ii Contents Must Know iii - x Chapter 1 Quadratic Functions and Equations in One Variable 1 – 11 NOTES 1 Paper 1 3 Paper 2 6 Chapter 2 Number Bases 12 – 20 NOTES 12 Paper 1 14 Paper 2 16 Chapter 3 Logical Reasoning 21 – 29 NOTES 21 Paper 1 22 Paper 2 24 Chapter 4 Operations on Sets 30 – 40 NOTES 30 Paper 1 31 Paper 2 35 Chapter 5 Network in Graph Theory 41 – 49 NOTES 41 Paper 1 42 Paper 2 46 Chapter 6 Linear Inequalities in Two Variables 50 – 60 NOTES 50 Paper 1 51 Paper 2 54 Chapter 7 Graphs of Motion 61 – 73 NOTES 61 Paper 1 62 Paper 2 66 Chapter 8 Measures of Dispersion for Ungrouped Data 74 – 86 NOTES 74 Paper 1 76 Paper 2 79 Chapter 9 Probability of Combined Events 87 – 95 NOTES 87 Paper 1 88 Paper 2 90 Chapter 10 Consumer Mathematics : Financial Management 96 – 102 NOTES 96 Paper 1 97 Paper 2 98 Form 4 Assessment 103 – 114 Answers 115 – 134 Content 1202QB Maths Form 4.indd 2 23/02/2022 6:09 PM
Important Facts (Chapter 1) 1 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 1) 7 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 1) 3 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 9 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 1) 5 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 11 @ Pan Asia Publications Sdn. Bhd. Sketching the Graph of a Quadratic Function Values of a, b and c on Quadratic Functions Roots of a Quadratic Equation Number Bases Conversion of Number Bases For example, f(x) = 2x2 + 5x – 12 Step 1: When a = 2 . 0, the shape Step 2: When c = –12, y-intercept = –12 Step 3: When f(x) = 0 2x2 + 5x – 12 = 0 (2x – 3)(x + 4) = 0 x = 3 2 or x = – 4 x 0 3 (–, 0) 2 (–4, 0) (0, –12) f(x) The value of a determines the shape of graph When a . 0, x y 0 b < 0 x y 0 b = 0 0 x y b > 0 When a , 0, x y 0 b < 0 x y 0 b = 0 x y 0 b > 0 The value of b determines the position of the axis of symmetry The value of c determines the position of y-intercept. When a . 0 When a , 0 x c y 0 x c y 0 1. To convert a number in a certain base to base ten 1728 = (1 × 82 ) + (7 × 81 ) + (2 × 80 ) = 64 + 56 + 2 = 12210 2. To convert a number in a base ten to certain base (a) 12310 to base five (b) 23410 to base eight 5 123 5 24 –3 5 4 –4 0 –4 8 234 8 29 –2 8 3 –5 0 –3 º 12310 = 4435 º 23410 = 3528 3. To convert a number in a certain base (not base ten) to another base (not base ten) Base p Base 10 Base q 1. Digits used in base two up to base ten: Number base Digit Base 2 0, 1 Base 3 0, 1, 2 Base 4 0, 1, 2, 3 Base 5 0, 1, 2, 3, 4 Base 6 0, 1, 2, 3, 4, 5 Base 7 0, 1, 2, 3, 4, 5, 6 Base 8 0, 1, 2, 3, 4, 5, 6, 7 Base 9 0, 1, 2, 3, 4, 5, 6, 7, 8 Base 10 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 2. For 24315 → Place value = 52 = 5 × 5 = 25 Digit value = 4 × 52 = 4 × 25 = 100 Number 2 4 3 1 Place value 53 52 51 50 Digit value 250 100 15 1 1. The roots of a quadratic equation ax2 + bx + c = 0 are the points of intersection of the graph and the x-axis. 2. The roots of the quadratic equation can be determined by: (a) Factorisation method: x2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 Thus, the roots are –3 and –2. (b) Graphical method: Step 1: Determine the shape of the graph by identifying the value of a. Step 2: Determine the y-intercept Step 3: Determine the x-intercept x y 0 –2 –1 –4 –2 2 Root 4 Root 1 2 3 4 1. The general form of the quadratic expression is ax2 + bx + c with a, b and c are constants and a ≠ 0. 2. The highest power of the quadratic expression is 2 and involved only one variable. 3. The general form of the quadratic functions is f(x) = ax2 + bx + c. 4. For a graph of quadratic function f(x) = ax2 + bx + c, When a . 0, x x = m y = ax2 + bx + c f(x) c 0 (m, n) n • Curved graph: . • Minimum point: (m, n) When a , 0, x x = m y = ax2 + bx + c f(x) 0 c (m, n) n • Curved graph: . • Maximum point, (m, n) Quadratic Expressions and Functions MUST KNOW Important Facts Must Know 1202QB Maths Form 4.indd 1 25/07/2022 4:33 PM
Common Mistakes (Chapter 2) 8 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 2 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 2) 10 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 4 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 2) 12 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 6 @ Pan Asia Publications Sdn. Bhd. Number Bases Quadratic Expression Conversion of Number Bases Conversion of Number Bases Quadratic Functions Factorisation Method Determine whether 7k2 + k 1 2 is a quadratic expression in one variable. Correct Wrong Not a quadratic expression in one variable because there is a variable with a power which is not a whole number. A quadratic expression in one variable. Determine whether 2x + 5 is a quadratic expression in one variable. Correct Wrong Not a quadratic expression in one variable because the highest power of the variable is not two. A quadratic expression in one variable. Calculate the sum of value of digit 3 in base 10 for the number 13026 and 23569 . Correct Wrong For 13026 ˜ Digit value = 3 × 62 = 10810 For 23569 ˜ Digit value = 3 × 92 = 24310 Sum of the value for digit 3 = 10810 + 24310 = 35110 Sum of the value for digit 3 = 62 + 92 = 11710 Diagram shows the graph of a quadratic function. Determine the quadratic function in the form of y = ax2 + bx + c. Correct Wrong y = a(x + 3)(x – 1) y = a(x2 + 3x – x – 3) y = a(x2 + 2x – 3)…a Substitute (–2, 6) into a: 6 = a[(–2)2 + 2(–2) – 3] a = –2 Thus, y = –2(x2 + 2x – 3) y = –2x2 – 4x + 6 y = (x + 3)(x – 1) y = x2 + 2x – 3 x y 0 (–2, 6) –3 1 Solve the quadratic equation (3x – 2)2 = 9. Correct Wrong (3x – 2)2 = 9 9x2 – 12x + 4 − 9 = 0 9x2 – 12x – 5 = 0 (3x – 5)(3x + 1) = 0 x = 5 3 or x = – 1 3 3x – 2 = 3 3x = 5 x = 5 3 Determine the roots of x2 + 7x + 6 = 14 by factorisation method. Correct Wrong x2 + 7x + 6 = 14 x2 + 7x – 8 = 0 (x + 8)(x – 1) = 0 x = –8 or x = 1 x2 + 7x + 6 = 14 (x + 6)(x + 1) = 14 x + 6 = 14 or x + 1 = 14 x = 8 x = 13 It is given 1m325 = 2478 , find the place value for m. Correct Wrong 2478 = 2 × 82 + 4 × 81 + 7 × 80 = 16710 = 11325 5 167 Remainder 5 33 – 2 5 6 – 3 5 1 – 1 0 1 º 2478 = 11325 Place value for m = 52 = 25 Place value for m = 1 State the value of the underlined digit. 5495 Correct Wrong Place value 52 51 50 Number 4 4 × 51 = 20 Place value 53 52 51 Number 4 4 × 52 = 100 Ensure the power of place value is start with 0. MUST KNOW Common Mistakes Must Know 1202QB Maths Form 4.indd 2 25/07/2022 4:33 PM
1 Chapter 1 Quadratic Functions and Equations in One Variable 1.1 Quadratic Functions and Equations 1. A quadratic expression is an expression in the form of ax2 + bx + c, where a, b and c are constants and a ≠ 0 while x is the variable. 2. A quadratic expression in one variable is an expression such as (a) involves one variable (b) the highest power for the variable is 2 3. General form for: (a) Quadratic expression: ax2 + bx + c (b) Quadratic function: f(x) = ax2 + bx + c (c) Quadratic equation : ax2 + bx + c = 0 4. The type of relation of a quadratic function is many-to-one relation. 5. Characteristics of quadratic functions: (a) When a . 0 • Has minimum point. • The axis of symmetry of the graph is parallel to the y-axis and passes through the minimum point. • The shape: Axis of symmetry Minimum point 0 f(x) y = f(x) x (b) When a , 0 • Has maximum point. • The axis of symmetry of the graph is parallel to the y-axis and passes through the maximum point. • The shape: Axis of symmetry Maximum point 0 f(x) y = f(x) x 6. Effect of changing the values of a, b and c on graphs of quadratic function, f(x) = ax2 + bx + c: (a) Changing the value of a • Affects the shape and width of the graph. y-intercept remain unchanged. • The width of the graph is decreasing when the value of a is increasing and vice versa. 0 2 –2 –2 2 f(x) y = x y = ax 2 2 , a > 1 y = ax2 , 0 < a < 1 y = ax2 , a < 0 x (b) Changing the value of b • Affects the position of the axis of symmetry. • The shape and y-intercept remain unchanged. For example, if a . 0: 0 2 –4 –2 2 4 f(x) y = x2 y = (x + b )2 y = (x – b )2 x (c) Changing the value of c • Affects the position of the y-intercept. • Shape of the graph unchanged. For example, if a . 0. x f(x) c y-intercept For example, if a , 0. x f(x) 0 y-intercept c NOTes C01 1202QB Maths Form 4.indd 1 22/07/2022 10:31 AM
3SOS TIP PAPER 1 1.1 Quadratic Functions and Equations 1. Which of the following is a quadratic expression in one variable? A 2x + 2y C x2 + y + 5 B 3y2 D 2 x2 + 4 2. Which of the following is a not quadratic expression in one variable? A x2 + 1 C 3x3 + 4 = 0 B 3x2 + 3x + 4 D x2 2 + 4 3. Which of the following is the values of a, b and c for f(x) = x2 – 2x + 6? A a = 1, b = – 2, c = 6 B a = 6, b = – 2, c = 1 C a = 0, b = – 2, c = 6 D a = 0, b = – 2, c = 1 4. Which of the following is the values of a, b and c for f(x) = 6x2 + x – 2? A a = 6, b = 0, c = –2 B a = –2, b = 0, c = 6 C a = –2, b = 1, c = 6 D a = 6, b = 1, c = –2 5. A quadratic function f(x) = 2x2 + 5x + c passes through point (−1, 2). What is the value of c? A −1 C 5 B 2 D 9 6. A quadratic function f(x) = –3x2 + 7x + c passes through point (–1, –12). What is the value of c? A –1 C –10 B –2 D 2 7. The diagram below shows a graph on Cartesian plane. x f(x) 1 0 (–5, 10) Which of the following is true? A Maximum point: (–5, 10) B Minimum point: (–5, 10) C Maximum point: (0, 1) D Minimum point: (0, 1) 8. Diagram below shows a graph of quadratic functions. 0 7 (5, –2) x f(x) Which of the following is true? A Maximum point: (5, −2) B Minimum point: (5, −2) C Maximum point: (0, 7) D Minimum point: (0, 7) 9. Diagram below shows two quadratic functions y = f(x) and y = g(x). HOTS Analysing f(x) = tx2 – 8 g(x) = 2x2 – 8 0 x y Which of the following is the range of value of t? A t , 0 C 0 , t , 2 B t , 2 D t . 2 10. Diagram below shows a graph on a Cartesian plane. 0 2 4 x f(x) Which of the following is the axis of symmetry for the graph? A 2 C 0 B 4 D 1 Question 2: Compare with the general form, f(x) = ax2 + bx + c. Question 10: The axis of symmetry of the graph of a quadratic function is parallel to the y-axis and passes through the maximum or minimum point. Each question has four answer choices A, B, C and D. Choose one answer for each question. C01 1202QB Maths Form 4.indd 3 22/07/2022 10:31 AM
6 SOS TIP PAPER 2 Section A 1. Determine whether each of the following expressions is a quadratic expression in one variable or not. (a) x2 – 3 (b) x2 + 3x–2 (c) y2 – x + 3 (d) –x2 (e) x 1 3 + x [5 marks] Answer: (a) (b) (c) (d) (e) 2. Determine whether the shapes of the following graphs of quadratic functions is or . (a) x2 – 2x (b) 2x – x2 (c) –2x2 + 2x + 4 (d) –5x + 2x2 – 3 [4 marks] Answer: (a) (b) (c) (d) 3. Determine the values of a, b and c for each of the following quadratic expressions. (a) 2x2 – 3x + 5 (b) x2 + 4x (c) 3x2 – 7 (d) 2 – 4x – 3x2 (e) 3y(y – 1) [5 marks] Answer: (a) (b) (c) (d) (e) 4. Determine the maximum point or minimum point and state the equation of the axis of symmetry for each graph of quadratic function below. (a) x f(x) 0 1 1 2 4 –3 –2 –1 2 3 3 (b) x f(x) 0 1 –1 –2 –4 –3 –2 –1 2 3 –3 [4 marks] Answer: (a) (b) 5. The quadratic function f(x) = 2x2 – 5x + c passes through point P as given below. Find the value of c for each of the following cases. (a) P(−1, 5) [2 marks] (b) P(3, 7) [2 marks] Answer: (a) (b) Question 2: Shape when a . 0, shape when a , 0. Question 5: Coordinate (x, f(x)). Substitute x and f(x) from coordinate to the equation. Hence, solve it by by making c as a subject. C01 1202QB Maths Form 4.indd 6 22/07/2022 10:32 AM
8 SOS TIP Question 10: Use the formula of Phytagoras theorom, a2 + b2 = c2 . Question 13: The reflection at x-axis will change the value of a. Section B 9. Form a quadratic expression based on the diagram below. HOTS Analysing [2 marks] Area = 66 cm2 (x + 7) cm (x + 2) cm Answer: 10. Form a quadratic expression based on the diagram below. HOTS Analysing [2 marks] p q r Answer: 11. Solve the following quadratic equation: x2 + 3x = –2(–3 – x) [4 marks] Answer: CLONE SPM 12. Solve the following quadratic equation: – 3 2x + 1 = x x – 2 [4 marks] Answer: 13. Diagram below shows a quadratic function f(x) = px2 + 5x + q with a minimum point (−2.5, −2.25). HOTS Analysing 0 4 x –4 –1 f(x) (a) Given p is an integer where –2 , p , 2, state the value of p. [1 mark] (b) Using the value of p from (a), find the value of q. Hence, state the axis of symmetry. [2 marks] (c) Is the quadratic function will change if the graph is reflected in the x-axis? If it changes, gives the answer in the form of f(x) = ax2 + bx + c. [2 marks] Answer : (a) (b) (c) CLONE SPM C01 1202QB Maths Form 4.indd 8 22/07/2022 10:32 AM
9SOS TIP Question 15: Volume = Length × Width × Height Question 16: Diameter is two times the radius. Section C 14. Sofea bought (8x + 14) text books with the price of RM5x each. She paid RM300 for all the text books. Calculate the number of text books that she bought. HOTS Analysing [4 marks] Answer: 15. A box has the length of (x + 5) cm, the width of x cm and the height of 30 cm. The total volume of the box is 4 500 cm3 . Calculate the value of x. HOTS Analysing [4 marks] Answer: CLONE SPM CLONE SPM 17. Razman wants to sell a rectangular cake during Entrepreneur’s Day. The length and width of the cake are (x + 6) cm and (x + 3) cm respectively. HOTS Analysing (a) Form an expression for the area of the cake, L cm2 , in terms of x. [2 marks] (b) Given that the area of the cake is 270 cm2 . Calculate the value of x. [3 marks] (c) Safi bought the cake and want to give to 30 friends at his birthday party. Will all his friends got the cake if he cut the cake 9 cm2 each? Explain your answer. [2 marks] 16. A box is filled with 12 same size balls as shown in the diagram below. 2x + 4 cm 3x + 3 cm Given the area of the box is 432 cm2 , find the diameter, in cm, of one ball. HOTS Analysing [4 marks] Answer: CLONE SPM Answer : (a) (b) (c) C01 1202QB Maths Form 4.indd 9 22/07/2022 10:32 AM
50 Chapter 6 Linear Inequalities in Two Variables 6.1 Linear Inequalities in Two Variables 1. Linear inequality in two variables is the inequality that involved two variables such that the highest power of both variables is 1. Table below shows the inequality that is suitable for certain situations. Situation Linear Inequality y is more than x y . x y is less than x y , x y is not more than x y < x y is not less than x y > x y is at least h times of x. y > hx Situation Linear Inequality y is at most h times of x y < hx The maximum value of y is h. y < h The minimum value of y is h. y > h The sum of x and y is at least k. x + y > k The ratio of x to y is not less than 2 3 . y < 3 2 x NOTes 2. Dotted line ( ) is used for the inequalities that has the sign . or , and solid line ( ) is used for the inequalities that has the sign > or <. 3. Linear inequality for the straight line on the graph can be determined using the general form of equation of the straight line, y = mx + c such that m is the gradient of the straight line and c is the y-intercept. 4. All the points located on the straight line y = mx + c satisfy the equation y = mx + c. 5. All the points in the region above the straight line y = mx + c satisfy the inequality y . mx + c. 6. All the points in the region below the straight line y = mx + c satisfy the inequality y , mx + c. 7. Diagram below shows a few regions that satisfy the certain inequalities. Linear Inequality y > mx + c, the region is above the solid line y = mx + c y , mx + c, the region is below the dashed line y = mx + c y < h, the region is below the solid line y = h x , k, the region is on the left side of the dashed line x = k x y y = mx + c 0 y > mx + c y < mx + c x y y = mx + c 0 y > mx + c y < mx + c x y y = h 0 y > h y < h x y x = k 0 x < k x > k 6.2 Systems of Linear Inequalities in Two Variables 1. A combination of two or more linear inequalities is known as a system of linear inequalities. 2. The region that satisfy a system of linear inequalities can be determined by using the following steps: I Determine and mark the region that is represented by each linear inequalities. II Determine the common region that satisfies all the linear inequalities. III Shade the region and make sure the region is being bounded by all the linear inequalities. 3. For example, the shaded R in the diagram on the right satisfies all the inequalities of y < 3x + 2, y > –x + 2 and x , 2. x y R x = 2 y = –x + 2 y = 3x + 2 0 2 2 2 – –3 C06 1202QB Maths Form 4.indd 50 29/07/2022 11:25 AM
51SOS TIP PAPER 1 6.1 Linear Inequalities in Two Variables 1. Which of the following inequalities is the linear inequality in two variables? A y . 3 B x + y < 4 C 2x – 3 , x – 2 D 3x + 5y , z – 2 2. Mr. Mark prepares x unit of tables and y unit of chairs for class 5A such that the number of tables is more than the number of chairs. Represents the situation above in the form of linear inequality. A x . y C y + x . 0 B y , x D y – x . 0 3. Diagram below shows a shaded region that satisfies the linear inequality P. x y 0 –2 2 4 –4 –2 2 Determine the linear inequality P. A y . 3 2 x + 3 C y , 3 2 x + 3 B y > 3 2 x + 3 D y < 3 2 x + 3 4. Diagram below shows the shaded region that satisfy the linear inequality R. x y 0 –2 2 4 –2 2 4 Determine the linear inequality R. HOTS Applying A y . – 1 2 x + 2 C y < – 1 2 x + 2 B y , – 1 2 x + 2 D y > – 1 2 x + 2 5. Diagram below shows the shaded region that satisfy the linear inequality Q. x y 0 –2 2 4 –6 –4 –2 2 Determine the linear inequality Q. HOTS Applying A 4y , 4 + x B y < 1 4 x + 1 C 4y – x , 1 D y – 4x + 1 , 0 6. Diagram below shows a shaded region which satisfy linear inequality P. –9 –3 x y 0 Determine the linear inequality P. HOTS Applying A y , –3x – 9 C y + 3x > –9 B 3y + x . –9 D y < – 1 3 x – 9 7. Which of the following region represents the inequality 4y , x – 8? HOTS Analysing A C –8 x y 0 2 2 x y 0 8 B D –2 x y 0 8 –8 x y –2 0 Question 1: A linear inequality in two variables is formed when symbols ,, <, >, or . used to relate two expressions involving two variables and the highest power of a variable is 1. Question 5: Equation of the straight line can be determined by using the equation y = mx + c such that m = y2 – y1 x2 – x1 and c is the y-intercept. Each question has four answer choices A, B, C and D. Choose one answer for each question. C06 1202QB Maths Form 4.indd 51 29/07/2022 11:25 AM
54SOS TIP Question 1: Equation of the straight line can be found using the equation y = mx + c where m = – y-intercept x-intercept and c is y-intercept. Question 3 (a): Point (5, 6) can be determined whether that satisfy the linear equation or the linear inequalities by the substituting the point into the equation of the straight line. PAPER 2 Section A 1. Diagram below shows a shaded region that satisfy the three inequalities. x y 0 –8 –6 –4 –2 2 4 6 6 4 2 –2 –4 –6 –8 –10 8 State all the linear inequalities. [4 marks] Answer: 3. (a) Determine whether the point (5, 6) satisfy 2y = x + 10, 2y . x + 10 or 2y , x + 10. [1 mark] (b) In the graph below, shade the region that satisfy 3y + x . 9, y > 1 2 x + 5 and x . –5. [3 marks] Answer: (a) (b) x y 0 2 4 2 4 6 8 –4 –2 2. In the graph in the diagram below, shade the region that satisfy the three inequalities y > –x + 6, y < x + 6 and x < 6. [3 marks] Answer: x y y = –x + 6 y = x + 6 0 6 6 4. In the graph below, shade the region that satisfy the three inequalities 2y < x + 12, y + x > –4 and y . 2x. [3 marks] Answer: y + x = –4 2y = x + 12 x y 0 2 4 6 –2 –4 2 4 6 8 10 –6 –4 –2 CLONE SPM CLONE SPM CLONE SPM CLONE SPM C06 1202QB Maths Form 4.indd 54 29/07/2022 11:25 AM
56SOS TIP Question 9: (a) The equation of straight line can be determined by the equation y = mx + c such that m = y2 – y1 x2 – x1 and c is y-intercept. (b) Draw the line y = 12 and determine the maximum value for x. 9. Diagram below shows the shaded region defined by a system of linear inequalities. x y 0 2 4 6 8 10 4 8 12 16 (a) State all the inequalities that satisfy the shaded region in the diagram above. (b) Hence, determine the maximum value of x if y = 12. HOTS Applying [5 marks] Answer: (a) (b) 10. Mr. Zakry has allocated RM4 800 to purchase x units of calculator and y units of geometrical set for school cooperation. The price of a unit of calculator and a unit of geometrical set is RM20 and RM16 respectively. The total of the calculator and geometrical set is at least 200 units and the number of unit of calculator purchased is not more than twice the number of unit of geometrical set. HOTS Applying (a) Write three inequalities, other than x > 0 and y > 0, that satisfy the situation above. [3 marks] (b) Using the scale of 1 cm to 50 units on both axes, construct and shade the region that satisfies the system of linear inequalities. [3 marks] (c) Determine the minimum and maximum units of calculator that can be purchased by Mr. Zakry if he purchased 100 units of geometrical set. [3 marks] Answer: (a) (b) (c) Section B C06 1202QB Maths Form 4.indd 56 29/07/2022 11:25 AM
58SOS TIP 13. Ramesh opens a restaurant that needs x waiters and y cooks. Total waiters and cooks employed by Ramesh is not more than 100 person and the number of waiters is not more than twice the number of cooks employed. HOTS Applying (a) Write two inequalities, other than x > 0 dan y > 0, that satisfy the situation above. [3 marks] (b) The third inequality is represented by the shaded region in the graph in Diagram (a). Write, in words, for the third inequality. [3 marks] (c) Using Diagram (a), construct and shade the region that satisfy the system of linear inequalities. [3 marks] (d) It is given that the monthly salary for each waiter and cook employed is RM2 000 and RM4 000 respectively. Calculate, in RM, the maximum salary allocation that need to be prepared by Ramesh if he employed 60 cooks employed. [3 marks] Answer: (a) (b) (c) x y 0 10 20 30 40 50 60 10 20 30 40 50 60 70 80 90 100 Diagram (a) (d) 14. Diagram below shows two types of hotel room that is available in a hotel in Kuala Lumpur. Type A room Type B room Type A room and Type B room can accommodate 2 and 3 person respectively. The hotel rental will be charged with service tax of 6%. (a) Mr. Rahman paid RM439.90 for a unit of Type A room and a unit of Type B room for a night stay. Madam Mariah paid RM1 293.20 for 4 units of Type A room and 2 units of Type B for a night stay. Calculate, in RM, the total payment payable by Mr. Siva if he would like to rent 3 units of Type A room and 2 units of Type B room for 2-night stay. [5 marks] Question 13(d): Draw the straight line y = 60 on the graph to get maximum value of x. Hence, determine the maximum salary of the allocation. Question 14 (a): The room rate of Type A room and Type B room can be obtained using simultaneous equation method. Section C C06 1202QB Maths Form 4.indd 58 29/07/2022 11:25 AM
103 1. Factorise 2x(2x – 7) – 3x – 15. A (4x + 3)(x – 5) B (4x – 3)(x – 5) C (4x + 3)(x + 5) D (4x – 3)(x + 5) 2. Given x2 = z2 + 2 y , express y in terms of x and z. A y = 1 2(x + z)(x – z) B y = 1 2(x – z) 2 C y = 2 (x + z)(x – z) D y = 2 (x – z) 2 3. Round off 0.08095 correctly to 3 significant figures. A 0.0810 B 0.0800 C 0.0890 D 0.0809 4. 1.85 × 106 + 1.49 × 105 = A 1.999 × 104 B 1.999 × 105 C 1.999 × 106 D 1.999 × 107 5. 7.89 × 106 5 000 = A 1.578 × 103 B 1.578 × 10–3 C 1.578 × 102 D 1.578 × 10–2 6. Express 15078 in base 10. A 6 663 C 839 B 840 D 832 7. Given x5 = 56 + (4 × 54 ) + (4 × 51 ) + 3, state the value of x. A 6 040 010 B 5 040 043 C 1 400 043 D 1 040 043 8. Determine the value of digit 3 in the number 130078 . A 512 C 192 B 12288 D 1536 9. 3215 + 1125 = A 4345 C 4435 B 4335 D 3345 10. Diagram 1 shows a regular hexagon ABCDEF and an equilateral triangle AGB. FAGH is a straight line. D C F x y E B A G H Diagram 1 Find the value of x + y. A 60° C 180° B 120° D 240° 11. In Diagram 2, ABC is a tangent to the circle centered at O, at B. y x O A B C 80° 70° Diagram 2 Calculate the value of x + y. A 50° C 90° B 70° D 110° 12. State the image coordinates of the point (–4, 3) under the rotation 90° anticlockwise at the center of (0, 1). A (–2, –3) B (–2, 5) C (–2, –5) D (–2, –4) PAPER 1 Time: 1 hour 30 minutes This question paper contains 40 questions. Answer all the questions. Form 4 Assessment APaper 1202QB Maths Form 4.indd 103 25/07/2022 11:53 AM
107 PAPER 2 1. On the graph in the answer space, shade the region which satisfies the three inequalites 2x + y > 8, y > x and y , 8. [3 marks] Answer: –2 2 –2 –4 4 6 8 10 12 2 4 6 8 10 12 14 16 y y = x 2x + y = 8 x 2. The Venn diagram below shows the set P, set Q and set R with the state of the universe set, ξ = P Q R. On the given diagram, shade the set (a) Q R Q R P [1 mark] (b) P (Q R) Q R P [2 marks] 3. Diagram 3 shows a solid combination of a hemisphere and a cylinder. Given that the height of cylinder is 13 cm and the volume of the solid combination is 1 282 59 60 cm3 . 13 cm Diagram 3 By using π = 22 7 , calculate the radius, in cm, for hemisphere and cylinder. [4 marks] Answer: 4. Diagram 4 shows the OPQ and OQRS sectors. OTQ is an right-angled triangle and TOS is a straight line. 14 cm S P Q T O R Diagram 4 Given that OS = 2OT. By using π = 22 7 , calculate (a) the perimeter, in cm, of the whole diagram. [2 marks] (b) the area, in cm2 , of the shaded region. [2 marks] Time: 2 hours 30 minutes Section A [40 marks] Answer all questions in this section. APaper 1202QB Maths Form 4.indd 107 25/07/2022 11:53 AM
110 11. (a) Diagram 11(a) shows a set of data. 21 20 31 25 26 Diagram 11(a) (i) Calculate the standard deviation of the data. (ii) Calculate the new variance if each value of the set of data is multiply by 3. [4 marks] Answer: (a) (i) (ii) (b) Table 11(b) shows the frequency for the number of goals scored by a team in football match. Score 0 1 2 3 4 5 Frequency 2 4 3 2 5 1 Table 11(b) (i) Calculate the interquartile range. (ii) Hence, draw a box plot by using the quartiles obtained. [5 marks] Answer: (b) (i) (ii) 12. Table 12 shows the distance that connecting 5 cities. City A – B B – C B – D C – E C – D D – E Distance (km) 52 38 75 55 60 28 Table 12 (a) Complete the non-directed graph in the answer space to represent the information in Table 12. [2 marks] (b) Calculate the nearest distance from City A to City E. [2 marks] (c) Calculate the longest distance from City E to City A such that all roads are only passed once. [2 marks] Answer: (a) A B C D E (b) (c) 13. (a) Complete Table 13 in the answer space for the equation y = 5x2 + 3x – 2 by writing values of y when x = –2.5 and x = 2. [2 marks] (b) For the subdivision of this question, use graph paper. You can use a flexible ruler. Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw a graph of y = 5x2 + 3x – 2 for –3 < x < 3. [4 marks] (c) From the graph in 13 (b), find (i) the value of y when x = 1.8. (ii) the values x when y = 2.2. [3 marks] Answer: (a) x –3 –2.5 –1 0 1 2 3 y 34 0 –2 6 52 Table 13 (b) Refer to the graph on page 111. (c) (i) (ii) Section B [45 marks] Answer all the questions in this section. APaper 1202QB Maths Form 4.indd 110 25/07/2022 11:53 AM
113 Graph for Question 15 Section C [15 marks] Answer any one questions from this section. 16. (a) A 5-digit number is written randomly using all digits 1, 2, 3, 4 and 5. Find the probability that the number is the even number. [4 marks] Answer: (a) (b) Teacher Ngu made a questionnaire about a hobby against one school with 100 students. It was found that 45 of his students’ hobby were reading, 21 students loved reading and listening to music, 10 loved listening to music and sporting, 22 students loved reading and sporting, 7 students loved reading only and 5 students had that three hobbies. Given the ratio of the number of students loved listening to music only to the number of students loved sporting only is 3 : 2. (i) Draw a Venn diagram based on the information provided. [5 marks] (ii) Calculate the number of pupils who are hobby to read or listen to music only. [2 marks] (iii) A student is selected at random, find the probability that the students loved to read and listen to music only. [2 marks] (iv) A student is randomly selected, find the probability that the students loved all three hobbies. [2 marks] APaper 1202QB Maths Form 4.indd 113 25/07/2022 11:53 AM
115 CHAPTER 1 Paper 1 1. B 2. C 3. A 4. D 5. C 6. B 7. A 8. B 9. D 10. A 11. B 12. A 13. D 14. D 15. B 16. A 17. C 18. C 19. D 20. D 21. C 22. A 23. D 24. A 25. B 26. A 27. D Paper 2 Section A 1. (a) Quadratic expression in one variable. (b) Not a quadratic expression in one variable because there is a variable with a power which is not a whole number. (c) Not a quadratic expression in one variable because there are two variables, x and y. (d) Quadratic expression in one variable. (e) Not a quadratic expression in one variable because there is a variable with a power which is not a whole number. 2. (a) (b) (c) (d) 3. (a) a = 2, b = –3, c = 5 (b) a = 1, b = 4, c = 0 (c) a = 3, b = 0, c = –7 (d) a = –3, b = –4, c = 2 (e) a = 3, b = –3, c = 0 4. (a) Maximum point : (–1, 4) Axis of symmetry, x = –1 (b) Minimum point : (0, –4) Axis of symmetry, x = 0 5. (a) Given f(x) = 2x2 – 5x + c. Substitute the values of x = –1 and f(x) = 5 into the quadratic function: 5 = 2(–1)2 – 5(–1) + c 5 = 2 + 5 + c c = –2 (b) Given f(x) = 2x2 – 5x + c. Substitute the values of x = 3 and f(x) = 7 into the quadratic function: 7 = 2(3)2 – 5(3) + c 7 = 18 – 15 + c c = 7 – 3 = 4 6. Let Sufi’s age = x – 2 x(x – 2) = 35 x2 – 2x = 35 x2 – 2x – 35 = 0 7. (a) f(x) = x2 – x – 6 Value of a = 1 > 0, shape Value of c = –6, y-intercept = –6 When f(x) = 0, x2 – x – 6 = 0 (x + 2)(x – 3) = 0 x = –2 or x = 3 The graph: –2 0 –6 3 f(x) x (b) Axis of symmetry at x = 3 + (–2) 2 = 1 2 f( 1 2) = ( 1 2) 2 – ( 1 2) – 6 = – 25 4 Thus, minimum point is ( 1 2 , – 25 4 ). (c) From the graph on (a), the roots for the graph function are x = –2 or x = 3. 8. (a) f(x) = x2 – 6x + 8 Value of a = 1 . 0, shape Value of c = 8, y-intercept = 8 When f(x) = 0, x2 – 6x + 8 = 0 (x – 4)(x – 2) = 0 x = 2 or x = 4 The graph: 0 8 2 4 f(x) x (b) f(x) = x2 – 4x + 4 Value of a = 1 . 0, shape Value of c = 4, y-intercept = 4 When f(x) = 0, x2 – 4x + 4 = 0 (x – 2)2 = 0 x = 2 The graph: 0 4 2 f(x) x (c) f(x) = x2 – 4 Value of a = 1 > 0, shape Value of b = 0, axis of symmetry is the y-axis Value of c = –4, y-intercept is –4 Thus, the minimum point is (0, –4). When f(x) = 0, x2 – 4 = 0 (x + 2)(x – 2) = 0 x = 2 or x = –2 The graph: 0 –4 2 f(x) x –2 (d) f(x) = –x2 + 2 Value of a = –1 . 0, shape Value of b = 0, axis of symmetry is the y-axis Value of c = 2, y-intercept is 2 Thus, the maximum point is (0, 2). When f(x) = 0, –x2 + 2 = 0 x = ±! 2 x = 1.4 or x = –1.4 The graph: 0 2 1.4 f(x) x –1.4 (e) f(x) = x2 – 2 Value of a = 1 . 0, shape Value of b = 0, axis of symmetry is the y-axis Value of c = –2, y-intercept is –2 Thus, the minimum point is (0, –2). When f(x) = 0, –x2 + 2 = 0 x = ±! 2 x = 1.4 or x = –1.4 The graph: 0 –2 1.4 f(x) x –1.4 Section B 9. Area of rectangle = 66 cm2 (x + 2)(x + 7) = 66 x2 + 2x + 7x + 14 = 66 x2 + 9x + 14 – 66 = 0 x2 + 9x – 52 = 0 10. By using Pythagoras Theorom, r2 = p2 + q2 11. x2 + 3x = –2(–3 – x) x2 + 3x = 6 + 2x x2 + 3x – 2x – 6 = 0 x2 + x – 6 = 0 (x – 2)(x + 3) = 0 x = 2, x = –3 12. – 3 2x + 1 = x x – 2 –3(x – 2) = x(2x + 1) –3x + 6 = 2x2 + x 2x2 + x + 3x – 6 = 0 2x2 + 4x – 6 = 0 x2 + 2x – 3 = 0 (x – 1)(x + 3) = 0 x = 1, x = –3 13. (a) p = 1 (b) f(x) = x2 + 5x + q From the graph, when x = 0, f(x) = 4. Substitute x = 0 and f(x) = 4 into quadratic function: 4 = 1(0)2 + 5(0) + q q = 4 CHAPTER 1 Answers Complete Answers (Paper 1) https://bit.ly/3KypArA Answers 1202QB Maths Form 4.indd 115 05/08/2022 3:58 PM
116 Given minimum point is (−2.5, −2.25), Axis of symmetry, x = –2.5. (c) When f(x) = x2 + 5x + 4 is reflected in the x-axis, the quadratic function change to f(x) = –x2 – 5x – 4 where the value of a, b and c will change to −a, –b and –c. 14. Price = RM300 (8x + 14)(5x) = 300 40x2 + 70x – 300 = 0 (x – 2)(4x + 15) = 0 x = 2, x = – 15 4 (Rejected) Number of text books bought by Sofea = 8(2) + 14 = 30 books 15. Volume of the box = 4 500 (x + 5)(x)(30) = 4 500 30x2 + 150x – 4 500 = 0 x2 + 5x – 150 = 0 (x – 10)(x + 15) = 0 x = 10, x = –15 (Rejected) Thus, the value of x is 10 cm. 16. Area of the box = 432 (2x + 4)(3x + 3) = 432 6x2 + 12x + 6x + 12 = 432 6x2 + 18x – 420 = 0 x2 + 3x – 70 = 0 (x + 10)(x – 7) = 0 x = 7, x = –10 (Rejected) Substitute x = 7, Diameter 4 balls = 3(7) + 3 = 21 + 3 = 24 cm So, diameter 1 ball = 24 ÷ 4 = 6 cm Substitute x = 7 into 2x + 4, Diameter 3 balls = 2(7) + 4 = 14 + 4 = 18 cm So, diameter 1 ball = 18 ÷ 3 = 6 cm Section C 17. (a) L = (x + 6)(x + 3) L = x2 + 6x + 3x + 18 L = x2 + 9x + 18 (b) Area of the cake = 270 x2 + 9x + 18 = 270 x2 + 9x + 18 – 270 = 0 x2 + 9x – 252 = 0 (x – 12)(x + 21) = 0 x = 12 cm (c) Number of small cakes = 270 cm2 ÷ 9 cm2 = 30 cakes Thus, the small cakes is enough to give to all Safi’s friend. 18. (a) Area of triangle, L = 1 2 (x + 3)(x – 1) = x2 + 2x – 3 2 (b) Area of triangle = 16 x2 + 2x – 3 = 32 x2 + 2x – 35 = 0 (x – 5)(x + 7) = 0 x = 5, x = –7 (Rejected) When x = 5, Height of triangle = 5 + 3 = 8 cm Side length of triangle = 5 – 1 = 4 cm 19. Area of square – Area of triangle = 40 2y(2y) – [ 1 2 (y + 2)(2y) ] = 40 4y2 – y2 – 2y – 40 = 0 3y2 – 2y – 40 = 0 (3y + 10)(y – 4) = 0 y = – 10 3 (Rejected) or y = 4 Perimeter of the combination objects = (4 + 6) + (4 + 2) + [3 × 2(4)] = 10 + 6 + 8 + 8 + 8 = 40 cm 20. (a) L = 1 2 (y)(y + 4) L = 1 2 (y2 + 4y) L = 1 2 y2 + 2y (b) Area of triangle = 48 1 2 y2 + 2y = 48 1 2 y2 + 2y – 48 = 0 y2 + 4y – 96 = 0 (y + 12)(y – 8) = 0 y = 8, y = –12 (Rejected) Thus, y = 8 cm. (c) Area of the polygon = 6 × 48 = 288 cm2 Thus, the area of polygon is 288 cm2 and its name is hexagon. 21. (a) L = (35 + y)(65 + y) L = 2 275 + 65y + 35y + y2 L = y2 + 100y + 2 275 (b) Area of the tile = 6 175 y2 + 100y + 2 275 = 6 175 y2 + 100y – 3 900 = 0 (y – 30)(y + 130) = 0 y = 30, y = –130 (Rejected) Thus, y = 30 cm. (c) The smallest part of tile represented by region DEFI. Area of DEFI = 0.3 m × 0.3 m = 0.09 m2 Number of tiles needed = 1.08 ÷ 0.09 = 12 tiles 22. (a) Area A – Area B = 12 cm2 2x(x + 3) – x(x + 5) = 12 2x2 + 6x – x2 – 5x – 12 = 0 x2 + x – 12 = 0 (x – 3)(x + 4) = 0 x = 3, x = –4 (Rejected) Thus, x = 3 cm. (b) Perimeter of new arrangement = 6 + 6 + 6 + 8 + 3 + 8 + 3 = 40 cm CHAPTER 2 Paper 1 1. A 2. D 3. D 4. A 5. C 6. C 7. A 8. B 9. B 10. D 11. C 12. D 13. B 14. D 15. C 16. A 17. C 18. C 19. B 20. A 21. C 22. B 23. B 24. D 25. A 26. C 27. D 28. A 29. B 30. D 31. C 32. C 33. A 34. B 35. D 36. D 37. B 38. C 39. D 40. C 41. B 42. B Paper 2 Section A 1. (a) 42 (b) 62 (c) 81 (d) 25 2. (a) 3 × 71 = 21 (b) 4 × 53 = 500 (c) 7 × 91 = 63 (d) 2 × 34 = 162 3. (a) 324 = 2 × 40 + 3 × 41 = 2 + 12 = 14 (b) 111012 = 1 × 20 + 0 × 21 + 1 × 22 + 1 × 23 + 1 × 24 = 1 + 4 + 8 + 16 = 29 (c) 4135 = 3 × 50 + 1 × 51 + 4 × 52 = 3 + 5 + 100 = 108 (d) 6247 = 4 × 70 + 2 × 71 + 6 × 72 = 4 + 14 + 294 = 312 4. (a) 53410 to base two 2 534 Remainder 2 267 0 2 133 1 2 66 1 2 33 0 2 16 1 2 8 0 2 4 0 2 2 0 2 1 0 0 1 53410 = 10000101102 (b) 53410 to base eight 8 534 Remainder 8 66 6 8 8 2 8 1 0 0 1 53410 = 10268 (c) 53410 to base five 5 534 Remainder 5 106 4 5 21 1 5 4 1 0 4 53410 = 41145 (d) 53410 to base four 4 534 Remainder 4 133 2 4 33 1 4 8 1 4 2 0 0 2 53410 = 201124 CHAPTER 1 – CHAPTER 2 Answers 1202QB Maths Form 4.indd 116 05/08/2022 3:58 PM