# Mathematics Grade 12 Paper 1 Eastern Cape Mock 2020 Memo > Grade 12 Mathematics Paper 1 memorandum from the Eastern Cape Mock Exam 2020, 16 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr-12-ec-maths-p1-memo-sep-2020/411f1d16-f302-46a4-88ab-36d8c0d25076 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2020 - Exam period: Eastern Cape Mock Exam - Paper: 1 - Curriculum: CAPS - Pages: 16 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 Paper 1 Eastern Cape Mock 2020 Memo](https://www.hlayiso.com/library/gr-12-ec-maths-p1-memo-sep-2020/411f1d16-f302-46a4-88ab-36d8c0d25076) ## About this document Mathematics · Grade 12 · Eastern Cape Mock Exam · 2020. Memorandum, 16 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## 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 SENIOR SERTIFIKAAT GRADE 12/GRAAD 12 SEPTEMBER 2020 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 16 pages./ Hierdie nasienriglyn bestaan uit 16 bladsye. ### Page 2 Downloaded from hlayiso.com 2 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2020) 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 applies in ALL aspects of the marking guideline. Volgehoue akkuraatheid 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 vir substitusie in die korrekte formule toegeken. QUESTION 1/VRAAG 1 1.1.1 2x2  x  3  0 (2 x  3)( x  1)  0  factorisation / faktorisering 3 x   or / of x 1 3 2  x  x 1 2 (3) 1.1.2 x(7 x  2)  1 7 x2  2 x 1  0  standard form / standaardvorm b  b 2  4ac x 2a 2  22  4(7)(1)  substitution / vervanging  2(7) 2  32  14  0, 26 or / of  0,55  0,26   0,55 (4) 1.1.3  x2  x  2  0 y x2  x  2  0  x2  x  2  0 ( x  2)( x  1)  0  x2  x  2  0  x  2 or / of x  1 2 1 x  x  2  x  1 OR/OF OR/OF y  x2  x  2  0  1 x (1  x)( x  2)  0  factorisation / faktorisering 2 1 x  x  2 or / of x 1  x  2  x 1 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) MATHEMATICS P1/WISKUNDE V1 3 1.1.4 17 2 x  22 x  2 2 x 2.2  2.2  17 x 23 2.2  x  17 x 2 2.2  17.2 x  8  0 2x  standard form / standaardvorm Let / Laat k  2 , x  2k 2  17 k  8  0  substitution / vervanging k  2 x (2k  1)(k  8)  0  factorisation / faktorisering 1 k or / of k  8 2  2 x  21 or / of 2 x  23  2x  21 and/en 2x  3 x  1 or / of x3  both answers / beide antwoorde OR/OF OR/OF 17 2 x  22 x  2 2 x 2.2  2.2  17 x 23 2.2 x   17 2x 2.22 x  17.2 x  8  0  standard form / standaardvorm (2.2 x  1)(2 x  8)  0  factorisation / faktorisering  2 x  21 or / of 2 x  23  2 x  2 1  2x  3 x  1 or / of x3  both answers / beide antwoorde (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com 4 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2020) 1.2 y ● B A ● ● O x ( x  2) 2  y 2  25 1 x  3  3y  0  2 x  3y  3  3  x  3y  3 (3 y  3  2) 2  y 2  25  substitution / vervanging (3 y  5) 2  y 2  25 9 y 2  30 y  25  y 2  25 standard form / standaardvorm 10 y 2  30 y  0  factorisation / faktorisering 10 y ( y  3)  0  y  0 or / of y3  x and y values / x en y waardes x  3 or / of x6  A(3;0) and/en B(6;3)  both sets of coordinates beide stelle van koördinate OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) MATHEMATICS P1/WISKUNDE V1 5 ( x  2) 2  y 2  25 1 x  3  3y  0  2 y  13 ( x  3)  3  y  13 ( x  3) 2 x   substitution / vervanging ( x  2)    1  25 2 3  x2 2x x  4x  4   2  1  25  0 9 3 9 x 2  36 x  36  x 2  6 x  216  0 10 x 2  30 x  180  0 standard form / standaardvorm x 2  3 x  18  0 ( x  3)( x  6)  0  factorisation / faktorisering  x  3 or / of x6  x and y values / x en y waardes y 0 or / of y3  A(3 ; 0) B(6 ; 3)  both sets of coordinates / beide stelle koördinate (6) 1.3 ( x  m)( x  n)  p 2 x 2  nx  mx  mn  p 2  0 x 2  (m  n) x  (mn  p 2 )  0  x 2  (m  n) x  (mn  p 2 )  0 For real roots / Vir reele wortels:  (m  n) 2  4(1)(mn  p 2 )  0 b 2  4ac  0  (m  n) 2  4(1)(mn  p 2 )  0 m 2  2mn  n 2  4mn  4 p 2  0 m 2  2mn  n 2  4 p 2  0  m  n  (2 p) 2  0 2  m  n   (2 p)2  0  explanation / verduideliking 2 Now/Maar: (m  n) 2  0 and/en (2 p) 2  0   0 (4) [26] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com 6 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2020) QUESTION 2/VRAAG 2 2.1 86 ; 119 ; 150 ; 179 ; 206 ; 231 33 31 29 27 25 2 2 2 2 2.1.1 206 ; 231  206 231 (2) 2.1.2 2 a  2  a  1  a  1 3a  b  33 3  b  33  b  36  b  36 a  b  c  86 1  36  c  86  c  51  c  51  Tn  n 2  36n  51 (4) Tn   n 2  36n  51 2.1.3 326  n 2  36n  51  substitution / vervanging n 2  36n  275  0 (n  11)(n  25)  0  method / metode  n  11 or / of n  25  answers / antwoorde (3) 2.1.4 86 +k ; 119 +k ; 150 +k ; 179 +k 33 31 29 2 2 2a  2  a  1 3a  b  33 3  b  33  b  36  a  1 and/en b  36 a  b  c  86  k 1  36  c  86  k  c  51  k  c  51  k Tn  k   n 2  36n  (51  k ) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) MATHEMATICS P1/WISKUNDE V1 7 2.2.1 2y 1 ; 4y 1 ; 6y 1 d  4 y  1  (2 y  1)  2y  d = 2y Tn  2 yn  1  2 yn  1 T30  2 y (30)  1  60 y  1  60 y  1 OR/OF OR/OF Tn  a  (n  1)d T30  (2 y  1)  (30  1)(2 y )  d = 2y  substitution / vervanging  2 y  1  58 y  answer / antwoord  60 y  1 (3) 2.2.2 Sn  n 2a  n  1d  2  d  2y S 30  30 2(2 y  1)  292 y   substitution into correct formula 2 vervanging in die korrekte formule  15(4 y  2  58 y )  15(62 y  2)  equating to  2820 /  2820  15(62 y  2) stel gelyk aan – 2820  188  62 y  2  186  62 y  y  3  answer / antwoord OR/OF OR/OF n Sn   a  l   d  2y 2 30  substituting a and l / S30   2 y  1  60 y  1 vervanging van a en l 2 2820  15  62 y  2  equating to  2820 188  62 y  2 stel gelyk aan – 2820 186  62 y  answer / antwoord  y  3 (4) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com 8 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2020) QUESTION 3/VRAAG 3 3.1 1  4  4 2  43  ...  4 n1 Tn  4 n1  Tn  4 n1 7 Sum/Som : 4 n 1 7 n 1   4 n1 For original sequence:/Vir oorspronklike ry: n 1 5 000 Sum/Som:  n 5000  n n 1 n 1 5 000 7 Sn :  n   4 n 1  answer / antwoord n 1 n 1 (4) 3.2 1 S  ar 1 1  x  x 2  x 3  ...  and/en 1 1 x  1 1 x 1  x  x 2  x 3  ...  1 1 x  1 x 1 1 Sum/Som :  1 x 1 x 1 x 1 x  (1  x)(1  x) 2  2 1  x2  1  x2 2  8  equating sum to 8 1  x2 stel som gelyk aan 8 8  8x 2  2  8 x 2  6 3 x2  4 3 3  x2  x 4 2  answer / antwoord (6) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) MATHEMATICS P1/WISKUNDE V1 9 QUESTION 4/VRAAG 4 4.1 a f ( x)   3, x 1 x  1; y 3  x 1  y3 (2) 4.2 a y 3 0 1  3 a  3a a 0 3  y0 x 1 a 3  x 1 3 x  3  a a 3 a a 3a x  1 or / of  x  1 or/of 3 3 3 3 (3) 4.3  asymptotes / asimptote  y-intercepts / y-afsnitte  x-intercepts / x-afsnitte  shape / vorm (4) 4.4 a 1  x + 2  +1 f ( x)   1 or/of f ( x)  1 x2 x2 (2) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com 10 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2020) QUESTION 5/VRAAG 5 5.1 (0 ;  7)  y  7 (1) 5.2 q  8  q  8 (1) 5.3 g ( x)  b x  8  5  b1  8  substituting (1 ; – 5) / vervanging van (1;  5) b  3 At turning point of f: / By die draaipunt van f: b 3 b x   x 2a 4 2a b 3  3  3   2a 4 2a 4  6a  12 b=3  simplifying / vereenvoudiging a2  5  2(1) 2  3(1)  c  f (1)  5 c  10 (6) 5.4 y  8 ; y  ℝ  y  8 (accuracy) / (akkuraatheid) (2) 5.5 y  9 x   28 y  9 x  28 y  2 x 2  3 x  10  2 x 2  3 x  10  9 x  28  2 x 2  3x  10  9 x  28 2 x 2  12 x  18  0 standard form / standaardvorm x 2  6x  9  0  factorisation / faktorisering ( x  3)( x  3)  0  x  3  x  3 y  2(3) 2  3(3)  10  y  1  1  T (3 ;  1) OR/OF OR/OF y  9 x  28  m  9  m  9 f ( x)  2 x 2  3 x  10  f ' x   4 x  3 f '  x   4 x  3  9  equating gradients /  4 x  12 stel gradiënte gelyk x  3  x  3 y  9(3)  28  1  y  1 T (3 ;  1) (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) MATHEMATICS P1/WISKUNDE V1 11 5.6 y  log 3 x   y  log 3 x (2) 5.7 p( x)  f ( x)  1  2 x 2  3 x  10  1  2 x 2  3x  9  p ( x)  2 x 2  3 x  9  (2 x  3)( x  3) when/wanneer: y  0, 3 x or / of  3  x-intercepts / x-afsnitte 2 3  x  3 or / of 0 x 2  x  3 3  0 x (accuracy/akkuraatheid) 2 (4) [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com 12 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2020) QUESTION 6/VRAAG 6 6.1 A  P (1  i ) n 1 1 x  x (1  i ) 4  x  x (1  i) 4 3 3 1  (1  i ) 4 3 1  3 4  1  i  0,24  i  i  0,24  i  0,24  r  24%  answer / antwoord (3) 6.2.1 x[(1  i )  1] n F 0,095 i  i  0,095 72  12 x 1    1  n  72  12    F  R 596 458,10 R 596 458,10  0,095  substitution into correct formula / 12 vervanging in die korrekte formule 0,095 R 596 458,10  x 12  0,095  72  1    1  12    answer / antwoord  R 6178,13 (5) 6.2.2 P  x 1  1  i  n  i   0,095  72   n  72 R 6178,131  1      12    substitution into correct formula /  vervanging in die korrekte formule 0,095 12  R 338 070,29 A  P (1  i ) n  0,095  5  A  R 338 070,29 R 338 070,29  P1    12   n5  substitution into correct formula / R 338 070,29 P  5 vervanging in die korrekte formule  0,095  1    answer / antwoord  12   R 325 000 (6) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) MATHEMATICS P1/WISKUNDE V1 13 QUESTION 7/VRAAG 7 𝑓 (𝑥+ℎ)−𝑓(𝑥) f ' x  = lim 7.1 ℎ→0 ℎ −2(𝑥+ℎ)2 −(−2𝑥2 ) = lim ℎ  substitution / vervanging ℎ→0 −2(𝑥 2 +2𝑥ℎ+ℎ2 )+2𝑥 2 = lim ℎ→0 ℎ  expansion / uitbreiding −2𝑥 2 −4𝑥ℎ−2ℎ2 +2𝑥 2 = lim ℎ→0 ℎ  −2𝑥 2 − 4𝑥ℎ − 2ℎ2 + 2𝑥 2 ℎ(−4𝑥−2ℎ) = lim  factorisation / faktorisering ℎ→0 ℎ = lim (−4𝑥 − 2ℎ) ℎ→0 = − 4𝑥   4x OR/OF OR/OF 𝑓(𝑥) = −2𝑥 2 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = −2(𝑥 + ℎ)2 − (−2𝑥 2 ) = −2𝑥 2 − 2𝑥ℎ − 2ℎ2 + 2𝑥 2  −2𝑥 2 − 2𝑥ℎ − 2ℎ2 + 2𝑥 2 = −4𝑥ℎ − 2ℎ2  −4𝑥ℎ − 2ℎ2 𝑓 (𝑥+ℎ)−𝑓(𝑥) f ' x  = lim ℎ→0 ℎ −4𝑥ℎ−2ℎ2 = lim  substitution / vervanging ℎ→0 ℎ ℎ(−4𝑥ℎ−2ℎ) = lim  factorisation / faktorisering ℎ→0 ℎ = lim (−4𝑥ℎ − 2ℎ) ℎ→0   4x = − 4𝑥 (5) 7.2 2 y  7x4  x3 3  3  7x  2x 4 2   2x 2 5 5 dy     28 x 3  3 x 2  28x 3   3x 2 (3) dx 7.3 1 5  Dt  gt 2   3g  2 t  1   5t 1  Dt  gt 2  5t 1  3g  2   gt  5t 2  gt   5t 2  Dt(3g) = 0 (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com 14 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2020) QUESTION 8/VRAAG 8 8.1 (0 ; 9)  (0 ; 9) (1) 8.2 f  x   2 x  x  12 x  9 3 2 f (1)  2(1)3  (1) 2  12(1)  9  method / metode 0  f  x   ( x  1)(2 x 2  3x  9)  2 x 2  3x  9  factorisation / faktorisering  0  ( x  1)( x  3)(2 x  3) 3  x  1 or / of x  3 or / of x  2 3  3  D(3;0), E(1;0), F  ;0   D(3; 0)  E(1; 0)  F ; 0  2  2  (6) 8.3 For concave down / Vir konkaaf afwaarts  f ''  x   0 f ' ' x   0 f '  x   6 x 2  2 x  12  f ' x   6 x 2  2 x  12  f ' ' x   12 x  2 f ' '  x   12 x  2 12 x  2  0 1 1  x x 6 6 (4) 8.4 6 x 2  2 x  12  0  b  b 2  4ac x 2a  2  2 2  4(6)(12)  substitution into correct formula  vervanging in die korrekte formule 2.6 2  292  x-values / x-waardes  12  1,59 or / of 1, 26  answer / antwoord   1,59  x  1, 26 (accuracy / akkuraatheid) (4) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com (EC/SEPTEMBER 2020) MATHEMATICS P1/WISKUNDE V1 15 QUESTION 9/VRAAG 9 1 1 Area of/van ∆PBA  bh   t  12  6t cm 2  Area of/van ∆PBA 2 2 1 7 t2 Area of/van ∆ACS  t (7  t )  t  cm 2  Area of/van ∆ACS 2 2 2 1 7t Area of/van ∆BQC  (12  t )7  42  cm 2  Area of/van ∆BQC 2 2 Area of rectangle/van reghoek: PBQS  12  7  84 cm 2 ∴ Area of/van ∆ABC:  subtracting from 84 / trek van 84 af 7t t 2 7t A(t)  84  6t    42  2 2 2 1  t 2  6t  42 2 a  0, so A  t  is a min graph / is ' n min grafiek A ' t   t  6  A ' (t )  0 At min / By min : A '(t )  0  t 6  0 t 6 Smallest area / Kleinste oppervlakte: A 6  6  6(6)  42 1 2 2  24 cm 2  smallest area / kleinste area (6) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com 16 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2020) QUESTION 10/VRAAG 10 10.1 S =135 Correct entries of: Korrekte waarde van: M P 12  12 y 2y+3  5 ; 12 ; 24 x 5 24  2y+3 ; y ; y y x 12 A (4) 10.2 x  y  5  24  60  x + y + 5 + 24 = 60 x  y  31 2 y  3  x  5  12  x  y  12  24  2 y  3  x  5  12  x  y  12  24 2 y  x  20  x  y  36  y  16 and/en x  15  y = 16 and/en  x = 15 (4) 10.3 2 y  3  5  12  x  24 P(M or/of (P and/en A))= 135  2 y  3  5  12  24 35  5  12  15  24  135 91 = or / of 0 , 67  answer / antwoord 135 (3) [11] QUESTION 11/VRAAG 11 11.1 7!= 5 040 ways / maniere   7! (2) 11.2 Sample space/Steekproefruimte: 5 4 2 3 1 2 1  arrangement / rangskikking Possible outcomes/Moontlike uitkomste: 3 2 2 2 1 1 3  arrangement / rangskikking 3  2  2  2 11 3 Probability/Waarskynlikheid = 5  4  2  3 1 2 1 72  3 2  2  2 11 3 and/en  5  4  2  3 1 2 1 240 3  answer / antwoord   0,3 10 (4) [6] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ## Related documents - [Gr 12 EC Maths P2 Memo Sep 2020 Errata](https://www.hlayiso.com/library/gr-12-ec-maths-p2-memo-sep-2020-errata/e1e6c6e0-9ebe-45f8-ba36-fe3147f0640a) — Mathematics · Memo · 2020 - [2020 GRADE 12 MATH TRIAL EXAM PAPER 1 KZN MEMO](https://www.hlayiso.com/library/2020-grade-12-math-trial-exam-paper-1-kzn-memo/d74b83c6-40ce-47f3-99b2-5422a6854104) — Mathematics · Memo · 2020 - [Mathematics P1 Nov 2020 Memo Afr Eng](https://www.hlayiso.com/library/mathematics-p1-nov-2020-memo-afr-eng/48e54b94-396f-4fda-b66d-100bed96bcc2) — Mathematics · Memo · 2020 · Eng - [MATHS P1 GR12 MEMO SEPT 2021 English IsiXhosa](https://www.hlayiso.com/library/maths-p1-gr12-memo-sept-2021-english-isixhosa/6e14ca2d-44f7-417b-b314-e40962f91676) — Mathematics · Memo · 2021 · Eng - [Maths P1 Memo](https://www.hlayiso.com/library/maths-p1-memo/7d83c9c7-adb9-4818-93b0-08773c9e1945) — Mathematics · Memo · 2016 - [2020 GRADE 12 MATH TRIAL EXAM PAPER 1 NW](https://www.hlayiso.com/library/2020-grade-12-math-trial-exam-paper-1-nw/29fe043e-9275-4a5d-9937-7bc72911b845) — Mathematics · QP · 2020 ## More from Grade 12 Mathematics - [MATHS P1 ANSWER BOOK GR12 SEPT 2025 English Afrikaans watermark](https://www.hlayiso.com/library/maths-p1-answer-book-gr12-sept-2025-english-afrikaans-watermark/d78c3fd3-dcd5-4f3f-911b-e5635b4686cc) — Addendum · 2025 · Afr - [Maths Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-gauteng/1d6bf7fd-d288-4a01-99f7-2c87e247760c) — QP · 2025 - [Maths Grade 12 NSC P2 QP September 2025 Afr Limpopo](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-afr-limpopo/f2f8d481-5d87-4e89-bb21-6077bbcf403d) — QP · 2025 · Afr - [Maths Grade 12 NSC P1 QP September 2025 Afr Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p1-qp-september-2025-afr-northern-cape/f5b3a4ce-fa91-4475-96e0-0654a87e1405) — QP · 2025 · Afr - [Maths Grade 12 NSC P2 QP September 2025 Eng Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-eng-northern-cape/4ede8207-ca82-4419-bae1-fae734498cc7) — QP · 2025 · Eng - [MATHS P2 ANSWER BOOK GR12 JUNE 2024 Afr+English](https://www.hlayiso.com/library/maths-p2-answer-book-gr12-june-2024-afr-english/4a71e975-4ac4-4f55-8e23-0b8155a5b0c6) — Addendum · 2024 · Eng ## Collection - [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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