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HKDSE Mathematics (Compulsory Part) Paper 1 Practice Paper | 15 questions, 105 marks, 2 hours 15 minutes
HKDSE Mathematics Paper 1
HKDSE Mathematics (Compulsory Part) Paper 1 practice paper: 15 long questions, 105 marks in 2 hours 15 minutes, with percentages, indices, quadratic equations, simultan
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Paper details Level: S6 (HKDSE) | Subject: Mathematics | Time: 2 hours 15 minutes | Marks: 105 | Questions: 15
Section A(1) (35 marks)
Answer ALL questions. Show your working.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 1. (4 marks) A jacket costs $480. Its marked price is 25% above the cost, and it is then sold at 15% discount on the marked price. (a) Find the marked price. (b) Find the selling price. (c) Find the profit percentage. |
Answer: (a) Marked price = 480 x 1.25 = $600 (1 mark) (b) Selling price = 600 x 0.85 = $510 (1 mark) (c) Profit = 510 – 480 = $30 (1 mark); profit percentage = 30 / 480 x 100% = 6.25% (1 mark) |
| 2 | 2. (4 marks) Simplify 4^(2n+1) / 8^(n-1), and hence find its value when n = 3. |
Answer: 4^(2n+1) = 2^(4n+2) (1 mark); 8^(n-1) = 2^(3n-3) (1 mark); quotient = 2^(4n+2-3n+3) = 2^(n+5) (1 mark); when n = 3 the value is 2^8 = 256 (1 mark) |
| 3 | 3. (4 marks) It is given that x + 1/x = 5 and x is not 0. (a) Find x^2 + 1/x^2. (b) Hence find x^4 + 1/x^4. |
Answer: (a) (x + 1/x)^2 = x^2 + 2 + 1/x^2 = 25 (1 mark), so x^2 + 1/x^2 = 23 (1 mark) (b) (x^2 + 1/x^2)^2 = x^4 + 2 + 1/x^4 = 529 (1 mark), so x^4 + 1/x^4 = 527 (1 mark) |
| 4 | 4. (5 marks) A rectangle has perimeter 34 cm and area 60 cm^2. (a) Let the width be w cm and form a quadratic equation in w. (2 marks) (b) Solve it and state the length and the width. (3 marks) |
Answer: (a) Length = (34 – 2w)/2 = 17 – w; w(17 – w) = 60, i.e. w^2 – 17w + 60 = 0 (2 marks) (b) Discriminant = 17^2 – 4×60 = 289 – 240 = 49 = 7^2 (1 mark); w = (17 +- 7)/2 = 5 or 12 (1 mark); so the length is 12 m and the width is 5 cm (1 mark) |
| 5 | 5. (4 marks) Solve the simultaneous equations 2x + 3y = 18 5x – 2y = 7 |
Answer: (1) x 2: 4x + 6y = 36 (1 mark); (2) x 3: 15x – 6y = 21 (1 mark); adding: 19x = 57, so x = 3 (1 mark); substituting: 2(3) + 3y = 18, so y = 4 (1 mark); hence x = 3, y = 4 |
| 6 | 6. (5 marks) A set of data is 4, 7, 7, 9, 13. Find (a) the mean; (b) the median; (c) the range; (d) the standard deviation, correct to 3 significant figures. |
Answer: (a) Mean = (4+7+7+9+13)/5 = 40/5 = 8 (1 mark) (b) Median = 7 (1 mark) (c) Range = 13 – 4 = 9 (1 mark) (d) Sum of squared deviations = 16+1+1+1+25 = 44 (1 mark); variance = 44/5 = 8.8, standard deviation = sqrt(8.8) = 2.97 (1 mark) |
| 7 | 7. (4 marks) A ladder 6.5 m long leans against a vertical wall. The foot of the ladder is 2.5 m from the wall. (a) Find the height of the top of the ladder above the ground. (b) Find the angle the ladder makes with the ground, correct to 0.1 degree. |
Answer: (a) By Pythagoras: h^2 = 6.5^2 – 2.5^2 = 42.25 – 6.25 = 36 (1 mark), so h = 6 m (1 mark) (b) tan(theta) = 6/2.5 = 2.4 (1 mark), so theta = 67.4 degrees (1 mark) |
| 8 | 8. (5 marks) A bag contains 5 red, 3 blue and 2 green balls. Two balls are drawn at random without replacement. (a) Find the probability that both are red. (b) Find the probability that at least one is red. |
Answer: (a) P(two red) = (5/10) x (4/9) = 20/90 = 2/9 (3 marks) (b) P(at least one red) = 1 – P(both not red) = 1 – (5/10 x 4/9) = 1 – 10/45 = 7/9 (2 marks) |
Section A(2) (35 marks)
Answer ALL questions. Show your working.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | 9. (8 marks) A solid is formed by a cylinder and a cone on the same base. The cylinder has radius 6 cm and height 10 cm; the cone has height 8 cm. (a) Find the volume of the solid in terms of pi. (3 marks) (b) Find the total surface area, including the base, in terms of pi. (3 marks) (c) Painting costs $0.05 per cm^2. Find the cost of painting the whole surface, correct to the nearest $0.1. (2 marks) |
Answer: (a) V = pi(6^2)(10) + (1/3)pi(6^2)(8) = 360pi + 96pi = 456pi cm^3, about 1432.6 cm^3 (3 marks) (b) Slant height = sqrt(36 + 64) = 10 cm; S = 120pi + 36pi + 60pi = 216pi cm^2, about 678.6 cm^2 (3 marks) (c) Cost = 216pi x 0.05 = 10.8pi, about $33.9 (2 marks) |
| 10 | 10. (9 marks) y varies partly as x and partly as x^2. When x = 2, y = 10; when x = 3, y = 24. (a) Express y in terms of x. (5 marks) (b) Find y when x = 5. (2 marks) (c) If y = 44 and x > 0, find x. (2 marks) |
Answer: (a) Let y = ax + bx^2. Then 2a + 4b = 10 and 3a + 9b = 24 (2 marks); so a + 2b = 5 and a + 3b = 8, giving b = 3 and a = -1 (2 marks); hence y = 3x^2 – x (1 mark) (b) At x = 5, y = 75 – 5 = 70 (2 marks) (c) 3x^2 – x – 44 = 0, discriminant = 1 + 528 = 529 = 23^2 (1 mark); x = (1 + 23)/6 = 4 (1 mark) |
| 11 | 11. (9 marks) It is given that A(1, 2) and B(7, 10). (a) Find the length of AB. (3 marks) (b) Find the equation of the perpendicular bisector of AB. (3 marks) (c) Find the equation of the circle with AB as a diameter. (3 marks) |
Answer: (a) AB = sqrt((7-1)^2 + (10-2)^2) = sqrt(36 + 64) = 10 (3 marks) (b) Midpoint = (4, 6); slope of AB = 8/6 = 4/3 (1 mark), so the perpendicular bisector has slope -3/4 (1 mark); y – 6 = -3/4 (x – 4), i.e. 3x + 4y = 36 (1 mark) (c) Centre (4, 6), radius 5 (1 mark); equation (x – 4)^2 + (y – 6)^2 = 25 (2 marks) |
| 12 | 12. (9 marks) $50,000 is deposited at 4% per annum compounded yearly for 3 years. (a) Find the amount and the interest after 3 years, correct to 2 decimal places. (4 marks) (b) How much less interest would simple interest give? (2 marks) (c) If interest is compounded half-yearly instead, find the amount after 3 years. (3 marks) |
Answer: (a) Amount = 50000 x 1.04^3 = $56243.20 (3 marks); interest = $6243.20 (1 mark) (b) Simple interest = 50000 x 4% x 3 = $6000; less by 6243.20 – 6000 = $243.20 (2 marks) (c) 50000 x 1.02^6 = $56308.12 (3 marks) |
Section B (35 marks)
Answer ALL questions. Show your working.
| # | Question | Answer and marking notes |
|---|---|---|
| 13 | 13. (11 marks) Let f(x) = x^2 – 6x + 5. (a) Complete the square and write f(x) in the form (x – h)^2 + k. (3 marks) (b) State the vertex and the axis of symmetry. (2 marks) (c) Solve f(x) = 0. (2 marks) (d) Find the range of x for which f(x) is negative. (2 marks) (e) If g(x) = f(x) – 5, solve g(x) = 0. (2 marks) |
Answer: (a) f(x) = (x^2 – 6x + 9) – 9 + 5 = (x – 3)^2 – 4 (3 marks) (b) Vertex (3, -4) (1 mark); axis of symmetry x = 3 (1 mark) (c) (x – 1)(x – 5) = 0, so x = 1 or 5 (2 marks) (d) The graph opens upwards, so x lies between 1 and 5 (2 marks) (e) g(x) = x^2 – 6x = x(x – 6), so x = 0 or 6 (2 marks) |
| 14 | 14. (12 marks) In the cuboid ABCDEFGH, ABCD is the base, AB = 8 cm, BC = 6 cm and AE = 5 cm. (a) Find AC. (3 marks) (b) Find the space diagonal AG in surd form and correct to 2 decimal places. (4 marks) (c) Find the angle between AG and the base ABCD, correct to 0.1 degree. (3 marks) (d) Find the angle between AG and the edge AE, correct to 0.1 degree. (2 marks) |
Answer: (a) AC = sqrt(64 + 36) = 10 cm (3 marks) (b) AG = sqrt(64 + 36 + 25) = sqrt(125) = 5sqrt(5), about 11.18 cm (4 marks) (c) If theta is the angle with the base, tan(theta) = 5/10 = 0.5 (2 marks), so theta = 26.6 degrees (1 mark) (d) cos(phi) = 5 / (5sqrt(5)) (1 mark), so phi = 63.4 degrees (1 mark) |
| 15 | 15. (12 marks) The table shows the monthly reading time (hours) of 30 students: 0-10: 4; 10-20: 8; 20-30: 10; 30-40: 6; 40-50: 2. (a) Find the mean. (4 marks) (b) Find the median class and the median. (4 marks) (c) Find the percentage of students reading less than 20 hours. (2 marks) (d) A student is chosen at random. Find the probability that the reading time is at least 30 hours. (2 marks) |
Answer: (a) Total = 5×4 + 15×8 + 25×10 + 35×6 + 45×2 = 20 + 120 + 250 + 210 + 90 = 690 (3 marks); mean = 690/30 = 23 hours (1 mark) (b) The median is the mean of the 15th and 16th values; cumulative frequencies are 4 and 12, so the median class is 20-30 (2 marks); median = 20 + (15 – 12)/10 x 10 = 23 hours (2 marks) (c) (4 + 8)/30 x 100% = 40% (2 marks) (d) (6 + 2)/30 = 4/15, about 0.267 (2 marks) |
A TutorZone practice paper; original questions and passages, not publisher material.
