# Mathematics Grade 12 Paper 1 NSC Supplementary 2015 Memo > Grade 12 Mathematics Paper 1 memorandum from the NSC Supplementary Exam 2015, 18 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/mathematics-p1-feb-march-2015-memo-afr-eng/150761f0-19b9-4722-8d45-5ab1ee701ffd - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2015 - Exam period: NSC Supplementary Exam - Paper: 1 - Curriculum: CAPS - Pages: 18 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 NSC Supplementary 2015 Memo](https://www.hlayiso.com/library/mathematics-p1-feb-march-2015-memo-afr-eng/150761f0-19b9-4722-8d45-5ab1ee701ffd) ## About this document Mathematics · Grade 12 · NSC Supplementary Exam · 2015. Memorandum, 18 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 GRADE/GRAAD 12 MATHEMATICS P1/WISKUNDE V1 FEBRUARY/MARCH/FEBRUARIE/MAART 2015 MEMORANDUM MARKS: 150 PUNTE: 150 This memorandum consists of 18 pages. Hierdie memorandum bestaan uit 18 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 2 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent accuracy applies in all aspects of the marking memorandum. LET WEL: • Indien ’n kandidaat ’n vraag TWEE keer beantwoord, merk slegs die EERSTE poging. • Volgehoue akkuraatheid is DEURGAANS op ALLE aspekte van die memorandum van toepassing. QUESTION/VRAAG 1 1.1.1 ( x + 4)( x − 5) = 0  factors/faktore ∴ x = − 4 or x = 5 answers/antwoorde (2) 1.1.2 2 x − 11x + 7 = 0 2 − b ± b 2 − 4ac x=  substitution into correct 2a formula/substitusie in − (− 11) ± (− 11)2 − 4(2)(7 ) korrekte formule = 2(2 ) 4,77 = 4,77 or 0,73 0,73 (3) OR/OF 2 x 2 − 11x + 7 = 0 11 7 x2 − x + = 0 2 2 11  1 11  2 7  1 11  2  correct completion of x2 − x+ .  + − .  =0 the square/korrekte 2 2 2  2 2 2  voltooiing van die 2  11  7 121 vierkant x −  + − =0  4 2 16 121 − 56 2  11  x −  =  4 16 11 65 x− =± 4 16 11 65 11 65 ∴x = + or x= − 4 4 4 4 x = 4,77 x = 0,73 4,77 0,73 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 3 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum 1.1.3 5 x 2 − 21x + 4 > 0  standard form/ standaardvorm (5 x − 1)( x − 4) > 0  factors/faktore 1 4 x < or/of x > 4 1 5 5 1 x< 5 or x>4 1 +  of + – 4 5 (5) 1.1.4 2 − 6.2 = 16 2x x 2 2 x − 6.2 x − 16 = 0 (2 − 8)(2 + 2) = 0 x x  factors/faktore 2 x = 23 or/of 2 x = −2 x=3 or/of No Solution or 2 x ≠ −2 no solution to/ geen oplossing 2 x = −2  2 x = 23  answer/antw. (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 4 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum 1.2 y = 2x − 1 y the subject/ x 2 − x(2 x − 1) + (2 x − 1) = 7 2 die onderwerp x 2 − 2x 2 + x + 4x 2 − 4x + 1 = 7 substitution/substitusie 3x − 3x − 6 = 0 2 simplification/vereenv. x2 − x − 2 = 0 (x − 2)(x + 1) = 0  factors/faktore x = 2 or/of x = −1 x-values/waardes y = 3 or/of y = −3 y-values/waardes (6) OR/OF y 1 x= +  x the subject/ 2 2 die onderwerp 2  y 1  y 1 substitution/substitusie  +  −  + y + y = 7 2  2 2  2 2 y2 y 1 y2 y simplification/vereenv. + + − − + y2 = 7 4 2 4 2 2 × 4 : y 2 + 2 y + 1 − 2 y 2 − 2 y + 4 y 2 − 28 = 0 3 y 2 − 27 = 0 y2 − 9 = 0 ( y − 3)( y + 3) = 0  factors/faktore ∴y =3 or y = −3 y-values/waardes 3 1 −3 1 ∴x = + x= + 2 2 2 2 x=2 x = −1 x-values/waardes (6) 1.3.1 k = – 2 or/of k = 2 answer/antw. (2) 1.3.2 k=–3 –3 (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 5 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum 1.4 7 2014 − 7 2012 12 = ( 7 2012 7 2 − 1 )  ( ) 7 2012 7 2 − 1 12 12 7 2012 .48 = 12  7 2012.4 = 7 2012 .4 = 2.7 1006  2.71006  a = 2; b = 1006 OR/OF a = 2 b = 1006 (4) [27] QUESTION/VRAAG 2 2.1 S n = a + (a + d ) + (a + 2d ) + ... + a + (n − 1)d  first series/eerste reeks S n = a + (n − 1)d + a + (n − 2 )d + a + (n − 3)d + ... + a series reversed/reeks omgekeer 2 S n = n(2a + (n − 1)d )  sum/som Sn = n [2a + (n − 1)d ]  division/deling 2 (4) 2.2 50 ∑ (100 − 3k ) = 97 + 94 + 91 + ... k =1 a = 97 T1 = a = 97 d = – 3 d = −3 n = 50 − 1 + 1 = 50 n = 50 n S n = [2a + (n − 1)d ] 2 50 = [2(97) + 49(−3)] 2 = 1175  answer/antwoord (4) OR/OF a = 97 T1 = a = 97  l = −50 l = 100 − 3(50 ) = −50 n = 50 − 1 + 1 = 50 n = 50 n S n = [a + l ] 2 50 = [97 − 50] 2 = 1175  answer/antwoord (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 6 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum 2.3.1 (a) T5 − T4 = 25  answer/antwoord (1) 2.3.1 (b) T70 − T69 = 7 + (69 − 1)(6 ) n = 69 = 415  7 + (69 − 1)(6 )  answer/antw. (3) 2.3.2 T89 − T69 = (T70 − T69 ) + (T71 − T70 ) + ... + (T89 − T88 )  expansion/uitbreiding = 415 + 421 + ...to 20 terms n = 20  method/metode = [2(415) + 19(6)] 20 a = 415 2 = 9440 T69 = T89 − (sum of the differences from/som van die verskille van T69 to T89 ) T69 = 23594 − 9440  answer/antwoord = 14154 (5) OR/OF 7 13 19 25 6 6 6 ∴ 2a = 6 a=3 3a + b = 7 b = −2  a and/en b T89 = 3(89) 2 − 2(89) + c = 23594  T89 (subst n = 89) ∴c = 9  Tn ∴Tn = 3n 2 − 2n + 9 ∴T69 = 3(69) 2 − 2(69) + 9 substitution/substitusie  answer/antwoord ∴T69 = 14154 (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 7 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum OR/OF 7 13 19 25 6 6 6 ∴ 2a = 6 a=3 7 − 6 =1 T1 − T0 = 1  a and/en b a + b + c − c =1 3+ b =1  T89 (subst n = 89) b = −2 T89 = 3(89) 2 − 2(89) + c = 23594  Tn ∴c = 9 substitution/substitusie ∴Tn = 3n 2 − 2n + 9  answer/antwoord ∴T69 = 3(69) 2 − 2(69) + 9 (5) [17] ∴T69 = 14154 OR/OF Tn+1 − Tn = 7 + 6(n − 1) 88 ∴ T89 − T1 = ∑ (Tn+1 − Tn ) formula/formule n =1 = n [2a + (n − 1)d ] 2 = [14 + 87 × 6] 88 2 value of/waarde van S 88 = 23584 ∴ T1 = 23594 − 23584 = 10 first term value/ 68 eerste term waarde ∴ T69 − 10 = ∑ (Tn+1 − Tn ) n =1 substitution/substitusie = 34(15 + 67 × 6) = 14144 answer/antwoord ∴ T69 = 14154 (5) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 8 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION 3 3.1 40,5 r = 0,9 r= = 0,9 45  substitution into T12 = 45(0,9 ) 12 −1 correct formula/substitusie in = 14,12147682... korrekte formule = 14,12  answer/antwoord (3) 3.2 r = 0,9 − 1 < 0,9 < 1 answer/antwoord (1) 3.3 45 S∞ =  substitution/substitusie 1 − 0,9 S ∞ = 450  450 (2) 3.4 S∞ − Sn < 1  450 − ( 45 1 − (0,9 ) n ) S ∞ − S n = 450 − ( 45 1 − (0,9) n ) 1 − 0,9 1 − 0,9 ( S ∞ − S n = 450 − 450 1 − (0,9) n ) 450(0,9) < 1 n  (0,9 ) = 1 (0,9)n < 1 n 450 450 log(0,9 ) < log n 1 450 n. log(0,9 ) < log 1 450  introducing/gebruik logs 1 log n> 450 log(0,9 ) making n the subject/maak n die onderwerp n > 57,98... Smallest value/Kleinste waarde: n = 58 n = 58 (5) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 9 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION/VRAAG 4 4.1 x = −2  x = −2 y = −1  y = −1 (2) 4.2.1 6 g (0) = −1 0+2 =2 y-intercept/afsnit (0 ; 2)  answer/antwoord (1) 4.2.2 6  equating to/stel gelyk aan 0 0= −1 x+2 6 1= x+2 x+2=6 x=4  answer/antwoord x-intercept/afsnit (4 ; 0) (2) 4.3  asymptotes/asimptote  intercepts/afsnitte  shape/vorm (3) 4.4 y + 1 = −( x + 2 ) m = – 1 y = −x − 3 substitution of (–2 ; –1) answer OR/OF (3) Using general formula/Gebruik algemene formule: y = −( x + p ) + q formula/formule substitution of p and q y = −( x + 2) − 1 values/substitusie van p- en y = −x − 3 q-waardes answer/antwoord (3) 4.5 x > −2  answer (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 10 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION/VRAAG 5 5.1 9 = a2  9 = a2 a=3 a =3 (2) OR/OF f −1 ( x) = log a x 2 = log a 9  9 = a2 a 2 = 9 = 32 a =3 (2) ∴a = 3 5.2 g ( x) = 3 − x  answer/antwoord (1) OR/OF x 1 g ( x) =    answer/antwoord 3 (1) 5.3 x≥9  answer/antwoord (2) OR/OF f −1 ( x) = log 3 x log 3 x = 2 x = 32 = 9 ∴x ≥ 9  answer/antwoord (2) OR/OF log 3 x ≥ 2 x ≥ 32 ∴x ≥ 9  answer/antwoord (2) 5.4 Yes/Ja. For every y-value there is only one x such that/Vir  Yes/Ja elke y-waarde is daar slegs een x sodanig dat y = f ( x ) .  Reason/Rede (2) OR/OF  Yes/Ja Yes/Ja. f is a one-to-one relation/is 'n een-tot-een-relasie.  Reason/Rede (2) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 11 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION/VRAAG 6 6.1 −3≤ x ≤ 2  critical values/ kritiese waardes  notation/notasie (2) 6.2 f : y = a( x − x1 )( x − x 2 ) y = a( x + 3)( x − 2)  y = a ( x + 3)( x − 2) − 8 = a(1 + 3)(1 − 2)  substitute/vervang (1 ; – 8) − 8 = −4a 2=a a = 2 y = 2( x + 3)( x − 2 ) y = 2 x 2 + 2 x − 12 b = 2 and/en c = – 12 b = 2 and/en OR/OF  c = – 12 (5) 2  1 y = a x +  + q  2 2  1 25 0 = a 2 +  + q → 0= a+q ....(1) equation/vergelyking 1  2 4 2  1 9 equation/vergelyking 2 − 8 = a 1 +  + q → −8= a+q .....(2)  2 4 (1) − (2) : 8 = 4a a = 2 a=2 25 25 q=0− (2) = − = −12,5 4 2 2  1 1  substitution/substitusie y = 2 x +  − 12  2 2  1 1 y = 2 x 2 + x +  − 12  4 2 1 1 y = 2 x 2 + 2 x + − 12 2 2 y = 2 x + 2 x − 12 2 b = 2 and/en ∴ b = 2 and c = −12  c = – 12 (5) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 12 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum f ′( x) = 2ax + b  1  1 f ′ −  = 2a −  + b = 0  2  2 ∴a = b (− 3;0) : 9a − 3b + c = 0 ∴ 6a + c = 0(1) equation/vergelyking 1 (1;− 8) : a + b + c = −8 ∴ 2a + c = −8(2) equation/vergelyking 2 (1) − (2) : 4a = 8 ∴a = 2 a = 2 ⇒b=2 ∴ c = −12 b=2 c = – 12 (5) 6.3 b x=− 2a 1 x=− 2 =− 1 x=− 2 2(2) 2 1 substitution/substitusie y = − 1 − 12 2 1 y = −12 2  1 1 y-value/waarde TP − ; − 12   2 2 OR/OF (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 13 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum y = 2[ x 2 + x − 6]  1  2 1   2 y = 2  x 2 + x +  .1 − 6 −  .1   2   2    method/metode  1 2  = 2  x +  − 6,25  2  2  1 = 2 x +  − 12,5  2 x-value/waarde  1  TP − ; − 12,5  y-value/waarde  2  (3) OR/OF −3+ 2 1 x= =− 2 2  1  1  method/metode y = 2 −  + 2 −  − 12 x-value/waarde  2  2 1 y = −12 2  1  TP − ; − 12,5  y-value/waarde  2  (3) OR/OF f ( x) = y = 2 x 2 + 2 x − 12 f / ( x) = 4 x + 2 4x + 2 = 0  method/metode 4 x = −2 1 x=− 2 2 x-value/waarde  1  1 25 ∴ y = 2 −  + 2 −  − 12 = −  2  2 2  1 25  y-value/waarde (3) TP − ; −   2 2  6.4 13 x=  answer/i 2 (2) 6.5 f ( x) = 4 x + 2 /  y = 4x + 2 / m = f ′(1) = 4(1) + 2  subst. x = 1  answer/antwoord m=6 (3) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 14 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION/VRAAG 7 7.1.1 R 400 × (44 × 12)  R 400 × (44 × 12) = R 211200 R211200 (2) 7.1.2 F= [ x (1 + i ) − 1 n ]  x = 400  n = 528 i 0,08  0,08  528  i = 400 1 +  − 1 12  12    substitution into correct = formula/substitusie in 0,08 korrekte formule 12 = R1 943 524 ,42 answer/antwoord (5) 7.1.3 −n x[1 − (1 + i ) ]  P = 2000000 P= i 0,1  n = 300 and/en i =   0,1   −300 12 x 1 − 1 +    substituting into correct   12   2000000 = formula/substitusie in 0,1 korrekte formule 12 x = R18 174 , 01  answer/antwoord (4) OR/OF   0,1  300  x   1 +  − 1 300   12   P = 2000000  0,1    20000001 +  =  12  0,1 0,1  n = 300 and/en i = 12 12 x = R18174 , 01 equating/stel gelyk  answer/antwoord (4) 7.2 Let P X and P Y be the populations of the two towns at the beginning of 2010./Laat P X en P Y die bevolkings wees van die twee dorpe aan die begin van 2010.  equating/stel gelyk AX = AY  A X = PX (1 − 0,08) 3 PX (1 − 0,08) 3 = PY (1 + 0,12) 3  AY = PY (1 + 0,12) 3 PX (1 + 0,12) 3 = PY (1 − 0,08) 3 1,404...  answer/antwoord = (4) 0,778... [15] = 1,8 : 1 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 15 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION/VRAAG 8 8.1 f ( x + h) = 2( x + h) 2 + 4  2 x 2 + 4 xh + 2h 2 + 4 = 2 x 2 + 4 xh + 2h 2 + 4 f ( x + h) − f ( x) = 2 x 2 + 4 xh + 2h 2 + 4 − 2 x 2 − 4 = 4 xh + 2h 2  4 xh + 2h 2 4 xh + 2h 2 f / ( x) = lim h →0 h h( 4 x + 2h) h( 4 x + 2h) = lim  lim h →0 h h →0 h = lim (4 x + 2h ) h →0 = 4x  4x (4) 8.2.1 f ( x) = −3x + 5 x 2 1  5x 2 1  − 6x f ( x) = −3x 2 + 5 x 2 1 5 − 5 − 1  x 2 f ( x) = −6 x + x 2 / 2 2 (3) 8.2.2  1  2 p( x) =  3 + 4 x  x  1 8 = 6 + 2 + 16 x 2  1 8 + + 16 x 2 x x x6 x2 = x + 8 x − 2 + 16 x 2 −6  x −6 + 8 x −2 + 16 x 2 p / ( x) = −6 x − 7 − 16 x −3 + 32 x  answer/antwoord (4) OR/OF p( x) = ( x −3 + 4 x) 2 by making use of the chain rule : p / ( x) = 2( x −3 + 4 x)(−3 x − 4 + 4)  2( x −3 + 4 x) p / ( x) = −6 x − 7 − 16 x −3 + 32 x  (−3 x −4 + 4) (4) 8.3.1 h / ( x) = 3 x 2 − 14 x + 14  finding/kry h / ( x) (1) 8.3.2 At/By B: h / ( x) = 0  derivative equal to/ afgeleide gelyk aan 0 3x 2 − 14 x + 14 = 0 14 ± (−14) 2 − 4(3)(14)  substitution into x= correct formula/substitusie 2(3) in korrekte formule = 1,45 or 3,22 x-value of/x-waarde n/a van 1,45 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 16 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum 8.3.3 x 3 − 7 x 2 + 14 x − 8 = ( x − 1)( x 2 − 6 x + 8)  ( x − 1) = (x − 1)(x − 2)(x − 4)  x2 − 6x + 8  (x − 2 )(x − 4 ) C(4 ; 0 ) coordinates of/koördinate van C (4) OR/OF x c > 3,22 h(4) = (4 ) − 7(4 ) + 14(4 ) − 8 = 0  xc > 3,22 3 2 ∴ xc = 4 substitution of/ substitusie van 4  h(4) = 0  xc (4) 8.3.4 h / ( x) = 3x 2 − 14 x + 14  h // ( x) = 6 x − 14 h // ( x) = 6 x − 14  6 x − 14 < 0 6 x − 14 < 0 6 x < 14 7 ∴x < 3 7 7 k = ∴k = 3 3 (3) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 17 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION/VRAAG 9 9.1 π r 2h = 6 6 6 h = (1) h= 2 πr 2 πr 9.2 ( S = 10 2π r 2 + 2π rh + 4π r 2 ) (  10 2π r 2 + 2π rh + 4π r 2 ) [ = 10 2π rh + 6π r 2 ]  20π rh + 60π r 2 = 20π rh + 60π r 2  6   substitution/substitusie = 20π r   + 60π r 2 2  π r  120 = 60π r 2 + (4) r OR/OF area of 10 spheres/ Area of/van 10 spheres/sfere = 10 × 4 × π × r = 40 π r 2 2 area van 10 sfere Area of/van 10 cylinders/silinders = 10(2 π r 2 + 2 π r h)  area of 10 cylinders/ area van 10 silinders 6 = 10(2 π r 2 + 2 π r )  substitution/substitusie π r2 120 = 20 π r 2 + r 120 Total area/Totale area = 40 π r 2 + 20 π r 2 + simplification/vereen- r voudiging 120 = 60 π r 2 + (4) r 9.3 S = 120πr − 120r −2 = 0 /  120πr − 120r −2 120 = 0 120πr − 2 = 0 r 120πr − 120 = 0 3 120 120 r3 =  r3 = 120π 120π 1 ∴ r = 1 = 0,68 cm π3 answer/antwoord (4) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 18 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION/VRAAG 10 10.1.1 d=5 d = 5 e=4 e = 4 f=7 f = 7 g=5 g = 5 (4) 10.1.2a 4 2 4 2 P(A and/en B and/en C) = =  = 54 27 54 27 (1) 10.1.2b 48 8 48 8 P(A or/of B or/of C) = =  = 54 9 54 9 (1) 10.1.2c 7 7 P(only/slegs C) =  54 54 (1) 10.1.2d P(that a country uses exactly two methods/dat 'n land 17  5 + 4 + 8 17 54 presies twee metodes gebruik) = = 54 54 (1) 10.2.1 1  answer/antwoord P(selects Midnight as drama/kies Midnight as drama) = (2) 5 10.2.2 Number of different selections of drama, romance and  product/produk comedy/Aantal verskillende keuses van drama, answer/antwoord (2) liefdesverhale en komedie = 5 × 4 × 3= 60 10.2.3 P(select Last Hero and Laughing Dragon/kies Last Hero en  product/produk 1 1 1 answer/antwoord Laughing Dragon) = × = (2) 5 3 15 OR/OF P(select Last Hero and Laughing Dragon/kies Last Hero en  product/produk 1× 4 ×1 1  answer/antwoord Laughing Dragon) = = (2) 60 15 [14] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou ## Related documents - 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