S5 Maths Paper: Indices, Logarithms, Sequences, Quadratics, Probability And Statistics (100 Marks)

S5 Maths Paper: Indices, Logarithms, Sequences, Quadratics, Probability and Statistics (100 marks)

2026-11-11

S5 Maths – sequences, probability and statistics

S5 Maths paper: indices and logarithms, geometric and arithmetic sequences, quadratics, probability, standard deviation and a savings plan, with full answers.

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Paper details　S5 Maths practice paper with six structured questions on indices and logarithms, geometric and arithmetic sequences, quadratic equations, probability, standard deviation and a savings plan, with full answers and marking points.

Question 1 Indices and logarithms (18 marks)

Show your working in all parts. Give approximations correct to 3 significant figures.

#	Question	Answer and marking notes
1	(a) Simplify (2^3 x 2^-5) / 2^-4.	Answer: add the indices: 2^(3 – 5 + 4) = 2^2 = 4.
2	(b) Solve log2 x + log2 (x – 2) = 3.	Answer: combine to get log2[x(x – 2)] = 3, so x(x – 2) = 8, giving x^2 – 2x – 8 = 0, x = 4 or x = -2; x must be positive, so x = 4.
3	(c) Solve 3^(2x) – 10 x 3^x + 9 = 0.	Answer: let y = 3^x, then y^2 – 10y + 9 = 0, so y = 1 or y = 9, i.e. 3^x = 1 or 3^x = 9, hence x = 0 or x = 2.
4	(d) Use the change of base formula to find log4 8.	Answer: log4 8 = log 8 / log 4 = 1.5.

Question 2 Geometric sequences (17 marks)

A geometric sequence has first term 5 and common ratio 3.

#	Question	Answer and marking notes
5	(a) Find the 6th term of the sequence.	Answer: T6 = a*r^5 = 5 x 3^5 = 1215.
6	(b) Find the sum of the first 6 terms.	Answer: S6 = a(r^6 – 1)/(r – 1) = 5 x 728 / 2 = 1820.
7	(c) Another geometric sequence has first term 5 and common ratio 1/3. Find its sum to infinity.	Answer: since \|r\| is less than 1, S = a/(1 – r) = 5 / (2/3) = 7.5.
8	(d) Determine whether 405 is a term of the sequence in this question, and state its position.	Answer: solve 5 x 3^n = 405, so 3^n = 81, n = 4, hence 405 is the 5th term.

Question 3 Quadratic equations and functions (16 marks)

The parts are independent.

#	Question	Answer and marking notes
9	(a) If x^2 + kx + 9 = 0 has a repeated root, find the positive value of k.	Answer: discriminant k^2 – 36 = 0, so k = 6 or k = -6; the positive value is k = 6.
10	(b) Solve 2x^2 – 7x + 3 = 0.	Answer: discriminant = 49 – 24 = 25, x = (7 +/- 5) / 4, so x = 3 or x = 0.5.
11	(c) Find the vertex and the axis of symmetry of y = x^2 – 6x + 5.	Answer: completing the square gives y = (x – 3)^2 -4, vertex (3, -4), axis x = 3.
12	(d) Two numbers have sum 14 and product 45. Find them.	Answer: they are roots of t^2 – 14t + 45 = 0, discriminant = 16, so the numbers are 9 and 5.

Question 4 Probability (17 marks)

A bag contains 4 red, 3 blue and 5 green balls, 12 in total, identical apart from colour.

#	Question	Answer and marking notes
13	(a) One ball is drawn at random. Find the probability that it is red.	Answer: P = 4/12 = 1/3 = 0.3333.
14	(b) Two balls are drawn without replacement. Find the probability that both are red.	Answer: P = (4/12) x (3/11) = 1/11 = 0.0909.
15	(c) Two balls are drawn without replacement. Find the probability that at least one is green.	Answer: P(no green) = (7/12) x (6/11), so P(at least one green) = 1 – 42/132 = 15/22 = 0.6818.
16	(d) Given that the first ball drawn is red, find the probability that the second is blue.	Answer: 11 balls remain, of which 3 are blue, P = 3/11 = 0.2727.

Question 5 Statistics (16 marks)

The test scores of five students are 4, 7, 7, 9 and 13.

#	Question	Answer and marking notes
17	(a) Find the mean.	Answer: total = 40, mean = 40 / 5 = 8.
18	(b) Find the median.	Answer: after arranging in order the middle value is 7.
19	(c) Find the standard deviation, correct to 3 significant figures.	Answer: sum of squared deviations = 44, variance = 44 / 5 = 8.8, standard deviation = 2.9665.
20	(d) If every score is increased by 3, state the new mean and the new standard deviation.	Answer: new mean = 11; the standard deviation is unchanged by a shift, still 2.9665.

Question 6 Arithmetic sequence and a savings plan (16 marks)

Chi Ming saves $500 in the first month and increases the amount by $80 each month.

#	Question	Answer and marking notes
21	(a) Find the amount saved in the 12th month.	Answer: T12 = 500 + 11 x 80 = $1380.
22	(b) Find the total saved in the first 12 months.	Answer: S12 = 12 x (500 + 1380) / 2 = $11280.
23	(c) If he wants to accumulate $20000, find the least number of months needed.	Answer: Sn = n(2 x 500 + (n – 1) x 80) / 2; n = 17 gives $19380 and n = 18 gives $21240, so at least 18 months.
24	(d) Explain why this arithmetic model is unsuitable for predicting savings ten years later.	Answer: the model assumes a fixed monthly increase forever, but income and expenses change, so long term projections would overestimate or underestimate.

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