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HKDSE Mathematics (Compulsory Part) Paper 2 MCQ Practice Paper | 45 questions, 45 marks, 75 minutes
HKDSE Mathematics Paper 2
HKDSE Mathematics Compulsory Part Paper 2 MCQ practice: 45 questions, 45 marks, 75 minutes on algebra, geometry, trigonometry, statistics and probability, with fully worked answers.
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Paper details Level: S6 (HKDSE) | Subject: Mathematics | Time: 75 minutes | Marks: 45 | Questions: 45
Section A: Foundation (Q1–Q30)
Choose the best answer. Each question carries 1 mark.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Expand (2x + 3)(x − 4). A. 2x² − 5x − 12 B. 2x² + 5x − 12 C. 2x² − 11x − 12 D. 2x² − 5x + 12 |
A — 2x·x + 2x·(−4) + 3·x + 3·(−4) = 2x² − 8x + 3x − 12 = 2x² − 5x − 12. |
| 2 | Solve 3x − 7 = 2x + 5. A. x = 2 B. x = 6 C. x = 12 D. x = −12 |
C — 3x − 2x = 5 + 7, so x = 12. |
| 3 | Factorise x² − 49. A. (x + 7)(x − 7) B. (x − 7)² C. (x + 7)² D. (x − 49)(x + 1) |
A — Difference of squares: a² − b² = (a + b)(a − b), giving (x + 7)(x − 7). |
| 4 | If a = 2 and b = −3, find a² − b². A. −5 B. 5 C. 13 D. −13 |
A — a² − b² = 4 − 9 = −5. |
| 5 | Increase 80 by 15%. The result is: A. 92 B. 95 C. 104 D. 68 |
A — 80 × (1 + 15%) = 80 × 1.15 = 92. |
| 6 | A car travels 240 km in 3 hours. Its average speed is: A. 60 km/h B. 72 km/h C. 80 km/h D. 120 km/h |
C — Speed = distance ÷ time = 240 ÷ 3 = 80 km/h. |
| 7 | The mean of 12, 15, 18, 20 and 25 is: A. 16 B. 18 C. 19 D. 20 |
B — Sum = 90, and 90 ÷ 5 = 18. |
| 8 | The median of 3, 7, 9, 12, 15, 20 is: A. 9 B. 10.5 C. 11 D. 12 |
B — There are 6 values; average the middle two: (9 + 12) ÷ 2 = 10.5. |
| 9 | A fair die is rolled. The probability of an even number is: A. 1/6 B. 1/3 C. 1/2 D. 2/3 |
C — Even outcomes are 2, 4, 6: 3 ÷ 6 = 1/2. |
| 10 | A circle has radius 7 cm. Its area is (take π ≈ 22/7): A. 44 cm² B. 154 cm² C. 308 cm² D. 49 cm² |
B — A = πr² = 22/7 × 49 = 154 cm². |
| 11 | An isosceles triangle has apex angle 40°. Each base angle is: A. 50° B. 60° C. 70° D. 80° |
C — (180 − 40) ÷ 2 = 70°. |
| 12 | A right-angled triangle has legs 3 and 4. Its hypotenuse is: A. 5 B. 6 C. 7 D. 12 |
A — √(3² + 4²) = √25 = 5. |
| 13 | The value of sin 30° is: A. 0 B. 1/2 C. √2/2 D. 1 |
B — Standard angle: sin 30° = 1/2. |
| 14 | The value of tan 45° is: A. 0 B. 1/2 C. 1 D. √3 |
C — tan 45° = 1. |
| 15 | For the line y = 2x + 3, the slope and y-intercept are: A. slope 3, intercept 2 B. slope 2, intercept 3 C. slope 2, intercept −3 D. slope −2, intercept 3 |
B — In y = mx + c, m = 2 and c = 3. |
| 16 | Solve the inequality 3x + 4 < 19. A. x < 5 B. x > 5 C. x < 7 D. x > 7 |
A — 3x < 15, so x < 5. |
| 17 | Simplify (a²)³. A. a⁵ B. a⁶ C. a⁸ D. a⁹ |
B — Multiply indices: 2 × 3 = 6, giving a⁶. |
| 18 | Simplify √72. A. 6√2 B. 8√2 C. 12√6 D. 36√2 |
A — √72 = √(36 × 2) = 6√2. |
| 19 | 0.000 42 in scientific notation is: A. 4.2 × 10⁻⁴ B. 4.2 × 10⁻³ C. 42 × 10⁻⁵ D. 4.2 × 10⁴ |
A — The point moves 4 places, giving 4.2 × 10⁻⁴. |
| 20 | A cuboid measures 4 by 5 by 6. Its total surface area is: A. 74 B. 120 C. 148 D. 296 |
C — 2(4×5 + 5×6 + 4×6) = 2 × 74 = 148; 120 is the volume trap. |
| 21 | $5 000 at 4% simple interest for 3 years earns: A. $200 B. $600 C. $640 D. $1 200 |
B — I = PRT = 5 000 × 0.04 × 3 = $600. |
| 22 | Solve x² − 5x + 6 = 0. A. x = 2 or 3 B. x = −2 or −3 C. x = 1 or 6 D. x = −1 or 6 |
A — (x − 2)(x − 3) = 0, so x = 2 or 3. |
| 23 | The 10th term of the arithmetic sequence 3, 7, 11, … is: A. 35 B. 39 C. 40 D. 43 |
B — a = 3, d = 4, T₁₀ = 3 + 9 × 4 = 39. |
| 24 | The distance between (1, 2) and (4, 6) is: A. 3 B. 5 C. 7 D. 25 |
B — √(3² + 4²) = 5. |
| 25 | The midpoint of (2, 3) and (6, 11) is: A. (4, 7) B. (8, 14) C. (3, 5) D. (4, 8) |
A — ((2 + 6)/2, (3 + 11)/2) = (4, 7). |
| 26 | An item marked $400 is discounted 20% then 10%. The selling price is: A. $280 B. $288 C. $300 D. $320 |
B — 400 × 0.8 × 0.9 = $288; the discounts do not add to 30%. |
| 27 | If a : b = 2 : 3 and b : c = 4 : 5, then a : c = A. 2 : 5 B. 8 : 15 C. 3 : 4 D. 5 : 6 |
B — Make b common: a : b = 8 : 12 and b : c = 12 : 15, so a : c = 8 : 15. |
| 28 | A sector has radius 6 cm and angle 60°. Its area is (π ≈ 3.1416): A. 9.42 cm² B. 18.85 cm² C. 37.70 cm² D. 113.1 cm² |
B — 60/360 × π × 6² = 6π ≈ 18.85 cm². |
| 29 | A cylinder has base radius 3 cm and height 10 cm. Its volume is (π ≈ 3.1416): A. 94.2 cm³ B. 188.5 cm³ C. 282.7 cm³ D. 565.5 cm³ |
C — V = πr²h = π × 9 × 10 ≈ 282.7 cm³. |
| 30 | An arc of a circle has radius 10 cm and angle 72° (π ≈ 3.1416). Its length is: A. 6.28 cm B. 12.57 cm C. 20.94 cm D. 62.83 cm |
B — 72/360 × 2π × 10 = 4π ≈ 12.57 cm. |
Section B: Advanced (Q31–Q45)
Choose the best answer. Each question carries 1 mark.
| # | Question | Answer and marking notes |
|---|---|---|
| 31 | Solve 2x² + 5x − 3 = 0. A. x = 0.5 or −3 B. x = −0.5 or 3 C. x = 1 or −1.5 D. x = 3 or 0.5 |
A — Δ = 25 + 24 = 49, x = (−5 ± 7)/4, giving 0.5 or −3. |
| 32 | The minimum value of f(x) = x² − 4x + 1 is: A. −4 B. −3 C. 1 D. 3 |
B — Completing the square: (x − 2)² − 3, minimum −3. |
| 33 | The vertex of the parabola y = x² − 6x + 5 is: A. (3, −4) B. (−3, −4) C. (3, 4) D. (6, 5) |
A — x = −b/2a = 3, y = 9 − 18 + 5 = −4. |
| 34 | The value of log₂ 32 is: A. 4 B. 5 C. 6 D. 16 |
B — 2⁵ = 32, so log₂ 32 = 5. |
| 35 | Solve 2^(x + 1) = 8^(x − 1). A. x = 1 B. x = 2 C. x = 3 D. x = 4 |
B — 8 = 2³, so x + 1 = 3x − 3, x = 2. |
| 36 | Find 1 + 2 + 3 + … + 50. A. 1 225 B. 1 275 C. 2 500 D. 2 550 |
B — n(n + 1)/2 = 50 × 51/2 = 1 275. |
| 37 | The 5th term of the geometric sequence 2, 6, 18, … is: A. 54 B. 108 C. 162 D. 486 |
C — Common ratio 3, T₅ = 2 × 3⁴ = 162. |
| 38 | The number of ways to choose 3 people from 8 is: A. 24 B. 56 C. 112 D. 336 |
B — C(8,3) = 8!/(3!5!) = 56. |
| 39 | A bag holds 3 red and 2 white balls. Two are drawn at random. P(both red) = A. 1/5 B. 3/10 C. 2/5 D. 3/5 |
B — C(3,2)/C(5,2) = 3/10. |
| 40 | In a triangle A = 60°, B = 45° and a = 10. Then b ≈ A. 7.07 B. 8.16 C. 10.0 D. 12.2 |
B — Sine rule: b = a sinB/sinA = 10 × 0.7071/0.8660 ≈ 8.16. |
| 41 | Ten numbers have mean 20. Adding 31 gives a new mean of: A. 20.5 B. 21 C. 21.5 D. 25.5 |
B — (200 + 31)/11 = 21. |
| 42 | The variance of 2, 4, 6, 8 is: A. 2 B. 4 C. 5 D. 10 |
C — Mean 5, sum of squared deviations 20, ÷4 = 5. |
| 43 | A cyclic quadrilateral has one angle 68°. The opposite angle is: A. 68° B. 102° C. 112° D. 132° |
C — Opposite angles are supplementary: 180 − 68 = 112°. |
| 44 | Two similar triangles have corresponding sides in ratio 3 : 4. Their areas are in ratio: A. 3 : 4 B. 6 : 8 C. 9 : 16 D. 27 : 64 |
C — Area ratio = (side ratio)² = 9 : 16. |
| 45 | $1 000 at 10% p.a. compounded yearly for 2 years amounts to: A. $1 100 B. $1 200 C. $1 210 D. $1 220 |
C — 1 000 × 1.1² = $1 210. |
A TutorZone practice paper; original questions and passages, not publisher material.
