Technical Mathematics Grade 10 P1 2026 Memo (Eng)

Mock 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 1.1.1 π ✓ answer A (1) √ 1.1.2 −25 ✓ answer A (1) √ √ √ 1.2 36 < 40 < 49 √ √ 6 < 40 < 7 ✓ 6 < 40 < 7 A √ ✓ −7 and −6 (correct −7 < − 40 < −6 Between −7 and −6 order/sign) CA (2) 1.3 1011012 ✓ 101101 A × 112 ✓ 1011010 (shifted row) A 101101 ✓ answer CA Answer only: full marks + 1011010 (3) = 100001112 (45 × 3 = 135) Copyright reserved Please turn over

Technical Mathematics/P1 3 Hlayiso/Mock 2026 NSC – Marking Guidelines 2,5 × 103 1.4 = 0,5 × 105 ✓ dividing / 0,5 × 105 M 5 × 10−2 ✓ answer in scientific = 5 × 104 pieces notation CA (2) 1.5 (3 − 2i)2 = 9 − 12i + 4i2 ✓ expansion A = 9 − 12i + 4(−1) ✓ i2 = −1 A = 5 − 12i ✓ answer CA (3) [12] QUESTION 2 2.1 (x − 3)(x2 + 3x + 9) = x3 − 27 ✓ x3 − 27 A x(x − 2)2 = x(x2 − 4x + 4) = x3 − 4x2 + 4x ✓ x2 − 4x + 4 A x3 − 27 − x3 + 4x2 − 4x ✓ −x3 + 4x2 − 4x CA = 4x2 − 4x − 27 ✓ answer CA (4) 2.2.1 2(x4 − 16) ✓ common factor A = 2(x2 − 4)(x2 + 4) ✓ (x2 − 4)(x2 + 4) CA = 2(x − 2)(x + 2)(x2 + 4) ✓ answer CA (3) 2.2.2 (3x − 4)(x + 3) ✓ (3x − 4) A ✓ (x + 3) A (2) 2.2.3 x2 (x − 2) − 9(x − 2) ✓ grouping A = (x − 2)(x2 − 9) ✓ (x − 2)(x2 − 9) CA = (x − 2)(x − 3)(x + 3) ✓ answer CA (3) 2.2.4 (3 + 2a)(9 − 6a + 4a2 ) ✓ (3 + 2a) A ✓ (9 − 6a + 4a2 ) A (2) Copyright reserved Please turn over

Technical Mathematics/P1 4 Hlayiso/Mock 2026 NSC – Marking Guidelines 2 8 2.3 = − ✓ factorising x2 − 4 A x − 2 (x − 2)(x + 2) ✓ LCD A 2(x + 2) − 8 ✓ 2x − 4 CA = (x − 2)(x + 2) ✓ answer CA 2x − 4 (4) = (x − 2)(x + 2) 2(x − 2) 2 = = (x − 2)(x + 2) x + 2 [18] QUESTION 3 2x+1 · 3x+1 · 2x−1 3.1.1 = ✓ prime bases (num.) A 22x · 3x ✓ 22x · 3x A =2 x+1+x−1−2x · 3x+1−x ✓ adding/subtracting = 20 · 31 exponents CA =3 ✓ answer CA (4) 2x · 22 − 2x 3.1.2 = ✓ 2x · 22 and 2x · 2−1 A 3 · 2x · 2−1 2x (4 − 1) ✓ common factor 2x A = ✓ answer CA 3 · 2x · 12 (3) 3 = 3 =2 2 3.2 4a2 · 4a−2 ✓ 4a2 and 4a−2 A = 16 ✓ answer CA (2) [9] Copyright reserved Please turn over

Technical Mathematics/P1 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1.1 x2 − 2x = 15 x2 − 2x − 15 = 0 ✓ standard form A (x − 5)(x + 3) = 0 ✓ factors CA x = 5 or x = −3 ✓ both answers CA (3) 4.1.2 2x+1 = 32 ✓ 2x+1 = 32 A 2x+1 = 25 ✓ 2 / equating exponents A 5 x+1 = 5 ✓ answer CA x=4 (3) 4.2 −5 < 3 − 2x ≤ 15 ✓ multiplying by 5 A −8 < −2x ≤ 12 ✓ subtracting 3 CA 4 > x ≥ −6 ✓ −6 ≤ x < 4 (inequality signs reversed) CA −6 ≤ x < 4 ✓ number line (closed at −6, −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 open at 4) CA (4) 4.3 4x + 3y = 174 …(1) ✓ equation (1) A 2x + 5y = 164 …(2) ✓ equation (2) A (2)×2: 4x + 10y = 328 …(3) ✓ elimination/substitution M (3)−(1): 7y = 154 ✓ y = 22 (R22) CA y = 22 4x + 3(22) = 174 ⇒ 4x = 108 ⇒ x = 27 ✓ x = 27 (R27) CA Light switch: R27; plug: R22 (5) 4.4 V 2 = PR ✓ V 2 = PR A √ √ V = ± PR (accept V = PR, V > 0) ✓ answer CA (2) Copyright reserved Please turn over

Technical Mathematics/P1 6 Hlayiso/Mock 2026 NSC – Marking Guidelines 4.5 Let t = time (hours) the motorcyclist travels. Van has then travelled for (t + 0,5) hours. ✓ t + 0,5 A 100t = 80(t + 0,5) ✓ equation (equal distances) M 100t = 80t + 40 20t = 40 t = 2 hours ✓t =2 CA 07:30 + 2 h = 09:30 ✓ 09:30 CA (4) [21] QUESTION 5 5.1 A = P(1 + i)n ✓ formula A 20 000 = P(1 + 0,075)4 ✓ SF A 20 000 P= ✓ answer CA (1,075)4 P = R14 976,01 (3) 5.2.1 R5 000 ✓ answer A (1) 5.2.2 Simple interest. ✓ simple A The graph is a straight line (the value increases by the ✓ reason A same amount every year). (2) 5.2.3 A = P(1 + in) 6 600 = 5 000(1 + 4i) ✓ SF A 1,32 = 1 + 4i ✓ 1,32 = 1 + 4i / i = 0,08 1 600 = 20 000i CA Rate = 8% per annum ✓ answer CA (3) 5.3.1 Deposit = 0,15 × 12 600 = R1 890 ✓ deposit A Loan = 12 600 − 1 890 = R10 710 ✓ loan CA (2) Copyright reserved Please turn over

Technical Mathematics/P1 7 Hlayiso/Mock 2026 NSC – Marking Guidelines 5.3.2 A = 10 710(1 + 0,14 × 1,5) ✓ SF (n = 1,5) A A = R12 959,10 ✓ R12 959,10 CA 12 959,10 ✓ instalment CA Instalment = = R719,95 (3) 18 5.3.3 1 890 + 12 959,10 = R14 849,10 ✓ answer CA (1) [15] QUESTION 6 x −4 −2 −1 1 2 4 6.1 ✓ f : 4 or more correct A f (x) 1 2 4 −4 −2 −1 ✓ g: 3 or more correct A ✓ both tables all correct A x −3 −2 −1 0 1 2 (3) g(x) 6 2 0 −1 −1,5 −1,75 6.2 y f: g ✓ shape (2nd and 4th quadrants) A ✓ asymptotes x = 0, y = 0 / points plotted CA (−2; 2) g: f x ✓ shape (decreasing) A −1 −1 f y = −2 ✓ asymptote y = −2 A ✓ intercepts (−1; 0) and (0; −1) CA (5) 6.3.1 y = −2 ✓ answer A (1) 6.3.2 y > −2 OR y ∈ (−2 ; ∞) ✓ answer CA (1) 6.3.3 x ∈ R , x ̸= 0 ✓ answer A (1) Copyright reserved Please turn over

Technical Mathematics/P1 8 Hlayiso/Mock 2026 NSC – Marking Guidelines ( 1 )x 6.4 2 −2 = 0 ✓ 2−x = 21 A −x 2 = 21 x = −1 i.e. (−1 ; 0) ✓ answer CA (2) 6.5 −2 < x < 0 OR x ∈ (−2 ; 0) ✓ −2 CA ✓ 0 and notation CA (2) [15] QUESTION 7 7.1 q = −8 ✓ answer A (1) 7.2 0 = a(2)2 − 8 ✓ substitution of B A 4a = 8 ⇒ a = 2 ✓ answer CA (2) 4−0 7.3 m= = −2 ✓ m = −2 A 0−2 ✓ equation (c = 4) CA g(x) = −2x + 4 (2) 7.4 2x2 − 8 = −2x + 4 ✓ equating f and g M 2x2 + 2x − 12 = 0 ⇒ x2 + x − 6 = 0 (x + 3)(x − 2) = 0 ✓ x = −3 CA ✓ y = 10 CA x = −3 (x = 2 at B) (3) y = −2(−3) + 4 = 10 E(−3 ; 10) 7.5 y ≥ −8 OR y ∈ [−8 ; ∞) ✓ answer CA (1) 7.6 CD = 4 − (−8) = 12 units ✓ answer A (1) [10] TOTAL: 100 Copyright reserved