WAEC General Mathematics (OBJ & Essay) Questions and Answers 2026
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Objective Answers (OBJ)
Essay & Theory Answers
Verified step-by-step detailed explanations
Number THEORY
*MATHS OBJECTIVE TYPED ANSWERS* *MATHS OBJ* 01. m = -1 and m = 5/3 02. sin p = 24/25 03. y = 2 years 04. 4a - b = 11 05. y = x/3 + 13 06. gradient = 44/3 07. height = 14.0 m 08. x = 4 09. V(x) = x(x-7)(9) 10. 435.6π cm³ 11. θ = 84° 12. d = -5/3 13. 6.03 14. √3 + 2 15. 4% 16. S₆ = 54 17. ∠SQR = 28° 18. If the weather is not cloudy, then it is not going to rain 19. x = 70° 20. h = 3V/πr² 21. q = 1 22. |XY| = 8 cm 23. 3 24. r = 14 cm 25. mean = 4.4 26. mode = 5 27. median = 5 28. (h+k)(h-k-p) 29. -4 ≤ x < 2 30. ∠ORQ = 53° 31. ∠QPR = 52° 32. area = 240 cm² 33. height = 32 cm 34. Z = 2 35. 50 balls 36. 17.5 ms⁻¹ 37. x = 1/3 38. 12x² + 11x + 2 = 0 39. age = 15 40. sum = 360° 41. p = 5, q = -7 42. x = 12 43. height = 5 cm 44. x = 2 45. ∠PRS = 34° 46. ₦70,110.00 47. 1.54 m 48. 15 minutes 49. 1/x 50. 343° *COMPLETED*✅ TRACE IT. Follow https://t.me/examgroupng *WAEC MATHEMATICS (PAPER 2 – ESSAY)* *COMPLETE ANSWERS – QUESTIONS 1, 2, 3, 4 & 5* =========================================================== QUESTION 1 A fun club donates to children's home annually. The amount given is increased by 8% of its value yearly. Donation in 2010 was $7,000.00. (a) Donation given in 2012 Increase rate = 8% = 0.08 per year Donation in 2011 = 7000 × (1 + 0.08) = 7000 × 1.08 = $7,560 Donation in 2012 = 7560 × 1.08 = $8,164.80 Answer: $8,164.80 (b) Total amount donated from 2010 to 2013 Donation in 2010 = $7,000.00 Donation in 2011 = $7,560.00 Donation in 2012 = $8,164.80 Donation in 2013 = 8164.80 × 1.08 = $8,817.98 Total = 7000 + 7560 + 8164.80 + 8817.98 = $31,542.78 Answer: $31,542.78 =========================================================== QUESTION 2 Two cars X and Y start from a common point and travel in opposite directions. X = 60 km/h, Y = 80 km/h. (a) How far apart after 3 hours Distance = speed × time Distance travelled by X = 60 × 3 = 180 km Distance travelled by Y = 80 × 3 = 240 km Since they travel in opposite directions, total distance apart = 180 + 240 = 420 km Answer: 420 km (b) If speed of X is increased by 20 km/h within the 3 hours, what would be the effect? New speed of X = 60 + 20 = 80 km/h Distance travelled by X = 80 × 3 = 240 km Distance travelled by Y = 240 km (unchanged) Total distance apart = 240 + 240 = 480 km Effect: The distance between them increases from 420 km to 480 km (an increase of 60 km). (c) Importance of considering speed when calculating total distance over the same time frame Speed determines how far each vehicle travels within a given time. Without considering speed, the total distance apart cannot be accurately calculated because different speeds produce different distances even when time is the same. =========================================================== QUESTION 3 Concrete blocks of length 14/25 m were laid end to end along the four sides of a rectangular lawn 28 m long and 21 m wide. (a) How many blocks were required to complete one layer? Length of block = 14/25 = 0.56 m Perimeter of rectangle = 2 × (length + width) = 2 × (28 + 21) = 2 × 49 = 98 m Number of blocks = Total perimeter ÷ Length of one block = 98 ÷ (14/25) = 98 × (25/14) = (98 ÷ 14) × 25 = 7 × 25 = 175 blocks Answer: 175 blocks =========================================================== QUESTION 4 (Circle geometry – based on the diagram described: centre O, triangle MNO with angle MNO = 90°, angle N = 120° – note: there is inconsistency. I will solve based on standard WAEC circle problem.) Assuming the diagram shows: · O is the centre · M, N, P are points on the circle · Angle MNO = 90° (triangle MNO with right angle at N) · Some angles given as 100° (likely at centre) Without the exact labelled diagram, here is the standard approach: If ON = radius = r In right triangle MNO, angle MNO = 90° If OM is radius, then OM = r Using Pythagoras or trigonometry, area of triangle MNO = ½ × base × height Since the diagram is not fully clear, the intended answers would be: Area of triangle MNO = ½ × r × r = ½ r² (if it is right-angled isosceles) For other parts, you would use: Area of sector = (θ/360) × πr² Area of triangle = ½ r² sinθ Note: To solve fully, please provide the missing lengths or radii from the diagram. =========================================================== QUESTION 5 A bowl contains red, green and black marbles. Let: Number of green marbles = G Number of black marbles = B Number of red marbles = R Given: (1) B = 2G (black is twice green) (2) G = R - 1 (green is 1 less than red) → R = G + 1 (3) B = R + 3 (black is 3 more than red) (4) P(red) = R/(R+G+B) = 1/4 From (1) and (3): 2G = (G+1) + 3 2G = G + 4 G = 4 Then R = G + 1 = 5 B = 2G = 8 Check: B = R + 3 → 8 = 5 + 3 ✓ (a) Number of red marbles = 5 (b) Probability that a marble drawn at random is red = R/(R+G+B) = 5/(5+4+8) = 5/17 Answer: 5/17 (c) Probability of drawing a black marble = B/(R+G+B) = 8/17 Note: Question says "probability that a marble drawn is red black" – likely means "red or black" P(red or black) = (5+8)/17 = 13/17 Answer: 13/17 =========================================================== *WAEC MATHEMATICS (PAPER 2 – ESSAY)* *COMPLETE ANSWERS – QUESTIONS 6 TO 13* =========================================================== QUESTION 6 In a remedial school of 13 teachers: n(G) = 8, n(Y) = 7, n(H) = 7 n(G∩Y only) = 2 n(G∩H only) = 2 n(Y∩H only) = 3 Let x = n(G∩Y∩H) Total teachers = 13 (a) Venn diagram description (draw in your answer booklet) Draw three intersecting circles labelled G, Y, H. Place x in the triple intersection. Place 2 in G∩Y only (not H) Place 2 in G∩H only (not Y) Place 3 in Y∩H only (not G) Number in G only = 8 - (2 + 2 + x) = 4 - x Number in Y only = 7 - (2 + 3 + x) = 2 - x Number in H only = 7 - (2 + 3 + x) = 2 - x Sum all regions = (4 - x) + (2 - x) + (2 - x) + 2 + 2 + 3 + x = 13 Simplify: (4+2+2) + (-x-x-x + x) + (2+2+3) = 13 8 + (-2x) + 7 = 13 15 - 2x = 13 -2x = -2 x = 1 (b)(i) n(G∩Y∩H) = 1 (ii) n(G∩Y) = n(G∩Y only) + n(G∩Y∩H) = 2 + 1 = 3 (iii) n(G∩H) ∩ G = n(G∩H) = 2 + 1 = 3 Answer: (i) 1, (ii) 3, (iii) 3 =========================================================== QUESTION 7 (a)(i) Walking speed Let walking speed = w km/h Cycling speed = 3w km/h Time cycling = 45 minutes = 45/60 = 0.75 h Time walking = 1 h Distance = speed × time Distance cycled = 3w × 0.75 = 2.25w km Distance walked = w × 1 = w km Total distance = 2.25w + w = 3.25w = 10.4 km w = 10.4 ÷ 3.25 = 3.2 km/h Answer: 3.2 km/h (to 2 s.f.) (ii) New cycling time Original cycling speed = 3w = 3 × 3.2 = 9.6 km/h New cycling speed = 9.6 × (1 + 20/100) = 9.6 × 1.2 = 11.52 km/h Same walking: distance walked = 3.2 × 1 = 3.2 km Distance to cycle = total distance - 3.2 = 10.4 - 3.2 = 7.2 km New cycling time = distance ÷ speed = 7.2 ÷ 11.52 = 0.625 h = 0.625 × 60 = 37.5 minutes Answer: 37.5 minutes (b) Equation of line l Line passes through K(-3, 5) and is parallel to 3x - 2y + 5 = 0 Gradient of given line: 3x - 2y + 5 = 0 → -2y = -3x - 5 → y = (3/2)x + 5/2 Gradient m = 3/2 Parallel lines have same gradient. Equation: y - y₁ = m(x - x₁) y - 5 = (3/2)(x + 3) y - 5 = (3/2)x + 9/2 Multiply by 2: 2y - 10 = 3x + 9 2y = 3x + 19 y = (3/2)x + 19/2 Answer: y = (3/2)x + 9.5 or 3x - 2y + 19 = 0 =========================================================== QUESTION 8 (a) Circle geometry (diagram described) Given: ∠DAB = 64°, ∠ADB = 10° (assuming ∠AD = 10° means ∠ADB = 10°) In triangle ABD: ∠ABD = 180° - (64° + 10°) = 180° - 74° = 106° Angle x is likely ∠ACB or related. Since points A, B, C, D are on circle, angles subtended by same chord are equal. If x = ∠ACB, then ∠ACB = ∠ADB = 10° (angles subtended by chord AB) Answer: x = 10° (b) Two consecutive even numbers Let smaller = n, larger = n + 2 One-third of larger exceeds one-fourth of smaller by 7: (1/3)(n + 2) = (1/4)n + 7 Multiply by 12: 4(n + 2) = 3n + 84 4n + 8 = 3n + 84 4n - 3n = 84 - 8 n = 76 Smaller = 76, larger = 78 Answer: 76 and 78 =========================================================== QUESTION 9 (a) Tower, ship and boat Tower height = 75 m Ship angle of elevation = 26.4° Boat angle of elevation = 57.4° (ii) Distance between ship and boat Let distance from tower to ship = d₁, to boat = d₂ tan θ = opposite/adjacent = height/distance For ship: tan 26.4° = 75/d₁ → d₁ = 75/tan 26.4° tan 26.4° = 0.4966 → d₁ = 75/0.4966 = 151.0 m For boat: tan 57.4° = 75/d₂ → d₂ = 75/tan 57.4° tan 57.4° = 1.562 → d₂ = 75/1.562 = 48.0 m Distance between ship and boat = d₁ - d₂ = 151.0 - 48.0 = 103 m Answer: 103 m (b) Area of sector Radius r = 18 cm, angle θ = 120°, π = 22/7 Area = (θ/360) × πr² = (120/360) × (22/7) × 18² = (1/3) × (22/7) × 324 = (22 × 324)/(21) = (7128)/21 = 339.42857... cm² To 5 significant figures = 339.43 cm² Answer: 339.43 cm² =========================================================== QUESTION 10 (a) Diagram description (draw in booklet) Draw circle with centre O. Points A, B, C on circumference. AB = BC = 15 cm, ∠ABC = 120° (angle at B on circumference) (b)(i) Radius of circle In triangle ABC, AB = BC = 15, ∠ABC = 120° By cosine rule: AC² = AB² + BC² - 2(AB)(BC)cos120° cos120° = -0.5 AC² = 225 + 225 - 2(15)(15)(-0.5) = 450 + 225 = 675 AC = √675 = 15√3 ≈ 25.98 cm ∠AOC (at centre) = 2 × ∠ABC = 240° (angle at centre is twice angle at circumference) Chord AC subtends 240° at centre, so minor arc subtends 120°. Using chord length formula: chord AC = 2R sin(θ/2) where θ at centre = 240°? No – careful. Better: In triangle AOC, OA = OC = R, AC = 15√3, ∠AOC = 120° (since reflex 240°, the smaller angle is 360-240=120°) By cosine rule: AC² = R² + R² - 2R² cos120° 675 = 2R²(1 - cos120°) = 2R²(1 - (-0.5)) = 2R²(1.5) = 3R² R² = 675/3 = 225 R = 15 cm Answer: Radius = 15 cm (ii) Perimeter of major segment cut off by chord AC Major segment is the larger part when chord AC divides circle. Major arc AC corresponds to angle at centre = 360° - 120° = 240° Length of major arc AC = (240/360) × 2πR = (2/3) × 2 × (22/7) × 15 = (2/3) × (660/7) = (1320/21) = 62.857 cm Perimeter of major segment = arc length + chord AC = 62.857 + 25.98 = 88.84 cm To nearest whole number = 89 cm Answer: 89 cm =========================================================== QUESTION 11 (a) Profit function given as P = (3y²)/18 - (9/2) – simplified P = (y²/6) - 4.5 (i) Profit = $16.00 16 = y²/6 - 4.5 y²/6 = 20.5 y² = 123 y = √123 = 11.09 ≈ 11 articles Answer: 11 articles (ii) No profit means P = 0 0 = y²/6 - 4.5 y²/6 = 4.5 y² = 27 y = √27 = 5.196 ≈ 5 articles Answer: 5 articles (b) Inequality (5 + x)/6 + (4x - 1)/3 ≥ (3x + 2)/5 Multiply by 30 (LCM of 6,3,5): 5(5 + x) + 10(4x - 1) ≥ 6(3x + 2) 25 + 5x + 40x - 10 ≥ 18x + 12 15 + 45x ≥ 18x + 12 45x - 18x ≥ 12 - 15 27x ≥ -3 x ≥ -3/27 = -1/9 Answer: x ≥ -1/9 =========================================================== QUESTION 12 (a) Temperature data: 24, 25, 23, 27, 26, 28, 29, 30, 27, 26, 25, 24 Sort: 23, 24, 24, 25, 25, 26, 26, 27, 27, 28, 29, 30 (i) Mean = sum ÷ 12 Sum = 23+24+24+25+25+26+26+27+27+28+29+30 = 314 Mean = 314/12 = 26.166... = 26.2°C (to 1 dp) (ii) Median: average of 6th and 7th values 6th = 26, 7th = 26 Median = 26 (iii) Mode: values that appear most often 24 appears twice, 25 twice, 26 twice, 27 twice Multimodal: 24, 25, 26, 27 (iv) Best measure – Median = 26 Because data is slightly skewed, median is not affected by extremes. (b) Two towns on same meridian: latitudes 15°N and 36°N Difference in latitude = 36° - 15° = 21° Distance = (θ/360) × 2πR = (21/360) × 2 × (22/7) × 6400 = (21/360) × (44/7) × 6400 = (21 × 44 × 6400) / (360 × 7) Simplify 21/7 = 3: = (3 × 44 × 6400)/360 = (844800)/360 = 2346.67 km Answer: 2350 km (to 3 s.f.) =========================================================== QUESTION 13 (a) Construction (steps – draw in booklet) 1. Draw line MN = 8.7 cm 2. At N, construct angle MNR = 75° using protractor or compasses 3. From M, draw arc of radius 10.5 cm to meet the ray from N at point R 4. Join MR (b) Construction of point S: · RS parallel to MN: through R, draw line parallel to MN · S equidistant from M and R: draw perpendicular bisector of MR · Intersection of the parallel line and perpendicular bisector = S (c) Measurement: (i) NR ≈ 10.2 cm (depends on accurate construction) (ii) ∠NRS ≈ 75° (alternate angles, RS∥MN, line RN transversal) (d) Shape of triangle MNR? It is scalene (all sides different: MN=8.7, MR=10.5, NR≈10.2) =========================================================== END OF ANSWERS – QUESTIONS 6 TO 13 =========================================================== END OF ANSWERS ===========================================================
Number Objective questions
OBJ
