IGCSE Mathematics Practice Paper: Algebra And Graphs

IGCSE Mathematics Practice Paper: Algebra and Graphs

2026-11-03

IGCSE Mathematics: Algebra and Graphs

IGCSE Mathematics practice paper: algebra and graphs, 12 questions, 100 marks, 75 minutes, covering expansion, factorising, quadratic equations and graphs of functions.

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Paper details　Level: S4-S5 | Subject: Mathematics | Time: 75 minutes | Marks: 100 | Questions: 12

Section A: Expansion, factorising and equations (Questions 1-4, 24 marks)

Answer all questions. Give answers to 3 significant figures unless stated otherwise, and show your working.

#	Question	Answer and marking notes
1	Expand (2x − 3)(x + 5) and simplify 6x⁴y² divided by 9x²y⁵. (6 marks)	(2x − 3)(x + 5) = 2x² + 10x − 3x − 15 = 2x² + 7x − 15 (3). 6x⁴y² divided by 9x²y⁵ = (2/3)x²y⁻³ = 2x² ÷ (3y³) (3).
2	Factorise (a) x² − 7x + 12, (b) 2x² + 5x − 3 and (c) 9x² − 16. (6 marks)	(a) (x − 3)(x − 4) (2). (b) 2x² + 6x − x − 3 = (2x − 1)(x + 3) (3). (c) (3x − 4)(3x + 4) (1).
3	Solve the simultaneous equations 3x + 2y = 13 and 5x − 2y = 11, then solve (2x − 1) ÷ 3 = (x + 4) ÷ 2. (6 marks)	Adding gives 8x = 24, so x = 3 (2); substituting gives y = 2 (2). Clearing denominators: 2(2x − 1) = 3(x + 4), so 4x − 2 = 3x + 12 and x = 14 (2).
4	Solve 2x² − 5x − 3 = 0 using the quadratic formula, giving exact answers, and hence factorise 2x² − 5x − 3. (6 marks)	With a = 2, b = −5 and c = −3 the discriminant is 25 + 24 = 49 (2); x = (5 ± 7) ÷ 4 gives x = 3 or x = −0.5 (2); hence 2x² − 5x − 3 = (2x + 1)(x − 3) (2).

Section B: Straight lines, quadratic graphs and functions (Questions 5-8, 32 marks)

Show the steps when solving equations, state the method used for an equation, and explain any rejected solution.

#	Question	Answer and marking notes
5	A(−1, 3) and B(3, 11) lie on the straight line L. (a) Find the gradient and the equation of L. (b) Find where L meets the x-axis. (c) Find the equation of the line through B perpendicular to L. (8 marks)	(a) gradient = (11 − 3) ÷ (3 + 1) = 2 (2); y − 3 = 2(x + 1) gives y = 2x + 5 (2). (b) y = 0 gives x = −2.5, so L meets the x-axis at (−2.5, 0) (1). (c) the perpendicular gradient is −0.5, so y − 11 = −0.5(x − 3) and y = −0.5x + 12.5 (3).
6	The curve y = x² − 4x + 1 is given. (a) Complete the square to find the vertex and the axis of symmetry. (b) Solve x² − 4x + 1 = 0, giving answers to 3 significant figures. (c) The line y = 2 meets the curve; find the x-coordinates of the points of intersection. (8 marks)	(a) y = (x − 2)² − 3, so the vertex is (2, −3) and the axis is x = 2 (3). (b) (x − 2)² = 3, x = 2 ± 1.73, so x ≈ 3.73 or 0.268 (3). (c) x² − 4x − 1 = 0, x = 2 ± 2.24, so x ≈ 4.24 or −0.236 (2).
7	Solve these inequalities: (a) 3 − 2x ≥ 7; (b) list every integer x for which 2x is greater than −3 and 2x ≤ 5; (c) find the range of x for which x² − 5x + 6 is negative. (8 marks)	(a) −2x ≥ 4; dividing by a negative reverses the sign, so x ≤ −2 (3). (b) x is greater than −1.5 and at most 2.5, so x = −1, 0, 1, 2 (3). (c) (x − 2)(x − 3) is negative, so x lies between 2 and 3 (2).
8	Given f(x) = 3x − 2 and g(x) = x² + 1, find (a) fg(2), (b) gf(2), (c) the inverse function f⁻¹(x), (d) the solution of f⁻¹(x) = 4. (8 marks)	(a) g(2) = 5 and f(5) = 13 (2). (b) f(2) = 4 and g(4) = 17 (2). (c) y = 3x − 2 gives x = (y + 2) ÷ 3, so the inverse is (x + 2) ÷ 3 (2). (d) (x + 2) ÷ 3 = 4 gives x = 10 (2).

Section C: Discriminants, exponential graphs and applications (Questions 9-12, 44 marks)

On graph questions label the coordinates of intersections and the vertex; in applied questions define the variables first.

#	Question	Answer and marking notes
9	(a) The quadratic equation x² + kx + 9 = 0 has equal roots; find the two possible values of k. (b) For k = 6 find the repeated root, and state where the curve y = x² + 6x + 9 touches the x-axis and its y-intercept. (11 marks)	(a) equal roots need discriminant k² − 36 = 0, so k = 6 or k = −6 (4). (b) for k = 6, x² + 6x + 9 = (x + 3)², so the repeated root is x = −3 (2); for k = −6 it is x = 3 (1); the curve touches the x-axis at (−3, 0) and its y-intercept is (0, 9) (4).
10	The curve y = a · b^x passes through (0, 3) and (2, 27). (a) Find a and b. (b) Find y when x = 4. (c) Find x when y = 1/9. (d) State the equation of the asymptote. (11 marks)	(a) at x = 0, y = a = 3 (2); at x = 2, 3b² = 27 so b² = 9 and b = 3 (2). (b) y = 3 × 3⁴ = 243 (2). (c) 3^(x + 1) = 3⁻², so x + 1 = −2 and x = −3 (3). (d) the asymptote is y = 0 (2).
11	(a) Make x the subject of y = (3x + 1) ÷ (x − 2). (b) Solve 3 ÷ (x − 1) + 2 ÷ (x + 2) = 1, giving answers to 3 significant figures. (c) State the values of x that must be excluded in (b). (11 marks)	(a) y(x − 2) = 3x + 1, so xy − 3x = 1 + 2y and x = (1 + 2y) ÷ (y − 3) (4). (b) multiplying by (x − 1)(x + 2) gives 3(x + 2) + 2(x − 1) = (x − 1)(x + 2), so 5x + 4 = x² + x − 2, x² − 4x − 6 = 0 and x = 2 ± √10 ≈ 5.16 or −1.16 (5). (c) x = 1 and x = −2 make a denominator zero (2).
12	A rectangle has perimeter 40 cm and sides x cm and (20 − x) cm. (a) Express the area A in terms of x. (b) Complete the square to find the greatest area and the value of x giving it. (c) Explain why x = 8 and x = 12 give the same area. (d) Find x when the area is 75 cm². (11 marks)	(a) A = x(20 − x) = 20x − x² (2). (b) A = −(x − 10)² + 100, so the greatest area is 100 cm² when x = 10 (4). (c) 8 and 12 are equally far from 10, so both give the same area of 96 cm² (2). (d) x² − 20x + 75 = 0 gives (x − 5)(x − 15) = 0, so x = 5 or 15 (3).

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