# Mathematics Grade 12 Paper 2 Gauteng Mock 2020 Memo > Grade 12 Mathematics Paper 2 memorandum from the Gauteng Mock Exam 2020, 21 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr-12-gauteng-maths-p2-sep-2020-memo/52194179-337d-4aeb-a0a1-aaf4b4ebbca4 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2020 - Exam period: Gauteng Mock Exam - Paper: 2 - Curriculum: CAPS - Pages: 21 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 2 Gauteng Mock 2020 Memo](https://www.hlayiso.com/library/gr-12-gauteng-maths-p2-sep-2020-memo/52194179-337d-4aeb-a0a1-aaf4b4ebbca4) ## About this document Mathematics · Grade 12 · Gauteng Mock Exam · 2020. Memorandum, 21 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 2020 MARKING GUIDELINES / NASIENRIGLYNE MATHEMATICS (PAPER 2) (10612) WISKUNDE (VRAESTEL 2) (10612) 21 pages / bladsye 1 ### Page 2 Downloaded from hlayiso.com 10612/20 NOTE:  If a candidate answers a question TWICE, mark only 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 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. LET WEL:  As ʼn kandidaat ʼn vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na.  As ʼn kandidaat ʼn antwoord van 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. 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 beide die bewering EN rede korrek is.) 2 ### Page 3 Downloaded from hlayiso.com 10612/20 QUESTION / VRAAG 1 1.1 1 581 1 581 x  31 31  51  51 OR / OF OR / OF x  51 (calculator method / sakrekenaar metode) 51 (2) 1.2  skewed to the right (positevely skewed)  skeef na regs (positief skeef)  answer / antwoord (1) 1.3 Physical Sciences performed better.  answer / antwoord Q1 is 40% in Physical Sciences and 28% in Mathematics which indicates the lower 25% of the class performed much better in Physical Sciences than in Mathematics. Fisiese Wetenskappe presteer beter.  reason / rede Q1 is 40% in Fisiese Wetenskappe en 28% in Wiskunde wat aandui dat die onderste 25% van die klas heelwat beter presteer in Fisiese Wetenskappe as in Wiskunde. (2) 1.4 Accept any mark between 40 – 50. Aanvaar enige punt tussen 40 – 50 .  answer / antwoord (2) 1.5 The greatest difference is 87%  71% 16%  87%  71% 16%  the Physical Sciences mark is 71% .  answer / antwoord Die grootste verskil is 87%  71% 16% answer only full marks /  die Fisiese Wetenskappe punt is 71% . antwoord alleenlik volpunte (2) [9] QUESTION / VRAAG 2 2.1 a  12,41  a  12,41 b  0,49  b  0,49 yˆ 12,41  0,49 x  yˆ 12,41  0,49 x (3) 2.2 yˆ 12,41  0,49 x 12,41  0,49150 substitution / vervanging  85,91  86%  answer / antwoord OR/OF OR/OF yˆ  85,17  yˆ  85,17 (2) 2.3 yˆ 12,41  0,49 x The 𝑦-intercept is 12,41 which means that a learner who did not begin the exam achieved 12,41%. This is clearly impossible. conclusion / gevolgtrekking Die 𝑦-afsnit is 12,41 wat beteken dat ʼn leerling wat die eksamen nie begin het nie, alreeds12,41% behaal het. Dit is onmoontlik. (1) 3 ### Page 4 Downloaded from hlayiso.com 10612/20 2.4 10,28  10,28 (2) 2.5 63,9    p  63,9    p 63,9    103,59  63,9    103,59 127,92  p  103,59 p  24,33  p  24,33 OR / OF OR / OF  103,59  63,96  39, 63  σ = 39,63 p  63,96  39, 63  63,96  39,63  24,33  p  24,33 (3) [11] QUESTION / VRAAG 3 𝑦 E O 𝑥 3.1  12 6  E ;   2 2 6 E(6 ; 3) 3 (2) 3.2 60 mBA  7 5  mBA  3 3 y  mx  c y  y1  m  x  x1  substitution of m and y  3x  c y  6  3 x  7 (7 ; 6) / (5 ; 0) 6  3(7)  c OR / OF Vervanging van m en y  3 x  21  6 c   15 (7 ; 6) / (5 ; 0) y  3 x 15 y  3 x 15  y  3x 15 OR / OF 4 ### Page 5 Downloaded from hlayiso.com 10612/20 y  mx  c y  3x  c y  y1  m  x  x1  0  3(5)  c OR / OF y  0  3( x  5) c   15 y  3x 15 y  3 x 15 (3) 3.3 rx  3 y  5  0 3 y   rx  5 r 5 standard form / y x 3 3 standaardvorm r r 3 3 3 3 r 9  r 9 (3) 3.4 Area Δ AOP 10 1  AO   h 10 2 1 1   5   h 10  5  h 10 2 2  h4   h4 butbut/ maar y  0 / maar y   4  y4   x  5   4  0  2 2 AP = BP and / en AP 2 = BP 2  x  7    4  6 2 2  x  5   4  0    x  7    4  6  2 2 2 2 equate the two lengths / gelykstel van die twee x 2 10 x  25 16  x 2 14 x  49 100 lengtes 4 x 108  4 x 108 x  27  x  27 P (27 ; 4 ) (7) [15] 5 ### Page 6 Downloaded from hlayiso.com 10612/20 QUESTION / VRAAG 4 𝑦 A (2; 4) E B(a ; b) R 45 D α P 𝑥 O S 4.1  135 ext  of  / buite  van  tan 135   m  tan 135  m m  1  m  1 y  mx  c y  y1  m  x  x1  y   1x  c y  4   1 x  2  substitute m and  2; 4  / 4   1 2   c OR / OF y   1x  2  4 vervang m en  2; 4  c2 y x2  equation / vergelyking y x2 (4) 4.2 P  4;0   a 4 a 4 mBA .mAS  1 mBA  1  mBA = 1 4b mBA  substitute / vervang 24 4b 1 2 2  4b b2 b2 (4) 6 ### Page 7 Downloaded from hlayiso.com 10612/20 4.3  x  a   y  b  r2 2 2 substitute midpoint in  x  4   y  2  r 2 2 2 correct formula / vervang  2  4    4  2   r 2 2 2 middelpunt in korrekte formule 4  4  r2 substitute / vervang  2; 4   x  4   y  2  8 2 2   x  4   y  2  8 2 2 (3) 4.4 x  2x  y  2 y  0 2 2   x 1 2  x  2 x  1   y  2 y  1  1  1 2 2   y 1 2  x  1   y  1  2 2 2 (3) 4.5 D (1 ; 1) answer / antwoord (1) 4.6 DE = 2 DE = 2 DA  (2  1) 2  (4  1) 2 correct substitution in correct formula /  99 korrekte vervanging in  18 OR/OF 3 2 korrekte formule DA = 18 OR/OF 3 2 ˆ  90 ˆ  90  DEA DEA radius  tangent / radius  raaklyn 2 2 2 AD = DE + AE pythagoras  18    2   AE  18    2   AE 2 2 2 2 2  2 18  2  AE 2  AE = 4 AE = 4 (6) [21] 7 ### Page 8 Downloaded from hlayiso.com 10612/20 QUESTION / VRAAG 5 5.1 1  4 sin 2 15  1  4 sin 2 (45  30)  1  4 sin (45  30) 2  sin 45 cos 30  cos 45 sin 30  1  4sin 45 cos 30  cos 45 sin 30 2  2 3 2 2 3 2 1 2 1  .  .  1  4 .  .  2 2 2 2  2 2 2 2 2  6  2  1  4   4  6  4 3  2 6  4 3  2  1  4   1  4   16   16  8  4 3   1  4   16  8  4 3   1    4   3 1  3 1 OR/OF OR/OF 1  4 sin 2 15  1  4 sin 2 (60  45)  1  4 sin 2 (60  45)  1  4sin 60 cos 45  cos 60 sin 45  sin 60 cos 45  cos 60 sin 45 2 2  3 2 1 2  1  4 .  .   3 2 1 2 .  .  2 2 2 2  2 2 2 2 2  6  2  1  4   4  6  4 3  2 6  4 3  2  1  4   1  4   16   16  8  4 3   1  4   16  8  4 3   1    4   3 1  3 1 (5) 8 ### Page 9 Downloaded from hlayiso.com 10612/20 5.2 3 sin x .sin 2 72  sin 2 198 . 3 cos  x  90  tan120 .sin x 3 sin x .sin 2  90 18   sin 2 180  18  . 3 sin x  sin x  tan 180  60  .sin x  cos2 18 3 sin x .cos 2 18  sin 2 18 . 3 sin x   sin 2 18  tan 60 .sin x   tan 60 3 sin x  cos 2 18  sin 2 18   factorise / faktoriseer  3 .sin x  1  1 OR/OF OR/OF 3 sin x. sin 2 72  sin 2 18. 3 sin x  sin 2 18  tan 60. sin x  sin x 3 sin x(sin 2 72  cos 2 72)   tan 60   3. sin x  factorise / faktoriseer 3 sin x(1)  cos 2 72   3. sin x  1  1 (6) 5.3 6 sin x. cos x  3 cos x  4 sin 2 x  2 sin x  0 grouping / groepeer 3 cos x(2 sin x  1)  2 sin x(2 sin x  1)  0 (2 sin x  1)(3 cos x  2 sin x)  0  factorise / faktoriseer 1 sin x   OR/OF 3 cos x  2 sin x 2  sin x   1 3 2 tan x  2  tan x  3 2 RA  30º RA  56,31º  210  k.360 x  210º  k .360º x  56,31º  k .180º  330  k.360 , x  330º  k .360º x  236,31º  k .180º ; k  Z  56,31  k.180 , k  Z OR/OF OR/OF 9 ### Page 10 Downloaded from hlayiso.com 10612/20 6 sin x. cos x  3 cos x  4 sin 2 x  2 sin x  0 (6 sin x. cos x  3 cos x)  (4 sin 2 x  2 sin x)  0 grouping / groepeer 3 cos x(2 sin x  1)  2 sin x(2 sin x  1)  0 (2 sin x  1)(3 cos x  2 sin x)  0  factorise / faktoriseer 1 sin x   OR/OF 3 cos x  2 sin x 2  sin x   1 3 2 tan x  2  tan x  3 2   30  k.360 x  30  k .360 x  56,31  k .180 ;k Z  210  k.360 x  210  k .360  56,31  k.180 ; k  Z (7) 5.4  cos A  1 1 tan A     cos 2A  cos A  sin A  cosA  LHS/ LK  (1  tan A)   cos2A   sin A  cos A  sin A  1      cos A  cos A  sin A  2 2 cos A  cos2 A  sin 2 A  cos A  sin A  cos A      cos A  (cos A  sin A)(cos A  sin A)  cos A  sin A 1   cos A cos A  sin A   cos A  sin A  cos A+sin A  LHS / LK = RHS / RK (4) 5.5.1 cos 2θ  1 k2  1  k 2 1 k OR/OF OR/OF cos 2 2  1  sin 2 2  cos 2 2  1  sin 2 2 1 k2 cos 2  1  k 2  1 k 2 (2) 10 ### Page 11 Downloaded from hlayiso.com 10612/20 5.5.2 sin 2 tan   2sin  .cos  2sin  .cos   sin  sin   cos  cos  cos   2sin  .cos  . sin   2 cos 2   2cos2  / maar cos 2  1 k 2 But/maar But 2cos 2  1 1 k 2  2cos2  1 2cos 2   1 k 2 1  1 k 2 1 sin 2 (5)  1 k2 1 tan [29] 11 ### Page 12 Downloaded from hlayiso.com 10612/20 QUESTION / VRAAG 6 6.1 a  1  a  1 d 2 d 2 (2) 6.2  1  1 D  150 ;    150 ;   2  2 (1) 6.3.1  90 x  90   90 and / en 90 OR / OF correct inequality / x 90 ;90 korrekte ongelykheid (2) 6.3.2 135 x   45  135 and / en 45 OR / OF  correct inequality / x 135 ;  45  korrekte ongelykheid (2) If learner gives all three answers, then [7] maximum 1/2 / Indien ʼn leerling al drie die antwoorde gee, dan maksimum 1/2 x  (135;45) or/of (0;45) or/of (135;180) 12 ### Page 13 Downloaded from hlayiso.com 10612/20 QUESTION / VRAAG 7 7.1 In  KLM KM correct trig ratio / korrekte trig = tan x verhouding LM KM = tan x r KM = r tan x  KM  r tan x (simplification / vereenvoudiging) In  KMN MN = tan x KM 2r 2r = tan x OR/OF KM =  2r  tan x KM tan x KM 2r 2r r tan x  2r = tan x tan x  r tan x r tan x 2 = tan2 x 2 = tan x  2  tan x x = 54,74°  x  54, 74 (6) 7.2 LN = LM + MN  2 LM.MN cos M 2 2 2 correct substitution in cos- LN 2   5   10   2  5  . 10  cos110 2 2 rule / korrekte vervanging in LN 2 159, 20 cos-reël LN  159, 20 LN 12, 62 m  12,62m (2) [8] 13 ### Page 14 Downloaded from hlayiso.com 10612/20 QUESTION / VRAAG 8 E A 1 67º B 2 3 G C 3 21 D 8.1.1 AD̂C  67 ext.  of cyclic quad / buite  van kvh S R OR / OF OR / OF ˆ +B B ˆ = 113° s straight line / e op reguit lyn 2 3 ˆ ADC= 67° opp s of cyclic quad / oorst e van kvh S R (2) 8.1.2 Ĉ  180º- 67º co-int  BC || AD / s  113º ko-binne e BC || AD S/R (1) 8.1.3 Â  67 opp  of cyclic quad / alt  BC||AD / s s S/R alt s EC || AD oorst e van kvh / verwisselende e BC||AD / verwis e EC || AD (1) 8.1.4 B̂ 2 67 s opposite = sides / e teenoor = sye S R D̂2  180 67  67 sum of s in  / som vd e v   46 S (3) 8.1.5 BD̂G 113 tan chord theorem / raaklyn koordstelling S R OR / OF OR / OF D̂1  67 tan chord theorem / raaklyn koordstelling R BD̂G 113 S (2) 14 ### Page 15 Downloaded from hlayiso.com 10612/20 8.2 ˆ D B ˆ  46 alt s BC||AD / verwisselende e BC||AD S 3 2 AB  CD s subtend = chords / e onderspan = R koorde OR/OF OR/OF AD̂C  Â both  67 S AB  CD sides opp equal angles in trapezium R ABCD /sye teenoor gelyke hoeke in (2) trapesium ABCD [11] QUESTION / VRAAG 9 9.1 D 1 A 21 R 1 2 2 B O 1 1= M C 9.1.1(a) Ô1  2 x  centre = 2   circumference / S/R middelpunts  = 2  omtreks  (1) 15 ### Page 16 Downloaded from hlayiso.com 10612/20 9.1.1(b) Â1  CD̂B  x s in the same segment / S/R e in dies. segment M̂ 2  90 line from centre to midpoint of chord / lyn S van middelpunt van sirkel na middelpunt R van koord ˆ  90  x ABO sum of s in  / ext  of a  / som vd e v  / buite  v  S OR/OF OR/OF Ô1  2 x proved/reeds bewys S M̂1  90 line from centre to midpoint of chord / lyn R van middelpunt van sirkel na middelpunt van koord Ĉ  90  2 x sum of s in  / som vd e v  S B̂1  90  2 x s in the same segment / S/R e in dies. segment (4) AB̂O  90  x 16 ### Page 17 Downloaded from hlayiso.com 10612/20 9.1.2 AD||OB midpoint theorem / middelpunt stelling S R Ô1  AD̂C  2 x corresponding s AD || OB / S ooreenkom e AD || OB  D̂1  x S Â1  x proved / reeds bewys  D̂1  Â1 S AB is a tangent / is ʼn raaklyn converse tan chord theorem / omgekeerde raaklyn koordstelling R OR / OF OR / OF  in a semi-circle /  in halwe sirkel S R Â 2  90 corr s are equal / ooreenk e gelyk R  AD||OB ˆ O ˆ  2x corr s DA || OB / ooreenk e DA || OB S CDA 1  D̂1  x D̂  Aˆ S 1 1 AB is a tangent / converse tan chord theorem / omgekeerde raaklyn koord R is ʼn raaklyn (6) 9.1.3 DC 2  AD2  AC2 Pythagoras S but /but/maar AC  2AM and/en DC  2DO maar but/maar AC  2AM enen DC  2DO and /and/ S 2DO2  AD2  2AM2 4DO2  AD2  4AM2 but/ but maarInInΔABM /maar but/ maar ΔABM Pythagoras S  2 2  2 2 AM  AB  MB2 AM AB MB 2 substitution /  4DO 2  AD 2  4  AB 2  MB 2  vervanging AD 2  4DO 2  4AB 2  4MB 2 (4) 17 ### Page 18 Downloaded from hlayiso.com 10612/20 9.2 Q P N 2 1 = U V 1 2 3 2 1 4 1 W L M 9.2.1 M̂ 2  Q̂ alternate s NQ || MP / S verwissel e NQ ||MP M̂ 4  Q̂  M̂ 2 tan chord theorem / raaklyn koordstelling S/R Ŵ1  Q̂  M̂ 2  in the same segment / tan chord s S/R theorem / e dieselfde sirkel segment / raaklyn koord stelling OR/OF Ŵ1  M̂ 4  M̂ 2 tan chord theorem / raaklyn koordstelling (3) 9.2.2 In ΔWMV and/en ΔQMN Ŵ1  Q̂ proved / reeds bewys S M̂1  M̂ 3 equal chords subtend equal s / gelyke S/R koorde onderspan gelyke e V̂1  N̂1  N̂ 2 sum of s in  / som vd e v  S  ΔWMV||| ΔQMN  OR / OF OR / OF In ΔWMV and/en ΔQMN Ŵ1  Q̂ proved/reeds bewys S M̂1  M̂ 3 equal chords subtend equal s / gelyke S/R koorde onderspan gelyke e  ΔWMV||| ΔQMN  R (3) 18 ### Page 19 Downloaded from hlayiso.com 10612/20 9.2.3 M V WV = S MN QN ΔWMV||| ΔQMN/||| s R MV MN  WV QN MV × QN = MN × WV given / gegee but / maar QN = PW S MV × PW = MN × WV MV MN (3) = WV PW [24] QUESTION / VRAAG 10 A h k D E B C 10.1 Construction: Join DC and BE and construction / altitudes k and h konstruksie Konstruksie: Verbind DC en BE asook die hoogtelyne k en h 1  AD  k S Area Δ ADE AD  2  Area Δ DEB 1  DB  k DB 2 1  AE  h Area Δ ADE 2 AE   S Area Δ DEC 1  EC  h EC 2 but// maar Area ΔDEB  Area ΔDEC but same base, same height / gelyke S R hoogte, dieselfde basis Area Δ ADE Area Δ ADE   S Area Δ DEB Area Δ DEC AD AE   DB EC (6) 19 ### Page 20 Downloaded from hlayiso.com 10612/20 10.2 10.2.1 ML BJ 2 line || one side ΔBCM OR prop theorem S   (a) LC JC 3 MB||JL / lyn || aan een sy van ΔBCM OF R eweredigheidsstelling MB||JL (2) 10.2.1 M C BC 5 line || one side ΔBMC OR prop theorem S (b)   M L BJ 2 MB||JL / lyn || aan een sy van ΔBMC OF eweredigheidsstelling MB || JL AM  M C diagonals of a parm bisect / hoeklyne van parm halveer S/R AM 5  ML 2 AK AM 5 line || one side ΔAJL OR prop theorem   KJ M L 2 MK||JL / lyn || aan een sy van ΔAJL OF S eweredigheidsstelling MK || JL (3) 10.2.2 AB || CD opposite sides of parm / oorst sye van parm S AB || QM In  ADC  QM || CD AM = MC proved / reeds bewys  AQ  QD line passing through the midpoint of 1 side S / R || to second side / lyn sny die middelpunt van 1 sy || aan tweede sy but AD = BC opposite sides of parm / oorst sye van parm S 1 AQ  AD 2 1  2 10     2  3  2  AQ  QD  10  2 3 10 S  units 3 20 ### Page 21 Downloaded from hlayiso.com 10612/20 OR / OF OR / OF In Δ ABD BM = MD diag of a parm / hoeklyne van parm S QM || AB given / gegee  AQ QD Line passing through the midpoint of S/R 1 side || 2nd side / lyn deur middelpunt aan 1 een sy || tweede sy OR/OF  QM  AB 2 midpoint theorem / middelpunt stelling S/R AQ = QD 1 2  AQ   10  2 3  10 S  units/eenhede 3 OR / OF OR / OF Let/stel BC = 2k en/and AB = 3k 3k  10 10 k 3 2 10 BC  S 3 line || one side ΔADC OR prop AQ AM 5   theorem QM||DC / lyn || aan een sy van AD AC 10 ΔADC OF eweredigheidsstelling S/R 1 QM||DC AQ  AD 2 2 10 opposite sides of parm / oorst sye van R AD  BC  parm 3 10 S AQ  3 (4) [15] TOTAL / TOTAAL: [150] 21 ## Related documents - [Gr 12 Gauteng Maths P2 Sep 2020](https://www.hlayiso.com/library/gr-12-gauteng-maths-p2-sep-2020/346151c4-588d-4f93-a0e8-ad2bbb142863) — Mathematics · QP · 2020 - [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 - [Gr 12 Gauteng June 2017 P2 Memo](https://www.hlayiso.com/library/gr-12-gauteng-june-2017-p2-memo/4aabd1d5-7a04-4e17-bff5-d1f3e3fa545a) — Mathematics · Memo · 2017 - [2020 GRADE 12 MATH TRIAL EXAM PAPER 2 KZN](https://www.hlayiso.com/library/2020-grade-12-math-trial-exam-paper-2-kzn/bc1658b3-f847-401c-bc59-8d5e8924854e) — Mathematics · QP · 2020 - [Maths Grade 12 NSC P2 QP September 2025 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