Matematik T4

Matematik Tingkatan 4 Jawapan J21 4 (a) {A, B, C, D, E, F, G, H} (b) {A, B, C, D, E, F, G} (c) {(A, B), (A, D), (B, C), (C, D), (D, E), (E, F), (E, H), (F, G), (G, H), (G, H)} (d) {(A, B), (A, G), (B, C), (B, F), (C, D), (C, F), (D, E), (E, F), (F, G)} (e) 8 (f) 7 (g) 2 (h) 2 (i) 2 (j) 3 (k) 2 (l) 3 (m) 3 (n) 2 (o) 3 (p) 2 (q) 2 (r) 4 (s) 3 (t) 2 (u) 3 (v) 20 (w) 18 5 (a) Q S T R P (b) 1 2 3 5 4 6 1 5 2 4 3 7 (a) 3 + 3 + 2 = 8, Boleh/ Can (b) 3 + 1 + 3 + 2 + 2 = 11, Tidak boleh/ Cannot (c) 2 + 2 + 2 + 2 = 8, Boleh/ Can 8 (a) (b) (c) (d) (e) 9 1 3 2 4 10 A B D C 11 (a) (b) (c) (d) (e) 12 Q R P 4 unit/ units 4 unit/ units 5 unit/ units 13 (a) (i) Mahkota Suria Pusing Damai Kati 60 km 28 km 55 km 75 km 52 km 38 km (ii) Perjalanan Encik Bakal/ Mr Bakal’s journey: Suria Mahkota Damai Pusing 52 + 38 + 55 = 145 km (iii) Perjalanan Encik Bakal/ Mr Bakal’s journey: Pusing Kati Damai Mahkota Suria 28 + 60 + 38 + 52 = 178 km (b) (i) H G F E 60 km 55 km 45 km 43 km 100 km 110 km (ii) G E F G H 60 + 55 + 43 + 110 = 268 km (iii) H G F G 100 + 45 + 43 = 188 km Beza jarak/ Difference in distance: 188 – 100 = 88 km CONTOH

Matematik Tingkatan 4 Jawapan J22 14 (a) (b) (c) (d) (e) 15 (a) (i) e6 e1 e4 e5 A B E D (ii) e4 e3 e2 D B C (b) (i) e7 e2 e e1 6 J O K L (ii) e6 e7 e5 J L N O 16 (a) (b) (c) (d) (e) 17 (a) (i) 6 (ii) 5 (b) (i) 7 (ii) 6 18 (a) Mei Li Luna Susie Kenny Janet 180 m 125 m 150 m 300 m 220 m (b) Badminton Sophia Pau Shan Melukis Drawing Berkebun Gardening Menyanyi Singing Elly David Zukri Shaufiq Shane 19 (a) Kereta kabel, kerana kereta kabel tidak menggunakan tenaga manusia berbanding dengan basikal dan masa perjalanannya bermula lebih awal daripada bas. Cable car, because cable car does not require human strength compared to bicycle and its depature time starts earlier than the bus. (b) RM2.50 × 2.22 = RM12.10 Praktis Kendiri Kertas 1 1 Rangkaian ialah satu graf yang mempunyai sekurang-kurangnya sepasang bucu yang berkait. Network is a graph which has at least a pair of linked vertices. Jawapan/ Answer: A 2 Σd(V) = 4 + 4 + 3 + 3 = 14 Σd(V) = 2 × n(E) 14 = 2 × n(E) n(E) = 7 Jawapan/ Answer: B 3 Graf dalam rajah mempunyai tepi yang menunjukkan arah, maka graf itu ialah graf terarah. The graph in the diagram has edges which show directions, thus the graph is a directed graph. Jawapan/ Answer: B 4 Σd(V) = 4 + 5 + 5 = 14 Jawapan/ Answer: C 5 Graf dalam rajah tiada bucu pada tengahnya. The graph in the diagram does not have a vertex at its centre. Jawapan/ Answer: A 6 n(E) = n(V) – 1 = 4 – 1 = 3 Jawapan/ Answer: D CONTOH

Matematik Tingkatan 4 Jawapan J23 Kertas 2 Bahagian A/ Section A 1 (a) AUH DUB LHR DXB 8.5 jam/ hours 1.5 jam/ hours 8.08 jam/ hours 7.33 jam/ hours 7.83 jam/ hours 7.17 jam/ hours 14.33 jam/ hours KUL (b) Masa transit yang paling lama: Longest transit time: 7.17 + 8.5 + 1.5 = 17.17 jam/ hours = 17 jam 10 minit = 17 hours 10 minutes Beza masa transit/ Difference in transit time: 17 jam 10 minit – 14 jam 20 minit 17 hours 10 minutes – 14 hours 20 minutes = 2 jam 50 minit/ 2 hours 50 minutes 2 (a) 15 × 2 = 30 elektron/ electrons (b) 4 Bahagian B/ Section B 3 (a) Laluan 1/ Route 1: Bandar A Town A Kereta sendiri Private car Rumah nenek Grandmother’s house Laluan 2/ Route 2: Bandar A Town A Bandar B Town B Kereta api Train Teksi/ Taxi Rumah nenek Grandmother’s house Laluan 3/ Route 3: Bandar A Town A Bas/ Bus Teksi/ Taxi Rumah nenek Grandmother’s house Bandar B Town B (b) Laluan 1/ Route 1: Jarak/ Distance: 200 km Jumlah perbelanjaan/ Total expenses: (200 × RM0.18) + RM5.90 = RM41.90 Masa/ Time: 2 jam 15 minit/ 2 hours 15 minutes Laluan 2/ Route 2: Jarak/ Distance: 190 + 15 = 205 km Jumlah perbelanjaan/ Total expenses: RM30 + RM20 = RM50 Masa: 2 jam 30 minit + 30 minit = 3 jam Time: 2 hours 30 minutes + 30 minutes = 3 hours Laluan 3/ Route 3: Jarak/ Distance: 190 + 15 = 205 km Jumlah perbelanjaan/ Total expenses: RM13 + RM20 = RM33 Masa: 3 jam + 30 minit = 3 jam 30 minit Time: 3 hours + 30 minutes = 3 hours 30 minutes (c) Laluan 1, kerana Laluan 2 terlalu mahal dan Laluan 3 mengambil masa yang agak lama. Route 1, because Route 2 is too expensive and Route 3 takes a very long time. Bab 6 6.1 Ketaksamaan Linear dalam Dua Pemboleh Ubah Linear Inequalities in Two Variables 1 (a) Tidak/ No (b) Ya/ Yes 2 (a) y < 10 (b) x + 5 > 70 (c) 2x + 3 N 20 3 (a) y = 5, 5 + 3 = 8 (5 < 8) Titik (5, 5) memuaskan y < x + 3. Point (5, 5) satisfies y < x + 3. (b) y = 8, 3 + 3 = 6 (8 > 6) Titik (3, 8) memuaskan y > x + 3. Point (3, 8) satisfies y > x + 3. (c) y = 2, –5 + 3 = –2 (2 > –2) Titik (–5, 2) memuaskan y > x + 3. Point (–5, 2) satisfies y > x + 3. (d) y = –2, –3 + 3 = 0 (–2 < 0) Titik (–3, –2) memuaskan y < x + 3. Point (–3, –2) satisfies y < x + 3. (e) y = 0, –3 + 3 = 0 (0 = 0) Titik (–3, 0) memuaskan y = x + 3. Point (–3, 0) satisfies y = x + 3. CONTOH

Matematik Tingkatan 4 Jawapan J24 4 (a) 3x + 2 > y y < 3x + 2 y = 3x + 2 y x 2 O – 2 3 (b) y M 3 y = 3 y x 3 O (c) y + 3x > 5 y > –3x + 5 y = –3x + 5 y x 5 O 5 3 (d) 2x + y < –4 y < –2x – 4 y = –2x – 4 y x –2 –4 O (e) x < –2 x = –2 y x –2 O 5 (a) y < –x + 3 x 0 3 y 3 0 x y y = –x + 3 3 3 O (b) 4x – 3y > 12 3y < 4x – 12 y < 4 3x – 4 x 0 3 y –4 0 x y y = 4 3x – 4 3 –4 O CONTOH

Matematik Tingkatan 4 Jawapan J25 (c) y M 3x + 9 x 0 –3 y 9 0 x y –3 9 O y = 3x + 9 6.2 Sistem Ketaksamaan Linear dalam Dua Pemboleh Ubah Systems of Linear Inequalities in Two Variables 1 (a) y N x (b) y – x M w (c) y N wx (d) y N w (e) y < x (f) y – x < w (g) x + y > w (h) y M w (i) y M x 2 (a) (i) x + y M 20 (ii) y – x > 3 (iii) x N 8 (b) (i) x + y N 40 (ii) y N 3x (iii) x – y N 10 (c) (i) 5m + 2b > 50 (ii) m + b M 20 (iii) m > b (d) (i) y – x M 6 (ii) x M 10 (iii) 1.5y – x < 2 3y – 2x < 4 3 (a) 6 + 2(6) = 18 18 M 10 6(6) – 6 = 30 30 > 8 (b) 3 + 2(2) = 7 7 N 10 6(3) – 2 = 16 16 > 8 (c) 1 + 2(3) = 7 7 N 10 6(1) – 3 = 3 3 < 8 (d) Rantau R/ Region R x + 1.5y N 25 2x + 3y N 50 4 (a) – 1 4 O 2 3 3 x = 2 5y – 4x = 1 x + y = 3 x y 1 5 (b) x y O 6 6 5 x + y = 6 y = 5 x = y (c) x y O 3 4 3 3x – 5y = –15 3x – y = 4 5 (a) 6 – 5y = x 2y = 3x – 5 5 3 O 6 x y 6 5 (b) 5x – y = 10 y = 2x – 10 y = 8 O 8 2 5 x CONTOH y

Matematik Tingkatan 4 Jawapan J26 (c) x – 5 = –y 2x – 2y = –4 O 2 5 5 x y 6 (a) x + y > 6, y N x dan/ and x N 6 (b) x + 2y N 8, 2y < x dan/ and y M 1 (c) x N 0, 2y – x < 4 dan/ and y > –x – 3 7 Persamaan / Equation : Kecerunan/ Gradient, m: 5 – (–1) 2 – 0 = 3 Maka/ Thus, y = 3x – 1 Persamaan / Equation : Kecerunan/ Gradient, m: – 6 12 = – 1 2 Maka/ Thus, y = – 1 2 x + 6 Ketaksamaan linear yang memuaskan rantau berlorek ialah y < 3x – 1 dan y N – 1 2 x + 6. The linear inequalities which satisfy the shaded region are y < 3x – 1 and y N – 1 2 x + 6. 8 (a) (i) ΔABE dan/ and ΔBCD (ii) Ketaksamaan linear dalam rantau ABE: Linear inequalities in region ABE: x M 0, 2y > 5x – 10 dan/ and 7x + 4y N 28 Ketaksamaan linear dalam rantau BCD: Linear inequalities in region BCD: y M 0, 7x + 4y N 28 dan/ and 2y < 5x – 10 (b) (i) x + y N 150 y N –x + 150 x 0 150 y 150 0 3x + 4y M 480 y M – 3 4 x + 120 x 0 160 y 120 0 (ii) x y 150 150 120 O 160 y = –x + 150 y = – 3 4x + 120 (c) Katakan bilangan oren = x dan bilangan lemon = y. Let the number of oranges = x and the number of lemons = y. (i) x + y N 20 y N 20 – x x – y M 2 y N x – 2 (ii) y N 20 – x x 0 20 y 20 0 y N x – 2 x 2 4 y 0 2 x y 20 O 2 20 y = 20 – x y = x – 2 (iii) Tidak, kerana y = 10 berada di luar rantau berlorek. No, because y = 10 is outside of the shaded region. CONTOH

Matematik Tingkatan 4 Jawapan J27 (d) Katakan x = bilangan lelaki dan y = bilangan perempuan. Let x = the number of men and y = the number of women. (i) x + y N 10 y – x > 2 y N 10 – x y > 2 + x (ii) y N 10 – x y > x + 2 x 0 10 y 10 0 x –2 0 2 y 0 2 4 x y 10 2 O 10 y = 10 – x y = x + 2 (iii) Bilangan maksimum lelaki yang boleh menyertai pasukan badminton ialah 4 orang. The maximum number of men who can join the badminton team is 4. Praktis Kendiri Kertas 1 1 Kecerunan/ Gradient = – 7 7 = –1 Pintasan-y/ y-intercept = 7 Persamaan garis lurus/ Straight line equation: y = –x + 7 x + y = 7 Rantau berlorek adalah di bawah garis padu x + y = 7, maka ketaksamaan linear ialah x + y N 7. The shaded region is below the solid line of x + y = 7, thus the linear inequality is x + y N 7. Jawapan/ Answer: A 2 Rantau berlorek adalah di bawah garis padu 3y = x dan di atas garis padu x + 2y = 10, maka ketaksamaan linear ialah 3y N x dan x + 2y M 10. The shaded region is below the solid line of 3y = x and above the solid line of x + 2y = 10, thus the linear inequalities are 3y N x and x + 2y M 10. Jawapan/ Answer: C 3 3x – y M 3 y N 3x – 3 x + 2y M 12 2y M 12 – x y M – 1 2 x + 6 Jawapan/ Answer: B 4 Rantau P berada di atas garis padu y = x – 2 dan di bawah garis padu 2x + y = 4, maka rantau P memuaskan y M x – 2 dan 2x + y N 4. Region P is above the solid line y = x – 2 and below the solid line 2x + y = 4, thus region P satisfy y M x – 2 and 2x + y N 4. Jawapan/ Answer: A 5 Rantau dilorek di atas garis padu 2y = x + 2, maka 2y M x + 2. The region is shaded above the solid line 2y = x + 2, thus 2y M x + 2. Rantau dilorek di bawah garis sempang y = 5, maka y < 5. The region is shaded below the dashed line y = 5, thus y < 5. Rantau dilorek di bawah garis sempang x + y = 5, maka x + y < 5. The region is shaded below the dashed line x + y = 5, thus x + y < 5. Jawapan/ Answer: D 6 Rantau berlorek hanya memuaskan x < 5, y – xN 4 dan 2x + y > 3. The shaded region only satisfy x < 5, y – x N 4 and 2x + y > 3. Jawapan/ Answer: B Kertas 2 Bahagian A/ Section A 1 (a) x y 3y = x 4x + 5y = 20 O CONTOH

Matematik Tingkatan 4 Jawapan J28 (b) Katakan x = bilangan guli merah dan y = bilangan guli hijau. Let x = the number of red marbles and y = the number of green marbles. x + y N 25, x > 2y dan/ and y M 5 2 x y O y = x – 2 4x – 9y = 36 y = –x 3 3 (6, 3) (–6, –3) –3 y x O 1 2 3 (a) Persamaan / Equation : Kecerunan/ Gradient, m: 0 ∴ y < 3 Persamaan ➁/ Equation ➁: Kecerunan/ Gradient, m: – 3 –3 = 1 ∴ y N x + 3 Persamaan ➂/ Equation ➂: Kecerunan/ Gradient, m: 3 – (–3) 6 – (–6) = 1 2 ∴ y M 1 2 x (b) –1 N x N 4 Bahagian B/ Section B 4 (a) Katakan x = bilangan kotak kertas 70 gram dan y = bilangan kotak kertas 80 gram. Let x = the number of boxes of 70 grams papers and y = the number of boxes of 80 grams papers. y M x   y M 30 (b) y N – 5 8 x + 60 x 0 80 96 y 60 10 0 x y 60 30 O 96 y = x y = 30 y = – 5 8x + 60 (c) Tidak, kerana nilai x = 80 berada di luar rantau berlorek. No, because the value of x = 80 is outside of the shaded region. Bab 7 7.1 Graf Jarak-Masa Distance-Time Graphs 1 (a) Jarak (m) Distance (m) Masa (saat) Time (second) 4 5 10 15 20 O 8 12 16 5x + 8y N 480 8y N –5x + 480 y N – 5 8 x + 60 CONTOH

Matematik Tingkatan 4 Jawapan J29 (b) Jarak (km) Distance (km) Masa (jam) Time (hour) 100 1 2 3 4 O 200 300 400 (c) Jarak (km) Distance (km) Masa (minit) Time (minute) 25 15 30 45 60 O 50 75 100 2 (a) Masa, t (jam) Time, t (hour) 0 1 2 3 Jarak, s (km) Distance, s (km) 0 50 100 150 Jarak (km) Distance (km) Masa (jam) Time (hour) 50 1 2 3 O 100 150 (b) Masa, t (saat) Time, t (second) 0 1 2 3 4 5 Jarak, s (mm) Distance, s (mm) 0 5 10 15 20 25 Jarak (mm) Distance (mm) Masa (saat) Time (second) 10 1 2 3 4 5 O 20 30 CONTOH

Matematik Tingkatan 4 Jawapan J30 3 (a) (i) (a) 25 – 5 = 20 minit/ minutes (b) Laju/ Speed = 7 – 4 5 – 0 60 = 36 km j–1/ km h–1 (c) Laju purata = Jumlah jarak Jumlah masa Average speed = Total distance Total time 60 = s 45 60 s = 60 × 45 60 = 45 km (ii) Laju/ Speed = 45 – 7 45 – 25 60 = 114 km j–1/ km h–1 Bas itu bergerak sejauh 38 km dengan kelajuan 114 km j–1 dalam tempoh 20 minit terakhir. The bus travelled 38 km with a speed of 114 km h–1 in the last 20 minutes. (b) Laju zarah A/ The speed of particle A: 186 – 36 25 = 6 m s–1 Laju zarah B/ The speed of particle B: 225 25 = 9 m s–1 Beza laju/ Difference of speed: 9 – 6 = 3 m s–1 (c) (i) Laju/ Speed = 60 – 35 20 60 = 75 km j–1/ km h–1 Bas itu bergerak sejauh 25 km dalam tempoh 20 minit dengan kelajuan 75 km j–1. The bus travelled 25 km in 20 minutes with a speed of 75 km h–1. (ii) Laju/ Speed = 100 – 60 80 – 50 60 = 80 km j–1/ km h–1 (d) (i) 30 – 25 = 5 minit/ minutes (ii) 9.20 a.m. (iii) Laju/ Speed = 6 20 60 = 18 km j–1/ km h–1 4 (a) Masa/ Time = 0714 – 0700 = 14 minit/ minutes Laju/ Speed = 0.28 14 60 = 1.2 km j–1/ km h–1 (b) Masa/ Time = 0719 – 0714 = 5 minit/ minutes (c) Laju purata = Jumlah jarak yang dilalui Jumlah masa yang diambil Average speed = Total distance travelled Total time taken = 0.56 28 60 = 1.2 km j–1/ km h–1 7.2 Graf Laju-Masa Speed-Time Graphs 1 (a) Laju (m s–1) Speed (m s–1) Masa (saat) Time (second) 2 2 4 6 8 O 4 6 8 (b) Laju (m s–1) Speed (m s–1) Masa (saat) Time (second) 5 10 20 30 40 O 10 15 20 CONTOH

Matematik Tingkatan 4 Jawapan J31 (c) Laju (m s–1) Speed (m s–1) Masa (saat) Time (second) 10 3 6 9 12 O 20 30 40 2 (a) Jarak/ Distance: = [ 1 2 × (8 + 10) × 9] + [ 1 2 × 8 × (12 – 9)] = 81 + 12 = 93 m (b) Jarak/ Distance: = [ 1 2 × (3 + 12) × 2] + (6 ×12) + [ 1 2 × 2 × 12] = 15 + 72 + 12 = 99 m (c) Jarak/ Distance: = [ 1 2 × (7 + 4) × 7] + [ 1 2 × (7 + 12) × 2] = 38.5 + 19 = 57.5 m (d) Jarak/ Distance: = [ 1 2 × (6 + 10) × 7] + [ 1 2 × 5 × 10] = 56 + 25 = 81 m 3 (a) [ 1 2 × (15 + 23) × t] + [(18 – t) × 15] = 298 19t + 270 – 15t = 298 19t – 15t = 298 – 270 4t = 28 t = 7 s (b) Laju purata = Jumlah jarak Jumlah masa Average speed = Total distance Total time = 298 18 = 16.56 m s–1 4 (a) (i) Kadar perubahan laju untuk tempoh 40 saat pertama The rate of change of speed for the first 40 seconds = Perubahan laju Perubahan masa = Change of speed Change of time = 70 – 0 40 – 0 60 = 105 m min–2 (ii) Kereta bergerak dengan laju seragam 70 m min–1 dari saat ke-40 hingga saat ke-80. The car moves at a uniform speed of 70 m min–1 from the 40th second to the 80th second. (iii) Jumlah jarak/ Total distance: [ 1 2 × ( 40 + 80 60 ) × 70] + [ 1 2 × (70 + 110) × 40 60] = 70 + 60 = 130 m (b) (i) Kadar perubahan laju/ Rate of change of speed: 21 – 7 35 – 0 60 = 24 m min–2 (ii) Kereta bergerak dengan laju seragam 21 m min–1 dari saat ke-35 hingga saat ke-70. The car moves with uniform speed of 21 m min–1 from the 35th second to the 70th second. (iii) Jumlah jarak/ Total distance: [ 1 2(7 + 21)( 35 60)] + [21( 70 – 35 60 )] + [1 2(21 + 35)( 90 – 70 60 )] = 8.167 + 12.25 + 9.333 = 29.75 m 5 (a) (i) (t – 40) s (ii) Kadar perubahan laju/ The rate of change of speed: = Perubahan laju Perubahan masa = Change of speed Change of time = 12 – 92 40 – 0 = –2 m s–2 (iii) [ 1 2 × (12 + 92) × 40] + [12(t – 40)] + [ 1 2 × 12 × (120 – t)] = 2 590 2 080 + (12t – 480) + (720 – 6t) = 2 590 2 320 + 6t = 2 590 6t = 270 t = 45 CONTOH

Matematik Tingkatan 4 Jawapan J32 (iv) Laju purata = Jumlah jarak Jumlah masa Average speed = Total distance Total time = 2 590 120 = 21.58 m s–1 (b) (i) Pecutan = Laju Masa Acceleration = Speed Time = 30 0.3 = 100 km j–2/ km h–2 (ii) Jarak/ Distance = 1 2 × (0.6 + 0.9) × 30 = 22.5 km (iii) Jumlah jarak/ Total distance: 22.5 + [ 1 2 × (t – 0.9) × 30] = 22.5 + 15t – 13.5 = 15t + 9 Laju purata/ Average speed = 20 km j–1/ km h–1 15t + 9 t = 20 15t + 9 = 20t 5t = 9 t = 1.8 (c) (i) Laju/ Speed = 600 – 150 50 = 9 m s–1 (ii) Jarak/ Distance = [ 1 2 (5 + 10)(25)] + [(10)(25)] = 187.5 + 250 = 437.5 m Laju purata/ Average speed = 437.5 50 = 8.75 m s–1 (iii) Jarak A/ Distance A = 600 – 150 = 450 m Jarak B/ Distance B = 437.5 m Beza jarak/ Difference in distance: 450 – 437.5 = 12.5 m Praktis Kendiri Kertas 1 1 Tempoh masa/ Duration of time = 0.7 – 0.3 = 0.4 jam/ hour Jawapan/ Answer: A 2 Jumlah jarak dilalui/ Total distance travelled: (30 – 20) + (45 – 30) + 45 = 10 + 15 + 45 = 70 km Laju purata/ Average speed = 70 110 60 = 38.18 km j–1/ km h–1 Jawapan/ Answer: C 3 Laju purata kereta A/ Average speed of car A: 120 28 = 30 7 m min–1 Laju purata kereta B/ Average speed of car B: 120 40 = 3 m min–1 Beza/ Difference = 30 7 – 3 = 9 7 m min–1 Jawapan/ Answer: C 4 Tempoh masa/ Duration of time = 9 – 3 = 6 s Jawapan/ Answer: C 5 Kadar perubahan laju/ Rate of change of speed: 0 – 25 10 = –2.5 km min–2 Jawapan/ Answer: D 6 Jumlah jarak = Luas di bawah graf Total distance = Area under the graph [ 1 2 (4 + 8)(3)] + [8(7 – 3)] + [ 1 2 (8)(t – 7)] = 62 18 + 32 + 4t – 28 = 62 22 + 4t = 62 4t = 40 t = 10 Jawapan/ Answer: C Kertas 2 Bahagian A/ Section A 1 (a) 30 – 20 = 10 s (b) Laju/ Speed = 25 20 = 1.25 m s–1 (c) Laju purata/ Average speed = 60 50 = 1.2 m s–1 Kereta itu bergerak dengan laju purata 1.2 m s–1 dalam tempoh 50 saat sejauh 60 m. The car moved with an average speed of 1.2 m s–1 in 50 seconds for a distance of 60 m. CONTOH

Matematik Tingkatan 4 Jawapan J33 2 (a) Masa (minit) Time (minutes) 75 k 250 Jarak (km) Distance (km) 211 O h (b) h = 211 – 80 = 131 km k = 1100 – 0900 = 2 jam/ hours = 120 minit/ minutes (c) Laju purata/ Average speed: 211 250 60 = 50.64 km j–1/ km h–1 Bahagian B/ Section B 3 (a) v = (0 – 7) km 3.5 min = –7 km 3.5 min = –7 km (3.5 ÷ 60) j/ h = –120 km j–1/ km h–1 (b) Jarak dari bandar R ke bandar Q: 7 – 3 = 4 km Distance from city R to city Q: Masa dari bandar R ke bandar Q: Time from city R to city Q: (4 ÷ 120 km j–1/ km h–1) × 60 min = 2 minit/ minutes Masa bertemu dengan kereta api A: The time met with train A: 0731 + 0002 = 0733 (c) Jarak yang dilalui oleh kereta api A: 3 km Distance travelled by train A: Jarak yang dilalui oleh kereta api B: Distance travelled by train B: 7 km – 3 km = 4 km Beza jarak/ Difference in distance: 4 – 3 = 1 km (d) Laju purata/ Average speed: 7 km (0737 ‒ 0730) ÷ 60 j/ h = 7 (7 ÷ 60) = 60 km/j / km/h 4 (a) 1.6 – 0.4 = 1.2 jam/ hours = 72 minit/ minutes (b) Laju purata/ Average speed: 100 2 = 50 km j–1/ km h–1 (c) (i) Di/ At S: 12 jam + 1.2 jam= 12 jam + 1 jam 12 minit 12 hours + 1.2 hours = 12 hours + 1 hour 12 minutes = 13 jam 12 minit = 13 hours 12 minutes = Jam 1312/ 1312 hour Di/ At R: 12 jam + 2 jam = 14 jam 12 hours + 2 hours = 14 hours = Jam 1400/ 1400 hour ∴ Kedua-dua kereta bertemu di S pada jam 1312 dan di R pada jam 1400. Both cars meet in S at 1312 hour and in R at 1400 hour. (ii) 1400 – 1312 = 48 minit/ minutes (d) Laju pada OP/ Speed at OP: 30 km 0.4 j/ h = 75 km/j/ km/h Laju pada QR/ Speed at QR: (100 – 30) km (2 – 1.6) j/ h = 70 0.4 = 175 km/j/ km/h Laju (km/j) Speed (km/h) Masa (jam) Time (hour) 175 75 0.4 1.6 2.0 O 5 (a) Jumlah jarak/ Total distance: [ 1 2 (4 + 8)(8)] + [8(10 – 8)] = 48 + 16 = 64 m Laju purata/ Average speed = 64 10 = 6.4 m s–1 (b) 1 2 (4 + 8)(8) = 2(8)(t – 8) 48 = 16t – 128 16t = 176 t = 11 s CONTOH

Matematik Tingkatan 4 Jawapan J34 (c) v – 8 13 – 11 = 2 v – 8 = 4 v = 12 m s–1 Bab 8 8.1 Serakan Dispersion 1 (a) Nilai tertinggi/ Highest value: 2.2 kg Nilai terendah/ Lowest value: 0.9 kg (b) Nilai tertinggi/ Highest value: 162 cm Nilai terendah/ Lowest value: 152 cm (c) Nilai tertinggi/ Highest value: 45 mangkuk/ bowls Nilai terendah/ Lowest value: 18 mangkuk/ bowls 2 3 4 5 6 7 8 9 10 Poin yang diperoleh Points obtained 3 3 5 6 4 8 9 5 0 6 7 7 5 8 2 Batang Stem Daun Leaf Kekunci: 3 | 5 bermaksud 35 markah Key: 3 | 5 means 35 marks 4 (a) 7 4 2 2 9 8 5 5 4 0 3 2 5 6 6 9 8 7 4 4 2 5 6 7 9 9 5 1 5 0 0 2 Kilang P Factory P Kilang Q Factory Q Kekunci: 2 | 2 | 9 bermaksud 22 tahun dan 29 tahun Key: 2 | 2 | 9 means 22 years old and 29 years old (b) Kilang P/ Factory P: Nilai tertinggi/ Highest value = 55 tahun/ years old Nilai terendah/ Lowest value = 22 tahun/ years old Beza/ Difference: 55 – 22 = 33 tahun/ years old Kilang Q/ Factory Q: Nilai tertinggi/ Highest value = 52 tahun/ years old Nilai terendah/ Lowest value = 29 tahun/ years old Beza/ Difference: 52 – 29 = 23 tahun/ years old ∴ Kilang P mempunyai serakan lebih besar kerana beza cerapan datanya adalah lebih tinggi. Factory P has higher dispersion because the difference in the observation data is higher. (c) Kilang Q mempunyai prestasi yang lebih tinggi kerana kilang tersebut mempunyai pekerja dengan umur yang lebih tinggi dan lebih berkemahiran. Factory Q has higher performance because the factory has workers with older age and more skilled. 8.2 Sukatan Serakan Measures of Dispersion 1 (a) Nilai tertinggi/ Highest value: 21 Nilai terendah/ Lowest value: 5 Julat/ Range: 21 – 5 = 16 (b) Nilai tertinggi/ Highest value: 10 Nilai terendah/ Lowest value: 2 Julat/ Range: 10 – 2 = 8 (c) Nilai tertinggi/ Highest value: 12 Nilai terendah/ Lowest value: 1 Julat/ Range: 12 – 1 = 11 2 (a) Julat antara kuartil/ Interquartile range: 37 + 42 2 – 17 + 24 2 = 39.5 – 20.5 = 19 (b) Julat antara kuartil/ Interquartile range: 20 + 20 2 – 18 + 18 2 = 20 – 18 = 2 (c) Julat antara kuartil/ Interquartile range: 69 – 34 = 35 3 (a) Q1 = cerapan ke-(10 4 )/ ( 10 4 ) th observation = cerapan ke-2.5/ 2.5th observation = 5 Q3 = cerapan ke-[ 3(10) 4 ]/ [3(10) 4 ]th observation = cerapan ke-7.5/ 7.5th observation = 7 + 8 2 = 7.5 CONTOH Julat antara kuartil/ Interquartile range: 7.5 – 5 = 2.5

Matematik Tingkatan 4 Jawapan J35 (b) Q1 = cerapan ke-(28 4 )/ ( 28 4 ) th observation = cerapan ke-7/ 7th observation = 4 Q3 = cerapan ke-[ 3(28) 4 ]/ [3(28) 4 ]th observation = cerapan ke-21/ 21th observation = 6 Julat antara kuartil/ Intequartile range: 6 – 4 = 2 (c) Q1 = cerapan ke-(12 4 )/ ( 12 4 ) th observation = cerapan ke-3/ 3th observation = 15 Q3 = cerapan ke-[ 3(12) 4 ]/ [3(12) 4 ]th observation = cerapan ke-9/ 9th observation = 23 Julat antara kuartil/ Interquartile range: 23 – 15 = 8 4 (a) Min/ Mean: 110 + 105 + 98 + 63 + 87 + 100 + 57 + 48 + 52 9 = 720 9 = 80 Varians/ Variance: (110 – 80)2 + (105 – 80)2 + (98 – 80)2 + (63 – 80)2 + (87 – 80)2 + (100 – 80)2 + (57 – 80)2 + (48 – 80)2 + (52 – 80)2 9 = 900 + 625 + 324 + 289 + 49 + 400 + 529 + 1 024 + 784 9 = 547.111 Sisihan piawai/ Standard deviation: √547.111 = 23.39 (b) Min/ Mean: 1(1) + 6(2) + 10(3) + 3(4) + 7(5) + 3(6) 1 + 6 + 10 + 3 + 7 + 3 = 108 30 = 3.6 Varians/ Variance: 1(1 – 3.6)2 + 6(2 – 3.6)2 + 10(3 – 3.6)2 + 3(4 – 3.6)2 + 7(5 – 3.6)2 + 3(6 – 3.6)2 30 = 6.76 + 15.36 + 3.6 + 0.48 + 13.72 + 17.28 30 = 1.907 Sisihan piawai/ Standard deviation: √1.907 = 1.381 5 (a) x 23 5 14 20 18 25 Σx = 105 x2 529 25 196 400 324 625 Σx2 = 2 099 Min/ Mean, x: 105 6 = 17.5 Varians/ Variance, σ2 : 2 099 6 – 17.52 = 43.583 Sisihan piawai/ Standard deviation, σ: √43.583 = 6.602 (b) x x2 10 100 22 484 30 900 17 289 18 324 25 625 32 1 024 8 64 16 256 22 484 Σx = 200 Σx2 = 4 550 Min/ Mean, x: 200 10 = 20 Varians/ Variance, σ2 : 4 550 10 – 202 = 455 – 400 = 55 Sisihan piawai/ Standard deviation, σ: √55 = 7.416 (c) x x2 60 3 600 41 1 681 53 2 809 44 1 936 54 2 916 45 2 025 39 1 521 42 1 764 Σx = 378 Σx2 = 18 252 Min/ Mean, x: 378 8 = 47.25 Varians/ Variance, σ2 : 18 252 8 – 47.252 = 2 281.5 – 2 232.5625 = 48.9375 Sisihan piawai/ Standard deviation, σ: √48.9375 = 6.996 CONTOH

Matematik Tingkatan 4 Jawapan J36 (d) x f x2 fx fx2 50 4 2 500 200 10 000 100 5 10 000 500 50 000 150 3 22 500 450 67 500 200 2 40 000 400 80 000 Σf = 14 Σfx = 1 550 Σfx2 = 207 500 Min/ Mean, x: 1 550 14 = 110.714 Varians/ Variance, σ2 : 207 500 14 – 110.7142 = 14 821.429 – 12 257.590 = 2 563.839 Sisihan piawai/ Standard deviation, σ: √2 563.839 = 50.634 6 (a) (i) Julat/ Range: 12 – 2 = 10 jam/ hours Q1 = Cerapan ke-(48 4 )/ ( 48 4 ) th observation = Cerapan ke-12/ 12th observation = 6 jam/ hours Q3 = Cerapan ke-[ 3(48) 4 ]/ [ 3(48) 4 ] th observation = Cerapan ke-36/ 36th observation = 8 jam/ hours Julat antara kuartil/ Interquartile range: 8 – 6 = 2 jam/ hours (ii) Julat kerana set data tiada pencilan, oleh itu julat dapat memberi gambaran serakan yang lebih jelas. Range because the set of data does not have outlier, thus the range can give a clearer image of dispersion. (b) (i) Sarah: x : 21.22 + 21.34 + 22.04 + 20.55 4 = 21.29 σ: 21.222 + 21.342 + 22.042 + 20.552 4 – 21.292 = 0.416 Sylvia: x : 20.33 + 21.45 + 21.04 + 20.58 4 = 20.85 σ: 20.332 + 21.452 + 21.042 + 20.582 4 – 20.852 = 0.43 (ii) Sarah ialah pelari dengan prestasi yang lebih konsisten kerana sisihan piawai masa lariannya adalah lebih rendah daripada Sylvia. Sarah is the runner with a more consistent performance because the standard deviation of her running time is lower the Sylvia. (iii) Kerana masa larian Sylvia adalah lebih pendek dan prestasinya adalah lebih kurang sama dengan Sarah. Because Sylvia’s running time is shorter and her performance is about the same as Sarah. 7 (a) Susun semula/ Rearrange: 23, 29, 32, 36, 40, 47, 50 Q1 Q3 Median 25 30 35 40 45 50 (b) Susun semula/ Rearrange: Bilangan cerapan genap Even number of observations 39, 42, 43, 43, 44, 44, 47, 48, 48, 49, 50, 51, 52, 53, 55, 59, 62, 66 9 cerapan 9 observations 9 cerapan 9 observations Median 48 + 49 2 = 48.5 Q1 Q3 Q1= Cerapan ke-5/ 5th observation = 44 Q3= Cerapan ke-14/ 14th observation = 53 40 45 50 55 60 65 (c) 28 cerapan (bilangan genap) 28 observations (even number) 14 cerapan 14 observations 14 cerapan 14 observations Median = Purata cerapan ke-14 dan ke-15 Median = Average of 14th and 15th observations = 103 + 104 2 = 103.5 Q1 = Purata cerapan ke-7 dan ke-8 = Average of 7th and 8th observations = 90 + 91 2 = 90.5 Q3 = Purata cerapan ke-21 dan ke-22 = Average of 21st and 22nd observations = 108 + 109 2 = 108.5 CONTOH

Matematik Tingkatan 4 Jawapan J37 90 95 100 105 110 115 8 (a) (i) Nilai minimum/ Minimum value: 20 Nilai maksimum/ Maximum value: 68 (ii) Julat/ Range: 68 – 20 = 48 (iii) Kuartil bawah/ Lower quartile: 34 Median/ Median: 42 Kuartil atas/ Upper quartile: 62 (iv) Julat antara kuartil/ Interquartile range: 62 – 34 = 28 (b) (i) Nilai minimum/ Minimum value: 5.6 Nilai maksimum/ Maximum value: 15.2 (ii) Julat/ Range: 15.2 – 5.6 = 9.6 (iii) Kuartil bawah/ Lower quartile: 9.2 Median/ Median: 11.2 Kuartil atas/ Upper quartile: 12.4 (iv) Julat antara kuartil/ Interquartile range: 12.4 – 9.2 = 3.2 9 (a) (i) Julat/ Range: 55 – 26 = 29 kg Q1 = Purata cerapan ke-5 dan ke-6 = Average of 5th and 6th observations = 30 + 31 2 = 30.5 kg Q3 = Purata cerapan ke-15 dan ke-16 = Average of 15th and 16th observations = 45 + 46 2 = 45.5 kg Julat antara kuartil/ Interquartile range: 45.5 – 30.5 = 15 kg (ii) Julat/ Range: 68 – 26 = 42 kg Q1 = Purata cerapan ke-6 dan ke-7 = Average of 6th and 7th observations = 31 + 31 2 = 31 kg Q3 = Purata cerapan ke-18 dan ke-19 = Average of 18th and 19th observations = 45 + 46 2 = 45.5 kg Julat antara kuartil/ Interquartile range: 45.5 – 31 = 14.5 kg (iii) Julat meningkat dengan banyak kerana terdapat satu nilai ekstrem. The range increased greatly because there is an extreme value. Julat antara kuartil masih kekal lebih kurang sama, iaitu 15 kg. The interquartile range still remains almost the same, which is 15 kg. (b) (i) x 4 8 12 16 20 f 3 7 6 4 2 Σf = 22 fx 12 56 72 64 40 Σfx = 244 x2 16 64 144 256 400 fx2 48 448 864 1 024 800 Σfx2 = 3 184 x: 244 22 = 11.091 Sisihan piawai/ Standard deviation: 3 184 11 – 11.0912 = 4.66 (ii) x 4 8 12 16 f 3 7 6 4 Σf = 20 fx 12 56 72 64 Σfx = 204 x2 16 64 144 256 fx2 48 448 864 1 024 Σfx2 = 2 384 x2 : 204 20 = 10.2 Sisihan piawai/ Standard deviation: 2 384 20 – 10.22 = 3.89 (iii)Serakan data menjadi kecil selepas jisim 20 kg dikeluarkan kerana sisihan piawai baharu menjadi kecil. The dispersion of data becomes smaller after the mass of 20 kg is removed because the new standard deviation has become smaller. 10 (a) Pencilan 21 kg dikeluarkan. The outlier 21 kg is removed. (b) Satu nilai yang jauh daripada min ditambah ke dalam set data tersebut. A value which is far from the mean was added to the set of data. 11 (a) (i) Kelas 2 Melati/ Class 2 Melati: Min/ Mean: 66 + 67 + 68 + 70 + 74 + 77 + 77 + 78 + 83 + 87 + 88 + 89 + 92 + 97 14 = 79.5 Sisihan piawai/ Standard deviation: 662 + 672 + 682 + 702 + 742 + 772 + 772 + 782 + 832 + 872 + 882 + 892 + 922 + 972 14 – 79.52 = 89 783 14 – 6 320.25 = √92.821 = 9.63 Kelas 2 Ros/ Class 2 Ros: Min/ Mean: 69 + 71 + 75 + 76 + 76 + 84 + 85 + 86 + 87 + 89 + 89 + 95 + 96 + 98 14 = 84 CONTOH

Matematik Tingkatan 4 Jawapan J38 Sisihan piawai/ Standard deviation: – 842 692 + 712 + 752 + 762 + 762 + 842 + 852 + 862 + 872 + 892 + 892 + 952 + 962 + 982 14 = 99 912 14 – 7 056 = √80.571 = 8.98 (ii) Markah murid di Kelas 2 Melati lebih berserak manakala markah murid di Kelas 2 Ros adalah lebih bertumpu kepada minnya. The marks of students in Class 2 Melati is more dispersed whereas the marks of students in Class 2 Ros is more converge towards its mean. (iii) Kelas 2 Ros kerana kebanyakan muridnya memperoleh markah yang lebih tinggi daripada Kelas 2 Melati. Class 2 Ros because most of its students obtained higher marks than Class 2 Melati. (b) (i) Jenis mentol Type of bulb Julat Range Julat antara kuartil Interquartile range A 75 – 50 = 25 hari/ days 71 – 57 = 14 hari/ days B 77 – 62 = 15 hari/ days 74 – 66 = 8 hari/ days C 75 – 57 = 18 hari/ days 72 – 60 = 12 hari/ days (ii) Taburan jangka hayat bagi mentol jenis C adalah lebih bertumpu ke kanan berbanding dengan mentol jenis A, maka penggunaan mentol jenis C adalah lebih tahan lama. The distribution of life span for type C bulb converged more to the right compared to type A bulb, thus the usage of type C bulb last longer. (iii) Mentol jenis B kerana antara tiga jenis mentol itu, median jangka hayat bagi mentol jenis B adalah paling tinggi dan taburan jangka hayatnya adalah agak konsisten dengan julat antara kuartil yang rendah. Type B bulb because among the three types of bulbs, the median of life span of type B bulb is the highest and the distribution of its life spans is very consistent with short interquartile range. (c) (i) Julat/ Range: 50 – 35 = 15 m Julat antara kuartil/ Interquartile range: 50 – 40 = 10 m (ii) Tinggi (m) Height (m) 30 35 40 45 50 65 Bilangan pokok Number of trees 2 5 13 8 11 1 Julat baharu/ New range: 65 – 30 = 35 m Q1 = Cerapan ke-(40 4 )/ ( 40 4 ) th observation = Cerapan ke-10/ 10th observation = 40 m Q3 = Cerapan ke-[ 3(40) 4 ]/ [ 3(40) 4 ] th observation = Cerapan ke-30/ 30th observation = 50 m Julat antara kuartil baharu/ New interquartile range: 50 – 40 = 10 m (iii) Tidak, kerana wujud pencilan. No, because there is an outlier. 12 (a) (i) σ = Σfx2 Σf – x2 7.331 = 16 200 Σf – 27.52 53.744 = 16 200 Σf – 756.25 809.994 = 16 200 Σf Σf = 20 orang penari/ dancers (ii) Bilangan penari yang tertinggal: Number of dancers left out: 20 – 18 = 2 Jumlah masa latihan bagi penari yang tertinggal: Total practice time for the left-out dancers: 2(15) + 3(20) + 6(25) + 4(30) + 35 + 2(40) + x = 20(27.5) 30 + 60 + 150 + 120 + 35 + 80 + x = 550 x = 75 75 – 5 2 = 35 jam/ hours ∴ Masa latihan bagi dua orang penari yang tertinggal ialah 35 jam dan 40 jam masing-masing. Practice times for the two left-out dancers are 35 hours and 40 hours respectively. (b) Set I: Σx N = x Σx 10 = 40.5 Σx = 405 Set II: Σx N = x Σx 15 = 44.3 Σx = 664.5 NI + NII = 10 + 15 = 25 Σx2 N – x2 = σ2 Σx2 10 – 40.52 = 1.2 Σx2 = 16 414.5 Σx2 N – x2 = σ2 Σx2 15 – 44.32 = 1.6 Σx2 = 29 461.35 CONTOH

Matematik Tingkatan 4 Jawapan J39 ΣxI + ΣxII = 405 + 664.5 = 1 069.5 Σx2 I + Σx2 II = 16 414.5 + 29 461.35 = 45 875.85 Min baharu/ New mean: 1 069.5 35 = 42.78 Sisihan piawai/ Standard deviation: 45 875.85 25 – 42.782 = √4.9056 = 2.215 (c) Σf = 13 + p + q Σfx = 59 + 3p + 5q Σfx2 = 311 + 9p + 25q x f fx x2 fx2 2 3 6 4 12 3 p 3p 9 9p 4 5 20 16 80 5 q 5q 25 25q 6 2 12 36 72 7 3 21 49 147 Min/ Mean: 59 + 3p + 5q 13 + p + q = 4.44 59 + 3p + 5q = 57.72 + 4.44p + 4.44q 1.44p = 1.28 + 0.56q p = 1.28 + 0.56q 1.44 ……… Sisihan piawai/ Standard deviation: 311 + 9p + 25q 13 + p + q – 4.442 = 1.47 311 + 9p + 25q 13 + p + q = 2.16 + 19.71 311 + 9p + 25q = 21.87(13 + p + q) 311 + 9p + 25q = 284.31 + 21.87p + 21.87q 12.87p – 3.13q = 26.69 ………➁ Gantikan ➀ ke dalam ➁: Replace ➀ into ➁ 12.87(1.28 + 0.56q 1.44 ) – 3.13q = 26.69 11.44 + 5q – 3.13q = 26.69 1.87q = 15.25 q = 8.16 ≈ 8 Apabila q = 8/ When q = 8, p = 1.28 + 0.56(8) 1.44 = 4 ∴ p = 4, q = 8 Praktis Kendiri Kertas 1 1 5 + 8 + p + q + 4 + 5 6 = 35 6 22 + p + q = 35 p + q = 13 Jawapan/ Answer: D 2 Min/ Mean: 100 + 96 + 42 + 87 + 50 + 60 6 = 435 6 = 145 2 Varians/ Variance: 1002 + 962 + 422 + 872 + 502 + 602 6 – (145 2 ) 2 = 34 649 6 – 21 025 4 = 518.58 Jawapan/ Answer: A 3 Σx = 90, σ2 = 15 Σx2 N – ( Σx N ) 2 = σ2 Σx2 6 – (90 6 ) 2 = 15 Σx2 6 – 225 = 15 Σx2 6 = 240 Σx2 = 1 440 Jawapan/ Answer: C 4 Varians tidak berubah dengan penambahan 2 pada setiap nombor. Variance does not change with the addition of 2 on each number. Jawapan/ Answer: B 5 x f fx x2 fx2 0 1 0 0 0 1 5 5 1 5 2 4 8 4 16 3 2 6 9 18 4 3 12 16 48 5 1 5 25 25 6 1 6 36 36 Σf = 17 Σfx = 42 Σfx2 = 148 CONTOH

Matematik Tingkatan 4 Jawapan J40 148 17 – ( 42 17) 2 = 148 17 – 1 764 289 = 752 289 = 1.61 Jawapan/ Answer: B 6 Set data baharu/ New set of data: 30, 36, 16, 40, 44, 56 N = 6 Σx = 30 + 36 + 16 + 40 + 44 + 56 = 222 Σx2 = 302 + 362 + 162 + 402 + 442 + 562 = 9 124 Sisihan piawai baharu/ New standard deviation: 9 124 6 – ( 222 6 ) 2 = 455 3 = 12.32 Jawapan/ Answer: B 7 Saiz kasut 42 merupakan pencilan kepada set data kelas tersebut, maka julat, varians dan sisihan piawai tidak sesuai mewakili taburan saiz kasut kelas. The shoe size of 42 is an outlier to the set of data, thus range, variance and standard deviation are not suitable to represent the distribution of shoe size of the class. Jawapan/ Answer: D Kertas 2 Bahagian A/ Section A 1 (a) (i) Susun semula/ Rearrange: Q1 Q Median 3 32, 35, 43, 44, 56, 59, 66, 78, 88 Julat antara kuartil/ Interquartile range: 66 + 78 2 – 35 + 43 2 = 72 – 39 = 33 (ii) Susun semula/ Rearrange: Q1 Q Median 3 0.2, 0.4, 0.8, 1.2, 1.35, 1.5, 1.6, 2.1, 2.2, 2.4, 2.4, 3.1 Julat antara kuartil/ Interquartile range: 2.2 + 2.4 2 – 0.8 + 1.2 2 = 2.3 – 1.0 = 1.3 (b) x : 5 + 3 + 7 + 8 + 8 + 6 + 7 + 5 + 7 + 9 10 = 65 10 = 6.5 Σx2 N : 52 + 32 + 72 + 82 + 82 + 62 + 72 + 52 + 72 + 92 10 = 451 10 = 45.1 Sisihan piawai/ Standard deviation: σ2 = 45.1 – 6.52 = 45.1 – 42.25 = 2.85 σ = 1.688 ∴ Dengan nilai sisihan piawai yang besar, cerapan tidak berserak dekat min. With a big value of standard deviation, the data are not distributed close to the mean. 2 (a) Julat/ Range: 22 – 10 = 12 kg Susun semula/ Rearrange: Q1 Q Median 3 10, 11, 14, 15, 16, 17, 20, 20, 21, 22 Julat antara kuartil/ Interquartile range: 20 – 14 = 6 kg (b) 10 14 16.5 20 22 Jisim/ Mass (kg) Bahagian B/ Section B 3 (a) 3 5 8 9 4 0 2 4 8 5 0 0 4 5 6 5 Batang Stem Daun Leaf Kekunci: 3 | 5 bermaksud RM35 Key: 3 | 5 means RM35 (b) 35, 38, 39, 40, 42, 44, 48, 50, 50, 54, 55, 65 Q1 Q Median 3 Julat antara kuartil/ Interquartile range: 50 + 54 2 – 39 + 40 2 = 52 – 39.5 = 12.5 CONTOH

Matematik Tingkatan 4 Jawapan J41 (c) 35 40 45 50 55 60 65 Pendapatan harian (RM) Daily income (RM) 4 (a) x f fx x2 fx2 40 4 160 1 600 6 400 45 6 270 2 025 12 150 50 8 400 2 500 20 000 55 7 385 3 025 21 175 60 15 900 3 600 54 000 Σf = 40 Σfx = 2 115 Σfx2 = 113 725 x = Σfx Σf = 2 115 40 = 52.875 σ2 = Σfx2 Σf – x2 = 113 725 40 – 52.8752 = 2 843.125 – 2 795.766 = 47.359 σ = √47.359 = 6.882 (b) x f fx x2 fx2 80 4 320 6 400 25 600 90 6 540 8 100 48 600 100 8 800 10 000 80 000 110 7 770 12 100 84 700 120 15 1 800 14 400 216 000 Σf = 40 Σfx = 4 230 Σfx2 = 454 900 x = Σfx Σf = 4 230 40 = 105.75 σ2 = Σfx2 Σf – x2 = 454 900 40 – 105.752 = 11 372.5 – 11 183.063 = 189.437 σ = √189.437 = 13.764 (c) σa : σb = 6.882 : 13.764 = 1 : 2 Apabila setiap cerapan dalam set data diganda dua, sisihan piawai baharu juga diganda dua. When each observation in the set of data is doubled, the standard deviation is also doubled. 5 (a) Ravi: Min/ Mean: 22.3 + 23.2 + 21.56 + 22.15 4 = 22.3025 s Sisihan piawai/ Standard deviation: 22.32 + 23.22 + 21.562 + 22.152 4 – 22.30252 = √0.345 = 0.5874 Anuar: Min/ Mean: 23.5 + 21.3 + 22.43 + 22.5 4 = 22.4325 s Sisihan piawai/ Standard deviation: 23.52 + 21.32 + 22.432 + 22.52 4 – 22.43252 = √0.6067 = 0.7789 ∴ Ravi, kerana sisihan piawai masa lariannya adalah lebih rendah. Ravi, because the standard deviation of his running time is lower. (b) Ravi: Min/ Mean: 22.3 + 21.56 + 22.15 3 = 66.01 3 = 22.003 s Sisihan piawai/ Standard deviation: 22.32 + 21.562 + 22.52 3 – ( 66.01 3 ) 2 = √0.102 = 0.319 Anuar: Min/ Mean: 21.3 + 22.43 + 22.5 3 = 66.23 3 = 22.077 s Sisihan piawai/ Standard deviation: 21.32 + 22.432 + 22.152 3 – (66.23 3 ) 2 = √0.3024 = 0.5499 ∴ Tidak, nilai min dan sisihan piawai Ravi masih lebih rendah daripada Anuar. No, Ravi’s mean value and standard deviation are CONTOH still lower than Anuar.

Matematik Tingkatan 4 Jawapan J42 Bab 9 9.1 Peristiwa Bergabung Combined Events 1 (a) S = {(Kepala, 1), (Kepala, 2), (Kepala, 3), (Kepala, 4), (Kepala, 5), (Kepala, 6), (Ekor, 1), (Ekor, 2), (Ekor, 3), (Ekor, 4), (Ekor, 5), (Ekor, 6)} S = {(Head, 1), (Head, 2), (Head, 3), (Head, 4), (Head, 5), (Head, 6), (Tail, 1), (Tail, 2), (Tail, 3), (Tail, 4), (Tail, 5), (Tail, 6)} (b) S = {(R, R), (R, A), (R, P), (R, I), (A, R), (A, A), (A, P), (A, I), (P, R), (P, A), (P, P), (P, I), (I, R), (I, A), (I, P), (I, I)} (c) S = {(Q, R), (Q, 5), (Q, 3), (Q, 2), (Q, 9), (R, Q), (R, 5), (R, 3), (R, 2), (R, 9), (5, Q), (5, R), (5, 3), (5, 2), (5, 9), (3, Q), (3, R), (3, 5), (3, 2), (3, 9), (2, Q), (2, R), (2, 5), (2, 3), (2, 9), (9, Q), (9, R), (9, 5), (9, 3), (9, 2)} (d) S = {(J, A), (J, M), (J, B), (J, U), (A, J), (A, M), (A, B), (A, U), (M, J), (M, A), (M, B), (M, U), (B, J), (B, A), (B, M), (B, U), (U, J), (U, A), (U, M), (U, B)} (e) S = {(Biru, Merah), (Biru, Hitam), (Merah, Biru), (Merah, Hitam), (Hitam, Biru), (Hitam, Merah)} S = {(Blue, Red), (Blue, Black), (Red, Blue), (Red, Black), (Black, Blue), (Black, Red)} 2 (a) (i) B = {2, 3, 5, 7} C = {2, 4, 6, 8, 10} S = {(2, 2), (2, 4), (2, 6), (2, 8), (2, 10), (3, 2), (3, 4), (3, 6), (3, 8), (3, 10), (5, 2), (5, 4), (5, 6), (5, 8), (5, 10), (7, 2), (7, 4), (7, 6), (7, 8), (7, 10)} (ii) n(S) = 20 (b) (i) P = {1, 3, 5} Q = {2, 3, 5} S = {(1, 2), (1, 3), (1, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)} (ii) n(S) = 9 (c) S = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2)} 9.2 Peristiwa Bersandar dan Peristiwa Tak Bersandar Dependent Events and Independent Events 1 (a) Peristiwa tak bersandar kerana kebarangkalian mendapat guli biru yang pertama tidak mempengaruhi kebarangkalian mendapat guli biru yang kedua. Independent event because the probability of obtaining the first blue marble does not affect the probability of obtaining the second blue marble. (b) Peristiwa bersandar kerana kebarangkalian mendapat butang putih mempengaruhi kebarangkalian mendapat butang kuning. Dependent event because the probability of obtaining the white button affects the probability of obtaining the yellow button. (c) Peristiwa tak bersandar kerana kebarangkalian memilih kad yang pertama tidak mempengaruhi kebarangkalian memilih kad yang kedua. Independent event because the probability of choosing the first card does not affect the probability of choosing the second card. (d) Peristiwa bersandar kerana kebarangkalian mendapat bola hijau yang pertama mempengaruhi kebarangkalian mendapat bola hijau yang kedua. Dependent event because the probability of obtaining the first green ball affects the probability of obtaining the second green ball. 2 (a) Kaedah I/ Method I: P(nombor perdana/ prime number) = 3 5 P(huruf vokal/ vowel letter) = 2 5 P(nombor perdana dan huruf vokal): P(a prime number and a vowel letter): 3 5 × 2 5 = 6 25 Kaedah II/ Method II: Kesudahan yang mungkin/ Possible outcomes: {(2, I), (2, A), (5, I), (5, A), (7, I), (7, A)} n(S) = 5 × 5 = 25 P(nombor perdana dan huruf vokal) = 6 25 P(a prime number and a vowel letter) ∴ Kedua-dua kaedah menghasilkan keputusan yang sama. Both methods produce the same result. (b) Kaedah I/ Method I: P(tuala krim pertama/ first cream towel) = 3 9 = 1 3 P(tuala krim kedua/ second cream towel) = 3 9 = 1 3 P(dua tuala krim/ two cream towels): 1 3 × 1 3 = 1 9 Kaedah II/ Method II: Kesudahan yang mungkin/ Possible outcomes: {(K1 , K1 ), (K1 , K2 ), (K1 , K3 ), (K2 , K1 ), (K2 , K2 ), (K2 , K3 ), (K3 , K1 ), (K3 , K2 ), (K3 , K3 )} n(S) = 9 × 9 = 81 P(dua tuala krim/ two cream towels): 9 81 = 1 9 ∴ Kedua-dua kaedah menghasilkan keputusan yang sama. Both methods produce the same result. 3 (a) P (nombor 4)/ P(number 4) = 1 6 P(nombor perdana)/ P( a prime number) = 3 6 = 1 2 P(nombor 4 dan nombor perdana): P(number 4 and a prime number): 1 6 × 1 2 = 1 12 CONTOH

Matematik Tingkatan 4 Jawapan J43 (b) P[(U, U) atau (B, B) atau (P, P)] P[(U, U) or (B, B) or (P, P)] = ( 3 12 × 2 11) + ( 4 12 × 3 11) + ( 5 12 × 4 11) = 6 + 12 + 20 132 = 38 132 = 19 66 (c) (i) P(B, B) = 15 25 × 14 24 = 210 600 = 7 20 (ii) P(B, M) = 15 25 × 10 24 = 150 600 = 1 4 (d) (i) P(P, B) = 25 45 × 20 44 = 500 1 980 = 25 99 (ii) P(P, P) = 25 45 × 24 44 = 600 1 980 = 10 33 (e) P(vokal dan konsonan) = 2 6 × 4 5 P(vowel and consonant) = 4 15 (f) Hari pertama First day Hari kedua Second day Kesudahan Outcomes 0.3 0.3 0.3 0.7 0.7 0.7 L L (L, L) (L, L�) (L�, L) (L�, L�) L L� L� L� L = Bangun lewat/ Wake up late L� = Tidak bangun lewat/ Does not wake up late P(L, L) = 0.3 × 0.3 = 0.09 (g) (i) P(huruf vokal dan nombor genap): P(a vowel letter and an even number): 2 4 × 2 3 = 1 3 (ii) P(huruf konsonan dan faktor bagi 9): P(a consonant letter and a factor of 9): 2 4 × 1 3 = 1 6 (iii) P(huruf konsonan dan gandaan 6): P(a consonant letter and a multiple of 6): 2 4 × 2 3 = 1 3 (h) Bola pertama First ball Bola kedua Second ball Kesudahan Outcomes 3 10 7 10 2 9 7 9 6 9 P P (P, P) (P, J) (J, P) (J, J) P J J J 3 9 P = Putih/ White J = Jingga/ Orange (i) P(J, P) = 7 10 × 3 9 = 7 30 (ii) P(J, J) = 7 10 × 6 9 = 42 90 = 7 15 (i) (i) P(kedua-dua gelas pecah)/ P(both glasses are broken): 3 17 × 2 16 = 6 272 = 3 136 (ii) P(tiada gelas pecah)/ P(none of the glasses are broken): 14 17 × 13 16 = 182 272 = 91 136 (iii) P(kedua-dua gelas pecah atau satu daripada gelas pecah): P(both glasses are broken or one of the glasses is broken): ( 3 17 × 2 16) + ( 3 17 × 14 16) = 6 + 42 272 = 48 272 = 3 17 CONTOH

Matematik Tingkatan 4 Jawapan J44 9.3 Peristiwa Saling Eksklusif dan Peristiwa Tidak Saling Eksklusif Mutually Exclusive Events and Non-Mutually Exclusive Events 1 (a) (i) Peristiwa tidak saling eksklusif kerana peristiwa D dan E boleh berlaku bersama. Non-mutually exclusive event because events D and E can occur together. (ii) Peristiwa saling eksklusif kerana peristiwa D dan F tidak boleh berlaku bersama. Mutually exclusive event because events D and F cannot occur together. (iii) Peristiwa saling eksklusif kerana peristiwa E dan F tidak boleh berlaku bersama. Mutually exclusive event because events E and F cannot occur together. (b) (i) Peristiwa tidak saling eksklusif kerana peristiwa F dan G boleh berlaku bersama. Non-mutually exclusive event because events F and G can occur together. (ii) Peristiwa tidak saling eksklusif kerana peristiwa F dan H boleh berlaku bersama. Non-mutually exclusive event because events F and H can occur together. (iii) Peristiwa saling eksklusif kerana peristiwa G dan H tidak boleh berlaku bersama. Mutually exclusive event because events G and H cannot occur together. (c) (i) Peristiwa saling eksklusif kerana peristiwa J dan K tidak boleh berlaku bersama. Mutually exclusive event because events J and K cannot occur together. (ii) Peristiwa tidak saling eksklusif kerana peristiwa J dan L boleh berlaku bersama. Non-mutually exclusive event because events J and L can occur together. (iii) Peristiwa saling eksklusif kerana peristiwa K dan L tidak boleh berlaku bersama. Mutually exclusive event because events K and L cannot occur together. 2 (a) (i) J = {G, M, B, R}, K = {E, I, A} J ∩ K = { } J ∪ K = {G, M, B, R, E, I, A} P(J ∪ K) = 7 7 = 1 ∴ P(J ∪ K) = P(J) + P(K) (ii) J = {G, M, B, R}, L = {R} J ∩ L = {R} J ∪ L = {G, M, B, R} P(J) + P(K) = 4 7 + 3 7 = 7 7 = 1 P(J ∪ L) = 4 7 P(J) + P(L) – P(J ∩ L) = 4 7 + 1 7 – 1 7 = 4 7 ∴ P(J ∪ L) = P(J) + P(L) – P(J ∩ L) (iii) K = {E, I, A}, L = {R} K ∩ L = { } K ∪ L = {E, I, A, R} P(K ∪ L) = 4 7 ∴ P(K ∪ L) = P(K) + P(L) (b) (i) F = {(Angka, 2), (Angka, 4), (Angka, 6)} F = {(Head, 2), (Head, 4), (Head, 6)} G = {(Gambar, 1), (Gambar, 3), (Gambar, 5)} G = {(Tail, 1), (Tail, 3), (Tail, 5)} F ∩ G = { } F ∪ G = {(Angka, 2), (Angka, 4), (Angka, 6), (Gambar, 1), (Gambar, 3), (Gambar, 5)} {(Head, 2), (Head, 4), (Head, 6), (Tail, 1), (Tail, 3), (Tail, 5)} P(F ∪ G) = 6 12 = 1 2 ∴ P(F ∪ G) = P(F) + P(G) (ii) F = {(Angka, 2), (Angka, 4), (Angka, 6)} F = {(Head, 2), (Head, 4), (Head, 6)} H = {(Angka, 3), (Angka, 6)} H = {(Head, 3), (Head, 6)} F ∩ H = {(Angka, 6)}/ {(Head, 6)} F ∪ H = {(Angka, 2), (Angka, 3), (Angka, 4), (Angka, 6)} {(Head, 2), (Head, 3), (Head, 4), (Head, 6)} P(F ∪ H) = 4 12 = 1 3 P(F) + P(H) – P(F ∩ H) = 3 12 + 2 12 – 1 12 = 4 12 = 1 3 ∴ P(F ∪ H) = P(F) + P(H) – P(F ∩ H) P(K) + P(L) = 3 7 + 1 7 = 4 7 P(F) + P(G) = 3 12 + 3 12 = 6 12 = 1 2 CONTOH

Matematik Tingkatan 4 Jawapan J45 (iii) G = {(Gambar, 1), (Gambar, 3), (Gambar, 5)} G = {(Tail, 1), (Tail, 3), (Tail, 5)} H = {(Angka, 3), (Angka, 6)} H = {(Head, 3), (Head, 6)} G Ç H = { } G ∪ H = {(Gambar, 1), (Gambar, 3), (Gambar, 5), (Angka, 3), (Angka, 6)} {(Tail, 1), (Tail, 3), (Tail, 5), (Head, 3), (Head, 6)} P(G ∪ H) = 5 12 ∴ P(G ∪ H) = P(G) + P(H) 3 (a) Nombor ganjil/ Odd numbers = {1, 3, 5} Nombor genap/ Even numbers = {2, 4, 6} P(nombor ganjil atau nombor genap): P(an odd number or an even number): P(A ∪ B) = P(A) + P(B) = 3 6 + 3 6 = 6 6 = 1 (b) P(gula-gula berperisa epal atau anggur): P(apple-flavoured sweet or grape-flavoured sweet): P(E È A) = P(E) + P(A) = 8 30 + 9 30 = 17 30 (c) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 1 4 + 1 3 – 1 5 = 15 + 20 – 12 60 = 23 60 (d) Kad nombor perdana/ Prime number cards = {5, 7} Kad huruf N/ Letter N card = {N} P(kad nombor perdana atau kad huruf N): P(a prime number card or letter N card): P(A ∪ B) = P(A) + P(B) = 2 4 + 1 5 = 10 + 4 20 = 14 20 = 7 10 P(G) + P(H) = 3 12 + 2 12 = 5 12 (e) Memakai cermin mata Wearing spectacles Tidak memakai cermin mata Does not wear spectacles Lelaki Boys 10 12 Perempuan Girls 7 11 P(murid lelaki atau murid perempuan yang tidak memakai cermin mata): P(a boy or a girl who does not wear spectacles): P(Ls , ∪ Ps ,) = P(Ls ,) + P(Ps ,) = 12 40 + 11 40 = 23 40 (f) (i) P(Muthu atau Mohan menang hadiah pertama): P(Muthu or Mohan wins the first prize): P(Mu ∪ Mo) = P(Mu) + (Mo) = 2 7 + 1 6 = 12 + 7 42 = 19 42 (ii) P(Muthu dan Mohan tidak memenangi hadiah pertama): P(Muthu and Mohan do not win the first prize): 1 – 19 42 = 23 42 (g) Danny Firdaus Kesudahan Outcomes M M (M, M) (M, M�) (M�, M) (M�, M�) M M� M� M� 3 5 3 8 3 8 2 5 5 8 5 8 M = Menyertai/ Join M� = Tidak menyertai/ Does not join P(salah seorang menyertai)/ P(one of them joins): P(M , M�) + P(M� , M) = ( 3 5 × 3 8 ) + ( 2 5 × 5 8 ) = 9 + 10 40 = 19 40 CONTOH

Matematik Tingkatan 4 Jawapan J46 9.4 Aplikasi Kebarangkalian Peristiwa Bergabung Application of Probability of Combined Events 1 (a) Kesudahan Outcomes 10 25 8 25 8 26 8 26 8 26 10 26 8 26 9 26 9 26 9 26 9 26 7 25 M M M M B B B (M, M) (M, B) (M, H) (B, M) (B, B) (B, H) (H, M) (H, B) (H, H) H H H H B M = Merah/ Red B = Biru/ Blue H = Hijau/ Green P(M, M) + P(B, B) + P(H, H) = (10 25 × 9 26) + ( 8 25 × 10 26) + ( 7 25 × 9 26) = 90 + 80 + 63 650 = 233 650 (b) (i) S = {(Kepala, 1), (Kepala, 2), (Kepala, 3), (Kepala, 4), (Kepala, 5), (Kepala, 6), (Ekor, 1), (Ekor, 2), (Ekor, 3), (Ekor, 4), (Ekor, 5), (Ekor, 6)} S = {(Head, 1), (Head, 2), (Head, 3), (Head, 4), (Head, 5), (Head, 6), (Tail, 1), (Tail, 2), (Tail, 3), (Tail, 4), (Tail, 5), (Tail, 6)} (ii) S = {(Ekor, 2), (Ekor, 4), (Ekor, 6)} S = {(Tail, 2), (Tail, 4), (Tail, 6)} P(ekor dan nombor genap)= 3 12 = 1 4 P(tail and an even number) (iii) S = {(Kepala, 3), (Ekor, 1)} S = {(Head, 3), (Tail, 1)} P[(Kepala, 3) ∪ (Ekor, 1)] = 1 12 + 1 12 P[(Head, 3) ∪ (Tail, 1)] = 2 12 = 1 6 (c) (i) 1 ✗ ✗ ✗ ✗ ✗ ✗ ✓ ✓ ✗ ✗ ✗ ✓ ✗ ✓ ✓ ✓ 1 3 2 4 2 3 4 B A Petunjuk/ Key: ✗ = Tidak melebihi 5/ Not exceed 5 ✓ = Melebihi 5/ Exceed 5 ✓ = Nombor ganjil dan melebihi 5 Odd numbers and exceed 5 P(nombor ganjil dan melebihi 5) = 2 16 = 1 8 P(an odd number and exceed 5) (ii) 1 ✗ ✗ ✗ ✗ ✗ ✗ ✓ ✓ ✗ ✗ ✗ ✓ ✗ ✓ ✓ ✓ 1 3 2 4 2 3 4 B A Petunjuk/ Key: ✗ = Tidak melebihi 5/ Not exceed 5 ✓ = Melebihi 5/ Exceed 5 = Nombor ganjil/ Odd numbers P(nombor ganjil atau melebihi 5): P(an odd number or exceed 5): P(✓ È ) = P(✓) + P( ) – P(✓ Ç ) = 6 16 + 8 16 – 2 16 = 12 16 = 3 4 (d) Tiro Titus Kesudahan Outcomes A A (A, A) (A, B) (B, A) (B, B) A B B B 5 7 5 7 2 7 1 6 5 6 2 7 A = Aiskrim/ Ice cream B = Biskut/ Biscuits (i) P(Tiro makan biskut)/ P(Tiro eats biscuits): 1 – 1 6 = 5 6 (ii) P(A, B) = 1 6 × 5 7 = 5 42 (iii) P(A, A) + P(B, B) = ( 1 6 × 2 7 ) + ( 5 6 × 5 7 ) = 2 + 25 42 = 27 42 = 9 14 CONTOH

Matematik Tingkatan 4 Jawapan J47 (e) P(Chloe) = 10 30 = 1 3 P(Cindy) = 14 30 = 7 15 P(S) = 6 30 = 1 5 (i) P(S, S, Chloe) + P(S, S, Cindy): = ( 1 5 × 1 5 × 1 3 ) + ( 1 5 × 1 5 × 7 15) = 1 75 + 7 375 = 4 125 (ii) P(Chloe, Cindy, Chloe) + P(Cindy, Chloe, Cindy): = ( 1 3 × 7 15 × 1 3 ) + ( 7 15 × 1 3 × 7 15) = 7 135 + 49 675 = 28 225 (iii) P(Cindy, Cindy, Chloe) + P(Cindy, Cindy, Cindy) + P(Cindy, Cindy, S): = ( 7 15 × 7 15 × 1 3 ) + ( 7 15 × 7 15 × 7 15) + ( 7 15 × 7 15 × 1 5 ) = 49 675 + 343 3 375 + 49 1 125 = 49 225 Praktis Kendiri Kertas 1 1 Nombor ganjil/ Odd numbers = {1, 3, 5} P(nombor ganjil dan nombor ganjil) = 3 6 × 3 6 P(odd numbers and odd numbers) = 1 4 Jawapan/ Answer: B 2 P(suka minum teh tarik) = 0.83 × 0.83 P(like to drink the tarik) = 0.6889 � 0.69 Jawapan/ Answer: C 3 P(murid lelaki dan murid lelaki) = 20 35 × 19 34 P(a boy and a boy) = 380 1 190 = 38 119 Jawapan/ Answer: B 4 P(E, O) atau/or P(O, E) = (10 18 × 8 17) + ( 8 18 × 10 17) = 80 + 80 306 = 160 306 = 80 153 Jawapan/ Answer: C 5 P(P È Q) = P(P) + P(Q) = 1 3 + 2 5 = 5 + 6 15 = 11 15 Jawapan/ Answer: D 6 P(pai epal È pai pisang)/ P(apple pie È banana pie): P(E) + P(P) – P(E Ç P) = 0.31 + 0.45 – 0.09 = 0.67 Jawapan/ Answer: C 7 A = {2, 4, 6} B = {Ekor}/ {Tail} P(A, B) = 3 6 × 1 2 = 1 4 Jawapan/ Answer: C 8 Kad merah/ Red card = 10 keping/ pieces P(Kad biru/ Blue card) = 2 5 P(Kad hijau/ Green card) = 4 15 P(Kad merah/ Red card) = 1 – ( 2 5 + 4 15) = 1 – 10 15 = 5 15 = 1 3 Kad biru/ Blue card = 2 5 × 3 × 10 = 12 keping/ pieces Jumlah bilangan kad/ Total number of cards: 3 × 10 = 30 keping/ pieces P(kad merah/ red card) = 10 30 + 5 = 10 35 = 2 7 Jawapan/ Answer: B CONTOH

Matematik Tingkatan 4 Jawapan J48 Kertas 2 Bahagian A/ Section A 1 (a) P(J, J) = 14 32 × 13 31 = 182 992 = 91 496 (b) P(P, P) = 6 32 × 5 31 = 30 992 = 15 496 2 Kesudahan Outcomes 18 40 22 40 21 39 18 39 22 39 17 39 P P (P, P) (P, L) (L, P) (L, L) P L L L P = Perempuan/ Girl L = Lelaki/ Boy P(L, L) = 22 40 × 21 39 = 462 1 560 = 77 260 ∴ Peristiwa bersandar/ Dependent event 3 (a) P(O ∩ T) = 3 7 × 2 3 = 2 7 P(O sahaja/ only) = 3 7 – 2 7 = 1 7 P(T sahaja/ only) = 2 3 – 2 7 = 14 – 6 21 = 8 21 ξ 4 21 O T 8 21 2 7 1 7 (b) P(Mary tidak membeli oren dan tembikai): P(Mary does not buy orange and watermelon): 1 – ( 1 7 + 8 21 + 2 7 ) = 1 – 17 21 = 4 21 Bahagian B/ Section B 4 (a) P(L, L, L) = 20 35 × 19 34 × 18 33 = 6 840 39 270 = 228 1 309 (b) P(L, L, L) = 20 35 × 19 34 × 18 33 = 6 840 39 270 = 2 28 1 309 (c) P(L, O, O) + P(O, L, O) + P(O, O, L) = (20 35 × 15 34 × 14 33) × (15 35 × 20 34 × 14 33) + (15 35 × 14 34 × 20 33) = 4 200 + 4 200 + 4 200 39 270 = 12 600 39 270 = 60 187 5 (a) Diana Ezzaty Kesudahan Outcomes L L (L, L) (L, G) (G, L) (G, G) L G G G 2 5 2 3 1 3 2 3 1 3 3 5 L = Lulus/ Passed G = Gagal/ Failed CONTOH

Matematik Tingkatan 4 Jawapan J49 (b) P(salah seorang gagal ujian): P(one of them failed in the test): P(L, G) + P(G, L) = ( 2 5 × 1 3 ) + ( 3 5 × 2 3 ) = 2 15 + 6 15 = 8 15 (c) P(sekurang-kurangnya seorang lulus ujian): P(at least one of them passed in the test): P(L, G) + P(G, L) + P(L, L) = ( 2 5 × 1 3 ) + ( 3 5 × 2 3 ) + ( 2 5 × 2 3 ) = 2 + 6 + 4 15 = 12 15 = 4 5 Bab 10 10.1 Perancangan dan Pengurusan Kewangan Financial Planning and Management 1 (a) • Khusus: Puan Siti ingin menyambut hari lahir anak perempuannya dalam tempoh setengah tahun kemudian. Specific: Mrs Siti wants to celebrate her daughter’s birthday half year later. • Boleh diukur: Puan Siti bercadang menyimpan RM200 sebulan daripada pendapatan bulanannya. Measurable: Mrs Siti plans to save RM200 every month from her monthly income. • Boleh dicapai: RM200 merupakan 8.33% daripada jumlah pendapatan bulanan Puan Siti, maka dia berkemampuan untuk menyimpannya. Attainable: RM200 is 8.33% of Mrs Siti’s monthly income, so she is able to save. • Realistik: Puan Siti mampu menyimpan RM200 setiap bulan dan dapat mencapai RM1 200 dalam setengah tahun. Realistic: Mrs Siti affords to save RM200 every month and able to achieve RM1 200 in half year. • Tempoh masa: Setengah tahun Time-bound: Half year Oleh itu, Puan Siti mengamalkan konsep SMART untuk mencapai matlamat tersebut. Therefore, Mrs Siti applied the SMART concept to achieve the goal. (b) • Khusus: Arjun merancang untuk bercuti ke Sabah dalam tempoh enam bulan. Specific: Arjun plans a trip to Sabah within six months. • Boleh diukur: Arjun bercadang menyimpan RM250 sebulan daripada gaji bulanannya. Measurable: Arjun plans to save RM250 every month from his monthly salary. • Boleh dicapai: Arjun perlu lapan bulan untuk mencapai matlamat kewangannya sebanyak RM2 000. Simpanan selama enam bulan hanya mencapai 75% matlamat kewangannya. Attainable: Arjun needs eight months to achieve his financial goal of RM2 000. Six months’ saving only reaches 75% of his financial goal. • Realistik: Matlamat kewangan Arjun untuk bercuti ke Sabah dalam tempoh enam bulan adalah tidak realistik dengan simpanan sebanyak RM250 sebulan. Beliau perlu menyimpan sebanyak RM334 sebulan untuk mencapai matlamatnya. Realistic: Arjun’s financial goal to go for a trip to Sabah within six months is not realistic with a saving of RM250 per month. He needs to save RM334 per month to achieve his goal. • Tempoh masa: Enam bulan Time-bound: Six months Oleh itu, Arjun tidak mengamalkan konsep SMART untuk mencapai matlamat tersebut. Therefore, Arjun did not apply the SMART concept to achieve the goal. 2 (a) RM3 500 + RM500 – RM1 200 – RM900 = RM1 900 (Aliran tunai positif/ Positive cash flow) Aliran tunai positif sebanyak RM1 900 adalah baik kerana Rachel boleh menyimpan wang tersebut di dalam bank dan menikmati faedah. Positive cash flow of RM1 900 is good as Rachel can save the money in a bank and receive interest. (b) RM1 900 + RM500 – RM1 700 – RM850 = –RM150 (Aliran tunai negatif/ Negative cash flow) Aliran tunai negatif sebanyak RM150 akan membebankan Encik Muthu dan mungkin menyebabkannya menggunakan kad kredit bagi mengatasi masalah kewangan tersebut. Negative cash flow of RM150 will burden Mr Muthu and he will probably use the credit card to facilitate his financial problems. (c) RM2 800 + RM750 – RM2 250 – RM1 050 = RM250 (Aliran tunai positif/ Positive cash flow) Aliran tunai positif sebanyak RM250 adalah baik kerana Yuan Xin boleh menggunakan lebihan pendapatan tersebut untuk melabur atau menghadapi situasi kecemasan jika perlu. Positive cash flow of RM250 is good as Yuan Xin can use the surplus of income for investment or spend it for emergency situations if necessary. (d) RM1 750 – RM1 100 – RM850 = –RM200 (Aliran tunai negatif/ Negative cash flow) Aliran tunai negatif sebanyak RM200 akan membebankan Encik Lim kerana dia mungkin menggunakan kad kredit untuk mengatasi masalah kewangan tersebut. Negative cash flow of RM200 will burden Mr Lim as he will probably use the credit card to facilitate his financial problems. CONTOH

Matematik Tingkatan 4 Jawapan J50 3 (a) Pendapatan lebihan/ Surplus of income: RM8 500 – RM3 000 = RM5 500 Jumlah simpanan dalam tempoh enam bulan: Total savings in six months: RM5 500 × 6 = RM33 000 Jumlah bulan yang diperlukan untuk membayar wang pendahuluan: Total number of months needed to make the down payment: RM100 000 RM5 500 = 18.18 bulan/ months ∴ Encik Philip dan isterinya tidak mungkin dapat membeli rumah teres tersebut dalam jangka masa enam bulan kerana jumlah wang yang berjaya dikumpulkan adalah kurang daripada wang pendahuluan. Mereka memerlukan 18.18 bulan untuk mencapai matlamat kewangan mereka. It is impossible for Mr Philip and his wife to buy the terrace house within six months as the total money saved is less than the down payment. They need to save for 18.18 months to achieve their financial goal. (b) (i) Simpanan bulanan yang diperlukan oleh Encik Ghafor: Monthly savings needed by Mr Ghafor: RM63 000 7 × 12 = RM63 000 84 = RM750 (ii) Tidak. Hal ini kerana walaupun Encik Ghafor mampu membayar wang pendahuluan sebanyak RM63 000, namun bayaran ansuran bulanan rumah akan membebankan beliau sekirannya jumlah perbelanjaannya adalah tinggi. No. It is because even though Mr Ghafor affords to pay the down payment of RM63 000, but his monthly housing loan instalments will burden him if his total expenses are high. (c) (i) Boleh. Simpanan setahun = RM550 × 12 = RM6 600. Puan Mary masih mempunyai lebihan RM600 selepas wang pendahuluan dibayar. Beliau juga mempunyai pendapatan lebihan sebanyak RM200 sebulan. Can. Annual savings = RM550 × 12 = RM6 600. Mrs Mary still has a surplus of RM600 after paying the down payment. She also has a surplus of income of RM200 every month. (ii) Ya. Hal ini kerana Puan Mary hanya berbelanja untuk keperluan asas, mempunyai simpanan bulanan dan insurans serta tidak memiliki hutang kad kredit. Yes. It is because Mrs Mary only spends her income on basic needs, has monthly savings and life insurance, and does not have any credit card debt. (iii) Ya/ Yes S Membeli sebuah kereta baharu yang berharga RM49 000. To buy a new car worth RM49 000. M Membayar wang pendahuluan sebanyak RM6 000. To make a down payment of RM6 000. A Wang pendahuluan sebanyak RM6 000 boleh dicapai dalam tempoh setahun. The down payment of RM6 000 can be achieved within a year. R Simpanan bulanan sebanyak RM550 adalah 10% daripada jumlah pendapatan bulanan, iaitu RM5 500. The monthly savings of RM550 is 10% of the total monthly income, which is RM5 500. T Dalam tempoh setahun. Within a year. (d) (i) Tidak kerana jumlah simpanan bulanan beliau ialah RM230, iaitu kurang 10% daripada pendapatan bulanan beliau. No because his monthly savings is RM230, which is less than 10% of his monthly income. (ii) RM230 × 12 × 8 = RM22 080 (<RM300 000) Beliau tidak boleh mencapai matlamat pelaburan RM300 000 dengan simpanan bulanan RM230. He cannot achieve his investment goal of RM300 000 with monthly savings of RM230. (e) (i) Deposit kereta/ Car’s deposit: 20 100 × RM90 000 = RM18 000 Jumlah pinjaman dari bank: Total loan from the bank: RM90 000 – RM18 000 = RM72 000 (ii) (a) Jumlah pinjaman/ Total loan amount: RM72 000 + (RM72 000 × 0.03 × 5) = RM72 000 + RM10 800 = RM82 800 (b) Jumlah pinjaman/ Total loan amount: RM72 000 + (RM72 000 × 0.03 × 7) = RM72 000 + RM15 120 = RM87 120 (c) Jumlah pinjaman/ Total loan amount: RM72 000 + (RM72 000 × 0.03 × 9) = RM72 000 + RM19 440 CONTOH = RM91 440

Matematik Tingkatan 4 Jawapan J51 (iii) (a) Bayaran ansuran bulanan: Payable monthly instalment: RM82 800 5 × 12 = RM1 380 (b) Bayaran ansuran bulanan: Payable monthly instalment: RM87 120 7 × 12 = RM1 037 (c) Bayaran ansuran bulanan: Payable monthly instalment: RM91 440 9 × 12 = RM847 (iv) S Membeli sebuah kereta dengan pinjaman bank sebanyak RM72 000. To buy a car with a loan of RM72 000 from the bank. M Jessica perlu membayar RM1 037 sebulan selama tujuh tahun. Jessica needs to pay RM1 037 every month for seven years. A RM1 037 merupakan 20.74% daripada jumlah pendapatan bulanan Jessica. Dia mampu membayarnya. RM1037 is 20.74% from Jessica’s monthly income. She affords to pay. R Jika perbelanjaan bulanan Jessica kurang daripada 79.26% sebulan, dia boleh membayarnya. If Jessica’s monthly expenses are less than 79.26%, she is able to pay. T Menjelaskan pinjaman dalam masa tujuh tahun. To settle the loan within seven years. (v) Pelan kewangan jangka panjang Long-term financial plan (f) (i) Pendapatan dan perbelanjaan Income and expenditure RM Gaji bersih/ Net salary Komisen/ Commission 3 600 650 Pendapatan pasif: Sewa rumah Passive income: House rental 700 Jumlah pendapatan bulanan Total monthly income 4 950 Tolak simpanan tetap bulanan Minus fixed monthly savings 360 Baki pendapatan Income balance 4 590 Pendapatan dan perbelanjaan Income and expenditure RM Tolak perbelanjaan tetap bulanan: Minus monthly fixed expenses Ansuran pinjaman rumah (1) Housing loan instalment (1) Ansuran pinjaman rumah (2) Housing loan instalment (2) Ansuran kereta/ Car instalment Takaful suami dan isteri Life insurance for husband and wife 750 550 560 200 Jumlah perbelanjaan tetap bulanan Total monthly fixed expenses 2 060 Tolak perbelanjaan tidak tetap bulanan: Minus monthly variable expenses: Perbelanjaan rumah Household expenses Perbelanjaan isteri/ Wife expenses Utiliti/ Utility Keperluan anak/ Children’s needs Perbelanjaan tol dan petrol Toll and petrol expenses Bil telefon/ Telephone bill Makan luar/ Eating out Percutian/ Holiday 800 300 150 400 200 100 300 400 Jumlah perbelanjaan tidak tetap bulanan Total monthly variable expenses 2 650 Lebihan/ Kurangan Surplus/ Deficit –120 (ii) Terdapat kurangan dalam pelan kewangan Encik Wafiy, iaitu aliran tunai negatif apabila jumlah perbelanjaan melebihi jumlah pendapatan. Bagi mencapai matlamat kewangannya, Encik Wafiy perlu mengurangkan perbelanjaan tidak tetap seperti percutian dan makan di luar di samping meningkatkan penjualan produk Y. There is a deficit in Mr Wafiy’s financial plan, which is negative cash flow as the total expenses is greater than the total income. To achieve his financial goal, Mr Wafiy needs to reduce his variable expenses such as holiday and eating out while increasing the sales of product Y. Praktis Kendiri Kertas 1 1 Jawapan/ Answer: D 2 Jawapan/ Answer: A CONTOH

Matematik Tingkatan 4 Jawapan J52 3 Jawapan/ Answer: D 4 RM3 500 + RM550 – RM2 200 – RM700 = RM1 150 Jawapan/ Answer: C 5 Jawapan/ Answer: D 6 Jawapan/ Answer: C 7 Simpanan bulanan/ Monthly savings: RM30 240 6 × 12 = RM420 Jawapan/ Answer: A 8 Simpanan tetap/ Fixed saving: 10 100 × RM3 680 = RM368 Jawapan/ Answer: A Kertas 2 Bahagian A/ Section A 1 • Khusus: Menyimpan RM600 dalam masa tiga bulan. Specific: Saves RM600 in three months. • Boleh diukur: Menyimpan RM200 dalam masa sebulan. Measurable: Saves RM200 a month. • Boleh dicapai: David mampu mengumpul RM600 dengan menyimpan sebanyak RM200 sebulan. Attainable: David is able to collect RM600 by saving RM200 every month. • Realistik: David mampu menyimpan RM200 sebulan jika perbezaan pendapatan dengan perbelanjaan lebih daripada RM200 sebulan. Realistic: David is able to save RM200 a month if the difference between his income and expenses is more than RM200 a month. • Tempoh masa: Dalam tempoh tiga bulan. Time-bound: Within three months. Bahagian B/ Section B 2 (a) Deposit kereta/ Car’s deposit: 10 100 × RM130 000 = RM13 000 Jumlah pinjaman dari bank: Total loan amount from the bank: RM130 000 – RM13 000 = RM117 000 (b) (i) Jumlah pinjaman/ Total loan amount: RM117 000 + (RM117 000 × 0.033 × 5) = RM117 000 + RM19 305 = RM136 305 (ii) Jumlah pinjaman/ Total loan amount: RM117 000 + (RM117 000 × 0.033 × 7) = RM117 000 + RM27 027 = RM144 027 (iii) Jumlah pinjaman/ Total loan amount: RM117 000 + (RM117 000 × 0.033 × 9) = RM117 000 + RM34 749 = RM151 749 (c) (i) Bayaran ansuran bulanan: Payable monthly instalment: RM136 305 5 × 12 = RM2 272 (ii) Bayaran ansuran bulanan: Payable monthly instalment: RM144 027 7 × 12 = RM1 715 (iii) Bayaran ansuran bulanan: Payable monthly instalment: RM151 749 9 × 12 = RM1 405 (d) S – Membeli sebuah kereta utiliti sukan dengan pinjaman bank sebanyak RM117 000. To buy a sport utility vehicle with a loan of RM117 000 from the bank. M – Edwin perlu membayar RM1 715 sebulan selama tujuh tahun. Edwin needs to pay RM1 715 every month for seven years. A – RM1 715 merupakan 28.58% daripada jumlah pendapatan bulanan Edwin. Dia mampu membayarnya. RM1 715 is 28.58% from Edwin’s monthly income. He affords to pay. R – Jika perbelanjaan bulanan Edwin kurang daripada 71.42% sebulan, dia boleh membayarnya. If Edwin’s monthly expenses are less than 71.42%, he is able to pay. T – Menjelaskan pinjaman dalam masa tujuh tahun. To settle the loan within seven years. Soalan SPM Kertas 2 Bahagian C 1 (a) (i) Lambungan dadu Chew Wai ialah suatu peristiwa tidak bersandar manakala lambungan dadu Lee Fang ialah suatu peristiwa bersandar. Chew Wai’s dice-tossing is an independent event whereas Lee Fang’s dice-tossing is a dependent event. (ii) Kebarangkalian Chew Wai mendapat nombor 6 pada dadu Q: 1 12 The probability of Chew Wai getting a number 6 on dice Q: 1 12 CONTOH

Matematik Tingkatan 4 Jawapan J53 Kebarangkalian Lee Fang mendapat nombor 6 pada dadu Q: The probability of Lee Fang getting a number 6 on dice Q: 1 6 × 1 12 = 1 72 (b) Panjang sisi dadu P/ Side length of dice P: √3.375 3 = 1.5 cm Katakan P sebagai jumlah luas permukaan dadu P dan Q sebagai jumlah luas permukaan dadu Q. Let P be the total surface area of dice P and Q be the total surface area of dice Q. P Q = 2 3 Q = 3P 2 = 3(6 × 1.52 ) 2 = 20.25 cm2 Luas bagi satu permukaan dadu Q: Area for one surface of dice Q: 20.25 12 = 1.6875 cm2 (c) Katakan P sebagai bilangan dadu P yang dibeli dan Q sebagai bilangan dadu Q yang dibeli. Let P be the number of dice P bought and Q be the number of dice Q bought. 7P + 3Q = 7.6, × 5 35P + 15Q = 38 ……➀ 9P + 5Q = 10.8, × 3 27P + 15Q = 32.4 ……➁ ➀ – ➁: 35P – 27P = 38 – 32.4 8P = 5.6 P = 0.7 P = 0.7 ↷ 7P + 3Q = 7.6, 7(0.7) + 3Q = 7.6 4.9 + 3Q = 7.6 3Q = 2.7 Q = 0.9 ∴ Harga bagi sebiji dadu P ialah RM0.70 dan sebiji dadu Q ialah RM0.90. The price of a dice P is RM0.70 and a dice Q is RM0.90. (d) Chew Wai dan Lee Fang perlu menyertai pertandingan itu. Ini kerana, jika bilangan peserta tidak mencapai 35 orang, maka Chew Wai dan Lee Fang akan turut menyertai pertandingan itu, dan bilangan peserta tidak mencapai 35 orang, iaitu hanya 30 orang telah mendaftar. Chew Wai and Lee Fang need to join the competition. Because if the number of participants does not reach 35, then Chew Wai and Lee Fang will have to join the competition, and the number of participants does not reach 35, that is only 30 people has registered. 2 (a) x�: 980 + 990 + 1 000 + 1 050 + 1 020 + 920 + 850 + 1 020 + 1 000 + 1 010 + 880 + 1 040 12 = 11 760 12 = 980 Σ(x ‒ x̅)2 = 44 000 Sisihan piawai/ Standard deviation: Σ(x ‒ x̅)2 N = 44 000 12 = √3666.6667 = ±60.55 x (x – x)2 980 0 990 100 1 000 400 1 050 4 900 1 020 1 600 920 3 600 850 16 900 1 020 1 600 1 000 400 1 010 900 880 10 000 1 040 3 600 ∴ Min jualan bulanan ialah RM980 dan sisihan piawainya ialah RM60.55. Berdasarkan sukatan serakan yang diperoleh, jualan nasi lemak Makcik Zuraini berserak antara julat RM980 ± RM60.55. The mean of monthly sales is RM980 and its standard deviation is RM60.55. Based on the measure of dispersion obtained, Auntie Zuraini’s nasi lemak sales dispersed between the range of RM980 ± RM60.55. (b) (i) Margin untung/ Profit margin = Jualan/ Sales ‒ Kos/ Cost Jualan/ Sales × 100% 75% = RM980 ‒ Kos/ Cost RM980 × 100% RM980 – Kos/ Cost = 0.75 × RM980 Kos/ Cost = RM980 – RM735 = RM245 Keuntungan bulanan/ Monthly profit: RM980 – RM245 = RM735 (ii) Pendapatan dan perbelanjaan Income and expenditure Amaun (RM) Amount (RM) Pendapatan aktif/ Active income Pendapatan pasif/ Passive income 735 500 Jumlah pendapatan/ Total income 1 235 Tolak simpanan/ Minus saving 0 Baki pendapatan/ Income balance 1 235 CONTOH

Matematik Tingkatan 4 Jawapan J54 Pendapatan dan perbelanjaan Income and expenditure Amaun (RM) Amount (RM) (‒) Perbelanjaan tetap bulanan Monthly fixed saving Sewa rumah/ Rent Premium insurans Insurance premium 500 250 Jumlah perbelanjaan tetap bulanan Total monthly fixed expenses 750 (‒) Perbelanjaan tidak tetap bulanan/ Monthly variable expenses Bil elektrik dan air Electric and water bills Barangan dapur/ Groceries 120 300 Jumlah perbelanjaan tidak tetap bulanan Total monthly variable expenses 420 Pendapatan lebihan Surplus of income 65 (iii) Jumlah simpanan dalam masa 6 bulan: Total saving in 6 months: 10% × RM65 × 6 = RM39 x 100 × RM65 × 6 > RM85 x > RM85 × 100 RM65 × 6 x > 21.79% ∴ Tidak munasabah, kerana RM39 < RM85, simpanan Makcik Zuraini adalah tidak cukup membeli set baju baharu itu. Makcik Zuraini perlu menyimpan sekurangkurangnya 22% daripada pendapatan lebihannya untuk membeli set baju itu. Not reasonable, because RM39 < RM85, Auntie Zuraini’s saving is not enough to buy the new set of clothes. Auntie Zuraini needs to save at least 22% of her surplus of income to buy the set of clothes. 3 (a) (i) Jarak, d (km) Distance, d (km) Masa, t (minit) Time, t (minute) 0 15 24 24 33 Motosikal Motorcycle Kereta Car (t, d) (ii) Laju/ Speed: 24 km (15 ÷ 60) j/ h = 24 0.25 = 96 km/j/ km/h (b) (i) Kereta/ Car: (15, 24), (33, 0) 0 ‒ 24 33 – 15 = d ‒ 24 t ‒ 15 t – 15 = 18(d ‒ 24) ‒24 = ‒ 3 4 d + 18 t = ‒ 3 4 d + 33 Motosikal/ Motorcycle: (0, 0), (24, 24) 24 ‒ 0 24 – 0 = t ‒ 0 d ‒ 0 t = d (ii) t = ‒ 3 4 d + 33 …… ➀ t = 3 4 d …… ➁ ➁ ↷ ➀: d = ‒ 3 4 d + 33 d + 3 4 d = 33 7 4 d = 33 d = 33 × 4 7 = 18.857 km t = 18.857 min (c) Jumlah masa yang diambil/ Total time taken: 24 + 20 + 24 = 68 minit/ minutes = 1 jam 8 minit/ 1 hour 8 minutes Masa tiba di bandar P/ Arrival time at town P: 11 jam 50 minit + 1 jam 8 minit = 12 jam 58 minit 11 hours minutes + 1 hour 8 minutes = 12 hours 58 minutes = 12.58 p.m. ∴ Penunggang motosikal itu akan pulang ke bandar P selepas menyiapkan kerjanya. The motorcyclist will return to town P after completing his work. 4 (a) ξ 3 M S B y x + y = 12 x + 2 6 8 3 2 (b) (i) x + y = 12 ……➀ x + 2 = y ……..➁ CONTOH

Matematik Tingkatan 4 Jawapan J55 (ii) ➁ ↷ ➀, x + (x + 2) = 12 2x = 10 x = 5 x = 5 ↷ ➁, 5 + 2 = y y = 7 Bilangan murid yang suka bahasa: Number of students who likes language: 3 + 8 + 2 + 7 = 20 Bilangan murid yang suka sains: Number of students who likes science: 6 + 8 + 2 + (5 + 2) = 23 (c) (i) ξ 3 M S B 6 7 8 3 12 2 n[(M È S) Ç B′] = 12 + 6 + 7 = 25 (ii) Kebarangkalian/ Probability: 25 48 × 24 47 = 25 94 (d) Jumlah bilangan ahli persatuan matematik dan sains, p: Total number of mathematics and science society members, p: 5 5 + 19 = 25 p p = 25 × 24 5 = 120 4 120 4 30 … 0 4 7 … 2 4 1 … 3 0 … 1 13204 ∴ p = 13204 Kertas Model Pra-SPM Kertas 1 1 A Persamaan ialah satu pernyataan yang menunjukkan dua nilai ungkapan yang sama. Equation is a statement which shows equal values of two expressions. 2 B 3 1 2 3 + 1 0 2 3 1 0 2 1 2 1 1 1 4 4 4 3 C Bentuk premis 1 ialah ‘Jika p, maka q’ dan bentuk premis 2 ialah ‘p adalah benar’. Oleh itu, bentuk kesimpulan ialah ‘q adalah benar’. The form of premise 1 is ‘If p, then q’ and the form of premise 2 is ‘p is true’. Therefore, the form of conclusion is ‘q is true’. 4 A Satu graf sekurang-kurangnya mempunyai sepasang bucu yang berkait. A graph has at least one pair of linked vertices. 5 D ξ = {70, 71, 72, 73, 74, 75, 76, 77, 78, 79} M = {70, 77} N = {71, 73, 75, 77, 79} (M ∪ N) = {70, 71, 73, 75, 77, 79} (M ∪ N)′ = (72, 74, 76, 78) 6 D 3x + 2 y A 3(1) + 2 = 5 2 (< 5) B 3(3) + 2 = 11 11 (= 11) C 3(5) + 2 = 17 7 (< 17) D 3(6) + 2 = 20 22 (> 20) 7 B Laju/ Speed: (1.2 – 0.6) × 1 000 m (1.8 – 0.2) × 60 min = 600 m 96 min = 6.25 m/min 8 C Kedua-dua peristiwa adalah saling eksklusif. Both events are mutually exclusive. 0.1 + 0.75 = 0.85 9 D Julat antara kuartil/ Interquartile range: Q3 – Q1 = 75 – 35 = 40 10 A (0.4 + 2 5) × 90 = 0.8 × 90 = 72 11 B Aliran tunai maksimum/ Maximum cash flow: RM3 500 + RM450 – RM2 000 = RM1 950 12 A PQ tiada perubahan laju, maka tiada pecutan. PQ does not have any change in speed, therefore there is no acceleration. CONTOH

Matematik Tingkatan 4 Jawapan J56 13 D X ialah baki pendapatan selepas menolak simpanan tetap bulanan. X is the balance of income after deducting fixed monthly savings. X = RM6 500 – RM650 = RM5 850 Y ialah baki selepas menolak semua perbelanjaan. Y is the balance after deducting all expenses. Y = RM5 850 – RM2 500 – RM3 180 = RM170 14 D Terdapat pencilan pada data di dalam plot batangdan-daun. Nilai julat, varians dan sisihan piawai berubah dengan kehadiran pencilan. There is an outlier in the data of the stem-and-leaf plot. The value of range, variance, and standard deviation change with the presence of outlier. 15 B 45267 = 4(73 ) + 5(72 ) + 2(7) + 6 = 163710 9 1637 9 181 8 9 20 1 9 2 2 0 2 ∴ 22189 16 D 17 D V = {G, H, I, J, K, L}, n(V) = 6 E = {(G, H), (G, K), (H, I), (H, L), (I, J), (J, K), (K, L), (L, J)} n(E) = 8 Σd(v) = 2(E) = 2(8) = 16 18 C 19 C x(x – 3) = 12 – x 2 x2 – 3x = 12 – x 2 2x2 – 6x = 12 – x 2x2 – 5x – 12 = 0 (2x + 3)(x – 4) = 0 x = –3 2 atau/ or 4 20 B 21 B c = –7, maka pintasan-y bagi fungsi ialah –7. c = –7, thus the y-intercept is –7. 22 B 1012 = 22 + 1 = 5 = Q 13558 = 83 + 3(82 ) + 5(8) + 5 = 74910 6 749 6 124 5 6 20 4 6 3 2 0 3 32456 ∴ P = 2, Q = 5 23 D Susunan ketinggian dalam tertib menurun: Arrangement of height in descending order: Rizal Mina Lina Nazri 24 A Perjalanan yang mungkin Possible route Jumlah jarak Total distance 250 m + 0.48 km 0.73 km 250 m + 250 m + 350 m 0.85 km 320 m + 250 m + 0.48 km 1.05 km 320 m + 350 m 0.67 km 0.4 km + 250 m + 0.2 km + 390 m 1.24 km Perbezaan/ Difference: 1.24 km – 0.67 km = 0.57 km 25 C Bilangan murid perempuan/ Number of girls: 42 – 23 = 19 P(dua orang yang dilantik ialah perempuan): P(two elected students are girls): 19 42 × 18 41 = 57 287 26 B (P ∪ Q)’ ∩ R = 3x + 1 (P ∩ R) ∪ Q = x + (x – 7) + (2x – 8) + (x – 6) + 2 = 5x – 21 + 2 = 5x – 19 3x + 1 = 5x – 19 2x = 20 x = 10 P ∩ Q ∩ R = 2x – 8 = 2(10) – 8 = 20 – 8 = 12 27 C Persamaan garis sempang/ Equation of dash line: m = – 4 –4 = 1 ∴ y = x + 4 y – x = 4 CONTOH

Matematik Tingkatan 4 Jawapan J57 Persamaan garis padu/ Equation of solid line: m = – 6 3 = –2 ∴ y = –2x + 6 Ketaksamaan linear yang memuaskan rantau berlorek: Linear inequalities which satisfy the shaded region: x M 0, y M 0, y – x < 4 dan/ and y N –2x + 6 28 C Jumlah bayaran balik/ Total repayment: RM60 000 + (RM60 000 × 3 100 × 4) = RM60 000 + RM7 200 = RM67 200 Ansuran bulanan/ Monthly instalment: RM67 200 ÷ (4 × 12) = RM1 400 RM1 400 – RM750 = RM650; Premium insurans dan ansuran kereta adalah perbelanjaan tetap bulanan, manakala perbelanjaan makanan tidak cukup dikurangkan dan adalah lebih penting daripada perbelanjaan melancong. Maka, hanya perbelanjaan melancong boleh dikurangkan. RM1 400 – RM750 = RM650; Insurance premium and car instalment are fixed monthly expenses, whereas food expense is not enough to be cut down and is more important than travel expense. Thus, only travel expense can be cut down. 29 C Luas di bawah graf laju-masa ialah jumlah jarak. The area under a speed-time graph is total distance. 1 2 × (3 + 8) × d = 198 11d = 396 d = 36 m s–1 30 D x 36 38 40 42 f 3 8 12 2 Σf = 25 fx 108 304 480 84 Σfx = 976 x2 1 296 1 444 1 600 1 764 fx2 3 888 11 552 19 200 3 528 Σfx2 = 38 168 x = 976 25 σ2 =38 168 25 – 39.042   σ = √2.598 = 39.04 = 2.598 = 1.612 31 C A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} B ={A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z} Nombor ganjil/ Odd number = {1, 3, 5, 7, 9} Vokal/ Vowel = {A, E, I, O, U} P(nombor ganjil, vokal/ odd number, vowel) 5 10 × 5 26 = 25 260 = 5 52 32 A Jumlah pendapatan/ Total income: RM4 000 + RM800 = RM4 800 Simpanan tetap bulanan/ Monthly fixed savings: RM4 800 × 10% = RM480 Jumlah bayaran balik kereta/ Total car repayment: RM48 000 + (RM48 000 × 4.2 100 × 4) = RM48 000 + RM8 064 = RM56 064 Ansuran kereta bulanan/ Monthly car instalment: RM56 064 ÷ (4 × 12) = RM1 168 Aliran tunai baharu/ New cash flow: RM4 000 + RM800 – RM480 – RM2 750 – RM1 168 = RM402 33 C Murid Student x� Sx2 σ2 σ Joshua 73.6 28 010 185.04 13.6 Lisa 73.0 27 723 215.6 14.68 Kamil 73.6 27 734 129.84 11.39 Fazia 75.2 28 768 98.56 9.93 34 B Biar H sebagai hari hujan dan U sebagai Cindy membawa payung. Let H be the rainy day and U be Cindy carrying an umbrella. P(U�) = P(H, U�) + P(H�, U�) = (0.4 × 0.2) + (0.6 × 0.7) = 0.08 + 0.42 = 0.5 35 C (x – 5)2 + (x + 2)2 = (x + 3)2 x2 – 10x + 25 + x2 + 4x + 4 = x2 + 6x + 9 2x2 – 6x + 29 = x2 + 6x + 9 x2 – 12x + 20 = 0 (x – 2)(x – 10) = 0 x = 2 atau/ or 10 Panjang tidak boleh bernilai negatif, maka x = 10. Length cannot be a negative value, thus x = 10. Perimeter: (10 – 5) + (10 + 2) + (10 + 3) = 5 + 12 + 13 = 30 cm 36 C Perlanggaran berlaku pada 2 s, maka laju baharu bagi zarah Q adalah pada 2 s < t < 8 s. Collision occurred at 2 s, thus the new speed of particle Q is at 2 s < t < 8 s. Laju bagi Q selepas perlanggaran: Speed of Q after collision: (140 – 0) m (8 – 2) s = 140 m 6 s = 23.33 m s–1 CONTOH

Matematik Tingkatan 4 Jawapan J58 37 C 445 = 4(5) + 4 = 2410 1327 = 72 + 3(7) + 2 = 7210 1439 = 92 + 4(9) + 3 = 12010 Pola/ Pattern: +24 Jujukan nombor dalam asas 10: Number sequence in base 10: 24, 48, 72, 96, 120 6 48 6 8 0 6 1 2 0 1 8 96 8 12 0 8 1 4 0 1 ∴ X6 = 1206 , Y8 = 1408 X = 120, Y = 140 38 B σ2 = Sx2 N – x� 2 24.75 = 15 000 24 – x� 2 x� 2 = 625 – 24.75 = 600.25 x�= 24.5 Min baharu/ New mean: 24(24.5) + 28.2 + 25.6 + 28.9 + 23.7 24 + 4 = 694.4 28 = 24.8 Hasil tambah kuasa dua baharu/ New sum of squares: 15 000 + 28.22 + 25.62 + 28.92 + 23.72 = 17 847.5 σ2 = 17 847.5 28 – 24.82 = 22.37 σ = 4.730 39 B n(S) = 8 + 6 + 2 = 16 P(KK, MM, HH): ( 8 16 × 7 15) + ( 6 16 × 5 15)+ ( 2 16 × 1 15) = 56 240 + 30 240 + 2 240 = 88 240 = 11 30 40 A (4y)2 = 320 8 – (√5x )2 16y2 = 40 – 5x 5x = 40 – 16y2 x = 40 – 16y2 5 Kertas 2 Bahagian A/ Section A 1 2(p2 – 6) p = –5 2p2 – 12 = –5p 2p2 + 5p – 12 = 0 (2p – 3)(p + 4) = 0 p = –4 atau/ or 3 2 2 (a) X Y Z (b) X Y Z 3 (a) 55556 (b) 55556 = 5(63 ) + 5(62 ) + 5(6) + 5 = 1080 + 180 + 30 + 5 = 129510 8 1295 8 161 7 8 20 1 8 2 4 0 2 ∴ 24178 4 (a) –x2 + 2x + 8 = 0 (–x – 2)(x – 4) = 0 x = –2 atau/ or 4 CONTOH

Matematik Tingkatan 4 Jawapan J59 f(x) x –2 O 4 8 (b) Paksi simetri/ Axis of symmetry: – 2 2(–1) = 1 Tinggi maksimum/ Maximum height: f(1) = –(1)2 + 2(1) + 8 = –1 + 2 + 8 = 9 m 5 (a) G(9, 9) (b) Bilangan tepi/ Number of edges: 9 – 1 = 8 6 (a) Aliran tunai/ Cash flow: RM6 000 + RM800 – RM2 850 – RM3 040 = RM910 (terbukti/ proven) ∴ Encik Syafiq mempunyai baki pendapatan sebanyak RM910. Mr Syafiq has an income balance of RM910. (b) Bayaran balik bulanan/ Monthly repayment: 74 700 5 × 12 = RM1 245 Minimum perbelanjaan tidak tetap bulanan yang perlu dikurangkan: The minimum of monthly variable expenses which needs to be reduce: RM1 245 – RM910 = RM335 7 2 –2 O –3 2 2y + 6 = –3x x = 2 x – y = –2 y x 8 (a) Peristiwa tidak saling eksklusif Non-mutually exclusive events (b) P(K ∪ T) = P(K) + P(T) – P(K ∩ T) = 5 12 + 3 8 – 5 24 = 10 + 9 – 5 24 = 14 24 = 7 12 9 Khusus Specific Matlamat Nancy adalah tetap iaitu membeli pakej pelancongan. Nancy’s goal is fixed, which is to buy the travel package. Boleh diukur Measurable Matlamat khusus Nancy boleh dihitung kerana harga pakej pelancongan tersebut adalah tetap pada RM2 100. Nancy’s specific goal can be measured because the price of the travel package is fixed at RM2 100. Boleh dicapai Attainable (15% × RM2 800) × 5 = RM420 × 5 = RM2 100 Matlamat kewangan Nancy boleh dicapai dengan menyimpan RM420 sebulan daripada pendapatannya. Nancy’s financial goal can be attained by saving RM420 a month from her income. Bersifat realistik Realistic Matlamat kewangan Nancy untuk mendapat tawaran pakej pelancongan dalam masa 5 bulan adalah bersifat realistik. Nancy’s financial goal to obtain the offer of the travel package in 5 months is realistic. CONTOH

Matematik Tingkatan 4 Jawapan J60 Tempoh masa Time-bound Matlamat kewangan Nancy mempunyai tempoh masa selama 5 bulan. Nancy’s financial goal has a time-limit of 5 months. \ Matlamat kewangan Nancy memenuhi konsep SMART. Nancy’s financial goal fulfils the SMART concept. 10 (a) 25 + 65 + 90 = 180 35 + 65 + 80 = 180 60 + 60 + 60 = 180 ∴ Hasil tambah nombor-nombor dalam setiap kumpulan ialah 180. The sum of the numbers in each group is 180. X + 10 + 50 = 180 X = 180 – 60 = 120 (b) Kumpulan-kumpulan nombor itu mewakili sudut-sudut pada sebuah segi tiga. The groups of numbers represent the angles on a triangle. Bahagian B/ Section B 11 (a) x + y N 100, y M 2x, x < 25 (b) O 25 25 50 75 100 50 75 100 x = 25 x + y = 100 y = 2x x y (c) (i) Bilangan maksimum/ Maximum number: 100 (ii) 25(RM1.20) + 75(RM1.50) = RM30 + RM112.50 = RM142.50 12 (a) Zarah H/ Particle H (b) (i) Pecutan/ Acceleration: (100 – 20) m/s (10 × 60) s = 80 m/s 600 s = 0.133 m/s2 (ii) Masa pada keadaan pegun hingga laju 20 m/s: Time from stationary state to speed of 20 m/s: (20 – 0) m/s t s = 0.133 m/s2 t = 150 s Jumlah masa yang diambil dari keadaan pegun hingga laju 100 m/s: Total time taken from stationary state up to speed of 100 m/s: 150 s + (10 × 60) s = 150 + 600 = 750 s Jumlah jarak/ Total distance: 1 2 × 100 × 750 = 37 500 m = 37.5 km (c) Pecutan/ Acceleration: (0 – 50) m/s (3 × 60) s = –50 m/s 180 s = –0.278 m/s2 (d) Jumlah jarak/ Total distance: 1 2 × 50 × 60(7 + 10) = 1 2 × 50 × 1 020 = 25 500 m Purata laju/ Average speed: 25 500 m 600 s = 42.5 m/s 13 (a) (i) Benar/ True (ii) Palsu/ False (b) 1. Jika x boleh dibahagi tepat dengan dua, maka x ialah satu nombor genap. If x is divisible by two, then x is an even number. 2. Jika x ialah satu nombor genap, maka x boleh dibahagi tepat dengan dua. If x is an even number, then x is divisible by two. (c) (i) Semua/ All (ii) Sebilangan/ Some (d) Jika daun tumbuhan tidak mempunyai klorofil, maka daun itu tidak berwarna hijau. If the leaves of the plant do not have chlorophyll, then the leaves are not green. (e) Jika semua sudut pedalaman bagi sebuah segi tiga ialah 60°, maka segi tiga tersebut ialah sebuah segi tiga sama sisi. If all the interior angles of a triangle is 60°, then the triangle is an equilateral triangle. 14 (a) Kumpulan A Group A Kumpulan B Group B 7 2 0 9 9 3 1 6 8 5 2 2 5 9 8 6 2 3 5 5 8 5 3 4 3 CONTOH

Matematik Tingkatan 4 Jawapan J61 (b) Kumpulan A/ Group A: 13 – 38, 38 – 13 = 25 Kumpulan B/ Group B: 18 – 35, 35 – 18 = 17 (c) x� A: 2 + 7 + 13 + 19 + 25 + 32 + 36 + 38 + 43 + 45 10 = 260 10 = 26 x� B: 9 + 16 + 18 + 22 + 25 + 29 + 35 + 35 + 38 + 43 10 = 270 10 = 27 (d) Kumpulan Group Hasil tambah kuasa dua Sum of squares A 22 + 72 + 132 + 192 + 252 + 322 + 362 + 382 + 432 + 452 = 8 846 B 92 + 162 + 182 + 222 + 252 + 292 + 352 + 352 + 382 + 432 = 8 354 σ2 A = 8 846 10 – 262 σ2 B = 8 354 10 – 272 = 208.6 = 106.4 σA = 14.44 σB = 10.32 ∴ Skor yang diperoleh kumpulan B adalah lebih konsisten kerana sisihan piawainya adalah lebih rendah berbanding dengan kumpulan A. The scores obtained by group B is more consistent because its standard deviation is lower compared to group A. 15 (a) (i) P(murid dari kelas yang sama): P(students from the same class): ( 3 18 × 2 14) + ( 3 12 × 2 21) = 6 252 + 6 252 = 1 21 (ii) P(lelaki dan perempuan dari kelas berlainan): P(boys and girls from different class): ( 3 18 × 2 21) + ( 3 12 × 2 14) = 1 63 + 1 28 = 4 252 + 9 252 = 13 252 (b) (i) 17 J T K 2x – 1 x2 + 1 7 15 6 y (ii) 15 + 7 + (x2 + 1) + (2x – 1) = 30 x2 + 2x + 22 = 30 x2 + 2x – 8 = 0 (x + 4)(x – 2) = 0 x = –4 atau/ or 2 ∴Bilangan orang tidak boleh bernilai negatif, maka x = 2. The number of people cannot be a negative value, hence x = 2. (iii) y + 6 + [2(2) – 1] + (22 + 1) = 33 y + 6 + 3 + 5 = 33 y = 33 – 14 = 19 Jumlah bilangan murid/ Total number of students: 30 + 19 + 6 + 17 = 72 Bahagian C/ Section C 16 (a) Pendapatan dan perbelanjaan Income and expenses RM Gaji bersih/ Net income Pendapatan pasif (sewa yang diterima) Passive income (rental received) 5 600 200 Jumlah pendapatan/ Total income 5 800 Tolak simpanan tetap bulanan Minus fixed monthly savings 1 160 Baki pendapatan/ Income balance 4 640 Tolak perbelanjaan tetap bulanan Minus fixed monthly expenses Pinjaman rumah/ Housing loan Premium insurans/ Insurance premium 1 850 250 Jumlah perbelanjaan tetap bulanan Total fixed monthly expenses 2 100 Tolak perbelanjaan tidak tetap bulanan Minus monthly variable expenses Perbelanjaan makanan/ Food expenses Belanja petrol/ Petrol expenses Utiliti rumah/ House utility Elaun ibu bapa/ Allowance for parents 550 120 220 1 000 Jumlah perbelanjaan tidak tetap bulanan Total monthly variable expenses 1 890 CONTOH Pendapatan lebihan/ Surplus of income 650

Matematik Tingkatan 4 Jawapan J62 (b) (i) Wang pendahuluan kereta: Down payment of car: RM127 000 × 10% = RM12 700 Tempoh simpanan/ Savings duration: RM12 700 ÷ RM1 160 = 10.95 ≈ 11 bulan/ months (ii) Pinjaman bank/ Bank loan: RM127 000 – RM12 700 = RM114 300 Bank Jumlah bayaran balik Total repayment Ansuran kereta bulanan Monthly car instalment A RM114 300 + (RM114 300 × 4 100 × 7) = RM114 300 + RM32 004 = RM146 304 RM146 304 ÷ (7 × 12) = RM1 742 B RM114 300 + (RM114 300 × 5 100 × 5) = RM114 300 + RM28 575 = RM142 875 RM142 875 ÷ (5 × 12) = RM2 381 Pendapatan lebihan selepas membayar wang pendahuluan kereta: Surplus of income after paying the down payment of car: RM1 160 + RM650 = RM1 810 ∴ Bank A. Pendapatan lebihan Sanjay hanya cukup membayar ansuran kereta bulanan yang ditawarkan oleh Bank A. Namun, jumlah bayaran balik Bank B kurang daripada jumlah bayaran balik Bank A. Bank A. Sanjay’s surplus of income is only enough to pay the monthly car instalment offered by Bank A. Although the total repayment amount of Bank B is less than the total repayment of Bank A. (c) Aliran tunai baharu/ New cash flow: RM5 600 + RM200 – (RM2 100 + RM1 742) – RM1 890 = RM68 17 (a) Pada permukaan air, h = 0 m, At the water surface, h = 0 m, t2 – 4t + 3 = 0 (t – 1)(t – 3) = 0 t = 1 atau/ or 3 ∴ Masa burung pekaka menembusi permukaan air ialah pada 1 s dan 3 s. The time that the kingfisher penetrates the water surface is at 1 s and 3 s. (b) t 0 1 2 3 4 5 h 3 0 –1 0 3 8 t = 2, h = 22 – 4(2) + 3 = 4 – 8 + 3 = –1 t = 1, h = 12 – 4(1) + 3 = 1 – 4 + 3 = 0 t = 4, h = 42 – 4(4) + 3 = 16 – 16 + 3 = 3 t = 3, h = 32 – 4(3) + 3 = 9 – 12 + 3 = 0 (c) –2 1 1 –1 2 3 4 5 3 2 4 h (m) t (s) 5 7 6 8 O (d) (i) 3 m dari aras sungai 3 m from the river level (ii) 1 m di bawah aras sungai 1 m below river level (e) Apabila/ When t = 10 s, h = 102 – 4(10) + 3 = 100 – 40 + 3 = 63 m ∴ 63 m atas aras sungai/ 63 m above river level CONTOH

CONTOH