Mathematics Grade 10 P1 November Mock Exam 2 2026 Memo (Eng)

Mock hlayiso mock exams Practice Series: Grade 10 Exam Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 10 MATHEMATICS P1 NOVEMBER EXAMINATION 2026 (MOCK) MARKING GUIDELINES MARKS: 100 These marking guidelines consist of 8 pages. Copyright reserved Please turn over

Mathematics/P1 2 Hlayiso/Mock 2026 NSC Mock – Marking Guidelines NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question 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. • Assuming answers/values in order to solve a problem is NOT acceptable. • The mark for substitution is awarded for substitution into the correct formula. QUESTION 1 1.1.1 = 9x2 − 12x + 4 − (x2 − 16) ✓ both expansions = 8x2 − 12x + 20 ✓ answer (2) 3 x − 10 1.1.2 = − ✓ (x − 2)(x + 2) x + 2 (x − 2)(x + 2) ✓ 3(x − 2) − (x − 10) 3(x − 2) − (x − 10) ✓ 2(x + 2) = (x − 2)(x + 2) 2x + 4 2(x + 2) ✓ answer (4) = = (x − 2)(x + 2) (x − 2)(x + 2) 2 = x−2 2x+1 .3x+1 .2x+1 1.1.3 = ✓ prime bases 22x .3x .3 2x+2 ✓ simplifying exponents 2 .3x+1 = 2x x+1 ✓ answer (3) 2 .3 2 =2 =4 ( ) 1 2 1.2.1 a− = 32 a ✓ a2 − 2 + a12 = 9 1 ✓ answer a2 − 2 + 2 = 9 (2) a 1 a2 + 2 = 11 a ( ) 1 2 1 1.2.2 2 a + 2 = a4 + 2 + 4 = 121 a a ✓ answer (CA from 1.2.1)(1) 1 a4 + 4 = 121 − 2 = 119 a Copyright reserved Please turn over

Mathematics/P1 3 Hlayiso/Mock 2026 NSC Mock – Marking Guidelines 1.3.1 = 3x(x2 − 4) ✓ 3x(x2 − 4) = 3x(x − 2)(x + 2) ✓ answer (2) 1.3.2 = 25 − (x2 − 6xy + 9y2 ) ✓ grouping = 25 − (x − 3y)2 −(x2 − 6xy + 9y2 ) = (5 − x + 3y)(5 + x − 3y) ✓ (x − 3y)2 ✓ answer (3) [17] QUESTION 2 2.1.1 3x2 − x − 10 = 0 ✓ standard form (3x + 5)(x − 2) = 0 ✓ factors 5 ✓ both answers (3) x=− or x = 2 3 2.1.2 2x .22 + 2x = 40 2x (4 + 1) = 40 ✓ common factor 2x (4 + 1) 2x = 8 = 23 ✓ 2x = 8 x=3 ✓ answer (3) 2.2.1 ×6: 2(2x + 1) − 3(x − 2) < 12 ✓ multiplying by LCD 6 4x + 2 − 3x + 6 < 12 ✓ expanding correctly x<4 ✓ answer (3) 2.2.2 x=3 ✓ answer (CA from 2.2.1)(1) 2.3 x + y = 160 …(1) 45x + 25y = 5 200 …(2) ✓ both equations From (1): y = 160 − x 45x + 25(160 − x) = 5 200 ✓ substitution / elimination 20x + 4 000 = 5 200 x = 60 and y = 160 − 60 = 100 ✓ x = 60 60 braai packs and 100 boerewors rolls were sold. ✓ y = 100 (4) OR elimination: (1) × 25: 25x + 25y = 4 000; subtract from (2): 20x = 1 200 … [14] Copyright reserved Please turn over

Mathematics/P1 4 Hlayiso/Mock 2026 NSC Mock – Marking Guidelines QUESTION 3 3.1.1 T2 − T1 = T3 − T2 ✓ equating the common (3x + 4) − (5x − 1) = (4x − 3) − (3x + 4) differences −2x + 5 = x − 7 −3x = −12 ✓ −2x + 5 = x − 7 x=4 ✓ −3x = −12 (x = 4 is given; no mark for the final line.) (3) 3.1.2 Terms: 19 ; 16 ; 13 d = −3 ✓ −3n Tn = 19 + (n − 1)(−3) ✓ +22 (2) Tn = −3n + 22 3.1.3 −3n + 22 = −131 ✓ equating −3n = −153 ✓ answer (2) n = 51 ∴ the 51st term 3.1.4 −3n + 22 < 0 ✓n=8 n > 7,33 ∴ n = 8 ✓ T8 = −2 (2) T8 = −3(8) + 22 = −2 3.2.1 The word has 7 letters. ✓ 7 × 35 + 5 (remainder 5) 250 = 7 × 35 + 5 ✓O (2) The 250th letter is the 5th letter of LIMPOPO: O 3.2.2 Each word has 2 O’s: 35 × 2 = 70 ✓ 70 Letters 246 to 250: L I M P O (1 more O) ✓ 71 (2) Total = 71 [13] Copyright reserved Please turn over

Mathematics/P1 5 Hlayiso/Mock 2026 NSC Mock – Marking Guidelines QUESTION 4 4.1.1 Deposit = 0,15 × 14 500 = R2 175 ✓ R2 175 Balance = 14 500 − 2 175 = R12 325 ✓ R12 325 (2) 4.1.2 A = P(1 + ni) A = 12 325(1 + 2 × 0,18) = R16 762 ✓ substitution 16 762 ✓ R16 762 Instalment = = R698,42 ✓ dividing by 24 24 Total monthly payment = 698,42 + 85 = R783,42 ✓ answer (4) 4.2 4 × $135 = $540 ✓ $540 540 × 17,85 = R9 639 ✓ R9 639 (2) 4.3 A = P(1 + i)n 23 000 = P(1 + 0,09)5 ✓ substitution 23 000 ✓ P the subject P= (1,09)5 ✓ answer (3) P = R14 948,42 [11] QUESTION 5 5.1 q = −2 ✓ answer (1) a 5.2 f (x) = − 2; subst. P(4 ; −1): x a ✓ substitution of P −1 = − 2 ✓ 4a = 1 (2) 4 a =1 4 a=4 4 5.3 0= −2 ✓ f (x) = 0 x ✓ A(2 ; 0) (2) x=2 A(2 ; 0) 0 − (−4) 5.4 m= =2 c = −4 ✓m=2 2−0 ✓ equation (2) g(x) = 2x − 4 Copyright reserved Please turn over

Mathematics/P1 6 Hlayiso/Mock 2026 NSC Mock – Marking Guidelines 4 5.5 − 2 = 2x − 4 ✓ equating x ×x: 4 − 2x = 2x2 − 4x ✓ standard form 2x2 − 2x − 4 = 0 ⇒ x2 − x − 2 = 0 (x − 2)(x + 1) = 0 ✓ x = −1 x = −1 (x = 2 at A) ✓ y = −6 (4) y = 2(−1) − 4 = −6 E(−1 ; −6) 5.6 x ≤ −1 or 0<x≤2 ✓ x ≤ −1 ✓0<x ✓x≤2 (Penalise 1 mark for 0 ≤ x.) (3) 4 5.7 k(x) = − + 2 ✓ answer (1) x [15] QUESTION 6 1 2 6.1 x −2 = 0 2 ✓ x2 = 4 x2 = 4 ✓ both intercepts (2) x = ±2 (−2 ; 0) and (2 ; 0) 6.2 y ✓ turning point (0 ; −2) and p x-intercepts ✓ shape (minimum, symmetric about the y-axis) (2) x (−2 ; 0) (2 ; 0) (0 ; −2) 6.3 y ≥ −2 OR p(x) ≥ −2 OR y ∈ [−2 ; ∞) ✓ answer (1) 6.4 k < −2 ✓ answer (1) Copyright reserved Please turn over

Mathematics/P1 7 Hlayiso/Mock 2026 NSC Mock – Marking Guidelines 6.5 Turning point (0 ; −2); positive x-intercept (2 ; 0) 0 − (−2) ✓m=1 m= =1 c = −2 ✓ y = x−2 (2) 2−0 y = x−2 [8] QUESTION 7 7.1 y < 4 OR h(x) < 4 OR y ∈ (−∞ ; 4) ✓ answer (1) 7.2 0 = −2x + 4 2x = 4 = 22 ✓ 2x = 4 x=2 (2 ; 0) ✓ (2 ; 0) (2) 7.3 y ✓ asymptote y = 4 ✓ both intercepts y=4 ✓ shape (decreasing, below the asymptote) (3) (0 ; 3) x (2 ; 0) h 7.4 −2x + 4 > 3 2x < 1 = 20 x<0 ✓x<0 (1) [7] QUESTION 8 8.1.1 P(A or B) = P(A) + P(B) − P(A and B) ✓ substitution = 0,45 + 0,3 − 0,12 = 0,63 ✓ answer (2) Copyright reserved Please turn over

Mathematics/P1 8 Hlayiso/Mock 2026 NSC Mock – Marking Guidelines 8.1.2 S ✓ 0,12 in the intersection A B ✓ 0,33 and 0,18 ✓ 0,37 outside (CA from 0,33 0,12 0,18 8.1.1) (3) 0,37 8.1.3 No. P(A and B) = 0,12 ̸= 0 ✓ No with reason (1) 8.2.1 Let x = number who use both: (28 − x) + x + (22 − x) + 7 = 50 ✓ equation 57 − x = 50 x=7 ✓x=7 (2) 8.2.2 U ✓ 7 in the intersection T W ✓ 21 and 15 ✓ 7 outside (3) 21 7 15 7 (U: the 50 learners.) 8.2.3 n(exactly one) = 21 + 15 = 36 ✓ 36 36 18 ✓ answer (2) P= = = 0,72 50 25 8.3 P(C) + P(D) = 1 ✓ P(C) + P(D) = 1 2P(D) + P(D) = 1 1 ✓ answer (2) P(D) = 3 [15] TOTAL: 100 Copyright reserved