Mathematics Grade 9 P1 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 P1 NOVEMBER 2026 (MOCK) MARKING GUIDELINES MARKS: 75 These marking guidelines consist of 7 pages. Copyright reserved Please turn over

Mathematics/P1 2 Hlayiso/Mock 2026 Senior Phase – Marking Guidelines NOTE: • If a learner answers a question TWICE, only mark the FIRST attempt. • If a learner has crossed out an attempt and not redone the question, mark the crossed-out version. • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines. Stop marking at the second calculation error. • Accept correct alternative methods. Answers only: award full marks ONLY where the guideline says so. • Assuming answers/values in order to solve a problem is NOT acceptable. SECTION A QUESTION 1 √ 1.1 C (12 is not a perfect square, so 12 is irrational; ✓C (1) √ 49 = 7, 0,3̇ = 13 and − 58 are rational) 1.2 A (4,72 × 10−5 ) ✓A (1) 1.3 B (d = −4; T1 = −4(1) + 13 = 9) ✓B (1) 1.4 D (− 23 × 32 = −1) ✓D (1) 1.5 C (4x2 − 20x + 25) ✓C (1) [5] SECTION B Copyright reserved Please turn over

Mathematics/P1 3 Hlayiso/Mock 2026 Senior Phase – Marking Guidelines QUESTION 2 2.1 252 = 2 × 2 × 3 × 3 × 7 ✓ prime factors 252 = 22 × 32 × 7 ✓ exponential form (2) 2.2 HCF = 22 × 32 = 36 ✓ HCF = 36 LCM = 22 × 32 × 5 × 7 = 1 260 ✓ LCM = 1 260 (CA from 2.1) (2) √ √ 2.3 = 64 + (−5) − 9 × 2 ✓ 64 = 8 = 8 − 5 − 18 ✓ −5 and −18 = −15 ✓ answer (3) 2.4 −7 + 12 = 5 ◦ C at 14:00 ✓ 5 ◦C 5 − 9 = −4 ◦ C ✓ −4 ◦ C (Answer only: full marks) (2) [9] QUESTION 3 3.1.1 Deposit = 10% × R8 400 = R840 ✓ R840 Balance = 8 400 − 840 = R7 560 ✓ R7 560 (2) 3.1.2 SI = P × n × i = 7 560 × 2 × 0,15 ✓ substitution SI = R2 268 ✓ R2 268 Total = 7 560 + 2 268 = R9 828 ✓ R9 828 OR A = 7 560(1 + 2 × 0,15) (CA from 3.1.1) A = 7 560 × 1,3 = R9 828 OR ✓ substitution into A = P(1 + ni) ✓ 1,3 ✓ R9 828 (3) 9 828 3.1.3 = R409,50 per month ✓ answer (CA) (1) 24 3.2 A = 12 000(1 + 0,08)3 ✓ substitution A = R15 116,54 ✓ answer (2) Copyright reserved Please turn over

Mathematics/P1 4 Hlayiso/Mock 2026 Senior Phase – Marking Guidelines 3.3 Distance = 100 × 3 = 300 km ✓ 300 km / 100 × 3 300 ✓ 3 h 45 min (2) Time = = 3,75 h = 3 h 45 min 80 OR inverse proportion: 100 × 3 = 80 × t ⇒ t = 3,75 h [10] QUESTION 4 −27a6 b3 × 2ab4 4.1 = ✓ −27a6 b3 6a5 b2 ✓ −54a7 b7 or −9 −54a7 b7 = ✓ a2 b5 (3) 6a5 b2 = −9a2 b5 2x+2 × (23 )x 4.2 = ✓ prime base 2 (22 )2x+1 ✓ 24x+2 in numerator and 2x+2+3x 24x+2 denominator = 4x+2 = 4x+2 2 2 ✓ answer (3) 0 =2 =1 4.3 = −3x4 + 1 ✓ −3x4 ✓ +1 (2) [8] QUESTION 5 5.1 = 2x2 + 8x − 3x − 12 − (x2 − 4x + 4) ✓ 2x2 + 5x − 12 = 2x2 + 5x − 12 − x2 + 4x − 4 ✓ x2 − 4x + 4 = x2 + 9x − 16 ✓ answer (3) 5.2.1 = 5x(x2 − 9) ✓ 5x(x2 − 9) = 5x(x − 3)(x + 3) ✓ answer (2) 5.2.2 = (x − 5)(x + 3) ✓ (x − 5) ✓ (x + 3) (2) Copyright reserved Please turn over

Mathematics/P1 5 Hlayiso/Mock 2026 Senior Phase – Marking Guidelines 5.2.3 = a(x − 2y) + 3(x − 2y) ✓ +3(x − 2y) = (x − 2y)(a + 3) ✓ answer (2) (x − 4)(x + 4) 5.3 = ✓ numerator factorised (x + 4)(x − 3) ✓ denominator factorised x−4 ✓ answer (3) = x−3 [12] QUESTION 6 6.1.1 ×12: 4(x − 1) − 3(x + 2) = 12 ✓ multiplying by LCD 12 4x − 4 − 3x − 6 = 12 ✓ removing brackets x − 10 = 12 correctly x = 22 ✓ answer (3) 6.1.2 3x−1 = 34 ✓ 34 x−1 = 4 ✓ answer (2) x=5 6.1.3 x2 − 3x − 10 = 0 ✓ standard form (x − 5)(x + 2) = 0 ✓ factors x = 5 or x = −2 ✓ both answers (3) 6.2 Naledi: x Sipho: 4x 4x + 6 = 3(x + 6) ✓ equation 4x + 6 = 3x + 18 x = 12 ✓ x = 12 Sipho is 4 × 12 = 48 years old. ✓ 48 years (3) [11] Copyright reserved Please turn over

Mathematics/P1 6 Hlayiso/Mock 2026 Senior Phase – Marking Guidelines QUESTION 7 7.1 24 ; 35 (add 3, 5, 7, 9, 11, … or Tn = n2 − 1) ✓ both terms correct for the mark (1) 7.2.1 p = 21 ✓p q = 26 ✓q (2) 7.2.2 Constant difference = 5 ✓ 5n Tn = 5n + 1 ✓ +1 (2) 7.2.3 5n + 1 = 156 ✓ = 156 5n = 155 ✓ Figure 31 (CA) (2) n = 31 ∴ Figure 31 7.2.4 5n + 1 = 200 ✓ n = 39,8 / 196 and 201 5n = 199 n = 39,8, which is not a whole number. ✓ No, with reason (2) No, he cannot use all 200 matchsticks for one figure. (Figure 39 uses 196 and Figure 40 uses 201 matchsticks.) [9] Copyright reserved

Mathematics/P1 7 Hlayiso/Mock 2026 Senior Phase – Marking Guidelines QUESTION 8 x −1 0 2 3 8.1 ✓ any two correct y 6 4 0 −2 ✓ all four correct (2) y (0 ; 4) (2 ; 0) x f 8.2 ✓ y-intercept (0 ; 4) ✓ x-intercept (2 ; 0) ✓ straight line with negative gradient through the intercepts, labelled f (3) 3−0 3 1 8.3.1 m= = = ✓ substitution 4 − (−2) 6 2 ✓ answer (2) 1 8.3.2 y = x + c; substitute A(−2 ; 0): ✓ substitute a point 2 1 ✓ equation (CA) (2) 0 = (−2) + c ⇒ c = 1 2 1 y = x+1 2 1 8.3.3 Perpendicular, because m f × mg = −2 × = −1 ✓ perpendicular with reason 2 (CA) (1) (Also accept: the gradients −2 and 21 are negative reciprocals of each other.) 8.3.4 y=3 ✓ answer (1) [11] TOTAL: 75 Copyright reserved