HKDSE Mathematics (Compulsory Part) Paper 1 Long Questions Practice Paper | 8 questions, 100 marks, 2 hours 15 minutes

2026-10-31

HKDSE Mathematics Paper 1

HKDSE Mathematics Paper 1 long questions practice paper: 8 questions, 100 marks, 2 hours 15 minutes, with full worked solutions and marks per part.

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Paper details　Level: S6 (HKDSE) | Subject: Mathematics | Time: 2 hours 15 minutes | Marks: 100 | Questions: 8

Section A (48 marks)

Long questions: show your working and give units in your answers.

#	Question	Answer and marking notes
1	(a) Simplify (4x³y⁻²)² ÷ (8x²y⁻³) and express the answer with positive indices. (2 marks) (b) Solve 3^2x−1 = 27^x−2. (3 marks) (c) Given 1/x + 1/y = 1/z, where x ≠ 0, y ≠ 0 and x ≠ z, express y in terms of x and z. (3 marks) (d) Solve 2x² − 7x − 15 = 0. (4 marks)	(a) (16x⁶y⁻⁴) ÷ (8x²y⁻³) = 2 x⁴ y⁻¹ = 2 x⁴ / y [2] (b) 3^2x−1 = 3^3x−6 ⟹ 2x − 1 = 3x − 6 ⟹ x = 5 [3] (c) 1/y = 1/z − 1/x = (x − z)/xz ⟹ y = xz/(x − z) [3] (d) (2x + 3)(x − 5) = 0 ⟹ x = −1.5 or x = 5 [4]
2	(a) A jacket marked at $640 is sold at a 25% discount. Find the selling price. (2 marks) (b) The cost of the jacket is $380. Find the percentage profit, correct to 3 significant figures. (3 marks) (c) $5,000 is deposited at 4% p.a. compounded yearly. Find the interest earned after 3 years. (3 marks) (d) A machine worth $80,000 depreciates by 15% each year. Find its value after 2 years. (4 marks)	(a) 640 × 0.75 = $480 [2] (b) (480 − 380)/380 × 100% = 26.3% [3] (c) 5,000 × 1.04³ − 5,000 = $624.32 [3] (d) 80,000 × 0.85² = $57,800 [4]
3	A(−2, 3) and B(6, 11) are two points on a rectangular coordinate plane, and O is the origin. (a) Find the length of AB, correct to 3 significant figures. (3 marks) (b) Find the coordinates of the mid-point M of AB. (2 marks) (c) Find the equation of the perpendicular bisector of AB. (4 marks) (d) Find the area of triangle OAB. (3 marks)	(a) √(8² + 8²) = √128 = 11.3 [3] (b) M = ((−2 + 6)/2, (3 + 11)/2) = (2, 7) [2] (c) slope of AB = 8/8 = 1, so the perpendicular bisector has slope −1; through M: y − 7 = −(x − 2) ⟹ y = −x + 9 [4] (d) ½·\|(−2)(11) − (6)(3)\| = ½·40 = 20 [3]
4	Let f(x) = 2x² − 12x + 13. (a) By completing the square, express f(x) in the form a(x − h)² + k, and write down the coordinates of the vertex. (4 marks) (b) Write down the axis of symmetry of y = f(x). (2 marks) (c) Solve f(x) = 0, correct to 3 significant figures. (3 marks) (d) Find the range of k such that f(x) = k has two distinct real roots. (3 marks)	(a) f(x) = 2(x² − 6x) + 13 = 2[(x − 3)² − 9] + 13 = 2(x − 3)² − 5; vertex (3, −5) [4] (b) axis of symmetry: x = 3 [2] (c) x = (12 ± √40)/4 = 3 ± √10/2 ⟹ x ≈ 4.58 or 1.42 [3] (d) the minimum value is −5, so f(x) = k has two distinct real roots when k > −5 [3]

Section B (52 marks)

Long questions: show your working and give units in your answers.

#	Question	Answer and marking notes
5	In triangle ABC, AB = 12 cm, AC = 9 cm and ∠BAC = 65°. (a) Find BC, correct to 3 significant figures. (3 marks) (b) Find the area of triangle ABC, correct to 3 significant figures. (3 marks) (c) Find ∠ABC, correct to the nearest 0.1°. (4 marks) (d) Find the shortest distance from A to BC, correct to 3 significant figures. (3 marks)	(a) BC² = 12² + 9² − 2(12)(9)cos 65° = 133.71 ⟹ BC = 11.6 cm [3] (b) area = ½(12)(9)sin 65° = 48.9 cm² [3] (c) sin∠ABC/9 = sin 65°/11.6 ⟹ ∠ABC = 44.9° [4] (d) shortest distance = 2 × area ÷ BC = 8.46 cm [3]
6	The test scores of nine students are 12, 15, 15, 18, 20, 22, 25, 28, 30. (a) Find the mean, correct to 3 significant figures. (2 marks) (b) Find the median. (2 marks) (c) Find the range and the inter-quartile range. (4 marks) (d) Find the standard deviation, correct to 3 significant figures. (3 marks) (e) If 3 is added to every score, write down the new mean and the new standard deviation. (2 marks)	(a) (185)/9 = 20.6 [2] (b) the 5th score in order = 20 [2] (c) range = 30 − 12 = 18; Q₁ = 15.0, Q₃ = 26.5 ⟹ IQR = 11.5 [4] (d) standard deviation = √(Σ(x − mean)²/9) = 5.85 [3] (e) new mean = 23.6; standard deviation = 5.85 (unchanged) [2]
7	A bag contains 5 red, 3 blue and 2 green balls, 10 balls in total. (a) A ball is drawn at random. Find the probability that it is red. (2 marks) (b) Two balls are drawn one after another without replacement. Find the probability that both are red. (3 marks) (c) Two balls are drawn one after another without replacement. Find the probability that at least one is blue. (4 marks) (d) Find the number of different arrangements of all nine letters T, U, T, O, R, Z, O, N, E. (4 marks)	(a) P(red) = 5/10 = 1/2 [2] (b) (5/10) × (4/9) = 2/9 [3] (c) 1 − C(7,2)/C(10,2) = 1 − 21/45 = 8/15 [4] (d) 9!/(2!2!) = 90,720 [4]
8	The first term of an arithmetic sequence is 7 and the common difference is 4. (a) Find the 10th term. (2 marks) (b) Find the sum of the first 20 terms. (3 marks) (c) Another geometric sequence has first term 3 and common ratio 2. Find its 6th term. (3 marks) (d) For the arithmetic sequence in (a), find the least n such that the sum of the first n terms exceeds 1000. (5 marks)	(a) T₁₀ = 7 + 9(4) = 43 [2] (b) S₂₀ = (20/2)[2(7) + 19(4)] = 900 [3] (c) T₆ = 3 × 2⁵ = 96 [3] (d) n(4n + 10)/2 > 1000 ⟹ 2n² + 5n − 1000 > 0; n = 21 gives 987 (not enough) while n = 22 gives 1078 ⟹ least n = 22 [5]

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