Mathematics Grade 9 Paper 2 GEC Test 2023 Memo (English)

Downloaded from hlayiso.com GEC PILOT STUDY MARKING GUIDELINE 2023 MATHEMATICS PAPER 2 GRADE 9 SECTION A  One mark per answer.  There are no half marks. No. Expected answer Clarification 1. C It is vertically opposite to 88° .  2. D 90°  Lines are perpendicular. 3. C Corresponding angles.  4. B 55°  5. A 54°  3𝑥 + 2𝑥 = 90° …compl ∠𝑠 5𝑥 = 90° 5𝑥 90° 5 = 5 𝑥 = 18° EÔ B = 3𝑥 = 3(18°) EÔ B = 54° 6. B 66°  68° + Ê + 46° = 180° …sum of ∠𝑠 of a ∆ ̂ = 180° − 68° − 46° E ̂ = 66° E ̂ = 𝑓 …alt ∠s; EB∥DC E ∴ 𝑓 = 66° OR ̂ + Ĉ = 180° …co-int ∠s; EB∥DC B 68° + 46° + 𝑓 = 180° 𝑓 = 180° − 114° ∴ 𝑓 = 66° 7. A 10°  20° + 55° + 2𝑎 − 15° = 180° …co-int ∠s ; MO∥PQ 2𝑎 + 60° = 180° 2𝑎 = 120° 𝑎 = 60° Grade 9 Mathematics P2 Marking Guideline 1

Downloaded from hlayiso.com No. Expected answer Clarification 2𝑏 + 2𝑎 + 20° + 𝑎 − 40° = 180° …sum of ∠s of a ∆ 2𝑏 + 2(60°) + 20° + 60° − 40° = 180° 2𝑏 + 120° + 20° + 60° − 40° = 180° 2𝑏 = 180° − 160° 2𝑏 = 20° 2𝑏 20° = 2 2 𝑏 = 10° 8. A one right angle and two acute  angles. 9. A  10. A The diagonals are  perpendicular, one diagonal bisects the other and one diagonal bisects the angles at the vertices. 11. D A square is a rectangle.  12. B 95°  ̂ A+ B̂ + Ĉ = 180° …sum of ∠s of a ∆ = 180° ̂ + 47° + 38° = 180° A ̂ = 180° − 47° − 38° A = 95° 13. A 162°  ̂ ̂ N1 = P1 = 35° …alt ∠s ; MN∥PQ ̂ NOQ = P ̂1 + PN̂ O …ext. ∠ = sum of 2 opp int ∠s = 35° + 127° ∴ 𝑦 = 162° OR 𝑦 = MN̂ O …alt ∠s ; MN∥PQ 𝑦 = 35° + 127° 𝑦 = 162° 14. A 4 cm  ∆ABC⦀∆EFG …corr. ∠s = AB BC AC EF = FG = EG …corr. sides are proportional 𝑦 6 = 8 12 Grade 9 Mathematics P2 Marking Guideline 2

Downloaded from hlayiso.com No. Expected answer Clarification 6×8 𝑦= 12 ∴AB = 4 cm 15. A 115°  EĈD = DE ̂C = 𝑦 + 15° … ∠s opp = sides 𝑦 + 15° + 𝑦 + 65° = 180° …∠s on a straight line 2𝑦 + 80° = 180° 2𝑦 = 100° 𝑦 = 50° ̂1 = 𝑦 + 65° E = 50° + 65° = 115° OR ̂B = D ̂ = 𝑦 …opp. ∠s of a rhombus = ̂1 = Ĉ1 + D E ̂ …ext. ∠ = sum of 2 opp int ∠s 𝑦 + 65° = 𝑦 + 15° + 𝑦 𝑦 = 50° ̂1 = 𝑦 + 65° E = 50° + 65° = 115° OR ̂B = D ̂ = 𝑦 …opp. ∠s of a rhombus = ̂ ECD = DE ̂C = 𝑦 + 15° … ∠s opp = sides EĈD + DE ̂C + D̂ = 180°… sum of ∠s of ∆ 𝑦 + 15° + 𝑦 + 15° + 𝑦 = 180° 3𝑦 = 150° 𝑦 = 50° ̂ E1 = 𝑦 + 65° = 50° + 65° = 115° 16. A 80°  ∆ RMS ≡ ∆TMS …given ∴ RM ̂ S = TM ̂ S …∆s congruent 𝑥 − 20° = 60° 𝑥 = 80° 17. D 16 cm  In ∆NRQ and ∆NOM: ̂1 = M Q ̂ 2 …corr..∠s; QR ∥ MO ̂ ̂ R1 = O1 … corr..∠s; QR ∥ MO ̂=N N ̂ …common. ∠ or sum ∠s of ∆ ∆NRQ⦀∆NOM ∠∠∠ NR RQ NQ NO = MO = NM … corr. sides are proportional 9 8 6 18 = MO = 12 ∴ MO = 16 cm OR Grade 9 Mathematics P2 Marking Guideline 3

Downloaded from hlayiso.com No. Expected answer Clarification ̂1 = O M ̂1 …alt.∠s ; NO ∥ MP ̂1 = O R ̂1 …corr.∠s; MO ∥ QR ̂ 1 = 𝑅̂1 ∴ M ̂=N P ̂ …opp. ∠s of a parm M̂2 = Ô 2 …alt.∠s ; NO ∥ MP M̂2 = Q̂ 1 …corr∠s MO ∥ QR ̂ ∴ O2 = Q1̂ ∆NRQ⦀∆PMO ∠∠∠ PO = QN + QM = 12 cm …opp.sides of a parm NR = RO = 9 cm …given In ∆PMO and ∆NRQ MO PM RQ = NR …corr. sides are proportional MO 18 = 8 9 MO = 2 × 8 MO = 16 cm 18. C  19. D Q and R  20. C Translated 2 units to the left  and 4 units up. 21. A P( −2 ; 7 )  22. C Q′(1 ; 5)  23. B (𝑥; 𝑦) → (−𝑥 ; 𝑦 − 1)  24. D A′(9 ; −5) and B′(2 ; 1)  25. D ∆KLM  26. B 15 cm  5 × 3 = 15 cm 27. B 𝐴 = 𝑙𝑏  Grade 9 Mathematics P2 Marking Guideline 4

Downloaded from hlayiso.com No. Expected answer Clarification 28. B the perimeter of a circle.  29. D 𝐴 = 𝜋𝑟 2  30. B 15,88 cm  𝐴 = 𝜋𝑟 2 198 = 𝜋𝑟 2 198 𝑟2 = 𝜋 198 𝑟= √ 𝜋 198 Diameter = √ × 2 = 15,88 𝜋 31. D 90  AB + BC + (FE + DC) + (AF + ED) = 20 + 25 + 20 + 25 = 90 32. C 48 cm2  TS = (30 ÷ 10) = 3 cm QS = 6 cm … PR bisects QS 1 Area of PQRS = 2 (PR × QS) 1 = 2 (16 × 6) = 48 cm2 = 48 cm2 33. B 530 cm2  Radius = 30 ÷ 2 = 15 cm 3 𝐴 = 𝜋𝑟 2 4 3 𝐴 = 𝜋(15)2 4 𝐴 = 530 cm2 34. B 20𝑥 cm  0,08𝑥 m = 8𝑥 cm 8𝑥 6𝑥 = 4𝑥 and = 3𝑥 …diagonals bisect at 2 2 right angles ∴ EF 2 = (3𝑥)2 + (4𝑥)2…Pyth theorem EF 2 = 9𝑥 2 + 16𝑥 2 EF 2 = 25𝑥 2 EF = √25𝑥 2 EF = 5𝑥 Perimeter = (5𝑥 × 4) Perimeter = 20𝑥 cm 35. D 13 times  𝑟 = 5 ÷ 2 = 2,5 cm Perimeter of wheel = 2𝜋𝑟 = 2𝜋(2,5) = 15,71 cm 2 m = 200 cm Number of times the wheel turns = 200 ÷ 15,71 = 12,73074475 ≈ 13 times Grade 9 Mathematics P2 Marking Guideline 5

Downloaded from hlayiso.com No. Expected answer Clarification 36. C 40 mm  𝑉 = s3 64 = s3 𝑠 = 4 cm = 40 mm 37. D 248 cm 2  𝑆𝐴 = 2[(6 × 4) + (10 × 4) + (10 × 6)] = 248 cm2 38. A 1120 cm2  1 𝑆𝐴 = 2 ( × 8 × 15) + (25 × 15) + (17 × 25) 2 + (25 × 8) = 120 + 375 + 425 + 200 = 1120 cm2 39. B SA = 2𝜋𝑟 2 + 2𝜋𝑟𝐻  40. C 250 cm3  0,05 m = 5 cm 𝑉 = 5 × 5 × 10 = 250 cm3 41. B 216 𝑙  400 mm = 40 cm and 600 mm = 60 cm 𝑉 = 90 × 40 × 60 = 216 000 cm3 = 216 𝑙 42. B 1,242 m3  90 cm = 0,9 m H of triangular prism = 1,4 – 0,9 = 0,5 𝑉 = (0,9 × 0,9 × 1,2) 1 + ( × 0,9 × 0,5 × 1,2) 2 = 0,972 + 0,27 = 1,242 m3 43. A 3,28 m2  Radius = 25 cm = 0,25 m Diameter = 25 x 2 = 50 cm = 0,5 m 𝑆𝐴 = (0,5)(2,4) + [𝜋(0,25)2 + 𝜋(0,25)(2,4)] = 1,2 + [0,20 + 1,88] = 3,28 m2 44. B 8,98%  𝑉 = 𝜋(90)2 (197) 𝑉 = 5 013 039,4 cm3 5 013 039,4 cm3 = 5 013 𝑙 Percentage of water in tank: 450 × 100 = 8,98% 5 013 45. B 150 mm  2 ℎ2 of triangle = (√4100) − (40)2 = 4100 − 1600 = 2500 ℎ = 50 mm 1 Area of base = × 80 × 50 2 Grade 9 Mathematics P2 Marking Guideline 6

Downloaded from hlayiso.com No. Expected answer Clarification = 2000 mm2 = 20 cm2 300 ml = 300 cm3 20 cm2 × 𝐻 = 300 cm3 𝐻 = 15 cm = 150 mm OR 2 ℎ2 of triangle = (√4100) − (40)2 = 4100 − 1600 = 2500 ℎ = 50 mm = 5 cm 1 Area of base = × 8 × 5 2 = 20 cm2 300 ml = 300 cm3 (1 ml = 1 cm3) 20 cm2 × 𝐻 = 300 cm3 𝐻 = 15 cm = 150 mm Section A total 45 SECTION B  Do not penalise the learner for the same mistake twice.  There are no half marks. MARKING CODES M Method mark A Accuracy mark CA Consistent Accuracy mark Grade 9 Mathematics P2 Marking Guideline 7

Downloaded from hlayiso.com No. Expected answer Clarification Mark 46. Translating 5 units to the left: 1 mark 3 Translating 1 unit up: 1 mark Correct quadrant: 1 mark M ̂1 = 𝑥 …vert opp ∠s M 47. O Statement and reason:1 mark 4 ̂ 3 = 3𝑥 O ̂5 = 𝑥 O Statement and reason:1 mark 3𝑥 + 3𝑥 + 𝑥 + 𝑥 + 𝑥 + 𝑥 = 360° …angles around point M Simplification:1 mark 10𝑥 = 360° M 10𝑥 360° = 10 10 𝑥 = 36° A Answer:1 mark OR OR Statement and reason:1 mark ̂ 3 = 3𝑥 …vert opp ∠s M O Statement and reason:1 mark 3𝑥 + 𝑥 + 𝑥 = 180° …∠s on a str line ADM 5𝑥 = 180° M Simplification:1 mark 5𝑥 180° = 5 5 𝑥 = 36° A Answer:1 mark Accept any other straight line used. 48. 40 mm = 4 cm A Correct conversion: 1 mark 3 1 𝐴 = 2 𝑏ℎ 1 100 = 2 × 4 × ℎ M Substitution into the correct formula: 1 100 = 2ℎ mark ℎ = 50 cm A Answer: 1 mark Grade 9 Mathematics P2 Marking Guideline 8

Downloaded from hlayiso.com No. Expected answer Clarification Mark ̂ + 38° + 75° = 180°…sum of ∠s of ∆ 49. P Statement and reason: 1 mark 5 M ̂ = 180° − 113° P ̂ = 67° CA P Answer: 1 mark ̂1 + 67° = 180°…co-int ∠s PQ∥ST T Statement and reason: 1 mark M ̂1 = 180° − 67° T ̂1 = 113° CA T Answer: 2 marks OR ̂ R1 = 75° …vert opp ∠s M Statement and reason: 1 mark Statement and reason: 1 mark Ŝ = 𝑄̂ …alt ∠s; PQ∥ST M Ŝ = 38° A Answer: 1 mark ̂ 1 + Ŝ …ext ∠ = sum of 2 opp ̂1 = R T Statement and reason: 1 mark int ∠s M ̂1 = 75° + 38° T ̂1 = 113° CA Answer: 1 mark T OR OR ̂+̂ ̂+Q P R 2 = 180° …sum of ∠s of ∆ Statement and reason: 1 mark M ̂ + 38° + 75° = 180° P ̂ = 180° − 113° P ̂ = 67° CA P Answer: 1 mark Statement and reason: 1 mark ̂2 = P T ̂ …alt ∠s; PQ∥ST M ̂2 = 67° T ̂1 + T T ̂2 = 180° …∠s on a str line M Statement and reason: 1 mark ̂1 + 67° = 180° T ̂1 = 180° − 67° T ̂1 = 113° CA T Answer: 1 mark Accept any other correct responses. Grade 9 Mathematics P2 Marking Guideline 9

Downloaded from hlayiso.com No. Expected answer Clarification Mark 50. OU = TO = 5 cm …diagonals of a Statement and reason: 1 mark 5 parm bisect each other M/A In ∆TON Statement and reason: 1 mark ON 2 = TO2 − TN2 …Pyth theorem M = (5)2 − (4)2 M Substitution: 1 mark = 25 − 16 M Simplification: 1 mark = 9 cm2 ON = 3 cm CA Answer: 1 mark 51. Volume of the room 4 𝑉 = 𝑙×𝑏×ℎ 𝑉 =5 ×3 ×2 𝑉 = 30 m3 A Answer: 1 mark Volume of the box 𝑉 = 𝑙×𝑏×ℎ 𝑉 = 10 × 6 × 4 Answer: 1 mark 𝑉 = 240 cm3 A Conversion: 1 mark 𝑉 = 0,00024 m3 A 30 Number of boxes = 0,00024 Number of boxes = 125 000 CA Answer: 1 mark OR OR Volume of the room 𝑉 = 𝑙×𝑏×ℎ 𝑉 =5 ×3 ×2 𝑉 = 30 m3 A Answer: 1 mark Conversion: 1 mark 𝑉 = 3000000 cm3 A Volume of the box 𝑉 = 𝑙×𝑏×ℎ 𝑉 = 10 × 6 × 4 Answer: 1 mark 𝑉 = 240 cm3 A 3000000 Number of boxes = 240 Number of boxes = 125 000 CA Answer: 1 mark Accept any other correct responses. Grade 9 Mathematics P2 Marking Guideline 10

Downloaded from hlayiso.com No. Expected answer Clarification Mark 52. Angle formed by the ray of light to the Identifying the common angle: 1 mark 6 top of the flag pole and the horizontal line from the light source to the pole, is common. M Angle between the wooden beam and Identifying the angles = 900 : 1 mark the horizontal line from the light source to the pole = angle formed by the flag pole and the horizontal line from the light source = 900 . M ∴ Angle formed by the ray of light and Concluding 3rd angles =: 1 mark wooden beam = angle formed by the ray of light and the flag pole. M The triangles formed by the light source, wooden beam and flag pole are similar. 10 18 110 = 𝑥 M Proportion: 1 mark 10𝑥 = 1980 10𝑥 1980 = 10 10 𝑥 = 198 from the height of the light source M Answer: 1 mark ℎ = 198 + 2 ℎ = 200 m A Answer: 1 mark Section B total 30 Grade 9 Mathematics P2 Marking Guideline 11