GCE Mathematics 2024 Theory Past Questions

Model answer

t ∝ V/P → t = kV/P. Given P=5, V=20, t=10: 10 = k(20)/5 = 4k → k = 2.5. (a) Relationship: t = 2.5V/P. (b) At V=50, P=2: t = 2.5×50/2 = 62.5 minutes. (c) At t=20 minutes, V=40, find P: 20 = 2.5×40/P = 100/P → P = 100/20 = 5 pumps.

Model answer

Time = Distance/Speed. Total time = 90 minutes = 1.5 hours. (2x+13)/67.5 + (5x−20)/72 = 1.5 Multiplying through by 540 (LCM of 67.5 and 72, since 540/67.5=8 and 540/72=7.5): 8(2x+13) + 7.5(5x−20) = 1.5×540 16x + 104 + 37.5x − 150 = 810 53.5x − 46 = 810 53.5x = 856 x = 16.

Model answer

Perimeter (circumference) = 66m = 2πr. r = 66/(2×22/7) = 66×7/44 = 10.5m. Area of floor = πr² = 22/7 × 10.5² = 22/7 × 110.25 = 346.5 m². Area of one tile = 40cm×40cm = 0.4m×0.4m = 0.16m². Number of tiles = 346.5/0.16 = 2165.625 ≈ 2166 tiles.

Model answer

Since OBCD is a rhombus and O is the centre of the circle passing through A, B, C, D, all sides OB, BC, CD, DO equal the radius r, and since C also lies on the circle, OC = r too. This means triangles OBC and OCD are each equilateral (all three sides = r), so ∠BOC = ∠COD = 60°, giving ∠BOD (measured through C) = 120°. (a) The angle t = ∠BAD is the inscribed angle standing on arc BD (the arc containing C), and the angle at the centre standing on the same arc is 120°. By the inscribed angle theorem (angle at centre = twice angle at circumference): t = 120°/2 = 60°. (b) Since OA=OB=r, triangle OAB is isosceles with ∠OAB=∠OBA=y. Similarly triangle OAD is isosceles with ∠OAD=∠ODA=y. Since OA bisects ∠BAD by symmetry: t = 2y → 60° = 2y → y = 30°. (c) In the rhombus OBCD, opposite angles are equal: ∠O=∠C=120°, so ∠B+∠D=360°−120°−120°=120°, and since ∠B=∠D, each = 60°. So ∠ODC=60°. ∠ADC = ∠ADO + ∠ODC = y + 60° = 30° + 60° = 90°.

Model answer

Let the number of silver marbles = s. Total marbles = 3+4+s = 7+s. P(both silver, no replacement) = [s/(7+s)] × [(s−1)/(6+s)] = 1/15. 15s(s−1) = (7+s)(6+s) 15s²−15s = s²+13s+42 14s²−28s−42 = 0 Dividing by 14: s²−2s−3 = 0 (s−3)(s+1) = 0 → s = 3 (rejecting the negative root). Number of silver marbles = 3.

Model answer

For consecutive G.P terms, the middle term squared equals the product of the outer terms: (4x+3)² = (x+2)(7x+24) 16x²+24x+9 = 7x²+38x+48 9x²−14x−39 = 0 Using the quadratic formula: x = [14±√(196+1404)]/18 = [14±√1600]/18 = [14±40]/18 x = 54/18 = 3 (taking the positive root). (a) x = 3. Check: (x+2)=5, (4x+3)=15, (7x+24)=45. 15/5=3, 45/15=3 ✓ confirms a valid G.P. (b) Common ratio = 3.

Model answer

Frequencies: Age5:2, Age6:(2x−1), Age7:(y+2), Age8:4, Age9:2, Age10:(y−1). Total=20. 2+(2x−1)+(y+2)+4+2+(y−1)=20 8+2x+2y=20 → 2x+2y=12 → x+y=6. Given y=2x: x+2x=6 → 3x=6 → x=2, y=4. (Check ratio x:y=2:4=1:2 ✓) Frequencies become: Age5:2, Age6:2(2)−1=3, Age7:4+2=6, Age8:4, Age9:2, Age10:4−1=3. Sum=2+3+6+4+2+3=20 ✓. (a) x=2, y=4. (b) Mean age = Σ(age×freq)/Σfreq = (5×2+6×3+7×6+8×4+9×2+10×3)/20 = (10+18+42+32+18+30)/20 = 150/20 = 7.5 years.

Model answer

(a) Let the height of the pole (to the board) be h. Since Badu's angle of elevation (48°) is greater than Abu's (40°), Badu is closer to the pole. Let Badu's horizontal distance from the foot of the pole = d, so Abu's distance = d+46 (Abu being further away). A simple diagram would show: the pole standing vertically at one end, with Badu at distance d from its foot and Abu at distance (d+46) from its foot, both on the same horizontal line, with lines of sight rising at 48° (from Badu) and 40° (from Abu) to the top of the pole. (b) tan48° = h/d → d = h/tan48° tan40° = h/(d+46) → d+46 = h/tan40° Subtracting: h/tan40° − h/tan48° = 46 h(1/tan40° − 1/tan48°) = 46 h(1.1918 − 0.9004) = 46 h(0.2914) = 46 h = 46/0.2914 ≈ 157.9m (correct to one decimal place).

Model answer

Radius r=13cm (half of diameter 26cm). Perpendicular distance from O to chord PQ = 5cm. Half-chord length = √(r²−d²) = √(169−25) = √144 = 12cm, so PQ=24cm. (a) In the right triangle formed by O, the midpoint of PQ, and P: cos(∠POQ/2) = 5/13 → ∠POQ/2 = cos⁻¹(5/13) = 67.38° → ∠POQ = 134.76° ≈ 135° (to the nearest whole number). (b) Area of sector POQ = (θ/360)×πr² = (134.76/360)×(22/7)×169 ≈ 0.3743×531.14 ≈ 198.8cm². Area of triangle POQ = (1/2)r²sinθ = 0.5×169×sin(134.76°) = 84.5×0.7106 ≈ 60.1cm². Area of minor segment = Sector − Triangle = 198.8−60.1 ≈ 138.7 ≈ 139cm² (to the nearest whole number).

Model answer

(a) Let total acres = T. Wife's share = 2/5T. Remainder = 3/5T. Of this remainder, 1/3 goes to family upkeep, leaving 2/3 × 3/5T = 2/5T for the three children, shared in ratio 3:5:2 (total 10 parts). Least share (2 parts) = 8 acres → 1 part = 4 acres → children's total share (10 parts) = 40 acres = 2/5T. (i) T = 40 × 5/2 = 100 acres. (ii) Wife's share = 2/5 × 100 = 40 acres. (b) Deposit = $400. Balance = $1600−$400 = $1200, paid over 12 months (1 year) at 25% p.a simple interest. Interest = PRT/100 = (1200×25×1)/100 = $300. Total instalment payments = $1200+$300 = $1500. Total cost = Deposit + Instalment total = $400+$1500 = $1900.

Model answer

(a) Solve x+2y=7 and x−y=4 simultaneously: from the second equation, x=y+4. Substituting: (y+4)+2y=7 → 3y+4=7 → 3y=3 → y=1, x=5. Intersection point = (5,1). Line through origin (0,0) and (5,1) has gradient = 1/5. Equation: y = (1/5)x, i.e. x−5y=0. (b) Interior:Exterior = 4:1. Since interior+exterior=180° (angles on a straight line): let exterior=e, interior=4e. 4e+e=180° → 5e=180° → e=36°. (i) Number of sides = 360°/e = 360/36 = 10 sides. (ii) Exterior angle = 36°. (iii) Sum of interior angles = (n−2)×180° = (10−2)×180° = 8×180° = 1440°.

Model answer

(a) Since V lies on line OP, ∠VOQ = ∠POQ = 56°. Using the exterior angle theorem on triangle OVQ, the exterior angle ∠SVP (=109°) equals the sum of the two opposite interior angles: ∠SVP = ∠VOQ + ∠VQO. So 109° = 56° + ∠VQO → ∠VQO = 53°. Since OQ is a radius and PR is a tangent to the circle at Q, OQ ⊥ PR, so ∠OQP = 90°. (i) ∠TQP = ∠OQP − ∠VQO = 90° − 53° = 37°. (ii) By the tangent-chord angle theorem, the angle between tangent QP and chord QT equals the angle in the alternate segment: ∠QTS = ∠TQP = 37°. (b) Factorising: 2n²−3n−2 = (2n+1)(n−2); 2n²+3n+1 = (2n+1)(n+1); n²−1 = (n−1)(n+1); n²−4 = (n−2)(n+2). Expression = [(2n+1)(n−2)/((2n+1)(n+1))] × [(n−1)(n+1)/((n−2)(n+2))] = [(n−2)/(n+1)] × [(n−1)(n+1)/((n−2)(n+2))] = (n−1)(n+1)/[(n+1)(n+2)] [the (n−2) terms cancel] = (n−1)/(n+2) [the (n+1) terms cancel].

Model answer

(a) By the tangent-chord angle theorem, ∠PTA (between tangent PT and chord TA) = angle in the alternate segment = ∠ABT = 60°. Since ABC is a straight line, in triangle ABT: ∠BAT+∠ABT+∠ATB=180° → 44°+60°+∠ATB=180° → ∠ATB=76°. Also by the tangent-chord angle theorem, ∠BTQ (between tangent TQ and chord TB) = angle in alternate segment = ∠BAT = 44°. Since TC bisects ∠BTQ: ∠BTC = ∠CTQ = 22° each. ∠ATC = ∠ATB + ∠BTC = 76° + 22° = 98°. Since ABC is a straight line, ∠TAC = ∠TAB = 44° (same ray). In triangle ATC: ∠ACT = 180° − ∠TAC − ∠ATC = 180° − 44° − 98° = 38°. (b) Circumference = 2πr = 11m → r = 11/(2×22/7) = 11×7/44 = 1.75m. (i) Radius = 1.75m. Height = 6r+3 = 6(1.75)+3 = 10.5+3 = 13.5m. (ii) Height = 13.5m. (iii) Volume = πr²h = 22/7×(1.75)²×13.5 = 22/7×3.0625×13.5 = 22/7×41.34375 ≈ 129.9m³.