Technical Mathematics Grade 10 P1 November Mock 2026 Memo

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

Technical Mathematics/P1 2 Hlayiso/Mock 2026 NSC – Marking Guidelines MARKING CODES A Accuracy CA Consistent accuracy M Method R Rounding NPR No penalty for rounding NPU No penalty for units omitted S Simplification SF Substitution in correct formula AO Answer only NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines where indicated. • No penalty for rounding (NPR) for ALL questions, unless stated otherwise. • AO = answer only. Where AO is indicated, the stated marks are awarded for a correct answer without working. • Where a question says WITHOUT using a calculator, the working must be shown to earn full marks. QUESTION 1 √ √ √ 1.1 64 < 75 < 81 ✓8 A √ 8 < 75 < 9 ✓9 A Between 8 and 9 (2) 1.2.1 Non-real ✓ answer A (1) 1.2.2 Irrational ✓ answer A (1) 1.3 45 = 32 + 8 + 4 + 1 ✓ powers of 2 / method M = 1 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 = 1011012 ✓ answer CA (2) 1.4 10112 ✓ method (carrying) M + 1102 = 100012 ✓ answer CA AO: full marks (11 + 6 = 17) (2) Copyright reserved Please turn over

Technical Mathematics/P1 3 Hlayiso/Mock 2026 NSC – Marking Guidelines 1.5 4,58 × 10−4 m ✓ answer A (1) 1.6 (2 − i)(3 + 4i) = 6 + 8i − 3i − 4i2 ✓ expansion A = 6 + 5i − 4(−1) ✓ i2 = −1 A = 10 + 5i ✓ answer CA (3) [12] QUESTION 2 2.1 2x2 + 10x − 3x − 15 − x2 + x ✓ 2x2 + 7x − 15 A = x2 + 8x − 15 ✓ −x2 + x A ✓ answer CA (3) 2.2.1 3(x2 − 9) ✓ common factor A = 3(x − 3)(x + 3) ✓ answer CA (2) 2.2.2 (2x + 1)(x − 3) ✓ (2x + 1) A ✓ (x − 3) A (2) 2.2.3 y(x − 2) + 3(x − 2) ✓ grouping A = (x − 2)(y + 3) ✓ answer CA (2) 2.2.4 (2m − 1)(4m2 + 2m + 1) ✓ (2m − 1) A ✓ (4m2 + 2m + 1) A (2) (x − 2)(x + 2) x + 3 2.3 = × ✓ numerator factors A (x + 3)(x − 2) x + 2 ✓ denominator factors A =1 ✓ ÷ → × (reciprocal) A ✓ answer CA (4) Copyright reserved Please turn over

Technical Mathematics/P1 4 Hlayiso/Mock 2026 NSC – Marking Guidelines 4a6 b2 · 3a−2 2.4 = ✓ 4a6 b2 A 6a4 b ✓ 12a4 b2 CA 12a4 b2 = ✓ answer CA 6a4 b (3) = 2b [18] QUESTION 3 5x · 5 − 5x 3.1.1 = ✓ 5x · 5 A 4 · 5x 5 (5 − 1) x ✓ common factor A = ✓ answer CA 4 · 5x =1 (3) 2x · 22x+2 3.1.2 = ✓ 22x+2 A 23x x+2x+2−3x ✓ 23x A =2 ✓ answer CA = 22 = 4 (3) 3.2.1 log2 32 = 5 ✓ answer A (1) 3.2.2 log3 34 + log 102 ✓ 4 and 2 A = 4+2 = 6 ✓ answer CA (2) [9] Copyright reserved Please turn over

Technical Mathematics/P1 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1.1 (x − 4)(x + 3) = 0 ✓ factors A x = 4 or x = −3 ✓x=4 CA ✓ x = −3 CA (3) 4.1.2 ×6: 3(x + 1) − 2(x − 2) = 12 ✓ multiplying by LCD M 3x + 3 − 2x + 4 = 12 ✓ removing brackets CA x=5 ✓ answer CA (3) 4.1.3 3x+2 = 3−2 ✓ 3−2 A x + 2 = −2 ✓ equating exponents CA x = −4 ✓ answer CA (3) 4.2 −4 ≤ 2x < 6 ✓ subtracting 1 A −2 ≤ x < 3 ✓ −2 ≤ x < 3 CA ✓ number line (closed at −2, −3 −2 −1 0 1 2 3 4 open at 3) CA (3) 4.3 y = 3x − 7 ✓ y as subject A x + 2(3x − 7) = 7 ✓ substitution CA 7x = 21 ✓x=3 CA x = 3 and y = 3(3) − 7 = 2 ✓y=2 CA (4) 4.4 at = v − u ✓ at = v − u A v−u ✓ answer CA a= t (2) 4.5 Let the shorter piece be x m. x + (x + 2,4) = 12 ✓ equation A 2x = 9,6 ✓ 4,8 m CA x = 4,8 ✓ 7,2 m CA The pieces are 4,8 m and 7,2 m. (3) [21] Copyright reserved Please turn over

Technical Mathematics/P1 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 5 5.1 A = P(1 + in) ✓ formula A A = 8 500(1 + 0,065 × 3) ✓ SF A A = R10 157,50 ✓ answer CA (3) 5.2.1 A = P(1 + i)n ✓ formula A A = 8 500(1 + 0,062)3 ✓ SF A A = R10 181,05 ✓ answer CA (3) 5.2.2 Lindiwe’s, because R10 181,05 > R10 157,50 ✓ answer with reason CA (compound interest earns interest on interest). (1) 5.3.1 0,1 × 14 800 = R1 480 ✓ answer A (1) 5.3.2 Loan = 14 800 − 1 480 = R13 320 ✓ loan amount CA A = 13 320(1 + 0,16 × 2) ✓ SF A A = R17 582,40 ✓ answer CA (3) 17 582,40 5.3.3 Instalment = ✓ dividing by 24 CA 24 = R732,60 ✓ answer CA (2) 5.4 450 × 18,40 = R8 280 ✓ multiplication A ✓ answer CA (2) [15] QUESTION 6 x −3 −2 −1 0 1 2 3 6.1 f (x) −5 0 3 4 3 0 −5 ✓ f : 4 or more correct A ✓ f : all correct A g(x) −1 0 1 2 3 4 5 ✓ g: all correct A (3) Copyright reserved Please turn over

Technical Mathematics/P1 7 Hlayiso/Mock 2026 NSC – Marking Guidelines 6.2 y f: g ✓ x-intercepts CA (0; 4) ✓ turning point (0; 4) CA (1; 3) ✓ shape (concave down) A 2 g: −2 ✓ intercepts (−2; 0) and x (0; 2) CA 2 ✓ straight line through (1; 3) CA (5) f 6.3.1 (0 ; 4) ✓ answer CA (1) 6.3.2 y ≤ 4 OR y ∈ (−∞ ; 4] ✓ answer CA (1) 6.3.3 x∈R ✓ answer A (1) 6.4.1 x = −2 or x=1 ✓ x = −2 CA ✓x=1 CA (2) 6.4.2 −2 < x < 1 OR x ∈ (−2 ; 1) ✓ critical values CA ✓ notation CA (2) [15] Copyright reserved Please turn over

Technical Mathematics/P1 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 k 7.1 3= ✓ substitution of P A 2 k=6 ✓ answer CA (2) 7.2 3 = b2 − 1 ✓ substitution of P A b2 = 4 b = 2 (b > 0) ✓ answer CA (2) 7.3 y = −1 ✓ answer A (1) 7.4 y ∈ R , y ̸= 0 ✓ answer A (1) 7.5 y > −1 OR y ∈ (−1 ; ∞) ✓ answer CA (1) 6 7.6 y= = −2 ✓ answer CA −3 (1) 7.7 x<0 ✓ answer A (1) 7.8 x>2 ✓ answer A (1) [10] TOTAL: 100 Copyright reserved