# Mathematics Grade 12 P1 NSC November 2025 Memo (Afrikaans) > Grade 12 Mathematics Paper 1 memorandum from the NSC November Exam 2025 (Afrikaans), 21 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/grade-12-mathematics-2025-nsc-november-exam-memorandum-paper-1-afrikaans/54753b66-b7e6-4677-99c8-6092dc2c7e96 - Updated: 2026-09-28 - Subject: Mathematics - Grade: Grade 12 - Language: Afrikaans - Document type: Memorandum - Year: 2025 - Exam period: NSC November Exam - Paper: 1 - 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 P1 NSC November 2025 Memo (Afrikaans)](https://www.hlayiso.com/library/grade-12-mathematics-2025-nsc-november-exam-memorandum-paper-1-afrikaans/54753b66-b7e6-4677-99c8-6092dc2c7e96) ## About this document Mathematics · Grade 12 · NSC November Exam · 2025 · Afrikaans. 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 GRADE 12/GRAAD 12 MATHEMATICS P1/WISKUNDE V1 NOVEMBER 2025 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 21 pages. Hierdie nasienriglyne bestaan uit 21 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Mathematics P1/Wiskunde V1 2 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent Accuracy applies in all aspects of the marking guidelines. LET WEL: • Indien 'n kandidaat 'n vraag TWEE keer beantwoord, sien slegs die EERSTE poging na. • Volgehoue akkuraatheid is DEURGAANS op ALLE aspekte van die nasienriglyne van toepassing. QUESTION 1/VRAAG 1 1.1.1 ( x + 5)( x − 2 ) = 0 x = −5 or x = 2 ✓ x = −5 ✓ x=2 (2) 1.1.2 5 x + 2 = −9 x 2 5x2 + 9 x + 2 = 0 ✓ standard form −9  ( 9 ) − 4 ( 5 )( 2 ) 2 ✓ substitution into the x= 2 ( 5) correct formula −9  41 x= 10 x = −0, 26 or x = −1, 54 ✓ answer ✓ answer (4) 1.1.3 8x  2x 2 8x2 − 2x  0 ✓ standard form 2 x ( 4 x − 1)  0 1 ✓ critical values/factors CV: 0 ; 4 0 1 x  0 or x  ✓✓ answer 4 (4) 1.1.4 2.2 − 9.2 + 4 = 0 2x x (2.2x − 1)(2x − 4) = 0 ✓factors 1 2x = or 2x = 4 2 2 x = 2 −1 2 x = 22 ✓ both equations x = −1 x=2 ✓ answer ✓ answer (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Mathematics P1/Wiskunde V1 3 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF OR/OF 2.2 − 9.2 + 4 = 0 2x x Let k = 2 x 2k 2 − 9k + 4 = 0 ( 2k − 1)( k − 4 ) = 0 ✓factors 1 k = or k = 4 2 1  2x = or 2 x = 4 ✓ both equations 2  x = −1 or x = 2 ✓ answer ✓ answer (4) 1.1.5 1 1 +2 = x x 2  1   1 2  +2 = ✓ squaring both sides  x   x    1 1 +2= x x 1 1 = −2 ✓ isolation of surd x x 2 2  1  1  ✓ squaring both sides only   =  x − 2  x   after isolation of surd 1 1 4 = − +4 x x2 x 1 5 − +4=0 x2 x 4x2 − 5x + 1 = 0 ✓ standard form (4 x − 1)( x − 1) = 0 1 x= or x  1 ✓ answer with selection 4 (5) OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Mathematics P1/Wiskunde V1 4 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 1.1.5 1 1 +2 = x x 1 Let =k x k +2 =k ✓ equation k + 2 = k2 ✓ squaring both sides k −k −2=0 2 ✓ standard form ( k − 2)( k + 1) = 0  k = 2 or k = −1 1 1 =2 or  −1 ✓ substitution x x 2 x =1 4x = 1 1 x = 4 ✓ answer with selection (5) 1.2 x= y+2 …..(1) ✓ x= y+2 5xy = x 2 + 6 …..(2) ✓ 5xy = x 2 + 6 5 ( y + 2) y = ( y + 2) + 6 2 ✓ substitution 5 y 2 + 10 y = y 2 + 4 y + 4 + 6 4 y 2 + 6 y − 10 = 0 2 y2 + 3 y − 5 = 0 ✓ standard form ( 2 y + 5)( y − 1) = 0 5 y=− or y = 1 ✓y-values 2 1 x = − or x = 3 ✓ x-values 2 (6) OR/OF OR/OF y = x−2 …..(1) ✓ y = x−2 5 xy − 6 = x 2 …..(2) ✓ 5 xy − 6 = x 2 5x( x − 2) − 6 = x 2 ✓ substitution 5 x 2 − 10 x − 6 = x 2 4 x 2 − 10 x − 6 = 0 ✓ standard form 2x2 − 5x − 3 = 0 (2 x + 1)( x − 3) = 0 1 ✓x-values x=− or x = 3 2 5 ✓ y-values y=− or y = 1 2 (6) [25] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Mathematics P1/Wiskunde V1 5 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1.1 (10 + t ) + ( t − 2 ) + ( t + 4 ) t −2 t +4 ✓ equating the ratios = t + 10 t − 2 ✓ cross multiplication ( t − 2 ) = ( t + 4 )( t + 10 ) 2 ✓ expansion t 2 − 4t + 4 = t 2 + 14t + 40 −18t = 36 (3)  t = −2 2.1.2 8 ; − 4 ;... ✓ 8 and − 4 −4 1 r= =− ✓r 8 2 24  1 T25 = 8  −   2 21 1 or T25 = ( 2 ) or T25 = ( 8 ) −21 −7 T25 =   or ✓ answer in exponential 2 form T25 = (128 ) or T25 = ( 2 097 152 ) −3 −1 (3) a 2.1.3 S = 1− r 8 S =  1 ✓substitution 1−  −   2 16 S = = 5,33 ✓ answer 3 (2) 2.2.1 T14 − T6 = 48 = 32 ✓✓answer (2) OR/OF OR/OF T14 − T6 = ( 4 (14 ) − 1) − ( 4 ( 6 ) − 1) ✓ subs = 55 − 23 = 32 ✓ answer (2) OR/OF OR/OF T14 − T6 = ( 4 ( k + 13) − 1) − ( 4 ( k + 5 ) − 1) ✓ subs = 4k + 51 − ( 4k + 19 ) = 32 ✓ answer (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Mathematics P1/Wiskunde V1 6 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 2.2.2 n = 118 − k ✓ number of terms T117 = 467 ✓ last term 118 − k S118− k =  4k − 1 + 467 ✓ substitution 2 26 675 = (118 − k )  2k + 233 26 675 = 236k + 27494 − 2k 2 − 233k ✓standard form 2 k 2 − 3k − 819 = 0 ( k − 21)(2 k + 39) = 0 39  k = 21 or k  − ✓answer with selection 2 (5) OR/OF OR/OF n = 118 − k ✓ number of terms 118 − k S118− k =  2(4k − 1) + (118 − k − 1)(4) ✓ substitution 2 118 − k 26 675 = 8k − 2 + 468 − 4k  2 ✓simplification 53 350 = (118 − k )  4k + 466 53 350 = 472k + 54 988 − 4k 2 − 466k 4k 2 − 6k − 1638 = 0 ✓ standard form 2 k 2 − 3k − 819 = 0 ( k − 21)(2 k + 39) = 0 39 ✓answer with selection  k = 21 or k  − 2 (5) OR/OF T1 = 3 T2 = 7 S117 − S k −1 = 26 675 n S117 =  2a + (n − 1)d  2 117 =  2(3) + (116)4  2 = 27 495 ✓ 27 495  S k −1 = 27 495 − 26 675 = 820 ✓ difference n 820 =  2(3) + 4 n − 4  ✓substitution 2 0 = 2n 2 + n − 820 ✓ standard form (2n + 41)( n − 20) = 0  n = 20  k − 1 = 20 k = 21 ✓answer with selection (5) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Mathematics P1/Wiskunde V1 7 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 3.1 36 ; 71 ; 104 ; 135 ; 164 35 33 31 29 ✓first differences –2 T5 = 164 ✓ answer (2) 3.2 2 a = −2 ✓ 2 a = −2 a = −1 3 ( −1) + b = 35 ✓ 3 ( −1) + b = 35 b = 38 −1 + 38 + c = 36 ✓ −1 + 38 + c = 36 c = −1 (3) Tn = −n2 + 38n − 1 3.3 −38 ✓ method n= = 19 2( −1) ✓ n T19 = −(19)2 + 38(19) − 1 = 360 ✓ answer (3) OR/OF OR/OF T / n = −2n + 38 = 0 ✓ method  n = 19 ✓ n T19 = −(19)2 + 38(19) − 1 = 360 ✓ answer (3) 3.4 n+3 ✓ method = 19 2 n + 3 = 38 n = 35 ✓ answer (2) OR/OF − n 2 + 38n − 1 = 104 ✓ method n 2 − 38n + 105 = 0 ( n − 35)( n − 3) = 0 ✓ answer n = 35 (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Mathematics P1/Wiskunde V1 8 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 y f x O A (3 ; t) 4.1 t = log 1 3 3 t = −1 ✓ answer (1) 4.2 A (1; 0 ) ✓ answer (1) 4.3 f ( x ) = log 1 x 3 y = log 1 x 3 x = log 1 y ✓ swopping 3 x 1 y =   = 3− x ✓ answer 3 (2) 4.4 y=0 ✓ answer (1) 4.5 y (−1 ; 3) f −1 (0 ; 1) ✓ decreasing exponential shape with asymptote O x y=0 ✓ y-intercept (0 ; 1) ✓ any point 4.6 (4 ; 3) is a point on h (3) 0 y3 or y  (0 ; 3) ✓✓ answer (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Mathematics P1/Wiskunde V1 9 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 y g P g f D(5 ; 6) x M O T 5.1 x  R, x  3 ✓ answer (1) 5.2 y8 ✓ answer (1) 5.3.1 3 x5 OR/OF x  ( 3;5 ✓✓ answer (2) 5.3.2 x  1 or x  5 OR/OF x  ( − ;1) or x  ( 5;  ) ✓ x 1 ✓ x5 (2) 5.4 y = a( x − 3) + 8 2 ✓ p and q values 6 = a(5 − 3)2 + 8 ✓ substitution (5 ; 6) −2 = 4a 1 a = − 2 1 y = − ( x − 3) 2 + 8 2 1 y = − ( x 2 − 6 x + 9) + 8 2 ✓ simplification 1 7 y = − x 2 + 3x + 2 2 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Mathematics P1/Wiskunde V1 10 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 5.5 1 7 y = − x 2 + 3x + 2 2 x2 − 6x − 7 = 0 ✓ solve for x ( x − 7)( x + 1) = 0 x=7 or x = −1 ✓ x-values M(– 1 ; 0) ✓ coordinates of M −4 0= +8 ✓ solve for x x−3 −8 x + 24 = −4 7 x= 2 ✓ x-value 7  T ; 0 2  7 9 MT = + 1 = = 4,5 ✓ MT 2 2 (6) 5.6 1 7 f ( x) = − x 2 + 3x + 2 2 f ( x) = − x + 3 / ✓ f / ( x) = − x + 3 m = f / (5) = −5 + 3 ✓ m = f / (5) = −2 6 = −2(5) + c c = 16  y = −2 x + 16 ✓ answer (3) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Mathematics P1/Wiskunde V1 11 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 y f g O x f 6.1 (–p ; 0) ✓ (–p ; 0) (1) 6.2 g ( x) = x + c q = 1+ c ✓ q = 1+ c a a ✓ f ( x) = +1+ c f ( x) = +1+ c x+c x+c Point (0 ; c) a c= +1+ c 0+c a −1 = c a = −c Point (3 ; 3 + c) a a 3+ c = +1+ c ✓ 3+ c = +1+ c 3+ c 3+ c −c 3+ c = +1+ c 3+ c 9 + 6c + c 2 = − c + 3 + 4c + c 2 3c = −6 c = −2 a=2 ✓a=2 q = −1 ✓q=–1 2 f ( x) = −1 x−2 (5) OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Mathematics P1/Wiskunde V1 12 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 0= x+c  x = −c ✓ q = 1+ c q = 1+ c a ✓ f ( x) = +1+ c a x+c f ( x) = +1+ c x+c ✓ 3+ c = a +1+ c a 3+ c 3+ c = +1+ c 3+ c a 2= 3+ c a = 6 + 2c 6 + 2c y= +1+ c x+c 6 + 2c c= +1+ c 0+c 6 + 2c −1 = c − c = 6 + 2c  c = −2 ✓a=2 a=2 ✓q=–1 q = −1 (5) 2  f ( x) = −1 x−2 6.3 Translation of 1 unit down ✓✓ answer (2) OR/OF OR/OF Translation of 1 unit to the right ✓✓ answer (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Mathematics P1/Wiskunde V1 13 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 7.1 A = P (1 + i ) n ✓ substitution into A = 40 000 (1 + 7,8% ) 5 correct formula A = R58 230, 94 ✓ answer (2) 7.2 x (1 + i ) − 1 n F=   i ✓ i  0, 058   24 ✓ substitution into 2 300  1 +  − 1 correct formula  4    0, 058  F=  1 +  0, 058  4  ✓ future value × (1 + i ) 1 4 F = R 66 411, 60 ✓ answer (4) OR/OF OR/OF  0, 058  25  ✓ i 2 300  1 +  − 1 ✓ substitution into  4   F= − 2300 correct formula 0, 058 4 ✓ – 2300 F = R66 411,60 ✓ answer (4) 7.3.1 A = P(1 + i) n 3  0, 068  A = 900 000  1 +   12  = R915 386,86 ✓answer x 1 − (1 + i )  −n P=   i   0, 068  − n  10 000 1 −  1 +     12   ✓ substitution into 915 386,86 = 0, 068 correct formula 12 −n  0, 068  1 +  = 0, 4812... ✓ correct use of logs  12  −n = log1,005... 0, 4812... n = 129, 419... months ✓ answer of n 132, 419 months since loan was granted 133 months since loan was granted ✓ final answer (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Mathematics P1/Wiskunde V1 14 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne   0, 068  −0,419...  10 000 1 −  1 +   ✓ substitution 7.3.2   12   ✓n P= 0, 068 12 = R4 173,55… ✓ balance 1  0, 068  Final payment = 4173, 55... 1 +  = R 4 197, 21 ✓ final payment  12  (4) OR/OF OR/OF 129  0, 068  A = 915 386,86  1 +   12  = R1 897 482,712 ✓answer  0, 068 129  10 000  1 +  − 1  12   F= 0, 068 12 = R1 893 309,16 ✓ answer Balance after 129 months = R4 173,552 ✓balance 1  0, 068  Final payment = 4 173, 552  1 +  = R 4 197, 21 ✓final payment  12  (4) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Mathematics P1/Wiskunde V1 15 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 8.1 f ( x ) = −2 x + 3 f (x + h ) − f (x ) f / ( x) = lim ✓ f ( x + h) h →0 h −2( x + h) + 3 − (−2 x + 3) ✓substitution = lim h →0 h −2 h = lim ✓simplification h→0 h = lim ( −2 ) h→ 0 ✓answer = −2 (4) OR/OF OR/OF f ( x ) = −2 x + 3 f ( x + h ) = −2( x + h ) + 3 ✓ f ( x + h) f ( x + h ) − f ( x ) = − 2 x − 2 h + 3 + 2 x − 3 = −2 h ✓simplification f (x + h ) − f (x ) f / ( x) = lim h →0 h −2 h = lim ✓substitution h→0 h = lim ( −2 ) h→ 0 = −2 ✓answer (4) 8.2.1 g ( x ) = −3x + 2 x 4 ✓ −12x 3 g ( x ) = −12 x + 2 / 3 ✓2 (2) 8.2.2 2x4 + 1 y= ✓ 2x2 x2 y = 2 x2 + x−2 ✓ x −2 dy ✓ derivative first = 4 x − 2 x −3 term dx ✓ derivative second term (4) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Mathematics P1/Wiskunde V1 16 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 y D x A O B C f E 9.1 f ( x) = x − 8x + 5x + 14 3 2 f / ( x) = 3x 2 − 16 x + 5 = 0 ✓ f / ( x) (3 x − 1)( x − 5) = 0 ✓ f / ( x) = 0 1 x = or x = 5 3 E(5 ; – 36) ✓ x-value ✓ y-value (4) 9.2 f ( x) = 6 x − 16  0 // ✓ f ( x) // x 8 ✓ f / / ( x)  0 3 ✓ answer (3) OR/OF OR/OF 1 +5 3 8 ✓ midpoint of TPs x= = 2 3 ✓ f / / ( x)  0 8 ✓ answer x 3 (3) x-intercepts of f: (– 1 ; 0) and (7 ; 0) ✓✓ x-intercepts 9.3 8 ✓ first interval −1  x  2 or x7 ✓ second interval 3 (4) 9.4 3 x − 16 x + 5 = −11 2 ✓ equating 3 x 2 − 16 x + 16 = 0 ( x − 4)(3 x − 4) = 0 4 x = 4 or x = ✓ x-values 3 At x = 4 : (4)3 − 8(4)2 + 5(4) + 14 = −11(4) + t t = 14 ✓ answer 3 2 4 4 4 4 4 At x = :   − 8   + 5   + 14 = −11   + t 3 3 3 3 3 634 13 ✓ answer t= = 23 = 23, 48 27 27 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Mathematics P1/Wiskunde V1 17 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 634 ✓✓ answer 14  t  (6) 27 OR/OF OR/OF x 3 − 8 x 2 + 5 x + 14 = −11x + t ✓ equating x 3 − 8 x 2 + 5 x + 14 + 11x = t x 3 − 8 x 2 + 16 x + 14 = t 3 x 2 − 16 x + 16 = 0 ( x − 4)(3 x − 4) = 0 4 ✓ x-values x = 4 or x = 3 At x = 4 : (4)3 − 8(4)2 + 5(4) + 14 = −11(4) + t ✓ answer t = 14 3 2 4 4 4 4 4 At x = :   − 8   + 5   + 14 = −11   + t 3 3 3 3 3 634 13 t= = 23 = 23, 48 ✓ answer 27 27 634 ✓✓ answer 14  t  27 (6) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 Mathematics P1/Wiskunde V1 18 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 h h x 10.1 2 x + 2h = 50 ✓ 2 x + 2h = 50 h = 25 − x 2 r = x x x r= ✓ r= 2 2 V =  r 2h 2  x  V =   ( 25 − x )  2   x2 V= ( 25 − x ) ✓ substitution 4 2 25 x 2 x 3 V= − (3) 4 4 10.2 50 x 3 x 2 V / ( x) = − 4 4 50 x 3 x 2 V / ( x) = − =0 50 x 3 x 2 4 4 ✓ − 4 4 50 x − 3 x 2 = 0 0 = x ( 50 − 3x ) ✓ V / ( x) = 0 50 x  0 or x = 3 50 x = = 16, 67 3 ✓ answer with selection (3) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 19 Mathematics P1/Wiskunde V1 19 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 11 ENERGY JUICE TOTAL DRINKS Females a 72 120 Males 36 54 90 Total 84 126 210 11.1.1 P ( Male and Prefer juice ) = P ( Male )  P ( Prefer juice ) ✓ P ( M and J ) = P ( M )  P ( J ) 36 90 e 36 =  ✓ 210 210 210 210 e = 84 90 ✓ P (Male) = 210 (3) 11.1.2 b P ( Female and like energy drink ) = 210 d = 210 − 84 = 126 ✓ d- value b = 126 − 54 = 72 ✓ b-value 72 12 P ( Female and like energy drink ) = = = 0,34 ✓ answer 210 35 (3) OR/OF OR/OF c = 120 ✓ c- value b = 72 ✓ b-value 72 12 P ( Female and like energy drink ) = = = 0,34 ✓ answer 210 35 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 20 Mathematics P1/Wiskunde V1 20 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 11.2 3 1 7 3 ( 3x ) + x = ✓ ( 3x ) 4 4 12 4 27 x + 3 x = 7 1 30 x = 7 ✓ x 4 7 7 x= ✓ = 30 12  28 cups ✓answer (4) OR/OF C OR/OF R 0,75 W C 0,25 S W 75  3 x  75  3 x  25  x  7 ✓    100  120  +  = 100  120  100  120  12 25  x  3 1 7 ✓   x+ x= 100  120  160 480 12 7 10 x = 280 ✓ = 12 x = 28 cups of coffee ✓answer (4) OR/OF OR/OF C R 0,75 W C 0,25 S W 75  3(120 − x )  ✓   75  3(120 − x)  25  120 − x  7 100  120   +  = 25  120 − x  100  120  100  120  12 ✓   9 1 100  120  270 − x + 30 − x = 70 7 4 4 ✓ = 5 12 − x = −230 2  x = 92 (bottles of water) ✓answer  there are 28 cups of coffee (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 21 Mathematics P1/Wiskunde V1 21 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 11.3.1 7.6! = 5 040 ✓✓ answer (2) OR/OF OR/OF 7! = 5 040 ✓✓ answer (2) 11.3.2 Possible outcomes: A×6×5×B×4×3×2×1 A×6×5×4×B×3×2×1 A × 6 × 5× 4 × 3 × B × 2 × 1 A×6×5×4×3×2×B×1 A×6×5×4×3 ×2×1×B 5  6! + 4  6! + 3  6! + 2  6! + 1 6! ✓✓ 6! (15) = 6!(5 + 4 + 3 + 2 + 1) = 6!(15) 6!(15) ✓ 8! in denominator P(two or more runners finishing after Andrew) = 6!(15) 15 8! ✓ or 15 8! 56 = (4) 56 = 0, 27 OR/OF OR/OF 8!− ( 7!.2 + 2.6.6!) ✓ 8! in numerator and denominator 8!.2 ✓✓ ( 7!.2 + 2.6.6! ) 15 = ✓ 15 56 56 (4) [16] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou ## Related documents - [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) — Mathematics · Addendum · 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) — Mathematics · QP · 2025 · Afr - [Gr12 Mathematics P1 (Afrikaans) SEP 2022 Possible Answers](https://www.hlayiso.com/library/gr12-mathematics-p1-afrikaans-sep-2022-possible-answers/38fb7466-deff-4faa-9c6c-825fcf932aae) — Mathematics · Memo · 2022 · 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 - [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 - [Mathematics P1 Nov 2019 Memo Afr Eng](https://www.hlayiso.com/library/mathematics-p1-nov-2019-memo-afr-eng/eef87169-da21-443f-af20-6207010a2376) — Mathematics · Memo · 2019 · Eng ## More from Grade 12 Mathematics - [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 P2 QP GR12 SEPT 2024 Afrikaans](https://www.hlayiso.com/library/maths-p2-qp-gr12-sept-2024-afrikaans/8572c34f-22e2-48bc-9d8a-f78c7e29781d) — QP · 2024 · 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 - [G12 Maths P2 Sept Free State 2024](https://www.hlayiso.com/library/g12-maths-p2-sept-free-state-2024/b81ee450-da0b-40b1-84a9-e4089f8e3d30) — QP · 2024 - [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) — QP · 2024 · Afr - [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 ## 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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