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Form 4 Mathematics Practice Paper: Trigonometry, Indices and Logarithms, Probability (12 questions, 100 marks, 75 minutes)
Form 4 Mathematics – 12 questions – 100 marks – 75 minutes
Form 4 Mathematics practice paper: 12 questions, 100 marks, 75 min; bearings, cosine rule, trig equations, 3D geometry, indices, logs, probability and standard deviation.
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Paper details Form 4 Mathematics practice paper with 12 structured questions, full worked solutions and a mark scheme (100 marks, 75 minutes).
Section A: Trigonometry
Answer all parts; show all working and units in calculations.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A boat leaves A on a bearing of 065° and sails 12 km to B, then turns and sails 9 km on a bearing of 155° to C. (a) Find the distance AC. (b) Find the bearing of C from A, to the nearest degree. (c) If the boat returns directly from C to A at 9 km/h, how long does the return journey take? |
(a) The two headings differ by 155° – 65° = 90°, so the angle at B is a right angle. AC² = 12² + 9² = 225, so AC = 15 km. (b) At A, tan θ = 9 / 12 = 0.75, giving θ ≈ 36.9°. The bearing of C is therefore 65° + 36.9° ≈ 102°. (c) Time = 15 / 9 = 1.667 h = 1 hour 40 minutes. |
| 2 | In triangle ABC, AB = 8 cm, BC = 11 cm and angle ABC = 118°. (a) Find AC, correct to 0.1 cm. (b) Find angle BAC, correct to 0.1°. (c) Find the area of triangle ABC, correct to 0.1 cm². |
(a) By the cosine rule, AC² = 8² + 11² – 2 × 8 × 11 × cos 118° = 185 + 82.6 = 267.6, so AC = 16.4 cm. (b) By the sine rule, sin BAC / 11 = sin 118° / 16.36, so sin BAC = 0.594 and angle BAC = 36.4°. (c) Area = ½ × 8 × 11 × sin 118° = 38.8 cm². |
| 3 | Consider the equation 2 sin²x + sin x – 1 = 0. (a) Factorise the left-hand side into two linear factors. (b) Solve the equation for 0° ≤ x ≤ 360°. (c) Solve the same equation for 0° ≤ x ≤ 720° and state how many solutions there are. |
(a) Let u = sin x, then 2u² + u – 1 = (2u – 1)(u + 1), so the equation becomes (2 sin x – 1)(sin x + 1) = 0. (b) 2 sin x – 1 = 0 gives sin x = 0.5, so x = 30° or 150°; sin x + 1 = 0 gives sin x = -1, so x = 270°. (c) Adding 360° to each solution gives 30°, 150°, 270°, 390°, 510° and 630°, that is 6 solutions in all. |
| 4 | A cuboid ABCDEFGH has length 8 cm, width 6 cm and height 5 cm, with ABCD as the base and CG as a vertical edge. (a) Find the base diagonal AC. (b) Find the space diagonal AG, correct to 0.1 cm. (c) Find the angle between AG and the base ABCD, correct to 0.1°, stating the triangle used. |
(a) AC² = 8² + 6² = 100, so AC = 10 cm. (b) CG is perpendicular to the base, so triangle ACG is right-angled at C: AG² = AC² + CG² = 100 + 25 = 125, giving AG = 11.2 cm. (c) In the right-angled triangle ACG, tan θ = CG / AC = 5 / 10 = 0.5, so θ = 26.6°. |
Section B: Indices and Logarithms
Answer all parts; give numerical answers to the stated precision.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | (a) Without a calculator, evaluate 8^(2/3), 16^(-3/4) and 27^(2/3). (b) Solve 3^(2x – 1) = 27^(x – 2). (c) Simplify (a³b⁻²)² ÷ (a⁻¹b³), giving your answer with positive indices. |
(a) 8^(2/3) = (cube root of 8)² = 4; 16^(-3/4) = 1 / 16^(3/4) = 1 / 8 = 0.125; 27^(2/3) = (cube root of 27)² = 9. (b) Since 27 = 3³, 27^(x – 2) = 3^(3x – 6). Equating indices gives 2x – 1 = 3x – 6, so x = 5. (c) The numerator is a⁶b⁻⁴; dividing by a⁻¹b³ gives a^(6+1)b^(-4-3) = a⁷b⁻⁷ = a⁷ / b⁷. |
| 6 | (a) Find the value of log₃ 54 – log₃ 2. (b) Solve log₂(x + 3) + log₂(x – 3) = 4. (c) Explain why the other algebraic root in (b) is rejected. |
(a) log₃ 54 – log₃ 2 = log₃(54 / 2) = log₃ 27 = 3, because 3³ = 27. (b) Combining the logarithms gives log₂((x + 3)(x – 3)) = 4, so x² – 9 = 2⁴ = 16, x² = 25 and x = 5 or x = -5. (c) The argument of a logarithm must be positive, so x + 3 and x – 3 must both exceed zero; x = -5 makes x + 3 negative, so it is rejected and the only solution is x = 5. |
| 7 | A culture of bacteria grows so that the number N after t hours is N = 500 × 2^(t/3). (a) Find the number of bacteria after 18 hours. (b) Find the time needed for the number to reach 100 000 for the first time, correct to 0.1 hour. (c) Find the first whole hour at which the number exceeds 1 000 000, and explain why a whole hour is used. |
(a) N = 500 × 2^(18 / 3) = 500 × 2⁶ = 32000. (b) We need 500 × 2^(t/3) = 100 000, that is 2^(t/3) = 200. Taking logarithms, t = 3 × log 200 / log 2 = 22.9 hours. (c) We need 2^(t/3) = 2000, so t = 3 × log 2000 / log 2 = 32.9 hours; counts are taken at whole hours, so the number first exceeds 1 000 000 in hour 33. |
| 8 | The sound intensity level L in decibels and the intensity I in W m⁻² are related by L = 10 log₁₀(I / I₀), where I₀ is a reference intensity. (a) Find the ratio of the intensities of sounds of 75 dB and 60 dB. (b) One machine produces 60 dB. How many identical machines are needed to reach 75 dB? (c) Explain what an increase of 3 dB means in terms of intensity. |
(a) The levels differ by 15 dB, so I₂ / I₁ = 10^(15 / 10) = 10^1.5 = 31.6, about 32 times. (b) Identical machines give equal intensities, so about 32 machines are needed, that is 32 in practice. (c) 10 log₁₀ 2 = 3.0, so doubling the intensity raises the level by about 3 dB; conversely an increase of 3 dB roughly doubles the intensity. |
Section C: Probability and Statistics
Answer all parts; support explanations with a formula or principle.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A bag contains 5 red balls and 3 blue balls, 8 balls in all. Two balls are drawn at random without replacement. (a) Find the probability that both balls are red. (b) Find the probability that exactly one ball is red. (c) Find the probability that at least one ball is blue, using your answer to (a). |
(a) P(red, red) = (5 / 8) × (4 / 7) = 20 / 56 = 5 / 14 ≈ 0.357. (b) There are two orders, red then blue and blue then red: P = 2 × (5 / 8) × (3 / 7) = 30 / 56 = 15 / 28 ≈ 0.536. (c) At least one blue is the complement of both red, so P = 1 – 5 / 14 = 9 / 14 ≈ 0.643. |
| 10 | Of 120 students in a school, 72 like mathematics, 45 like physics and 30 like both subjects. One student is chosen at random. (a) Find the probability that the student likes mathematics. (b) Given that the student likes mathematics, find the probability that the student likes physics. (c) Decide whether the two events are independent, supporting your answer with a calculation. |
(a) P(mathematics) = 72 / 120 = 0.6. (b) P(physics | mathematics) = 30 / 72 = 5 / 12 ≈ 0.417. (c) P(mathematics) × P(physics) = 0.6 × 0.375 = 0.225, while P(both) = 30 / 120 = 0.25; the two are not equal, so the events are not independent (students who like mathematics are more likely to like physics). |
| 11 | A class has 5 boys and 4 girls, 9 students in all. (a) How many ways are there to choose a committee of 3 from the 9 students? (b) How many committees of 3 contain at least one girl? (c) How many ways are there to choose 4 students containing exactly 2 girls? |
(a) C(9, 3) = 84. (b) Subtract the all-boy committees from all committees: 84 – C(5, 3) = 84 – 10 = 74. (c) Choose 2 girls and 2 boys: C(4, 2) × C(5, 2) = 6 × 10 = 60. |
| 12 | The test scores of nine students are 10, 12, 14, 16, 18, 20, 22, 24 and 26. (a) Find the mean and the median. (b) Find the standard deviation, dividing by n, correct to 0.01. (c) A tenth student scores 28. Find the new mean and standard deviation, correct to 0.01, and compare the spread. |
(a) Sum = 162, mean = 162 / 9 = 18; the median (5th value) is 18. (b) The sum of squared deviations is 240.0, so the variance is 240.0 / 9 = 26.67 and the standard deviation is 5.16. (c) The new sum is 190, so the mean is 19; the sum of squares rises from 3156 to 3940, giving a variance of 33 and a standard deviation of 5.74, so the spread has increased slightly. |
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