# Mathematics Grade 11 Paper 1 Eastern Cape November 2025 Memo > Grade 11 Mathematics Paper 1 memorandum from the Eastern Cape November Exam 2025, 21 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/maths-p1-memo-gr11-nov-2025-e-a-watermark/5af2bd74-cc9d-4651-895a-310a59d8ce25 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Document type: Memorandum - Year: 2025 - Exam period: Eastern Cape November Exam - Paper: 1 - Curriculum: CAPS - Pages: 21 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 Paper 1 Eastern Cape November 2025 Memo](https://www.hlayiso.com/library/maths-p1-memo-gr11-nov-2025-e-a-watermark/5af2bd74-cc9d-4651-895a-310a59d8ce25) ## About this document Mathematics · Grade 11 · Eastern Cape November Exam · 2025. 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 NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT SENIOR SERTIFIKAAT GRADE/GRAAD 11 NOVEMBER 2025 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 21 pages. Hierdie nasienriglyn bestaan uit 21 bladsye. ### Page 2 2 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2025) 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 toegeken vir substitusie in die korrekte formule. QUESTION/VRAAG 1 1.1.1 x ( 2x + 6) = 0 x = 0 or/of 2 x + 6 = 0 ✓ x=0 ✓ x = −3 x = 0 or/of x = −3 (2) 1.1.2 4 x + 3x − 9 = 0 2 Penalise 1 mark for −b  b − 4ac 2 incorrect rounding off./  standard form / standaardvorm x= Penaliseer 1 punt vir 2a verkeerde afronding. −3  (3) 2 − 4(4)( −9) x= 2(4)  substitution / vervanging −3  153 x= 8  x = 1,17 or / of x = −1,92  x = 1,17 or/of  x = − 1, 92 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 (EC/NOVEMBER 2025) MATHEMATICS P1/WISKUNDE V1 3 1.1.3 x ( x − 4) + 2 ( 4 − x )  0 x ( x − 4) − 2 ( x − 4)  0 ✓ −2 ( x − 4 ) ( x − 4 )( x − 2 )  0 critical values / kritieke waardes ✓ critical values / kritieke waardes x = 4 or / of x = 2 2 4 x  2 or / of x  4 ✓✓ x  2 or x  4 (accuracy / akkuraatheid) (4) 1.1.4 x + 14  0 ; x  −2 ✓ x  −14 (a) x  −14 ✓ x  −2 (2) (b) x + 14 ✓ isolating surd / isoleer wortelvorm =1 x+2 ✓ square both sides / x + 14 = x + 2 kwadreer beide kante x + 14 = x 2 + 4 x + 4 ✓ factors / faktore x 2 + 3 x − 10 = 0 ( x + 5 )( x − 2 ) = 0 ✓ answer with choice / x  −5 or / of x = 2 antwoord met keuse (–5 excluded / uitgesluit) (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 4 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2025) 1.2 x − 3 y = 1......................................... (1) x 2 − 2 xy + 9 y 2 = 17..........................(2) From / Vanaf (1) : x = 3 y + 1 ..........(3) ✓ x = 3y +1 (3) into/in (2) : ( 3 y + 1) − 2 y ( 3 y + 1) + 9 y 2 = 17 2 ✓ substitution / vervanging 9 y 2 + 6 y + 1 − 6 y 2 − 2 y + 9 y 2 − 17 = 0 12 y 2 + 4 y − 16 = 0 ✓ standard form / standaardvorm 3y2 + y − 4 = 0 ✓ factors / faktore ( y − 1)( 3 y + 4 ) = 0 4 y = 1 or/of y = − ✓ y values / y-waardes 3 x = 4 or/of x = −3 ✓ x values / x-waardes (6) 1.3 10 1+ =7 1 1+ 1 1+ x 10 1+ =7 1 1 ✓ 1+ x +1 x +1 x x 10 1+ =7 1+ x x +1 ✓ x +1 x 10 1+ =7 x +1+ x x +1 10 1+ =7 2x +1 2x + 1 ✓ x +1 x +1 10 ( x + 1) 1+ =7 2x + 1 2 x + 1 + 10 x + 10 2 x + 1 + 10 x + 10 =7 ✓ 2x + 1 2x +1 12 x + 11 =7 2x + 1 12 x + 11 = 14 x + 7 2x = 4 ✓ answer / antwoord x=2 (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 (EC/NOVEMBER 2025) MATHEMATICS P1/WISKUNDE V1 5 QUESTION/VRAAG 2 2.1 125 x. 54 x − 2 3 12 x + 6 x ✓ 125 x = 33 x 5 .5 3 x 2 x −1 ✓ 54 x −2 = 52 x −1 and/en 5 .5 = 4 x+2 x 3 512 x +6 = 54 x + 2 5 .5 = 53 x − x + 2 x −1− 4 x − 2 ✓53x+2x-1-4x-2-x −3 =5 1 = ✓ answer / antwoord 125 (4) 2.2.1 3 6 x = 482 3 ✓ dividing by 6 / deel deur 6 x =82 2 ✓ raising by reciprocal x = (2 ) 3 3 verhef tot omgekeerde/resiprook x = 22 ✓ answer / antwoord x=4 (3) 2.2.2  14  12   x − 3 x 4 +2=0 2    14  1 ✓ x    let/laat : k = x4 k 2 − 3x + 2 = 0  1  1  ( k − 2 )( k − 1) = 0 OR /OF  x 4 − 2  x 4 − 1    ✓ factors / faktore k = 2 or/of k = 1 1 1 1 1 x = 2 or/of x = 1 4 4 ✓ x = 2 or/of x = 1 4 4 x = 16 or/of x = 1 ✓ answers / antwoord (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 6 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2025) 2.3 A = L B 16 x − 100 = L  ( 4 x + 10 ) ✓ substitution into area formula L= ( 4 ) − 100 x 2 vervanging in oppervlakte formule 4 x + 10 ✓ 16 x = ( 4 x ) 2 L= ( 4 − 10 )( 4 + 10 ) x x ✓ factors of difference of 2 squares 4 + 10 x faktore van verskil tussen 2 L = 4 − 10 x kwadrate  Length = 4 x − 10 ✓ answer / antwoord (4) 2.4 Smaller integer / Kleiner heelgetal = x next consecutive integer / opeenvolgende heelgetal = x + 1 ✓ difference / verskil Difference / Verskil : ( x + 1) − x 2 = 45 2 x 2 + 2 x + 1 − x 2 = 45 ✓ equating the difference to 45 stel verskil gelyk aan 45 2 x = 44 x = 22 ✓ x = 22  ( 22 + 23) = 2025 2 ✓ (sum)2 / (som)2 = 2025 (4) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 (EC/NOVEMBER 2025) MATHEMATICS P1/WISKUNDE V1 7 QUESTION/VRAAG 3 3.1.1 d = − 0,5 ✓ 4 4 ; 3,5 ✓ 3,5 (2) 3.1.2 Tn = −0,5n + 6 ✓ −0, 5n ✓ +6 (2) 3.1.3 1 − n + 6  −113,5 2 1 ✓ − n + 6  −113,5 −0,5n  119,5 2 n  239 ✓ answer / antwoord n = 240 (2) 3.2.1 d =5 Tn = 5n + c ✓ Tn = 5n + c but / maar: T38 = 192 ✓ substitution / vervanging 192 = 5 ( 38 ) + c c=2 T1 = 5 (1) + 2 = 7  First term is 7 / Eerste term is 7 ✓ answer / antwoord (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 8 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2025) QUESTION/VRAAG 4 4.1.1 Tn = n 2 − 12n + k ✓ substitution / vervanging 3132 = ( 62 ) − 12 ( 62 ) + k 2 k = 32 ✓ answer / antwoord (2) 4.1.2 Terms of the quadratic number pattern: Terme van die kwadratiese patroon: T1 = (1) − 12 (1) + 32 = 21 2 T2 = ( 2 ) − 12 ( 2 ) + 32 = 12 2 ✓ terms of quadratic number pattern / T3 = ( 3) − 12 ( 3) + 32 = 5 2 terme van kwadratiese getalpatroon Terms of the first difference: Terme van eerste verskille: −9 ; −7 ; −5 ✓ answer / antwoord (2) 4.1.3 Tn = n − 12n + 32 2 ✓ method / metode − ( −12 ) n= =6 ✓ answer / antwoord 2 (1) (2) Tn −1 = ( n − 1) − 12 ( n − 1) + 32 ✓ simplification / vereenvoudiging 2 4.1.4 = n 2 − 2n + 1 − 12n + 12 + 32 = n 2 − 14n + 45 ✓ answer / antwoord (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 (EC/NOVEMBER 2025) MATHEMATICS P1/WISKUNDE V1 9 4.1.5 Tn − Tn −1 = 149 n 2 − 12n + 32 − n 2 + 14n − 45 = 149 ✓ Tn − Tn−1 = 149 149 = 2n − 13 ✓ simplification / n = 81 vereenvoudiging consecutive terms are 80 and 81 / opeenvolgende terme is 80 en 81 ✓ answer / antwoord (3) OR / OF OR/OF Tn = 2n − 11 ✓ Tn = 2n − 11 149 = 2n − 11 160 = 2n ✓ equating Tn to 149 n = 80 consecutive terms are 80 and 81 / opeenvolgende terme is 80 en 81 ✓ answer / antwoord (3) 4.2 x 4 3x 18 4− x 3x − 4 18 − 3 x ✓ 2nd diff i.t.o x / 2de versk. i.t.v x 4x − 8 − 6 x + 22 4 x − 8 = − 6 x + 22 ✓ equating / gelykstelling 10 x = 30 x=3 ✓ answer / antwoord  T3 = 9 (3) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 10 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2025) QUESTION/VRAAG 5 5.1 p = −3 ✓ answer / antwoord (1) a a 5.2 f ( x) = −1 ✓ f ( x) = −1 x −3 x −3 Subst. / Vervang: ( 4; 2 ) a ✓ substitution of ( 4;2 ) / 2= −1 4−3 vervanging van ( 4;2 ) a=3 3 f ( x) = −1 ✓ equation of f / vergelyking van f x −3 (3) x 5.3 g ( x) = +c 3 3 −1 = +c 3 c = −2  B (0 ; − 2) x ✓ coordinates of B / koördinate van B 0= −2 3 ✓ g ( x) = 0 x 2= 3 x=6  A (6 ; 0) ✓ coordinates of A / koördinate van A (3) OR/OF OR/OF f ( 0 ) = −2 B (0 ; − 2) ✓ coordinates of B / koördinate van B 3 0= −1 x−3 3 ✓ f ( x) = 0 1= x−3 x−3= 3 x=6  A (6 ; 0) ✓ coordinates of A / koördinate van A (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 (EC/NOVEMBER 2025) MATHEMATICS P1/WISKUNDE V1 11 5.4 − ( x − 4) + 3 = 0 ✓ replacing x by − ( x − 4 ) −x + 4 − 3 = 0 vervang x met – (x – 4) −x + 1 = 0 ✓ x −1 = 0 / x = 1 x −1 = 0 V.A: x = 1 ✓ answer / antwoord Domain / Gebied : x  R ; x  1 (3) OR / OF OR/OF 3 h ( x) = −1 − ( x − 4) − 3 ✓ replacing x by − ( x − 4 ) 3 vervang x met – (x – 4) = −x + 4 − 3 3 = −1 − ( x − 1) ✓ denominator / noemer: ( x − 1) −3 = −1 x −1 Domain / Gebied : x  R ; x  1 ✓ answer / antwoord (3) 5.5 x  3 ,x  6 OR / OF ✓✓ answer / antwoord 3  x  6 or/of x  6 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 12 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2025) 5.6.1 1 mAB = 3  = 18, 43 ˆ = 18, 43 (Vert. opp 's / Regoorstaande e)  BOA 0−2 mEA = = −1 ✓ mAB and/en mEA 6−4 ˆ = 45 Ref.  / Verwys..  : EAO  EABˆ = 63, 43 ✓ Length of EA / Lengte van EA EA = ( 4 − 6) + ( 2 − 0) 2 2 ✓ 18, 43 and/en 45 =2 2 1 ˆ Area of  ABE =  EA  AB  sin EAB ✓ substitution in the area formula / 2 vervanging in die oppervlakte 1 ( ) ( ) =  2 2  2 10 sin 63, 43 2 formule = 8 units 2 / eenhede2 ✓ answer / antwoord (5) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 (EC/NOVEMBER 2025) MATHEMATICS P1/WISKUNDE V1 13 QUESTION/VRAAG 6 6.1 a = −1 6 ✓ value of a / waarde van a x= = −3 −2 ✓ x = −3 and / en y = 9 f ( −3) = − (1) − 6 ( −3) = 9 2 ✓ equation in T.P-form / f ( x ) = − ( x + 3) + 9 Vergelyking in DP-vorm 2 OR / OF (3) OR/OF f ( x ) = − x2 − 6x ✓ common factor –1 / = − ( x2 + 6 x ) gemene faktor –1 ( = − x 2 + 6 x + ( 3) − ( 3) 2 2 ) ✓ completing the square  ( 3) / 2 = − ( ( x + 3) − 9 ) voltooing van vierkant  ( 3) 2 2 = − ( x + 3) + 9 ✓ equation in T.P-form / 2 vergelyking in DP-vorm (3) 6.2 xTP = −3 ✓ x = −3 xB + x A = xTP 2 −5 + x A = −3 2 −6 = − 5 + x A x A = −1 ✓ xA = −1 f ( −1) = − ( −1) − 6 ( −1) = 5 2 ✓ f ( −1)  A ( −1;5 ) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 14 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2025) 6.3 g ( x ) = x2 + 6x ✓ g ( x) CD = g ( x ) − f ( x ) 14 = x 2 + 6 x − ( − x 2 − 6 x ) ✓ g ( x) − f ( x) 14 = 2 x 2 + 12 x 2 x 2 + 12 x − 14 = 0 x2 + 6x − 7 = 0 ✓ standard form / standaardvorm ( x + 7 )( x − 1) = 0 ✓ both x − values / x = −7 or / of x = 1 beide x-waardes  x = −7  OE = 7 units / eenhede ✓ answer / antwoord (5) 6.4.1 k  −9 ✓✓ answer / antwoord (2) 6.4.2 −x − 6x = 0 2 x2 + 6x = 0 x ( x + 6) = 0 x = 0 or / of x = −6 ✓ x − values / x-waardes  −6 k  0 ✓✓ answer / antwoord (3) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 (EC/NOVEMBER 2025) MATHEMATICS P1/WISKUNDE V1 15 QUESTION/VRAAG 7 7.1.1 y = −3 ✓ answer / antwoord (1) 7.1.2 x +1 1   −3= 0 ✓ f ( x) = 0 3 x +1 1 3=  3 −1 x +1 1 1   =  OR / OF 31 = 3− x −1 3  3 x + 1 = −1 x = −2 ✓ x = −2 0 +1 1 y =   −3 3 8 8 = −  −2, 7 ✓ y=− 3 3 (3) 7.1.3 ✓ shape / vorm ✓ x and y intercepts / x- en y-afsnitte ✓ asymptote / asimptoot Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 16 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2025) 7.1.4 x  −2 ✓✓ answer / antwoord (2) 7.2 y = a.k + rx y = a.k x − 2 ✓ y = a.k x − 2 Subst./ Vervang: ( 0 ; 0 ) 0 = a.k 0 − 2 2=a ✓ a=2 6 = 2k 2 − 2 8 = 2k 2 ✓ substitution of ( 2;6 ) / 4 = k2 vervanging van ( 2;6 ) k = 2 k = 2 ;k  0 h ( x ) = 2.2 x − 2 ✓ equation h / vergelyking h (4) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 (EC/NOVEMBER 2025) MATHEMATICS P1/WISKUNDE V1 17 QUESTION/VRAAG 8 8.1 A = P (1 − i ) n ✓ i=6 = 250 000 (1 − 13% ) 6 ✓ substitution / vervanging = R108 406,55 ✓ answer / antwoord (3) n  i  8.2 1 + 7,5% = 1 + nom   4  ✓ substitution / vervanging 43  inom  4 = 1 +  40  4  43 i 4 − 1 = nom 40 4 ✓ inom subject of equation /  43  inom = 4  4 − 1 inom onderwerp van formule  40  inom = 0, 07297..... irate = 7,30% ✓ irate (3)  24   7%  6 2 ✓ n = 24 ; 6 and / en 2  4%   7%  8.3 A =  25 000 1 +  + 10 000  1 +  − 5000 1 +    24   12   2   2   4%  ✓ 25 000 1 +  A = R 40 222,91  12  24  4%  ✓ +10000 to 25 000 1 +   12  ✓ subtracting 5000 with its interest / aftrekking van 5000 met rente ✓ answer / antwoord (5) OR / OF OR/OF ✓ n = 24 ; 4 and/en 2   4%    7%  24 4   7%  2 24 A =   25 000 1 + + 10000  1 + − 5000  1 +  4%        ✓ 25 000 1 +    12   2   2   12  A = R 40 222,91 24  4%  ✓ +10000 to 25 000 1 +   12  2  7%  ✓ P 1 +   2  ✓ answer / antwoord (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 18 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2025) 8.4 Option 1 : annual compounding * Amount after 2 years: x (1 + 0,12 ) ✓ x (1 + 0,12) − 5 000 2 2 *Brian pays 5 000, leaving : x (1 + 0,12 ) − 5 000 and/en x (1 + 0,06) − 3 000 2 2 ( *Amount after another 2 years: x (1 + 0,12 ) − 5 000 (1 + 0,12 ) 2 ) 2 ( ) ✓ x (1 + 0,12 )2 − 5 000 (1 + 0,12 )2 Option 2 : semi - annual compounding * Amount after 1 year: x (1 + 0,06 ) 2 *Brian pays 3 000, leaving : x (1 + 0,06 ) − 3 000 2 ( ) ✓ x (1 + 0, 06 )2 − 3 000 (1 + 0, 06 )4 ( *Amount after another 2 years: x (1 + 0,06 ) − 3 000 (1 + 0,06 ) 2 ) 4 Times lines for both options ( x (1 + 0,12) − 5000) (1 + 0,12) = ( x (1 + 0,06) − 3000 ) (1 + 0,06) 2 2 2 4 ✓ equating both options / gelykstel van opsies ( x (1,12) − 5000) (1,12) = ( x (1,06) − 3000 ) (1,06 ) 2 2 2 4 ✓ distribution by / uitbreiding (1,12) and / en (1,06) 2 2 x (1.12 ) − 5000 (1,12 ) = x (1,06 ) − 3000 (1,06 ) 4 2 6 4 −5000 (1,12 ) + 3000 (1,06 ) = x (1,06 ) − x (1,12 ) 2 4 6 4 −2484,56912 = −0,1550002477 x ✓ answer /antwoord (6) x = R16029, 45 [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 19 (EC/NOVEMBER 2025) MATHEMATICS P1/WISKUNDE V1 19 Opsie 1 : jaarliks saamgestel * Bedrag na 2 jaar: x (1 + 0,12 ) ✓ x (1 + 0,12) − 5 000 2 2 *Brian betaal 5 000, dit laat: x (1 + 0,12 ) − 5 000 and/en x (1 + 0,06 ) − 3 000 2 2 ( *Bedrag na 'n verdere 2 jaar: x (1 + 0,12 ) − 5 000 (1 + 0,12 ) 2 ) 2 ( ) ✓ x (1 + 0,12 )2 − 5 000 (1 + 0,12 )2 Opsie 2 : half - jaarliks saamgestel * Bedrag na 1 jaar: x (1 + 0, 06 ) 2 *Brian betaal 3 000, dit laat : x (1 + 0, 06 ) − 3 000 2 ( ) ✓ x (1 + 0, 06 )2 − 3 000 (1 + 0, 06 )4 ( *Bedrag na 'n verdere 2 jaar: x (1 + 0, 06 ) − 3 000 (1 + 0, 06 ) 2 ) 4 Tydlyne vir beide opsies: ( x (1 + 0,12) − 5000) (1 + 0,12) = ( x (1 + 0,06) − 3000 ) (1 + 0,06) 2 2 2 4 ✓ equating both options / gelykstel van opsies ( x (1,12) − 5000) (1,12) = ( x (1,06) − 3000 ) (1,06 ) 2 2 2 4 ✓ distribution by / uitbreiding (1,12) and / en (1,06) 2 2 x (1.12 ) − 5000 (1,12 ) = x (1,06 ) − 3000 (1,06 ) 4 2 6 4 −5000 (1,12 ) + 3000 (1,06 ) = x (1,06 ) − x (1,12 ) 2 4 6 4 −2484,56912 = −0,1550002477 x ✓ answer /antwoord x = R16029, 45 (6) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 20 20 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2025) QUESTION/VRAAG 9 9.1.1 NO / NEE ✓ NO / NEE P ( A ) ×P ( B ) = 0,3  0, 6 ✓ valid reason / = 0,18 geldige rede  P ( A and / en B ) = 0,18  0 (2) 9.1.2 ✓ 0,18 ✓ 0,12 and/en 0,42 ✓ 0,28 (3) 9.1.3 P ( not A and not B) = P not ( A or B) = 0, 28 ✓✓ answer / antwoord P(nie A en nie B) = [nie(A of B)] = 0,28 (2) 9.2.1 42 21 = 106 53 ✓ answer / antwoord (1) 56 9.2.2 P ( Enjoy jogging/Geniet draf ) = 106 42 P ( Male/Manlik ) = 106 56 42 P ( Enjoy jogging/Geniet draf )  P ( Male/Manlik ) =  106 106 ✓ = 0, 20932... P ( Enjoy jogging )  P ( Male ) / 26 P ( Enjoy jogging and Male/Geniet draf en Manlik ) = P (Geniet draf) x P (Manlik) 106 = 0, 24528... ✓ answer / antwoord  P ( Enjoy jogging and Male/Geniet draf en Manlik ) ✓  P ( Enjoy jogging/Geniet draf )  P ( Male/Manlik ) P ( Enjoy jogging and Male ) /  Events are not independent. / Gebeurtenisse is nie P (Geniet draf en Manlik) onafhanklik nie. ✓ conclusion / slotsom (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 21 (EC/NOVEMBER 2025) MATHEMATICS P1/WISKUNDE V1 21 9.3.1 Sunny / Sonnig -S Not Sunny / Nie Sonnig nie - S Use Sunscreen / Gebruik Sonskerm - U Not using Sunscreen / Gebruik nie sonskerm nie - U ✓ First branches i.e. S and S / S Eerste takke dws S en S ✓ Second branches i.e. U and U / Tweede takke dws U en U ✓ outcomes / uitkomste U (3) 9.3.2 0, 7 x + 0, 06 x = 0, 62 ✓ 0, 7 x + 0, 06 x 0, 76 x = 0, 62 ✓ = 0, 62 x = 0,82 ✓ answer / antwoord (3) [18] 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 - [Final Marking Guideline Grade 11 Mathematics P1 2017 Common Examination](https://www.hlayiso.com/library/final-marking-guideline-grade-11-mathematics-p1-2017-common-examination/67badd2d-178b-4a32-8ead-5f13604b50a3) — Mathematics · Memo · 2017 - [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) — Mathematics · QP · 2025 - [MATHS P1 QP GR11 NOV2024 ENG D watermark](https://www.hlayiso.com/library/maths-p1-qp-gr11-nov2024-eng-d-watermark/739068f1-b085-4977-8a45-18f9bb927644) — Mathematics · QP · 2024 · Eng - [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) — Mathematics · QP · 2023 · Afr - [MATHS P1 QP GR11 NOV 2023 English](https://www.hlayiso.com/library/maths-p1-qp-gr11-nov-2023-english/11b74f81-704a-47bc-b151-d2eea1003c3c) — Mathematics · QP · 2023 · Eng ## More from Grade 11 Mathematics - [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 - [NW 11 MATHEMATICS P2 AB ENGAFR NOV 2024](https://www.hlayiso.com/library/nw-11-mathematics-p2-ab-engafr-nov-2024/8ecc44b8-11a8-4044-8a89-cabd7fafe08c) — Addendum · 2024 ## 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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