# Mathematics Grade 11 P1 Eastern Cape November 2023 Memo > Grade 11 Mathematics Paper 1 memorandum from the Eastern Cape November Exam 2023 (Afrikaans), 15 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/maths-p1-gr11-memo-nov-2023-english-afrikaans/84e969c7-880d-4ed8-8946-551447d298f0 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Language: Afrikaans - Document type: Memorandum - Year: 2023 - Exam period: Eastern Cape November Exam - Paper: 1 - Curriculum: CAPS - Pages: 15 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Mathematics](https://www.hlayiso.com/library/grade-11/mathematics) › [Mathematics Grade 11 P1 Eastern Cape November 2023 Memo](https://www.hlayiso.com/library/maths-p1-gr11-memo-nov-2023-english-afrikaans/84e969c7-880d-4ed8-8946-551447d298f0) ## About this document Mathematics · Grade 11 · Eastern Cape November Exam · 2023 · Afrikaans. Memorandum, 15 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Matching paper and memo - [MATHS P1 QP GR11 NOV 2023 Afrikaans D](https://www.hlayiso.com/library/maths-p1-qp-gr11-nov-2023-afrikaans-d/c2f240dc-7fd6-4ceb-81c1-a511d86b6035) — QP · 2023 · Afr ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIORSERTIFIKAAT GRADE/GRAAD 11 NOVEMBER 2023 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 15 pages./ Hierdie nasienriglyn bestaan uit 15 bladsye. ### Page 2 Downloaded from hlayiso.com 2 MATHEMATICS P1 / WISKUNDE V1 (EC/NOVEMBER 2023) NOTE/LET WEL: • If a candidate answers a question TWICE, mark the FIRST attempt ONLY. Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord, merk SLEGS die EERSTE poging. • Consistent accuracy (CA) applies in ALL aspects of the marking guideline. Volgehoue akkuraatheid (VA) geld deurgaans in ALLE aspekte van die nasienriglyn. • If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt. Indien ʼn kandidaat ʼn poging vir ʼn vraag deurgetrek het en nie die vraag weer beantwoord het nie, merk die poging wat deurgetrek is. • The mark for substitution is awarded for substitution into the correct formula. Die punt vir substitusie word toegeken vir substitusie in die korrekte formule. QUESTION 1/VRAAG 1 1.1.1 x 2 − 3x = 0 x( x − 3) = 0  factorisation / faktorisering  x = 0 or / of x = 3  answers / antwoorde (2) 1.1.2 x(3x + 1) = 5 3x 2 + x − 5 = 0  standard form / standaardvorm −b  b 2 − 4ac x= 2a −1  12 − 4(3)(−5)  substitution / vervanging = 2(3) −1  61 = 6 = 1,14 or / of − 1, 47  answers / antwoorde (4) 1.1.3 2 x − 5x + 3  0 2 (2 x − 3)( x − 1)  0  factors / faktore 1  x  1 1 1 112 2   answer / antwoord (A) (3) 1.1.4 2 x + 2 = x −1 ( 2 x + 2 ) = ( x − 1) 2 2  squaring / kwadreer 4( x + 2) = x 2 − 2 x + 1 4x + 8 = x2 − 2x + 1 x2 − 6 x − 7 = 0  standard form / standaardvorm ( x − 7)( x + 1) = 0  factors / faktore  x = 7 or / of x  −1  both answers / beide antwoorde  selection / seleksie (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com (EC/NOVEMBER 2023) MATHEMATICS P1 / WISKUNDE V1 3 1.2 x + 3y = 2 (1) x 2 − 3 xy = 4 (2) x = 2 − 3y (3)  x = 2 − 3y (2 − 3 y ) 2 − 3 y (2 − 3 y) = 4  substitution / vervanging 4 − 12 y + 9 y 2 − 6 y + 9 y 2 = 4  standard form / standaardvorm 18 y − 18 y = 0 2  method/factors / metode/faktore 18 y ( y − 1) = 0  both y-values / beide y-waardes  y = 0 or / of y = 1  both x-values / beide x-waardes  x = 2 − 3(0) or / of x = 2 − 3(1) =2 x = −1 OR/OF OR/OF x + 3y = 2 (1) x − 3 xy = 4 2 (2) 2− x  y= 3 2− x y= (3) 3  substitution / vervanging  2− x x 2 − 3x  =4  3  x 2 − x(2 − x) = 4  standard form / standaardvorm x2 − 2 x + x2 − 4 = 0 2x2 − 2 x − 4 = 0  factors / faktore x2 − x − 2 = 0  both x-values / beide x-waardes ( x − 2)( x + 1) = 0  x = 2 or / of x = −1 2−2 2 − ( −1)  both y-values / beide y-waardes y = or / of y= 3 3 =0 =1 (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com 4 MATHEMATICS P1 / WISKUNDE V1 (EC/NOVEMBER 2023) 1.3 ( x − 3) = p 2 − 4 2 ( x − 3) =  p 2 − 4 2  square root / vierkantswortel  x − 3 =  p2 − 4  x = 3  p2 − 4  x = 3  p2 − 4 For non-real roots: Vir nie - reëele wortels: -2 2  p2 − 4  0 p2 − 4  0  factors / faktore ( p − 2)( p + 2)  0  answer / antwoord −2  p  2 OR/OF OR/OF ( x − 3) = p 2 − 4 2  expansion / uitbreiding x − 6x + 9 = p − 4 2 2  standard form / standaardvorm x − 6 x + 13 − p = 0 2 2 For non-real roots: Vir nie - reëele wortels  b2 − 4ac  0 b − 4ac  0 2 ( −6 ) − 4(1)(13 − p 2 )  0 2 36 − 52 + 4 p 2  0  4 p 2 − 16  0  factors / faktore (2 p + 4)(2 p − 4)  0  answer / antwoord −2  p  2 (5) [25] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com (EC/NOVEMBER 2023) MATHEMATICS P1 / WISKUNDE V1 5 QUESTION 2/VRAAG 2 2.1 2n +1 − 8.2n −3 2n.2 − 8.2n.2−3  numerator / teller =  denominator / noemer 2n − 2 2n.2− 2 2n (2 − 8.2−3 ) =  factorisation / faktoriseering 2n.2− 2 2 −1 = −2 2 =4  answer / antwoord (4) 2.2.1 x 27 = 2187 1  27 x = 2187 1 27 = 2187x 1 (3 ) = 3 1  ( 33 ) x = 37 OR/OF 27 x = 37 −1 27 x = 37 3 x 7 −1 OR /OF 3 −1 3 x = 37 33 x = 37 3  =7  3 x −1 = 7  equating exponents x gelyk stel van eksponente 3 7 x= x −1 =  answer / antwoord 7 3 3 x= 7 (4) 2.2.2 4 − 16 = 6.2 x x ( 2 ) − 6.2 − 16 = 0 x 2 x  standard form / standaardvorm (2 x − 8)(2 x + 2) = 0  factors / faktore  2 x = 8 or / of 2 x  −2  selection / seleksie  2 x = 23  2 x = 23 x=3  answer / antwoord OR/OF OR/OF 4x – 16 = 6.2x  standard form / standaardvorm (2x)2 = 6.2x –16 = 0 Let/Laat k = 2x, ∴ k2 – 6k – 16 = 0  factors / faktore (k – 8)(k + 2) = 0 ∴ k = 8 or/of k = -2  selection / seleksie ∴ 2x = 8 or/of 2x ≠ -2  2 x = 23 2x = 23  answer / antwoord ∴x=3 (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com 6 MATHEMATICS P1 / WISKUNDE V1 (EC/NOVEMBER 2023) x + 1 ( 3 − 2) + 1 2.3 2 2 =  substitution / vervanging x −5 ( 3 − 2) − 5 2 2 3 − 4 3 + 4 +1 = 3− 4 3 + 4−5 8−4 3 8−4 3 =  2−4 3 2−4 3 = (8 − 4 3 )( 2 + 4 3 )  rationalisation / rasionalisering ( 2 − 4 3 )( 2 + 4 3 ) 16 + 32 3 − 8 3 − 16.3  simplification / vereenvoudiging = 4 − 16.3 24 3 − 32 = −44 8−6 3  answer / antwoord = 11 (5) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com (EC/NOVEMBER 2023) MATHEMATICS P1 / WISKUNDE V1 7 QUESTION 3/VRAAG 3 3.1.1 17 ; 14 ; 11 ; ... ; − 247 8;5 8 5 (2) 3.1.2 Tn = 20 − 3n  20  − 3n (2) 3.1.3 Tn = 20 − 3n T17 = 20 − 3(17)  substitution / vervanging = −31  answer / antwoord (2) 3.1.4 Tn = 20 − 3n −247 = 20 − 3n  Tn = −247 −267 = −3n  n = 89  answer / antwoord (2) 3.2 2 x + 11 ; 2 ; T3 ; 2 x − 4 2x − 4 − 2 2 + 2x − 4 T3 = +2 OR / OF T3 =  method / metode 2 2 2x − 6 2x − 2 = +2 =  simplifying / vereenvoudiging 2 2 = x −3+ 2 = x −1 = x −1  T3 = x − 1  2 − (2 x + 11) = ( x − 1) − 2  equating / gelykstel − 2x − 9 = x − 3 − 3x = 6  x = −2  answer / antwoord OR/OF OR/OF d = 2 − (2 x + 11) = −2 x − 9  d = −2 x − 9 2d = 2 x − 4 − 2 ✓ 2d = 2 x − 4 − 2 d = x−3 ✓ d = x −3  − 2x − 9 = x − 3  equating / gelykstel − 3x = 6 x = −2  answer / antwoord (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com 8 MATHEMATICS P1 / WISKUNDE V1 (EC/NOVEMBER 2023) QUESTION 4/VRAAG 4 4.1.1 94 ; 90 ; 82 ; 70 -4 -8 -12 -4 -4 54 ; 34 54  and/en 34  (2) 4.1.2 2a = −4 3a + b = −4 a + b + c = 94  a = −2  a = −2 3(−2) + b = −4 2 − 2 + c = 94 b = 2  c = 94 b=2 c = 94 Tn = −2n 2 + 2n + 94  Tn = −2n 2 + 2n + 94 (4) 4.1.3 First differences / Eerste verskille: t n = − 4n  method / metode −136 = −4n  n = 34  n = 34 Tn = −2n 2 + 2n + 94 T34 = −2(34) 2 + 2(34) + 94 = −2150  T34 = −2150 T35 = −2(35) 2 + 2(35) + 94 = −2286  T35 = −2286 OR/OF OR/OF Tn +1 − Tn = −136 −2 ( n + 1) + 2(n + 1) + 94 − ( −2n 2 + 2n + 94 ) = −136 2  method / metode −2(n 2 + 2n + 1) + 2n + 2 + 94 + 2n 2 − 2n − 94 = −136 −2n 2 − 4n − 2 + 2n + 2 + 94 + 2n 2 − 2n − 94 = −136 −4n = −136  n = 34 n = 34 n + 1 = 35 Tn = −2n 2 + 2n + 94 T34 = −2(34) 2 + 2(34) + 94  T34 = −2150 = −2150 T35 = −2(35) 2 + 2(35) + 94 = −2286  T35 = −2286 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com (EC/NOVEMBER 2023) MATHEMATICS P1 / WISKUNDE V1 9 4.2 Tn = an 2 + bn − 15 T1 = a + b − 15 T2 = 4a + 2b − 15  𝑇1 and/en 𝑇2 and/en T3 T2 − T1 = 3a + b = 3  3a + b = 3 T3 = 9a + 3b − 15 T3 − T2 = 5a + b = 7  5a + b = 7  5a + b = 7 3a + b = 3 2a = 4 a = 2  value for a / waarde van a b = −3  value for b / waarde van b (5) [15] QUESTION 5/VRAAG 5 5.1 p=3  p =3 q = −1  q = −1 (2) 5.2 a f ( x) = +q x+ p a = −1  substituting for p and q x+3 vervanging vir p en q a 0= −1  substituting for x and y −5 + 3 vervanging vir x en y a 1= −2  a = −2  answer / antwoord (3) 5.3 a f ( x) = +q x+ p −2 = −1 x+3 −2 y= −1  substituting x = 0 0+3 vervang x = 0 5  answer / antwoord =− 2 (2) 5.4 𝑥 ∈ ℝ, but/maar 𝑥 ≠ -3  𝑥 ∈ ℝ  x  −3 (2) 5.5 y = −( x + 3) − 1  y = −( x + 3) −1 = −x − 3 −1  answer / antwoord = −x − 4 (2) 5.6 −5  x  −3  answer / antwoord (A) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com 10 MATHEMATICS P1 / WISKUNDE V1 (EC/NOVEMBER 2023) 5.7 f is reflected in the x-axis and then shifted 4 units  reflected / gereflekteer to the right.  x-axis / x-as f is gereflekteer in die x-as en dan 4 eenhede na  4 units / 4 eenhede regs geskuif.  right / regs (4) [17] QUESTION 6/VRAAG 6 6.1 (0;3)  answer / antwoord (1) 6.2 y =1  answer / antwoord (2) 6.3  y-intercept / y-afsnit  asymptote / asimptoot  shape (must be increasing) vorm (moet stygend wees) 1 (3) 6.4 y  −5 answer / antwoord (A) (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com (EC/NOVEMBER 2023) MATHEMATICS P1 / WISKUNDE V1 11 QUESTION 7/VRAAG 7 7.1 x = 12  answer / antwoord (1) 7.2 x  12  answer / antwoord (1) 7.3 Average gradient / Gemiddelde gradiënt 4−6  method / metode = −1 − 0  answer / antwoord =2 (2) 7.4 g ( x) = mx + q y2 − y1 m= x2 − x1 6−4 = 1 − (−1) =1  m =1 y = x+q 6 = 1 + q or / of 4 = −1 + q  substituting a point vervanging van ’n punt q = 5  q=5  g ( x) = x + 5 (3) 7.5 f ( x) = ax + bx + c 2  c=6 c=6  f ( x) = ax 2 + bx + 6 4 = a(−1) 2 + b(−1) + 6  both substitutions / 6 = a(1) 2 + b(1) + 6 beide vervangings −2 = a −b 0 = a+b  method / metode 2a = −2  a = −1 b =1  values of a and b  f ( x) = − x 2 + x + 6 waardes van a en b (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com 12 MATHEMATICS P1 / WISKUNDE V1 (EC/NOVEMBER 2023) 7.6 g ( x) = x + 5 0 = x+5  substitution / vervanging x = −4 S(0; − 4)  S(0 ; –4) f ( x) = − x 2 + x + 6 0 = − x2 + x + 6 x2 − x − 6 = 0 ( x − 3)( x + 2) = 0  factors / faktore  x = −2 or / of x = 3  both x-intercepts  U(3;0) beide x-afsnitte  SU = 3 − (−4) = 7 units/eenhede  answer / antwoord (5) 7.7 x  −1 or / of x  1  x  −1  x  1 (2) 7.8 yV − yW = f ( x) − g ( x) = (− x 2 + x + 6) − ( x + 5)  f ( x) − g ( x) = − x2 + x + 5 − x − 5 = − x2 + 1  answer / antwoord  Max.length of VW is 1 unit  interpretation / interpretasie Maks. lengte van VW is 1 eenheid (3) [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com (EC/NOVEMBER 2023) MATHEMATICS P1 / WISKUNDE V1 13 QUESTION 8/VRAAG 8 m 8.1  i  ieff + 1 = 1 + nom   formula / formule  m  12  0, 093   substitution / vervanging = 1 +  −1  12  = 0, 09707 = 9, 71%  answer / antwoord (3) 8.2 A = P(1 + i ) n = R312000 (1 + 0, 0691)  n = 5 substitution / vervanging 5 = R 435 758,88  answer / antwoord (3) 0,0925 8.3.1   0, 0925  12   0, 082  24 i = and/en 𝑛 = 12 A =  23000 1 +  + 13500  1 +  4   4   12  i = 0,082 and/en 𝑛 = 24   12 = R51 530,18    0, 0925  12   23000 1 +  + 13500    4   24   0,082  1 +   12   answer / antwoord OR/OF (5) OR/OF   0, 0925  12  A =  23000 1 +  + 13500    4   0,0925   i= and/en 𝑛 = 12 4 = R 43760, 23 i= 0,082 and/en 𝑛 = 24 24 12  0, 082   R51530,18 A = R 43760, 23 1 +   12   0,082  24 = R51530,18  R43760, 23 1 +   12   answer / antwoord (5) 8.3.2 A = P (1 + i ) n i 36  12 and/en n = 36  i  64 487, 24 = 51530,18 1 +   12   substituting / vervang A = R64487, 24  64 487, 24   i =  36 − 1  12  51530,18   substituting into correct formula = 0, 075 vervanging in korrekte formule rate/koers = 7,5%  answer / antwoord (4) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com 14 MATHEMATICS P1 / WISKUNDE V1 (EC/NOVEMBER 2023) QUESTION 9/VRAAG 9 9.1.1 For mutually exclusive events: Vir onderling uitsluitende gebeurtenisse: P(A or B) = P(A) + P(B) formula / formule 0, 61 = 0, 35 + P(B)  substitution / vervanging  P( B) = 0, 61 − 0,35 = 0, 26  answer / antwoord (3) 9.1.2 For independent events: Vir onafhanklike gebeurtenisse: P(A or/of B) = P(A) + P(B) − P(A and/en B)  formula / formule 0, 61 = 0,35 + P(B) − P(A).P(B)  substitution / vervanging 0, 61 = 0,35 + P(B) − 0,35  P(B)  0,61 = 0,35 + P(B) − 0,35  P(B) 0, 61 = 0,35 + 0, 65  P(B)  0, 65  P(B) = 0, 26 0, 26  P(B) = 0, 65 = 0, 4  answer / antwoord (4) 9.2.1 a = 28, b = 86, c = 216  a = 28  b = 86  c = 216 (3) 216 9 9.2.2 = or / of 0,36  answer / antwoord 600 25 (1) 9.2.3 P(not defective) + P(Axisphones and defective) P(nie foutief) + P(Axis Phones en foutief) 514 58 = + 514 58 600 600  + 600 600 572 143 = = or/of 0,95  answer / antwoord 600 150 (3) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com (EC/NOVEMBER 2023) MATHEMATICS P1 / WISKUNDE V1 15 QUESTION 10/VRAAG 10 10.1 4 R x −1 5 R x x −5 x −1 G 5 R x −1 x −5 x G x−6 CA only if proper fractions are used and -1 for x −1 G 2nd pick VA slegs as egte breuke gebruik word en –1 vir tweede keuse. P(GG) = P(G)  P(G) x −5 x −6 =  x −5 x −6 x x −1   x −5 x −6 3 x x −1   = 3 3 x x − 1 11  equating to / stel gelyk aan 11( x − 5)( x − 6) = 3x( x − 1) 11 11 11( x 2 − 11x + 30) = 3x 2 − 3x  getting rid of fractions raak ontslae van breuke 11x 2 − 121x + 330 = 3x 2 − 3x 8 x 2 − 118 x + 330 = 0 4 x 2 − 59 x + 165 = 0  standard form / standaardvorm [4] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ## Related documents - [MATHS P1 GR11 MEMO NOV 2019 English+Afrikaans2](https://www.hlayiso.com/library/maths-p1-gr11-memo-nov-2019-english-afrikaans2/c5737a2c-1e54-41a6-8970-5bbb54f884f3) — Mathematics · Memo · 2019 · Afr - [MATHS P2 QP GR11 NOV 2023 Afrikaans](https://www.hlayiso.com/library/maths-p2-qp-gr11-nov-2023-afrikaans/17065526-5498-4c8a-b44d-031e45899cfc) — Mathematics · QP · 2023 · Afr - [Gr 11 Math P1 (Afrikaans) June 2023 Question Paper](https://www.hlayiso.com/library/gr-11-math-p1-afrikaans-june-2023-question-paper/3a6596d1-46af-4766-9815-0901dee27aaa) — Mathematics · QP · 2023 · Afr - [MATHS P1 GR11 MEMO NOV 2020 English+Afrikaans](https://www.hlayiso.com/library/maths-p1-gr11-memo-nov-2020-english-afrikaans/89544f20-6bf0-4ca5-b1c6-bb6abeb0c3cd) — Mathematics · Memo · 2020 · Afr - [Gr11 Math P1 (Afrikaans and English) November 2019 Possible Answers](https://www.hlayiso.com/library/gr11-math-p1-afrikaans-and-english-november-2019-possible-answers/e4962b19-d0cb-4874-a7fa-58cd8e95ab68) — Mathematics · Memo · 2019 · Afr - [Gr11 Math P1 (Afrikaans) June 2017 Possible Answers](https://www.hlayiso.com/library/gr11-math-p1-afrikaans-june-2017-possible-answers/b8504619-0cb6-476d-944d-f53d2ea22f72) — Mathematics · Memo · 2017 · Afr ## More from Grade 11 Mathematics - [FS Maths Grade 11 June 2025 P1 only](https://www.hlayiso.com/library/fs-maths-grade-11-june-2025-p1-only/82015d8f-bcc0-478e-ad47-d3b295dabb01) — QP · 2025 - [FS Maths Grade 11 June 2025 P2 only](https://www.hlayiso.com/library/fs-maths-grade-11-june-2025-p2-only/5d0e7dd8-cb37-419c-8fe2-0aee27fbdc84) — QP · 2025 - [KZN Maths Grade 11 November 2025 P2 and Memo](https://www.hlayiso.com/library/kzn-maths-grade-11-november-2025-p2-and-memo/a2aac931-a969-41c8-bde2-85d23035d042) — QP + Memo · 2025 - [MATHS P2 ANSWER BOOK GR11 2024 ENG AFR D watermark](https://www.hlayiso.com/library/maths-p2-answer-book-gr11-2024-eng-afr-d-watermark/91fd4f6b-a252-4d30-b6b1-f8f546015336) — Addendum · 2024 · Afr - [Gr 11 Mathematics P1 (Afrikaans) June 2024 Question Paper](https://www.hlayiso.com/library/gr-11-mathematics-p1-afrikaans-june-2024-question-paper/1a0f4a9d-f18a-4aca-af77-5a81acfb8e0c) — QP · 2024 · Afr - [Gr 11 Mathematics P2 (Afrikaans) June 2024 Question Paper](https://www.hlayiso.com/library/gr-11-mathematics-p2-afrikaans-june-2024-question-paper/4934c5d3-e2a1-4d6b-acf6-d00e0623d46d) — QP · 2024 · Afr ## Collection - [Grade 11 past papers](https://www.hlayiso.com/grade-11/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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