Mathematics Grade 9 P2 November Mock 2026 Memo (English)

Mock hlayiso mock exams Practice Series: Grade 9 Exam Preparation WWW.HLAYISO.COM SENIOR PHASE END-OF-YEAR EXAMINATION GRADE 9 MATHEMATICS P2 NOVEMBER 2026 (MOCK) MARKING GUIDELINES MARKS: 75 These marking guidelines consist of 7 pages. Copyright reserved Please turn over

Mathematics/P2 2 Hlayiso/Mock 2026 Senior Phase – Marking Guidelines 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. QUESTION 1 1.1 B (2x + 30◦ = 110◦ , vert opp ∠s, so x = 40◦ ) ✓B (1) 1.2 C (square) ✓C (1) 1.3 A (∠∠∠ gives similar, not necessarily congruent, ✓A (1) triangles) 1.4 B P′ (−3 ; −4) ✓B (1) 1.5 D 2,4 kℓ = 2,4 m3 = 2 400 000 cm3 ✓D (1) [5] QUESTION 2 2.1.1 B̂1 + B̂2 + B̂3 = 180◦ ∠s on a str line ✓ 180◦ ✓ reason x + 2x + 3x − 30◦ = 180◦ Substitution ✓ 210◦ 6x = 210◦ ✓ x = 35◦ (4) ∴ x = 35◦ Copyright reserved Please turn over

Mathematics/P2 3 Hlayiso/Mock 2026 Senior Phase – Marking Guidelines 2.1.2 B̂3 = 3(35◦ ) − 30◦ = 75◦ ✓ answer (CA from 2.1.1)(1) 2.2 a = GĤK = 68◦ [corresp ∠s; AB ∥ CD] ✓S ✓R (a) b = GK̂H = 40◦ [alt ∠s; AB ∥ CD] ✓S ✓R (b) ◦ ◦ ◦ ◦ ✓S ✓R (c) (6) c = HĜK = 180 − 68 − 40 = 72 [sum of ∠s in △] OR BĜH = 180◦ − 68◦ = 112◦ [∠s on a str line]; c = 112◦ − 40◦ = 72◦ 2.3.1 3x + 40◦ + 5x − 20◦ = 180◦ [∠s on a str line] ✓S 8x + 20◦ = 180◦ ✓R 8x = 160◦ ✓ x = 20◦ (3) x = 20◦ 2.3.2 PV̂W = 3(20◦ ) + 40◦ = 100◦ and ✓ both angles = 100◦ VŴS = 4(20◦ ) + 20◦ = 100◦ ✓R (2) ∴ PV̂W = VŴS ∴ PQ ∥ RS [alt ∠s equal] OR QV̂W + VŴS = 80◦ + 100◦ = 180◦ ; PQ ∥ RS [co-int ∠s suppl] [16] Copyright reserved Please turn over

Mathematics/P2 4 Hlayiso/Mock 2026 Senior Phase – Marking Guidelines QUESTION 3 3.1.1 Trapezium ✓ answer (1) 3.1.2 Kite ✓ answer (1) 3.1.3 Rhombus ✓ answer (1) 3.2 L̂ = M̂1 [∠s opp = sides] ✓ S/R L̂ = M̂1 ◦ ◦ ◦ M̂1 = 180 − (3x − 20 ) = 200 − 3x [∠s on a str line] ✓ S/R M̂1 = 200◦ − 3x 3x − 20◦ = x + 10◦ + (200◦ − 3x) [ext ∠ of △] ✓ S/R ext ∠ / sum of ∠s 5x = 230 ◦ ✓ x = 46◦ (4) x = 46◦ OR x + 10◦ + 2(200◦ − 3x) = 180◦ [sum of ∠s in △] ⇒ 5x = 230◦ , x = 46◦ (Check: K̂ = 56◦ , L̂ = M̂1 = 62◦ , KM̂N = 118◦ ) 3.3 Â + B̂ = 180◦ [co-int ∠s; AD ∥ BC] ✓S 3y − 10◦ + 2y + 15◦ = 180◦ ✓R 5y = 175◦ ✓ y = 35◦ (3) y = 35◦ 3.4.1 In △AEC and △BED: ✓ AE = EB AE = EB [given] ✓ CE = ED CE = ED [given] ✓ S/R Ê1 = Ê2 [vert opp ∠s] ✓ R (SAS) (4) ∴ △AEC ≡ △BED [SAS / s∠s] 3.4.2 BD = AC [≡ △s] ✓ BD with reason (1) 3.5.1 In △MNP and △RST: ✓ both pairs of angles M̂ = R̂ [given] ✓ S/R P̂ = T̂ N̂ = Ŝ [given] ✓ R (∠∠∠) (3) P̂ = T̂ [sum of ∠s in △ / 3rd ∠] ∴ △MNP ||| △RST [∠∠∠] Copyright reserved Please turn over

Mathematics/P2 5 Hlayiso/Mock 2026 Senior Phase – Marking Guidelines ST RS 3.5.2 = [||| △s] ✓ correct proportion NP MN ST 8 ✓ ST = 12 cm (2) = 9 6 ST = 12 cm [20] QUESTION 4 4.1 y ✓ correct position and shape of image C′ ✓ labels ✓ all three coordinates (3) B′ A′ x A′ (−1 ; 2), B′ (−4 ; 2), C′ (−1 ; 5) 4.2 (x ; y) → (−x ; y) ✓ answer (1) 4.3.1 (x ; y) → (x − 2 ; y − 6) ✓ answer (1) 4.3.2 B′′ (2 ; −4) ✓ answer (1) 4.4 C′′′ (5 ; 1) ((x ; y) → (y ; x)) ✓ answer (1) 4.5.1 D(2 ; 4), E(8 ; 4), F(2 ; 10) ✓ any two correct ✓ all three correct (2) 4.5.2 4 times (22 = 4; area of △ABC = 4,5 units2 , area of ✓ answer (1) △DEF = 18 units2 ) [10] Copyright reserved Please turn over

Mathematics/P2 6 Hlayiso/Mock 2026 Senior Phase – Marking Guidelines QUESTION 5 5.1.1 EC = 15 − 9 = 6 cm and BE = AD = 8 cm ✓ EC = 6 and BE = 8 BC2 = BE2 + EC2 [Pythagoras] ✓ substitution in Pythagoras BC2 = 82 + 62 = 100 ✓ BC = 10 cm (3) BC = 10 cm 5.1.2 P = 9 + 10 + 15 + 8 = 42 cm ✓ answer (CA from 5.1.1)(1) 5.1.3 A = 12 (AB + DC) × AD = 12 (9 + 15) × 8 ✓ substitution = 96 cm2 ✓ answer (2) OR rectangle 9 × 8 = 72 plus triangle 12 × 6 × 8 = 24; total 96 cm2 5.2.1 Two semicircles = one full circle with d = 40 m ✓ 2 × 60 P = 2 × 60 + π d = 120 + 3,14 × 40 ✓ 3,14 × 40 = 120 + 125,6 = 245,6 m ✓ 245,6 m (3) 5.2.2 Field area = 60 × 40 + π r2 = 2 400 + 3,14 × 202 ✓ 60 × 40 = 2 400 + 1 256 = 3 656 m2 ✓ 3,14 × 202 (1 256) Grass = 3 656 − 22 × 3 = 3 656 − 66 = 3 590 m2 ✓ 3 590 m2 (3) [12] Copyright reserved Please turn over

Mathematics/P2 7 Hlayiso/Mock 2026 Senior Phase – Marking Guidelines QUESTION 6 6.1.1 h2 = 52 + 122 = 169 [Pythagoras] ✓ substitution h = 13 cm ✓ 13 cm (2) 6.1.2 V = area of base × height = 12 × 5 × 12 × 20 ✓ 21 × 5 × 12 (= 30) = 600 cm3 ✓ 600 cm3 (2) 6.1.3 TSA = 2 × 12 × 5 × 12 + (5 + 12 + 13) × 20 ✓ two triangles (60) = 60 + 600 ✓ three rectangles (600) (CA = 660 cm2 from 6.1.1) ✓ 660 cm2 (3) 6.2.1 r = 0,7 m ✓ r = 0,7 V = π r2 h = 3,14 × 0,72 × 1,5 ✓ substitution = 2,31 m3 (2,3079) ✓ answer (3) 6.2.2 2,3079 × 1 000 = 2 307,9 ℓ ✓ answer (CA from 6.2.1)(1) Accept 2 307,9 ℓ / 2 308 ℓ / 2 310 ℓ (CA from 6.2.1) 6.2.3 4 times bigger (π (2r)2 h = 4π r2 h) ✓ answer (1) [12] TOTAL: 75 Copyright reserved