S5 Mathematics (Compulsory Part) Practice Paper: Indices, Logarithms, Sequences, Quadratic Functions And Probability (12 Questions, 100 Marks, 75 Minutes, Answers Included)

S5 Mathematics (Compulsory Part) Practice Paper: Indices, Logarithms, Sequences, Quadratic Functions and Probability (12 Questions, 100 Marks, 75 Minutes, Answers Included)

2026-11-06

HKDSE Mathematics (Compulsory Part): Indices, Logarithms, Sequences, Quadratics and Probability

S5 Maths (Compulsory Part) paper: 12 questions, 100 marks, 75 minutes on indices, logarithms, sequences, quadratic graphs, circles, probability and standard deviation.

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Section A: Indices, Logarithms and Sequences (24 marks)

Each question carries 6 marks; show your working or reasoning.

#	Question	Answer and marking notes
1	Solve the following equations: (a) 3^(2x-1) = 81; (b) log_5 (2x+3) = 2. State the index or logarithm law used at each step.	(a) 81 = 3^4, so 2x – 1 = 4 and x = 2.5. (b) log_5 (2x+3) = 2 means 2x + 3 = 5^2 = 25, so x = 11. Marking points: same base comparison, conversion between index and logarithm, answers.
2	Given log 2 = a and log 3 = b. (a) Express log 72 in terms of a and b. (b) Express log 4.5 in terms of a and b. (c) Explain why the coefficients of log 72 come from the prime factorisation of 72.	(a) 72 = 2^3 x 3^2, so log 72 = 3a + 2b, the coefficients following the powers 3 and 2. (b) log 4.5 = log(9/2) = 2 log 3 – log 2 = 2b – a. (c) A logarithm turns multiplication into addition, so each prime power becomes a coefficient. Marking points: factorisation, both expressions, explanation.
3	An arithmetic sequence has first term 7 and common difference 4. (a) Find the 20th term. (b) Find the sum of the first 20 terms. (c) Find the least n for which the sum of the first n terms exceeds 900.	(a) 20th term = 7 + 19 x 4 = 83. (b) Sum of first 20 terms = 20(7 + 83)/2 = 900. (c) The sum of the first n terms is n(2 x 7 + (n-1) x 4)/2: n = 15 gives 570, n = 18 gives 792, n = 19 gives 874 and n = 20 gives 900, so the least n is 20. Marking points: general term, sum formula, comparison.
4	A geometric sequence has first term 3 and common ratio 2. (a) Find the sum of the first 10 terms. (b) Find the least n for which the nth term exceeds 1000, and state that term.	(a) Sum of first 10 terms = 3(2^10 – 1)/(2 – 1) = 3069. (b) The nth term is 3 x 2^(n-1): 2^8 = 256 gives 768, which does not exceed 1000, while 2^9 = 512 gives 1536, so the least n is 10 and that term is 1536. Marking points: geometric sum, successive comparison, answer.

Section B: Quadratic Functions, Lines and Circles (36 marks)

Each question carries 9 marks; note the marks for each part.

#	Question	Answer and marking notes
5	The curve is y = x^2 – 6x + 5. (a) Express it in the form (x – h)^2 + k. (b) State the vertex and the minimum value. (c) Find the x-intercepts and the y-intercept.	(a) x^2 – 6x + 5 = (x – 3)^2 – 4. (b) The vertex is (3, -4) and the minimum value is -4. (c) The x-intercepts are x = 1 and x = 5; the y-intercept is y = 5. Marking points: completing the square, vertex, intercepts.
6	The line L is 3x + 4y – 12 = 0. (a) Find its slope. (b) Find its x-intercept and y-intercept. (c) Find the perpendicular distance from the origin to L.	(a) 4y = -3x + 12, so y = -(3/4)x + 3 and the slope is -3/4. (b) The x-intercept is (4, 0) and the y-intercept is (0, 3). (c) Distance = 12/sqrt(3^2 + 4^2) = 12/5 = 2.4. Marking points: slope-intercept form, both intercepts, point-line distance formula.
7	The circle C has equation x^2 + y^2 – 4x + 6y – 12 = 0. (a) Find its centre and radius. (b) Show that the point (6, 0) lies on C. (c) Find the length of the chord that C cuts on the x-axis.	(a) x^2 – 4x + y^2 + 6y = 12, so (x – 2)^2 + (y + 3)^2 = 4 + 9 + 12 = 25; the centre is (2, -3) and the radius is 5. (b) Substituting (6, 0): 36 + 0 – 24 + 0 – 12 = 0, so the point lies on C. (c) Putting y = 0 gives x^2 – 4x – 12 = 0, so x = 6 or x = -2 and the chord length is 8. Marking points: completing the square, substitution check, chord length.
8	The equation is x^2 – 8x + k = 0. (a) Find the range of k for two distinct real roots. (b) If both roots are positive, find the range of k. (c) Solve the equation when k = 12.	(a) The discriminant is 64 – 4k, which must be positive, so k is less than 16. (b) The sum of the roots is 8, which is positive, and the product is k, which must be positive, so 0 is less than k and k is less than 16. (c) x^2 – 8x + 12 = 0 gives (x – 2)(x – 6) = 0, so x = 2 or x = 6. Marking points: discriminant, sum and product of roots, factorisation.

Section C: Probability and Statistics (40 marks)

Each question carries 10 marks; give full working or explanations.

#	Question	Answer and marking notes
9	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 they are of the same colour. (c) Find the probability that at least one is red.	(a) Both red = (5/10) x (4/9) = 2/9. (b) Same colour = (5 x 4 + 3 x 2 + 2 x 1)/(10 x 9) = 14/45. (c) At least one red = 1 – both non-red = 1 – (5 x 4)/(10 x 9) = 7/9. Marking points: without replacement, adding mutually exclusive cases, complement.
10	The data set is 2, 4, 6, 8, 10. (a) Find the mean, median and mode. (b) Find the range. (c) Find the variance and standard deviation. (d) If 5 is added to every value, describe how the mean, variance and standard deviation change.	(a) Mean = 30/5 = 6, median = 6, and there is no mode because each value occurs once. (b) Range = 10 – 2 = 8. (c) Variance = 40/5 = 8 and standard deviation = sqrt(8) = 2.83. (d) Adding 5 to every value makes the mean 11; the variance and standard deviation are unchanged because a shift does not change spread. Marking points: three averages, range, variance and standard deviation, shift property.
11	The marks of an examination are normally distributed with mean 60 and standard deviation 10. (a) State the percentages covered by one and two standard deviations from the mean. (b) Estimate the percentage of candidates scoring above 80. (c) The top 16% gain grade A; estimate the boundary mark.	(a) One standard deviation from the mean covers about 68%, two cover about 95% and three cover about 99.7%. (b) 80 is two standard deviations above the mean, so about (100% – 95%)/2 = 2.5% score above 80. (c) About 84% score below 70, one standard deviation above the mean, so the grade A boundary is about 70 marks. Marking points: empirical rule, tail percentage, boundary.
12	A team of 3 is chosen from 6 boys and 4 girls. (a) Find the number of possible teams. (b) Find the number of teams containing at least one girl. (c) Find the probability that a team contains exactly two girls.	(a) Number of teams = C(10, 3) = 120. (b) All-boy teams = C(6, 3) = 20, so teams with at least one girl = 120 – 20 = 100. (c) Exactly two girls and one boy = C(4, 2) x C(6, 1) = 6 x 6 = 36, so the probability is 36/120 = 3/10. Marking points: combination formula, complement, multiplication principle.

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