 Jadual 3/Table 3 (b) (c) 14 9 Firasat SPM Maths_Set1 K2_vim3p(2023).indd 19 19/6/2023 9:08:00 AM PENERBIT ILMU BAKTI SDN. BHD.

• Firasat SPM 2023: Matematik – Jawapan J1• Jawapan Kertas Model 1 Kertas 1 1 B 2.422 + 4.26 × 8.5 = 2.422 + 36.21 = 42.0664 = 42.1 (3 angka bererti/3 signifi cant fi gures) Nombor Number 1 angka bererti 1 signifi cant fi gure 2 angka bererti 2 signifi cant fi gure 3 angka bererti 3 signifi cant fi gure 4 angka bererti 4 signifi cant fi gure 42.0664 40 42 42.1 42.07 2 B Pernyataan ialah suatu ayat yang dapat ditentukan nilai kebenarannya, iaitu sama ada benar atau palsu, tetapi bukan kedua-duanya. 32 = –9 adalah satu pernyataan kerana nilainya adalah palsu. A statement is a sentence which truth value can be determined, that is, either true or false, but not both. 32 = –9 is a statement because the value is false. 3 A Nilai digit 4 dalam nombor 334029 : Value of digit 4 in 334029 : Nombor Number 3 3 4 0 2 Nilai tempat Place value 94 93 92 91 90 Nilai digit Digit value 3 × 94 = 19 683 3 × 93 = 2 187 4 × 92 = 324 0 × 91 = 9 2 × 90 = 2 Nilai digit 4 dalam nombor 41256 : Value of digit 4 in 41256 : Nombor Number 4 1 2 5 Nilai tempat Place value 63 62 61 60 Nilai digit Digit value 4 × 63 = 864 1 × 62 = 36 2 × 61 = 12 5 × 60 = 5 Beza nilai digit 4 dalam nombor 334029 dengan 41256 : Diff erence between the values of digit 4 in 334029 and 41256 : 864 – 324 = 540 4 A 43215 = (4 × 53 ) + (3 × 52 ) + (2 × 51 ) + (1 × 50 ) = 58610 12345 = (1 × 53 ) + (2 × 52 ) + (3 × 51 ) + (4 × 50 ) = 19410 58610 – 19410 = 39210 5 A 3g + 1 = 2g – 2 2h 6gh + 2h = 2g – 2 6gh – 2g = –2h – 2 2g(3h – 1) = –2(h + 1) g = –2 (h + 1) 2 (3h – 1) g = –h – 1 3h – 1 g = h + 1 1 – 3h 6 A 1 mn = m–n 1 34 = 3–4 ∴ m = 3, n = 4 7 B y = 180° – 47° – 47° = 86° ∠AED = (5 – 2) × 180° 5 = 108° x = 108° – [(180° – 108°) ÷ 2] – 47° = 25° x + y = 86° + 25° = 111° 8 D (4x – 30°) + 2x = 180° 6x – 30° = 180° 6x = 180° + 30° 6x = 210° x = 35° 4x – 30° = 2y + 16° 4(35°) – 30° = 2y + 16° 2y = 94° 2y – x = 94° – 35° = 59° 9 C K L • 52° N J M x O 52° 38° 104° ∠JKN = ∠NMK = 52° ∠NOK = 52° × 2 = 104° Firasat SPM Maths_Jaw_vim3p(2023).indd 1 19/6/2023 9:12:36 AM PENERBIT ILMU BAKTI SDN. BHD.

© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J2• ∠KNM = 180° – 52° 2 = 64° ∠KNO = 180° – 104° 2 = 38° x = 64° – 38° = 26° 10 A P b K L X a M N Q R 145° Y 55° 35° 55° ∠MLQ = 360°– 90° – 90°– 145° = 35° ∠MLQ = ∠PLK a = 35° ∠PLX = ∠XPL = 90° – 35° = 55° ∠PXL = 180° – 55° – 55° b = 70° a + b = 35° + 70° = 105° 11 C x + y = 180° – 52° = 128° z = 22° × 2 = 44° x + y + z = 128° + 44° = 172° 12 D 2 7 × jumlah kad/Total card = 12 Jumlah kad/Total card = 42 Kad merah/Red card, p = 1 6 × 42 = 7 Kad biru/Blue card, q = 42 – 7 – 12 = 23 13 C 2m + n 8m – 6n ÷ 4m2 – n2 4mn – 2n2 = 2m + n 8m – 6n × 4mn – 2n2 4m2 – n2 = 2m + n 2(4m – 3n) × 2n(2m – n) (2m + n)(2m – n) = n 4m – 3n 14 A (2p – 2)2 – (p + 1)(p + 4) = (4p2 – 8p + 4) – (p2 + 5p + 4) = 3p2 – 13p 15 D 16 B Saiz/Size XS S M L XL Bilangan murid Number of students 5 6 3 2 4 Kekerapan longgokan Cumulative frequency 5 11 14 16 20 Median = 1 2 × 20 = 10 = Cerapan ke-10/10th value 17 A 3 cm 4 cm 13 cm x 12 cm 5 cm 32 + 42 = 9 + 16 = 5 cm 132 – 52 = 169 – 25 = 12 cm tan x = 5 12 18 B 0.25 km = 250 m [ 1 2 × (5 + 20) × 4] + [ 1 2 × [(k – 4) + (18 – 4)] × 20] = 250 50 + [10(k + 10)] = 250 50 + [10k + 100 ] = 250 10k = 100 k = 10 19 B 20 C 21 B Z Y X 6 26 4 17 53 – 36 = 17 36 – 26 – 6 = 4 n(X) = 36 n(Y) = 26 n(ξ) = 53 Bahagian berlorek/Shaded region = [Y'∩ (X ∩ Z')] n[Y'∩ (X ∩ Z')] = 17 + 4 = 21 Cerapan ke-6 hingga ke-11 6th value to 11th value Firasat SPM Maths_Jaw_vim3p(2023).indd 2 19/6/2023 9:12:36 AM PENERBIT ILMU BAKTI SDN. BHD.

•J3• Firasat SPM 2023: Matematik – Jawapan 22 C 23 B 24 C n = 1 2 + n(n + 1) = 2 + 1 (1 + 1) = 2 + 2 = 4 n = 2 2 + n(n + 1) = 2 + 2(2 + 1) = 2 + 6 = 8 25 D a K ∝ L2 b K ∝ l M Gabungan/Combined a & b: K ∝ L2 M 26 C 90° 180° 270° 360° y x 3 2 1 0 –1 2 y = 2 kos/cos x + 1 y = 2 kos/cos x y = kos/cos x y = a kos/cos bx + c a: amplitud (Nilai maksimum/minimum berubah) a: amplitude (Maximum/minimum value change) b: tempoh/period c: kedudukan graf (sekiranya c . 0, graf beranjak menegak ke atas paksi-x) c: position of graph (if c . 0, the graph shifts vertically on the x-axis) 27 C MV = 3 200 (1 + 0.035 2 ) 2(6) = RM3 940.61 28 D 130 – 42 = 88 29 A Faktor skala/Scale factor, k = Jarak imej/Image distance Jarak objek/Object distance = AB PQ = 2 6 = 1 3 30 C 31 D Jumaat/Friday: 132546 = (1 × 64 ) + (3 × 63 ) + (2 × 62 ) + (5 × 61 ) + (4 × 60 ) = 2 050 Sabtu/Saturday: 329410 = 3 294 Ahad/Sunday: 3100314 = (3 × 45 ) + (1 × 44 ) + (0 × 43 ) + (0 × 42 ) + (3 × 41 ) + (1 × 40 ) = 3 341 Maka/Hence: 2 050 + 3 294 + 3 341 3 = 2 895 32 B KO JO = 5 3 KO 2.4 = 5 3 KO = 4 cm MK = MO – KO = 8.5 – 4 = 4.5 m 33 B 4.5 m = 450 cm [450 – (68 × 2) – (84 × 2)] 2 = 73 34 C A B C D V 2 O –2 –4 –8 –6 –4 –2 2 4 6 8 10 y x 4 ✗ ✗ V' V' = Pantulan pada paksi-x Reflection on the x-axis = (3, –1) C = Putaran 90° ikut arah jam pada pusat (–1, 1) Rotation of 90° clockwise at point (–1, 1) = (–4, –3) 35 D A–1 = 1 ad – bc [ d –c –b a ] [ 8 –4 5 2] = 1 (8 × 2) – (5 × –4) [ 2 4 –5 8 ] = 1 16 – (–20) [ 2 4 –5 8 ] = 1 36 [ 2 4 –5 8 ] ∴ m = 36, n = 2 36 C Kos perubatan/Medical cost = RMx (x – RM350) × 80 100 = RM12 232 x = RM15 640 Deduktibel ditanggung oleh Puan Yatt = RM15 290 Deductible covered by Puan Yatt = RM15 290 (80% daripada syarikat insuran from insurance company) (20% daripada pemegang polisi from policy holder) RM15 290 × 80 100 = RM12 232 RM15 290 × 20 100 = RM3 058 + RM 350 RM3 408 RM15 290 Firasat SPM Maths_Jaw_vim3p(2023).indd 3 19/6/2023 9:12:37 AM PENERBIT ILMU BAKTI SDN. BHD.

© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J4• 37 B Penyakit kritikal/Critical illness 30 100 × RM200 000 = RM60= 000 RM60 000 RM1 000 × RM2.70 = RM162 38 A Pendapatan bercukai/Chargeable income = Jumlah pendapatan tahunan/Total annual income – Pengecualian cukai/Tax exemption – pelepasan cukai/tax relief = RM57 630 – RM400 – (RM9 000 + RM2 600 + RM3 400) = RM42 230 39 B x Burung P Burung Q Azim 50° y 12 m tan 50° = 12 x x = 10.07 tan y = 12 (10.07 + 5) y = 38.53° 40 B K2 – 29 = 47 K2 = 76 83 – K1 = 76 K1 = 7 Kertas 2 Bahagian A 1 60° x 40 cm tan 60° = x 40 [K1] h = 40 × tan 60° = 69.28 cm [N1] Luas/Area = 1 2 × 69.28 × 80 [K1] = 2 771.2 cm2 [N1] 2 Kereta A/Car A 2223768 = (2 × 85 ) + (2 × 84 ) + (2 × 83 ) + (3 × 82 ) + (7 × 81 ) + (6 × 80 ) [K1] = RM75 00610 Kereta B/Car B RM43243015 = (4 × 56 ) + (3 × 55 ) + (2 × 54 ) + (4 × 53 ) + (3 × 52 ) + (0 × 51 ) + (1 × 50 ) = RM73 70110 RM75 00610 – RM73 70110 = RM1 30510 [N1] Beza kereta A dengan kereta B dalam asas lima Difference between car A and B in base five = RM202105 [N1] 5 1 305 5 261 – 0 5 52 – 1 5 10 – 2 5 2 – 0 0 – 2 3 (a) (RM6 200 – RM2 350 – RM1 700) × 6 [K1] = RM2 150 × 6 = RM12 900 RM12 900 . RM12 500 Luqman dapat mencapai matlamat kewangannya Luqman can achieve his financial goal. [N1] (b) Jumlah pendapatan – jumlah perbelanjaan = +(Aliran tunai positif) Total income – total expenses = +(Positive cash flow) RM3 800 + x – RM2 935 = RM1 720 [K1] x = RM855 [N1] 4 (2x + 3) + 6 + (x – 2) = (2x + 3) + 3 + (3x – 7) [K1] 3x + 7 = 5x – 1 2x = 8 x = 4 [N1] n[Y' ∩ X] = (4 – 2) + 4 [K1] = 6 [N1] 5 Isi padu (a)/Volume (a) (20 × 14 × 15) + 1 2 ( 22 7 × 72 × 15) = 4 200 + 1 155 [K1] = 5 355 m2 Isi padu(b)/Volume (b): 22 7 × 3.52 × 10 = 385 m3 Bilangan bekas/Number of container: 5 355 385 = 13.9 [K1] = 14 bekas/containers [N1] 6 (a) Implikasi 1: Jika sin θ = 1, maka θ = 90° [K1] Implication 1: If sin θ = 1, thus θ = 90° Implikasi 2: Jika θ = 90°, maka sin θ = 1 [K1] Implication 2: If θ = 90°, thus sin θ = 1 (b) 21 tidak boleh dibahagi tepat dengan 9. [K1] 21 cannot be divided exactly by 9. (c) 3n² + n, n = 1, 2, 3, 4, … [K1] 7 [(5x) × (5(5x) + 4)] – 3(x2 ) – (x + 2)(2x – 3) = 4 440 (125x2 + 20x) – 3x2 – (2x2 + x – 6) = 4 440 120x2 + 19x – 4 434 = 0 [K1] (x – 6)(120x + 739) = 0 [K1] x = 6, x = – 739 120 [N1] Luas pintu/Area of the door = (6 + 2) × (2(6) – 3) [K1] = 8 × 9 = 72 [N1] 8 2 5 = p + q p + q + 7 + 11 …① [K1] 2(18 + p + q) = 5(p + q) 3p + 3q = 36 p = 2q…② Firasat SPM Maths_Jaw_vim3p(2023).indd 4 19/6/2023 9:12:37 AM PENERBIT ILMU BAKTI SDN. BHD.

•J5• Firasat SPM 2023: Matematik – Jawapan 3(2q) + 3q = 36 9q = 36 [K1] q = 4 [N1] p = 8 [N1] 9 (a) RM63 000 – RM500 – (RM9 000 + RMk + RM3 000) = RM46 140 [K1] k = 4 360 [N1] (b) RM2 940 – RM1 500 = RM1 440 [K1] 100 20 × RM1 440 = RM7 200 [K1] RM7 200 + RM1 500 = RM8 700 [N1] 10 (a), (b), (c) Titik persilangan [K1] Point of intersection Lokus P [K1] Locus of P Lokus Q [K1] Locus of Q ⊗ A B C D Bahagian B 11 (a) (1, 3) [N1] (b) (32 + 42 ) = 5 km [N1] Jarak di antara rumah Aina dengan padang = 5 km The distance between Aina’s house and the field m = 9 – 6 1 – (–3) = 3 4 [K1] y = 3 4 x + c [K1] 6 = 3 4 (–3) + c [K1] c = 33 4 y = 3 4 x + 33 4 atau/or 4y = 3x + 33 [N1] (c) Restoran/Restaurant (1, 15) [N1] m = 15 – 6 1 – (–3) = 9 4 [K1] y = 9 4 x + c 6 = 9 4 (–3) + c c = 51 4 y = 9 4 x + 51 4 atau/or 4y = 9x + 51 [N1] 12 (a) (i) 7 4 – 3 4 = 4 4 = 1 minit/minute Kadar perubahan laju/Rate of change of speed = Laju/Speed Masa/Time [K1] = 10 60 = 1 6 m s–2 [N1] (ii) 1 2 = [(1 3 × 60) × 10] + [1 2 × (10 + 30) × (t – 20)] [(1 3 × 60) × 10] + [1 2 × (10 + 30) × (t – 20)] + [ 1 2 × (10 + 30) × [(3 4 × 60) – t ] + ( 1 2 × 10 × 60) [K1] 1 2 = (200) + [20(t – 20)] (200) + [20(t – 20)] + [20(45 – t)] + (300) [K1] t = 35 saat/seconds [N1] (iii) 1 2 × (10 + 30) × 15 atau/or 1 2 × (10 + 30) × 10 [K1] Laju purata/Average speed [K1] = Jumlah jarak/Total distance Jumlah masa/Total time = 200 + 300 + 200 + 300 7 4 × 60 = 200 + 300 + 200 + 300 105 [K1] = 9.52 [N1] (b) Laju purata/Average speed = Jumlah jarak/Total distance Jumlah masa/Total time 96 = 142 + x + 26 (95 + 97 + 18) 60 [K1] 96 = 142 + x + 26 3.5 336 = 142 + x + 26 x = 168 km [N1] Maka, jarak dari bandar B ke bandar C ialah 168 km. Thus, the distance from town B to town C is 168 km. 13 (a) PQRS dan/and PUTS [N1] ABCD dan/and AFED [N1] (b) Y: Pantulan pada garis y = 2 [K1][K1] Reflection on line y = 2 X: Pembesaran pada pusat (4, 2) dengan faktor skala 2 [K1] Enlargement at point (4, 2) with the scale factor 2 (c) Luas imej = Luas objek × k2 Area of image = Area of object × k2 = 35 × 22 [K1] = 140 m2 Luas kawasan berlorek/Area of shaded region = 2 × (140 – 35) [K1] = 210 m2 [N1] 14 (a) x –3 –2 –1 0 1 2 3 y 18 2 –2 0 2 –2 –18 Firasat SPM Maths_Jaw_vim3p(2023).indd 5 19/6/2023 9:12:37 AM PENERBIT ILMU BAKTI SDN. BHD.

© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J6• (b) x y 0 –2–3 –1 1 3 2 –5 –10 –15 –20 20 15 10 5 ✗ ✗ ✗ ✗ ✗ ✗ [K4] (c) –1.4, –1.4, 1.8 [N3][N1][N3] 15 (a) Masa (min) Time (min) Kekerapan longgokan Cumulative frequency Kekerapan Frequency Titik Tengah Midpoint 0 – 10 4 4 5 10 – 20 11 7 15 20 – 30 22 11 25 30 – 40 35 13 35 40 – 50 45 10 45 50 – 60 50 5 55 [N1] [N1] [N1] (b) (i) Min/Mean = 4(5) + 7(15) + 11(25) + 13(35) + 10(45) + 5(55) 50 [K1][K1] = 31.6 [N1] (ii) Sisihan piawai/Standard deviation = 4(5)2 + 7(15)2 + 11(25)2 + 13(35)2 + 10(45)2 + 5(55)2 50 [K1] = (31.6)2 [K1] = 14.09 [N1] Bahagian C 16 (a) (i) Perimeter empat kawasan kolam Perimeter of four swimming pool = 2 × 22 7 × 70 [K1] = (2 × 22 7 × 70) + (8 × 70) [K1] = 1 000 m [N1] (ii) Luas empat kawasan kolam Area of four swimming pool = 22 7 × 702 [K1] Luas taman/Area of garden = (140 × 280) – ( 22 7 × 702 ) [K1] = 23 800 m2 [N1] (b) t α p w t = k p w 7 = k 210 6 [K1] k = 0.2 t = 0.2 p w [K1] 8 = 0.2 1000 w [K1] w = 25 [N1] (c) x – dewasa/adult y – kanak-kanak/children 2x + 3y = 121 atau/or 3x + 28y = 628 [K1] [ 2 3 3 28 ][ x y ] = [ 121 628] [K1] [ x y ] = 1 2(28) – 3(3) [ 28 –3 –3 2 ][121 628] [K1] [ x y ] = 1 47 [ 1 504 893 ] [ x y ] = [ 32 19] x = RM32, y = RM19 [N1] [N1] 17 (a) (i) I: x + y < 3200 [K1] II: y > 900 [K1] III: x > 1 3 y atau/or y < 3x [K1] (ii) y 500 1000 1500 2000 2500 3000 3500 500 1000 1500 2000 2500 3000 0 x 3500 ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ x = 1 3 y y > 900 x + y < 3200 [K4] (iii) 1700 [N1] (b) Jenis A/Type A: Min/Mean = 9(16500) +10(19500) + 18(22500) + 8(25500) + 5(28500) 50 = 1095000 50 [N1] = 21900 Firasat SPM Maths_Jaw_vim3p(2023).indd 6 19/6/2023 9:12:38 AM PENERBIT ILMU BAKTI SDN. BHD.

•J7• Firasat SPM 2023: Matematik – Jawapan Sisihan piawai/Standard deviation = 9(16500)2 + 10(19500)2 + 18(22500) + 8(25500)2 + 5(28500)2 50 – (21900)2 [K1] = 24628500000 50 – (21900)2 = 3600 Jenis B/Type B: Min/Mean = 10(16500) + 9(19500) + 12(22500) + 10(25500) + 9(28500) 50 [K1] = 1122000 50 = 22440 Sisihan piawai/Standard deviation = 10(16500)2 + 9(19500)2 + 12(22500)2 + 10(25500)2 + 9(28500)2 50 – (22440)2 = 26032500000 50 – (22440)2 = 4134.78 [N1] Mentol jenis A/Bulb type A Sisihan piawainya lebih rendah, maka lebih konsisten. The standard deviation is lower, thus its more consistent. [N1] (c) (200 × RM0.263) + (100 × RM0.327) + (230 × RM0.468) + (230 × RM0.468 × 6%) [K1] = RM199.40 [N1] Kertas Model 2 Kertas 1 1 D 0 . 0 6 6 9 9 = 0 . 0 6 7 0 (3 angka bererti/signifi cant fi gures) 2 C Nilai tempat Place value 74 73 72 71 70 Digit 1 0 3 0 2 3 C Kemungkinan jawapan/Possible answers: B, C atau/or D 2256 = (2 × 62 ) + (2 × 61 ) + (5 × 6°) = RM8910 11214 = (1 × 43 ) + (1 × 42 ) + (2 × 41 ) + (1 × 4°) = RM8910 4 B 0.001084 – 2.003 × 10–5 = 0.001084 – 0.00002003 = 0.00106397 = 1.06 × 10–3 5 B 243 × 1 3 = 3n 81 = 3n 34 = 3n n = 4 6 D 5.4 ÷ 700 × 25 = 0.193 cm 0.193 cm = 0.00193 m = 1.93 × 10–3 m +1 7 A 2 p + 3 = 3 p + 5 2(p + 5) = 3(p + 3) 2p + 10 = 3p + 9 p = 1 8 C 9m – mn2 mn ÷ 3m – mn 6n = m(9 – n2 ) mn × 6n m(3 – n) = m(3 + n)(3 – n) mn × 6n m(3 – n) = 6(3 + n) m 9 A p q = 30 p 5 = 30q 5 p 5 = 6q 1 5 = 6q p 6q p = 1 5 10 B ∠HNM = 4 × 180° 6 = 120° 5x = 120° x = 24° 2y = 180° – 120° – 24° 2y = 36° y = 18° 11 D ∠RQO = 90° ∠QOR = 180° – 90° – 50° = 40° ∠OST = 40° 2 = 20° ∠SOT = 180° – 40° = 140° ∠OTS = 180° – 40° – 20° = 20° y = 180° – 20° = 160° 12 D s = 1 3 + t 5 15s = 5 + 3t 3t = 15s – 5 t = 5s – 5 3 13 C POQ (Diameter) = 34 cm QO = OP = 34 2 = 17 cm OS = 17 – 9 = 8 cm RS = 172 –82 = 15 cm 8 O R 17 S 15 cm RT = 15 × 2 = 30 cm a2 – b2 = (a + b)(a – b) H Q M P N 2y° 120° 24° • 50° T y S P Q O 140° 20° 20° 40° R Firasat SPM Maths_Jaw_vim3p(2023).indd 7 19/6/2023 9:12:39 AM PENERBIT ILMU BAKTI SDN. BHD.

© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J8• 14 D y = –3x2 + 2x + 8 –3x2 + 2x + 8 = 0 3x2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0 x = – 4 3 atau/or = 2 koordinat/coordinate V(– 4 3 , 0) koordinat/coordinate W(2, 0) 15 A –3x + 2x2 = – 1 3 (4 – x) –9x + 6x2 = –4 + x 6x2 – 10x + 4 = 0 3x2 – 5x + 2 = 0 (3x – 2)(x – 1) = 0 x = 2 3 , x = 1 16 B W X Z Y B 5 cm 17 C Beli/Buy = RM0.80 × 7 500 = RM6 000 Jual/Sell = RM1.20 × 7 500 = RM9 000 9 000 – 6 000 + ( 6 100 × 6 000) 6 000 × 100% = 56% 18 D y = a sin bx + c a = Amplitud/Amplitude = 2 b = Bilangan kitaran/Number of cycles = 2 c = pintasan-y/y-intercept c = 0 y = 2 sin 2x 19 C Luas lukisan/Scale drawing = Luas poster/Area of poster Luas rajah/Area of picture 1 1 2 = Luas poster/Area of poster (9.6 × 17.5) Luas poster/Area of poster= 336 20 B P ∩ Q ∩ R = {h, k, l} (P ∩ Q ∩ R)' = {g, i, j} 21 D Julat asal/Initial range = 10 – 4 = 6 Julat baharu/New range = 6 × 3 = 18 22 D 0.6 × 0.7 = 0.42 23 B RM4 700 × 5.2%= RM244.40 Cukai pintu setiap setengah tahun Property assessment tax for each half-year = RM244.40 2 = RM122.20 24 B Gred/Grade 1 2 3 4 5 6 Bilangan murid Number of students 4 3 6 7 3 2 ↓ Mod/Mode Mod/Mode = 4 Bilangan murid yang memperoleh gred lebih baik daripada gred mod (gred 5 dan gred 6) Number of students who achieve better grades than the mode grade (grade 5 and grade 6) = 3 + 2 = 5 25 A Matriks songsang yang tidak wujud: Inverse matrices that do not exists: ad – bc = 0 (–2)(2) – (–4)(1) = –4 – (–4) = 0 26 A Q(0, 3) P(–4, 0) 5 units O(0, 0) 3 units 4 units mPQ = 3 4 Q(0, 3) PQ = 32 + 42 = 5 units 10 units 8 units Q(0, 3) 6 units R(6, 11) QR = 5 × 2 = 10 units mQR = 4 3 R(6, 11) 27 C mOM = 5 4 tan p = 5 4 p = 51.34° 28 A 72 m – 1.7 m = 70.3 m tan 51° = 70.3 z z = 56.93 51° 70.3 z Firasat SPM Maths_Jaw_vim3p(2023).indd 8 19/6/2023 9:12:39 AM PENERBIT ILMU BAKTI SDN. BHD.

•J9• Firasat SPM 2023: Matematik – Jawapan 29 C Akas: Jika q, maka p Inverse: If ~p, then ~q. Songsangan: Jika ~p, maka ~q Converse: If ~p, then ~q. Akas: Jika g ialah gandaan 4 q maka, g boleh dibahagi tepat dengan 2 p . If g is a multiple of 4 q then, g can be divided exactly by 2 p . Songsangan/Inverse: Jika g tidak boleh dibahagi tepat dengan 2, maka g bukan gandaan 4. If g cannot be divided exactly by 2, then g is not a multiple of 4. 30 C (A) 90 + 100 + 120 = 310 (B) 40 + 30 + 50 310 × 100% = 38.7% (C) P : 190 L: 120 P . 2L (Tidak benar/Not true) (D) P(Ahad/Sunday) = 70 L: (Jumaat/Friday) + (Sabtu/Satuday) = 40 + 30 = 70 P(Ahad/Sunday) = L(Jumaat/Friday) + L(Sabtu/Satuday) 31 A n(K' ∩ B) = n(F ∩ K) 8 + (2x + 1) = (x + 4) + (2x) 2x + 9 = 3x + 4 x = 5 Jumlah murid/Total students = 6 + (5 + 4) + 2(5) + 8 + 12 + 7 + (2(5) + 1) = 63 32 D Q & R tidak bersilang Q & R are not intersect Q ∩ R = ϕ Q R P ⊂ Q Q P 33 A n(S) = 10 A = {hasil tambah digit/sum of digits . 6} = {17, 29, 25} P(A) = 3 10 B = {nombor perdana/prime number} = {17, 29, 3, 11} P(B) = 4 10 A ∩ B = {17, 29} P(A ∩ B) = 2 10 P(A atau/or B) = P(A ∪ B) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 3 10 + 4 10 – 2 10 = 5 10 = 1 2 34 C Titik G ialah imej/Point G is an image (4, 1) Q G(1, 4) Objek Imej Object     Image (1, 3) Objek translasi ( –3 2 ) Object translation R (4, 1) Imej Image 35 B 9 100 × RM350 000 = RM315 000 315 000 + (315 000 × p% × 30) 30 × 12 = 1 715 p = 3.2 36 B Pendapatan pasif/Passive income: • sewa yang diterima/rent received • faedah/interest • dividen/divided 37 B x° 1 2 3 2 1 y° 3 2 1 2 1 tan x = ( 1 2 ) ( 3 2 ) = 30° kos x° + sin y° = kos 30° + sin 60° = 1.73 38 D Pokok/Tree: → Graf mudah tanpa gelung atau berbilang tepi A simple graph without loops and multiple edges → Semua bucu mesti berkait/All the vertices are connected → n(E) = n(V) – 1 39 C R ∝ = p d3 R = k P d3 8 = k 108 33 k = 2 R = 2( p d3) 40 B 80 100 × RM280 000 = RM238 000 5 = 2( 140 d3 ) 5 2 = 140 d3 d3 = 140 ( 5 2 ) d = 3.83 Firasat SPM Maths_Jaw_vim3p(2023).indd 9 19/6/2023 9:12:39 AM PENERBIT ILMU BAKTI SDN. BHD.

© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J10• Kertas 2 Bahagian A 1 p + q = 180° …① [K1] p = q – 42° …② [K1] (q – 42°) + q = 180° 2q = 222° q = 111° [N1] p = 111° – 42° p = 69° [N1] 2 (a) ℎ + (ℎ + 1) + (2ℎ) + (2ℎ + 2) + (3ℎ − 1) 5 = 7.6 [K1] ℎ + (ℎ + 1) + (2ℎ) + (2ℎ + 2) + (3ℎ − 1) = 38 9ℎ + 2 = 38 ℎ = 4 [N1] (b) Set nombor/Set of numbers = 4, 5, 8, 10, 11 Julat/Range = 11 – 4 [K1] = 7 [N1] 3 g(5x² + 3x) – 6 = 0 5gx² + 3gx – 6 = 0 5g(–1)² + 3g(–1) – 6 = 0 [K1] 5g – 3g – 6 = 0 2g = 6 g = 3 [N1] 3(5x² + 3x) – 6 = 0 15x² + 9x – 6 = 0 5x² + 3x – 2 = 0 [K1] (5x – 2)(x + 1) = 0 [K1] x = 2 5 atau/or x = –1 Punca yang lain/The other root = 2 5 [N1] 4 (a) 2y – 3x = 6 2(0) – 3x = 6 [K1] x = –2 [N1] (b) mPQ = mRS = 3 2 [N1] P(–6, 0) Persamaan garis lurus/The equation of straight line PQ: 0 = 3 2 (–6) + c [K1] c = 9 y = 3 2 x + 9 [N1] 5 (a) Benar/True [N1] (b) Implikasi 1/Implication 1: Jika M ⊂ N, maka M ∩ N = M [K1] If M ⊂ N, thus M ∩ N = M Implikasi 2/Implication 2: Jika M ∩ N = M, maka M ⊂ N [K1] If M ∩ N = M, thus M ⊂ N (c) 4(n)2 – n [K1] n = 1, 2, 3, … [N1] 6 OU = OS = OR = OP = 29 cm PQ = OP2 – OQ2 = 292 – 212 [K1] = 20 cm [N1] PQR = 20 + 20 = 40 cm [N1] 7 (a) J N O M L K 8 4 5 7 2 [K1] [N1] (b) 7 + 2 + 4 + 5 + 8 = 26 [N1] 8 1 2 (4 3 × 22 7 × 1.53 ) + ( 22 7 × 1.52 × h) = 99 [K1] [K1] h = 13 [N1] h = 13 + 1.5 = 14.5 [N1] 9 (2 × 22 7 × 14) + ( 300 360 × 2 × 22 7 × 3 + 3 + 3) + ( 300 360 × 2 × 22 7 × 3.5 + 3.5 + 3.5) [K1] = 88 + 21 5 7 + 25 1 3 [K1] = 135 1 25 [N1] 10 a = Amplitud/Amplitude = 50 [N1] b = Bilangan kitaran/Number of cycles [N1] = 2 c = Pintasan y/y-intercept = 20 [N1] Bahagian B 11 (a) A = {15, 20} [N1] B = {17, 19, 23} [N1] C = {15, 17, 19, 21, 23} [N1] (b) • 20 • 22 • 24 • 15 • 17 • 19 • 23 • 16 • 18 • 21 B C A ξ [K1] [N1][N1] (c) (i) 9 [N1] (ii) 4 [N1] 12 (a) Simpanan bulanan Monthly savings Kekerapan Frequency Titik tengah Midpoint RM10 – RM14 4 12 RM15 – RM19 7 17 RM20 – RM24 11 22 RM25 – RM29 14 27 RM30 – RM34 9 32 [K1] RM35 – RM39 5 37 [N1] (b) 37 y 0 5 10 15 20 x 12 17 22 3227 42 ✗ ✗ ✗ ✗ ✗ ✗ – Kedua-dua paksi betul dengan skala seragam bagi 7 < x < 42 dan 0 < y < 50 [K1] Both axis are correct with uniform scale of 7 < x < 42 and 0 < y < 50 Firasat SPM Maths_Jaw_vim3p(2023).indd 10 19/6/2023 9:12:40 AM PENERBIT ILMU BAKTI SDN. BHD.