Mathematics Grade 8 Paper 2 November Mock 2026 Memo

Copyright reserved Please turn over Mock hlayiso mock exams Practice Series: Grade 8 Exam Preparation WWW.HLAYISO.COM SENIOR PHASE END-OF-YEAR EXAMINATION GRADE 8 MATHEMATICS P2 NOVEMBER 2026 (MOCK) MARKING GUIDELINES MARKS: 75 These marking guidelines consist of 7 pages.

Mathematics/P2 2 Senior Phase – Marking Guidelines Hlayiso/Mock 2026 NOTE: • If a learner answers a question TWICE, only mark the FIRST attempt. • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines, unless stated otherwise. Stop marking at the second calculation error. • Accept answers rounded correctly to TWO decimal places. Penalise units only where the question asks for them. • Accept any other correct method and award marks accordingly. GEOMETRY S A mark for a correct statement (a statement mark is independent of a reason). R A mark for the correct reason (a reason mark may only be awarded if the statement is correct). S/R Award a mark if the statement AND reason are both correct. Accept the usual abbreviations, e.g. ∠s on a str line; vert opp ∠s; corresp ∠s, AB∥CD; alt ∠s; co-int ∠s; sum of ∠s in △; ext ∠of △; ∠s opp = sides; opp ∠s of ∥m; sides of rhombus. QUESTION 1 1.1 C (reflex: between 180◦and 360◦) ✓C (1) 1.2 C (x = 180◦−115◦= 65◦; co-int ∠s, KL ∥MN) ✓C (1) 1.3 A (360◦−100◦−80◦−95◦= 85◦) ✓A (1) 1.4 D (rhombus) ✓D (1) 1.5 A (SAS / s∠s) ✓A (1) 1.6 B (52 +122 = 169 = 132) ✓B (1) 1.7 C (3,14×102 = 314 cm2) ✓C (1) 1.8 D (100×100 = 10000 cm2) ✓D (1) Copyright reserved Please turn over

Mathematics/P2 3 Senior Phase – Marking Guidelines Hlayiso/Mock 2026 1.9 A ( 5 10 = 1 2) ✓A (1) 1.10 B (x; y) →(−x; y) gives (−4; −3) ✓B (1) [10] QUESTION 2 2.1.1 3x+10◦+2x+20◦= 180◦ ∠s on a str line 5x = 150◦ ∴x = 30◦ ✓180◦ ✓reason ✓x = 30◦ (3) 2.1.2 DˆEB = AˆEC = 3(30◦)+10◦= 100◦ [vert opp ∠s] OR CˆEB = 80◦, so DˆEB = 180◦−80◦= 100◦[∠s on a str line] ✓100◦(reason required; CA from 2.1.1) (1) 2.2 a = V ˆWX = 64◦ [corresp ∠s; PQ ∥RS] b = V ˆXW = 47◦ [alt ∠s; PQ ∥RS] c = W ˆVX = 180◦−64◦−47◦= 69◦[sum of ∠s in △] OR Q ˆVW = 180◦−64◦= 116◦[∠s on a str line]; c = 116◦−47◦= 69◦ ✓S ✓R (a) ✓S ✓R (b) ✓S ✓R (c) (6) 2.3.1 2x+15◦+3x−5◦= 180◦ [co-int ∠s; AB ∥CD] 5x+10◦= 180◦ 5x = 170◦ x = 34◦ ✓S ✓R ✓x = 34◦ (3) 2.3.2 B ˆGH = 2(34◦)+15◦= 83◦ A ˆGH = 180◦−83◦= 97◦ [∠s on a str line] OR A ˆGH = G ˆHD = 3(34◦)−5◦= 97◦[alt ∠s] ✓97◦(CA from 2.3.1) (1) [14] Copyright reserved Please turn over

Mathematics/P2 4 Senior Phase – Marking Guidelines Hlayiso/Mock 2026 QUESTION 3 3.1.1 Rectangle ✓answer (1) 3.1.2 Kite ✓answer (1) 3.1.3 Parallelogram ✓answer (1) 3.2 ˆQ = ˆR1 [∠s opp = sides] ˆR1 = 180◦−(4x+10◦) = 170◦−4x [∠s on a str line] 4x+10◦= 2x−10◦+(170◦−4x) [ext ∠of △] 6x = 150◦ x = 25◦ OR 2x−10◦+2(170◦−4x) = 180◦[sum of ∠s in △] ⇒6x = 150◦, x = 25◦ (Check: ˆP = 40◦, ˆQ = ˆR1 = 70◦, P ˆRS = 110◦) ✓S/R ˆQ = ˆR1 ✓S/R ˆR1 = 170◦−4x ✓S/R ext ∠/ sum of ∠s ✓x = 25◦ (4) 3.3 AB = BC [sides of rhombus] ∴ˆA1 = B ˆCA [∠s opp = sides] ˆA1 = 180◦−116◦ 2 = 32◦ [sum of ∠s in △] ✓S/R ˆA1 = B ˆCA (AB = BC) ✓R sum of ∠s in △ ✓32◦ (3) 3.4 AB = DE (= 5 cm) ∴△ABC ≡△DEF [SAS / s∠s] ✓DE ✓SAS (2) 3.5 YZ LM = XY KL [||| △s] YZ 6 = 10 4 (scale factor 2,5) YZ = 15 cm ✓correct proportion / scale factor 2,5 ✓YZ = 15 cm (2) [14] Copyright reserved Please turn over

Mathematics/P2 5 Senior Phase – Marking Guidelines Hlayiso/Mock 2026 QUESTION 4 4.1 BC2 = AC2−AB2 [Pythagoras] BC2 = 262 −102 = 676−100 = 576 BC = 24 cm ✓correct use of Pythagoras ✓576 ✓BC = 24 cm (3) 4.2 PQ2+ QR2 = 92 +402 = 81+1600 = 1681 PR2 = 412 = 1681 PQ2+ QR2 = PR2, so △PQR is right-angled (at Q) [converse of Pythagoras] ✓1681 (sum of squares) ✓412 = 1681 ✓conclusion (3) 4.3 brace2 = 1,62 +1,22 = 2,56+1,44 = 4 brace = 2 m ✓substitution ✓2 m (2) [8] QUESTION 5 5.1.1 CE2 = BC2+ BE2 = 62 +82 = 100 [Pythagoras] CE = 10 cm ✓substitution ✓10 cm (2) 5.1.2 P = AE + EC + CD + DA = 22+10+14+6 = 52 cm ✓52 cm (CA from 5.1.1) (1) 5.1.3 A = 14×6+ 1 2 ×8×6 = 84+24 = 108 cm2 ✓84 and 24 ✓108 cm2 (2) 5.2.1 C = πd = 3,14×14 = 43,96 m ✓43,96 m (1) 5.2.2 r = 7 m Pond = πr2 = 3,14×72 = 153,86 m2 Lawn = 20×20 = 400 m2 Grass = 400−153,86 = 246,14 m2 ✓3,14×72 (153,86) ✓400 ✓246,14 m2 (3) [9] Copyright reserved Please turn over

Mathematics/P2 6 Senior Phase – Marking Guidelines Hlayiso/Mock 2026 QUESTION 6 6.1 V = l ×b×h = 30×20×15 = 9000 cm3 ✓substitution ✓9 000 cm3 (2) 6.2 TSA = 2(30×20)+2(30×15)+2(20×15) = 1200+900+600 = 2700 cm2 ✓any two pairs of faces correct ✓all three pairs ✓2 700 cm2 (3) 6.3 9000 cm3 = 9000 mℓ = 9 ℓ ✓9 000 mℓ(CA from 6.1) ✓9 ℓ (2) 6.4 5 faces (2 triangles and 3 rectangles) ✓5 (1) [8] QUESTION 7 7.1 A′(1; −1), B′(4; −1), C′(1; −3) [(x; y) →(x; −y)] ✓any two correct ✓all three correct (2) 7.2 (x; y) →(x−5; y+2) ✓answer (1) 7.3 B′′(−4; −1) [(x; y) →(−x; −y)] ✓answer (1) [4] Copyright reserved Please turn over

Mathematics/P2 7 Senior Phase – Marking Guidelines Hlayiso/Mock 2026 QUESTION 8 8.1 7; 7; 11; 12; 13; 14; 15; 15; 15; 16; 18 ✓correct order (1) 8.2 Range = 18−7 = 11 ✓11 (1) 8.3 Mean = 7+7+11+12+13+14+15+15+15+16+18 11 = 143 11 = 13 ✓143 ✓13 (2) 8.4 Median = 6th value = 14 ✓14 (CA from 8.1) (1) 8.5 Mode = 15 ✓15 (1) 8.6 Marks more than 14: 15; 15; 15; 16; 18 (5 learners) P(more than 14) = 5 11 (≈0,45 or 45,45%) ✓5 favourable outcomes ✓5 11 (2) [8] TOTAL: 75 Copyright reserved