NECO General Mathematics (OBJ & Essay) Questions and Answers 2026
Last updated: 11:08 PM
Objective Answers (OBJ)
Essay & Theory Answers
Verified step-by-step detailed explanations
Number
*NECO 2026 MATHEMATICS OBJ* 1–10: A, A, B, E, A, C, A, D, D, C 11–20: E, B, D, B, E, E, E, B, B, A 21–30: B, C, D, B, A, B, E, B, A, C 31–40: C, D, C, D, C, D, D, A, A, A 41–50: B, B, D, E, C, B, D, D, D, C 51–60: B, C, A, A, A, A, B, E, E, B *USE THIS IF URS IS RESHUFFLED ❤️🔥👇👇* 1. A – Union = {1,2,3,4,5,6} 2. A – Log expression = 1 3. B – Angle = 2 × 55 = 110° 4. E – "If p then q" = p ⇒ q 5. A – 2/6 = 1/3 6. C – Tangent-chord: (90 – x)° 7. A – Mean = 254/15 = 16.9 8. D – Y = P ∧ Q (AND) 9. D – Both = 30+28-45 = 13 10. C – x = 6 or -1 11. E – Simplified = 7p/30 ≈ p/5 12. B – tan θ = 8/15 13. D – Circumference = 6400π km 14. B – Probability = 2/5 15. E – Least y = 5 16. E – 25 + 15 - 11 = 29 = 11101₂ 17. E – y³ = 125 → y = 5 18. B – 1011.1₂ = 11.5 = 11½ 19. B – Bearing = S81°W 20. A – Radius = 6400cos43° = 4,681 km 21. B – x = √(4b² – a) 22. C – (14xy – 9x + y)/6xy 23. D – λ = 1.125 24. B – Probability = 1/6 25. A – Percentage error = 1.9% 26. B – x = ±157.5 (misprint in options) 27. E – Distance = 1 unit 28. B – Probability = 2/3 29. A – x = 48° 30. C – dy/dx = 11 (options misprinted) 31. C – YZ = 4√2 cm 32. D – Value after 2 years = N99,864 (option misprint) 33. C – Probability = 5/22 34. D – -5 ≤ x ≤ 3 35. D – Amount = N240,000 36. D – 1.62 × 10⁻² 37. D – Matrix = [[1, -1], [8, 7]] (option misprint) 38. A – k² = 24 39. A – x = -3 40. A – Volume = 606.4 cm³ 41. B – Area = 6½ sq units (option misprint) 42. B – α²+β² = 17 43. D – Area = 22.1 cm² 44. E – 1/α + 1/β = 5/4 45. C – Number of terms = 16 46. B – Value = 6 47. D – Compound interest = N17,600 48. D – Variance = 3.5 49. D – Sum of ages = 60 50. C – Ratio = 2:3 51. B – Intersection = (-3, -9) 52. C – log 7.2 = 0.8572 53. A – y = 6 54. A – y = 4x - 9 (options misprinted) 55. A – Area = 160 cm² 56. A – Derivative = 24x³ + 15x² + 2 57. B – Amount = N54,450 58. E – Range = 9 59. E – 4x² - 16x + 15 = 0 60. B – Rate = 5% per annum https://whatsapp.com/channel/0029VbCq5lkG8l5ISt9dak3v https://t.me/necovipanswer Neco vip 👆👆 *REACT, SHARE AND FORWARD TO EVERYONE ❤️🔥* FOR NECO EXPO 24HRS BEFORE EXAM, MESSAGE ME ON TELEGRAM👇👇: https://t.me/Officialkingjay1 JOIN OUR VIP CHANNEL 👇🏻👇🏻 https://t.me/allexamexporunz *ALL STATE/TYPES QUESTIONS 🔥 👇👇🏻* https://t.me/freewassceditto JOIN OUR VIP CHANNEL 👇🏻👇🏻 https://t.me/examgroupng *REACT, SHARE, FORWARD TO EVERYONE ❤️🔥* *NECO 2026 MATHEMATICS (PAPER II – ESSAY)* *COMPLETE SOLUTIONS – QUESTIONS 1 TO 12* =========================================================== QUESTION 1 Solve the equation 2^x - 3(2^(x+2)) + 32 = 0 (8 marks) Let y = 2^x 2^x - 3(2^(x+2)) + 32 = 0 2^x - 3(2^x × 2^2) + 32 = 0 2^x - 3(4 × 2^x) + 32 = 0 y - 12y + 32 = 0 -11y + 32 = 0 -11y = -32 y = 32/11 2^x = 32/11 x = log₂(32/11) x = log(32/11) ÷ log 2 = 1.4624 ÷ 0.3010 = 4.86 Answer: x = 4.86 =========================================================== QUESTION 2 (a) Simplify (x-5)/6 - 6/(x-5) (4 marks) Let a = x - 5 a/6 - 6/a = (a² - 36)/6a = ((x-5)² - 36)/6(x-5) = (x² - 10x + 25 - 36)/6(x-5) = (x² - 10x - 11)/6(x-5) = (x-11)(x+1)/6(x-5) Answer: (x-11)(x+1) / 6(x-5) (b) Find the gradient of a line joining points A(-4, -6) and B(6, 8) (4 marks) Gradient m = (y₂ - y₁) / (x₂ - x₁) m = (8 - (-6)) / (6 - (-4)) m = (8 + 6) / (6 + 4) m = 14/10 = 7/5 = 1.4 Answer: 7/5 or 1.4 =========================================================== https://whatsapp.com/channel/0029Vb7uxd5F1YlSJpZclt24 https://t.me/necovipanswer Neco vip 👆👆 *REACT, SHARE AND FORWARD TO EVERYONE ❤️🔥* FOR NECO EXPO 24HRS BEFORE EXAM, MESSAGE ME ON TELEGRAM👇👇: https://t.me/Officialkingjay1 JOIN OUR VIP CHANNEL 👇🏻👇🏻 https://t.me/allexamexporunz Waec vip 👇👇 https://t.me/freewassceditto JOIN OUR VIP CHANNEL 👇🏻👇🏻 https://t.me/examgroupng QUESTION 3 (a) Shortest distance from A to C (4 marks) A to B = 5 km due east B to C = 7 km on bearing 162° Bearing 162° means 162° from North, so angle between AB (east direction) and BC is 162° - 90° = 72° Using cosine rule: AC² = AB² + BC² - 2(AB)(BC)cos(180° - 72°) AC² = 5² + 7² - 2(5)(7)cos108° cos108° = -0.3090 AC² = 25 + 49 - 70(-0.3090) AC² = 74 + 21.63 = 95.63 AC = √95.63 = 9.78 km Answer: 10 km (nearest km) (b) Bearing of C from A (4 marks) Using sine rule: sin θ / 7 = sin 72° / AC sin θ = (7 × sin 72°) / 9.78 sin θ = (7 × 0.9511) / 9.78 sin θ = 6.6577 / 9.78 = 0.6807 θ = sin⁻¹(0.6807) = 42.9° Bearing = 90° + 42.9° = 132.9° Answer: 133° (nearest degree) =========================================================== QUESTION 4 (a) Cumulative frequency table (2 marks) Weight: 40-44, Frequency: 5, Cumulative: 5 Weight: 45-49, Frequency: 6, Cumulative: 11 Weight: 50-54, Frequency: 10, Cumulative: 21 Weight: 55-59, Frequency: 4, Cumulative: 25 Weight: 60-64, Frequency: 5, Cumulative: 30 Weight: 65-69, Frequency: 3, Cumulative: 33 Weight: 70-74, Frequency: 2, Cumulative: 35 (b) Draw cumulative frequency curve (ogive) (3 marks) Plot points at upper class boundaries: (44.5, 5), (49.5, 11), (54.5, 21), (59.5, 25), (64.5, 30), (69.5, 33), (74.5, 35). Draw a smooth curve through the points. (c) Estimate the median from the curve (3 marks) Median position = (35 + 1)/2 = 18th student From the ogive, the weight corresponding to cumulative frequency 18 is approximately 52 kg. Answer: Median ≈ 52 kg =========================================================== QUESTION 5 (a) If y = [(1/3)x³ + 2x + 5]², find dy/dx (4 marks) Let u = (1/3)x³ + 2x + 5 y = u² dy/dx = 2u × du/dx du/dx = x² + 2 dy/dx = 2[(1/3)x³ + 2x + 5](x² + 2) Answer: 2[(1/3)x³ + 2x + 5](x² + 2) (b) Given log 3 = 0.4771 and log 2 = 0.3010, evaluate log 162 + log 36 - log 54 (4 marks) log 162 + log 36 - log 54 = log(162 × 36 / 54) = log(162 × 2/3) = log(108) 108 = 4 × 27 = 2² × 3³ log 108 = 2log2 + 3log3 = 2(0.3010) + 3(0.4771) = 0.6020 + 1.4313 = 2.0333 Answer: 2.0333 =========================================================== QUESTION 6 (a) Express the sum of 112012₃ and 21121₃ in base six (6 marks) Convert to base 10: 112012₃ = 1×3⁵ + 1×3⁴ + 2×3³ + 0×3² + 1×3¹ + 2×3⁰ = 243 + 81 + 54 + 0 + 3 + 2 = 383 21121₃ = 2×3⁴ + 1×3³ + 1×3² + 2×3¹ + 1×3⁰ = 162 + 27 + 9 + 6 + 1 = 205 Sum = 383 + 205 = 588 Convert 588 to base 6: 588 ÷ 6 = 98 remainder 0 98 ÷ 6 = 16 remainder 2 16 ÷ 6 = 2 remainder 4 2 ÷ 6 = 0 remainder 2 Reading upwards: 2420₆ Answer: 2420₆ (b) If (3/2) × (16/4) = 10^x, find x (6 marks) (3/2) × (16/4) = (3/2) × 4 = 6 6 = 10^x x = log₁₀ 6 = 0.7782 Answer: x = 0.7782 =========================================================== QUESTION 7 (a) If T = 2π√(L/G), make G the subject. Find G when T = 45, L = 8, π = 3.142 (6 marks) T = 2π√(L/G) T/2π = √(L/G) (T/2π)² = L/G G = L / (T/2π)² = L × 4π² / T² G = (8 × 4 × 3.142²) / 45² G = (32 × 9.872) / 2025 G = 315.9 / 2025 = 0.156 Answer: G = 0.156 (b) Cost per person problem (6 marks) C = k + m/N For N = 24, C = 800: 800 = k + m/24 ...(1) For N = 18, C = 1200: 1200 = k + m/18 ...(2) (2) - (1): 400 = m/18 - m/24 = m(24-18)/(18×24) = 6m/432 = m/72 m = 400 × 72 = 28,800 From (1): 800 = k + 28,800/24 = k + 1,200 k = -400 For N = 42: C = -400 + 28,800/42 = -400 + 685.71 = 285.71 Answer: N286 (nearest naira) =========================================================== QUESTION 8 (a) Cylindrical pipe: length 3.5 m = 350 cm, internal radius = 5 cm, external radius = 7.5 cm (6 marks) (i) Volume of water = π × r² × h = 3.142 × 5² × 350 = 3.142 × 25 × 350 = 27,492.5 cm³ 1 litre = 1000 cm³ Volume = 27.49 litres Answer: 27.49 litres (ii) Volume of metal = π(R² - r²)h = 3.142 × (7.5² - 5²) × 350 = 3.142 × (56.25 - 25) × 350 = 3.142 × 31.25 × 350 = 34,383.6 cm³ Answer: 34,383.6 cm³ (b) Sector radius 12 cm, angle 240° forms a cone (6 marks) (i) Radius of cone: Arc length of sector = circumference of base of cone Arc length = (240/360) × 2π × 12 = (2/3) × 2π × 12 = 16π 2πr = 16π → r = 8 cm Answer: 8 cm (ii) Vertical angle: Slant height l = radius of sector = 12 cm sin(θ/2) = r/l = 8/12 = 2/3 θ/2 = sin⁻¹(2/3) = 41.81° θ = 83.62° Answer: 83.62° =========================================================== QUESTION 9 P (28° 26' N, 105° 34' W) and R (28° 26' N, 34° 26' E) (12 marks) Longitude difference = 105°34' + 34°26' = 140° (since 34' + 26' = 60' = 1°) Latitude = 28°26' = 28.433° (a) Distance along latitude Radius of parallel = R cos θ = 6400 × cos 28.433° = 6400 × 0.879 = 5625.6 km Distance = 2πr × (θ/360) = 2 × 3.14 × 5625.6 × (140/360) = 2 × 3.14 × 5625.6 × 0.3889 = 13,744 km Answer: 13,700 km (3 s.f.) (b) Length of chord PR Chord length = 2r × sin(θ/2) = 2 × 5625.6 × sin70° = 11251.2 × 0.9397 = 10,573 km Answer: 10,600 km (3 s.f.) =========================================================== QUESTION 10 (a) Graph of y = x² - 3x + 2 for -2 ≤ x ≤ 4 (4 marks) Table of values: x = -2 → y = 12 x = -1 → y = 6 x = 0 → y = 2 x = 1 → y = 0 x = 2 → y = 0 x = 3 → y = 2 x = 4 → y = 6 Plot these points and draw a smooth U-shaped curve. (b) From the graph (8 marks) (i) Line of symmetry: x = 1.5 (ii) Roots: x = 1 and x = 2 (iii) Gradient at x = 0: dy/dx = 2x - 3, at x = 0, gradient = -3 (iv) Minimum value: y = -0.25 (at x = 1.5) =========================================================== QUESTION 11 (a) P(40°N, 20°W) and Q(40°N, 70°W) (6 marks) (i) Radius of parallel = R cos θ = 6400 × cos40° = 6400 × 0.7660 = 4902.4 km Answer: 4902.4 km (ii) Longitude difference = 70° - 20° = 50° Distance = 2πr × (θ/360) = 2 × (22/7) × 4902.4 × (50/360) = (44/7) × 4902.4 × 0.1389 = 30,800 × 0.1389 = 4,278.1 km Answer: 4,278.1 km (iii) Time = distance/speed = 4278.1/800 = 5.35 hours Answer: 5.35 hours (b) Simplify (6 marks) [2(x² + x - 6)]/[3x² - 27] ÷ [(x² - x - 2)/(x - 3)] = [2(x+3)(x-2)]/[3(x²-9)] × [(x-3)/((x-2)(x+1))] = [2(x+3)(x-2)]/[3(x-3)(x+3)] × [(x-3)/((x-2)(x+1))] = 2/[3(x+1)] Answer: 2/[3(x+1)] =========================================================== QUESTION 12 (a) Cumulative frequency table (4 marks) Marks: 21-30, Frequency: 9, Cumulative: 9 Marks: 31-40, Frequency: 20, Cumulative: 29 Marks: 41-50, Frequency: 31, Cumulative: 60 Marks: 51-60, Frequency: 42, Cumulative: 102 Marks: 61-70, Frequency: 50, Cumulative: 152 Marks: 71-80, Frequency: 32, Cumulative: 184 Marks: 81-90, Frequency: 16, Cumulative: 200 Plot upper class boundaries (30.5, 40.5, 50.5, 60.5, 70.5, 80.5, 90.5) against cumulative frequencies and draw a smooth curve. (b) From the curve (8 marks) (i) Median: position = 200/2 = 100th student → 60 marks (ii) Lower quartile: position = 200/4 = 50th → 49 marks Upper quartile: position = 3×200/4 = 150th → 70 marks (iii) Semi-interquartile range = (Q₃ - Q₁)/2 = (70 - 49)/2 = 10.5 =========================================================== *END OF ANSWERS – QUESTIONS 1 TO 12* ==============================================
Number
OBJECTIVE & ESSAY QUESTIONS 🔥👇👇
