S5 Mathematics Practice Paper: Quadratic Equations, Indices And Logarithms, Sequences, Circle Properties, Probability And Statistics, Polynomials

S5 Mathematics Practice Paper: Quadratic Equations, Indices and Logarithms, Sequences, Circle Properties, Probability and Statistics, Polynomials

2026-11-12

S5 Mathematics – Compulsory Part

S5 Mathematics practice paper on quadratic discriminants, indices and logarithms, sequences, circle properties, probability and statistics. Show full working.

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Paper details　S5 Mathematics practice paper with model answers on quadratic equations, indices and logarithms, sequences, circle properties, probability and polynomials.

Part 1: Quadratic equations (16 marks)

Four questions. State the formula or theorem used and show your steps.

#	Question	Answer and marking notes
1	Solve x2 – 5x + 6 = 0 and state the value of the discriminant.	The discriminant is D = b2 – 4ac = 25 – 24 = 1, which is positive, so there are two distinct real roots. Using x = (5 +/- sqrt(1)) / 2 gives x = 3 or x = 2. Check: substituting x = 3 gives 9 – 15 + 6 = 0.
2	Find the values of k for which 2×2 + kx + 8 = 0 has a repeated root.	A repeated root requires a zero discriminant, so k2 – 4 x 2 x 8 = 0 and k2 = 64. Hence k = 8 or k = -8; both values give a repeated root.
3	Let a and b be the roots of 3×2 – 7x – 2 = 0. Find a + b and ab.	By the sum and product of roots, a + b = -b / a = 7/3 and ab = c / a = -2/3. A check with the quadratic formula, roots (7 +/- sqrt(73)) / 6, gives the same sum.
4	Find a quadratic equation with roots 3 and -5 and with coefficient 1 for x2.	The sum of the roots is -2 and the product is -15. The equation is x2 – (sum)x + (product) = 0, that is x2 + 2x – 15 = 0.

Part 2: Indices and logarithms (16 marks)

Four questions. Logarithms are base 10 unless stated; give answers to 3 significant figures.

#	Question	Answer and marking notes
5	Evaluate log2 32 and log3 (1/9).	Since 2 to the power 5 equals 32, log2 32 = 5. Since 3 to the power -2 equals 1/9, log3 (1/9) = -2.
6	Solve 2^(x+1) = 32.	Taking log base 2 of both sides gives x + 1 = 5, so x = 4. Check: 2 to the power 5 equals 32, as required.
7	Solve 3^(2x) = 27.	Since 27 = 3 to the power 3, 2x = 3 and x = 1.5. Check: 2 x 1.5 = 3, so the exponent matches.
8	Simplify log 8 + log 125 and state the logarithm law used.	Using log m + log n = log (mn), log 8 + log 125 = log 1000 = 3. Since 10 to the power 3 equals 1000, the value is 3.

Part 3: Sequences (18 marks)

Four questions. Write the general term before substituting.

#	Question	Answer and marking notes
9	An arithmetic sequence has first term 7 and common difference 4. Find the 10th term.	The general term is T(n) = a + (n – 1)d = 7 + 9 x 4 = 43.
10	Find the sum of the first 15 terms of that arithmetic sequence.	Using S(n) = n/2 x [2a + (n – 1)d], S(15) = 7.5 x [14 + 56] = 525. A check with the last term 63 gives 15 x (7 + 63) / 2 = 525.
11	A geometric sequence has first term 3 and common ratio 2. Find the 8th term and the sum of the first 10 terms.	The 8th term is 3 x 2^7 = 384. The sum of the first 10 terms is 3 x (2^10 – 1) / (2 – 1) = 3069.
12	A geometric sequence has first term 24 and common ratio 0.6. Find the sum to infinity.	Since the modulus of 0.6 is less than 1, the sum to infinity is a / (1 – r) = 24 / 0.4 = 60.

Part 4: Circle properties (16 marks)

Four questions. State the theorem used and show the calculation.

#	Question	Answer and marking notes
13	A circle has radius 13 cm and a chord is 5 cm from the centre. Find the length of the chord.	The perpendicular from the centre bisects the chord. With half chord h, h2 = 13^2 – 5^2 = 144, so h = 12 cm and the chord is 2 x 12 = 24 cm.
14	A circle has radius 8 cm and an external point P is 17 cm from the centre. Find the tangent length from P.	The tangent is perpendicular to the radius, so by Pythagoras the tangent length is sqrt(17^2 – 8^2) = sqrt(225) = 15 cm.
15	An angle at the circumference subtended by an arc is 35 degrees. Find the angle at the centre subtended by the same arc.	The angle at the centre is twice the angle at the circumference on the same arc, so it is 2 x 35 = 70 degrees.
16	What is the size of an angle in a semicircle? State the theorem and give a brief reason.	An angle in a semicircle is 90 degrees. Reason: the angle at the centre on a diameter is 180 degrees and the angle at the centre is twice the angle at the circumference, so the angle is 180 / 2 = 90 degrees.

Part 5: Probability and statistics (18 marks)

Four questions. Give probabilities as fractions in lowest terms and statistics to 3 significant figures.

#	Question	Answer and marking notes
17	A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability that both are red.	The first draw is red with probability 5/8, and the second with probability 4/7, so the probability is 5/8 x 4/7 = 5/14.
18	Continuing from the previous question, find the probability of at least one blue ball.	Using the complement, the probability of at least one blue is 1 – 5/14 = 9/14.
19	For the data 4, 7, 7, 9, 13, find the mean, median and variance (population form).	The mean is 40 / 5 = 8. The data are ordered, so the median is 7. The variance is [16 + 1 + 1 + 1 + 25] / 5 = 44/5 = 8.8.
20	Find the standard deviation of the data and the standard score of 13, to 3 significant figures.	The standard deviation is sqrt(8.8) = 2.966, that is 2.97 to 3 significant figures. The standard score is (13 – 8) / 2.9665 = 1.685, about 1.69, so 13 lies about 1.69 standard deviations above the mean.

Part 6: Polynomials (16 marks)

Four questions with f(x) = 2×3 – 3×2 – 11x + 6. Show the substitution.

#	Question	Answer and marking notes
21	Find f(2) and f(-1).	f(2) = 16 – 12 – 22 + 6 = -12 and f(-1) = -2 – 3 + 11 + 6 = 12.
22	Use the remainder theorem to find the remainder when f(x) is divided by (x – 3).	The remainder is f(3) = 54 – 27 – 33 + 6 = 0, which is zero, so (x – 3) is a factor of f(x).
23	Factorise f(x) completely into three linear factors.	From the previous part, f(x) = (x – 3)(2×2 + 3x – 2). Factorising the quadratic gives (2x – 1)(x + 2), so f(x) = (x – 3)(2x – 1)(x + 2).
24	Solve f(x) = 0.	Setting each factor to zero: x = 3, x = 0.5 and x = -2. These three values satisfy f(x) = 0.

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