ii CONTENTS Learning Area: Numbers and Operations Chapter 1 Patterns and Sequences 1.1 Patterns .......................................................... 1 1.2 Sequences ................................................... 2 1.3 Patterns and Sequences ............................. 3 Mastery Challenge 1 ............................................. 5 Learning Area: Relationship and Algebra Chapter 2 Factorisation and Algebraic Fractions 2.1 Expansion .................................................... 8 2.2 Factorisation ............................................... 11 2.3 Algebraic Expressions and Laws of Basic Arithmetic Operations ................................. 14 Mastery Challenge 2 ........................................... 17 Chapter 3 Algebraic Formulae 3.1 Algebraic Formulae .................................... 22 Mastery Challenge 3 ........................................... 26 Learning Area: Measurement and Geometry Chapter 4 Polygons 4.1 Regular Polygons ....................................... 30 4.2 Interior Angles and Exterior Angles of Polygons ..................................................... 31 Mastery Challenge 4 ........................................... 36 Chapter 5 Circles 5.1 Properties of Circles .................................. 40 5.2 Symmetrical Properties of Chords ............. 42 5.3 Circumference and Area of a Circle .......... 44 Mastery Challenge 5 ........................................... 49 Chapter 6 Three-Dimensional Geometrical Shapes 6.1 Geometric Properties of ThreeDimensional Shapes ................................... 55 6.2 Nets of Three-Dimensional Shapes ........... 58 6.3 Surface Area of Three-Dimensional Shapes ...................................................... 59 6.4 Volume of Three-Dimensional Shapes ...... 62 Mastery Challenge 6 ........................................... 66 Learning Area: Relationship and Algebra Chapter 7 Coordinates 7.1 Distance in the Cartesian Coordinate System ....................................................... 72 7.2 Midpoint in the Cartesian Coordinate System ......................................................... 75 7.3 The Cartesian Coordinate System.............. 77 Mastery Challenge 7 ........................................... 79 Chapter 8 Graphs of Functions 8.1 Functions .................................................... 84 8.2 Graphs of Functions .................................. 86 Mastery Challenge 8 ........................................... 93 Chapter 9 Speed and Acceleration 9.1 Speed ......................................................... 98 9.2 Acceleration ............................................. 102 Mastery Challenge 9 ......................................... 105 Chapter 10 Gradient of a Straight Line 10.1 Gradient ................................................... 110 Mastery Challenge 10 ....................................... 119 Learning Area: Measurement and Geometry Chapter 11 Isometric Transformations 11.1 Transformations ....................................... 124 11.2 Translation ................................................ 126 11.3 Reflection ................................................. 131 11.4 Rotation .................................................... 136 11.5 Translation, Reflection and Rotation as Isometries ............................................. 140 11.6 Rotational Symmetry ................................ 142 Mastery Challenge 11 ....................................... 145 Learning Area: Statistics and Probability Chapter 12 Measures of Central Tendencies 12.1 Measures of Central Tendencies ............. 152 Mastery Challenge 12 ....................................... 162 Chapter 13 Simple Probability 13.1 Experimental Probability ........................... 167 13.2 Probability Theory Involving Equally Likely Outcomes......................................... 168 13.3 Probability of the Complement of an Event ................................................... 172 13.4 Simple Probability .................................... 174 Mastery Challenge 13 ....................................... 175 Pentaksiran Sumatif Ujian Akhir Sesi Akademik (UASA) ......................................................................................182 Answers ................................................................................. 195 Contents Spot A+ Maths Form1-3.indd 2 12/04/2023 4:55 PM
1.1 Patterns 1.1.1 Recognise and describe patterns of various number sets and objects based on real life situations, and hence make generalisation on patterns Pattern of a set of numbers or objects is an arrangement of the numbers or objects according to a certain pattern. Example 1 Identify and describe the pattern for each of the following sets of numbers. (a) 2, 4, 6, 8, 10, …. (b) 1, 1, 2, 3, 5, 8, 13, …… (c) 1, 3, 12, 60, 360 Solution: (a) 2 , 4 , 6 , 8 , 10 , ... +2 +2 +2 +2 Each number is obtained by adding 2 to its previous number. (b) 1 , 1 , 2 , 3 , 5 , 8 , 13 , ... ↑ ↑ ↑ ↑ ↑ 1+1 1+2 2+3 3+5 5+8 Each number after the second is obtained by adding two of its previous numbers. (c) 1 , 3 , 12 , 60 , 360 × 3 × 4 × 5 × 6 Each number is obtained by multiplying 3, 4, 5, 6 to its previous number successively. ➡ Diagnostic Test 1.1: Question 1 Example 2 The diagram shows a Pascal’s Triangle. 1 1 1 2 1 1 3 a 1 b c d 4 1 Find the values of a, b, c and d. Solution: 1 1 1 2 1 1 3 a 1 b c d 4 1 a = 2 + 1 = 3 b = 1 c = 1 + 3 = 4 d = 3 + a = 3 + 3 = 6 ➡ Diagnostic Test 1.1: Question 2 Diagnostic Test 1.1 1. Identify and describe the pattern for each of the following sets of numbers. (a) 1, 3, 5, 7, 9, …. (b) 108, 36, 12, 4, …. (c) 2, 7, 13, 20, 28 2. Find the values of a, b, c and d as shown in the following Pascal’s Triangle. 1 1 1 2 1 1 3 3 1 1 4 6 4 1 a 5 b c d 1 1 CHAPTER 1 Patterns and Sequences 1 1 Patterns and Sequences CHAPTER C01 SpotlightA+ Maths Form2.indd 1 18/04/2023 9:09 AM
1.2 Sequences 1.2.1 Explain the meaning of sequence Sequence is a list of objects or numbers which are arranged according to a certain pattern. Example 3 Describe each of the following sequences. (a) 2, 10, 50, 250, ... (b) 7, 15, 25, 37, 51 Solution: (a) 2 , 10 , 50 , 250 , ... ×5 ×5 ×5 2, 10, 50, 250, ... is a sequence formed by multiplying 5 to its previous number. (b) 7 , 15 , 25 , 37 , 51 +8 +10 +12 +14 7, 15, 25, 37, 51 is a sequence formed by adding 8, 10, 12, 14 to its previous number successively. ➡ Diagnostic Test 1.2: Question 1 TIPS Corner Infinite number of terms Finite number of terms, 5 Infinite sequence Finite sequence 2, 10, 50, 250, ... 7, 15, 25, 37, 51 1.2.2 Identify and describe the pattern of a sequence, and hence complete and extend the sequence Example 4 Describe the pattern for each of the following sequences and hence complete the sequence. (a) 3, 8, 13, , , 28 (b) , 24, 12, 6, , ... Solution: (a) 3 , 8 , 13 , 18 , 23 , 28 +5 +5 +5 +5 +5 Each number is obtained by adding 5 to its previous number. TIPS Corner 13 + 5 = 18 18 + 5 = 23 (b) 48 , 24 , 12 , 6 , 3 , ... ÷2 ÷2 ÷2 ÷2 Each number is obtained by dividing 2 to its previous number. TIPS Corner ÷ 2 = 24 = 24 × 2 = 48 6 ÷ 2 = 3 ➡ Diagnostic Test 1.2: Question 2 Example 5 Describe the pattern for each of the following sequences and hence extend the sequence with the next two numbers. (a) 13, 21, 29, 37, ... (b) 5, 5, 10, 30, 120, ... Solution: (a) 13, 21, 29, 37, 45, 53, ... +8 +8 +8 +8 +8 Each number is obtained by adding 8 to its previous number. The next two numbers are 45, 53. (b) 5, 5, 10, 30, 120, 600, 3 600, . . . ×1 ×2 ×3 ×4 ×5 ×6 Each number is obtained by multiplying 1, 2, 3, 4, 5, 6, ... to its previous number successively. The next two numbers are 600, 3 600. ➡ Diagnostic Test 1.2: Question 3 2 Learning Area: Numbers and Operations 1 C01 SpotlightA+ Maths Form2.indd 2 12/04/2023 4:57 PM
Diagnostic Test 1.2 1. Describe each of the following sequences. (a) 1, 3, 9, 27, 81, ... (b) 55, 48, 41, 34, 27, 20 (c) 1, 1, 3, 15, 105, 945, … 2. Describe the pattern for each of the following sequences and hence complete the sequence. (a) 7, 8.5, 10, 11.5, , , 16 (b) 2, 8, , 128, , 2 048 (c) 9, , 12, 15, , , ... 3. Describe the pattern for each of the following sequences and hence extend the sequence with the next two numbers. (a) 15, 28, 41, 54, … (b) 1, 4, 9, 16, 25, … (c) 6 400, 1 600, 400, 100, … 1.3 Patterns and Sequences 1.3.1 Make generalisation about the pattern of a sequence using numbers, words and algebraic expressions Example 6 Make a generalisation about the pattern for each of the following sequences. (a) 5, 10, 15, 20, … (b) 11, 14, 19, 26, … Solution: (a) 5 = 5 × 1 10 = 5 × 2 15 = 5 × 3 20 = 5 × 4 . . . Each number of the sequence 5, 10, 15, 20, ... can be stated in the form 5n, n = 1, 2, 3, 4, ... (b) 11 = 12 + 10 14 = 22 + 10 19 = 32 + 10 26 = 42 + 10 . . . Each number of the sequence 11, 14, 19, 26, ... can be stated in the form n2 + 10, n = 1, 2, 3, 4, ... ➡ Diagnostic Test 1.3: Question 1 1.3.2 Determine specific terms of a sequence Example 7 Determine the 20th term of the sequence 4, 11, 18, 25, ... Solution: First term, 4 = 7 × 1 – 3 Second term, 11 = 7 × 2 – 3 Third term, 18 = 7 × 3 – 3 Fourth term, 25 = 7 × 4 – 3 . . . The 20th term = 7 × 20 – 3 = 137 ➡ Diagnostic Test 1.3: Question 2 Example 8 The nth term of the sequence 1, 13, 33, 61, ... is given by an2 + b, n = 1, 2, 3, 4, ... (a) Determine the values of a and b. (b) Find the 35th term. Solution: (a) First term, 1 = 4 × 12 – 3 Second term, 13 = 4 × 22 – 3 Third term, 33 = 4 × 32 – 3 Fourth term, 61 = 4 × 42 – 3 . . . The nth term = 4n2 – 3 a = 4 ; b = −3 3 CHAPTER 1 Patterns and Sequences 1 C01 SpotlightA+ Maths Form2.indd 3 12/04/2023 4:57 PM
(b) The 35th term = 4 × 352 – 3 = 4 897 Calculator Press 4 × 3 5 x2 – 3 = 4 897 ➡ Diagnostic Test 1.3: Question 3 1.3.3 Solve problems involving sequences Example 9 A car distributor sold 50 cars in January, 80 cars in February and its subsequent sales increase 30 cars for every month. (a) Determine the number of cars sold in the nth month. (b) Hence, find (i) the number of cars sold in December of the same year, (ii) the value of n when the number of cars sold exceeded 600 cars for the first time. Solution: (a) 50, 80, 110, 140, ... First term, 50 = 30 × 1 + 20 Second term, 80 = 30 × 2 + 20 Third term, 110 = 30 × 3 + 20 Fourth term, 140 = 30 × 4 + 20 The number of cars sold in the nth month is 30n + 20. (b) (i) When n = 12, 30n + 20 = 30(12) + 20 = 380 The number of cars sold in December of the same year is 380. (ii) 30n + 20 > 600 30n > 580 n > 19.3 n = 20 ➡ Diagnostic Test 1.3: Question 4 Diagnostic Test 1.3 1. Make a generalisation about the pattern for each of the following sequences. (a) 3, 6, 9, 12, … (b) 7, 8, 9, 10, … (c) 8, 20, 32, 44, … (d) 4, 7, 12, 19, … 2. Given the sequence 11, 13, 15, 17, ... , determine (a) the 10th term, (b) the 33rd term. 3. (a) Complete the following. 3 = 8 × 1 – 11 = 8 × 2 – 19 = 8 × 3 – 27 = 8 × 4 – (b) Hence, find the 17th and the 70th terms of the sequence 3, 11, 19, 27, … 4. Starting with one lorry in the year 2000, Syarikat Pengangkutan Maju added another 3 lorries in 2001. In 2002, the number of lorries acquired increased to 9. The number of lorries in 2003 is 7 units more than the number of lorries in the previous year. By assuming the number of lorries acquired by the company increases every year according to the pattern stated, (a) write the information about the number of lorries as a sequence, (b) determine the number of lorries acquired by the company in 2017. HOTS Analysing 4 Learning Area: Numbers and Operations 1 C01 SpotlightA+ Maths Form2.indd 4 12/04/2023 4:57 PM
MASTERY CHALLENGE 1 1. The sum of the next two consecutive numbers in the sequence of numbers 480, 240, 120, 60, … is A 30 B 45 C 75 D 90 2. 41, P, Q, R, S, 59 is a sequence of prime numbers. Which of the following is the number 47? A P B Q C R D S 3. 8 , 5 , H , K , –4 The diagram above shows a sequence of five numbers. Determine the value of H and of K. A H = 1, K = –3 B H = 2, K = –1 C H = 2, K = 3 D H = 3, K = 1 4. The first five terms of the Fibonacci numbers are 1, 1, 2, 3, 5. Find the 12th term. A 89 B 138 C 144 D 233 5. 1 k 6 4 1 1 5 m m 5 1 1 6 15 n 15 6 1 The diagram above shows a part of the Pascal’s Triangle. Find the value of k + m + n. A 34 B 36 C 44 D 46 6. I : 7, 13, 21, p, 43, 57, … II : 8, 24, 72, q, 648, … Based on the two sequences of numbers I and II given, determine the values of p and q. A p = 31, q = 216 B p = 31, q = 243 C p = 32, q = 218 D p = 33, q = 216 7. 1, 1, 5, 45, m, n, … Which of the following shows the relation between m and n in the sequence of numbers above? A m × 13 = n B n × 13 = m C n × 17 = m D m × 17 = n 8. Determine the nth term of the sequence 27, 22, 17, 12, … such that n = 1, 2, 3, 4, … A 35 – 8n B 34 – 7n C 32 – 5n D 30 – 3n 9. Find the sum of the 5th term and the 20th term of the sequence 4, 9, 16, 25, … A 397 B 477 C 490 D 578 10. Given the nth term of the sequence 21, 34, 47, 60, … is 190. The value of n is A 11 B 12 C 13 D 14 11. Match each of the following. (a) 31, 28, 25, 22, ... • • Add 3 to its previous number. (b) 5, 10, 20, 40, ... • • Divide its previous number by 2. (c) 1, 4, 7, 10, ... • • Subtract 3 from its previous number. (d) 200, 100, 50, 25, ... • • Multiply its previous number by 2. 12. Identify and describe the following patterns. (a) 3, 9, 15, 21, ... (b) 81, 76, 71, 66, 61, 56, ... (c) 7, 28, 112, 448, 1 792, ... (d) 360, 120, 40, 13 1 — 3 , ... 5 CHAPTER 1 Patterns and Sequences 1 C01 SpotlightA+ Maths Form2.indd 5 12/04/2023 4:57 PM
2 2.1 Expansion 2.1.1 Explain the meaning of the expansion of two algebraic expressions Expansion by the representation of algebraic tiles: 1. Draw a rectangle by using two algebraic expressions as its length and width. 2. Arrange the algebraic tiles to form the rectangle. 3. The expansion of the two algebraic expressions is the sum of the area of each of the algebraic tiles. Example 1 (a) Complete the area of each of the following algebraic tiles. x x 1 x x2 x + 2 1 x 1 2x + 1 (b) Hence, expand (2x + 1)(x + 2). Solution: (a) x x 1 x x2 x2 x x + 2 1 x x 1 1 x x 1 2x + 1 (b) (2x + 1)(x + 2) = x2 + x2 + x + x + x + x + x + 1 + 1 = 2x2 + 5x + 2 ➡ Diagnostic Test 2.1: Question 1 Example 2 By using algebraic tiles, expand each of the following. (a) (x + 1)(2y + 3) (b) (a + 2b)(c + d) Solution: (a) x 1 y xy y 2y + 3 y xy y 1 x 1 1 x 1 1 x 1 x + 1 (x + 1)(2y + 3) = xy + xy + x + x + x + y + y + 1 + 1 + 1 = 2xy + 3x + 2y + 3 TIPS Corner Area of rectangle = Sum of area of each algebraic tiles (b) a b b c ac bc bc c + d d ad bd bd a + 2b (a + 2b)(c + d) = ac + bc + bc + ad + bd + bd = ac + 2bc + ad + 2bd ➡ Diagnostic Test 2.1: Question 2 2 Factorisation and Algebraic Fractions CHAPTER 8 C02 SpotlightA+ Maths Form2.indd 8 12/04/2023 4:59 PM
2 2.1.2 Expand two algebraic expressions Example 3 Expand each of the following. (a) (7x + 3)(x + 4) (b) (4h + 5k)(2h – k) Solution: (a) (7x + 3)(x + 4) = 7x(x) + 7x(4) + 3(x) + 3(4) = 7x2 + 28x + 3x + 12 = 7x2 + 31x + 12 (b) (4h + 5k)(2h – k) = 4h(2h) + 4h(–k) + 5k(2h) + 5k(–k) = 8h2 – 4hk + 10hk – 5k2 = 8h2 + 6hk – 5k2 ➡ Diagnostic Test 2.1: Question 3 TIPS Corner The expansion of the forms (a + b)(a – b), (a + b) 2 and (a – b) 2 are easier to perform by using the following formulae. (a + b)(a – b) = a2 – b2 (a + b) 2 = a2 + 2ab + b2 (a – b) 2 = a2 – 2ab + b2 Example 4 Expand. (a) (2m + 3n)(2m – 3n) (b) (p – 4r)2 (c) (6x + 11y)2 Solution: (a) (2m + 3n)(2m – 3n) = (2m) 2 – (3n)2 = 4m2 – 9n2 (b) (p – 4r)2 = p2 – 2p(4r) + (4r) 2 = p2 – 8pr + 16r 2 (c) (6x + 11y)2 = (6x)2 + 2(6x)(11y) + (11y)2 = 36x2 + 132xy + 121y2 Alternative Method (p – 4r) 2 = (p – 4r)(p – 4r) = p(p) + p(–4r) – 4r(p) – 4r(–4r) = p2 – 4pr – 4pr + 16r2 = p2 – 8pr + 16r2 ➡ Diagnostic Test 2.1: Question 4 Learn it right (a) (2m + 3n)(2m – 3n) = 2m2 – 3n2 ✗ (b) (p – 4r) 2 = p2 – 2p(4r) + 4r2 = p2 – 8pr + 4r2 ✗ (c) (6x + 11y) 2 = 6x2 + 2(6x)(11y) + 11y2 = 6x2 + 132xy + 11y2 ✗ 2.1.3 Simplify algebraic expressions involving combined operations, including expansion Example 5 Simplify each of the following. (a) (3x – 2)(x + 6) + 4(2x + 5) (b) (r – 2t) 2 – (8t – 5r)(t + 2r) Solution: (a) (3x – 2)(x + 6) + 4(2x + 5) = 3x2 + 18x – 2x – 12 + 8x + 20 = 3x2 + 18x – 2x + 8x – 12 + 20 = 3x2 + 24x + 8 (b) (r – 2t) 2 – (8t – 5r)(t + 2r) = r 2 – 4rt + 4t 2 – (8t 2 + 16rt – 5rt – 10r 2 ) = r 2 – 4rt + 4t 2 – (8t 2 + 11rt – 10r 2 ) = r 2 – 4rt + 4t 2 – 8t 2 – 11rt + 10r 2 = 11r 2 – 15rt – 4t 2 ➡ Diagnostic Test 2.1: Question 5 Learn it right (r – 2t) 2 – (8t – 5r)(t + 2r) = r 2 – 4rt + 4t 2 – 8t 2 + 16rt – 5rt – 10r2 = –9r2 + 7rt – 4t2 ✗ 9 CHAPTER 2 Factorisation and Algebraic Fractions C02 SpotlightA+ Maths Form2.indd 9 12/04/2023 4:59 PM
2 6. Simplify each of the following. (a) 5c —– a × b —– 2d (b) 3x —– 8y × 4w—–9 (c) n —– 6m ÷ mn —– 12 (d) 15v —– u ÷ 10v —– 3w 7. Simplify each of the following. (a) 2h + 6 ———— 2k2 – 7k × 6k – 21 ———— (h + 3)2 (b) v + 2w ———— 12v – 9w ÷ 4v2 – 16w2 ————— 4v – 3w 8. Simplify each of the following. (a) x – 1 ——— 2x + 1 × 2x2 + 3x + 1 —————– (x – 1)2 + 2 ——— x2 – 1 (b) (y + 1)(y + 7) + 3y – 6y2 ———— y + 5 ÷ 6 – 12y ———— y2 – 25 MASTERY CHALLENGE 2 1. (4x + 7)(x – 3) = A 4x2 – 5x – 21 B 4x2 – 9x – 21 C 4x2 – 14x – 21 D 4x2 – 19x – 21 2. 2(v – 2)2 + 13v = A 2v2 – 9v – 4 B 2v2 + 9v + 8 C 2v2 + 5v + 8 D 2v2 + 9v + 4 3. Which of the following is not correct? A 6y – 18 = 6(y – 3) B 4 – 9p2 = (2 + 3p)(2 – 3p) C 64k2 + 25 = (8k + 5)(8k – 5) D 16m2 n2 + 2mn2 = 2mn2 (8m + 1) 4. Factorise 12p2 – 16p + 5. A (2p – 5)(6p – 1) B (2p – 1)(6p – 5) C (4p – 1)(3p + 5) D (4p – 5)(3p – 1) 5. 2ad – 2bc – 4bd + ac = A (2a – b)(c + 2d) B (2a + b)(d – 2c) C (a + 2b)(c – 2d) D (a – 2b)(c + 2d) 6. m —2 – 2m – 3 —–——8 = A 2m – 3 —–——8 B 2m + 3 —–——8 C 2m + 3 —–—— 10 D 3 – 2m —–—— 16 7. 4k + 9 —––—9k + 5 – k —–—2k = A 63 – k —–—— 18k B 7k + 54 —–—— 18k C 3k + 13 —–—— 9k D 53 – 5k —–—— 9k 8. p – 1 —–———— 2p2 – 4p + 2 + 7 —p = A 8p – 14 —–—— p(p – 1) B 14p – 7 —–—— p(p – 1) C 15p – 14 —–——– 2p(p – 1) D 3p + 7 —–——– 2p(p – 1) 9. 4a2 + 16a —–——–— 6a2 + a – 1 × 3a – 1 —–—— 6a + 24 = A a —–——– 6(2a – 1) B 2a —–——– 3(2a – 1) C 2 —–——– 3(2a – 1) D 2a —–——– 3(2a + 1) 17 CHAPTER 2 Factorisation and Algebraic Fractions C02 SpotlightA+ Maths Form2.indd 17 12/04/2023 4:59 PM
Learning Area: Relationship and Algebra 9 9.1 Speed 9.1.1 Explain the meaning of speed as a rate involving distance and time 1. Speed is the rate of change of distance with respect to time. Example 1 Explain the meaning of speed in each of the following situations. (a) A horse runs at a speed of 40 km/h. (b) An ostrich travels at a speed of 25 m/s. Solution: (a) The speed 40 km/h means the horse runs over a distance of 40 km in 1 hour. (b) The speed 25 m/s means the ostrich travels over a distance of 25 m in 1 second. ➡ Diagnostic Test 9.1: Question 1 9.1.2 Describe the differences between uniform and non-uniform speed uniform Speed non-uniform distances travelled by the object are the same in equal time intervals distances travelled by the object are different in equal time intervals 14243 123 123 Example 2 The diagram shows the positions of a bicycle travelling on a road in time intervals of 5 seconds. Determine whether the bicycle is travelling with uniform speed or non-uniform speed. Give your justification. (a) 10 m 10 m 10 m 10 m 10 m (b) 10 m 5 m 15 m 20 m Solution: (a) The bicycle is travelling with uniform speed because it covers the same distance, 10 m in equal time intervals of 5 seconds. (b) The bicycle is travelling with non-uniform speed because it covers different distances in equal time intervals of 5 seconds. ➡ Diagnostic Test 9.1: Question 2 Example 3 The table shows the distance travelled by a car in a period of 5 minutes. State whether the car is moving with uniform speed or non-uniform speed. Give your reason. (a) Time (minute) 0 1 2 3 4 5 Distance (km) 0 2 4 6 8 10 (b) Time (minute) 0 1 2 3 4 5 Distance (km) 0 1.6 2.8 4.8 6 8 Solution: (a) The car is moving with uniform speed because it covers the same distance, 2 km in equal time intervals of 1 minute. (b) The car is moving with non-uniform speed because it covers different distances in equal time intervals of 1 minute. ➡ Diagnostic Test 9.1: Question 3 TIPS Corner 0 1 2 4 2 3 4 5 6 8 10 12 Time (minute) Distance (km) If a car moves at a uniform speed, then the distance-time graph is a straight line. 0 1 2 4 2 3 4 5 6 8 10 12 Time (minute) Distance (km) If a car moves at a non-uniform speed, then the distance-time graph is not a straight line. 9 Speed and Acceleration CHAPTER 98 C09 SpotlightA+ Maths Form2.indd 98 12/04/2023 5:17 PM
CHAPTER 9 Speed and Acceleration 9 9.1.3 Perform calculation involving speed and average speed including unit conversion 1. If an object moves with uniform speed, then its speed can be found from the formula speed = distance time Example 4 Find the speed in each of the following situations. (a) A taxi moves a distance of 270 km in 3 hours. State the unit in km/h. (b) Asril runs a distance of 360 m in 1.2 minutes. State the unit in m/s. Solution: (a) Speed = distance time = 270 3 = 90 km/h (b) 1.2 minutes = 1.2 × 60 s = 72 s Speed = 360 72 = 5 m/s ➡ Diagnostic Test 9.1: Question 4 Example 5 (a) A lorry moves at a speed of 80 km/h. Find the distance travelled in 1 1 2 hours. (b) An aeroplane flies at a speed of 100 m/s. How long does it take for the aeroplane to fly 420 km? Give the unit in minute. Solution: (a) Distance = speed × time = 80 × 1 1 2 = 120 km (b) 420 km = 420 000 m Time = distance speed = 420 000 100 = 4 200 s = 70 minutes ➡ Diagnostic Test 9.1: Question 5 Example 6 Convert (a) 86 cm/s to m/minute, (b) 234 km/h to m/s. Solution: (a) 86 cm/s = 86 cm 1 s = 86 100 m 1 60 minute = 86 100 × 60 1 m/minute = 51.6 m/minute TIPS Corner 86 cm 1 s 86 100 m 1 60 minute Conversion of unit (b) 234 km/h = 234 km 1 h = 234 × 1 000 m 1 × 60 × 60 s = 234 000 m 3 600 s = 65 m/s TIPS Corner 234 km 234 000 m 1 h 3 600 s Conversion of unit ➡ Diagnostic Test 9.1: Question 6 2. If an object moves with non-uniform speed, then its average speed can be found from the formula average speed = total distance travelled total time taken S = speed D = distance T = time S = D T T = D S D = S × T U seful A cronyms D S T 99 C09 SpotlightA+ Maths Form2.indd 99 18/04/2023 9:09 AM
Learning Area: Relationship and Algebra 9 Example 7 Nordin walked at a speed of 6 km/h for 20 minutes and cycled at a speed of 15 km/h for 40 minutes. Calculate the average speed of Nordin for the whole journey. Solution: Journey by walk: 20 minutes = 20 60 h Distance = 6 × 20 60 = 2 km Journey by bicycle: 40 minutes = 40 60 h Distance = 15 × 40 60 = 10 km Total distance travelled = 2 + 10 = 12 km Total time taken = 20 + 40 = 60 minutes = 1 h Average speed = total distance travelled total time taken = 12 1 = 12 km/h Average speed = 6 + 15 2 = 10.5 km/h ✗ Learn it right ➡ Diagnostic Test 9.1: Question 7 9.1.4 Solve problems involving speed Example 8 A and B are two towns which are 150 km apart. A van which departed from A travelled to B at a speed of 100 km/h. The van stopped at B for 1 1 4 hours before resuming its journey from B to A. The time of journey from B to A is 15 minutes less than the time of journey from A to B. Find (a) the time of the van journey from A to B, (b) the speed of the van from B to A, (c) the average speed of the van for the whole journey. Solution: A 150 km B 100 km/h (a) Time = distance speed = 150 100 = 1.5 h ∴ The time of the van journey from A to B is 1.5 hours. (b) The time of the van journey from B to A = 1.5 – 0.25 = 1.25 h Speed = distance time = 150 1.25 = 120 ∴ The speed of the van from B to A is 120 km/h. (c) Total distance travelled = 150 + 150 = 300 km Total time taken = 1.5 + 1 1 4 + 1.25 = 4 h Average speed = 300 4 = 75 ∴ The average speed of the van for the whole journey is 75 km/h. The time for rest has to be taken into account in determining the average speed. Total time taken = 1.5 + 1.25 = 2.75 h Average speed = 300 2.75 = 109.1 km/h ✗ Learn it right ➡ Diagnostic Test 9.1: Question 8 Diagnostic Test 9.1 1. Explain the meaning of speed in each of the following situations. (a) Osman walks at a speed of 13 m/minute. (b) Sharks swim at a speed of 64 km/h. (c) A deer runs at a speed of 9 m/s. (d) A tortoise moves at a speed of 28 cm/minute. TIPS Corner 15 minutes = 15 60 h = 0.25 h 100 C09 SpotlightA+ Maths Form2.indd 100 12/04/2023 5:17 PM
Learning Area: Relationship and Algebra 9 9.2 Acceleration 9.2.1 Explain the meaning of acceleration and deceleration as a rate involving speed and time 1. Acceleration is the rate of increase of speed with respect to time. TIPS Corner Two quantities involved in acceleration are speed and time. 2. Deceleration is the rate of decrease of speed with respect to time. TIPS Corner Deceleration is the negative acceleration. Example 9 Explain the meaning of acceleration or deceleration in each of the following situations. (a) A racing car moves with an acceleration of 4 m/s2 . (b) The deceleration of a trailer while ascending a hill is 3 km/h per second. Solution: (a) The acceleration 4 m/s2 means the speed of the racing car increases 4 m/s in 1 s. (b) The deceleration 3 km/h per second means the speed of the trailer decreases 3 km/h in 1 second. ➡ Diagnostic Test 9.2: Question 1 9.2.2 Perform calculations involving acceleration including unit conversion The acceleration of a moving object is found from the formula acceleration = change of speed time taken TIPS Corner Change of speed = final initial speed – speed final speed initial speed final speed initial speed 1442443 123 123 positive acceleration negative deceleration Example 10 Find the acceleration or deceleration in each of the following situations. (a) A boat travelling at a speed of 12 m/s attained a speed of 30 m/s in 5 seconds. State the unit in m/s2 . (b) A train travelling at 90 km/h took a time of 1 minute to stop at the station. State the unit in km/h2 . Solution: (a) Change of speed = final speed – initial speed = 30 – 12 = 18 m/s Acceleration = change of speed time taken = 18 5 = 3.6 m/s2 (b) Change of speed = final speed – initial speed = 0 – 90 = –90 1 minute = 1 60 hour Acceleration = change of speed time taken = –90 1 60 = –5 400 km/h2 or deceleration = 5 400 km/h2 ➡ Diagnostic Test 9.2: Question 2 102 C09 SpotlightA+ Maths Form2.indd 102 12/04/2023 5:17 PM
PENTAKSIRAN SUMATIF 182 PENTAKSIRAN SUMATIF UJIAN AKHIR SESI AKADEMIK (UASA) 1. Given 3, 13, 23, 33, … is a sequence of numbers. Which of the following is not a possible sum for the five consecutive terms of the sequence? A 180 B 215 C 248 D 515 2. (3x + 4)2 – (5x – 2)(x – 2) = A 4x2 + 12x + 12 B 4x2 + 12x + 16 C 4x2 + 36x + 12 D 4x2 + 36x + 16 3. m (m – 1)2 + 1 4(m – 1) = A 5m – 1 (m – 1)2 B m + 4 (m – 1)2 C 2m – 1 4(m – 1)2 D 5m – 1 4(m – 1)2 4. Given p + 3 3p = 2 – 3q r , express p in terms of q and r. A p = r 5r – 3q B p = r 7r – 6q C p = 3r 5r – 9q D p = 6r 5r – 9q 5. In Diagram 1, PQRST is a regular pentagon. STV and TPW are straight lines. R Q W T S P x y V 32° Diagram 1 SECTION A [20 marks] Answer all questions. Find the value of x + y. A 102° B 108° C 112° D 148° 6. In Diagram 2, OHK is a sector of a circle with centre O and radius 24 cm. M is the midpoint of OK. OMN is a quadrant of a circle with centre O. O N M K H 100° Diagram 2 Find the area, in cm2 , of the whole diagram. A 163p B 196p C 232p D 268p 7. In Diagram 3, OPQ is a sector of a circle with centre O and radius 12 cm. O P Q Diagram 3 The perimeter of the sector OPQ is (4p + 24) cm. Find the area, in cm2 , of the sector OPQ. A 24p B 28p C 32p D 44p P.Sumatif Spot A+1 Maths Form2.indd 182 12/04/2023 5:34 PM
Pentaksiran Sumatif PENTAKSIRAN SUMATIF 183 8. Diagram 4 shows a hemispherical bowl of radius 7 cm and a container in the shape of a cuboid. 14 cm 11 cm Diagram 4 The bowl is completely filled up with water. All the water is poured into the container. Find the height, in cm, of water in the container. A 42 3 B 71 2 C 82 3 D 91 3 9. Diagram 5 shows a composite solid consisting of a cuboid ABCDHEFG and a pyramid EFGHK such that CGK is a straight line. 8 cm 8 cm 5 cm K A B D C H E F G Diagram 5 If FK = HK = 10 cm and the volume of the pyramid is 128 cm3 , calculate the surface area, in cm2 , of the composite solid. A 224 B 352 C 400 D 416 10. Diagram 6 shows a straight line PQ that is drawn on a Cartesian plane. S(0, –2) Q(4, 5) P(–2, 1) y x O Diagram 6 M is the midpoint of PQ. Find the distance between M and S. A 1.41 units B 2.45 units C 4.78 units D 5.10 units 11. Based on the following arrow diagrams, which of the relation between set X and set Y is not a function? A X Y 1 • 2 • 3 • • 1 • 2 • 3 B X Y 0 • 2 • 5 • 7 • • 3 • 5 • 7 C X Y p • q • r • • 4 • 8 D X Y m • n • • 2 • 6 • 9 12. Alias stayed 105 km away from Batu Pahat. One day, Alias departed from his house by car at 9:30 a.m. and arrived at Batu Pahat at 10.45 a.m.. Find the average speed, in km/h, of Alias’s car. A 72 B 80 C 84 D 95 13. Diagram 7 shows two straight lines, PQ and QR that are drawn on a Cartesian plane. Q(3, 1) R(x, y) P(–3, –3) y x O Diagram 7 If the gradient of PQ is 1 2 time the gradient of QR, find a relation between x and y. A 2x – 3y = 12 B 4x – 3y = 9 C 5x – 3y = 12 D 5x – 4y = 10 P.Sumatif Spot A+1 Maths Form2.indd 183 12/04/2023 5:34 PM
Pentaksiran Sumatif PENTAKSIRAN SUMATIF 185 SECTION B [20 marks] Answer all questions. 21. (a) Diagram 1 shows four numbers. –15 –7 5 9 Diagram 1 Fill in the empty boxes in the answer space with a suitable number so as to form a sequence of numbers. [2 marks] Answer: –13, , –1, , 11 (b) Match the sequence of numbers with the correct pattern. [2 marks] Answer: Sequence of numbers Pattern (i) 5, 10, 15, 20, 25, … • • 4n + 1, n = 1, 2, 3, 4, … • n3 , n = 1, 2, 3, 4, … (ii) 1, 8, 27, 64, 125, … • • 5n, n = 1, 2, 3, 4, … 22. (a) Mark “✓” on the empty box for the correct answer. [2 marks] Answer: (i) 21r – 14 = 7(3r – 2) 3(7r – 5) (ii) 64 – 9w2 = (3w + 8)(3w – 8) (8 + 3w)(8 – 3w) (b) Fill in the following empty boxes. [2 marks] Answer: 6y – 12 3y + 13 ÷ 4 – 2y y = (y – ) 3y + 13 × y (y – ) = 3y + 13 P.Sumatif Spot A+1 Maths Form2.indd 185 12/04/2023 5:34 PM
Pentaksiran Sumatif PENTAKSIRAN SUMATIF 187 25. Box P contains four number cards 7, 11, 13 and 17. Box Q contains three number cards 6, 10 and 14. In an experiment, Mazlan picked a number card at random from box P and another number card from box Q. (a) Complete the possible outcomes of the sample space S in the answer space. [2 marks] Answer: (a) S = {( , 6), ( , 10), ( , 14), (11, ), (11, ), (11, ), ( , 6), (13, 10) (13, ), (17, 6), (17, ), ( , )} (b) A represents the event of getting the sum of two numbers divisible by 3. B represents the event of getting the product of two numbers greater than 100. Circle the correct answer. [2 marks] Answer: (i) P(A) = 1 12 1 4 1 3 7 12 (ii) P(B) = 1 12 1 4 1 3 7 12 SECTION C [60 marks] Answer all questions. 26. (a) In Diagram 3, ABCDEFGH is a regular octagon. BAK and DHK are straight lines. K A B C D F E G x y H Diagram 3 Find the values of x and y. [5 marks] Answer: P.Sumatif Spot A+1 Maths Form2.indd 187 12/04/2023 5:34 PM
195 ANSWERS Chapter 1 Diagnostic Test 1.1 1. (a) Each number is obtained by adding 2 to its previous number. (b) Each number is obtained by dividing its previous number by 3. (c) Each number is obtained by adding 5, 6, 7, 8 to the previous number successively. 2. a = 1 , b = 10 , c = 10 , d = 5 Diagnostic Test 1.2 1. (a) Multiply its previous number by 3. (b) Subtract 7 from its previous number. (c) Multiply its previous number by 1, 3, 5, 7, 9 successively. 2. (a) Add 1.5 to its previous number; 13, 14.5 (b) Multiply its previous number by 4; 32, 512 (c) Add 1, 2, 3, 4, 5, ... to its previous number successively; 10, 19, 24 3. (a) Add 13 to its previous number; 67, 80 (b) Add 3, 5, 7, 9, 11, 13, ... to its previous number successively; 36, 49 (c) Divide its previous number by 4; 25, 6.25 Diagnostic Test 1.3 1. (a) Each number of the sequence 3, 6, 9, 12, ... can be stated in the form 3n, n = 1, 2, 3, 4, ... (b) Each number of the sequence 7, 8, 9, 10, ... can be stated in the form n + 6, n = 1, 2, 3, 4, ... (c) Each number of the sequence 8, 20, 32, 44, ... can be stated in the form 12n – 4, n = 1, 2, 3, 4, ... (d) Each number of the sequence 4, 7, 12, 19, ... can be stated in the form n2 + 3, n = 1, 2, 3, 4, ... 2. (a) 29 (b) 75 3. (a) 5; 5; 5; 5 (b) 131, 555 4. (a) 1, 4, 9, 16, ... (b) 324 Mastery Challenge 1 1. B 2. B 3. B 4. C 5. A 6. A 7. D 8. C 9. B 10. D 11. (a) Subtract 3 from its previous number. (b) Multiply its previous number by 2. (c) Add 3 to its previous number. (d) Divide its previous number by 2. 12. (a) Add 6 to its previous number. (b) Subtract 5 from its previous number. (c) Multiply its previous number by 4. (d) Divide its previous number by 3. 13. (a) Each number is obtained by adding 6 to its previous number. (b) Each number is obtained by adding 3, 4, 5, 6, ... to its previous number successively. (c) Each number is obtained by multiplying its previous number by 2, 3, 4, 5, ... successively. (d) Each number is obtained by dividing its previous number by 2. 14. (a) 15 (b) 51 (c) 64 (d) 125 15. (a) 17, 21 (b) 33 16. p = 17, q = 9, r = 243 17. a = 13, b = 31, c = 57 18. (a) Multiply the previous number by 1, 2, 3, 4, 5, 6, 7, . . . successively. (b) 120 ; 720 ; 5 040 19. 4n + 6 ; 106 20. (a) 21 = 12 × 1 + 9 33 = 12 × 2 + 9 45 = 12 × 3 + 9 57 = 12 × 4 + 9 (b) Each number of the sequence 21, 33, 45, 57, ... can be stated in the form 12n + 9, n = 1, 2, 3, 4, . . . (c) 585 21. (a) p = 3 , q = –1 (b) 1 199 22. (a) 108 000, 112 500, 117 000, 121 500, ... (b) 4 500n + 103 500 (c) RM193 500 23. (a) 12 , 12 , 22 , 32 , 52 , 82 , 132 , ... (b) a = 2, b = 5, c = 13, d = 34, e = 89, f = 233 (c) The new sequence is the 3rd, 5th, 7th, 9th, 11th, 13th, ... of the sequence 1, 1, 2, 3, 5, 8, 13, ... 24. (a) 33 pieces (b) 63 – 3n (c) n = 20 Chapter 2 Diagnostic Test 2.1 1. (a) (i) x 1 x x2 x x + 3 1 x 1 1 x 1 1 x 1 x + 1 ANSWERS Ans SpotlightA+ Maths Form2.indd 195 11/04/2023 1:15 PM
196 ANSWERS (ii) a 1 1 b ab b b 3b + 1 b ab b b b ab b b 1 a 1 1 a + 2 (b) (i) x2 + 4x + 3 (ii) 3ab + a + 6b + 2 2. (a) 2xy + 8x + y + 4 (b) 2a2 + 3ab + b2 (c) 9m2 + 12mn + 4n2 (d) 8pr + 4ps + 2qr + qs 3. (a) 10x2 + 39x + 14 (b) 3y2 + 5y – 8 (c) 4m2 – 25mn + 6n2 (d) 2p2 + 3pq – 9q2 4. (a) 25x2 – 81 (b) 4h2 – k2 (c) 16x2 – 56x + 49 (d) 9a2 + 12ac + 4c2 5. (a) 16p – 10p2 – 9 (b) 11 – 25y – 8y2 (c) 9k2 – 20km – 9m2 (d) 6a2 – 12ab + 33b2 6. (8k2 – 30k + 24) cm2 Diagnostic Test 2.2 1. (a) (y – 5)(2y – 3) (b) y – 5, 2y – 3 2. (a) 2(3w + 4) (b) 4v(w – 2v) (c) 6hk(2h + 3k) (d) 3a2 (5c – 4b) 3. (a) (k + 5)(k – 5) (b) 5(2 + 3p)(2 – 3p) (c) (2x + 1)2 (d) 3(3y – 4)2 4. (a) (x + 1)(x + 6) (b) (x + 3)(x – 5) (c) (2x – 1)(x + 4) (d) (4x + 3)(2x – 3) 5. (a) (a + 4)(6b – c) (b) (5h – 2k)(3 – 2m) (c) (2p + 9r)(s – 2q) (d) (6x – z)(2w – 7y) 6. (x + 4)(x + 3) 7. (a) (2x + 3)(x + 2) (b) (3x + 2)(x + 4) (c) (2x + 3)2 (d) (5x + 1)(x + 2) 8. (a) RM(4x + 20) (b) RM4(x + 3)(x + 6) Diagnostic Test 2.3 1. (a) 2a 3 (b) 3c – 2 7 (c) 4p 5r (d) 13k – 25 3k 2. (a) 6r2 + q2 6qr (b) 3b – 4a ab (c) 5k2 + 2h2 2hk (d) 45v + 8w 5vw 3. (a) 8m2 – 3n2 (m + n)(2m – 3n) (b) x 2 + 29 (x – 4)(x + 1) 4. (a) 2x + 1 x – 1 (b) –2x (2 + x)(2 – x) 5. (a) 30w + 1 3(3w + 1)(w – 1) (b) k2 – 8 4(k – 4)(2k – 5) 6. (a) 5bc 2ad (b) xw 6y (c) 2 m2 (d) 9w 2u 7. (a) 6 k(h + 3) (b) 1 12(v – 2w) 8. (a) x2 + 2x + 3 (x + 1)(x – 1) (b) 3y2 + 11y + 14 2 Mastery Challenge 2 1. A 2. C 3. C 4. B 5. D 6. B 7. A 8. C 9. D 10. A 11. (a) x 1 1 1 x x2 x x x x + 1 1 x 1 1 1 x + 3 (b) (x + 3)(x + 1) = x2 + 4x + 3 12. (a) (x + 2)(x + 3) = x2 + 5x + 6 (b) (x + 4)(2x + 1) = 2x2 + 9x + 4 (c) (3x + 2)(x + 1) = 3x2 + 5x + 2 13. (a) a a b b b a b b (b) (2a + 3b)(a + 2b) = 2a2 + 7ab + 6b2 14. (2x + 3)(x + 2) 15. (a) x2 + 9x + 18 (b) 2x2 + 7x – 4 (c) 2x2 – 5x – 25 (d) x2 – 10x + 16 16. (a) ✗ (b) ✓ (c) ✗ (d) ✗ 17. (a) 6x2 – 17 (b) 16x – 9 (c) 14x + 10 18. (a) 17h2 – 14hk + 10k2 (b) 3m2 + 8mn – 3n2 19. (a) (5x – 3) cm (b) (36x2 – 36x + 9) cm2 (c) (3x2 + x) cm2 (d) (33x2 – 37x + 9) cm2 20. (a) 4a + 20 = 4(a + 5) (b) (x + 2)(x – 1) = x2 + x – 2 (c) y2 – 4y + 4 = (y – 2)2 21. (a) (i) Expansion (ii) Factorisation (b) a = 4 , b = 3 , c = –3 (c) 4m – 3n ; 2m – 3n 22. (a) 2 ; k (b) 4 ; 4 (c) 3 ; 5 23. (a) ✓ (b) ✗ (c) ✗ (d) ✓ 24. 2x –5 x +4 –5x +8x 2x2 –20 +3x 2x2 + 3x – 20 = (2x – 5)(x + 4) 25. (a) (i) 12x2 – 16x – 3 (ii) 12x2 – 5x – 3 (iii) 12x2 + 35x – 3 (b) (i) 4x ; 3x (ii) 6x ; 2x (iii) 12x ; x 26. AC2 = (2x + 6)2 + [(3x + (–4)]2 = 4x2 + 24x + 36 + 9x2 – 24x + 16 = 13x2 + 52 = 13(x2 + 4) 27. (a) – ; − (b) 1.99 ; 3.96 ; 6.04 (c) 13.98 ; 2.08 (d) 160.6 28. (a) 2x 3 (b) 10 3x (c) x2 – 4 2x (d) x – 16 4x Ans SpotlightA+ Maths Form2.indd 196 11/04/2023 1:15 PM
210 ANSWERS 2. (a) G A G A G A G A G A G A G A {GGG, GGA, GAG, GAA, AGG, AGA, AAG, AAA} (b) (i) {GAA, AGA, AAG} (ii) {GGG, GGA, GAG, AGG} 3. (a) (i) 1 2 (ii) 1 3 (iii) 1 6 (b) (i) → 1 2 (ii) → 1 3 (iii) → 1 6 4. (a) {(20, 10), (20, 30), (20, 50), (20, 70), (20, 80), (40, 10), (40, 30), (40, 50), (40, 70), (40, 80), (50, 10), (50, 30), (50, 50), (50, 70), (50, 80), (60, 10), (60, 30), (60, 50), (60, 70), (60, 80), (90, 10), (90, 30), (90, 50), (90, 70), (90, 80)} (b) (i) 3 25 (ii) 13 25 5. (a) 3 13 (b) 20 39 (c) 5 52 Diagnostic Test 13.3 1. (a) Aʹ = Event that Mei Fong will fail in her music test. (b) Bʹ = Event that the afternoon of Sunday in this week will not rain. (c) Cʹ = Event that Naga team will lose in the dancing competition. (d) Dʹ = Event that the essay of Zaini contains not more than five spelling mistakes. 2. Aʹ = {V, E, T, O, R}, Bʹ = {E, O}, Cʹ = {V, K, T, R} 3. (a) 8 9 (b) 11 15 (c) 11 18 Diagnostic Test 13.4 1. (a) 10 (b) 1 3 2. (a) 3 20 (b) 1 2 (c) 2 5 Mastery Challenge 13 1. C 2. C 3. B 4. C 5. A 6. C 7. B 8. D 9. C 10. (a) (i) 3 (ii) 3 20 (b) (i) 4 (ii) 4 20 = 1 5 (c) (i) 5 (ii) 5 20 = 1 4 11. (a) ✗ (b) ✓ (c) ✓ 12. (a) (i) 1 4 (ii) 3 5 (iii) 3 20 (b) (i) 3 10 (ii) 1 2 (iii) 1 5 13. (a) 59 300; 8 25 ; 13 60 ; 4 15 (b) Ice cream: 1 6 ; Cake: 1 3 ; Chocolate: 2 9 ; Drinks: 5 18 14. (a) {(orange, orange), (orange, guava), (orange, duku), (guava, orange), (guava, guava), (guava, duku), (duku, orange), (duku, guava), (duku, duku)} (b) (i) {(guava, orange), (guava, guava), (guava, duku)} (ii) {(guava, orange), (orange, guava)} (iii) {(orange, orange), (guava, guava), (duku, duku)} 15. (a) 2 3 4 5 3 4 5 2 4 5 2 3 5 2 3 4 (b) {(2, 3), (2, 4), (2, 5), (3, 2), (3, 4), (3, 5), (4, 2), (4, 3), (4, 5), (5, 2), (5, 3), (5, 4)} (c) (i) ✓ (ii) ✗ (iii) ✗ 16. (a) (i) 1 3 (ii) 1 2 (b) (i) approaches 1 3 (ii) approaches 1 2 17. (a) 3 10 (b) 1 5 (c) 7 20 (d) 3 20 18. (a) (i) 1 10 (ii) 2 5 (b) 31 19. (a) Aʹ = Event of choosing not a multiple of 3 (b) Bʹ = Event of choosing not a perfect square = {10, 15, 20, 30, 35, 40} (c) Cʹ = {10, 15, 20, 25} 20. (a) 3 4 (b) 7 12 (c) 2 3 21. (a) 1 4 (b) 1 2 (c) 7 8 22. P(Aʹ) = 8 9 ; P(Bʹ) = 17 22 ; P(Cʹ) = 33 46 23. (a) (i) 1 10 (ii) 50 eggs (b) (i) 9 14 (ii) 5 Ans SpotlightA+ Maths Form2.indd 210 11/04/2023 1:16 PM
211 ANSWERS Pentaksiran Sumatif Ujian Akhir Sesi Akademik (UASA) Section A 1. n, n + 10, n + 20, n + 30, n + 40 are five consecutive terms of the sequence. n + (n + 10) + (n + 20) + (n + 30) + (n + 40) = 5n + 100 C: 5n + 100 = 248 5n = 148 n = 29.6 º 248 is not a possible sum for the five consecutive terms of the sequence. Answer: C 2. (3x + 4)2 – (5x – 2)(x – 2) = 9x2 + 24x + 16 – (5x2 – 12x + 4) = 9x2 + 24x + 16 – 5x2 + 12x – 4 = 4x2 + 36x + 12 Answer: C 3. m (m – 1)2 + 1 4(m – 1) = 4m + (m – 1) 4(m – 1)2 = 5m – 1 4(m – 1)2 Answer: D 4. p + 3 3p = 2 – 3q r p + 3 3p = 2r – 3q r (p + 3)r = 3p(2r – 3q) pr + 3r = 6pr – 9pq 5pr – 9pq = 3r p(5r – 9q) = 3r p = 3r 5r – 9q Answer: C 5. Interior angle of pentagon PQRST = (5 – 2) × 180° 5 = 108° ˙PQR = 108° x = 1 2 × (180° – 108°) = 36° y + 32° = 108° y = 76° x + y = 36° + 76° = 112° Answer: C 6. Area of the whole diagram = 100° 360° × p × 242 + 1 4 × p × 122 = 160p + 36p = 196p cm2 Answer: B 7. Let ˙POQ = x x 360° × 2p × 12 + 12 + 12 = 4p + 24 x 15° × p = 4p x = 60° Area of the sector OPQ = 60° 360° × p × 122 = 24p cm2 Answer: A 8. Let h = the height of water in the container 2 3 × 22 7 × 73 = 14 × 11 × h 2 3 × 7 = h h = 14 3 = 4 2 3 Answer: A 9. 1 3 × 82 × GK = 128 1 3 × GK = 2 GK = 6 cm Surface area of the composite solid = 8 × 8 + 4(8 × 5) + 2( 1 2 × 8 × 10) + 2( 1 2 × 8 × 6) = 64 + 160 + 80 + 48 = 352 cm2 Answer: B 10. Midpoint of PQ = M( –2 + 4 2 , 1 + 5 2 ) = M(1, 3) Distance of MS = (1 – 0)2 + (3 + 2)2 = 1 + 25 = 26 = 5.10 units Answer: D 11. D: The element n in set X is mapped to two elements, 6 and 9 in set Y. º The relation is not a function. Answer: D 12. Distance travelled = 105 km Time taken = 1 hour 15 minutes = 1.25 hours Average speed = 105 1.25 = 84 km/h Answer: C 13. mPQ = 1 2 × mQR 1 – (–3) 3 – (–3) = 1 2 × y – 1 x – 3 4 6 = 1 2 × y – 1 x – 3 4(x – 3) = 3(y – 1) 4x – 12 = 3y – 3 4x – 3y = 9 Answer: B Ans SpotlightA+ Maths Form2.indd 211 11/04/2023 1:16 PM
212 ANSWERS 14. Let c = y-intercept of TW c – (–3) 0 – 5 = –2 c + 3 = 10 c = 7 Answer: B 15. y P T Q R S x –2 O 2 –2 2 4 –6 –4 4 6 8 10 º The coordinates of point S are (9, –2). Answer: C 16. 2 4 6 R 8 y P W L M N Q x –2 O 2 –2 –6 –4 4 6 8 –4 10 º The coordinates of W are (0, –2). Answer: A 17. a + 7 + 15 + 4 + 27 + 7 6 = 12 a + 60 6 = 12 a + 60 = 72 a = 12 4, 7, 7, 12, 15, 27 Median = 7 + 12 2 = 9.5 Answer: B 18. The data 45 has the highest frequency. º The mode of the data is 45 seconds. Answer: C 19. x 5 + 8 + x + 12 = 2 7 x x + 25 = 2 7 7x = 2x + 50 5x = 50 x = 10 Answer: B 20. Number of cards labelled with vowel = 3 Total number of cards = 7 + 8 = 15 P(a card that is labelled with vowel is picked) = 3 15 = 1 5 Answer: B Section B 21. (a) –13, –7, –1, 5, 11 (b) (i) 5n, n = 1, 2, 3, 4, … (ii) n3 , n = 1, 2, 3, 4, … 22. (a) (i) 7(3r – 2) (ii) (8 + 3w)(8 – 3w) (b) 6y – 12 3y + 13 ÷ 4 – 2y y = 6(y – 2) 3y + 13 × y –2(y – 2) = –3y 3y + 13 23. (a) a = 4(2b + 1 c – d ) (b) b = 1 2 ( a 4 – 1 c + d ) (c) 1 c = a 4 – 2b + d = a – 8b + 4d 4 c = 4 a – 8b + 4d (d) d = 2b + 1 c – a 4 d = (2b + 1 c – a 4 ) 2 24. (a) Distance of QR = 52 + 122 = 13 units (b) (6, –6) (c) (12, –3) (d) (6, 5 2 ) 25. (a) S = {(7, 6), (7, 10), (7, 14), (11, 6), (11, 10), (11, 14), (13, 6), (13, 10), (13, 14), (17, 6), (17, 10), (17, 14)} (b) A = {(7, 14), (11, 10), (13, 14), (17, 10)} B = {(11, 10), (11, 14), (13, 10), (13, 14), (17, 6), (17, 10), (17, 14)} n(A) = 4 n(B) = 7 n(S) = 12 (i) P(A) = n(A) n(S) = 4 12 = 1 3 (ii) P(B) = n(B) n(S) = 7 12 Section C 26. (a) Interior angle of octagon ABCDEFGH = (8 – 2) × 180° 8 = 135° º x = 135° Ans SpotlightA+ Maths Form2.indd 212 11/04/2023 1:16 PM