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IGCSE Mathematics 0580 Paper 4 (Extended) Mock Paper (10 questions, 100 marks)
IGCSE Mathematics
IGCSE Mathematics 0580 Paper 4 (Extended) mock paper: 10 questions, 100 marks, 2 hours 30 minutes, covering standard form, algebra, ratio, geometry, trigonometry, statistics, probability, sequences, quadratic functions and mensuration, with full worked answers and marking notes.
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Paper details Level: IGCSE (Year 10-11) | Subject: Mathematics | Time: 2 hours 30 minutes | Marks: 100 | Questions: 10
All questions (100 marks) — Questions 1 to 10
Answer ALL questions. Write your answers in the spaces provided and show all working; give units where appropriate. Use π = 3.142 unless told otherwise.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (a) Write 0.000472 in standard form (a × 10n, 1 ≤ a < 10). [1] (b) Calculate (3.6 × 105) × (2.5 × 10-3), giving your answer in standard form. [2] (c) A car costs $186,000. Its price is increased by 7.5% and then reduced by 4%. Find the final price. [2] (d) Express 45 minutes as a fraction of 2 hours in its simplest form. [1] (e) $4,500 is invested at a simple interest rate of 3.2% per year for 2.5 years. Find the interest earned. [2] | (a) 4.72 × 10-4. [1] (b) 3.6 × 2.5 = 9.0 and 105 × 10-3 = 102, so 9.0 × 102 (= 900). [2] (c) 186,000 × 1.075 = $199,950; 199,950 × 0.96 = $191,952. [2] (d) 45 / 120 = 3/8. [1] (e) 4500 × 3.2% × 2.5 = $360. [2] |
| 2 | (a) Expand (2x − 3)(x + 5). [2] (b) Factorise 6x2 − 9x. [2] (c) Solve 5(x − 3) = 2x + 6. [2] (d) Given V = (1/3)πr2h, make r the subject. [2] (e) Solve x2 − 7x + 10 = 0. [2] | (a) 2x2 + 10x − 3x − 15 = 2x2 + 7x − 15. [2] (b) 3x(2x − 3). [2] (c) 5x − 15 = 2x + 6, so 3x = 21 and x = 7. [2] (d) r2 = 3V / (πh), so r = √(3V / (πh)) taking the positive root. [2] (e) (x − 2)(x − 5) = 0, so x = 2 or x = 5. [2] |
| 3 | A biscuit recipe uses 250 g of flour, 100 g of sugar and 150 g of butter to make 24 biscuits. (a) Find the mass of flour, sugar and butter needed to make 60 biscuits. [4] (b) y is directly proportional to x2 and y = 48 when x = 4. Find y when x = 7. [2] (c) 6 workers take 8 days to finish a job. How long would 4 workers take, working at the same rate? [2] (d) A map has a scale of 1 : 25,000 and a distance of 6.4 cm is measured on it. Find the real distance in km. [2] | (a) Scale factor = 60 / 24 = 2.5; flour 250 × 2.5 = 625 g; sugar 100 × 2.5 = 250 g; butter 150 × 2.5 = 375 g. [4] (b) y = kx2, so 48 = 16k and k = 3; when x = 7, y = 3 × 49 = 147. [2] (c) Total work = 6 × 8 = 48 worker-days, so 4 workers need 48 / 4 = 12 days. [2] (d) 6.4 × 25,000 = 160,000 cm = 1,600 m = 1.6 km. [2] |
| 4 | (a) The exterior angle of a regular polygon is 24°. Find the number of sides and the size of each interior angle. [2] (b) The angles of a triangle are (2x + 10)°, (3x − 10)° and x°. Find x and the three angles. [2] (c) The angle at the centre AOB and the angle at the circumference ACB stand on the same arc. Given ∠ACB = 68°, find ∠AOB. [2] (d) In a cyclic quadrilateral ABCD, ∠A = 115°. Find ∠C. [1] (e) The bearing of B from A is 118°. Find the bearing of A from B. [3] | (a) Number of sides = 360 / 24 = 15; each interior angle = 180 − 24 = 156°. [2] (b) (2x + 10) + (3x − 10) + x = 180, so 6x = 180 and x = 30; the angles are 70°, 80° and 30°. [2] (c) Angle at centre = 2 × angle at circumference = 2 × 68 = 136°. [2] (d) Opposite angles of a cyclic quadrilateral add to 180°, so ∠C = 180 − 115 = 65°. [1] (e) Back bearing = 118 + 180 = 298°. [3] |
| 5 | (a) In a right-angled triangle the hypotenuse is 13 cm and the opposite side is 5 cm. Find the acute angle θ to 1 decimal place. [2] (b) A ladder 7.5 m long makes an angle of 68° with the ground. Find the height of the top of the ladder, to 3 significant figures. [2] (c) A ship sails 12 km east and then 9 km north. Find its distance from the start and the bearing of the ship from the start. [3] (d) Two sides of a triangle are 8 cm and 11 cm and the included angle is 60°. Find the third side to 3 significant figures. [3] | (a) sin θ = 5 / 13, so θ = 22.6°. [2] (b) Height = 7.5 × sin 68° = 6.95 m. [2] (c) Distance = √(122 + 92) = √225 = 15 km; bearing = tan-1(12 / 9) = 53.1°, that is 053°. [3] (d) a2 = 64 + 121 − 2 × 8 × 11 × cos 60° = 185 − 88 = 97, so a = 9.85 cm. [3] |
| 6 | The daily reading times, in minutes, of ten students are: 12, 15, 11, 18, 15, 20, 14, 15, 17, 13. (a) Find the mean. [2] (b) Find the median. [1] (c) Find the mode and the range. [2] (d) Another class of 15 students has a mean of 18 minutes. Find the mean of all 25 students. [3] (e) Which average, mean or median, better describes a typical reading time here? Give a reason. [2] | (a) Total = 150, so mean = 150 / 10 = 15 minutes. [2] (b) In order: 11, 12, 13, 14, 15, 15, 15, 17, 18, 20; median = (15 + 15) / 2 = 15 minutes. [1] (c) Mode = 15 minutes; range = 20 − 11 = 9 minutes. [2] (d) Combined total = 150 + 15 × 18 = 420, so mean = 420 / 25 = 16.8 minutes. [3] (e) Both are 15 minutes and there are no extreme values, so both work; the mode (15 minutes, occurring 3 times) also describes it, but the mean uses all the data. [2] |
| 7 | A bag contains 5 red, 3 blue and 2 green balls. One ball is taken at random. (a) Find the probability that it is red. [1] (b) Two balls are taken without replacement. Find the probability that both are red. [2] (c) Two balls are taken without replacement. Find the probability that at least one is red. [3] (d) A ball is taken, its colour noted, and it is replaced; then a second ball is taken. Find the probability of blue then green. [2] (e) The ball is drawn 200 times with replacement. How many red balls are expected? [2] | (a) 5 / 10 = 1/2. [1] (b) 5/10 × 4/9 = 20/90 = 2/9. [2] (c) P(no red) = 5/10 × 4/9 = 2/9, so P(at least one red) = 1 − 2/9 = 7/9. [3] (d) 3/10 × 2/10 = 6/100 = 0.06. [2] (e) 200 × 1/2 = 100 times. [2] |
| 8 | (a) Find the n-th term of the sequence 7, 12, 17, 22, … [2] (b) Find the 10th term of this sequence. [1] (c) Is 103 a term of the sequence? Explain. [2] (d) The sequence 2, 7, 14, 23, 34, … has a constant second difference. Find its n-th term. [3] (e) Find the sum of the first 20 terms of 7, 12, 17, … [2] | (a) Common difference 5 and first term 7, so the n-th term is 5n + 2. [2] (b) 10th term = 5 × 10 + 2 = 52. [1] (c) 5n + 2 = 103 gives n = 20.2, which is not an integer, so 103 is not a term. [2] (d) First differences 5, 7, 9, 11; the second difference is 2, so the n-th term is n2 + bn + c. From n = 1: 1 + b + c = 2 and n = 2: 4 + 2b + c = 7, giving b = 2 and c = −1, so the n-th term is n2 + 2n − 1 (check: 5th term = 34). [3] (e) Sum = 20/2 × (2 × 7 + 19 × 5) = 10 × 109 = 1,090. [2] |
| 9 | Given that f(x) = x2 − 4x + 3. (a) Write f(x) in the form (x − h)2 + k and state the minimum point of the graph. [3] (b) Solve f(x) = 0. [2] (c) Find f(−2). [1] (d) Solve f(x) = 8. [3] (e) State the line of symmetry and the y-intercept of the graph. [2] | (a) x2 − 4x + 3 = (x − 2)2 − 4 + 3 = (x − 2)2 − 1; the minimum point is (2, −1). [3] (b) (x − 1)(x − 3) = 0, so x = 1 or x = 3. [2] (c) f(−2) = 4 + 8 + 3 = 15. [1] (d) x2 − 4x + 3 = 8, so x2 − 4x − 5 = 0 and (x − 5)(x + 1) = 0, giving x = 5 or x = −1. [3] (e) Line of symmetry x = 2; y-intercept 3, that is the point (0, 3). [2] |
| 10 | Take π = 3.142. (a) A cylinder has radius 7 cm and height 10 cm. Find its volume in cm3 to 3 significant figures. [2] (b) A cone has radius 7 cm and height 24 cm. Find its slant height, its volume and its curved surface area. [5] (c) A sphere has radius 6 cm. Find its surface area and volume to 3 significant figures. [2] (d) A solid hemisphere has radius 6 cm. Find its total surface area. [2] | (a) V = π × 72 × 10 = 1539.58 ≈ 1,540 cm3. [2] (b) Slant height l = √(72 + 242) = √625 = 25 cm; V = (1/3) × π × 49 × 24 = 1231.66 ≈ 1,230 cm3; curved area = π × 7 × 25 = 549.85 ≈ 550 cm2. [5] (c) Surface area = 4 × π × 62 = 452.45 ≈ 452 cm2; volume = (4/3) × π × 63 = 904.90 ≈ 905 cm3. [2] (d) Total surface area = 3 × π × 62 = 339.34 ≈ 339 cm2 (curved surface plus the flat circle). [2] |
A TutorZone practice paper; original questions and passages, not publisher material.
