Mathematics Grade 9 GEC Test 2022 Memo (English)

Downloaded from hlayiso.com GEC PILOT STUDY MARKING GUIDELINE 2022 MATHEMATICS: ENGLISH GRADE 9 SECTION A  One mark per answer.  There are no half marks.  Tick () only the correct answer and underline the incorrect one.  No marks should be credited for multiple answers. Underline them all.  If there is no answer indicated, skip the question Option No. Expected answer 1. An irrational number D 2. 14 = 2 × 7 B 16 = 2 × 2 × 2 × 2 ∴LCM = 2 × 2 × 2 × 2 × 7 = 112 3. 𝑑 D 𝑠= 𝑡 180 𝑠= = 90 km/h 2 210 𝑡 = 60 = 3,5 h 𝑑 =𝑠×𝑡 𝑑 = 90 × 3,5 = 315 km 4. = 3[−(14)] + 4 × 2 D = −42 + 8 = −34 5. 3(−2(2) × 3) + 6(2)2 (3)2 B = 2×3 3(−12) + 216 = 6 180 = 6 = 30 Grade 9 Mathematics: English Marking Guideline 1

Downloaded from hlayiso.com Option No. Expected answer 6. −64 8 A = + −4 2×2 2 = −16 + 4 − 4 = −16 7. = 𝑎12−3 B = 𝑎9 8. 1 1 11 D + 11 30 = 5 6 = 30 = × = 11 1 1 1 30 1 × 5 6 30 9. = 23 (𝑥 5 )6 D = 8𝑥 30 10. = 1 − 1 + 56 ÷ 56 A =1−1+1=1 11. 25 C Figure 1 2 3 4 Dots 7 13 19 25 12. 𝑇5 𝑇4 𝑇3 𝑇2 C = = = 𝑇4 𝑇3 𝑇2 𝑇1 1 𝑎 = 16 and 𝑏 = 16 13. B Arrangement 1 2 3 4 5 6 7 No of tables 1 2 3 4 5 6 7 No of people 4 8 8 16 12 24 16 Sequence 1: Arrangement 1; Arrangement 3 Arrangement 5; … Sequence 2: Arrangement 2; Arrangement 4; Arrangement 6;… The rule for the 7th arrangement is 4𝑛. 14. 10 handshakes A 15. 1 C − 4 16. 7𝑥𝑎𝑦 + 3𝑎𝑥𝑦 D 17. 𝑥 2 + 7𝑥𝑦 − 15𝑦 2 A Grade 9 Mathematics: English Marking Guideline 2

Downloaded from hlayiso.com Option No. Expected answer 18. 16𝑥 3 + 8𝑥 2 D = −2 𝑥 2 2 8𝑥 (2𝑥 + 1) = −2𝑥 2 = −4 (2𝑥 + 1) = −8𝑥 − 4 19. = (1 − 3𝑎)(1 + 3𝑎) B 20. = 3𝑥(𝑥 2 + 3𝑥 − 10) D = 3𝑥(𝑥 + 5)(𝑥 − 2) 21. L = (𝑥 + 1) and W = (𝑥 − 2) D 22. 6(𝑥 2 + 3𝑥 − 10) 8𝑥 C = × 2 48𝑥 𝑥 −4 6(𝑥 + 5)(𝑥 − 2) 8𝑥 = × 48𝑥 (𝑥 − 2)(𝑥 + 2) (𝑥 + 5) = (𝑥 + 2) 23. = 15(𝑠 4 − 𝑧 4 ) C = 15(𝑠 2 − 𝑧 2 )(𝑠 2 + 𝑧 2 ) = 15(𝑠 − 𝑧)(𝑠 + 𝑧)(𝑠 2 + 𝑧 2 ) 24. 𝑥 − 5 = −7 C 𝑥 = −2 25. 3𝑥 + 1 = 10 D 3𝑥 = 9 𝑥=3 26. 𝑥 = 0 or 𝑥 = 4 B 27. 6𝑥−1 = 62 B ∴ 𝑥−1=2 𝑥=3 28. −6 = 12𝑥 − 18 B 12 = 12𝑥 𝑥=1 29. 𝑦 = 𝑥2 + 𝑐 A 𝑎 = (−1)2 + 2 𝑎=3 30. 3𝑥(6𝑥 − 8) = 0 D 3𝑥 = 0 or 6𝑥 − 8 = 0 4 𝑥 = 0 or 𝑥 = 3 31. 4𝑥 2 − 4𝑥 − 48 = 0 … (÷ 4) D 𝑥 2 − 𝑥 − 12 = 0 (𝑥 + 3)(𝑥 − 4) = 0 𝑥 = −3 or 𝑥 = 4 Grade 9 Mathematics: English Marking Guideline 3

Downloaded from hlayiso.com Option No. Expected answer 32. 𝑑2 = 𝑙 2 + 𝑤 2 …Theor. of Pyth. A (𝑤 + 9)2 = (𝑤 + 7)2 + 𝑤 2 𝑤 2 + 18𝑤 + 81 = 𝑤 2 + 14𝑤 + 49 + 𝑤² 𝑤 2 − 4𝑤 − 32 = 0 (𝑤 − 8)(𝑤 + 4) = 0 𝑤 ≠ −4 or 𝑤 = 8 𝑙 = 8 + 7 = 15 A= 𝑙 ×𝑏 = 15 × 8 = 120 cm2 33. 𝑦 = −2𝑥 − 4 D 34. D 1 𝑦 =− 𝑥−1 3 𝑐 = −1 1 𝑚=− 3 35. Rule: 𝑦 = 𝑥 2 + 2 C 258 = 𝑧 2 + 2 𝑧 2 = 256 𝑧 = 16 36. A 𝑦-values are all equal to 4, which implies that the horizontal line is above the X-axis. Grade 9 Mathematics: English Marking Guideline 4

Downloaded from hlayiso.com Option No. Expected answer 37. 𝑦 = 𝑚𝑥 + 𝑐 C 𝑦 = 𝑚𝑥 + 4 Substitute 𝑐 = 4 from the graph 0 = −3𝑚 + 4 Substitute (−3; 0) from the graph 3𝑚 = 4 4 𝑚=3 4 ∴ 𝑦 = 3𝑥 + 4 4 𝑦 = 3 (−2) + 4 −8 𝑦= +4 3 4 𝑦= 3 4 ∴ (−2; 3) lies on the graph (The learner may test different values to check which points lie on the graph. 𝑦 −intercept eliminates options B and D) 38. 𝑦 = 2𝑥 − 8 (Line AB) D 1 1 ∴ 𝑦 = −2𝑥 + 𝑐 m ⊥ AB= − 2 1 4 = − 2 (−2) + 𝑐 Substitute (−2; 4) 𝑐=3 1 ∴ 𝑦 = −2𝑥 + 3 39. 𝐴′(−3; −5) B 40. (𝑥; 𝑦) (−𝑥; 𝑦) A 41. Translated 5 units up A 42. 90° B 43. 80° A 44. Alternate angles AB‖CD A 45. AÔ C = BÔ D Vertically Opp. ∠𝑠 = D 3𝑥 − 5 = 𝑥 + 250 0 3𝑥 − 𝑥 = 250 + 50 2𝑥 = 300 𝑥 = 150 ∴ AÔ C = 3𝑥 − 50 = 3(15°) − 5° = 45° − 5° = 40° 46. ̂ D = 90° AD⊥BE, BO D 𝑥 +5° +35° = 90° …∠𝑠 on a str. line ∴ 𝑥 =50° Grade 9 Mathematics: English Marking Guideline 5

Downloaded from hlayiso.com Option No. Expected answer 47. Two angles opposite equal sides are equal. D 48. ̂H = GM MG ̂ H = GH ̂ M = 60° ∠𝑠 of OR equilat. Δ ̂M = GM 𝐴G ̂ H = 60° Alt. ∠𝑠 =, AB//CD ̂ D + GH GH ̂ M = 180° ∠𝑠 on a str. line ̂H = AG AG ̂M + MG ̂H° ̂ D = 120° GH = 60° + 60° ̂B = GH EG ̂ D Corr. ∠𝑠, AB//CD = 120° 4𝑦 + 20° = 120° ̂H = EG AG ̂B Vert. opp. ∠𝑠 4𝑦 = 100° 4𝑦 + 20° = 120° 100 4𝑦 = 100° ∴𝑦= 4 100 ∴𝑦= = 25° 4 OR = 25° ̂H = GM MG ̂ H = GH ̂ M = 60° ∠𝑠 of equilat. Δ ̂H = GM BG ̂ H = 60° … Alt. ∠𝑠, AB//CD 4𝑦 + 20° + 60° = 180° … ∠𝑠 on a str 4𝑦 + 20° = 120° 4𝑦 = 100° 100 ∴𝑦= 4 = 25° Grade 9 Mathematics: English Marking Guideline 6

Downloaded from hlayiso.com Option No. Expected answer 49. In ΔEFG and ΔABC OR A EF FG EG In ΔEFG and ΔABC = = AB BC AC EF FG EG 12 cm 4 cm 8 cm = = = = =2 AB BC AC 6 cm 2 cm 4 cm 12 cm 4 cm 8 cm ∴ ΔEFG||| ΔABC Corr. Sides are 6 cm = 2 cm = 4 cm = 2 proportional ∴ ΔEFG||| ΔABC Corr. Sides are 𝐴𝐶̂ 𝐵 = 𝐸𝐺̂ 𝐹 = 86° proportional 𝐵̂1 = ̂F = 58° 𝐵̂1 = F̂ corr. … ∠𝑠 of similar △ 𝑠 ̂ ∴A =E ̂2 corr. 𝐵̂1 = 58° ∠𝑠 of similar △ = ̂ + 𝐵̂1 + Ĉ = 180° … sum of ∠𝑠 of △ A = 36° ̂ A + 58° + 86° = 180° 𝐸̂1 = ̂A alter. ∠𝑠 DE∥AC ̂ A = 180° − 144° =36° ̂ A = 36° 𝐸̂1 = ̂ A alter…. ∠𝑠 = DE∥AC 𝐸̂1 =36° 50. 𝐵̂2 = 𝐵̂1 = 420 diagonals of rhombus bisect ∠𝑠 C 𝐴𝐵̂𝐶 = 840 𝐴𝐵̂𝐶 + 𝐵𝐶̂ 𝐷 = 1800 Co-int ∠𝑠 AB∥CD 0 ̂ 84 + 𝐵𝐶 𝐷 = 180 0 𝐵𝐶̂ 𝐷 = 960 𝐵𝐶̂ 𝐷 = 𝐵𝐶̂ 𝐴 + 𝐴𝐶̂ 𝐷 960 = 𝐵𝐶̂ 𝐴 + 𝐴𝐶̂ 𝐷 But 𝐵𝐶̂ 𝐴 = 𝐴𝐶̂ 𝐷 diagonals of rhombus bisect ∠𝑠 ∴ 96 = 2𝐴𝐶̂ 𝐷 0 𝐴𝐶̂ 𝐷 = 480 51. 𝐹̂2 = 𝐵̂1 = 680 Alt ∠𝑠 AB∥CD OR A 𝐸̂2 + 𝐵̂1 = 180 0 Co-int ∠𝑠 ED∥BF 𝐵̂1 = 𝐸̂1 = 680 … Corr. ∠𝑠 ED∥BF 𝐸̂2 + 680 = 1800 𝐴̂ + 𝐷̂1 + 𝐸̂1 = 1800 …. ∠𝑠 of ∆ 𝐸̂2 = 1120 𝐴̂ + 𝐷̂1 + 680 = 1800 𝐴̂ + 𝐷 ̂1 = 𝐸̂2 Ext ∠ of ∆ 𝐴̂ + 𝐷 ̂1 = 1120 ̂ ̂ 𝐴 + 𝐷1 = 112 0 But 𝐴̂ = 𝐷 ̂1 …∠𝑠 opp equal sides But 𝐴̂ = 𝐷 ̂1 ∠𝑠 opp equal sides ∴ 𝐴̂ = 560 ∴ 2𝐴̂ = 1120 𝐴̂ = 560 52. AC 2 = AB 2 + BC 2 … Theor. Of Pyth. C = (15 cm)2 + (8 cm)2 = 225 cm2 + 64 cm2 = 289 cm2 AC = 17 cm Grade 9 Mathematics: English Marking Guideline 7

Downloaded from hlayiso.com Option No. Expected answer 53. 𝑃 = 2(𝑙 + 𝑏) = 40 C 𝑙 + 𝑏 = 20 For maximum area: 𝑙 = 𝑏 = 10 ∴ 𝐴 = 𝑙2 = 102 = 100 Using different combinations other 10 gives smaller area. e.g. 𝐴 = 9 × 11 = 99 cm2 𝐴 = 8 × 12 = 96 cm2 54. 1 B For 2 smaller rectangles AB = (5 ÷ 10 = ) cm 2 BC = 4 cm 1 ∴area = 2 × 4 = 2 cm2 55. 𝐴 = 𝜋𝑟 2 = 36𝜋 A ∴ 𝑟 2 = 36 𝑟=6 𝐶 = 2𝜋𝑟 =2×6×𝜋 = 12𝜋 56. 𝐶 = 2𝜋𝑟 B =2×𝜋×3 = 6𝜋 2 For 2 out of 3 children: 𝐶 = × 6𝜋 3 = 12,57 cm 57. 1 A V= 2 × 𝑏 × ℎ × 𝑙 1 = ( × 8 × 6 × 14) cm3 2 = (4 × 6 × 14) cm3 = 336 cm3 58. 𝑆𝐴 = 2𝜋𝑟ℎ + 2𝜋𝑟 2 D 22 22 = (2 × × 7 × 16 + 2 × × 72 ) cm2 7 7 = (2 × 22 × 16 + 2 × 22 × 7) cm2 = (704 + 308) cm2 = 1012 cm2 Grade 9 Mathematics: English Marking Guideline 8

Downloaded from hlayiso.com Option No. Expected answer 59. 300 ml = 300 cm3 B 𝑉 = 𝜋𝑟 2 ℎ 22 300 cm3 = × 𝑟 2 × 13 cm 7 300 cm3 7 × = 𝑟2 13 cm 22 1050 cm2 = 𝑟 2 143 𝑟 = 2,71 cm 𝑑 = 2 × 2,71 cm = 5,42 𝑐m 60. 𝑆𝐴 = 2(𝑙𝑏 + 𝑏ℎ + 𝑙ℎ) C ℎ = 140 mm= 14 cm 𝑏 = 2ℎ 𝑏 = 2(14) cm= 28 cm OR 3640 = 2(𝑙(28) + 28(14) + 𝑙(14)) 3640 = 28𝑙 + 392 + 14𝑙 3640 = 2(28𝑙 + 392 + 14𝑙) 2 3640 = 2(42𝑙 + 392) 1820 − 392 = 42𝑙 3640 = 84𝑙 + 784 1428 = 42𝑙 1428 84𝑙 =784 =𝑙 𝑙 = 34 cm 42 34 cm = 𝑙 Grade 9 Mathematics: English Marking Guideline 9

Downloaded from hlayiso.com SECTION B Marking guideline for Section B  Do not penalise the learner for the same mistake more than once.  There are no half marks.  Underline errors committed by learners do not place a cross (X).  In instances where learners have used different but mathematically sound strategies to solve problems, they (learners) must be credited.  Credit learners for providing the correct answer without showing calculations. M a mark for a correct method A a mark for accurate calculation key CA a mark for consistent accuracy No. Expected answer Rationale/Clarification Mark 61. 2𝑦 = 2𝑥 + 4 1 mark for the 𝑦 = 𝑥 + 2 A standard form 𝑦 −intercept if 𝑥 = 0 𝑦 =0+2 1 mark for plotting both 𝑦=2 intercepts 𝑥 −intercept if 𝑦 = 0 0=𝑥+2 1 mark for the 𝑥 = −2 shape/gradient Full marks for the correct graph plotted OR 3 1 mark for plotting 𝑥 −intercept 1 mark for plotting 𝑦 −intercept 1 mark for the shape/gradient NB: 1 mark if the learner has plotted incorrect CACA intercepts but the gradient of the graph is positive. Grade 9 Mathematics: English Marking Guideline 10

Downloaded from hlayiso.com No. Expected answer Rationale/Clarification Mark 62. A (1 ; 2), B (−2 ; 3) and C (−4 ; −1) 1 mark for correct plotting 𝐴′ (1 ; −2) of 𝐴′ . 𝐵′ (−2 ; −1) 1 mark for correct plotting 𝐶 ′ (−4 ; −5) of 𝐵′ . 1 mark for correct plotting of 𝐶 ′ . Full marks for the correct plotted image 3 AAA 63. MN ̂ P = MP̂N = 54°  M ∠𝑠 opp equal sides (MN=MP) 1 mark for statement and ̂ + 54° + 54° = 180° M sum of int ∠𝑠 of△ reason ̂ = 72°CA M ̂ +Q M ̂ = 180° co-int ∠𝑠 (MN∥RQ) ∴Q ̂ = 108° CA 1 mark for 𝑄̂ OR ̂P1 = 54° ∠𝑠 opp equal sides (MN=MP) OR ̂ N2 = 54° Alt ∠𝑠 (NR∥MQ) ̂ = 108° CA 1 mark for 𝑄̂ ∴Q Opp ∠𝑠 of ∥gram In ∆QPR ∴Q̂+R ̂2 + P̂3 = 180° sum of int ∠𝑠 of△ ̂ ̂ But R 2 = P3 ∠𝑠 opp equal sides (PQ=RQ) 108°+2P ̂3 = 180° 2P̂3 = 72° ̂3 1 mark for P ̂3 = 36°  CA P ̂ ̂ P1 + 𝑦 + P3 = 180° but P̂1 = 54° ∠𝑠 opp equal sides (MN=MP) 1 mark for 𝑦 54° + 𝑦 + 36° = 180° ∠𝑠 on a straight line 𝑦 = 90°CA OR ̂P1 = N ̂ 2 = 54° … alt ∠𝑠 (NR∥MQ) ̂2 1 mark for N ̂P3 = R ̂ 1 = 36° … alt ∠𝑠 (NR∥MQ) ̂ 1 mark for R1 Grade 9 Mathematics: English Marking Guideline 11

Downloaded from hlayiso.com No. Expected answer Rationale/Clarification Mark In ∆NPR 𝑦+N ̂2 + R ̂ 1 = 180° … sum of int ∠𝑠 of △ 𝑦 + 54° + 36° = 180° 𝑦 = 90°CA 1 mark for 𝑦 Full marks for answer 4 only. 64. Angle at B + 45° + 45° = 180° 1 mark for angle at B ∴ Angle at B = 90° A AB 2 = (Home to 𝐴)2 − (Home to 𝐵)2 … Theorem of 1 mark for statement and PythM reason = (2 km)2 − (1 km)2 = 4 km2 − 1 km2 = 3 km2 AB = √3 km2 = 1,7 km Angle at the shop = angle at school given = 45° B to school = B to shop … sides opp. equal ∠𝑠 1 mark for distance from B to the school = 3 km A (Shop to school)2 = (B to shop)2 + (B to school)2 …Theorem of Pyth = ( 3 km)2 + (3 km)2 = 9 km2 + 9 km2 5 = 18 km2 Shop to school = √18 km2 = 4,2 km (School to Home)2 = (Home to B)2 + (B to school)2 Theorem of Pyth = ( 1 km)2 + (3 km)2 = 1 km2 + 9 km2 = 10 km2 1 mark for distance from School to home = √10 km2 school to home = 3,2 km CA 1 mark for the answer The total distance = 2 km + 1,7 km+ 3 km +4,2 km+3,2 km = 14,1 km CA Full marks for answer only. Grade 9 Mathematics: English Marking Guideline 12