Form 4 CHAPTER 5 48 Solution Calculate the number of terms first. Tn = 5 1 16 ar n – 1 = 81 16 2 3 1 3 2 2 n – 1 = 81 16 r = 1 2 3 = 3 2 1 3 2 2 n – 1 = 81 16 × 3 2 1 3 2 2 n – 1 = 243 32 1 3 2 2 n – 1 = 1 3 2 2 5 Make the base to be the same, i.e. 3 2 . n – 1 = 5 n = 6 Next, find the sum of the first 6 terms. Sn = a(r n – 1) r – 1 Since r 1, the formula Sn = a(r n – 1) r – 1 is used. S6 = 2 3 31 3 2 2 6 – 14 3 2 – 1 S6 = 1341 48 Example 15 Given that 6k + 8, 16 and 2k are the first three terms of a geometric progression, find (a) the values of k, (b) the sum of the first n terms, using the positive value of k. Solution (a) Since 6k + 8, 16 and 2k are three consecutive terms of a geometric progression, use the idea that the common ratio is always the same. 16 6k + 8 = 2k 16 2k(6k + 8) = 256 12k2 + 16k = 256 3k2 + 4k = 64 3k2 + 4k – 64 = 0 (3k + 16)(k – 4) = 0 Solve by factorisation k = – 16 3 or 4 (b) When k = 4, the first three terms are 32, 16 and 8. Sn = a(1 – r n ) 1 – r = 3231 – 1 1 2 2 n 4 1 – 1 2 = 6431 – 1 1 2 2 n 4 Example 16 The 3rd term of a geometric progression is 24 and the sum of the first two terms is 288. If all the terms are positive, find (a) the first term and the common ratio, (b) the sum of all the terms from the 4th term to the 8th term. Solution (a) T3 = 24 ar 2 = 24 … ➀ S2 = 288 a(r 2 – 1) r – 1 = 288 a(r + 1) (r – 1) r – 1 = 288 a(r + 1) = 288 … ➁ Note that S2 is simplified to the simplest form by factorisation and elimination. ➁ ➀ : a(r + 1) ar 2 = 288 24 Note that for geometric progression, the simultaneous equations formed is solved using division. r + 1 r 2 = 12 12r 2 = r + 1 12r 2 – r – 1 = 0 (3r – 1)(4r + 1) = 0 r = 1 3 or – 1 4 r = – 1 4 is not accepted because the all the terms are positive. ∴r = 1 3 Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 48 16-Feb-23 7:22:20 PM PENERBIT ILMU BAKTI SDN. BHD.
Form 4 CHAPTER 5 49 From ➀, when r = 1 3 , a1 1 3 2 2 = 24 ⇒ a = 216 (b) The sum of all the terms from the 4th term to the 8th term S8 – S3 = 2163 1 – 1 1 3 2 8 4 1 – 1 3 – 2163 1 – 1 1 3 2 3 4 1 – 1 3 = 323 77 81 – 312 = 11 77 81 SMART TIP The sum of all the terms from the 4th term to the 8th term = S8 – S3 S8 S3 S8 – S3 T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 Example 17 The price of a shop at a commercial centre is RM220 000. Its value increases by 5% each year. Calculate the minimum of years required for its value to exceed RM400 000 for the first time. Solution Number of years 0 1 2 … Tn T1 T2 T3 … Value (RM) 220 000 231 000 242 550 … × 1.05 × 1.05 Tn 400 000 arn – 1 400 000 220 000(1.05)n – 1 400 000 1.05n – 1 400 000 220 000 1.05n – 1 1.8182 lg 1.05n – 1 lg 1.8182 Take lg of both sides of the equation. (n – 1)lg 1.05 lg 1.8182 (n – 1)0.02119 0.2596 n – 1 0.2596 0.02119 n – 1 12.25 n 13.25 Smallest integer value of n = 14 Hence, the minimum of years required for its value to exceed RM400 000 for the first time is 13 years. Caution! n = 14. The minimum of years required for its value to exceeds RM400 000 for the first time is 14 years. Incorrect Example 18 Ibrahim saves 5 sen on the first day, 10 sen on the second day, 20 sen on the third day and so on such that the amount of money he saves each day is two times of the preceding day. Find the minimum number of days for the sum of his money to exceed RM1 000 for the first time. Solution 5 sen, 10 sen, 20 sen, … a = 5, r = 2 Sn 100 000 RM1 000 = 100 000 sen a(r n – 1) r – 1 100 000 5(2n – 1) 2 – 1 100 000 5(2n – 1) 100 000 2n – 1 100 000 5 2n – 1 20 000 2n 20 001 lg 2n lg 20 001 Take lg of both sides of the n equation. lg 2 lg 20 001 n(0.3010) 4.3011 Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 49 16-Feb-23 7:22:20 PM PENERBIT ILMU BAKTI SDN. BHD.
Form 4 CHAPTER 5 50 n 4.3011 0.3010 n 14.29 Smallest integer value of n = 15 Hence, the minimum number of days for the sum of money to exceed RM1 000 for the first time is 15 days. Example 19 Calculate the sum to infinity of the geometric progression 24, 12, 6, … Solution S∞ = a 1 – r = 24 1 – 1 2 r = T2 T1 = 12 24 = 1 2 = 48 Example 20 Write the recurring decimal 0.181818… as a single fraction in its lowest terms. Solution 0.181818… = 0.18 + 0.0018 + 0.000018 + … = 0.18 1 – 0.01 S∞ = a 1 – r = 0.18 0.99 r = 0.0018 0.18 = 0.01 = 18 99 = 2 11 Example 21 The first term of a geometric progression is 81. The sum to infinity of the geometric progression is 121 1 2 . Find (a) the common ratio, (b) the term that is less than 1 2 for the first time. Solution (a) S∞ = 121 1 2 a 1 – r = 243 2 81 1 – r = 243 2 243 – 243r = 162 243r = 81 r = 81 243 = 1 3 (b) Tn 1 2 arn – 1 1 2 (81)1 1 3 2 n – 1 1 2 1 1 3 2 n – 1 1 162 lg 1 1 3 2 n – 1 lg 1 1 162 2 (n – 1)lg 1 1 3 2 lg 1 1 162 2 ** (n – 1)(-0.4771) –2.2095 n – 1 . –2.2095 –0.4771 SMART TIP Reverse the inequality sign each time a division by a negative number is carried out. n – 1 . 4.63 n . 5.63 Smallest integer value of n = 6 Hence, the term that is less than 1 2 for the first time is the 6th term. Caution! From the step ** (n – 1)lg 1 1 3 2 lg 1 1 1622 n – 1 lg 1 1 1622 lg 1 1 3 2 n – 1 4.63 n 5.63 ∴n = 5 Incorrect because the inequality sign is not reversed when a division by a negative number is carried out. Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 50 16-Feb-23 7:22:21 PM PENERBIT ILMU BAKTI SDN. BHD.
Form 4 CHAPTER 5 51 A piece of wire with a length of 78π cm is cut to form 6 circles as shown in the diagram below. The radii of the circles differ from each other successively by 1 cm. Find (a) the radius of the smallest circle, (b) the number of complete circles that can be formed if the original length of the wire is 200π cm. HOTS Applying HOTS Analysing HOTS Evaluating Solution First of all, prepare a table as follows. Radius r r + 1 r + 2 … Circumference 2πr 2π(r + 1) 2π(r + 2) … Term T1 T2 T3 … (a) d = T2 – T1 = 2π(r + 1) – 2πr = 2π Given that the length of the wire is 78π cm, thus S6 = 78π n 2 32a + (n – 1)d4 = 78π 6 2 32(2πr) + 5(2π)4 = 78π 3(4πr + 10π) = 78π 4πr + 10π = 26π 4πr = 16π r = 16π 4π = 4 Hence, the radius of the smallest circle is 4 cm. (b) a = 2πr = 2π(4) = 8π Given that Sn = 200π, n 2 32(8π) + (n – 1)(2π)4 = 200π 8nπ + n(n – 1)π = 200π 8nπ + n2 π – nπ = 200π 7nπ + n2 π = 200π 7n + n2 = 200 The equation is divided throughout by π. n2 + 7n – 200 = 0 n = –7 ± ! 72 – 4(1)(–200) (2)(1) x = –b ± ! b2 – 4ac 2a = –7 ± ! 849 2 = –18.07 or 11.07 n = –18.07 is not accepted. ∴n = 11.07 Hence, the number of complete circles that can be formed is 11. HOTS Zone Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 51 16-Feb-23 7:22:21 PM PENERBIT ILMU BAKTI SDN. BHD.
Form 4 CHAPTER 5 52 5.1 Arithmetic Progressions 1 Calculate the 16th term of the arithmetic progression 1 3 , 1 2 3 , 3, … 2 Find the nth term of the arithmetic progression 4, 6 1 2 , 9, … 3 Find the number of terms of the arithmetic progression 4, 1, –2, …, –23. 4 Find the number of the multiples of 3 between 20 and 110. 5 For the arithmetic progression 5, 8, 11, …, which term is equal to 320? 6 Given that 4k, 12 and 2k2 + 4k are three consecutive terms of an arithmetic progression, find the possible values of k. HOTS Applying 7 The 12th term and the 17th term of an arithmetic progression are 38 and 53 respectively. Find the common difference. 8 For the arithmetic progression 185, 182, 179, …, find the value of the first negative term. HOTS Applying 9 Find the sum of the first 12 terms of the arithmetic progression –10, –7, –4, … 10 Find the sum of the first n terms of the arithmetic progression 11, 7, 3, … 11 For the arithmetic progression –4, 1, 6, ... , find the sum of all the terms from the 6th term to the 14th term. 12 Find the sum of all the terms of the arithmetic progression 1 4 , 1 1 2 , 2 3 4 , …, 14. 13 How many terms of the arithmetic progression 12, 16, 20, … that has to be taken for its sum to be equal to 132? 14 The first term and the last term of an arithmetic progression are –12 and 36 respectively. If the sum of all the terms is 240, find the number of terms. 15 The second term of an arithmetic progression is 8 while the sum of the first six terms is 30. Find the common difference. 16 The 3rd term and the 7th term of an arithmetic progression are 6 and 26 respectively. Find the 11th term. 17 The sum of the first six terms of an arithmetic progression is 72 while the sum of the first nine terms is 81. Find the common difference. 18 The 3rd term and the 9th term of an arithmetic progression are 7 and 31 respectively. Find the sum of the first 12 terms. 19 If the sum of the first n terms of an arithmetic progression is given by Sn = n(4n – 7), find the 4th term. 20 If the sum of the first n terms of an arithmetic progression is given by Sn = n(2n + 3), find the common difference. 21 A long distance runner takes 3 minutes 45 seconds for the first kilometre but his speed decreases at a constant rate such that for each of the following kilometres, he needs 12 seconds more than the preceding kilometre. Find the time needed for the 11th kilometre (in minutes and seconds). 22 Find the number of months needed to settle a loan of RM5 800 by monthly installments of RM100 for the first month and for the each of the following months, it is increased by RM20 successively. 23 The bricks are arranged in a straight row such that their lengths increase successively in an arithmetic progression. Given that the shortest brick has a length of 0.8 m and the longest brick has a length of 1.25 m. If the sum of the lengths of all the bricks is 20.5 m, calculate the number of bricks required. SPM Practice 5 Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 52 16-Feb-23 7:22:21 PM PENERBIT ILMU BAKTI SDN. BHD.
Form 4 CHAPTER 5 53 24 The perimeter of a hexagon is 36 cm. The lengths of the sides of the hexagon are in an arithmetic progression. The length of the longest side is five times of the length of the shortest side. Find the length of the shortest side. HOTS Applying 25 A wire with a length of 176 cm is cut into 20 parts such that the lengths of the parts form an arithmetic progression. It is given that the sum of the first five shorter parts is 14 cm. Find the length of the longest part. HOTS Applying 26 A string with a length of 62 m is cut into n parts such that the lengths of the parts form an arithmetic progression with a common difference of d m. It is given that the shortest part is 0.5 m and the longest part is 3.5 m. Find the value of n and of d. HOTS Applying 27 (a) Given that x + 2, 2x and 2x + 3 are three first terms in an arithmetic progression, find (i) the value of x, (ii) the sum of 20 following terms. (b) In an arithmetic progression, the sum of the first n terms is given by Sn = 6n2 + 3n. Find (i) the nth term, (ii) the common difference. 28 The first term of an arithmetic progression is 24. It consists of 23 terms. The sum of the last three terms is 5 times of the sum of the first three terms. Find (a) the common difference, (b) the sum of the first 11 terms. HOTS Applying 29 The sum of the first 6 terms of an arithmetic progression is 39 and the sum of the 6 following terms is –69. Find (a) the first term and the common difference, (b) the sum of all the terms from the 15th term to the 25th term. HOTS Applying HOTS Analysing 30 An arithmetic progression has 14 terms. The sum of the first 14 terms is 301 while the sum of all the odd terms (first, third, fifth and so on) is 140. Find (a) the first term and the common difference, (b) the last term. HOTS Applying HOTS Analysing 31 An arithmetic progression has n terms. The sum of all the terms is 300. If the common difference and the last term are 2 and 36 respectively, find the possible values of n and the corresponding values of the first term. HOTS Applying HOTS Analysing 32 The following diagram shows a series of cone. The base-radius of each cone is a constant at 6 cm. The height of the first cone is h cm. The height of the second cone is (h + 1) cm. The height of the third cone is (h + 2) cm and so on such that the height of each cone is 1 cm more than its preceding cone successively. 6 cm 6 cm 6 cm h cm (h + 1) cm (h + 2) cm Determine whether the volumes (in cm3 ) of the cones are in an arithmetic progression or a geometric progression. Hence, state its common difference or common ratio. 33 A circle is divided into 8 sectors such that the angles subtended at the centre of the circle forms an arithmetic progression. Given that the angle of the smallest sector is 10°, find (a) the common difference and the angle of the largest sector, (b) the sum of the angles of the first 6 sectors, (c) the area of the fourth sector if the area of the whole circle is 36π cm2 . HOTS Applying HOTS Analysing Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 53 16-Feb-23 7:22:22 PM PENERBIT ILMU BAKTI SDN. BHD.
Form 4 CHAPTER 5 54 34 The diagram below shows a few sectors of circles with centre O. The angle subtended at the centre of the circle is π 3 radians. The length of each arc increases by π cm successively. O 15 cm π 3 rad. Given that the radius of the nth sector is 84 cm, find (a) the length of arc of the nth sector, in terms of π, (b) the value of n, (c) the sum of the lengths of arcs of the first 10 sectors, in terms of π. HOTS Applying HOTS Analysing 35 There are two pieces of wires. One of the wires is 110 cm longer than the other wire. Each wire is cut such that the length of the first part is a cm while the lengths of the following parts decrease by d cm successively. (a) If the shorter wire is cut into 15 parts while the longer wire is cut into 25 parts, express a in terms of d. (b) If d = 2, calculate the value of a. Hence, find the original length of each of the two wires. (c) For the longer wire, which part has a length of 12 cm? HOTS Applying HOTS Analysing 36 A piece of wire with a length of 156 cm is cut to form 8 equilateral triangles as shown in the following diagram. The length of each side of the triangles increases by 1 cm successively. Find (a) the perimeter of the smallest triangle, (b) the number of complete triangle that can be formed if the original length of the wire used is 756 cm. HOTS Applying HOTS Analysing 37 A piece of wire with a length of 100π cm is bent to form 8 semicircles as shown in the diagram below. The radius of each semicircle increases by 1 cm successively. Find (a) the radius of the smallest semicircle, (b) the number of complete semicircles that can be formed if the original length of the wire used is 500π cm. HOTS Applying HOTS Analysing 38 (a) Four angles of a quadrilateral are in an arithmetic progression. Given that the largest angle is four times of the smallest angle, find the size of each of the four angles. (b) Aminah wants to construct a wall in the form of a triangle such that the highest row has 1 brick, the second row has 2 bricks, the third row has three bricks and so on. If she has 250 bricks, how many rows that can be constructed and how many bricks that are left? HOTS Applying HOTS Analysing 39 The following diagram shows the side view of a stairs. The height of each block is 15 cm. The length of the lowest block is 130 cm. The length of each of the following blocks is 4 cm less than its preceding block. 130 cm 4 cm 4 cm 15 cm Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 54 16-Feb-23 7:22:22 PM PENERBIT ILMU BAKTI SDN. BHD.
Form 4 CHAPTER 5 55 (a) If the height of the whole stairs is 3 m, find (i) the length of the highest block, (ii) the sum of the lengths of all the blocks. (b) Find the possible maximum height of the whole stairs. HOTS Applying HOTS Analysing 40 Two companies, Epsilon and Theta, start to sell motorcycles at the same time. (a) Epsilon sells h motorcycles in the first month and its sales increase by k motorcycles successively in the following months. It sells 320 motorcycles in the 5th month and the sum of sales in the first 8 months is 2 440. Find the value of h and of k. (b) Theta sells 160 motorcycles in the first month and its sales increase by 35 successively in the following months. If both companies sell the same number of motorcycles in the nth month, find the value of n. 41 The following diagram shows part of the construction model of a building using cubes of the same size, of side 3 cm. Each level of the cubes represents a floor of the building. The highest floor consists of 1 cube. 3 cm (a) If the lowest floor of the model consists of 105 cubes, calculate the height of the model. (b) If the cost of each cube is 90 sen, calculate the total cost to construct the model. 42 Four angles in a quadrilateral form an arithmetic progression. Given that the largest angle is three times the smallest angle, find the value of each angle. HOTS Applying HOTS Analysing 43 The diagram below shows the arrangements of a series of cylinders that have a constant base radius of 3 cm. The height of the first cylinder is 5 cm, the height of the second cylinder is 7 cm and so on such that the height of each of the following cylinders increases by 2 cm successively. 3 cm 3 cm 3 cm 5 cm 7 cm 9 cm (a) Find the volume, in cm3 , of the 15th cylinder, in terms of π. (b) Given that the sum of the first n cylinders is 1 260π cm3 , find the value of n. 5.2 Geometric Progressions 44 Calculate the 8th term of the geometric progression 3, –6, 12, –24, … 45 Find the nth term of the geometric progression 1 2 , 1 4 , 1 8 , … 46 Calculate the number of terms of the geometric progression 1 4 , 1 12, 1 36, …., 1 324 . 47 Given that x + 1, 12 and 4x + 4 are three consecutive terms of a geometric progression, find the possible values of x. HOTS Applying 48 For the geometric progression 500, 100, 20, …, find the term that is less than 1 10 for the first time. HOTS Applying 49 Calculate the sum of the first six terms of the geometric progression 3, 2, 4 3 , … 50 Calculate the sum of all the terms of the geometric progression 2 3 , 2, 6, …, 486. Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 55 16-Feb-23 7:22:22 PM PENERBIT ILMU BAKTI SDN. BHD.
Form 4 CHAPTER 5 56 51 Given the geometric progression 27, 9, 3, … Find the sum of all the terms from the 5th term to the 8th term. 52 Find the number of terms for the geometric progression 48, 24, 12, … that have to be taken such that its sum is equal to 95 1 4 . HOTS Applying 53 For the geometric progression 3, 6, 12, ... find the minimum number of terms that have to be taken such that its sum exceeds 1 000. HOTS Applying 54 The 2nd term and 7th term of a geometric progression are 3 and 32 81 respectively, find the first term. 55 The 3rd term and the 6th term of a geometric progression are 27 and 8 respectively. Find the 2nd term. 56 The 2nd term and the 3rd term of a geometric progression are 1 2 and 1 respectively. Find the sum of the first three terms. 57 For a geometric progression, the sum of the 2nd term and the 3rd term is 12 while the sum of the 3rd term and the 4th term is 6. Find the common ratio. HOTS Applying HOTS Analysing 58 The first term of a geometric progression is 18 and the sum of the first three terms is 38. Find the possible values of the common ratio. HOTS Applying 59 The lengths of the sides of a triangle are in a geometric progression. The length of the longest side is 36 cm. If the perimeter of the triangle is 76 cm, find the positive value of the common ratio. HOTS Applying 60 The value of the insurance coverage of a new imported car bought in the year 2011 is RM120 000. The value of the insurance for each of the following years is 5% less than its preceding year. Calculate the value of the insurance in the year 2019. HOTS Applying 61 The 3rd term of a geometric progression exceeds the 2nd by 6 while the 4th term exceeds the 3rd term by 2. Find the sum of the first 5 terms. 62 The 2nd term and the 3rd terms of a geometric progression are m3 4 and m5 16 respectively. (a) Find the values that cannot be taken by m other than 0. (b) If m = 6, find (i) the first term and the common ratio, (ii) the sum of all the terms from the 2nd term to the 5th term. 63 An arithmetic progression and a geometric progression have the same first term. Both progressions also have the same values of common difference and common ratio i.e. 1 3 . The 10th term of the arithmetic progression is 12. The 13th term of the arithmetic progression is equal to the sum of the first n terms of the geometric progression. Find (a) the first term of both progressions, (b) the value of n. HOTS Applying HOTS Analysing 64 The first term and the common difference of an arithmetic progression are 8 and d respectively. Given that the first term, 5th term and 21st term of the arithmetic progression form three consecutive terms of a geometric progression, find the of d. HOTS Applying HOTS Analysing HOTS Evaluating 65 Kien Heng arranges 50 sen coins as shown in the diagram below. The number of coins in each pile forms a geometric progression. 50 sen 50 sen 50 sen Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 56 16-Feb-23 7:22:23 PM PENERBIT ILMU BAKTI SDN. BHD.
Form 4 CHAPTER 5 57 Find (a) the number of coins in the 7th pile, (b) the number of complete piles obtained if Kien Heng has 1500 coins. HOTS Applying HOTS Analysing 66 The diagram below shows five squares drawn successively such that the length of each side of the squares forms a geometric progression. The length of each side of the first square is x cm and the ratio of each side of the fourth square to the length of each side of the first square is 125 : 64. x cm (a) Find the ratio of the length of each side of the second square to the length of each side of the first square. (b) If 5 squares are formed from a piece of wire with a length of 131 5 16 cm, find the value of x. HOTS Applying HOTS Analysing 67 On 1 January 2020, Encik Shamsul opened a fixed deposit account with an amount of money of RM10 000. The annual interest is 5% of the amount of money left at the end of each year. The interest is added into the account at the beginning of each year commencing 1 January 2021. No withdrawal is done. Find (a) the ratio of the amount of money on 1 January 2021 to the amount of money on 1 January 2020, (b) the year when there is more than RM25 000 for the first time in the fixed deposit account after the interest has been added to it. HOTS Applying HOTS Analysing HOTS Evaluating 68 A gardener is given a task to cut the grass on an area of 1 000 m2 . On the first day, he cuts the grass on an area of 16 m2 . On each of the following days, the area of grass he cuts is 1.1 times of the area of grass he cuts on the preceding day until the day his task is completed. Find (a) the area of grass he cuts on the 10th day, (b) the minimum number of days required to complete his task. HOTS Applying HOTS Analysing 69 A welfare body gives financial assistance to an orphan’s home beginning from 2017. The amount of financial assistance given in the year 2017 is RM10 000 and after that, it is 95% of the financial assistance given in the preceding year. Find (a) the year when financial assistance given is RM8 573.75, (b) the number of years taken for the sum of money given to be more than RM70 000 for the first time. HOTS Applying HOTS Analysing 70 Shahril starts to work at a company on 1 January 2015 with an annual salary of RM24 000. Each January in each of the following years, the company increases his salary by 5% of his salary in the preceding year. Find (a) his annual salary, correct to the nearest RM, in the year 2020, (b) the minimum value of n such that his annual salary in the nth year exceeds RM35 000, (c) his total salary, correct to the nearest RM, from the year 2015 to the year 2020. HOTS Applying 71 The following diagram shows a series of similar right-angled triangles. A B C E G b D F h The first triangle is ∆ABC (with base b and height h), followed by ∆ADE, ∆AFG and so on. The base and the height of each triangle are two times of the base and the height of the preceding triangle. Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 57 16-Feb-23 7:22:23 PM PENERBIT ILMU BAKTI SDN. BHD.
Form 4 CHAPTER 5 58 (a) Show that the areas of the triangles form a geometric progression. State its common ratio. (b) If b = 6 cm and h = 3 cm, find (i) the area of the eight triangle, (ii) the sum of the areas of the first four triangles. 72 A piece of wire is cut into n parts. The length of each part increases and forms a geometric progression. Given that the length of the seventh part of the wire is 8 times of the length of the fourth part of the wire. (a) Calculate the common ratio. (b) If the sum of the lengths of all the parts of the wire is 3 069 cm and the length of the first part of the wire is 3 cm, find (i) the value of n, (ii) the length, in cm, of the last part of the wire. HOTS Applying HOTS Analysing 73 Given that …, 729, k, 6 561, … is part of a geometric progression and the sum of the first five terms is 1 089, find (a) the common ratio, (b) the first term, (c) the smallest value of n such that the nth exceeds 12 000 HOTS Applying 5.2b Sum to Infinity 74 If the sum to infinity of a geometric progression is 64 and the first term is 16, find the common ratio. 75 Write the recurring decimal 0.459459459… as a single fraction in its lowest terms. 76 The 2nd term and the 4th term of a geometric progression are 20 and 12 4 5 respectively. If all the terms are positive, find the sum to infinity of the progression. HOTS Applying 77 The 2nd term and the sum to infinity of a geometric progression are 6 and 32 respectively. Find the possible values of the common ratio. HOTS Applying 78 Given that 5x – 12, x and x – 4 are three consecutive terms of a geometric progression, find (a) the value of x if all the terms are positive, (b) the sum to infinity of the progression, (c) the value of the term that is less than 1 10 for the first time. 79 P and Q are two geometric progressions with common ratios of 1 2 and 1 3 respectively. The sum to infinity of both geometric progressions are the same. If the third term of the geometric progression P exceeds the third term of the geometric progression Q by 11 6 , find the first term of each geometric progression. HOTS Applying HOTS Analysing 80 Stannum is extracted from tin ore. During the first year of operation, the tin ores produce 20 000 kg of stannum. The production of stannum in each of the following years is reduced by 8% of the production in the preceding year. (a) Find the maximum total of stannum that can be extracted if the mining of tin ores and the extraction of stannum is continued in the same manner for a very long period of time. (b) Due to economic reasons, the mining of tin ores has to be stopped if the total of stannum produced is less than 2 000 kg. Find the number of years the mining activity can be operated. 81 The following diagram shows a set of trapeziums PQSR, RSUT, TUWV, … The base of each trapezium is 2 cm. The length of SR is 9 10 times of the length of QP. The length of UT is 9 10 times of the length of SR. The length of WV is 9 10 times of the length of UT. Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 58 16-Feb-23 7:22:23 PM PENERBIT ILMU BAKTI SDN. BHD.
Form 4 CHAPTER 5 59 2 cm 2 cm 2 cm P Q R S T U V W 100 cm If QP = 100 cm, (a) show that the areas of the trapeziums PQSR, RSUT, TUWV, … (in cm2 ) form a geometric progression and state its common ratio, (b) find the sum to infinity of the areas of the set of trapeziums. HOTS Applying HOTS Analysing 82 Syamala drops a rubber ball from a height of h cm above the floor. After the first bounce, the ball reaches a height of h1 cm, such that h1 = 0.9h. After the second bounce, the ball reaches a height of h2 cm, such that h2 = 0.9h1. The ball continues to bounce in the same manner until it stops bouncing. This is shown in the diagram below. h h2 h1 Given that h = 100 cm, find (a) the number of bounces when the maximum height of the ball from the floor is less than 25 cm for the first time, (b) the total distance, in cm, travelled by the ball until it stops bouncing. HOTS Applying HOTS Analysing 83 An oil well produced 100 000 barrels of oil in its first year of operation but the output fell by 5% each year thereafter. (a) Find the amount of oil, in barrels, extracted in the fifth year of its operation. (b) What is the maximum amount of oil, in barrels, that could be extracted? (c) If the well is closed down after 12 years, what was the total amount of oil, in barrels, extracted during that period of time? HOTS Applying HOTS Analysing 84 The following diagram shows part of several squares which are drawn in sequence. x cm The length of each side of the squares form a geometric progression. Given that the length of the side of the first square is x cm and the ratio of the length of the side of the third square to that of the first square is 4 9 . (a) Find the ratio of the length of the side of the second square to the length of the side of the first square. (b) If the sum of the area of the first three squares is 532 cm2 , find the value of x, (c) Calculate the sum to infinity of the areas of the squares. HOTS Applying HOTS Analysing HOTS Evaluating Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 59 16-Feb-23 7:22:23 PM PENERBIT ILMU BAKTI SDN. BHD.
Form 4 CHAPTER 5 60 1 20 1 3 2 1 2(5n + 3) 3 10 4 30 5 T106 6 k = 2 or –6 7 3 8 –1 9 78 10 13n – 2n2 11 369 12 85 1 2 13 6 14 20 15 –2 16 46 17 –2 18 252 19 21 20 4 21 5 minutes 45 seconds 22 20 months 23 20 bricks 24 2 cm 25 16.4 cm 26 n = 31, d = 0.1 27 (a) (i) x = 5 (ii) 890 (b) (i) 12n – 3 (ii) 12 28 (a) 6 (b) 594 29 (a) a = 14, d = –3 (b) –473 30 (a) a = 2, d = 3 (b) 41 31 n = 25, a = –12 n = 12, a = 14 32 Arithmetic progression, d = 12π 33 (a) d = 10°, l = 80° (b) 210° (c) 12.57 cm2 34 (a) 28π cm (b) n = 24 (c) 95π cm 35 (a) a = 11 + 19.5d (b) a = 50 ; 540 cm, 650 cm (c) 20th portion 36 (a) 9 cm (b) 20 37 (a) 9 cm (b) 24 38 (a) 36°, 72°, 108°, 144° (b) 21 rows, 19 bricks 39 (a) (i) 54 cm (ii) 1 840 cm (b) 495 cm 40 (a) h = 200, k = 30 (b) n = 9 41 (a) 81 cm (b) RM1 287.90 42 45°, 75°, 105°, 135° 43 (a) 297π cm3 (b) 10 44 –384 45 1 2n 46 5 47 x = 5 or –7 48 T7 49 8 17 81 50 7282 3 51 40 81 52 7 53 9 54 4 1 2 55 401 2 56 1 3 4 57 1 2 58 2 3 or –12 3 59 1 1 2 60 RM79 610.45 61 –401 3 62 (a) 2 or –2 (b) (i) a = 6, r = 9 (ii) 44 280 63 (a) 9 (b) n = 3 64 d = 6 65 (a) 128 (b) 9 66 (a) 5 : 4 (b) x = 4 67 (a) 1.05 (b) Year 2039 68 (a) 37.73 m2 (b) 21 days 69 (a) Year 2020 (b) 9 years 70 (a) RM30 631 (b) n = 9 (c) RM163 246 71 (a) r = 4 (b) (i) 147 456 cm2 (ii) 765 cm2 72 (a) r = 2 (b) (i) n = 10 (ii) 1 536 cm 73 (a) r = 3 (b) a = 9 (c) n = 8 74 r = 3 4 75 17 37 76 125 77 1 4 or 3 4 78 (a) x = 6 (b) 27 (c) 2 27 79 ap = 18, aQ = 24 80 (a) 250 000 kg (b) 28 years 81 (a) 9 10 (b) 1 900 cm2 82 (a) 14 (b) 1 900 cm 83 (a) 81 451 barrels (b) 2 000 000 barrels (c) 919 280 barrels 84 (a) 2 : 3 (b) x = 18 (c) 583.2 cm2 Answers Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 60 16-Feb-23 7:22:23 PM PENERBIT ILMU BAKTI SDN. BHD.
260 Paper 1 Section A Answer all questions in this section. [64 marks] Time: 2 hours [80 marks] 1 It is given that f(x) = n – mx, where m and n are constants. Find (a) f –1(x) in terms of m and n. [2 marks] (b) the value of m and of n such that f –1(1) = 1 and f(2) = –3. [4 marks] 2 Given that k 3 and k 5 are the roots of the quadratic equation 15x2 – 16x + m = 0, find the value of m and of k. [5 marks] 3 Express the quadratic function f(x) = 2x – 3 – 2x2 in the vertex form f(x) = a(x – h) 2 + k, where a, h and k are constants. Hence, sketch the graph of f(x) = 2x – 3 – 2x2 and state the equation of the axis of symmetry of the graph. [6 marks] 4 Solve the equation: ! x – ! x – 2 = 1 [4 marks] 5 Solve the simultaneous equations: 4x 2y = 2 log10 (2x + 2y) = 1 [5 marks] 6 The coordinates of three points, A, B and C, are (6, 10), (4, 4) and (9, 6) respectively. (a) (i) Find the area of triangle ABC. (ii) Calculate the distance of AB. Hence, determine the perpendicular distance from C to AB. [3 marks] (b) R(x, y) is a moving point such that its distance from point A is always twice its distance from point C. Find the equation of the locus of point R. [2 marks] 7 Given that k – 1, 2k – 2 and 3k – 1 are the first three terms of a geometric progression, find (a) the value of k such that k . 1, [3 marks] (b) the sum of all the terms from the fourth term to the ninth term. [3 marks] SPM Model Test Ana&Tip(BM) SPM Add Maths(Eng)-MT 3rd.indd 260 16-Feb-23 7:34:23 PM PENERBIT ILMU BAKTI SDN. BHD.
261 Model Test SPM 8 It is given that r ~ = (h + 1)i ~ + 3j ~ and s ~ = 8i ~ + (h + 3)j ~ . Find the value (or values) of h for each of the following cases: (a) r ~ is parallel to s ~, [4 marks] (b) the vector s ~ is parallel to the straight line 2y = x – 4. [2 marks] 9 Diagram 1 shows a sector AOB of a circle with centre O. PQ is a straight line. A B Q O P θ 20 cm Diagram 1 Given OB = 20 cm, OQ : QB = 3 : 2 and the area of triangle OPQ = 42.42 cm2 , find KBAT Mengaplikasi (a) the value of θ, in radians, [4 marks] (b) area, in cm2 , of the shaded region. [2 marks] [Use π = 3.142] 10 It is given that sec x = 1 p, such that x is an acute angle. Find (a) cot x, [2 marks] (b) sin2 x 2 . [3 marks] 11 Given that e10 4 f(x) dx = 7, find (a) e 5 4 f(x) dx + e 7 5 f(x) dx + e10 7 f(x) dx, [2 marks] (b) the value of h if e 4 10 [10f(x) – hx] dx = 182. [3 marks] 12 The straight line 4y + x = q is a normal to the curve y = (x – 1)2 – 3 at point A. Find (a) the coordinates of point A, [4 marks] (b) the value of q. [1 mark] Section B Answer any two questions from this section. [16 marks] 13 Puan Junaidah bought 3 tins of baby milk powder, 2 packets of baby diapers and 4 baby rompers. Puan Ooi bought 2 tins of baby milk powder, 4 packets of baby diapers and 3 baby rompers. Puan Kamala bought 4 tins of baby milk powder, 3 packets of baby diapers and 2 baby rompers. The amount payable by Puan Junaidah, Puan Ooi and Puan Kamala are RM480, RM405 and RM600 respectively. Find the price of a tin of baby milk powder, a packet of baby diapers and a baby rompers. [8 marks] Ana&Tip(BM) SPM Add Maths(Eng)-MT 3rd.indd 261 16-Feb-23 7:34:23 PM PENERBIT ILMU BAKTI SDN. BHD.
262 Model Test SPM 14 In Diagram 2, the straight lines AB and CD intersect at point M. The equations of the straight lines AB and CD are x + 2y – 7 = 0 and x – y + 2 = 0 respectively. The point N(0, –3) divides the line segment PQ in the ratio 2 : 1. The coordinates of point P are (–4 , –6). y x C A M Q 0 D x – y + 2 = 0 B x + 2y – 7 = 0 P(–4, –6) N(0, –3) Diagram 2 (a) Find the equation of the straight line MQ. [5 marks] (b) Point R(x, y) moves such that its distance from point M and point Q are always the same. (i) Describe the locus of the moving point R. (ii) Hence, find the equation of the locus of moving point R. [3 marks] 15 Diagram 3 shows a triangle OAB. O B P A R Q 2u ∼ 3v ∼ Diagram 3 The straight line BR intersects the straight line OP at Q. It is given that O → R = 3v ~ , O → B = 2u ~, O → A = 3O → R and B → P = 1 4 B → A. (a) Express each of the following vectors in terms of u ~ and v ~. (i) B → R, (ii) B → P, (iii) O → P. [3 marks] (b) (i) Given that B → Q = hB → R, express B → Q in terms of h, u~ and v ~. (ii) Given that Q → P = kO→ P, express Q → P in terms of k, u~ and v ~. [2 marks] (c) Using B → Q and Q → P in 15(b), find the value of h and of k. [3 marks] Ana&Tip(BM) SPM Add Maths(Eng)-MT 3rd.indd 262 16-Feb-23 7:34:23 PM PENERBIT ILMU BAKTI SDN. BHD.
263 Model Test SPM Paper 2 Section A Answer all questions in this section. [50 marks] Time: 2 hours 30 minutes [100 marks] 1 Diagram 1 shows two points, A and B, on a straight line such that AB = 8.5 m. X Y A 8.5 m B Diagram 1 Two particles, X and Y, move simultaneously from point A and point B respectively. Particle X travels 52 cm during the 1st second, 50 cm during the 2nd second, 48 cm during the 3rd second and so on. Particle Y travels 19 cm during the 1st second, 18 cm during the 2nd second, 17 cm during the 3rd second and so on. Find HOTS Applying (a) the distance, in cm, travelled by particle X during the 10th second, [2 marks] (b) the total distance, in cm, travelled by particle Y in the first 20 seconds, [2 marks] (c) the time, in seconds, taken by particle X and particle Y to collide. [4 marks] 2 (a) It is given that y = x 2x – 4 and dy dx = 4f(x), such that f(x) is a function in terms of x. Find the value of e 1 0 f(x) dx. [3 marks] (b) Diagram 2 shows the curves y = x(x – 2)(x – 3) and y = x(x – 2). y x 0 y = x(x – 2)(x – 3) y = x(x – 2) Diagram 2 Calculate the area of the shaded region. [4 marks] 3 (a) Sketch the graph of y =|sin x| + 1 for 0 < x < 2π. [3 marks] (b) Hence, by sketching a suitable straight line on the same axes, determine the number of solutions that satisfy the equation π(|sin x| + 1) – 2π + x = 0 for 0 < x < 2π. [4 marks] 4 A Science Society committee that consists of 6 members are to be selected from 5 girls and 4 boys. (a) Calculate the number of different committees that can be formed if (i) there are no restrictions, [2 marks] (ii) the number of girls must be more than the number of boys. [4 marks] (b) Find the probability that the committee formed has more number of girls than boys. [1 mark] Ana&Tip(BM) SPM Add Maths(Eng)-MT 3rd.indd 263 16-Feb-23 7:34:24 PM PENERBIT ILMU BAKTI SDN. BHD.
264 Model Test SPM 5 In Diagram 3, X represents the number of long LED light bulbs which have a lifespan of less than nine months. P(X = x) x 0 1 2 3 k 9 64 1 64 0 Diagram 3 It is given that the probability of the lifespan of the long light bulbs which are less than nine months is p. A sample of 3 long light bulbs is chosen at random. HOTS Applying (a) Find the value of p. [3 marks] (b) Calculate the value of k. [2 marks] (c) Calculate the number of long light bulbs which are still functioning after nine months if 20 long light bulbs of the same brand are used. [2 marks] 6 (a) Calculate e 2 1 3x3 – 2x5 + 2 x3 dx. [3 marks] (b) Diagram 4 shows the curve 3y2 = x – 2, the straight lines y = –1 and y = 3. y 3y2 = x – 2 x 3 –1 0 2 Diagram 4 Find the area of the shaded region. [4 marks] 7 Given that y = x2 + 2 x2 , show that x2 d2 y dx2 + 4x dy dx + 2y – 12x2 = 0. [7 marks] Ana&Tip(BM) SPM Add Maths(Eng)-MT 3rd.indd 264 16-Feb-23 7:34:24 PM PENERBIT ILMU BAKTI SDN. BHD.
265 Model Test SPM Section B Answer any three questions from this section. [30 marks] 8 Use graph paper to answer this question. The variables x and y are related by the equation py = p2 x + q x, where p and q are constants. The corresponding values of x and y obtained from an experiment are shown in Table 1. x 2 4 6 8 10 11 y 12.5 17.5 25.0 32.0 40.0 41.0 Table 1 One of the values of y was incorrectly recorded. (a) Plot xy against x2 , using a scale of 2 cm to 20 units on the x2 -axis and 2 cm to 50 units on the xy-axis. Hence, draw the line of best fit. [5 marks] (b) Use the graph in 8(a) to answer the following questions. (i) Determine the value of y that is incorrectly recorded and state its actual value. (ii) Find the value of p and of q. [5 marks] 9 Diagram 5 shows a straight line DP which is a normal to a curve at point P(4, 12). y x D O P(4, 12) Diagram 5 The gradient function of the curve is – 1 2 x. (a) Find (i) the coordinates of point D, (ii) the equation of the curve. [5 marks] (b) If the volume generated when the region bounded by the straight line y = k, y-axis and the curve is revolved through 360° about the y-axis is 50π units3 , find the value of k. [5 marks] Ana&Tip(BM) SPM Add Maths(Eng)-MT 3rd.indd 265 16-Feb-23 7:34:24 PM PENERBIT ILMU BAKTI SDN. BHD.
266 Model Test SPM 10 Diagram 6 shows a sector AOB with centre O and a radius of 8 cm. PBCD is a semicircle with centre P and a radius of 3 cm. A B P D O C Diagram 6 It is given that the straight line OA is a tangent to the semicircle PBCD at point C. Calculate (a) ∠AOB, in radians, [2 marks] (b) the perimeter, in cm, of the shaded region, [4 marks] (c) the area, in cm2 , of the shaded region. [4 marks] [Use π = 3.142] 11 (a) In Taman Cempaka, termites were found in 3 out of 5 houses. If 8 houses are chosen at random, calculate the probability that (i) exactly 2 houses are infested with termites, (ii) more than 2 houses are infested with termites. [5 marks] (b) The masses of the students in a school are normally distributed with a mean of µ kg and a standard deviation of 12 kg. (i) A student is chosen at random from the school. The probability that the student has a mass of less than 45 kg is 0.2266. Find the value of μ. (ii) Hence, calculate the probability that a student chosen at random has a mass of between 42 kg and 45 kg. [5 marks] Ana&Tip(BM) SPM Add Maths(Eng)-MT 3rd.indd 266 16-Feb-23 7:34:24 PM PENERBIT ILMU BAKTI SDN. BHD.
268 Model Test SPM 14 Solution by scale drawing is not accepted. Diagram 7 shows a quadrilateral PQRS. P S R Q 40° 10.3 cm 15.1 cm 8.5 cm 11.7 cm Diagram 7 (a) Find (i) ∠PQR, (ii) ∠PRS, (iii) the area, in cm2 , of the quadrilateral PQRS. [8 marks] (b) Sketch ∆P'Q'R' that has a different shape from ∆PQR such that P'R' = PR, Q'R' = QR and ∠Q'P'R' = ∠QPR. Hence, state ∠P'Q'R'. [2 marks] 15 Use graph paper to answer this question. In a certain week, Beauty Boutique bought x blouses and y skirts based on the following constraints: I The total number of blouses and skirts bought cannot be more than 10. II The ratio of the number of skirts to the number of blouses bought cannot be more than 2 : 1. III The cost prices of a blouse and a skirt are RM60 and RM40 respectively. The allocated capital is RM480. (a) Write three inequalities, other than x > 0 and y > 0 that satisfy all the above constraints. [3 marks] (b) Using a scale of 2 cm to 1 unit on both axes, construct and shade the region R which satisfies all the above constraints. [3 marks] (c) The profits from the sale of a blouse and a skirt are RM12 and RM9 respectively. Based on the graph constructed in 15(b), answer the following questions: (i) If the boutique bought 4 skirts, find the minimum and maximum number of blouses bought. (ii) Find the maximum profit obtained if all the blouses and skirts purchased are sold. [4 marks] Ana&Tip(BM) SPM Add Maths(Eng)-MT 3rd.indd 268 16-Feb-23 7:34:24 PM PENERBIT ILMU BAKTI SDN. BHD.
269 Model Test SPM Paper 1 Section A 1 (a) f –1(x) = n – x m (b) n = 5; m = 4 2 k = –2; m = 4 3 –21 x – 1 2 2 2 – 5 2 , x = 1 2 4 x = 9 4 5 x = 2; y = 3 6 (a) (i) 13 units2 (ii) 4.111 units (b) 3x2 + 3y2 – 60x – 28y + 332 = 0 7 (a) k = 3. (b) 1 008 8 (a) h = 3 or h = –7 (b) h = 1 9 (a) θ = 0.6301 rad (b) 83.60 cm2 10 (a) p 1 – p2 ! (b) 1 – p 2 11 (a) 7 (b) h = 6 12 (a) (3, 1) (b) q = 7 Section B 13 RM120, RM30 and RM15 respectively. 14 (a) 2y = –9x + 15 (b) (i) R is the perpendicular bisector of MQ. (ii) 36y = 8x + 15 15 (a) (i) –2u ~ + 3v ~ (ii) – 1 2 u ~ + 9 4 v ~ (iii) 3 2 u ~ + 9 4 v ~ (b) (i) 3hv ~ – 2hu ~ (ii) 3 2 ku ~ + 9 4 kv ~ (c) h = 1 2 ; k = 1 3 Paper 2 Section A 1 (a) 34 cm (b) 190 cm (c) 20 seconds 2 (a) – 1 8 (b) 4 units2 3 (a), (b) y = |sin x| + 1 y = 2 – x p y x 2 1 0 –1 Number of solutions = 2 4 (a) (i) 84 (ii) 34 (b) 17 42 5 (a) p = 3 4 (b) 27 64 (c) 5 long light bulbs 6 (a) – 11 12 (b) 36 units3 7 y = x2 + 2 x2 = x2 + 2x–2 dy dx = 2x – 4x–3 = 2x – 4 x3 d2 y dx2 = 2 + 12x–4 = 2 + 12 x4 x2 d2 y dx2 + 4x dy dx + 2y – 12x2 = x2 12 + 12 x4 2 + 4x12x – 4 x3 2 + 21x2 + 2 x2 2 – 12x2 = 2x2 + 12 x2 + 8x2 – 16 x2 + 2x2 + 4 x2 – 12x2 = 0 [Shown] Answers Ana&Tip(BM) SPM Add Maths(Eng)-MT 3rd.indd 269 16-Feb-23 7:34:24 PM PENERBIT ILMU BAKTI SDN. BHD.
270 Model Test SPM Section B 8 (a) x 2 4 6 8 10 11 y 12.5 17.5 25.0 32.0 40.0 41.0 xy 25.0 70.0 150.0 256.0 400.0 451.0 x2 4 16 36 64 100 121 20 40 60 80 100 120 140 500 450 400 350 300 250 200 150 100 50 10 0 400 – 10 = 390 100 x2 xy Incorrect Actual (b) (i) The incorrect value of y recorded is 41.0. The actual value of xy is 480. y = 43.6 (actual value) (ii) p = 3.9; q = 39 9 (a) (i) (0, 10). (ii) y = – 1 4 x2 + 16 (b) k = 11 10 (a) 0.6436 rad (b) 15.79 cm (c) 4.630 cm2 11 (a) (i) 0.04129 (ii) 0.9502 (b) (i) μ = 54 (ii) 0.0679 Section C 12 (a) (i) The initial velocity of the particle is 20 m s–1. The initial acceleration of the particle is –12 m s–2. (ii) 2 , t , 10. (iii) 0 , t , 6. (b) 104 m 13 (a) x = 5.40; y = 96; z = 4.80 (b) 124.26 (c) RM2.58 14 (a) (i) 56.06° (ii) 40.85° (iii) 129.8 cm2 (b) P' Q' Q R' 40° 56.06° 56.06° 15.1 cm 11.7 cm ∠P’Q’R’ = 123.94° 15 (a) I x + y , 10 II y , 2x III 3x + 2y , 24 (b) 1 2 3 4 5 6 7 8 12 11 10 9 8 7 6 5 4 3 2 1 0 4.5 R x y Maximum (4, 6) y = 2x 3x + 2y = 24 12x + 9y = 54 x + y = 10 (c) (i) xminimum = 2 xmaximum = 5 (ii) RM102 Ana&Tip(BM) SPM Add Maths(Eng)-MT 3rd.indd 270 16-Feb-23 7:34:25 PM PENERBIT ILMU BAKTI SDN. BHD.
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