Matematik T5

Matematik Tingkatan 5 Jawapan J17 (e) Pendapatan bercukai/ Chargeable income: RM111 284 – RM12 100 = RM99 184 Cukai pendapatan/ Income tax: RM4 400 + (RM99 184 – RM70 000) × 21% − RM1 300 = RM4 400 + RM6 128.64 – RM1 300 = RM9 228.64 (f) Pendapatan bercukai/ Chargeable income: RM75 000 – RM10 200 – RM130 = RM64 670 Cukai pendapatan/ Income tax: RM1 800 + (RM64 670 – RM50 000) × 13% = RM1 800 + RM1 907.10 = RM3 707.10 Jumlah PCB yang dipotong: Total PCB deducted: RM250 × 12 = RM3 000 Cukai yang perlu dibayar > PCB Tax payable > PCB Baki cukai pendapatan yang perlu dibayar: Balance of payable income tax: RM3 707.10 − RM3 000 = RM707.10 ∴ Dia perlu membayar lagi RM707.10 kepada LHDN. She needs to pay another RM707.10 to IRB. (g) Pendapatan bercukai/ Chargeable income: RM190 000 – RM27 000 = RM163 000 Cukai pendapatan/ Income tax: RM10 700 + (RM163 000 – RM100 000) × 24% = RM10 700 + RM15 120 = RM25 820 Jumlah PCB yang dipotong: Total PCB deducted: RM2 500 × 12 = RM30 000 PCB > cukai pendapatan PCB > income tax Lebihan PCB/ Excess monthly tax deduction: RM30 000 − RM25 820 = RM4 180 ∴ Tidak, LHDN akan memulangkan RM4 180 kepada Xavier. No, IRB will refund RM4 180 to Xavier. (h) Pendapatan bercukai/ Chargeable income: (RM9 975 × 12) – RM9 000 – RM200 – RM140 = RM119 700 – RM9 340 = RM110 360 Cukai pendapatan/ Income tax: RM10 700 + (RM110 360 – RM100 000) × 24% = RM10 700 + RM2 486.40 = RM13 186.40 Jumlah PCB yang dipotong: Total PCB deducted: RM1 200 × 12 = RM14 400 PCB > cukai pendapatan PCB > income tax Lebihan PCB/ Excess monthly tax deduction: RM14 400 − RM13 186.40 = RM1 213.60 ∴ Tidak, LHDN perlu memulangkan RM1 213.60 kepada Josh. No, IRB needs to refund RM1213.60 to Josh. (i) Pendapatan bercukai/ Chargeable income: RM232 000 − RM9 000 − RM7 000 – RM3 000 – RM2 900 – RM500 = RM209 600 Cukai pendapatan/ Income tax: RM10 700 + (RM209 600 – RM100 000) × 24% = RM10 700 + RM26 304 = RM37 004 CONTOH

Matematik Tingkatan 5 Jawapan J18 (j) (i) Perkara Item Taksiran cukai bersama Joint tax assessment Taksiran cukai berasingan Separate tax assessment Suami dan isteri Husband and wife Suami Husband Isteri Wife Jumlah pendapatan Total income RM62 000 + RM42 000 = RM104 000 RM62 000 RM42 000 Jumlah pengecualian (derma) Total exemption (donation) – RM1 500 – RM600 – RM900 Pelepasan Tax relief – Individu Individual – RM9 000 – RM9 000 – RM9 000 – Insurans perubatan (had RM3 000) Medical insurance (limited to RM3 000) – RM3 000 – RM1 930 – RM1 500 Pendapatan bercukai Chargeable income RM90 500 RM50 470 RM30 600 Cukai dasar Base tax RM4 400 RM1 800 RM150 Cukai atas baki Tax on the next balance Baki/ Balance: RM90 500 – RM70 000 = RM20 500 RM20 500 × 21% = RM4 305 Baki/ Balance: RM50 470 – RM50 000 = RM470 RM470 × 13% = RM61.10 Baki/ Balance: RM30 600 – RM20 000 = RM10 600 RM10 600 × 3% = RM318 Rebat cukai Tax rebate – RM0 – RM0 – RM400 (Pendapatan bercukai N RM35 000) – RM400 (Chargeable income N RM35 000) Cukai pendapatan yang perlu dibayar Payable income tax RM4 400 + RM4 305 = RM8 705 RM1 800 + RM61.10 = RM1 861.10 RM150 + RM318 – RM400 = RM68 RM1 929.10 (ii) Taksiran cukai berasingan kerana cukai pendapatan yang perlu dibayar jauh lebih rendah, iaitu RM1 929.10 berbanding dengan RM8 705 melalui taksiran cukai bersama. Separate tax assessment because the payable income tax is much lower, which is RM1929.10 as compared to RM8705 for joint tax assessment. 4 (a) RM30 – RM9 = RM21 (b) (i) Cukai jalan/ Road tax: RM880 + (2 700 − 2 500) × RM2.50 = RM880 +RM500 = RM1 380 (ii) Cukai jalan/ Road tax: RM2 130 + (3 300 − 3 000) × RM4.50 = RM2 130 + RM1 350 = RM3 480 5 (a) Jumlah cukai pintu/ Property assessment tax: 3 100 × RM7 945 = RM238.35 setahun/ per year Cukai pintu setiap setengah tahun: Property assessment tax for each half-year: RM238.35 2 = RM119.18 CONTOH

Matematik Tingkatan 5 Jawapan J19 (b) Nilai tahunan/ Annual value: RM1 280 × 12 = RM15 360 Kadar cukai pintu: Property assessment tax rate: RM768 RM15 360 × 100% = 0.05 × 100% = 5% (c) Jumlah cukai pintu/ Property assessment tax: RM600 × 2 = RM1 200 Nilai tahunan/ Annual value: RM1 200 4% = RM1 200 0.04 = RM30 000 6 (a) Kadar cukai tanah/ Quit rent rate: RM132.30 210 = RM0.63 seunit/ per unit (b) Cukai perkhidmatan/ Service tax: RM150 × 3 × 6% = RM27 (c) Jumlah bayaran bagi setiap individu: Total fee for each person: 106 100 × RM63.90 = RM67.73 Jumlah wang yang perlu dibayar: Total money that needs to be paid: RM67.73 × 6 = RM406.38 7 (a) (i) Oleh sebab pendapatan bercukai kurang daripada RM35 000, dia boleh menuntut rebat cukai sebanyak RM400. Beacause chargeable income is less than RM35 000, she can claim a tax rebate of RM400. Jumlah rebat cukai Subaidah: Subaidah’s total tax rebate: RM400 + RM230 = RM630 (ii) Cukai bagi RM20 000 pertama = RM150 Tax on the first RM20 000 Cukai atas baki berikutnya: Tax on the next balance: (RM30 600 – RM20 000) × 3% = RM10 600 × 3% = RM318 Cukai pendapatan yang perlu dibayar: Payable income tax: RM150 + RM318 – RM630 = (–RM162) = RM0 ∴ –RM162 < 0, maka Subaidah tidak perlu membayar cukai pendapatan. –RM162 < 0, thus Subaidah does not need to pay the income tax. (b) (i) Pendapatan bercukai Zul: Zul’s chargeable income: RM270 000 – RM61 000 = RM209 000 (ii) Cukai bagi RM100 000 pertama Tax on the first RM100 000 = RM10 700 Cukai atas baki berikutnya: Tax on the next balance: (RM209 000 – RM100 000) × 24% = RM109 000 × 24% = RM26 160 Cukai pendapatan yang perlu dibayar: Payable income tax: RM10 700 + RM26 160 – RM4 200 = RM32 660 (iii) Jumlah PCB yang dipotong: Total PCB deducted: RM5 000 × 12 = RM60 000 Cukai yang perlu dibayar < PCB Payable tax < PCB Lebihan PCB: Excess monthly tax deduction: RM60 000 – RM32 860 = RM27 340 ∴ Tidak, LHDN perlu memulangkan RM27 340 kepada Zul No, IRB needs to refund RM27340 to Zul. (c) (i) Pendapatan bercukai: Chargeable income: RM237 400 – RM9 000 – RM7 000 – RM2 400 – RM2 800 – RM2 100 = RM214 100 (ii) Cukai bagi RM100 000 pertama Tax on the first RM100 000 = RM10 700 Cukai atas baki berikutnya: Tax on the next balance: (RM214 100 – RM100 000) × 24% = RM114 100 × 24% = RM27 384 Cukai pendapatan yang perlu dibayar: Payable income tax: RM10 700 + RM27 384 = RM38 084 (iii) Jumlah PCB yang dipotong: Total PCB deducted: RM3 500 × 12 = RM42 000 Oleh sebab PCB > cukai yang perlu dibayar, maka Puan Chew tidak perlu membuat pembayaran cukai pendapatan tambahan. Because PCB > payable tax, hence Madam Chew does not need to make an additional income tax payment. CONTOH

Matematik Tingkatan 5 Jawapan J20 Lebihan PCB: Excess monthly tax deduction: RM42 000 – RM38 084 = RM3 916 ∴ Tidak, LHDN perlu memulangkan RM3 916 kepada Puan Chew. No, IRB needs to refund RM3 916 to Madam Chew. Praktis Kendiri Kertas 1 1 A 2 A 3 B Cukai pendapatan/ Income tax: RM10 900 + (RM167 900 − RM100 000) × 24% = RM27 196 4 B Cukai tanah yang dibayar oleh Jin: Quit rent payable by Jin: RM0.32 × 135 = RM43.20 setahun/ RM43.20 per year 5 C Nisbah/ Ratio = RM9 : RM9 : RM12 = RM9 RM3 : RM9 RM3 : RM12 RM3 = 3 : 3 : 4 6 A Anggaran sewa bulanan/ Estimate monthly rental: RM33 600 12 = RM2 800 7 B Jumlah cukai pintu yang perlu dibayar: Property assessment tax payable: 2 100 × RM6 000 = RM120 Kertas 2 Bahagian A/ Section A 1 Perkara Item Sean dan isteri Sean and wife Jumlah pendapatan Total income RM53 000 + RM39 000 = RM92 000 Jumlah pengecualian (Derma) Total exemption (Donation) – RM1 500 Pelepasan cukai Tax relief – Individu Individual – RM9 000 – Insurans hayat (had RM7 000) Life insurance (limited to RM7 000) – RM7 000 Pendapatan bercukai Chargeable income RM74 500 Cukai dasar Base tax RM4 400 Cukai atas baki Tax on the next balance (RM74 500 – RM70 000) × 21% = RM945 Rebat cukai Tax rebate – RM0 Cukai pendapatan yang perlu dibayar Payable income tax RM4 400 + RM945 = RM5 345 2 Nilai tahunan/ Annual value: RM3 000 × 12 = RM36 000 Jumlah cukai pintu yang perlu dibayar: Property assessment tax payable: RM36 000 × 7% = RM2 520 setahun/ per year Cukai pintu setiap setengah tahun: Property assessment tax for each half-year: RM2 520 2 = RM1 260 3 (a) Cukai jalan/ Road tax: RM380 + (2 400 – 2 000) × RM1.00 = RM380 + RM400 = RM780 (b) Cukai jalan/ Road tax: RM2 130 + (4 000 – 3 000) × RM4.50 = RM2 130 + RM4 500 = RM6 630 Bahagian B/ Section B 4 (a) k = 100 × 0.334 = 33.40 l = 300 × 0.516 = 154.80 m = 150 × 0.546 = 81.90 (b) Jumlah bayaran yang tidak dikenakan cukai perkhidmatan: Total amount of payment which is not subjected to service tax: RM43.60 + RM33.40 + RM154.80 = RM231.80 (c) Cukai perkhidmatan yang dikenakan: Service tax charged: 6 100 × RM81.90 = RM4.91 CONTOH

Matematik Tingkatan 5 Jawapan J21 Bab 5 5.1 Kekongruenan Congruency 1 (a) Bukan, kerana bukan semua sudut sepadan adalah sama. No, because not all corresponding angles are equal. (b) Bukan, kerana bukan semua sisi sepadan adalah sama. No, because not all corresponding sides are equal. (c) Ya, kerana panjang jejari sepadan adalah sama. Yes, because the corresponding radius is equal. (d) Ya, kerana semua panjang sisi dan sudut sepadan adalah sama. Yes, because all corresponding sides and angles are equal. (e) Bukan, kerana kedua-dua bentuk adalah tidak sama. No, because both shapes are not equal. 2 (a) Sudut-Sudut-Sisi (AAS) Angle-Angle-Side (AAS) (b) Sisi-Sudut-Sisi (SAS) Side-Angle-Side (SAS) (c) Sudut-Sisi-Sudut (ASA) Angle-Side-Angle (ASA) (d) Sudut-Sudut-Sudut (AAA) Angle-Angle-Angle (AAA) (e) Sisi-Sisi-Sudut (SSA) Side-Side-Angle (SSA) 3 (a) AB = EF, AC = DF, BC = DE Segi tiga ABC dan segi tiga DEF adalah kongruen Triangles, ABC and DEF are congruent (b) AB = √52 – 42 = 3 cm = EF AC = DE DF = √42 + 32 = 5 cm = BC Segi tiga ABC dan segi tiga DEF adalah kongruen Triangles, ABC and DEF are congruent (c) AC = √32 + 42 = 5 cm = EF DF = √52 + 122 = 13 cm AB ≠ DE, AC ≠ DF, BC ≠ EF Segi tiga ABC dan segi tiga DEF adalah tidak kongruen Triangles, ABC and DEF are not congruent 4 (a) ∠A = 180° – 2(50°) = 80° = ∠E AB = DE, AC = EF Segi tiga ABC dan segi tiga DEF adalah kongruen Triangles, ABC and DEF are congruent (b) AB = EF, ∠B = ∠E, BC = DE Segi tiga ABC dan segi tiga DEF adalah kongruen Triangles, ABC and DEF are congruent (c) ∠A = 60° = ∠D (segi tiga sama sisi/ equilateral triangle) AB = DE, AC = DF Segi tiga ABC dan segi tiga DEF adalah kongruen Triangles, ABC and DEF are congruent 5 (a) ∠A = ∠D, AB = DF, ∠F = 180° – 80° – 45° = 55° = ∠B Segi tiga ABC dan segi tiga DEF adalah kongruen Triangles, ABC and DEF are congruent (b) ∠C = ∠F, BC = DF, ∠D = 180° – 50° – 70° = 60° = ∠B Segi tiga ABC dan segi tiga DEF adalah kongruen Triangles, ABC and DEF are congruent (c) ∠D = 180° – 85° – 65° = 30° ≠ ∠A AC = DF ∠C = 180° – 35° – 85° = 60° ≠ ∠F Segi tiga ABC dan segi tiga DEF adalah tidak kongruen Triangles, ABC and DEF are not congruent 6 (a) ∠A = ∠D, ∠B = ∠E, BC = EF Segi tiga ABC dan segi tiga DEF adalah kongruen Triangles, ABC and DEF are congruent (b) ∠F = 180° – 50° – 60° = 70° = ∠C ∠B = ∠E, AB = DE Segi tiga ABC dan segi tiga DEF adalah kongruen Triangles, ABC and DEF are congruent (c) ∠A = ∠D, ∠C = ∠F, BC ≠ EF Segi tiga ABC dan segi tiga DEF adalah tidak kongruen Triangles, ABC and DEF are not congruent 7 (a) ∠A = 180° – 2(70°) = 40° = ∠D AB = DE, AC = DF Segi tiga ABC dan segi tiga DEF adalah kongruen kerana mematuhi sifat SisiSudut-Sisi (SAS) Triangles, ABC and DEF are congruent because they satisfy the Side-Angle-Side (SAS) rule CONTOH

Matematik Tingkatan 5 Jawapan J22 (b) ∠A = ∠D, AC = DE ∠C = 180° – 100° – 30° = 50° = ∠E Segi tiga ABC dan segi tiga DEF adalah kongruen kerana mematuhi sifat SudutSisi-Sudut (ASA) Triangles, ABC and DEF are congruent because they satisfy the Angle-Side-Angle (ASA) rule (c) ∠AB ≠ DE, BC ≠ EF, AC ≠ DF Segi tiga ABC dan segi tiga DEF adalah tidak kongruen kerana tidak mematuhi sifat Sisi-Sisi-Sisi (SSS) Triangles, ABC and DEF are not congruent because they do not satisfy the Side-Side-Side (SSS) rule (d) AC = DE, AB = DF, ∠A ≠ ∠D Segi tiga ABC dan segi tiga DEF adalah tidak kongruen kerana tidak mematuhi sifat Sisi-Sudut-Sisi (SAS) Triangles, ABC and DEF are not congruent because they do not satisfy the Side-Angle-Side (SAS) rule (e) ∠B = ∠E, BC = DE, ∠C = ∠D Segi tiga ABC dan segi tiga DEF adalah kongruen kerana mematuhi sifat SudutSisi-Sudut (ASA) Triangles, ABC and DEF are congruent because they satisfy the Angle-Side-Angle (ASA) rule (f) ∠A = ∠D, AC ≠ DF, ∠C = ∠F Segi tiga ABC dan segi tiga DEF adalah tidak kongruen kerana satu panjang sisi sepadan adalah tidak sama Triangles, ABC and DEF are not congruent because the corresponding sides are not the same (g) Segi tiga ABC dan segi tiga DEF adalah tidak kongruen walaupun mematuhi sifat Sisi-Sisi-Sudut (SSA) kerana luas keduadua segi tiga adalah tidak sama Triangles, ABC and DEF are not congruent although they satisfy the Side-Side-Angle (SSA) rule because the areas of both triangles are not the same 8 (a) ∠EAD = 180° – 90° – ∠AED = 90° – ∠AED ∠AED = ∠BEF, maka/ thus ∠EAD = ∠EBF ∠EBF = 180° – 90° – ∠BEF = 90° – ∠BEF Segi tiga ACF dan segi tiga BEF memenuhi sifat Sudut-Sisi-Sudut (ASA), maka keduadua segi tiga itu adalah kongruen. Oleh itu, BE = AC. Triangle ACF and triangle BEF satisfy the property of Angle-Side-Angle (ASA), thus both triangles are congruent. Therefore, BE = AC. (b) RS = TU, SU = US, UR = ST Segi tiga RSU dan segi tiga TUS adalah kongruen kerana mematuhi sifat Sisi-SisiSisi (SSS) Triangles, RSU and TUS are congruent because they satisfy the Side-Side-Side (SSS) rule Oleh sebab segi tiga RSU adalah kongruen dengan segi tiga TUS, maka semua sudut sepadan adalah sama. Since triangle RSU is congruent to triangle TUS, thus all their corresponding angles are equal. ∴ ∠SRU = ∠STU (c) (i) XZ = XY, ∠ZXW = ∠YXW, XW = XW Segi tiga XWZ dan segi tiga XWY adalah kongruen kerana mematuhi sifat SisiSudut-Sisi (SAS) Triangles, XWZ and XWY are congruent because they satisfy the Side-Angle-Side (SAS) rule (ii) ∠ZXW + ∠YXW + ∠XZY + ∠XYZ = 180° 2∠ZXW + 2∠XZY = 180° ∠ZXW + ∠XZY = 90° ∠ZXW + ∠XZY + ∠XWZ = 180° 90° + ∠XWZ = 180° ∠XWZ = 90° ∴ XW ∟ YZ 5.2 Pembesaran Enlargement 1 (a) AB PQ = AC PR = 3 4.5 = 2 3 , BC QR = 2 3 Serupa, kerana semua sudut sepadan dan nisbah sisi sepadan adalah sama Similar, because all corresponding angles and the ratios of corresponding sides are equal (b) Tidak serupa, kerana semua sudut sepadan adalah tidak sama Not similar, because all the corresponding angles are not equal (c) AB PQ = 4 10 = 2 5 , BC QR = 5 10 = 1 2 , CD RS = 8 18 = 4 9 , AD PS = 3 6 = 1 2 Tidak serupa, kerana nisbah semua sisi sepadan adalah tidak sama Not similar, because the ratios of all corresponding sides are not equal 2 (a) Q Q� k = 3 5 (1, 1) (b) R R� k = –3 3 = –1 (0, –1) (c) S S� k = –1 2 (0, –1) CONTOH

Matematik Tingkatan 5 Jawapan J23 3 (a) k = 2 4 = 1 2 x O 2 4 y 4 2 A� A (2, 5) A� ialah imej bagi A di bawah suatu pembesaran pada pusat (2, 5) dengan faktor skala 1 2 . A� is the image of A under an enlargement at centre (2, 5) with a scale factor of 1 2 . (b) k = 4 8 = 1 2 x O 2 4 y 2 A� (–6, 1) –2 A –4–6 A� ialah imej bagi A di bawah suatu pembesaran pada pusat (–6, 1) dengan faktor skala 1 2 . A� is the image of A under an enlargement at centre (–6, 1) with a scale factor of 1 2 . (c) O 2 4 2 A� k = – 1 x 2 4 (1, 3) –2 6 A –2 y A� ialah imej bagi A di bawah suatu pembesaran pada pusat (1, 3) dengan faktor skala – 1 2 . A� is the image of A under an enlargement at centre (1, 3) with a scale factor of – 1 2 . (d) k = 1 3 x y –2 O 2 4 4 2 6 A� A (–2, 0) A� ialah imej bagi A di bawah suatu pembesaran pada pusat (–2, 0) dengan faktor skala 1 3 . A� is the image of A under an enlargement at centre (–2, 0) with a scale factor of 1 3 . (e) k = –1 x y O 2 4 2 • (2, 3) A� A A� ialah imej bagi A di bawah suatu pembesaran pada pusat (2, 3) dengan faktor skala –1. A� is the image of A under an enlargement at centre (2, 3) with a scale factor of –1. 4 (a) A A� O (b) A O • A� CONTOH

Matematik Tingkatan 5 Jawapan J24 (c) • A� O A (d) A� O• A (e) • A A� O 5 (a) AO A� • (b) • O A A� (c) • A� A O (d) A� A CONTOH • O

Matematik Tingkatan 5 Jawapan J25 (e) • A� A O 6 (a) 1 2 4 mm2 4 × (1 2) 2 = 1 mm2 (b) –2 64 ÷ (–2)2 = 16 unit2 64 unit2 (c) – 1 3 3 ÷ (– 1 3) 2 = 27 cm2 3 cm2 (d) 60.5 2 = √30.25 = ±5.5 2 cm2 60.5 cm2 (e) 6 96 = 1 16 = ±1 4 96 m2 6 m2 7 (a) Luas objek/ Area of object: 1 × 2 = 2 unit2 Luas imej/ Area of image: 2 × 32 = 18 unit2 (b) Luas objek/ Area of object: 1 2 × (4 + 6) × 4 = 20 unit2 Luas imej/ Area of image: 20 × (1 2) 2 = 5 unit2 (c) Luas objek/ Area of object: 1 2 × 4 × 6 = 12 unit2 Luas imej/ Area of image: 12 × (–1)2 = 12 unit2 8 (a) (i) Faktor skala/ Scale factor: 1 2 (ii) Luas ABCD/ Area of ABCD: 40 × 22 = 160 cm2 Luas BCDGFE/ Area of BCDGFE: 160 – 40 = 120 cm2 (b) Faktor skala/ Scale factor: Luas A�/ Area of A� Luas A/ Area of A = 1 4 = ± 1 2 (c) Kolam ikan B ialah imej bagi kolam ikan A. Fish pond B is the image of fish pond A. k2 = Luas permukaan B/ Surface area of B Luas Permukaan A/ Surface area of A 42 = 256 Luas A/ Area of A Luas A/ Area of A = 256 16 = 16 m2 Beza/ Difference: 256 – 16 = 240 m2 (d) Luas kolam/ Area of the pool: 22 7 × (2.8 2 ) 2 = 6.16 m2 Faktor skala/ Scale factor: 1.54 6.16 = 1 4 = 1 2 Luas permukaan air kolam: Surface area of the pool water: 6.16 – 1.54 = 4.62 m2 (e) 2 O y 4 6 • –2 –2 2 4 x 6 8 A A� (–2, –3) Kemungkinan 1/ Possibility 1: Pembesaran pada titik (–2, –3) dengan faktor skala 1 2 Enlargement at point (–2, –3) with a scale factor of 1 CONTOH 2

Matematik Tingkatan 5 Jawapan J26 2 O 2 4 y x 6 4 6 8 A (4, 3) • A� Kemungkinan 2/ Possibility 2: Pembesaran pada titik (4, 3) dengan faktor skala – 1 2 Enlargement at point (4, 3) with a scale factor of – 1 2 5.3 Gabungan Transformasi Combined Transformation 1 (a) x p(3, –1) y O 2 –2 –4 2 4 p��(2, 0) • 6 • –2 • p�(0, 0) x = 1 (b) x y O 2 –2 –4 2 4 y = x 4 –6 G G�� G� –2 (c) x y U O –2 –4 –6 2 4 2 –2 (3, 2) U�� U� 2 (a) x y 3 2 O 1 –1 1 2 3 –1 –2 –3 –3 –2 Q(0, –1) Q�(2, –2) Q��(2, 2) (b) x y O 4 R� 2 –2 –4 –2 6 R R�� x = 1 CONTOH

Matematik Tingkatan 5 Jawapan J27 (c) S O 2 S�� (5, 8) y x 4 6 8 10 2 4 6 8 • S� •(7, 6) 3 (a) (i) A�� O –2 y x –2 2 4 –4 2 4 –4 –6 –8 A A� (ii) A�� O –2 y x –2 2 4 –4 2 4 –4 –6 –8 A� A Imej di bawah gabungan transformasi TS dan ST adalah tidak sama, maka gabungan transformasi TS dan ST tidak mematuhi sifat kalis tukar tertib. Images under the combined transformations TS and ST are not the same, therefore the combined transformations TS and ST do not satisfy the commutative law. (b) (i) –6 –2 O 2 4 y x –4 6 2 4 6 A�� A� A CONTOH

Matematik Tingkatan 5 Jawapan J28 (ii) –6 –2 O 2 4 y x –4 6 2 4 6 A�� A� A Imej di bawah gabungan transformasi UV dan VU adalah sama, maka gabungan transformasi UV dan VU mematuhi sifat kalis tukar tertib. Images under the combined transformations UV and VU are the same, therefore the combined transformations UV and VU satisfy the commutative law. (c) (i) A�� A� A –6 –2 O 2 4 y x –4 6 2 4 6 (ii) A�� A� A –6 –2 O 2 4 y x –4 6 2 4 6 Imej di bawah gabungan transformasi JK dan KJ adalah tidak sama, maka gabungan transformasi JK dan KJ tidak mematuhi sifat kalis tukar tertib. Images under the combined transformations JK and KJ are not the same, therefore the combined transformations JK and KJ do not satisfy the commutative law. 4 (a) Pantulan pada paksi-x Reflection on the x-axis Pembesaran pada titik (3, –1) dengan faktor skala 2 Enlargement at point (3, –1) with a scale factor of 2 Pembesaran pada titik (3, –5) dengan faktor skala 2 Enlargement at point (3, –5) with a scale factor of 2 (b) Pembesaran pada asalan dengan faktor skala 2 Enlargement at origin with a scale factor of 2 Putaran 180° pada asalan Rotation 180° at origin Pembesaran pada asalan dengan faktor skala –2 Enlargement at origin with a scale factor of –2 (c) Pembesaran pada asalan dengan faktor skala –1 Enlargement at origin with a scale factor of –1 Pantulan pada paksi-x Reflection on the x-axis Pantulan pada paksi-y Reflection on the y-axis 5 (a) (i) Transformasi X ialah pembesaran pada titik (–3, 3) dengan faktor skala 2 dan transformasi Y ialah putaran 180° pada titik (0, 3). Transformation X is an enlargement at point (–3, 3) with a scale factor of 2 and transformation Y is a rotation of 180° at point (0, 3). (ii) Transformasi tunggal yang setara dengan gabungan transformasi XY ialah pembesaran pada titik (1, 3) dengan faktor skala –2. The single transformation which is equivalent to combined transformation XY is an enlargement at point (1, 3) with a scale factor of –2. (iii) Biar x sebagai luas ABCD dan y sebagai luas IJKL Let x be the area of ABCD and y be the area of IJKL y = 36 + x … ① = x × (–2)2 y = 4x……… ② Gantikan ① ke dalam ② : Substitute ① into ②: 36 + x = 4x 3x = 36 x = 12 m2 CONTOH

Matematik Tingkatan 5 Jawapan J29 (b) (i) Transformasi S ialah pembesaran pada titik (4, 6) dengan faktor skala 1 2 dan transformasi T ialah putaran 90° ikut arah jam pada titik (4, 6). Transformation S is an enlargement at point (4, 6) with a scale factor of 1 2 and transformation T is a rotation of 90° clockwise at point (4, 6). (ii) Gabungan transformasi ST mematuhi sifat kalis tukar tertib kerana imej di bawah gabungan transformasi ST dan gabungan transformasi TS adalah sama. Combined transformation ST satisfies the commutative law because the images under combined transformation ST and combined transformation TS are the same. (iii) Luas EFGH/ Area of EFGH Luas ABCD/ Area of ABCD = (1 2) 2 = 1 4 ∴ Nisbah luas EFGH kepada luas ABCD ialah 1 : 4. The ratio of the area of EFGH to the area of ABCD is 1 : 4. 5.4 Teselasi Tessellation 1 (a) Suatu teselasi yang terdiri daripada suatu bentuk yang berulang A tessellation which is made up of a repeating shape (b) Bukan teselasi kerana tiada bentuk yang tetap dan berulang Not a tessellation because there is no fixed and repeating shape (c) Suatu teselasi yang terdiri daripada suatu corak yang berulang tanpa bertindih A tessellation which is made up of repeating patterns without overlapping 2 3 (a) Putaran 180° pada pusat O Rotation of 180° at centre O (b) Pantulan di garis CD/ Reflection on line CD Praktis Kendiri Kertas 1 1 B Bentuk pada rajah adalah sebuah pentagon sekata seperti di B The shape on the diagram is a regular pentagon as in B 2 B – AB = CD – BEC ialah sebuah segi tiga sama kaki, maka ∠EBC = ∠ECB, BEC is an isosceles triangle, thus ∠EBC = ∠ECB, – Segi tiga ABC dan BCD berkongsi satu tapak BC Triangles ABC and BCD share one base BC Berdasarkan sifat Sisi-Sisi-Sudut (SSA), segi tiga ABC dan BCD adalah kongruen Based on the Side-Side-Angle (SSA) rule, triangles ABC and BCD are congruent 3 D 4 B 2.5 cm 5 cm = x cm 6 cm x cm = 15 5 = 3 cm 5 C Luas ABCD/ Area of ABCD Luas AEFG/ Area of AEFG = 1.52 45 Luas AEFG/ Area of AEFG = 2.25 Luas AEFG/ Area of AEFG = 20 cm2 Luas kawasan berlorek/ Area of shaded region: 45 – 20 = 25 cm2 6 C Faktor skala/ Scale factor: – 1.1 cm 1.1 m = – 1.1 cm 110 cm = – 1 100 Jarak objek dari lubang jarum, x: Distance of object from pinhole, x: – 5 cm x = – 1 100 x = 500 cm = 5 m 7 C 8 A 9 C Kertas 2 Bahagian A/ Section A 1 (a) Transformasi T ialah putaran 90° lawan arah jam pada titik (5, 2). Transformation T is a rotation of 90° anticlockwise at point (5, 2). Transformasi S ialah pembesaran pada titik F dengan faktor skala 2. Transformation S is an enlargement at point F with a scale factor of 2. CONTOH

Matematik Tingkatan 5 Jawapan J30 (b) k2 = Luas imej/ Area of image Luas objek/ Area of object Luas ABCD / Area of ABCD = 60 22 = 15 cm2 Bahagian B/ Section B 2 (a) BD dikongsi oleh segi tiga ABD dan BCD. BD is shared by triangles ABD and BCD. AB dan AD adalah selari dengan DC dan BC masing-masing, maka ∠ADB = ∠CBD dan ∠ABD = ∠BDC. AB and AD are parallel to DC and BC respectively, hence ∠ADB = ∠CBD and ∠ABD = ∠BDC. ∴ Segi tiga ABD dan BCD mematuhi sifat Sudut-Sisi-Sudut (ASA), oleh itu AB dan AD masing-masing adalah sama panjang dengan DC dan BC. Triangles ABD and BCD satisfy the AngleSide-Angle (ASA) rule, therefore AB and AD are respectively equal to DC and BC. (b) ∠ADB = ∠CBD = 67.5° ∠BDC + ∠CBD + ∠BCD = 180° ∠BDC + 67.5° + 60° = 180° ∠BDC = 180° – 127.5° = 52.5° (c) Luas/ Area = Tinggi/ Height × AB 660 = 22 × AB AB = 30 mm 5AB = 6AD 5(30) = 150 � 6 = 25 mm Perimeter = 2AD + 2AB = 2(25) + 2(30) = 50 + 60 = 110 mm Bab 6 6.1 Nilai Sinus, Kosinus dan Tangen bagi Sudut θ, 0° N θ N 360° The Value of Sine, Cosine and Tangent for Angle θ, 0° N θ N 360° 1 (a) 101° Sukuan II Quadrant II 180° – 101° = 79° (b) 250° Sukuan III Quadrant III 250° – 180° = 70° (c) 345° Sukuan IV Quadrant IV 360° – 345° = 15° (d) 89° Sukuan I Quadrant I 89° (e) 154° Sukuan II Quadrant II 180° – 154° = 26° 2 (a) kos/ cos 300° = kos/ cos (360° – 300°) = kos/ cos 60° (b) tan 240° = tan (240° – 180°) = tan 60° (c) kos/ cos 120° = –kos/ cos (180° – 120°) = –kos/ cos 60° (d) sin 280° = –sin (360° – 280°) = –sin 80° (e) tan 210° = tan (210° – 180°) = tan 30° (f) kos/ cos 158° = –kos/ cos (180° – 158°) = –kos/ cos 22° (g) sin 95° = sin (180° – 95°) = sin 85° 3 (a) sin θ = 0.9659 kos/ cos θ = –0.2588 tan θ = 0.9659 –0.2588 = –3.7322 (b) sin θ = –0.5736 kos/ cos θ = –0.8192 tan θ = –0.5736 –0.8192 = 0.7 (c) sin θ = –0.9659 kos/ cos θ = 0.2588 tan θ = –0.9659 0.2588 = –3.7322 (d) sin θ = 0.5 kos/ cos θ = –0.866 tan θ = 0.5 –0.866 = –0.577 (e) sin θ = 0.766 kos/ cos θ = –0.6428 tan θ = 0.766 –0.6428 = –1.1917 4 (a) sin 359° = –sin (360° – 359°) = –sin 1° = –0.017 (b) tan 234° 21ʹ = tan (234° 21ʹ – 180°) = tan 54° 21ʹ = 1.394 (c) sin 115° = sin (180° – 115°) = sin 65° = 0.906 (d) tan 165° = –tan (180° – 165°) = –tan 15° = –0.268 (e) kos/ cos 325° = kos/ cos (360° – 325°) = kos/ cos 35° = 0.819 5 (a) sin 135° = sin (180° – 135°) = sin 45° = 1 2 CONTOH

Matematik Tingkatan 5 Jawapan J31 (b) kos/ cos 315° = kos/ cos (360° – 315°) = kos/ cos 45° = 1 2 (c) tan 120° = –tan(180° – 120°) = –tan 60° = – 3 (d) sin 240° = –sin (240° – 180°) = –sin 60° = – 3 2 (e) kos/ cos 300° = kos/ cos (360° – 300°) = kos/ cos 60° = 1 2 (f) tan 225° = tan (225° – 180°) = tan 45° = 1 (g) sin 330° = –sin (360° – 330°) = –sin 30° = –1 2 (h) kos/ cos 150° = –kos/ cos (180° – 150°) = –kos/ cos 30° = – 3 2 6 (a) tan–1 5.6713 = 80° θ = 80° atau/ or (180° + 80°) = 80° atau/ or 260° (b) tan–1 0.1228 = 7° θ = (180° – 7°) atau/ or (360° – 7°) = 173° atau/ or 353° (c) kos–1/ cos–1 0.7193 = 44° θ = (180° – 44°) atau/ or (180° + 44°) = 136° atau/ or 224° (d) sin–1 0.9063 = 65° θ = 65° atau/ or (180° – 65°) = 65° atau/ or 115° (e) kos–1/ cos–1 0.9659 = 15° θ = 15° atau/ or (360° – 15°) = 15° atau/ or 345° (f) sin–1 0.1736 = 10° θ = (180° + 10°) atau/ or (360° – 10°) = 190° atau/ or 350° (g) kos–1/ cos–1 0 = 90° θ = 90° atau/ or (360° – 90°) = 90° atau/ or 270° 7 (a) AB AC = 1 1 + 1.5 AB 20 cm = 1 2.5 AB = 8 cm BD2 = AB2 + AD2 = 82 + 62 = 100 BD = 100 = 10 cm (i) sin z = 6 10 = 3 5 (ii) kos/ cos z = 8 10 = 4 5 kos/ cos x = –kos/ cos z = – 4 5 (iii) 270° < y < 360°, maka nilai tan y adalah negatif. 270° < y < 360°, thus the value of tan y is negative. tan y = – 8 6 = – 4 3 (b) 4 cm 4 cm P Q R S 5 cm QR2 = PR2 – PQ2 = 52 – 42 = 9 QR = 9 = 3 cm (i) tan ∠PRQ = 4 3 (ii) PS2 = PQ2 + QS2 PS = 65 = 42 + (3 + 4)2 = 8.06 cm = 65 kos/ cos ∠QSP = 7 8.06 (iii) kos–1/ cos–1 7 8.06 = 29.72° atau/ or 29° 43ʹ CONTOH

Matematik Tingkatan 5 Jawapan J32 (c) (i) tan θ = sin θ kos/cos θ 0.5774 = 0.5 kos/cos θ kos/ cos θ = 0.5 0.5774 = 0.8660 (ii) (0.8660, 0.5) y x O θ (iii) Sukuan Quadrant Nilai kos θ The value of cos θ II –0.8660 III –0.8660 IV 0.8660 6.2 Graf Fungsi Sinus, Kosinus dan Tangen The Graphs of Sine, Cosine and Tangent Functions 1 (a) Graf kosinus/ Cosine graph (b) Graf tangen/ Tangent graph (c) Graf sinus/ Sine graph 2 (a) 0 1 –1 90° 180° y = kos/ cos x y x (b) 90° 180° 270° y x 0 y = tan x (c) 0 1 30° 90° –0.5 0.5 –1 180° 270° 330° y = sin x y x 3 (a) 90°, 270° (e) 1 (b) 0°, 180°, 360° (f) ∞ (c) 0 (g) –1 (d) 1 4 (a) y = 2 kos/ cos x 2 360° (b) y = tan 1 3 x 1 540° (c) y = 2 kos/ cos 2x – 4 2 180° (d) y = 1 7 tan 3x + 1 1 7 60° (e) y = 4 sin x 4 360° (f) y = 2 sin 2x 3 2 540° (g) y = kos/ cos 2x + 3 1 180° 5 (a) p = 180° 20° + q = 90° q = 70° (b) p = 270° p + q = 360° 270° + q = 360° q = 90° (c) q = 270° q – p = 180° 270° – p = 180° p = 90° 6 (a) y = 3 sin x y x –3 3 90° 180° 270° 360° Amplitud telah menjadi 3 The amplitude has become 3 (b) y = sin x + 2 y x 3 2 1 0 90° 180° 270° 360° Graf telah anjak 2 unit ke atas The graph has shifted up by 2 units CONTOH

Matematik Tingkatan 5 Jawapan J33 (c) y = 1 2 sin x + 3 y x 3.5 2.5 0 90° 180° 270° 360° 3 Graf telah anjak 3 unit ke atas dan amplitud telah menjadi 1 2 The graph has shifted up by 3 units and the amplitude has become 1 2 7 (a) y = kos/ cos x – 1 0 180° 360° –2 y x Graf telah anjak 1 unit ke bawah The graph shifted down by 1 unit (b) y = kos/ cos 2x y x 1 0 –1 45° 90° 135° 180° 225° 270° 315° 360° Tempoh fungsi telah menjadi 180° The period of the function has become 180° (c) y = 3 kos/ cos 2x y 3 0 45° 90° 135° 180° 225° 270° 315° 360° –3 x Amplitud telah menjadi 3 dan tempoh fungsi telah menjadi 180° The amplitude has become 3 and the period of the function has become 180° 8 (a) y = tan 2x y 0° 45° 90° 135° 180° 225° 270° 315° 360° x Tempoh fungsi telah menjadi 90° The period of the function has become 90° (b) y = 4 tan x y x 0° 90° 180° 270° 360° Tiada perubahan kerana fungsi tangen tiada amplitud There is no change because tangent function does not have amplitude (c) y x y = 2 tan 2x – 2 –2 0° 45° 90° 135°180° 225° 270° 315° 360° Graf telah anjak 2 unit ke bawah dan tempoh fungsi telah menjadi 90° The graph has shifted down by 2 units and the period of the function has become 90° 9 (a) Daripada graf/ From the graph, graf telah anjak 1 unit ke bawah, maka c = –1; the graph has moved down by 1 unit, hence c = –1; Amplitud fungsi ialah 2, maka a = 2 The amplitude of the function is 2, hence a = 2 tempoh fungsi ialah 180°, maka the period of the function is 180°, hence 360° b = 180° b = 2 CONTOH

Matematik Tingkatan 5 Jawapan J34 ∴ Bentuk graf merupakan graf kosinus, maka fungsi graf ialah y = 2 kos 2x – 1. The shape of the graph shows a cosine graph; hence the function of the graph is y = 2 cos 2x – 1. (b) (i) Berdasarkan bentuk graf, graf mewakili fungsi sinus, maka h = a sin bt + c. Based on the shape of the graph, the graph represents a sine function, hence h = a sin bt + c. Daripada graf, amplitud adalah 1, graf menunjukkan dua tempoh (b = 2) dan graf telah anjak 4 unit ke atas (c = +4). From the graph, the amplitude is 1, the graph shows two periods (b = 2) and has moved up by 4 units (c = +4). ∴ Fungsi yang memodelkan aras air ialah h = sin 2t + 4. The function that models the water level is h = sin 2t + 4. (ii) Pada pukul 3 pagi, aras air adalah 3 m. At 3 a.m., the water level is 3 m. (iii) Beza/ Difference: 5 – 3 = 2 m Praktis Kendiri Kertas 1 1 B 2 B 3 C Sukuan II/ Quadrant II: 180° ‒ 50° = 130° Sukuan III/ Quadrant III: 180° + 50° = 230° Sukuan IV/ Quadrant IV: 360° ‒ 50° = 310° 4 A tan 320° = ‒tan (360° ‒ 320°) = ‒ tan 40° = ‒0.8391 5 D A: sin‒1 0.8661 = 60° B: kos‒1/ cos–1 0.5 = 60° C: tan‒1 1.7321 = 60° D: sin‒1 0.9848 = 80° 6 A Apabila/ When y = 0, 2 kos/ cos x = 0 kos/ cos x = 0 kos‒1/ cos–1 0 = 90° atau/ or (360° ‒ 90°) = 90° atau/ or 270° 7 A sin θ = 0.8984, kos/ cos θ = 0.4391 tan θ = 0.8984 0.4391 = 2.046 8 C Pada sukuan IV/ At quadrant IV, 360° ‒ 45° = 315° 9 C kos/ cos θ = ‒0.9457 kos‒1/ cos–1 0.9457 = 18° 58′ sin 18° 58′ = 0.325 Sebab titik berada di sukuan III, maka Because the point is at quadrant III, hence sin (180° + 18° 58′) = sin 198° 58′ = ‒0.325 10 B a = 2, maka amplitud graf fungsi ialah 2. a = 2, hence the amplitude of the graph of function is 2. Kertas 2 Bahagian A/ Section A 1 (a) PQ = TS = 3 mm tan ÐRTS = RS TS tan 53.13° = RS 3 RS = 4 mm QR = RS = 4 mm ⸫ QS = 8 mm (b) ÐPRS = ÐQRT tan ÐQRT = ‒ 3 4 ÐQRT = ‒tan‒1 3 4 = 180° ‒ 36.87° = 143.13° Bahagian B/ Section B 2 (a) (i), (ii) (b) (i) y 2 1 0 –1 y = sin 2x + 1 x 45° 90° 135° 180° 225° 270° 315° 360° y = sin 2x Tempoh/ Period: 360° 2 = 180° ∴ Amplitud bagi y = sin 2x ialah 1 dan tempoh fungsi ialah 180°. The amplitude for y = sin 2x is 1 and the period of the function is 180°. (ii) c = +1, maka fungsi baharu ialah y = sin 2x + 1. c = +1, hence the new function is y = sin 2x + 1. CONTOH

Matematik Tingkatan 5 Jawapan J35 3 (a), (b)(i) y 0 –2 45° 90° 135° 180° 225° 270° 315° 360° x y = tan 2x y = tan 2x – 2 Tempoh graf ialah/ The period of the graph is 180° 2 = 90°. Daripada graf, amplitud bagi y = tan 2x adalah tidak tertakrif dan tempoh fungsi ialah 90°. From the graph, the amplitude of y = tan 2x is undefined and the period of the function is 90°. (b) (ii) c = –2 apabila graf anjak 2 unit ke bawah, maka fungsi bagi graf baharu ialah y = tan 2x – 2. c = –2 when the graph moves down by 2 units, hence the function of the new graph is y = tan 2x – 2. Bab 7 7.1 Serakan Dispersion 1 Saiz selang kelas/ Size of class intervals: 105 ‒ 81 5 = 4.8 (≈ 5) Jisim (g) Mass (g) Kekerapan Frequency Had bawah Lower limit Had atas Upper limit Titik tengah Midpoint Sempadan bawah Lower boundary Sempadan atas Upper boundary 81 – 85 1 81 85 83 80.5 85.5 86 – 90 2 86 90 88 85.5 90.5 91 – 95 4 91 95 93 90.5 95.5 96 – 100 8 96 100 98 95.5 100.5 101 – 105 5 101 105 103 100.5 105.5 2 (a) Kekerapan longgokan Cumulative frequency 3 7 19 31 32 33 (b) Kekerapan longgokan Cumulative frequency 13 23 33 37 39 40 (c) Kekerapan longgokan Cumulative frequency 1 11 19 26 29 35 3 (a) Titik tengah Midpoint Sempadan bawah Lower boundary Sempadan atas Upper boundary 44.5 39.5 49.5 54.5 49.5 59.5 64.5 59.5 69.5 74.5 69.5 79.5 84.5 79.5 89.5 CONTOH

Matematik Tingkatan 5 Jawapan J36 0 5 10 15 34.5 44.5 54.5 64.5 74.5 84.5 94.5 Bilangan murid/ Number of students Jisim (g) Mass (g) (b) Titik tengah Midpoint Sempadan bawah Lower boundary Sempadan atas Upper boundary 148 145.5 150.5 153 150.5 155.5 158 155.5 160.5 163 160.5 165.5 168 165.5 170.5 0 5 10 15 145.5 150.5 155.5 160.5 165.5 170.5 Bilangan murid/ Number of students Tinggi (cm) Height (cm) 4 (a) (i) Bentuk taburan histogram bagi markah ujian Kelas 5P1 ialah bentuk loceng manakala bagi markah ujian Kelas 5S1 ialah pencong ke kiri. The histogram for the test marks by Class 5P1 shows a bell-shaped distribution whereas the distribution shape of the histogram for the test marks by Class 5S1 is skewed to the left. (ii) Markah ujian yang diperoleh muridmurid di Kelas 5P1 mempunyai serakan yang lebih luas kerana beza markah yang diperoleh adalah lebih besar, iaitu 80 markah (90.5 – 10.5). The test marks obtained by the students in Class 5P1 has a wider dispersion because the difference in the marks obtained is larger, which is 80 marks (90.5 – 10.5). (iii) Kelas 5S1 menunjukkan keputusan yang lebih cemerlang kerana kebanyakan markah murid adalah lebih tinggi. Class 5S1 shows a more excellent result because most of the student’s marks are higher. (b) (i) Bentuk taburan poligon kekerapan bagi harga jualan rumah di kawasan A ialah pencong ke kanan manakala bagi harga jualan rumah di kawasan B ialah pencong ke kiri. The distribution shape of frequency polygon for the house selling price in area A is skewed to the right whereas for the house selling price in area B is skewed to the left. (ii) Harga jualan bagi rumah di kawasan A dan kawasan B mempunyai serakan yang lebih kurang sama. The selling price of the houses in area A and area B has approximately the same dispersion. (iii) Kawasan A merupakan kawasan luar bandar kerana harga jualan rumah adalah lebih rendah manakala kawasan B merupakan kawasan bandar kerana harga jualan rumah adalah lebih tinggi. Area A is a rural area because the selling price of houses is lower whereas area B is an urban area because the selling price of houses is higher. CONTOH

Matematik Tingkatan 5 Jawapan J37 5 (a) (i) Pengambilan gula harian (g) Daily sugar intake (g) 1 – 10 11 – 20 21 – 30 31 – 40 41 – 50 51 – 60 61 – 70 71 – 80 Kekerapan/ Frequency 0 1 1 5 7 3 2 1 Sempadan atas Upper boundary 10.5 20.5 30.5 40.5 50.5 60.5 70.5 80.5 Kekerapan longgokan Cumulative frequency 0 1 2 7 14 17 19 20 0 14 16 18 Kekerapan longgokan Cumulative frequency 20 12 10 8 6 Pengambilan gula harian (g)/ Daily sugar intake (g) 10.5 20.5 30.5 40.5 50.5 60.5 70.5 80.5 4 2 Q1 Q2 Q3 (ii) (a) kuartil pertama = 37 first quartile (b) median = 45 median (c) kuartil ketiga = 53 third quartile (b) (i) Jisim betik (g) Mass of papaya(g) Kekerapan Frequency Sempadan atas Upper boundary Kekerapan longgokan Cumulative frequency 401 – 500 0 500.5 0 501 – 600 7 600.5 7 601 – 700 12 700.5 19 701 – 800 17 800.5 36 801 – 900 14 900.5 50 901 – 1 000 CONTOH 10 1 000.5 60

Matematik Tingkatan 5 Jawapan J38 0 30 40 50 Kekerapan longgokan Cumulative frequency 60 20 10 500.5 600.5 700.5 800.5 900.5 1 000.5 Q1 Q2 Q3 Jisim (g) Mass (g) (ii) (a) kuartil pertama = 670.5 first quartile (b) median = 765.5 median (c) kuartil ketiga = 860.5 third quartile (c) (i) Kandungan karbon/ Carbon content (%) 3.1 – 4.0 4.1 – 5.0 5.1 – 6.0 6.1 – 7.0 7.1 – 8.0 Kekerapan/ Frequency 8 12 23 5 2 Sempadan bawah/ Lower boundary 3.05 4.05 5.05 6.05 7.05 Sempadan atas/ Upper boundary 4.05 5.05 6.05 7.05 8.05 Kekerapan longgokan/ Cumulative frequency 8 20 43 48 50 (ii) (a) kuartil pertama first quartile = 4.5 (b) median = 5.25 median (c) kuartil ketiga third quartile = 5.75 0 30 40 50 Kekerapan longgokan Cumulative frequency 20 10 3.05 4.05 5.05 6.05 7.05 8.05 Q1 Q2 Q3 Kandungan karbon (%) CONTOH Carbon content (%)

Matematik Tingkatan 5 Jawapan J39 6 (a) 0 50 Kekerapan longgokan Cumulative frequency 40 30 20 10 Skor Scores 20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5 100.5 74.5 59 31.5 (i) 15% daripada jumlah kekerapan/ 15% of the total frequency = 15 100 � 40 = 6 Daripada ogif/ From the ogive, P15 = 40.5 50% daripada jumlah kekerapan/ 50% of the total frequency = 50 100 � 40 = 20 Daripada ogif/ From the ogive, P50 = 59 (ii) 85% daripada jumlah kekerapan/ 85% of the total frequency = 85 100 � 40 = 34 Daripada ogif/ From the ogive, P85 = 74.5 ∴ Skor minimum untuk mendapat sebuah hadiah ialah 74.5. The minimum scores to get a present is 74.5. (iii) Peratusan murid yang memperoleh skor sekurang-kurangnya 70: The percentage of students who scored at least 70: 40 – 31.5 40 � 100% = 21.25% CONTOH

Matematik Tingkatan 5 Jawapan J40 (b) 0 Kekerapan longgokan Cumulative frequency 40 30 20 10 Gaji (RM) Salary (RM) 1 000.5 1 500.5 2 000.5 2 500.5 3 000.5 3 500.5 4 000.5 4 500.5 2 575.5 1 925.5 17 (i) 25% daripada jumlah kekerapan: 25% of the total frequency: 25 100 × 36 = 9 Daripada ogif/ From the ogive, P25 = RM1 925.50 50% daripada jumlah kekerapan: 50% of the total frequency: 50 100 × 36 = 18 Daripada ogif/ From the ogive, P50 = RM2 575.50 (ii) Peratusan pekerja yang mendapat gaji kurang daripada RM2 500: The percentage of workers who have salary below RM2 500: 17 36 � 100% = 47.22% 7.2 Sukatan Serakan Measures of Dispersion 1 (a) 0 10 20 30 40 Kekerapan longgokan Cumulative frequency 35.5 40.5 45.5 50.5 55.5 60.5 Jisim/ Mass (kg) 44.5 50.5 Q1 Q3 Julat/ Range = 55.5 + 60.5 2 – 35.5 + 40.5 2 = 58 – 38 = 20 Julat antara kuartil/ Interquartile range: 50.5 – 44.5 = 6 (b) 0 20 40 60 80 Kekerapan longgokan Cumulative frequency 0.5 3.5 6.5 9.5 12.5 15.5 Bilangan buku/ Number of books 5.75 9.35 Q1 Q3 Julat/ Range = 12.5 + 15.5 2 – 0.5 + 3.5 2 = 14 – 2 = 12 Julat antara kuartil/ Interquartile range: CONTOH 9.35 – 5.75 = 3.6

Matematik Tingkatan 5 Jawapan J41 (c) 0 20 40 60 80 Kekerapan longgokan Cumulative frequency 4.5 9.5 14.5 19.5 24.5 29.5 Jisim tin aluminium yang dikumpulkan (kg) The mass of aluminium cans collected (kg) 20.5 28.0 Q1 Q3 100 34.5 39.5 Julat/ Range = 34.5 + 39.5 2 – 4.5 + 9.5 2 = 37 – 7 = 30 Julat antara kuartil/ Interquartile range: 28.0 – 20.5 = 7.5 2 (a) Bilangan murid, f The number of students, f Titik tengah, x Midpoint, x fx x 2 fx 2 5 5.5 27.5 30.25 151.25 14 15.5 217 240.25 3 363.5 8 25.5 204 650.25 5 202 6 35.5 213 1 260.25 7 561.5 7 45.5 318.5 2 070.25 14 491.75 ∑f = 40 ∑fx = 980 ∑fx2 = 30 770 Min/ Mean, x̅: 980 40 = 24.5 Varians/ Variance: 30 770 40 – (24.5)2 = 769.25 ‒ 600.25 = 169 Sisihan piawai/ Standard deviation: 169 = 13 (b) Bilangan murid, f The number of students, f Titik tengah, x Midpoint, x fx x 2 fx 2 3 13 39 169 507 5 18 90 324 1 620 6 23 138 529 3 174 8 28 224 784 6 272 8 33 264 1 089 8 712 ∑f = 30 ∑fx = 755 ∑fx2 = 20 285 Min/ Mean, x̅: 755 30 = RM25.17 Varians/ Variance: 20 285 30 – 755 30 2 = 676.17 ‒ 633.36 = RM42.81 Sisihan piawai/ Standard deviation: 42.81 = RM6.54 3 0 20 30 40 Kekerapan longgokan Cumulative frequency 10 2 4 6 8 Masa (min) Time (min) Q1 Q2 Q3 5.8 4.9 Nilai minimum/ Minimum value: 2 Nilai maksimum/ Maximum value: 8 Kedudukan/ Position of Q1 : 1 4 � 40 = 10 Kedudukan/ Position of Q2 : 1 2 � 40 = 20 Kedudukan/ Position of Q3 : 3 4 � 40 = 30 2 4 6 8 Masa (min)/ Time (min) Data ini mempunyai taburan pencong ke kanan kerana bahagian kanan plot kotak lebih besar daripada bahagian kiri plot kotak. The distribution of the data is skewed to the right because the right half of the box plot is longer than the left half of the box plot. CONTOH

Matematik Tingkatan 5 Jawapan J42 4 Bagi kategori A/ For category A, Bilangan soalan Number of questions Kekerapan, f Frequency, f Titik tengah, x Midpoint, x fx x 2 fx 2 1 – 2 5 1.5 7.5 2.25 11.25 3 – 4 7 3.5 24.5 12.25 85.75 5 – 6 5 5.5 27.5 30.25 151.25 7 – 8 4 7.5 30 56.25 225 9 – 10 9 9.5 85.5 90.25 812.25 ∑f = 30 ∑fx = 175 ∑fx2 = 1 285.5 Min/ Mean: Varians/ Variance: Sisihan piawai: 175 30 = 5.8333 1 285.5 30 – 175 30 2 = 8.8222 Standard deviation: 8.8222 = 2.9702 Bagi kategori B/ For category B, Bilangan soalan Number of questions Kekerapan, f Frequency, f Titik tengah, x Midpoint, x fx x 2 fx 2 1 – 2 2 1.5 3 2.25 4.5 3 – 4 3 3.5 10.5 12.25 36.75 5 – 6 5 5.5 27.5 30.25 151.25 7 – 8 12 7.5 90 56.25 675 9 – 10 8 9.5 76 90.25 722 ∑f = 30 ∑fx = 207 ∑fx2 = 1 589.5 Min/ Mean: Varians/ Variance: Sisihan piawai: 207 30 = 6.9 1 589.5 30 – (6.9)2 = 5.3733 Standard deviation: 5.3733 = 2.3180 Murid daripada kategori B mempunyai prestasi yang lebih baik kerana minnya adalah lebih tinggi daripada kategori A (6.9 > 5.8333). Students from category B has a better performance because the mean is higher than category A (6.9 > 5.8333). 5 (a) (i) Masa yang diambil untuk mengulang kaji (minit) Time taken to do revision (minutes) Kekerapan, f Frequency, f Titik tengah, x Midpoint, x fx x 2 fx 2 60 – 119 18 89.5 1 611 8 010.25 144 184.5 120 – 179 28 149.5 4 186 22 350.25 625 807 180 – 239 34 209.5 7 123 43 890.25 1 492 268.5 240 – 299 20 269.5 5 390 72 630.25 1 452 605 ∑f = 100 ∑fx = 18 310 ∑fx2 = 3 714 865 (ii) Min/ Mean: Varians/ Variance: Sisihan piawai: 18 310 100 = 183.1 3 714 865 100 – (183.1)2 = 3 623.04 Standard deviation: 3 623.04 = 60.1917 CONTOH

Matematik Tingkatan 5 Jawapan J43 (b) (i) Jisim (kg) Mass (kg) Kekerapan, f Frequency, f Titik tengah, x Midpoint, x fx x 2 fx 2 40 – 49 5 44.5 222.5 1 980.25 9 901.25 50 – 59 11 54.5 599.5 2 970.25 32 672.75 60 – 69 22 64.5 1 419 4 160.25 91 525.5 70 – 79 18 74.5 1 341 5 550.25 99 904.5 80 – 89 4 84.5 338 7 140.25 28 561 ∑f = 60 ∑fx = 3 920 ∑fx2 = 262 565 (ii) Min/ Mean: Varians/ Variance: Sisihan piawai: 3 920 60 = 65.3333 262 565 60 – 3 920 60 2 = 107.6389 Standard deviation: 107.6389= 10.37 (c) (i) Bermain permainan komputer Play computer games Mengulang kaji Do revision Sempadan atas Upper limit Kekerapan Frequency Kekerapan longgokan Cumulative frequency Kekerapan Frequency Kekerapan longgokan Cumulative frequency 3.5 0 0 – – 8.5 3 3 0 0 13.5 7 10 6 6 18.5 15 25 24 30 23.5 10 35 8 38 28.5 4 39 2 40 33.5 1 40 (ii) 0 Bilangan murid Number of students 40 30 Masa (jam) 3.5 8.5 13.5 18.5 23.5 28.5 33.5 Time (hours) 20 10 Mengulang kaji Do revision Bermain permainan komputer Playing computer games 20.75 Murid-murid di Kelas 5 Cempaka lebih cenderung bermain permainan komputer daripada mengulang kaji. The students in Class 5 Cempaka is more inclined on playing computer games than do revision. CONTOH

Matematik Tingkatan 5 Jawapan J44 Praktis Kendiri Kertas 1 1 C Kekerapan longgokan/ Cumulative frequency: 4 + 5 + 6 + 9 = 24 2 B Saiz selang kelas/ Size of class interval: 134.5 – 129.5 = 5 3 D 4 D 5 D Julat antara kuartil/ Interquartile range: 5.8 – 3.5 = 2.3 kg 6 C Titik tengah, x Midpoint, x Kekerapan, f Frequency, f fx 0.8 4 3.2 1.3 13 16.9 1.8 5 9 2.3 2 4.6 2.8 1 2.8 Σf = 25 Σfx = 36.5 Min/ Mean: Σfx Σf = 36.5 25 = 1.46 7 C σ2 = Σfx2 Σf ‒ x2 1.28282 = 1491.5 30 ‒ x2 x2 = 49.7167 ‒ 1.6456 x ̅= 48.0710 = 6.93 Kertas 2 Bahagian A/ Section A 1 (a) Titik tengah Midpoint fA fAx fB fBx 0.2 4 0.8 10 2 0.7 6 4.2 15 10.5 1.2 8 9.6 13 15.6 1.7 15 25.5 3 5.1 2.2 8 17.6 6 13.2 2.7 9 24.3 3 8.1 ∑ 50 82 50 54.5 x̅ A = 82 50 = 1.64 tahun/ years x̅ B = 54.5 50 = 1.09 tahun/ years (b) Beza/ Difference: 1.64 – 1.09 = 0.55 tahun/ years (c) Bateri A boleh tahan lebih lama kerana min jangka hayat adalah lebih panjang daripada bateri B. Battery A can last longer because the mean of lifespan is longer than battery B. Bahagian B/ Section B 2 (a) Kekerapan, f Frequency, f Titik tengah, x Midpoint, x fx x 2 fx 2 Kekerapan longgokan Cumulative frequency 3 45.5 136.5 2 070.25 6 210.75 3 8 55.5 444 3 080.25 24 642 11 6 65.5 393 4 290.25 25 741.5 17 6 75.5 453 5 700.25 34 201.5 23 6 85.5 513 7 310.25 43 861.5 29 3 95.5 286.5 9 120.25 27 360.75 32 ∑f = 32 ∑fx = 2 226 ∑fx2 = 162 018 (b) Min/ Mean: Varians/ Variance: Sisihan piawai/ Standard deviation: 2 226 32 = 69.5625 162 018 32 – (69.5625)2 224.1211= 14.9707 = 224.1211 CONTOH

Matematik Tingkatan 5 Jawapan J45 (c) (i) 0 35 Kekerapan longgokan Cumulative frequency 30 25 20 15 Markah/ Marks 40.5 50.5 60.5 70.5 80.5 90.5 100.5 10 5 ≈ 20 (ii) Daripada ogif, bilangan murid yang memperoleh lebih daripada 75 markah ialah 12 orang (32 – 20). From the ogive, the number of students who obtained more than 75 marks is 12 (32 – 20). Peratusan/ Percentage: 12 32 � 100% = 37.5% 3 (a) Bilangan orang, f Number of people, f Titik tengah, x Midpoint, x fx x 2 fx 2 Kekerapan longgokan Cumulative frequency 7 5 35 25 175 7 5 8 40 64 320 12 4 11 44 121 484 16 3 14 42 196 588 19 1 17 17 289 289 20 ∑f = 20 ∑fx = 178 ∑fx2 CONTOH = 1 856

Matematik Tingkatan 5 Jawapan J46 (b), (c) 0 20 Kekerapan longgokan Cumulative frequency 15 10 5 Bilangan telur (biji) Number of eggs 3.5 6.5 9.5 12.5 15.5 18.5 Q1 Q2 Q3 Data ini mempunyai taburan pencong ke kanan kerana bahagian kanan plot kotak lebih besar daripada bahagian kiri plot kotak. The distribution of the data is skewed to the right because the right half of the box plot is longer than the left half of the box plot. Bab 8 8.1 Pemodelan Matematik Mathematical Modelling 1 (a) Masalah/ Problem: Kita mengetahui amaun prinsipal dan kadar faedah. Faedah mesti dimasukkan ke dalam prinsipal Jason untuk membayar wang pendahuluan sebuah rumah. Kita perlu mencari tempoh simpanan Jason di dalam bank. We know about the principal’s amount and the interest rate. The interest must be included into Jason’s principal to pay for the down payment of a house. We need to determine how long Jason needs to keep his saving in the bank. Andaian/ Assumptions: • Kadar faedah tidak berubah sepanjang tempoh simpanan. The interest rate does not change during the period of saving. • Harga rumah tidak berubah sehingga Jason mempunyai wang yang mencukupi untuk membayar wang pendahuluan rumah. The price of the house does not change until Jason has enough money to pay for the down payment of the house. Pemboleh ubah: Faedah, prinsipal sebanyak RM50 000, kadar faedah sebanyak 1.6% dan masa Variables: Interest, principal of RM50 000, interest rate of 1.6% and time (b) Masalah/ Problem: Kita mengetahui kelajuan arus sungai dan tempoh perjalanan antara H dan K. Kita perlu mencari kelajuan sampan di atas air tenang. We know about the speed of the river current and the travel time between H and K. We need to find the speed of the canoe on calm water. Andaian/ Assumptions: • Kelajuan arus sungai dan kelajuan sampan tidak berubah sepanjang masa. The speed of the river current and the speed of the canoe do not change throughout the time. • Kesan geseran antara permukaan sampan dengan air sungai dan rintangan angin melawan sampan diabaikan. The effect of friction between the surface of the canoe and the river as well as the wind resistance against the canoe are neglected. CONTOH

Matematik Tingkatan 5 Jawapan J47 Pemboleh ubah: v untuk kelajuan sampan, t untuk tempoh perjalanan dan d untuk jarak antara H dengan K Variables: v for the speed of the canoe, t for travel time and d for the distance between H and K (c) Masalah/ Problem: Kita mengetahui prinsipal dan kadar faedah. Harga telefon baharu ialah prinsipal simpanan termasuk faedah. Kita perlu mencari tempoh simpanan Jamie di bank. We know about the principal and the interest rate. The price of the new phone is the saving’s principal including interest. We need to determine the duration of Jamie’s saving in the bank. Andaian/ Assumptions: • Kadar faedah tidak akan berubah sepanjang tempoh simpanan. The interest rate does not change during the period of saving. • Harga telefon tidak berubah sehingga Jamie berjaya mengumpul amaun wang yang diperlukan. The price of the phone does not change until Jamie has enough money to buy it. Pemboleh ubah: Faedah, prinsipal sebanyak RM3 000, kadar faedah sebanyak 1.3% dan masa Variables: Interest, principal of RM3 000, interest rate of 1.3% and time 2 (a) Masalah/ Problem: • Tentukan bilangan soalan yang boleh diselesaikan dalam masa 235 minit. Determine the number of questions which can be solved in 235 minutes. • Diketahui bahawa semakin banyak masa diambil, semakin banyak soalan Matematik boleh diselesaikan. Oleh itu, bilangan soalan berubah secara langsung dengan masa. We know that the more time taken, the more Mathematics questions can be solved. Thus, the number of questions varies directly as the time. Andaian/ Assumptions: • Andaikan tahap kesukaran bagi semua soalan Matematik adalah sama Assume that the difficulty level for all Mathematics questions is the same • Charles tidak mempunyai masalah dalam penyelesaian soalan Matematik Charles does not have any problems in solving Mathematics questions Pemboleh ubah/ Variables: • Katakan t ialah masa yang diambil, dalam minit dan n ialah bilangan soalan Matematik Let t be the time taken, in minutes and n be the number of Mathematics questions • n berubah secara langsung dengan t, maka n = kt dengan keadaan k ialah pemalar n varies directly as t, hence n = kt where k is a constant Mengaplikasi matematik Applying mathematics: Gantikan t = 120 dan n = 24 ke dalam n = kt, Substitute t = 120 and n = 24 into n = kt, 24 = k(120) k = 24 120 = 1 5 Maka/ Hence, n = 1 5 t Persamaan ini menghuraikan hubungan antara bilangan soalan Matematik yang dapat diselesaikan dengan masa. This equation describes the relationship between the number of Mathematics questions which can be solved with the time. Apabila/ When t = 235, n = 1 5 × 235 = 47 soalan/ questions Maka, 47 soalan Matematik boleh diselesaikan oleh Charles dalam 235 minit. Hence, 47 Mathematics questions can be solved by Charles in 235 minutes. Menentusahkan dan mentafsir penyelesaian dalam konteks masalah berkenaan Verifying and interpreting solutions in the context of the problem • Model fungsi linear n = 1 5 t yang diperoleh mungkin tidak dapat digunakan dalam semua situasi penyelesaian soalan kerana tahap kesukaran soalan Matematik adalah tidak sama. The linear function model n = 1 5 t obtained may not be used in all question solving situations because the difficulty levels of Mathematics questions are not the same. n O t n = 1 5 CONTOH t

Matematik Tingkatan 5 Jawapan J48 • Charles mungkin perlu mengambil lebih masa untuk memahami dan menyelesaikan soalan Matematik yang lebih sukar dan dia mungkin mengambil masa yang lebih singkat untuk menyelesaikan soalan Matematik yang lebih mudah. Charles may need to take more time to understand and solving a more difficult Mathematics question and he may take a shorter time to solve an easier Mathematics question. • Apabila diterjemah kembali ke dunia sebenar, model fungsi linear yang diperoleh tidak sesuai digunakan untuk menangani masalah berkenaan. When this is reflected to the real-world situation, the linear function model obtained is not suitable to solve this problem. Memurnikan model matematik: Refining the mathematical model: • Dalam masalah ini, kita tidak dapat memurnikan model memandangkan maklumat yang diberi adalah terhad. In this problem, we are not able to refine the model due to limited information given. (b) Masalah/ Problem: • Bagaimana menentukan kedalaman kolam pada 7.5 m dari tengah kolam How to determine the depth of the pond at 7.5 m away from the centre of the pond Andaian dan pemboleh ubah: Assumptions and variables: • Andaikan bentuk kolam ialah hemisfera. Assume that the shape of the pond is hemispherical. • Kolam adalah paling dalam di bahagian tengah The pond is the deepest in the centre. • Pemboleh ubah yang terlibat dalam kajian ini ialah kedalaman kolam, y dan jarak dari tengah kolam, x The variables involved in this study is the depth of the pond, y and the distance from the centre of the pond, x Mengaplikasi matematik: Applying mathematics: • Tulis jarak dari tengah kolam dan kedalaman kolam sebagai set pasangan tertib (x, y) dan lukis satu graf bagi data tersebut. Write the distance from the centre of the pond and the depth of the pond as a set of ordered pairs (x, y) and draw a graph for the data. (7.5, 1.0) y x 10 20 30 1.4 1.2 1 0.8 0.6 0.4 0.2 0 • Graf yang dilukis menunjukkan lengkung penyuaian terbaik dan menyerupai graf fungsi kuadratik. The graph drawn shows the curve of best fit and resembles the graph of a quadratic function. • Nilai anggaran digunakan dalam pemodelan matematik ini untuk mewakili situasi sebenar. Daripada graf, kedalaman kolam ialah 1.0 m. An approximate value is used in this mathematical modelling to represent the actual situation. From the graph, the depth of the pond is 1.0 m. Menentusahkan dan mentafsir penyelesaian: Verifying and interpreting solutions: Tentukan pemalar fungsi kuadratik yang mempunyai bentuk y = ax2 + bx + c dengan menggantikan sebarang tiga data [misalnya (0, 1.2), (15, 0.6) dan (20, 0.1)]. Determine the constants of the quadratic function in the form of y = ax2 + bx + c by substituting any three data [for example (0, 1.2), (15, 0.6) and (20, 0.1)]. 1.2 = a(0)2 + b(0) + c → 1.2 = c 0.6 = a(15)2 + b(15) + c → 0.6 = 225a + 15b + c 0.1 = a(20)2 + b(20) + c → 0.1 = 400a + 20b + c Oleh sebab c ialah pemalar, sistem bagi dua persamaan linear dalam dua pemboleh ubah ialah Since c is a constant, the system for the two linear equations in two variables is 0.6 = 225a + 15b + 1.2 –0.04 = 15a + b ………① 0.1 = 400a + 20b + 1.2 –0.055 = 20a + b ………② CONTOH

Matematik Tingkatan 5 Jawapan J49 ② – ①: –0.015 = 5a a = –0.003 Gantikan a = –0.003 ke dalam ①, Substitute a = –0.003 into ①, –0.04 = 15(–0.003) + b b = 0.005 Maka fungsi kuadratik yang mungkin ialah Hence, the possible quadratic function is y = –0.003x2 + 0.005x + 1.2. Apabila/ When x = 7.5, y = –0.003(7.5)2 + 0.005(7.5) + 1.2 = 1.07 m (hampir dengan jawapan yang diperoleh daripada graf/ approximate to the answer obtained from the graph) Memurnikan model matematik: Refining the mathematical model: • Dalam model ini, kita andaikan kedalaman kolam adalah paling dalam di bahagian tengah dan bentuk kolam ialah hemisfera. Model baharu diperlukan untuk andaian baharu. In this model, we assume the depth of the pond is the deepest at the centre and the shape of the pond is hemispherical. A new model is required for new assumptions. • Kejituan jawapan boleh diperbaik jika lebih banyak data dikumpulkan. The accuracy of the answer can be improved if more data are collected. Praktis Kendiri Kertas 1 1 A Masa bagi setiap individu makan 2 mangkuk mi adalah tidak setara. Oleh itu, masalah ini tidak boleh diterjemah sebagai model matematik. The time for each individual to eat 2 bowls of noodles are not equal. Therefore, this problem cannot be translated into mathematical model. 2 C Jisim karbon-14 (m) menjadi setengah pada setiap tempoh 5 730 tahun (t). The mass of carbon-14 (m) becomes half at every period of 5 730 years (t). Tempoh Period Perhitungan Calculation 1t 10.5 × 1 2 2t (10.5 × 1 2 ) × 1 2 3t [(10.5 × 1 2 ) × 1 2 ] × 1 2 nt m × ( 1 2 ) n = m 2n Model fungsi yang diperoleh ialah fungsi eksponen. The function model obtained is exponential function. 3 B Kita perlu menentusahkan dan mentafsir penyelesaian masalah untuk menentukan sama ada model matematik adalah sah. We need to verify and interpret the solutions of the problem to determine whether the mathematical model is valid or not. 4 A 5 D Kita perlu membuat andaian dan menentukan pemboleh ubah sebelum dapat membentukkan model matematik dan mengaplikasikannya dalam penyelesaian masalah. We need to make assumptions and determine the variables before we can form a mathematical model and apply it to solving problems. 6 C Harga beras dan jarak pasar raya perlu dipertimbangkan untuk menentukan pasar raya manakah yang menunjukkan pembelian beras yang lebih berbaloi. The price of rice and the distance of supermarket need to be considered to determine which supermarket shows the more worthy purchase of rice. 7 B Pilihan warna tidak boleh menjadi pemboleh ubah dalam model matematik kerana pilihan warna adalah berbeza bagi setiap individu. The choice of colour cannot be a variable in the mathematical model because the choice of colour differs for each individual. 8 D Kertas 2 Bahagian A/ Section A 1 (a) Masalah 1: Sama ada bajet Mak Cik Minah cukup membeli kerepek kentang yang mencukupi untuk satu kelas murid. Problem 1: Whether Aunt Minah’s budget is enough to buy enough potato chips for one class of students. Masalah 2: Bilangan kerepek kentang yang perlu dibeli oleh Mak Cik Minah. Problem 2: The number of potato chips which need to be bought by Aunt Minah. CONTOH

Matematik Tingkatan 5 Jawapan J50 (b) Andaian/ Assumptions: • Bilangan maksimum murid ialah 30 orang. The maximum number of students is 30. • Mak Cik Minah hanya mempunyai RM75 untuk membeli kerepek kentang. Aunt Minah only has RM75 to buy the potato chips. 2 (a) Had masa untuk menghantar semua bungkusan kepada semua pelanggan. The time limit to send all the parcels to all the customers. (b) Andaian/ Assumptions: • Perjalanan penghantarannya tidak mengalami sesak jalan His delivery trip is not caught in traffic jam • Jarakantara setiappasangpelanggannya adalah sama The distance between each pair of customers is the same • Pelanggannya menerima bungkusan sebaik sahaja dia sampai His customers receive the parcel right when he arrives 3 (a) Jualan nasi lemak/ Sales of nasi lemak: 25, 36, 49, …, n2 → Fungsi kuadratik/ Quadratic function Jualan mi goreng/ Sales of fried noodles: 20, 30, 40, …, 10n → Fungsi linear/ Linear function (b) Andaian/ Assumptions: • Penjualan makanan tidak mengalami masalah kekurangan pelanggan The selling of food does not face the lack of customers • Kualiti makanan kekal sama dan tidak menyebabkan bilangan pelanggan berkurang The food quality remains the same and does not cause a decrease in the number of customers 4 (a) 1200 – 1000 = 2 jam/ hours 205.2 km ÷ 2 jam/ hours = 102.6 km/ j/ km/h (b) Encik Harun andaikan laju pemanduannya adalah kekal sama sepanjang perjalanannya. Perkara ini tidak dapat menentusahkan penyelesaian masalahnya sepenuhnya kerana apabila diterjemah kembali ke dunia sebenar, sesetengah perjalanan mempunyai had laju yang tetap. Mr Harun assumes his driving speed remains the same along his journey. This matter cannot fully verify the solution to his problem when translated to the real-world situation, some routes have fixed speed limits. Bahagian B/ Section B 5 Masalah/ Problem: • Tentukan isi padu petrol yang diperlukan untuk suatu jarak sejauh 234 km Determine the volume of petrol required for a distance of 234 km • Diketahui bahawa semakin jauh jarak, semakin banyak petrol digunakan. Oleh itu, isi padu petrol berubah secara langsung dengan jarak perjalanan. We know that the further the distance, the more petrol is consumed. Thus, the volume of petrol varies directly as the distance travelled. Andaian dan pemboleh ubah: Assumptions and variables: • Andaikan kelajuan memandu bagi keduadua perjalanan adalah sama. Assume that the driving speed for both routes are the same. • Danella tidak menghadapi sebarang masalah sepanjang perjalanannya. Danella does not face any problems throughout his travel. • Katakan d mewakili jarak perjalanan dan V mewakili isi padu petrol yang digunakan. Let d represents the distance travelled and V represents the volume of petrol consumed. • V berubah secara langsung dengan d, maka V = kd dengan keadaan k ialah pemalar. V varies directly as d, hence V = kd where k is a constant. Mengaplikasi matematik/ Applying mathematics: Gantikan V = 34 dan d = 306 ke dalam V = kd, Substitute V = 34 and d = 306 into V = kd, 34 = k(306) k = 34 306 = 1 9 Maka/ Therefore, V = 1 9 d Persamaan ini menghuraikan hubungan antara isi padu petrol yang digunakan dengan jarak perjalanan. This equation describes the relationship between the volume of petrol consumed with the distance travelled. Apabila/ When d = 234, V = 1 9 (234) = 26 liter/ litres ∴ 26 liter petrol akan digunakan untuk suatu jarak sejauh 234 km. 26 litres of petrol will be consumed for a distance of 234 km. Menentusahkan dan mentafsir penyelesaian: Verifying and interpreting solutions: • Model fungsi linear V = 1 9 d yang diperoleh mungkin tidak dapat digunakan untuk semua situasi perjalanan. The linear function model V = 1 9 d obtained may not be used in all situations. V O V = 1 9 d d CONTOH

Matematik Tingkatan 5 Jawapan J51 • Hal ini kerana bukan semua perjalanan berada dalam keadaan yang sama. Misalnya, penggunaan petrol akan dipengaruhi sekiranya perjalanan ke Syarikat A sering menghadapi masalah kesesakan lalu lintas. This is because not all routes are in the same conditions. For example, the consumption of petrol will be affected if the route to Company A is always facing traffic jam problems. • Oleh itu, model fungsi linear yang diperoleh tidak sesuai digunakan untuk menangani masalah berkenaan di dunia sebenar. Therefore, the linear function model obtained is not suitable to solve this type of problem in the real-world situation. Memurnikan model matematik: Refining the mathematical model: Dalam masalah ini, kita tidak dapat memurnikan model memandangkan maklumat yang diberi adalah terhad. In this problem, we are not able to refine the model due to limited information given. 6 Masalah/ Problem: • Encik Chang diberikan faedah kompaun dengan pengkompaunan sekali setahun. Mr Chang is given compound interest which compounds once per annum. • Prinsipal Encik Chang ialah RM100 000. Mr Chang’s principal is RM100 000. • Kadar faedah tahunan ialah 4.5%. The interest rate is 4.5% per annum. • Tertibkan satu model matematik bagi jumlah simpanan Encik Chang pada akhir tahun ke-t. Derive a mathematical model for Mr Chang’s total savings at the end of tth year. Andaian dan pemboleh ubah: Assumptions and variables: • AndaikanEncikChang tidakmengeluarkan atau menambah wang simpanannya sepanjang tempoh penyimpanannya. Assumes Mr Chang did not withdraw or raise his savings throughout the period of saving. • Pemboleh ubah yang terlibat ialah prinsipal, RMP, kadar faedah tahunan, r = 0.045, bilangan kali faedah dikompaun, n dan masa, t tahun. The variables involved are the principal, RMP, the annual interest rate, r, the number of times the interest is compounded, n and the time, t years. Mengaplikasi matematik untuk menyelesaikan masalah: Applying mathematics to solve the problem: Faedah kompaun ialah faedah yang dihitung berdasarkan hasil tambah prinsipal asal dan faedah yang terkumpul daripada penyimpanan sebelumnya, maka hasil simpanan boleh dihitungkan seperti jadual berikut: Compound interest is an interest which is calculated based on the sum of original principal and the accumulated interest from previous periods of savings, thus the amount of savings can be calculated as the following table: Tahun Year Prinsipal (RM) Principal (RM) Faedah yang diterima (RM) Collected interest (RM) Prinsipal + Faedah (RM) Principal + Interest (RM) Hasil simpanan (RM) Amount of savings (RM) 1 100 000 100 000 × 0.045 100 000 + (100 000 × 0.045) = 100 000(1 + 0.045) 100 000(1.045) 2 100 000(1.045) 100 000(1.045) × 0.045 100 000(1.045) + [100 000(1.045) × 0.045] = 100 000(1.045)(1 + 0.045) 100 000(1.045)2 3 100 000(1.045)2 100 000(1.045)2 × 0.045 100 000(1.045)2 + [100 000(1.045)2 × 0.045] = 100 000(1.045)2 (1 + 0.045) 100 000(1.045)3 4 100 000(1.045)3 100 000(1.045)3 × 0.045 100 000(1.045)3 + [100 000(1.045)3 × 0.045] = 100 000(1.045)3 (1 + 0.045) 100 000(1.045)4 5 100 000(1.045)4 100 000(1.045)4 × 0.045 100 000(1.045)4 + [100 000(1.045)4 × 0.045] =100 000(1.045)4 (1 + 0.045) 100 000(1.045)5 CONTOH

Matematik Tingkatan 5 Jawapan J52 Berdasarkan jadual, hasil simpanan ialah suatu kuasa 1.045 didarab dengan prinsipal simpanan sebanyak RM100 000 dan kuasa adalah sepadan dengan bilangan tahun. Based on the table, the amount of savings is a power of 1.045 times the principal, RM100000 and the power corresponds to the number of years. Oleh sebab 1.045 = 1 + r, dengan menggunakan pemboleh ubah yang ditetapkan, model matematik bagi masalah ini boleh ditulis sebagai A(t) = P(1 + r)t dengan keadaan A(t) sebagai hasil simpanan dan t sebagai bilangan tahun. As 1.045 = 1 + r, using the set variables, the mathematical model for this problem can be written as A(t) = P(1 + r)t where A(t) as the amount of savings and t as the number of years. Hasil simpanan, A(t) (RM) Amount of saving, A(t) (RM) Masa, t (tahun) Time, t (years) 0 10 20 30 40 50 200 000 400 000 600 000 800 000 1 000 000 A(t) = 100 000(1.045)t Graf bagi model matematik ini menunjukkan suatu fungsi eksponen dengan pertumbuhan eksponen bagi hasil simpanan, A(t), apabila masa, t, bertambah. The graph for this mathematical model shows an exponent function with an exponential growth on the amount of savings, A(t), as time, t, increases. Memurnikan model matematik: Refining the mathematical model: Model matematik ini, A(t) = P(1 + r)t , digunakan untuk menghitung faedah kompaun dengan kekerapan pengkompaunan sekali setahun. Sekiranya pengkompaunan berlaku beberapa kali, n, setahun, maka model matematik ini boleh ditambah baik seperti berikut: This mathematical model, A(t) = P(1 + r)t , is used to calculate the annual compounding. If the compounding occurs few times, n, annually, thus this mathematical model can be improved as follows: A(t) = P(1 + r n ) t Soalan SPM Kertas 2 Bahagian C Kertas 2 Bahagian C/ Section C 1 (a) (i) Perimeter laluan/ Perimeter of the route = AB + BC + CD + DO + OA = ( 60 360 × 2 × 22 7 × 7) + 3 + ( 120 360 × 2 × 22 7 × 4) + 4 + 7 = 29.71 km (ii) Target jarak/ Distance target Jarak laluan/ Distance route = 65 29.71 = 2.19 � 3 ⸫ 3 pusingan lengkap/ complete routes (b) (i) Laju (m s–1) Speed (m s–1) Masa (saat) Time (second) O 42 8 14 70 120 (ii) Laju/ Speed Masa/ Time = 0 – 14 120 – 70 = – 0.28 m s−2 (iii) Jumlah jarak yang dilalui: Total distance travelled: = Luas di bawah graf/ Area under the graph = (42 × 8) + [ 1 2 × (8 + 14) × 28] + ( 1 2 × 50 × 14) = 336 + 308 + 350 = 994 m (c) (i) Kos perubatan selepas deduktibel: Medical cost after deductible: = RM15 300 – RM400 = RM14 900 (ii) Kos ditanggung oleh Arifi: Cost borne by Arifi: = ( 20 100 × RM14 900) + RM400 CONTOH = RM3 380

Matematik Tingkatan 5 Jawapan J53 2 (a) (i) p q [r s] = 21124 13005 12206 11117 p q [r s] = [11x] (ii) p q [r s] = pr qr ps qs = 21124 13005 12206 11117 = 150 200 300 400 pr qr = 150 200 p r = 3 4 Apabila/ When p = 3, 3r = 150 r = 50 Apabila/ When p = 3, 3s = 300 s = 100 ⸫ p = 3, q = 4, r = 50, s = 100 (iii) p q [r s] = [pr + qs] = [3(50) + 4(100) = [150 + 400] = [550] = [11x] 11x = 550 x = 50 (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) 86 = 60 + 50 – n(S ∩ M) n(S ∩ M) = 110 – 86 = 24 (ii) ξ S M 36 24 26 3 (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 CONTOH

Matematik Tingkatan 5 Jawapan J54 ⸫ 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 (a) Jumlah pendapatan/ Total income: RM90 000 + 12(RM1 800) = RM90 000 + RM21 600 = RM111 600 Pendapatan bercukai/ Chargeable income: RM111 600 – RM16 750 – RM1 800 = RM93 050 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 RM1559.50 (RM10 800 – RM9 240.50) to Ms Zuraini through her bank account. (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 (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 360000 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 360000 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 (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 (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 CONTOH

Matematik Tingkatan 5 Jawapan J55 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 ⸫ 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 (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 CONTOH

Matematik Tingkatan 5 Jawapan J56 (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 6 (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 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 765 = 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

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