# Mathematics Grade 12 P2 Gauteng Prelim 2024 Memo (Afrikaans) > Grade 12 Mathematics Paper 2 memorandum from the Gauteng Prelim Exam 2024 (Afrikaans), 20 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr-12-mathematics-p2-english-and-afrikaans-2024-possible-answers/62b34748-9b6e-41cd-9210-ea13ad628fe1 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Language: Afrikaans - Document type: Memorandum - Year: 2024 - Exam period: Gauteng Prelim Exam - Paper: 2 - Curriculum: CAPS - Pages: 20 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 P2 Gauteng Prelim 2024 Memo (Afrikaans)](https://www.hlayiso.com/library/gr-12-mathematics-p2-english-and-afrikaans-2024-possible-answers/62b34748-9b6e-41cd-9210-ea13ad628fe1) ## About this document Mathematics · Grade 12 · Gauteng Prelim Exam · 2024 · Afrikaans. Memorandum, 20 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 PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN 2024 OFFICIAL/ AMPTELIKE MARKING GUIDELINES/NASIENRIGLYNE MATHEMATICS/WISKUNDE (PAPER/VRAESTEL 2) (10612) 20 pages/bladsye 1 ### Page 2 Downloaded from hlayiso.com 10612/24 NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempted question and not redone the question, mark the crossed-out version. • Consistent accuracy applies in ALL aspects of the marking guidelines. Stop marking at the second calculation error. • Assuming answers/values to solve a problem is NOT acceptable. • Question 7.3 will be the only question in this marking guideline to have a rounding off penalty. LET WEL: • As ʼn kandidaat ʼn vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na. • As ʼn kandidaat ʼn antwoord tot ʼn vraag doodtrek en nie oordoen nie, sien die doodgetrekte poging na. • Volgehoue akkuraatheid word in ALLE aspekte van die nasienriglyne toegepas. Hou op nasien by die tweede berekeningsfout. • Aannames van antwoorde/waardes om ʼn probleem op te los, word NIE toegelaat nie. • Vraag 7.3 is die enigste vraag in die nasienriglyne waar afronding gepenaliseer sal word. GEOMETRY/MEETKUNDE A mark for a correct statement (A statement mark is independent of a reason.) S ʼn Punt vir ʼn korrekte bewering (ʼn Punt vir ʼn bewering is onafhanklik van die rede.) A mark for a correct reason (A reason mark may only be awarded if the statement is correct.) R ʼn Punt vir ʼn korrekte rede (ʼn Punt word slegs vir die rede toegeken as die bewering korrek is.) Award a mark if the statement AND reason are both correct. S/R Ken ʼn punt toe as die bewering EN rede beide korrek is. 2 ### Page 3 Downloaded from hlayiso.com 10612/24 QUESTION/VRAAG 1 Monthly salary (in thousand Rand) Number of employees Maandelikse salaris (in duisend Rand) Aantal werknemers R0  x  R10 3 R10  x  R20 4 R20  x  R30 13 R30  x  R40 20 R40  x  R50 21 R50  x  R60 12 R60  x  R70 12 R70  x  R80 8 R80  x  R90 5 R90  x  R100 2 1.1 Cumulative Frequency/Kumulatiewe Frekwensie Ogive/Ogief ✓ (0 ; 0) ✓ Upper limit/Boonste limiet ✓ Cumulative frequency/ Kumulatiewe frekwensie ✓ Shape/Vorm Monthly salaries in thousand Rand/Maandelikse salaris in duisend Rand (4) 1.2 30 employees/werknemers ✓ answer/antwoord Accept/aanvaar 29-31 (1) 1.3 Median/Mediaan = R45 000 ✓ answer/antwoord Accept/aanvaar R44 000 – R46 000 (1) 1.4.1 Mean/Gemiddeld = R85 857,14 ✓ answer/antwoord (1) 1.4.2 Standard deviation/Standaardafwyking = R2 294,63 ✓ answer/antwoord (1) 1.4.3 (83 562,51 ; 88 151,77 ) ✓ interval 4 ✓4  100 ✓ answer/antwoord 7 (3) = 57,14% [11] 3 ### Page 4 Downloaded from hlayiso.com 10612/24 QUESTION/VRAAG 2 Stall Number of Amount of money Stalletjie customers/ spent (R)/ Aantal klante Bedrag spandeer (R) 1 10 163 2 7 138 3 9 185 4 12 220 5 8 147 6 13 245 SCATTER PLOT/SPREIDIAGRAM 270 Amount/Bedrag (R) 250 230 210 190 170 150 130 110 5 6 7 8 9 10 11 12 13 14 Number of customers/Aantal klante 2.1 a = 12,59627329 ANSWER ONLY: full marks ✓a b = 17,32919255 SLEGS ANTWOORD: volpunte ✓b ✓ equation/vergelyking yˆ = 12, 60 + 17,33 x (3) 2.2 yˆ = 12, 60 + 17,33(11) ✓ substitution/substitusie ANSWER ONLY: full marks ✓ R203,23 yˆ = 203, 23 SLEGS ANTWOORD: volpunte (2) OR/OF yˆ = 203, 22 (with calculator/met sakrekenaar) ✓✓answer/antwoord 2.3 r = 0,949501084 ✓ value of 𝑟/waarde van r r = 0,95 (1) 2.4 No/Nee ✓ no, strong positive The correlation shows a positive strong correlation between correlation/nee sterk the money spent and the number of customers./Daar is ‘n positiewe korrelasie sterk positiewe korrelasie tussen die geld spandeer en die NO marks for NO only/ aantal klante. Geen punte vir slegs NEE nie (1) 4 ### Page 5 Downloaded from hlayiso.com 10612/24 2.5 The gradient will decrease as the line will become less steep. ✓ decrease/afneem The point is far away from the line. Or the point is above the ✓ point is above the data/far data./Die gradiënt sal afneem want die lyn word minder steil. away from the line/punt is Die punt is ver van die lyn af. Of die punt lê bokant die data. bokant die lyn/punt is ver weg van die lyn af (2) [9] QUESTION/VRAAG 3 𝑦 𝜃 𝑥 O M R 3.1 2 y − 3x + 9 = 0 ✓y=0 2(0) − 3 x = −9  x=3 x=3  M(3; 0) (2) 3.2 12 − 0  correct substitution of P mPM = −9 − 3 and M into gradient = −1 formula/korrekte Substitute/Vervang P (−9;12) substitusie van P en M in die gradiënt formule y − 12 = − ( x + 9 ) ✓ mPM = −1 y = − x − 9 + 12 ✓sub of point P or M/ y = −x + 3 vervang punt P of M ✓equation/vergelyking (4) 5 ### Page 6 Downloaded from hlayiso.com 10612/24 3.3 3 ✓ tan  = 3 tan  = 2 2 ✓answer/antwoord  = 56,31 (2) 3.4 Substitute/vervang (a ; b) into/in y = − x + 3 :  b = −a + 3  substitution/substitusie b = 3 − a (1) 3.5 (5 5 ) = (a − 9) + (b − 9) 2 2 2  substitution into correct formula/substitusie in die Substitute/vervang b = 3 − a : korrekte formule 125 = (a − 9) 2 + (3 − a − 9) 2 125 = a 2 − 18a + 81 + (−a − 6) 2 ✓subs of 𝑎/vervang a 125 = a 2 − 18a + 81 + a 2 + 12a + 36 125 = 2a 2 − 6a + 117  0 = 2a 2 − 6a − 8  0 = a 2 − 3a − 4  standard form/ standaardvorm  0 = (a − 4)(a + 1)  a = 4 or a = −1 Since N is in the second quadrant, we choose a = −1 . Omdat N in die tweede kwadrant is, kies a = −1 . ✓ correct value of a/  b = 3 − (−1) korrekte waarde van a b = 4 ✓ correct value of b/ korrekte waarde van b (5) 3.6 9+x 9+ y 3= 0= 2 2 ✓ xR = −3 ✓ yR = −9 R(−3 ; − 9) (−3)2 + (−9)2 = r 2 ✓substitution of R/ x + y = 90 2 2 vervang R ✓ x + y = 90 2 2 (4) 3.7 Let 𝛽 be the inclination angle of the line/Laat 𝛽 die inklinasiehoek wees van die lyn y = mx + 4 ✓  = 11, 31  = 56,31 − 45 or/of / of  = 56,31 + 45 ✓  = 101,31  = 11,31 or/of / of  = 101,31 ✓ m = 0, 20 m = tan11,31 or/of / of m = tan101.31 ✓ m = −5, 0 m = 0, 20 or/of or / of m = −5,00 (4) See diagram for Q3.7 on the next page./ Sien die diagram vir V3.7 op die volgende bladsy. [22] 6 ### Page 7 Downloaded from hlayiso.com 10612/24 3.7 𝑦 45° 𝛽 56,31° 𝑥 O M R 𝑦 𝛽 56,31° 𝑥 O M 45° R 7 ### Page 8 Downloaded from hlayiso.com 10612/24 QUESTION/VRAAG 4 𝑦 F C A 𝜃 B 𝑥 𝜃 O D 4.1 B(−1;1) ✓answer/antwoord (1) 4.2 BC = 6 − 1 = 5 ✓BC = 5 AB = 20 ˆ = 90 tan ⊥ rad/rklyn ⊥ rad CAB ✓ CÂB = 90 AC2 = ( 5 ) − ( 20 ) 2 2 ✓ subs in Pythagoras/ AC = 5 vervang in Pythagoras (3) 4.3 tan  = AC = 5 = 1 ✓ tan  = 1 AB 2 20 2 (1) 4.4 mDC  mAB = −1 1 ✓ mDC = 1 mDC = tan  = 2 2 mAB = −2 ✓ mAB = −2 Substitute/Vervang B( −1;1) ✓substitution of 𝑚 and B/ y − 1 = −2( x + 1) substitusie van m en B y = −2 x − 1 (3) OR/OF 8 ### Page 9 Downloaded from hlayiso.com 10612/24 m AB = tan (90 + 26,57) ✓ mAB = tan (90 + 26,57) m AB = − 2 ✓ m AB = − 2 Substitute/ Vervang B(− 1 ; 1) y − 1 = − 2(x + 1) ✓substitution of 𝑚 and B/ substitusie van m en B y = − 2x − 1 (3) 𝑦 F A 26,67° 90° + 26,67° 𝑥 D B O 4.5 (x + 1)2 + (− 2 x − 1 − 1)2 = 20 ✓substitution into correct formula/substitusie in x 2 + 2 x + 1 + 4 x 2 + 8 x + 4 = 20 die korrekte formule 5 x 2 + 10 x − 15 = 0 x 2 + 2x − 3 = 0 (x + 3)(x − 1) = 0 ✓standard form/ x = − 3 or/of x = 1 standaardvorm y = − 2(− 3) − 1 ✓ x = −3 y= 5 A(− 3 ; 5) ✓y =5 OR/OF (4) 9 ### Page 10 Downloaded from hlayiso.com 10612/24 Eq of DC/Verg van DC: ✓equation of DC/ 1 vergelyking van DC y − 6 = ( x + 1) 2 1 y = x + 6,5 ✓equate/gelykstel 2 1 x + 6,5 = −2 x − 1 2 ✓ x = −3 x + 13 = −4 x − 2 5 x = −15 ✓y =5 x = −3 y = −2(−3) − 1 y=5  A(−3 ; 5) (4) 4.6 Area of/van ΔABC = 1 2 ( )( ) 20 5 sin 26,57 ✓1 2 ( 20 )( ) 5 sin 26,57 =5 ✓ area of / van ΔABC = 5 Area of/van ΔODF = (13) (14,53) sin 26,57 1 ✓ 13 and/en 14,53 2 ✓ sin 26,57 = 42,25 ✓ Area of/van ΔABC : Area of ΔODF = 20 : 169 area of / van ΔODF = 42,25 ✓answer/antwoord (6) OR/OF Area Δ ABC = 1  5  20 = 5 ✓ 1  5  20 2 2 Equation of DC/Vergelyking van DC: 1 y − 6 = ( x + 1) 2 1 ✓ y = 1 x + 13 y = x + 6,5 2 2 2 ✓ OF = 13 1 13 Equation of/verg van DC is y = x + 2 2 2 ✓OD = 13 13 OF = and/en OD = 13 2 1  13  169 ✓ 1  13 13 Area of Δ ODF =    13 = 2 2 2 2 4 ✓answer/antwoord Area Δ ABC : Area Δ ODF = 20 :169 (6) [18] 10 ### Page 11 Downloaded from hlayiso.com 10612/24 QUESTION/VRAAG 5 5.1 sin 2 (180  + x ) .sin (− x ) −1 − sin (90 + x ). tan x ✓ (− sin x ) 2 = (− sin x )2 . − sin x − 1 sin x ✓ − sin x − cos x. ✓ − cos x cos x ✓ sin x = sin 2 x − 1 cos x ( = −1 1 − sin x 2 ) ✓ sin x − 1 2 = − cos2 x ✓ − cos2 x (6) 5.2.1 sin (A + B) = cos90 − (A + B) ✓ cos 90 − ( A + B )  = cos (90 − A ) − B ✓ cos ( 90 − A ) − B = cos(90 − A ) cos B + sin(90 − A ) sin B ✓expansion/uitbreiding = sin A cos B + cos A sin B (3) 5.2.2 cos 420  cos15 + sin 300  cos105  ✓ cos60 = cos 60 cos15 + (− sin 60)(− sin 15) ✓ − sin 60 ✓ − sin 15 = cos(60 − 15) ✓ cos(60 −15) = cos 45 2 2 = ✓ 2 2 (5) 5.3.1  1  LHS/LK = tan 2 x  − 1  tan x  2 ✓ 1 − tan x 2 = 1 − tan x 2 sin 2 x =1− sin 2 x cos 2 x ✓ cos 2 x − sin 2 cos2 x = cos2 x − sin 2 cos 2 x ✓ cos 2 x cos2 x = cos 2 x = RHS/RK (3) 5.3.2 x = 90 ✓ answer/antwoord (1) 11 ### Page 12 Downloaded from hlayiso.com 10612/24 5.4 cos 2 x = cos x 2 cos2 x − 1 = cos x ✓ cos 2 x = 2 cos x − 1 2 2 cos2 x − cos x − 1 = 0 ✓standard form (2 cos x + 1)(cos x − 1)= 0 standaardvorm 1 cos x = − or / of cos x = 1 2 x = 180 − 60 + k 360 ; k   or x = k 360 ; k   ✓both answers for cos x x = 120 + k 360 ; k   albei antwoorde vir or/of cos x ✓x = k 360 ; k   x = 180 + 60 + k 360 ; k   ✓ x = 240 + k 360 ; k   x =120 and / en x = 240 ✓ x = + k 360 ; k   (6) 5.5 sin 3 = sin (2 +  ) ✓ sin 3 = sin (2 +  ) = sin 2 cos + cos 2 sin  ✓ compound expansion/ ( ) = (2 sin  cos )cos + 1 − 2 sin 2  sin  saamgestelde uitbreiding = 2 sin  cos  + sin  − 2 sin  2 3 ✓ sin 2 = 2 sin  cos ( ) = 2 sin  1 − sin 2  + sin  − 2 sin 3  ✓ cos 2 = 1 − 2 sin  2 = 2 sin  − 2 sin 3  + sin  − 2 sin 3  ✓ cos  = 1 − sin  2 2 = 3 sin  − 4 sin 3  3 1 1 = 3  − 4  2 2 =1 ✓sub/vervang sin  = 1 2 ✓answer/antwoord OR/OF (7) 1 sin  = 2  = 30 or  = 150  ✓  = 30 sin 3 sin 3 ✓  = 150  = sin 3(30) = sin 3(150 ) ✓ sin 3(30) = sin 90 = sin 450  ✓ sin 3(150) =1 = sin 90 ✓ sin 450 =1 ✓ sin 450 = sin 90 ✓ sin 90 = 1 (7) [31] 12 ### Page 13 Downloaded from hlayiso.com 10612/24 QUESTION/VRAAG 6 6.1 3 ✓ shape/vorm ✓ x-intercepts/x-afsnitte 2 ✓ turning points/draaipunte 𝒇 1 𝒈 -180° -135° -90° -45° 0° 45° 90° 135° 180° -1 -2 (3) -3 6.2 180 ✓ 180 (1) 6.3 x  0 ; 45 ✓ x  0 ; 45 x  90 ; 135 ✓ x  90 ; 135 (2) 6.4 Maximum value of/Maksimum waarde van ✓3 f (x)− g (x) = 2 − (− 1) = 3 (1) 6.5 𝑦 = 22𝑐𝑜𝑠𝑥+2 Range of / Waardevers van 2cos x : −2 y  2 Range of / Waardevers van 2cos x + 2 : 0  y  4 Range of / Waardevers van 22cos x + 2 20  y  2 4  1  y  16 ✓✓ 1  y  16 (2) [9] 13 ### Page 14 Downloaded from hlayiso.com 10612/24 QUESTION/VRAAG 7 7.1 y ✓correct ratio/ sin  = korrekte verhouding BE y BE = ✓ BE = y sin  sin  (2) 7.2 In  ABE : ˆ = 90 +  . ˆ = 90 +  ✓ BAE BAE AB BE ✓ correct use of sine rule/ = sin 2 sin(90 +  ) korrekte gebruik van BE  sin 2 sin reël AB = cos  ✓ double angle/dubbelhoek y  2sin  cos  ✓co-function/ko-funksie = sin  ✓ substitution of BE/ cos  = 2y vervang BE (5) 7.3 In  BAF is: 7y ✓correct ratio in terms of AF y/korrekte verhouding in ˆ cos BAF= = 4 terme van y AB 2 y 7 ✓ cosBÂF = 7 = 8 8 ˆ = 28,955.. BAF ✓29° / PENALTY for = 29 incorrect rounding off/ Penaliseer vir foutiewe afronding (3) [10] 14 ### Page 15 Downloaded from hlayiso.com 10612/24 QUESTION/VRAAG 8 8. A 1 23 G 2 B 1 2 1 3 = H 21 D =C 8.1.1 B̂2 = 90 ∠ in semi-circle/∠ in ʼn halwe sirkel S R (2) 8.1.2 Bˆ 1 + Bˆ 2 + Gˆ = 180 opposite ∠s of cyclic quad/teenoorst ∠e van R kvh B̂1 = 32 S (2) 8.1.3 ˆ =B A ˆ = 32 ∠s in the same segment/ ✓S R 2 1 (2) ∠e in dieselfde segment 8.2 Ĉ1 = Â 3 = 29 ∠s opposite equal sides/∠e teenoor S/R gelyke sye D̂ = Ĉ1 = 29 ∠s in the same segment/∠e in dieselfde S/R segment  Â 3 = D̂ = 29 ∴AB is a tangent to the converse tan chord theorem/ circle/AB is ‘n raaklyn aan omgekeerde raaklyn koordstelling R (3) die sirkel [9] 15 ### Page 16 Downloaded from hlayiso.com 10612/24 QUESTION/VRAAG 9 9.1 Construction/Konstruksie: ✓construction/ Draw diameter KS and join konstruksie KR/Trek middellyn KS en trek KR ✓S ✓R ˆ = 90 QSK radius ⊥ tangent/raaklyn ˆ = 90 ✓S/R SRK ∠ in semicircle/halwe sirkel ✓S ✓R ˆ = TRK but/maar TSK ˆ ∠’s in the same segment/ ˆ = Rˆ ∠e in dieselfde segment (6)  QST 16 ### Page 17 Downloaded from hlayiso.com 10612/24 9.2 F 1 2 Q E G 3 23 1 2 1 D 2 1 C B A 9.2.1 ˆ =x E  centre = 2   at R circumference/ (1) midpts∠=2 x omtreks∠ 9.2.2 Fˆ1 = Eˆ = x tan chord theorem/ S/R rklyn koordstelling S/R ˆ = Fˆ = x C ext. ∠ of cyclic quad/ 1 1 ˆ  C = Eˆ buite ∠ van kvh 1 R (3) corresponding ∠s are equal/  GC|| FE ooreenk ∠e is gelyk 9.2.3 AB AG prop. theorem/eweredigheid- S R = BD GF stelling; GB || FD AG AC = prop. theorem/eweredigheid- S GF CE AB AC stelling; GC || FE  = BD CE (3) 17 ### Page 18 Downloaded from hlayiso.com 10612/24 9.2.4 AB 4 AC 4 = and/en = ✓substitution/ BD 5 CE 5 12 4 AC 4 substitusie = and/en = BD 5 22, 5 5 ✓BD =15 BD = 15 and/en AC = 18 BC = AC − AB ✓AC =18 = 18 − 12 =6 ✓BC = 6 CD = BD − BC = 15 − 6 =9 ✓CD = 9 (5) [18] 18 ### Page 19 Downloaded from hlayiso.com 10612/24 QUESTION/VRAAG 10 10. C K 1 H D G 1 2 E A . F B 10.1 Tangents from the same point R (1) 10.2 In Δ DCE and Δ HEK is : ˆ = Eˆ 1. C alt. ∠s./verw. ∠e ,DC || EH S /R 2 ˆ 2. D = E ˆ ∠s opp equal sides/∠e teenoor gelyke sye S 1 ˆ = 180 − Cˆ 2D sum of ∠s in triangle/ binne ∠e van ∆ S ˆ =H K ˆ ∠s opp radii/ /∠e teenoor gelyke radii S/R 1 ˆ = 180 − Eˆ 2H sum of ∠s in triangle / binne ∠e van ∆ S 2 ˆ ˆ but C = E proved 2 Dˆ =Hˆ   CDE|||  EHK ∠∠∠ R (6) 19 ### Page 20 Downloaded from hlayiso.com 10612/24 ˆ = 90 tan 10.3 FGE tan ⊥radius / rkln⊥radius S R  DG = GE line from centre ⊥ to chord/ lyn vanuit R midpt loodreg op koord (3) 10.4 DC DE Δ CDE|||ΔEHK S/R = HE HK ✓substitute DC×HK = DE×HE DE=2GE DC×HK = 2GE×HE radii/radiusse but GE = HE ✓S  DC×HK = 2HE 2 (3) [13] TOTAL/TOTAAL: 150 20 ## Related documents - [MATHS P2 QP GR12 SEPT 2024 Afrikaans](https://www.hlayiso.com/library/maths-p2-qp-gr12-sept-2024-afrikaans/8572c34f-22e2-48bc-9d8a-f78c7e29781d) — Mathematics · QP · 2024 · Afr - [Gr 12 Mathematics P1 (Afrikaans) June 2024 Question Paper](https://www.hlayiso.com/library/gr-12-mathematics-p1-afrikaans-june-2024-question-paper/141815e5-df0c-4d8f-a354-4ef5b9b2a2e2) — Mathematics · QP · 2024 · Afr - [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) — Mathematics · QP · 2025 · Afr - [MATHS P2 QP GR12 JUNE 2022 Afrikaans](https://www.hlayiso.com/library/maths-p2-qp-gr12-june-2022-afrikaans/49feb536-3d0c-4e63-b857-c88def50a8c1) — Mathematics · QP · 2022 · Afr - [G12 Maths P2 Sept Free State 2024](https://www.hlayiso.com/library/g12-maths-p2-sept-free-state-2024/b81ee450-da0b-40b1-84a9-e4089f8e3d30) — Mathematics · QP · 2024 - [KZN GR 12 MATHS P2 SEPT 2024](https://www.hlayiso.com/library/kzn-gr-12-maths-p2-sept-2024/6db62b2f-3c0b-4c51-8d6b-e875f4b5d67a) — Mathematics · QP · 2024 ## 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 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 - [FS Gr 12 Sept 2024 Afr en Eng P1](https://www.hlayiso.com/library/fs-gr-12-sept-2024-afr-en-eng-p1/4d7be3dc-5f53-4da3-a833-8fa9c4d73be1) — QP · 2024 · Eng - [KZN Maths Grade 12 March 2024 QP and Memo](https://www.hlayiso.com/library/kzn-maths-grade-12-march-2024-qp-and-memo/49c516f7-1294-466c-8f55-9d573aae2c13) — QP + Memo · 2024 - [Mathematics Maths NSC Prac P1 QP May June 2024 KZN](https://www.hlayiso.com/library/mathematics-maths-nsc-prac-p1-qp-may-june-2024-kzn/f4a136af-f532-49fe-878e-dea9f5333405) — QP · 2024 ## 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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