Mathematics Grade 10 P2 November Mock 2026 Memo (English)

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 9 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 1 448 1.1 x̄ = = 29,87 ✓ 448 15 ✓ answer (2) 1.2 Minimum = 14; Q1 = 23; Median = 30; Q3 = 36; ✓ min and max Maximum = 47 ✓ Q1 and Q3 ✓ median (3) 1.3 ✓ box with Q1 , median and Q3 ✓ whiskers to 14 and 47 (2) 10 20 30 40 50 1.4 IQR = 36 − 23 = 13 ✓ answer (1) [8] QUESTION 2 2.1 20 ≤ t < 30 ✓ answer (1) 2.2 Median is between the 20th and 21st values. ✓ 20th/21st value or Cumulative frequencies: 4 ; 13 ; 27 ; . . . cumulative frequencies Median class: 20 ≤ t < 30 ✓ answer (2) Copyright reserved Please turn over

Mathematics/P2 3 Hlayiso/Mock 2026 NSC – Marking Guidelines 2.3 Midpoints: 5 ; 15 ; 25 ; 35 ; 45 ✓ midpoints 4(5) + 9(15) + 14(25) + 8(35) + 5(45) ✓ ∑ f x products x̄ = ✓ 1 010 ÷ 40 40 20 + 135 + 350 + 280 + 225 1 010 ✓ answer (4) = = 40 40 = 25,25 minutes 8+5 13 2.4 × 100 = × 100 = 32,5% ✓ 13 40 40 ✓ answer (2) [9] QUESTION 3 √ 3.1 AB = (2 − (−2))2 + (5 − 3)2 ✓ substitution √ √ √ = 16 + 4 = 20 = 2 5 ✓ answer (2) 1 − 5 −4 3.2 mBC = = = −2 ✓ substitution 4−2 2 ✓ answer (2) 5−3 1 3.3 mAB = = ✓ mAB = 12 2 − (−2) 2 ✓ product = −1 (2) 1 mAB × mBC = × (−2) = −1 2 ∴ AB ⊥ BC ( ) −2 + 4 3 + 1 3.4 M ; = M(1 ; 2) ✓x=1 2 2 ✓y=2 (2) 1−3 1 3.5 mAC = =− ✓ mAC = − 31 4 − (−2) 3 ✓ substitution of B 1 ✓ equation y − 5 = − (x − 2) (3) 3 1 17 y = − x+ 3 3 OR 5 = − 13 (2) + c ⇒ c = 17 3 Copyright reserved Please turn over

Mathematics/P2 4 Hlayiso/Mock 2026 NSC – Marking Guidelines √ √ √ 3.6 BC = (4 − 2)2 + (1 − 5)2 = 20 ✓ BC = 20 1 ✓ answer (2) Area △ABC = × AB × BC (AB ⊥ BC) 2 1 √ √ = × 20 × 20 = 10 square units 2 3.7 Diagonals of a square bisect each other, so M is the midpoint of BD: ✓ xD = 0 2 + xD 5 + yD ✓ yD = −1 (2) = 1 ⇒ xD = 0; = 2 ⇒ yD = −1 2 2 D(0 ; −1) OR (translation B→C: x + 2, y − 4; apply to A) D(−2 + 2 ; 3 − 4) [15] QUESTION 4 4.1.1 1,51 ✓ answer (1) 2 4.1.2 = = 2,61 ✓ cos240◦ cos 40◦ ✓ answer (2) (√ ) 2 (√ )2 2× 1 3 √ 4.2 = 1 + 2 ✓ 21 and ( 3)2 √ × √1 2 2 ✓ √12 × √12 3 = 21 + 2 ✓ sec2 45◦ = 2 2 ✓ answer (4) = 3+2 = 5 4.3.1 cos θ = 0,65 ✓ cos θ = 0,65 θ = 49,46◦ ✓ answer (2) 4.3.2 2θ + 10◦ = 72,646 . . .◦ ✓ 72,65◦ 2θ = 62,646 . . .◦ ✓ subtracting 10◦ θ = 31,32◦ ✓ answer (3) [12] Copyright reserved Please turn over

Mathematics/P2 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 5 12 5.1 tan β = − ; β in the 2nd quadrant: x = −5, y = 12 ✓ x = −5 and y = 12 in 5 correct quadrant y (−5 ; 12) 13 ✓ r = 13 ✓ answer (3) β x r2 = (−5)2 + 122 = 169 r = 13 12 sin β = 13 ( ) ( )2 5 2 12 5.2 = − − ✓ substitution 13 13 ✓ answer (2) 25 144 119 = − =− 169 169 169 13 12 5.3 = × ✓ sec β = − 13 −5 13 5 ✓ answer (2) 12 =− 5 [7] Copyright reserved Please turn over

Mathematics/P2 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 6 6.1 y f: ✓ x-intercepts / end points (90◦ ; 2) 2 (360◦ ; 2) ✓ turning points ✓ shape g 1 g: ✓ y-intercept and end points x (0◦ ; 2), (360◦ ; 2) 90◦ 180◦ 270◦ 360◦ ✓ minimum (180◦ ; 0) −1 f ✓ shape (6) −2 (270◦ ; −2) f (solid): x-intercepts 0◦ , 180◦ , 360◦ ; TPs (90◦ ; 2), (270◦ ; −2). g (dashed): (0◦ ; 2), minimum (180◦ ; 0), end point (360◦ ; 2). 6.2 360◦ ✓ answer (1) 6.3 2 ✓ answer (1) 6.4 0 ≤ y ≤ 2 OR y ∈ [0 ; 2] ✓ critical values ✓ notation (2) 6.5 53,13◦ ≤ x ≤ 180◦ ✓ critical values ✓ notation (2) [12] Copyright reserved Please turn over

Mathematics/P2 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 TB 7.1 tan 38◦ = ✓ correct ratio 50 ✓ TB = 50 tan 38◦ TB = 50 tan 38◦ ✓ answer (3) TB = 39,06 m 50 7.2 cos 38◦ = ✓ correct ratio TA 50 ✓ answer (2) TA = ◦ = 63,45 m cos 38 √ OR TA = 502 + 39,062 = 63,45 m TB 7.3 tan 24◦ = ✓ correct ratio CB ✓ CB as subject 39,06 CB = ✓ CB = 87,74 tan 24◦ CB = 87,74 m ✓ answer (4) AC = 87,74 − 50 = 37,74 m Accept 37,73 m – 37,75 m (rounding). [9] QUESTION 8 8.1 V = π (1,2)2 (2,5) ✓ substitution V = 11,31 m3 ✓ answer (2) 8.2 TSA = 2π (1,2)2 + 2π (1,2)(2,5) ✓ substitution = 9,05 + 18,85 = 27,90 m2 ✓ answer (2) ( ) h h ( ) 8.3 Vnew = π (2r)2 = π . 4r2 . = 2π r2 h ✓ π (2r)2 h2 2 2 ✓ volume doubles (2) The volume is doubled. (Vnew = π (2,4)2 (1,25) = 22,62 m3 = 2 × 11,31) [6] Copyright reserved Please turn over

Mathematics/P2 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 9 9.1 In △PTS and △RUQ: PS = RQ (opp sides of parm) ✓S ✓R PŜT = RQ̂U (alt ∠s; PS ∥ QR) ✓S ✓R PT̂S = RÛQ = 90◦ (given) ∴ △PTS ≡ △RUQ (∠∠S / AAS) ✓ 90◦ and conclusion with reason (5) 9.2 PT = RU (≡ △s) ✓ S/R PT ∥ RU (both ⊥ QS / alt ∠s = 90◦ ) ✓ S/R ∴ PTRU is a parallelogram (one pair of opp sides = ✓R (3) and ∥) [8] QUESTION 10 10.1.1 Midpoint theorem: the line joining the midpoints of ✓R (1) two sides of a triangle is parallel to the third side. 1 10.1.2 DE = BC (midpoint theorem) ✓ S/R 2 1 ✓ equation 3x − 2 = (4x + 6) 2 3x − 2 = 2x + 3 ✓ answer (3) x=5 10.1.3 Ĉ = 180◦ − 50◦ − 70◦ = 60◦ (sum of ∠s of △) ✓S ✓R AÊD = Ĉ = 60◦ (corresp ∠s; DE ∥ BC) ✓ S/R (3) 10.2.1 BÂD = 180◦ − 116◦ = 64◦ (co-int ∠s; AD ∥ BC) ✓S ✓R (2) 1 10.2.2 BÂC = (64◦ ) = 32◦ (diag of rhombus bisects ∠) ✓S ✓R (2) 2 OR AB = BC (sides of rhombus); ◦ ◦ BÂC = BĈA = 180 −116 2 = 32◦ (∠s opp = sides) 10.2.3 AÔB = 90◦ (diags of rhombus ⊥) ✓ S/R (1) Copyright reserved Please turn over

Mathematics/P2 9 Hlayiso/Mock 2026 NSC – Marking Guidelines 1 10.2.4 AB̂D = (116◦ ) = 58◦ (diag of rhombus bisects ∠) ✓S ✓R (2) 2 OR 180◦ − 90◦ − 32◦ = 58◦ (sum of ∠s of △AOB) [14] TOTAL: 100 Copyright reserved