Spotlight A+1 Mathematics Form 3

ii Learning Area: Numbers and Operations Chapter 1 Indices 1.1 Index Notation .............................................. 1 1.2 Law of Indices .............................................. 2 Mastery Challenge 1 ............................................. 6 Chapter 2 Standard Form 2.1 Significant Figures ....................................... 9 2.2 Standard Form ........................................... 11 Mastery Challenge 2 ........................................... 13 Chapter 3 Consumer Mathematics: Savings and Investments, Credit and Debt 3.1 Savings and Investments ........................... 17 3.2 Credit and Debt Management ................... 25 Mastery Challenge 3 ........................................... 30 Learning Area: Measurement and Geometry Chapter 4 Scale Drawings 4.1 Scale Drawings .......................................... 37 Mastery Challenge 4 ........................................... 44 Chapter 5 Trigonometric Ratios 5.1 Sine, Cosine and Tangent of Acute Angles in Right-angled Triangles ............................... 49 Mastery Challenge 5 ........................................... 56 Chapter 6 Angles and Tangents of Circles 6.1 Angle at the Circumference and Central Angle Subtended by an Arc ................................. 61 6.2 Cyclic Quadrilaterals .................................. 64 6.3 Tangents to Circles .................................... 67 6.4 Angles and Tangents of Circle ................... 71 Mastery Challenge 6 ........................................... 72 Chapter 7 Plans and Elevations 7.1 Orthogonal Projections .............................. 80 7.2 Plans and Elevations ................................. 83 Mastery Challenge 7 ........................................... 90 Chapter 8 Loci in Two Dimensions 8.1 Loci ............................................................. 97 8.2 Loci in Two Dimensions ............................. 98 Mastery Challenge 8 ......................................... 104 Learning Area: Relationship and Algebra Chapter 9 Straight Lines 9.1 Straight Lines ........................................... 110 Mastery Challenge 9 ......................................... 120 Pentaksiran Sumatif Ujian Akhir Sesi Akademik (UASA)......................................................................................125 Answers ..................................................................................137 CONTENTS Contents Spot A+1 Maths Form3.indd 2 08/05/2023 12:58 PM

3 17 3.1.1 Recognise various types of savings and investments Deposits and investments Types of savings Types of investments Savings account Shares Fixed deposit account Unit trust Current account Property Example 1 Complete each of the following by the word “saving” or “investment”. (a) Share is a type of . (b) National Education Saving scheme is a type of . (c) Residential house is a type of . (d) BSN Premium Saving certificate is a type of . Solution: (a) investment (b) saving (c) investment (d) saving ➡ Diagnostic Test 3.1: Question 1 3.1.2 Perform calculations involving simple interest and compound interest for savings, and hence explain the impact of changes in period, rate of interest or return and compounding frequency on the future value of savings 1. Deposit that gives simple interest is found by the formula I = Prt. 3.1 Savings and Investments I = interest P = principal r = annual interest rate t = time in year Example 2 (a) Find the simple interest for a deposit of RM4000 with an interest rate of 5% per annum based on the following terms. (i) 1 year (ii) 4 years (iii) 6 years (b) Hence, explain the effect of change of term on the future deposit value. Solution: (a) (i) Interest, I = 4 000 × 0.05 × 1 = RM200 (ii) Interest, I = 4 000 × 0.05 × 4 = RM800 (iii) Interest, I = 4 000 × 0.05 × 6 = RM1 200 (b) RM(4 000 + 200) = RM4 200 RM(4 000 + 800) = RM4 800 RM(4 000 + 1 200) = RM5 200 Future deposit value Term of 1 year Term of 4 years Term of 6 years The future deposit value increases when the term of deposit increases. TIPS Corner In the calculations of simple interest, the value of r has to be changed from percentage to decimal. ➡ Diagnostic Test 3.1: Question 2 Consumer Mathematics: Savings and 3 Investments, Credit and Debt CHAPTER C03 Spotlight A+1 Maths Form3.indd 17 10/05/2023 5:19 PM

Learning Area: Numbers and Operations 3 22 Example 12 The following table shows the number of units and price per unit of VCT unit trust invested by Jui Min. Number of units Price per unit (sen) 20 000 63 70 000 72 50 000 58 60 000 66 (a) Determine the average cost per unit of the VCT unit trust bought. (b) Jui Min sold all the units of the unit trust at the price of 69 sen per unit. Find the profit obtained. Solution: (a) Total investment of VCT unit trust = 0.63 × 20 000 + 0.72 × 70 000 + 0.58 × 50 000 + 0.66 × 60 000 = RM131 600 Total number of units of the unit trust = 20 000 + 70 000 + 50 000 + 60 000 = 200 000 Average cost per unit of the unit trust = RM131 600 200 000 = RM0.658 (b) Profit = (0.69 – 0.658) × 200 000 = RM6 400 ➡ Diagnostic Test 3.1: Question 12 3.1.6 Solve problems involving savings and investments Example 13 A sum of money of RM80 000 was invested for a term of 6 years 10 months. The investment was offered an interest rate of 7% per annum compounded half-yearly for the first 2 years, 10% per annum compounded quarterly for the next 1 1 2 years and 6% per annum compounded every 4 months for the remaining duration. Calculate the matured value of the investment. Solution: For the first 2 years, r = 0.07, n = 2 and t = 2. The matured value at the end of the first 2 years = RM80 000 1 + 0.07 2  2 × 2 = RM91 801.84 For the next 1 1 2 years, r = 0.1, n = 4 and t = 1 1 2 . The matured value at the end of the first 3 1 2 years = RM91 801.84 1 + 0.1 4  4 × 3 2 = RM106 461.99 For the next 3 1 3 years, r = 0.06, n = 3 and t = 3 1 3 . The matured value at the end of the first 6 5 6 years = RM106 461.991 + 0.06 3  3 × 10 3 = RM129 776.57 ➡ Diagnostic Test 3.1: Question 13 Learn it right The value P = 80 000 cannot be used to calculate the matured value at the end of 3 1 2 years and 6 5 6 years. The matured value at the end of the first 3 1 2 years = RM80 000 1 + 0.1 4  4 × 3 2 = RM92 775.47 ✗ The matured value at the end of the first 6 5 6 years = RM80 000 1 + 0.06 3  3 × 10 3 = RM97 519.55 ✗ Example 14 Kok Liang invested in BCS shares. He bought 3 000 units of the shares at a price of RM2.12 per unit, 2 000 units of the shares at the price of RM1.85 per unit and 5 000 units of the shares at the price of RMx per unit over a period of time. The average cost per share unit for the BCS shares bought is RM1.876. (a) Determine the value of x. (b) Kok Liang intended to sell part of the BCS shares at the price of RM2.65. How many units of BCS shares need to be sold to obtain a profit of RM5 418? C03 Spotlight A+1 Maths Form3.indd 22 10/05/2023 5:19 PM

5 49 5.1.1 Identify the opposite side and adjacent side based on an acute angle in a right-angled triangle C A B θ Hypotenuse Opposite side Adjacent side 1. Opposite side is the side that lies opposite to the acute angle θ in a right-angled triangle. 2. Adjacent side is the side that lies adjacent to the acute angle θ in a right-angled triangle. Example 1 R Q P θ Based on the angle θ marked on the right-angled triangle PQR in the diagram above, name (a) the opposite side, (b) the adjacent side. Solution: (a) The opposite side to the angle θ is QR. (b) The adjacent side to the angle θ is PR. ➡ Diagnostic Test 5.1: Question 1 5.1.2 Make and verify the conjecture about the relationship between acute angles and the ratios of the sides of right-angled triangles, and hence define sine, cosine and tangent C A B θ 1. sin θ = opposite side hypotenuse = BC AC 2. cos θ = adjacent side hypotenuse = AB AC 3. tan θ = opposite side adjacent side = BC AB TIPS Corner sin θ cos θ tan θ sine of θ cosine of θ tangent of θ Abbreviation Example 2 R Q P S β α Based on the diagram above, determine (a) (i) sin α, (ii) cos α, (iii) tan α, (b) (i) sin β, (ii) cos β, (iii) tan β. 5.1 Sine, Cosine and Tangent of Acute Angles in Right-angled Triangles 5 Trigonometric Ratios CHAPTER C05 Spotlight A+1 Maths Form3.indd 49 10/05/2023 2:57 PM

CHAPTER 5 Trigonometric Ratios 5 51 5.1.4 Determine the values of sine, cosine and tangent of acute angles Example 4 16 cm 12 cm B A C θ The diagram above shows a right-angled triangle, ABC. (a) Calculate the length of AB correct to four significant figures. (b) Determine the values of sin θ, cos θ and tan θ. Solution: (a) AB2 = 162 – 122 = 112 AB =  112 = 10.58 cm (b) sin θ = 12 16 = 0.75 cos θ = 10.58 16 = 0.6613 tan θ = 12 10.58 = 1.134 ➡ Diagnostic Test 5.1: Question 4 5.1.5 Determine the values of sine, cosine and tangent of angles 30°, 45° and 60° without using a calculator Example 5 2 cm 2 cm 1 cm 1 cm 60° A D B C 30° In the diagram above, ABC is an equilateral triangle. D is the midpoint of AB. (a) Find the length of CD, in the form a. (b) Hence, find the value of (i) sin 60°, (ii) tan 30°, (iii) cos 60°. Solution: (a) CD2 = AC2 – AD2 = 22 – 12 = 3 CD = 3 cm (b) (i) sin 60° = CD AC (ii) tan 30° = AD CD = 3 2 = 1 3 (iii) cos 60°= AD AC = 1 2 ➡ Diagnostic Test 5.1: Question 5 TIPS Corner The values of sine, cosine and tangent for the angles 30°, 45° and 60° can be obtained from the following two right-angled triangles. 2 45° 60° √ — 2 3 1 1 1 √ — 30° 45° θ sin θ cos θ tan θ 30° 1 2 3 2 1 3 45° 1 2 1 2 1 60° 3 2 1 2 3 5.1.6 Perform calculations involving sine, cosine and tangent Example 6 The following diagram shows a right-angled triangle, UVW, with sin θ = 2 3 . 36 cm U V W θ (a) Determine the length of UW. (b) Find the value of tan θ correct to four significant figures. Solution: (a) sin θ = UW VW 2 3 = UW 36 UW = 2 3 × 36 = 24 cm TIPS Corner sin θ cos θ = 0.75 0.6613 = 1.134 ∴ tan θ = sin θ cos θ C05 Spotlight A+1 Maths Form3.indd 51 10/05/2023 2:57 PM

Learning Area: Measurement and Geometry 5 52 (b) UV2 = 362 – 242 = 720 UV =  720 = 26.83 cm tan θ = UW UV = 24 26.83 = 0.8945 ➡ Diagnostic Test 5.1: Question 6 Example 7 Given sin θ = 2 3 , determine the values of cos θ and tan θ. Give the answers in surd form. Solution: AB2 = 32 – 22 = 5 AB = 5 cos θ = AB AC = 5 3 tan θ = BC AB = 2 5 ➡ Diagnostic Test 5.1: Question 7 Example 8 Use a calculator to find the values, correct to four significant figures, for each of the following. (a) sin 35° (b) tan 72.6° (c) cos 48° 23ʹ Solution: (a) sin 35° = 0.5736 (b) tan 72.6° = 3.191 (c) cos 48° 23ʹ = 0.6641 Calculator Press (a) sin 3 5 °ʹ ʺ = (b) tan 7 2 . 6 °ʹ ʺ = (c) cos 4 8 °ʹ ʺ 2 3 °ʹ ʺ = ➡ Diagnostic Test 5.1: Question 8 TIPS Corner sin θ = BC AC = 2 3 Label BC = 2 and AC = 3 Example 9 Find the acute angle θ for each of the following. (a) sin θ = 0.78 (b) cos θ = 0.2713 (c) tan θ = 9 5 Solution: (a) sin θ = 0.78 (b) cos θ = 0.2713 θ = sin–1 0.78 θ = cos–1 0.2713 = 51° 16ʹ = 74° 16ʹ (c) tan θ = 9 5 θ = tan–1  9 5  = 60° 57ʹ Calculator Press (a) SHIFT sin 0 . 7 8 = °ʹ ʺ (b) SHIFT cos 0 . 2 7 1 3 = °ʹ ʺ (c) SHIFT tan 9 5 ➤ = °ʹ ʺ ➡ Diagnostic Test 5.1: Question 9 TIPS Corner θ = sin–1  a b  θ = cos–1  a b  θ = tan–1  a b  Find angle sin θ = a b cos θ = a b tan θ = a b Example 10 Determine the values of θ for each of the following. (a) (b) 15 cm 8 cm θ 1.3 m 3.6 m θ (c) 102 mm 79 mm θ 2 3 C A B θ C05 Spotlight A+1 Maths Form3.indd 52 10/05/2023 2:57 PM

6 61 6.1.1 Make and verify conjectures about the relationships between (i) angles at the circumference, (ii) angles at the circumference and central angle subtended by particular arcs, and hence use the relationships to determine the values of angles in circles 1. The angle at circumference subtended by an arc AB of a circle is the angle formed by the line joining points A and B to point C at the circumference. 2. The angle at the centre of a circle subtended by an arc AB is the angle formed by the radii OA and OB. 3. The angles at the circumference subtended by the same arc are equal. Example 1 Determine the value of x in each of the following diagrams. (a) 24° x (b) 106° x Solution: (a) x = 24° (b) x = 106° ➡ Diagnostic Test 6.1: Question 1 4. The angles subtended at the circumference by arcs of the same length are equal. 5. The angles subtended at the centre of a circle by arcs of the same length are equal. Example 2 In the following diagram, O is the centre of the circle PQR. Find the values of x and y. (a) 25° x C B A E D (b) 133° y R P Q O Solution: (a) ∠ACB = ∠DAE x = 25° (b) ∠POR = ∠QOR y = 133° ➡ Diagnostic Test 6.1: Question 2 6. The angle at the centre of a circle is twice the angle at the circumference subtended by the same arc. O x 2x O y 2y 6.1 Angle at the Circumference and Central Angle Subtended by an Arc C A B O A B x x x x x x x O 6 Angles and Tangents of Circles CHAPTER C06 Spotlight A+1 Maths Form3.indd 61 10/05/2023 2:58 PM

CHAPTER 6 Angles and Tangents of Circles 6 69 3. The angle between tangent and chord is equal to the angle in the alternate segment subtended by the chord. C B E A F TIPS Corner (a) The angles between tangent EF and chords AC and AB are ∠CAE and ∠BAF respectively. (b) The angle in the alternate segment subtended by the chords AC and AB are ∠ABC and ∠ACB respectively. (a) ∠CAE = ∠ABC (b) ∠BAF = ∠ACB Example 17 In the following diagram, MN is the tangent to the circle. Determine the values of x and y. 75° 3y 32° C A N M x B Solution: ∠BAM = ∠ACB ∠CAN = ∠ABC x = 32° 75° = 3y y = 25° ➡ Diagnostic Test 6.3: Question 4 6.3.3 Solve problems involving tangents to circles Example 18 In the following diagram, LM is the tangent to the circle at point P. P L M Q R S x y 30° 50° Given PR = SR, find the values of x and y. Solution: ∠PRS = ∠MPS = 50° ∠PSR = 1 2 × (180° – 50°) = 65° ∠LPR = ∠PSR x + 30° = 65° x = 65° – 30° = 35° ∠PQR + ∠PSR = 180° y + 65° = 180° y = 180° – 65° = 115° Alternative Method ∠PRQ = x x + y + 30° = 180° 35° + y + 30° = 180° y = 180° – 35° – 30° = 115° ➡ Diagnostic Test 6.3: Question 5 Example 19 In the following diagram, A and B are the centres of two circles with radii 15 cm and 5 cm respectively. CD is the common tangent to the two circles touching at point H. θ A H B C D E Determine (a) the length of CD, (b) the angle θ, (c) the area of the shaded region. Solution: (a) AB = 15 + 5 = 20 cm AE = 15 – 5 = 10 cm BE 2 = 202 – 102 = 300 BE =  300 = 17.32 cm ∴CD = 17.32 cm TIPS Corner The common tangent to two circles is the straight line touching both circles at only one point. C06 Spotlight A+1 Maths Form3.indd 69 10/05/2023 2:59 PM

Learning Area: Measurement and Geometry 6 70 (b) cos θ = AE AB = 10 20 = 1 2 ∴ θ = 60° (c) Area of trapezium ABCD = 1 2 × (15 + 5) × 17.32 = 173.2 cm2 Area of sector ADH = 60° 360° × 22 7 × 152 = 117.86 cm2 ∠ABC = 180° – 60° = 120° Area of sector BCH = 120° 360° × 22 7 × 52 = 26.19 cm2 Area of the shaded region = 173.2 – 117.86 – 26.19 = 29.15 cm2 ➡ Diagnostic Test 6.3: Question 6 TIPS Corner Area of sector OAB = θ 360° × πr 2 θ r r B O A Diagnostic Test 6.3 1. The following diagram shows a circle and four straight lines, AB, CD, EF and GH. H D F B C E A G Identify the straight lines that are tangents to the circle. 2. In the following diagram, PQ is the tangent to a circle with centre O and radius 8 cm at point R. OSP is a straight line. P 12 cm R Q O S Calculate ∠OPR. 3. In the following diagram, HK and HL are tangents to the circle with centre O. 40° (x + 8) cm (3x – 4) cm H L K y O Determine the values of x and y. 4. In the following diagram, PQ is the tangent to the circle. x 21° Q L P M N y 4y 43° Find the values of x and y. 5. In the following diagram, HK is the tangent to a circle at point P. 63° S T H P K Q R y x 60° 116° 40° Determine the values of x and y. C06 Spotlight A+1 Maths Form3.indd 70 10/05/2023 2:59 PM

9 110 9.1.1 Make connection between the equation, y = mx + c, and the gradient and y-intercept, and hence make generalisation about the equation of a straight line 1. Equation y = mx + c with gradient, m and y-intercept, c is the equation of a straight line. TIPS Corner Equation of straight line y = mx + c Gradient y-intercept Example 1 In the following diagram, A(–2, –4) and B(3, 11) are two points that lie on the graph y = 3x + 2. x B (3, 11) A (–2, –4) –2 O 2 4 –2 2 4 6 8 10 –4 y (a) Determine the gradient and y-intercept of the straight line AB. (b) State the connection between the equation y = 3x + 2 and the gradient and y-intercept of the straight line AB. Solution: (a) Gradient of straight line AB = 11 – (–4) 3 – (–2) = 15 5 = 3 The y-intercept of the straight line AB = 2 (b) The gradient of the straight line AB is the coefficient of x in the equation y = 3x + 2. The y-intercept of the straight line AB is the constant term in the equation y = 3x + 2. ➡ Diagnostic Test 9.1: Question 1 2. Parallel to the x-axis with y-intercept, k Parallel to the y-axis with x-intercept, h y = k x = h Equation of straight line TIPS Corner x x = h y = k O y k h Example 2 x A B D C –2 O 2 4 –2 2 4 y –4 Based on the above diagram, write down the equation of the straight line (a) AB, (b) BC, (c) CD, (d) AD. 9.1 Straight Lines 9 Straight Lines CHAPTER C09 Spotlight A+1 Maths Form3.indd 110 08/05/2023 5:26 PM

Learning Area: Relationship and Algebra 9 120 MASTERY CHALLENGE 9 19. In the following diagram, HK is parallel to the x-axis and the area of triangle HKL is 25 units2 . x O L (4, 0) H (–3, 5) K y Find (a) the coordinates of point K, (b) the equation of the straight line KL in the form ax + by = c. 20. The following diagram shows a parallelogram PQRS. x O R (2, 2) S (0, –2) P (–3, 1) Q y Determine (a) the equation of the straight line QR, (b) the x-intercept of PQ, (c) the coordinates of point Q. 1. Which of the following equations has gradient 2 3 ? A 3x – 2y = 10 B 3y = 2x + 12 C x 6 + y 4 = 1 D x 4 – y 6 = 1 2. Determine the equation of the straight line that does not have y-intercept = –4. A y = x – 4 B x + y + 4 = 0 C x – y 4 = 1 D – x 2 + y 4 = 1 3. Find the y-intercept of the straight line –5x – 3y = 2. A – 2 3 B – 3 2 C – 5 2 D – 2 5 4. The straight line x 6 + y k = 1 passes through the point (–6, 8). Find the value of k. A –4 B –2 C 2 D 4 5. Which of the following points does not lie on the straight line –7x + 2y = 14? A (–2, 0) B (2, 14) C (–1, 7) D (–4, –7) 6. Determine the equation of the straight line that is parallel to the straight line x 4 – y 8 = 1. A y = –2x + 9 B 1 2 x + y = 4 C x + y 2 = 1 D x 2 – y 4 = 1 7. Find the equation of the straight line that passes through the points (1, 15) and (–3, –5). A y = – 5 2 x – 25 2 B y = 5x + 10 C y = –5x + 20 D y = –10x – 35 C09 Spotlight A+1 Maths Form3.indd 120 08/05/2023 5:26 PM

CHAPTER 9 Straight Lines 9 121 8. The following diagram shows a straight line drawn on a Cartesian plane. x O 8 10 y Determine the equation of the straight line. A y = – 5 4 x + 8 B 5y = –4x + 40 C 4x + 5y = 8 D x 10 – y 8 = 1 9. Determine the point of intersection of the two straight lines 5x – 2y = 16 and 3x + y = 3. A (2, 9) B (3, 9) C (2, –3) D (3, –3) 10. In the following diagram, PQ and MN are two straight lines drawn on a Cartesian plane. x O M Q (4, 5) P (–2, 1) N (8, –1) y M is the midpoint of PQ. Find the x-intercept of MN. A 27 8 B 25 4 C 10 3 D 50 7 11. Fill in each of the following empty boxes with the correct answer. Equation of straight line y = mx + c (a) and are variables (b) m = = coefficient of (c) c = -intercept = (d) Coefficient of = 1 (e) Power of x = Power of y = 12. Complete the following table. Equation of straight line Gradient y-intercept (a) y = 4x (b) y = x – 8 (c) y = –2x + 5 (d) y = 14 13. Match the two forms of equation of straight line representing the same straight line. (a) 3x – 2y = 12 • • (i) y = – 3 2 x + 3 (b) x 2 + y 3 = 1 • • (ii) y = 2 3 x – 2 (c) 2x + 3y + 12 = 0 • • (iii) y = 3 2 x – 6 (d) x 3 – y 2 = 1 • • (iv) y = – 2 3 x – 4 C09 Spotlight A+1 Maths Form3.indd 121 08/05/2023 5:26 PM

Learning Area: Relationship and Algebra 9 122 14. Complete the following diagram. (b) y = x – (c) y = –3 Gradient = y-intercept = 10 Gradient = – 1 2 y-intercept = –4 Gradient = y-intercept = Equation of straight line y = mx + c (a) y = 3x + 15. Mark “✓” or “✗” for the equation of straight line that is equivalent to 5x + 2y = 20. (a) y = – 5 2 x + 20 (b) x 4 + y 10 = 1 16. Complete the following table. Form y = mx + c Form ax + by = c Form x a + y b = 1 (a) y = x – 2 (b) 3x – y = –9 (c) – x 12 – y 4 = 1 17. Circle the correct gradient and y-intercept for each of the following straight lines. (a) 4x – 7y – 14 = 0 (i) Gradient 4 7 – 4 7 (ii) y-intercept –14 –2 (b) x 9 + y 6 = 1 (i) Gradient – 3 2 – 2 3 (ii) y-intercept 6 9 18. Mark “✓” or “✗” for each of the following. (a) Point (3, 7) lies on the straight line y = 5x – 8. (b) Point (–2, 1) lies on the straight line 3x – 6y = 10. (c) Point (8, –2) lies on the straight line x 4 + y 2 = 1. 19. (2, 4) (–4, 1) (1, 3) (–1, 2) (–5, 3) (6, 5) Based on the above list of points, complete the following diagram. (a) Points that lie on the straight line (b) Points that do not lie on the straight line Straight line –2x + 5y = 13 20. Circle the correct answer. (a) The straight line y = 2x + c passes through the point (–3, 1). c = –5 7 (b) The point (–5, p) lies on the straight line x 5 – y 2 = 1. p = –4 –2 21. Mark “✓” or “✗” for each of the following. (a) Point (2, 4) does not lie on the straight line kx + 3y = 4. k ≠ –4 (b) Point (6, k) does not lie on the straight line x 3 – y 6 = 1. k ≠ 1 C09 Spotlight A+1 Maths Form3.indd 122 08/05/2023 5:26 PM

PENTAKSIRAN SUMATIF 125 PENTAKSIRAN SUMATIF UJIAN AKHIR SESI AKADEMIK (UASA) 1. 125 2 3 = A 5 B 15 C 25 D 30 2. (pmr–3) × (2p–2rn ) = 2p3 r –5 Find the value of mn. A –10 B –3 C 3 D 10 3. State 4 574 000 in the standard form correct to two significant figures. A 4.5 × 106 B 4.6 × 106 C 4.5 × 108 D 4.6 × 108 4. 3 × 107 × 7.7 × 1010 = A 2.31 × 1016 B 2.31 × 1017 C 2.31 × 1018 D 2.31 × 1019 5. Which of the following gives the lowest simple interest on a total savings of RM5 000 with the given interest rate and term? Interest rate Term A 2% per annum 3 years B 1.5% per annum 30 months C 1% per month 15 months D 0.6% per month 2 years 6. Winson invested in Boulex share since 2022. The following table shows the price and number of shares bought. Price per share (RM) Number of shares 1.25 1 000 1.40 2 000 0.96 2 000 0.75 5 000 Calculate the average cost per share for the Boulex shares bought by Winson. A RM0.843 B RM0.945 C RM0.972 D RM1.08 SECTION A [20 marks] Answer all questions. 7. The East Coast Rail Line project with a distance of 665 km will be built to connect the states in the East Coast and Klang Valley. If the scale of 1 : 2 000 000 is used on the map to draw the rail line, find the length of the rail line on the map. A 33.25 cm B 35.45 cm C 38.15 cm D 39.75 cm 8. The diagram shows the drawing of a space rocket with a height of 25 cm. Given the actual height of the space rocket is 100 m, find the scale used in the scale drawing. A 1 : 3 B 1 : 4 C 1 : 300 D 1 : 400 9. The diagram shows straight lines FS and HP which intersect at point M. 9 cm 15 cm M P S H F 6 cm Calculate the length, in cm, of MS. A 6 B 8 C 10 D 12 P.Sumatif Spot A+1 Maths Form3.indd 125 08/05/2023 5:31 PM

Pentaksiran Sumatif PENTAKSIRAN SUMATIF 126 10. The diagram shows two right-angled triangles, HLM and LMP. 9 cm M y x H L P 3 cm Given tan x = 3 4 , calculate the value of sin y. A 1 3 B 4 9 C 5 9 D 1 3 11. In the diagram, HST and PMS are straight lines. M is the midpoint of PS. M x H T 4 cm S 8 cm P Given cos x = 4 5 , find the length, in cm, of PT. A 73 B 10 C 12 D 14 12. In the diagram, HPT and SQT are tangents to the circle PQR. PRS is a straight line. x T S 56° 100° P H Q R Find the value of x. A 24° B 28° C 32° D 50° 13. The diagram shows a circle PQRS with centre O. x S 110° 88° P Q O R Find the value of x. A 22° B 24° C 35° D 44° 14. The diagram shows a circle HKLM. PHK and QKLR are straight lines. 130° 60° P Q K L R y x M H Determine the value of x + y. A 170° B 190° C 210° D 250° 15. The diagram shows a solid right prism that lies on a horizontal plane. 5 cm 4 cm 2 cm 6 cm Which of the following is the plan of the solid? A 6 cm 2 cm 3 cm P.Sumatif Spot A+1 Maths Form3.indd 126 08/05/2023 5:31 PM

Pentaksiran Sumatif PENTAKSIRAN SUMATIF 128 SECTION B [20 marks] Answer all questions. 21. On the diagram in the answer space, write four index numbers in the form (10m)n , such that m and n are positive integers not equal to 1 to represent 1028. 1028 [4 marks] 22. Given a = 4 × 108 , mark “✓” for the correct value or “✗” for the wrong value. (a) 3a = 1.2 × 109 (b) 1 a = 2.5 × 10–7 (c) a2 = 1.6 × 1018 (d) a = 2 × 104 [4 marks] 23. Match the simple interest with the correct savings. Savings Simple interest (a) RM2 000 at an interest rate of 4% per annum for a term of 1 year. • • RM56.25 (b) RM800 at an interest rate of 7% per annum for a term of 3 years. • • RM80 (c) RM1 500 at an interest rate of 2.5% per annum for a term of 1 1 2 years. • • RM168 (d) RM2 400 at an interest rate of 1% per month for a term of 6 months. • • RM144 [4 marks] P.Sumatif Spot A+1 Maths Form3.indd 128 08/05/2023 5:31 PM

Pentaksiran Sumatif PENTAKSIRAN SUMATIF 130 SECTION C [60 marks] Answer all questions. 26. (a) Rani invested RM40 000 on a joint venture scheme offering an interest rate of 9% per annum for a term of 4 years. The interest is compounded every four months. Calculate the matured value obtained. [2 marks] Answer: (b) Jamal wanted to buy a car at a price of RM100 000. He managed to settle 5% of the car price as down payment and the balance amount is to be paid through a bank loan. A bank has approved his loan at the rate of 2.4% per annum for a term of 8 years. Determine the total loan required to be repaid. [4 marks] Answer: (c) Soo Teng bought a condominium with his own savings and its balance amount is obtained from a bank loan. He rented the condominium and later on sold it when the price is reasonable. Buying price: RM500 000 Down payment: RM50 000 Legal fee charge: RM12 000 Stamp duty paid: RM10 000 Total monthly instalment settled: RM60 000 Balance of bank loan required to settle: RM360 000 Total amount of rental received: RM28 800 Selling price: RM620 000 Based on the information above, calculate the return on investment obtained by Soo Teng. [4 marks] Answer: P.Sumatif Spot A+1 Maths Form3.indd 130 08/05/2023 5:31 PM

137 ANSWERS Chapter 1 Diagnostic Test 1.1 1. (a) 36 (b) 28 (c)  2 5  5 (d) (–6)4 2. (a) 2 × 2 × 2 × 2 (b) 4.8 × 4.8 × 4.8 (c) 1 4 × 1 4 × 1 4 × 1 4 × 1 4 (d) (–3) × (–3) × (–3) × (–3) × (–3) × (–3) 3. (a) 25 (b) 33 (c) 64 (d) 104 4. (a) 1 000 (b) 64 (c) 15 5 8 (d) 7 58 81 Diagnostic Test 1.2 1. (a) (i) 24 × 25 = (2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2) = 29 (ii) 24 + 5 = 29 (iii) 24 × 25 = 24 + 5 (b) 2m × 2n = 2m + n 2. (a) (i) 35 ÷ 33 = 3 × 3 × 3 × 3 × 3 3 × 3 × 3 = 32 (ii) 35 – 3 = 32 (iii) 35 ÷ 33 = 35 – 3 (b) 3m ÷ 3n = 3m – n 3. (a) (i) (53 )4 = 53 × 53 × 53 × 53 = 53 + 3 + 3 + 3 = 512 (ii) 53 x 4 = 512 (iii) (53 )4 = 53 × 4 (b) (5m)n = 5m × n 5. (a) 4 2 (b) 3 5 (c) 5 10 (d) 9  313 6. (a) 4 1 5 (b) 21 1 8 (c) 209 1 12 (d) 424 1 18 7. (a) 10 4 7 = (104 ) 1 7 = (10 1 7 ) 4 (b) 26 5 9 = 9  265 = (9 26)5 8. (a) 514 (b) 212 (c) 317 (d) 621 9. (a) 27 × 32 (b) 56 × 75 (c) 67 × 1112 (d) 316 × 109 10. (a) 210 (b) 316 11. (a) 4 27 (b) 625 (c) 27 32 (d) 3 4 12. (a) 4 (b) 125 (c) 81 (d) 3 7 Mastery Challenge 1 1. A 2. C 3. B 4. B 5. D 6. B 7. D 8. D 9. B 10. B 11. (a) ✓ (b) ✗ (c) ✓ (d) ✗ 12. (a) 73 ; 3; 7 (b) 3.24 ; 4; 3.2 (c) (–6)5 ; 5; –6 (d)  1 2  6 ; 6; 1 2 13. (a) 3 243 (b) 243 = 35 3 81 3 27 3 9 3 3 1 14. 84 , 46 , 212 15. (a) 512 (b) 125 (c) 729 (d) 625 16. (a) 29 (b) 38 (c) 521 (d) 102 17. (a) 2p + 4 × 23p + 2 = 2(p + 4) + (3p + 2) = 24p + 6 (b) 52q + 1 ÷ 53 – q = 5(2q + 1) – (3 – q) = 53q – 2 (c) (113 )4r – 3 = 113 × (4r – 3) = 1112r – 9 18. (a) ✗ (b) ✓ (c) ✗ (d) ✓ 19. (a) 5 3 (b) 7 1 4 (c) 8 10 20. (a) ✗ (b) ✓ (c) ✓ (d) ✗ 21. (a) 35 × 27n = 314 (b) 163 × 26 ÷ 2n = 28 35 × (33 )n = 314 (24 )3 × 26 ÷ 2n = 28 35 × 33n = 314 212 × 26 ÷ 2n = 28 35 + 3n = 314 212 + 6 – n = 28 5 + 3n = 14 12 + 6 – n = 8 3n = 9 18 – n = 8 n = 3 n = 10 22. (a) 11 (b) 5 (c) 3 (d) 4 23. (a) 16 (b) 8 9 24. (a) (22 ) 1 5 , (2 1 5 ) 2 , 5 22 , (5 2)2 (b) (34 ) 1 7 , (3 1 7 ) 4 , 7 34 , (7 3)4 (c) (53 ) 1 8 , (5 1 8 ) 3 , 8 53 , (8 5)3 (d) (75 ) 1 9 , (7 1 9 ) 5 , 9 75 , (9 7)5 25. (a) 6 (b) 64 (c) 16 (d) 32 26. (a) 3n + 11 (b) n = 7 27. m = 2 28. 36 29. (a) 8 125 (b) 216 30. 1 Chapter 2 Diagnostic Test 2.1 1. (a) 4 (b) 1 (c) 5 (d) 2 (e) 5 (f) 3 (g) 3 (h) 6 2. (a) 1.0 (b) 3 620 (c) 2.48 (d) 0.3 (e) 5 002 400 (f) 8 000 (g) 0.00462 (h) 43.00 3. (a) 30 000; 25 000; 25 200 (b) 30; 30; 30.5 (c) 0.08; 0.075; 0.0755 (d) 0.0007; 0.00070; 0.000698 Diagnostic Test 2.2 1. 3.1 × 109 , 4.04 × 1017, 6 × 10–3, 5.92 × 10–2 2. (a) 5.726 × 103 (b) 2.14 × 105 (c) 3.2 × 10–4 (d) 4.8 × 10–7 3. (a) 4.6 × 10–7 (b) 2.08 × 108 (c) 7 × 10–8 (d) 3.800 × 1010 4. (a) 4.8 × 107 (b) 7.73 × 10–11 (c) 3.7 × 104 (d) 1.875 × 105 5. (a) 1.014 × 109 m2 (b) RM6 per m2 ANSWERS ANS Spotlight A+1 Maths Form3.indd 137 08/05/2023 5:45 PM

147 ANSWERS 16. (a) –2 2 4 2 4 6 –2 O 2y = x + 6 y = 3x – 2 y x (b) (2, 4) 17. (a) (2, 5) (b) (4, –2) (c) (–1, 4) (d) (–3, –2) 18. (a) (1, –1) (b) (–2, 6) (c) (4, 7) (d) (–3, –5) 19. (a) K(7, 5) (b) 5x – 3y = 20 20. (a) y = –x + 4 (b) – 7 2 (c) (–1, 5) Mastery Challenge 9 1. B 2. D 3. A 4. D 5. C 6. D 7. B 8. B 9. C 10. B 11. (a) x and y are variables (b) m = gradient = coefficient of x (c) c = y-intercept = constant (d) coefficient of y = 1 (e) Power of x = 1, power of y = 1 12. (a) 4; 0 (b) 1; –8 (c) –2; 5 (d) 0; 14 13. (a) iii (b) i (c) iv (d) ii 14. (a) y = 3x + 10, gradient = 3 (b) y = – 1 2 x – 4 (c) Gradient = 0, y-intercept = –3 15. (a) ✗ (b) ✓ 16. (a) x – y = 2; x 2 – y 2 = 1 (b) y = 3x + 9; – x 3 + y 9 = 1 (c) y = – 1 3 x – 4; 1 3 x + y = –4 17. (a) (i) 4 7 (ii) –2 (b) (i) – 2 3 (ii) 6 18. (a) ✓ (b) ✗ (c) ✓ 19. (a) (–4, 1), (1, 3), (6, 5) (b) (2, 4), (–1, 2), (–5, 3) 20. (a) 7 (b) –4 21. (a) ✓ (b) ✗ 22. y = 1 2 x + 1, 2x – 4y = 1, – x 6 + y 3 = 1, x 5 – 2y 5 = 1 23. (a) y = 3x + 6 (b) y = –8x + 24 (c) y = – 7 2 x – 14 24. (a) y = 8 (b) x = 11 (c) y = 2x – 5 25. (a) 2 3 ; y = 2 3 x – 5 (b) –2; y = –2x + 2 26. (a) k – 1 = 1 2 k = 3 2 (b) –3 =  3 2 – 1(4) + r –3 = 2 + r r = –5 27. (a) 2 O –4 –2 4 2 4 6 (1, 5) y = 2x + 3 y = 5 y x (b) –2 O 2 6 8 10 2 4 (4, 2) x + 3y = 10 y x x = 4 4 (c) 2 2 4 6 –8 –6 –4 –2 O (–2, 3) 2y = x + 8 x + y = 1 y x 28. (a) (3, 1) (b) (2, –2) (c) (3, 4) (d) (–1, –3) 29. (a) ✓ (b) ✗ (c) ✗ (d) ✓ 30. (a) 1 (b) 3 (c) y = 2x + 6 (d) x + 3y = –10 (e) (–4, –2) Pentaksiran Sumatif Ujian Akhir Sesi Akademik (UASA) Section A 1. 125 2 3 = (53 ) 2 3 = 53 × 2 3 = 52 = 25 Answer: C 2. (pmr–3) × (2p–2rn ) = 2p3 r−5 (pm × 2p−2) × (r−3 × rn ) = 2p3 r−5 2pm−2r−3+n = 2p3 r−5 m − 2 = 3 –3 + n = –5 m = 5 n = −2 mn = 5(−2) = −10 Answer: A 3. 4 574 000 ≈ 4 600 000 = 4.6 × 106 Answer: B ANS Spotlight A+1 Maths Form3.indd 147 08/05/2023 5:45 PM