Mathematics Grade 10 P2 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 P2 NOVEMBER EXAMINATION 2026 (MOCK) MARKING GUIDELINES MARKS: 100 These marking guidelines consist of 10 pages. Copyright reserved Please turn over

Mathematics/P2 2 Hlayiso/Mock 2026 NSC – 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. 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. QUESTION 1 1.1 5 + 3 + 1 = 9 households ✓ answer (1) 1.2 Mode = 4 people ✓ answer (1) 1.3 n = 30: median is the average of the 15th and 16th ✓ 15th and 16th values / values. cumulative frequencies Cumulative frequencies: 2 ; 6 ; 13 ; 21 ; 26 ; 29 ; 30 ✓ answer (2) 15th and 16th values are both 4, so median 4+4 = =4 2 1.4 Q1 = 8th value = 3 ✓ Q1 = 3 Q3 = 23rd value = 5 ✓ Q3 = 5 (2) 1.5 ✓ box with Q1 , median and Q3 (CA from 1.3, 1.4) ✓ whiskers to 1 and 7 (2) 0 1 2 3 4 5 6 7 8 Minimum = 1; Q1 = 3; Median = 4; Q3 = 5; Maximum = 7 [8] Copyright reserved Please turn over

Mathematics/P2 3 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 2 2.1 14 ✓ answer (1) 2.2 Midpoints: 3 ; 5 ; 7 ; 9 ; 11 ✓ midpoints 6(3) + 14(5) + 19(7) + 13(9) + 8(11) ✓ ∑ f x products x̄ = ✓ 426 ÷ 60 60 18 + 70 + 133 + 117 + 88 426 ✓ answer (4) = = 60 60 = 7,1 kg 2.3 Q3 is the 34 (60) = 45th value. ✓ 45th value / cumulative Cumulative frequencies: 6 ; 20 ; 39 ; 52 ; 60 frequencies Q3 lies in 8 ≤ m < 10 ✓ answer (2) 13 + 8 21 2.4 × 100 = × 100 = 35% ✓ 21 60 60 ✓ answer (2) [9] Copyright reserved Please turn over

Mathematics/P2 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 √ 3.1 AB = (1 − (−2))2 + (3 − (−1))2 ✓ substitution √ √ = 9 + 16 = 25 = 5 units ✓ answer (2) ( ) −2 + 6 −1 + 3 3.2 M ; = M(2 ; 1) ✓x=2 2 2 ✓y=1 (2) 3.3 M is the midpoint of BD: ✓ midpoint equations 1 + xD ✓ xD = 3 = 2 ⇒ xD = 3 2 ✓ yD = −1 (3) 3 + yD = 1 ⇒ yD = −1 2 D(3 ; −1) 3 − (−1) 4 1 3.4 mAC = = = ✓ mAC = 12 6 − (−2) 8 2 ✓ mBD = −2 −1 − 3 −4 ✓ product = −1 mBD = = = −2 (3) 3−1 2 1 mAC × mBD = × (−2) = −1 ∴ AC ⊥ BD 2 3.5 The diagonals bisect each other (M is the midpoint of ✓ parallelogram (diagonals AC and of BD), so ABCD is a parallelogram. bisect) The diagonals of the parallelogram are perpendicular ✓ diagonals ⊥ / adjacent (3.4), so ABCD is a rhombus. sides = (2) OR parallelogram with AB = BC = 5 (adjacent sides equal) √ √ √ 3.6 AC = 82 + 42 = 80 = 4 5; BD ✓ AC and BD (or base and √ √ √ = 22 + 42 = 20 = 2 5 height) 1 1 √ √ ✓ area formula Area = × AC × BD = × 4 5 × 2 5 = 20 square ✓ answer (3) 2 2 units OR AD = 5 (horizontal, y = −1); BC on y = 3, so height = 4; Area = 5 × 4 = 20 √ √ OR 4× area △AMB = 4 × 12 × 20 × 5 = 20 [15] Copyright reserved Please turn over

Mathematics/P2 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1.1 cos(56,2◦ + 43,4◦ ) = cos 99,6◦ = −0,17 ✓ answer (1) 1 4.1.2 = = 1,45 ✓ sin2 156,2◦ sin 56,2◦ 2 ✓ answer (2) ( √ )2 ( )2 2 3 − √1 2 (√ )2 4.2 = + 3 ✓ 43 and 12 in numerator 1 × 12 ✓ denominator 12 4−2 3 1 = 1 +3 ✓ tan2 60◦ = 3 2 ✓ answer (4) 1 1 7 = 41 + 3 = + 3 = (or 3,5) 2 2 2 8 5 4.3.1 sec θ = ⇒ cos θ = = 0,625 ✓ cos θ = 85 5 8 ✓ answer (2) θ = 51,32◦ θ 4.3.2 + 25◦ = 60,945 . . .◦ ✓ 60,95◦ 2 θ ✓ subtracting 25◦ = 35,945 . . .◦ ✓ answer (3) 2 θ = 71,89◦ [12] Copyright reserved Please turn over

Mathematics/P2 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 5 8 5.1 sin α = − ; α in the 3rd quadrant: y = −8, r = 17 ✓ y = −8 and r = 17 in 17 correct quadrant y α x 17 (−15 ; −8) ✓ x = −15 ✓ answer (3) x2 = 172 − (−8)2 = 225 x = −15 15 cos α = − 17 −8 17 8 17 5.2 = + = − ✓ substitution of tan α and −15 −15 15 15 sec α 9 3 =− =− ✓ answer (2) 15 5 ( )2 ( ) 17 −15 2 5.3 = − ✓ substitution −8 −8 ✓ answer (2) 289 225 64 = − = =1 64 64 64 Note: answer 1 only (no substitution, no identity shown): 1/2. Correct use of the identity cosec2 α − cot2 α = 1 with working: 2/2. [7] Copyright reserved Please turn over

Mathematics/P2 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 6 1 − (−3) 6.1 Amplitude: = 2, so a = 2 ✓a=2 2 q = 1 − 2 = −1 ✓ q = −1 (2) ( f (x) = 2 cos x − 1) 6.2 y ✓ y-intercept and end points (0◦ ; −1), (360◦ ; −1) 1 (90◦ ; 0) ✓ turning points (90◦ ; 0) and x (270◦ ; −2) 90◦ 180◦ 270◦ 360◦ ✓ shape (3) −1 (360◦ ; −1) g −2 (270◦ ; −2) −3 f g (dashed): y-intercept (0◦ ; −1); maximum and x-intercept (90◦ ; 0); minimum (270◦ ; −2); end point (360◦ ; −1). 6.3 2 ✓ answer (1) 6.4 −3 ≤ y ≤ 1 OR y ∈ [−3 ; 1] ✓ critical values ✓ notation (2) 6.5 63,43◦ < x < 243,43◦ OR x ∈ (63,43◦ ; 243,43◦ ) ✓ critical values ✓ notation (strict inequality) (2) 6.6 Reflection: y = −(2 cos x − 1) = −2 cos x + 1 ✓ −2 cos x h(x) = −2 cos x + 4 ✓ +4 (2) [12] Copyright reserved Please turn over

Mathematics/P2 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 DF 7.1 sin 32◦ = ✓ correct ratio 85 ✓ answer (2) DF = 85 sin 32◦ = 45,04 m AF 7.2 cos 32◦ = ✓ correct ratio 85 ✓ answer (2) AF = 85 cos 32◦ = 72,08 m √ OR AF = 852 − 45,042 = 72,08 m DF 7.3 tan 47◦ = ✓ correct ratio FB 45,04 ✓ FB as subject FB = = 42,00 m ✓ FB = 42,00 tan 47◦ AB = 72,08 + 42,00 = 114,09 m (using unrounded ✓ answer (4) values) Accept 114,08 m – 114,09 m (rounding). 7.4 47◦ (alternate angles with the horizontal through D) ✓ answer (1) [9] QUESTION 8 8.1 s2 = 52 + 122 = 169 ∴ s = 13 cm ✓ Pythagoras substitution (1) 8.2 TSA = π (5)2 + π (5)(13) ✓ substitution = 25π + 65π = 90π = 282,74 cm2 ✓ answer (2) 1 8.3 Vcone = π (5)2 (12) = 100π (= 314,16 cm3 ) ✓ volume of cone 3 4 3 ✓ equating volumes π r = 100π 3 ✓ answer (3) r3 =√75 r = 3 75 = 4,22 cm [6] Copyright reserved Please turn over

Mathematics/P2 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 9 9.1 In △PQS and △RSQ: Q̂1 = Ŝ1 (alt ∠s; PQ ∥ SR) ✓ S/R Ŝ2 = Q̂2 (alt ∠s; PS ∥ QR) ✓ S/R QS = SQ (common) ✓ S/R common side ∴ △PQS ≡ △RSQ (∠∠S / AAS) ✓ ≡ (AAS) and conclusion ∴ PQ = SR and PS = QR (≡ △s) (4) 9.2.1 Â1 = Â2 (AE bisects DÂB) ✓ S/R (bisector) Â2 = AÊD (alt ∠s; AB ∥ DC) ✓ S/R (alt ∠s; AB ∥ DC) ∴ Â1 = AÊD ∴ AD = DE (sides opp equal ∠s) ✓ R (sides opp = ∠s) (3) 9.2.2 DC = AB = 12 cm (opp sides of parm) ✓ answer (1) EC = 12 − 7 = 5 cm [8] QUESTION 10 10.1.1 OK = OL (diags of rectangle are equal and bisect each ✓ S/R (OK = OL) other) ✓ answer with reason (2) 180◦ − 124◦ OL̂K = OK̂L = = 28◦ (∠s opp = sides; 2 sum of ∠s of △) 10.1.2 KL̂M = 90◦ (∠ of rectangle) ✓ S/R OL̂M = 90◦ − 28◦ = 62◦ ✓ answer (2) 1 1 10.1.3 ON = LN = KM (diags of rect equal and bisect) ✓ equation with reason 2 2 1 y + 4 = (3y + 2) ✓ answer (2) 2 2y + 8 = 3y + 2 y=6 1 10.2.1 EG = BC (midpoint theorem in △ABC) ✓ S/R 2 1 ✓ answer (2) EG = (14) = 7 cm 2 Copyright reserved Please turn over

Mathematics/P2 10 Hlayiso/Mock 2026 NSC – Marking Guidelines 10.2.2 EG ∥ BC (midpoint theorem) ✓ S/R EG ∥ BC AD ∥ BC (given) ∴ GF ∥ AD ✓ GF ∥ AD In △ACD: G is the midpoint of AC and GF ∥ AD ∴ F is the midpoint of DC (line through midpt of one ✓ R (converse of midpoint side ∥ to second side) theorem) (3) 1 10.2.3 GF = AD (midpoint theorem in △ACD) ✓ S/R 2 1 ✓ GF = 4 GF = (8) = 4 cm ✓ answer (3) 2 EF = EG + GF = 7 + 4 = 11 cm [14] TOTAL: 100 Copyright reserved