# Mathematics Grade 11 P1 National Exemplar November 2015 Memo > Grade 11 Mathematics Paper 1 memorandum from the National Exemplar November Exam 2015, 19 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/mathematics-p1-grade-11-nov-2015-memo-afr-eng/50c02c3a-1d85-4d37-8b9f-b1c52711e135 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Document type: Memorandum - Year: 2015 - Exam period: National Exemplar November Exam - Paper: 1 - Curriculum: CAPS - Pages: 19 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 P1 National Exemplar November 2015 Memo](https://www.hlayiso.com/library/mathematics-p1-grade-11-nov-2015-memo-afr-eng/50c02c3a-1d85-4d37-8b9f-b1c52711e135) ## About this document Mathematics · Grade 11 · National Exemplar November Exam · 2015. Memorandum, 19 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 NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 MATHEMATICS P1/WISKUNDE V1 NOVEMBER 2015 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 19 pages. Hierdie memorandum bestaan uit 19 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 2 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum NOTE: • If a candidate answered a question TWICE, mark only the FIRST attempt. • If a candidate crossed out an answer and did not redo it, mark the crossed-out answer. • Consistent accuracy applies to ALL aspects of the marking memorandum. • Assuming values/answers in order to solve a problem is unacceptable. LET WEL: • As 'n kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na. • As 'n kandidaat 'n antwoord deurgehaal en nie oorgedoen het nie, sien die deurgehaalde antwoord na. • Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing. • Dit is onaanvaarbaar om waardes/antwoordde aan te neem om 'n probleem op te los. QUESTION/VRAAG 1 1.1.1 x 2 + x − 12 = 0 factors/faktore (x + 4)(x − 3) = 0 answer/antwoord x = −4 or x=3 answer/antwoord (3) 1.1.2 2x +1 = x −1 2 x + 1 = (x − 1) 2 squaring both sides/ 2 2x +1 = x − 2x +1 kwadreer aan albei kante 2 x − 4x = 0 standard form/stand vorm x(x − 4) = 0 factors/faktore x = 0 or x=4 answers/antwoord n/a x = 4 (correct selection) (korrekte keuse) (5) 1.1.3 2 x x = 2 27 3  2x 2 3 2x = 2 27 2 3 3  x 2 = 27 x = 27 2 2 raise both sides to 2 3 x = (27 ) 3 answer/antwoord (4) x=9 1.1.4 x 2 − 2x − 8 < 0 (x − 4)(x + 2) < 0  ( x + 4 )( x + 2 ) < 0 OR/ -2 4 -2 4 OF critical values/krit wrdes − 2 < x < 4 OR / OF x ∈ (− 2 ; 4) inequalities/ongelykh (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 3 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum 1.2.1 5x 2 + 6 x − 7 = 0 formula/e − b ± b 2 − 4ac x= 2a substitution −6± (6)2 − 4(5)(− 7 ) = 2(5) answers/antwoord (4) = 0 ,73 or − 1,93 1.2.2 2 5x + 6 x − d = 0 − b ± b 2 − 4ac x= 2a substitution −6± (6)2 − 4(5)(− d ) = 2(5)  36 + 20d = 0 36 + 20d = 0 9 answer/antwoord (3) d =− 5 OR/OF For equal roots/ vir gelyke wortels : ∆=0 2 ∆ = b − 4ac = (6 ) − 4(5)(− d ) 2  substitution 36 + 20d = 0  36 + 20d = 0 9 d =− 5 answer/antwoord (3) OR/OF 5x 2 + 6 x − d = 0 6x d x2 + = 5 5 completing the square/  3 d 9 2 voltooi die kwadraat x +  = +  5 5 25 5d + 9 = 25 5d + 9 For equal roots =0  5d + 9 = 0 25 ∴ 5d + 9 = 0 answer/antwoord (3) −9 d= 5 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 4 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum 1.3 x = 2y − 3 ... (1) making x the subject/ xy = 20 ... (2) maak x die o/w substitute (1) into (2) : (2 y − 3) y = 20 substitution 2 2 y − 3 y − 20 = 0 standard form/stand vorm factors/faktore (2 y + 5)( y − 4) = 0 5 y-values/wrdes y=− or y=4 2 x = −8 or x=5 x-values/wrdes (6) OR/OF x + 3 = 2y x+3 y= ... ( 1 ) 2 xy = 20 ... ( 2 ) making y the subject/ substitute ( 1 ) into ( 2 ) : maak y die o/w  x + 3 subst x  = 20  2  x 2 + 3 x = 40 x 2 + 3 x − 40 = 0 standard form/stand vorm (x + 8)(x − 5) = 0 factors/faktore x = −8 or x=5 x-values/wrdes 5 y-values/wrdes (6) y=− or y=4 2 OR/OF x − 2 y = −3 ... ( 1 ) 20 y= ... ( 2 ) x making y the subject/ substitute ( 2 ) into ( 1 ) : maak y die o/w  20  x − 2  = −3  x  substitution 2 x − 40 = −3 x x 2 + 3 x − 40 = 0 standard form/stand vorm (x + 8)(x − 5) = 0 factors/faktore x = −8 or x=5 x values/wrdes 5 y=− or y=4 2 y values/wrdes (6) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 5 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum x − 2 y = −3 ... (1) 20 x= ... (2) y making x the subject/ substitute (2) into (1) : maak x die o/w 20 − 2 y = −3 y subst 20 − 2 y 2 = −3 y 0 = 2 y 2 − 3 y − 20 0 = (2 y + 5)( y − 4 ) standard form/stand vorm 5 factors/faktore y=− or y = 4 2 x = −8 or x=5 y values/wrdes x values/wrdes (6) [28] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 6 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 2 2.1.1 2 n + 2. 4 n +1 8 n −1 writing as prime bases/ 2 n + 2.2 2 n + 2 skryf as priembasisse = 2 3 n −3 applying exponential laws/ = 2 n + 2 + 2 n + 2 − (3 n − 3 ) pas ekspon.wette toe = 27 = 128 answer/antwoord (3) OR/OF 2 n.2 2.4 n.4 8 n.8 −1 writing as separate bases/ 8 n.2 2.2 2 = n −3 skryf as aparte basisse 8 .2 applying exponential laws/ = 27 pas ekspon.wette toe = 128 answer/antwoord (3) 2.1.2 x + 2x − 1 . x − 2x − 1 writing as one surd/ = (x + 2 x − 1)(x − 2 x − 1) skryf as een wortel = x 2 − 2x + 1  x2 − 2x +1  ( x − 1)2 = (x − 1)2 answer/antwoord (4) = x −1 OR/OF Let 2 x − 1 = k 2x − 1 = k 2 writing as one surd/ x + k. x −k skryf as een wortel = (x − k ) 2 2  x2 − 2x +1 = x 2 − 2x + 1  ( x − 1)2 = (x − 1)2 answer/antwoord (4) = x −1 2.2.1 5 For P to be a real number, we cannot divide by 0 and x+2 must be positive, i.e. we must have: Vir P om reëel te wees, kan ons nie deur 0 deel nie en 5 moet positief wees, dus moet ons die vlg hê: x+2 x+2>0 x+2>0 x > −2 answer/antwoord (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 7 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum 2.2.2 5 x P= + x+2 3 5 3 substitution x = 3 = + 3+ 2 3 = 1+1 =2 answer/antwoord (2) 2.3 2 2015 ×5 2019 2015 2015 4  5 4 × 2 2015 × 5 2015 =2 ×5 .5 = 625 × (2 ⋅ 5) 2015 = 625 × 10 2015  10 2015  625 000...0 = 625 000 000...0 ∴ Sum of digits will be/Som van die getalle 6 + 2 + 5 + 0 + 0 + 0... + 0 answer/antwoord (4) = 13 [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 8 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 3 3.1.1 5 –2 –9 + –7 –7 First difference/Eerste verskil d = –7  answer/antwoord (1) 3.1.2 T4 = −16  answer/antwoord (1) 3.1.3 Tn = −7n + 12  Tn = −7 n + 12 − 289 = −7 n + 12  Tn = −289 7 n = 301 n = 43  answer/antwoord (3) 3.2.1 T22 = 64 + 3 + 3 = 70  answer/antwoord (1) 3.2.2 T20 = 3n + b 64 = 3(20) + b  64 = 3(20 ) + b b=4 Tn = 3n + 4 3T5 − 2 = 3[3(5) + 4] − 2  3[3(5) + 4] − 2 = 55 3n + 4 = 55  3n + 4 = 55 3n = 51 n = 17  n = 17 (4) OR/OF T20 = 3n + b 64 = 3(20 ) + b  64 = 3(20 ) + b b=4 T5 = 64 − 15(3) = 19 19 Tn = 3T5 − 2 3n + 4 = 3(19 ) − 2  3n + 4 = 3(19 ) − 2 3n = 51 n = 17  n = 17 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 9 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum 3.3.1 5 12 29 56 7 17 27 10 10 93 ; 140  93  140 (2) 3.3.2 7 17 27 10 10 Tn = 10n − 3  Tn = 10n − 3 = (10n − 2 ) − 1  10n − 2 = 2(5n − 1) = 2(5n − 1) − 1 Since 10n − 2 = 2(5n − 1) , 10n − 2 is even for any value of n. Thus Tn is always odd, since for any value of n, Tn is always one explanation/verduidlkng less than an even number (3) Aangesien 10n − 2 = 2(5n − 1) , is 10n − 2 dus ewe vir enige waarde van n. Dus Tn is altyd onewe, want vir enige waarde van n, is Tn altyd een minder as 'n ewe getal OR/OF 7 17 27 10 10 Tn = 10n − 3  Tn = 10n − 3 n 10n T n = 10n –3 or 4 10 x 4 = 40 37 T n = 10n –3 5 10 x 5 = 50 47 7; 17; 27; 37; 47; 57..... 6 10 x 6 = 60 57 7 10 x 7 = 70 67 8 10 x 8 = 80 77 The sequence of first differences, for all values of n, will always end all answers end in 7/ in a 7, since they are all multiples of 10 (10n) less three. alle antwoord eindig in 7 Any number ending in 7 is odd. explanation/verduidlkng Therefore all the terms in the sequence of first differences will be (any logical explanation/ odd numbers. enige logiese verduidlkng) (3) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 10 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum Term 1 is odd and if the first difference of 10(even number) is added to an odd number, the answer remains odd. Die ry van eerste verskille, vir alle waardes van n, sal altyd in 7 eindig want hulle is almal veelvoude van 10 (10n) min drie. Enige getal wat in 7 eindig, is onewe. Daarom sal alle getalle in die ry van eerste verskille onewe getalle wees. OF Term 1 is 'n onewe getal en as die eerste verskil van 10(ewe getal) by 'n onewe getal getel word, bly die antwoordoord 'n onewe getal. [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 11 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 4 4.1 3 5 8 12 2 3 4 1 1 2a = 1 1 a= 2 1 a = 1 2 3  + b = 2 2 1 1 b= b = 2 2 a+b+c =3 1 1 + +c =3 2 2 c = 2 c=2 n2 n n2 n Tn = + +2  Tn = + +2 2 2 2 2 26 2 26 T26 = + +2 2 2 subst n = 26 = 353 answer/antwoord (6) OR/OF (n − 1)(n − 2)d 2 Tn = T1 + (n − 1)d1 +  formula/e 2 (n − 1)(n − 2)(1)  substitution = 3 + (n − 1)(2) + 2 2 n − 3n + 2 = 3 + 2n − 2 + 2 simplifying/vereenv 1 3 = 2n + 1 + n 2 − n + 1 2 2 1 1 = n2 + n + 2 2 2 n 2 n n2 n Tn = + +2  Tn = + +2 2 2 2 2 26 2 26 T26 = + +2 substitution n = 26 2 2 = 353 answer/antwoord (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 12 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum 4.2 p 18 28 4p 18 – p 10 4p – 28  T3 = 28 c c  T4 = 4 p 10 − (18 − p ) = 4 p − 28 − 10  10 − (18 − p ) 10 − 18 + p = 4 p − 28 − 10  4 p − 28 − 10 3 p = 30 equating p = 10 answer/antwoord (6) [12] QUESTION/VRAAG 5 5.1.1 (1 ; − 2)  for/vir 1 for/vir – 2 (2) 5.1.2 For x-intercept/Vir x-afsnit: −9 0= −2 y = 0 x −1 −9 2= x −1 2(x − 1) = −9  simplification/vereenv 2 x = −7 7  7  x=−  − ; 0  answer/antwoord 2  2  For y-intercept/Vir y-afsnit: −9 x = 0 y= −2 0 −1 =9−2  answer/antwoord (5) =7 (0; 7 ) 5.1.3 y = −x − 1 – x  –1 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 13 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum 5.1.4 Closest point is a point of intersection between the axis of symmetry and the hyperbola/Naaste punt is 'n snypunt tussen die simmetrie-as en die hiperbool: −9 − x −1 = −2  equating/vgl x −1 −9 − x +1 = x −1 9 x −1 = x −1 (x − 1) = 9  ( x − 1) = 9 2 2 x − 1 = 3 or x − 1 = −3 x=4 x = −2  answers for/antwoord vir x in the fourth quadrant, x > 0, hence x = 4 only  selects x = 4 only/ kies slegs x = 4 y = −4 − 1 y = −5  answer for/antwoord vir y Point/Punt is (4 ; − 5) (5) OR/OF Closest point is a point of intersection between the axis of symmetry and the hyperbola/Naaste punt is 'n snypunt tussen die simmetrie-as en die hiperbool: −9 − x −1 = −2  equating/vgl x −1 (− x − 1)(x − 1) = −9 − 2(x − 1) − x 2 + 1 = −9 − 2 x + 2 0 = x2 − 2x − 8 0 = (x − 4 )(x + 2 )  0 = x2 − 2x − 8 x=4 x = −2  answers for/antwoord vir x in the fourth quadrant, x > 0, hence x = 4 only  selects x = 4 only/ y = −4 − 1 kies slegs x = 4 y = −5 Point is (4 ; − 5)  answer for/antwoord vir y OR/OF (5) −9 −9 y= → f (x ) = −2 x translate 1 right and 2 down x −1 points in 4th quad stay in 4th quad Under translation 1 right and 2 down, points in the fourth  origin becomes A quadrant will stay in the fourth quadrant. closest point to origin on Since the origin becomes A under the translation 1 right parent function is (3 ; − 3) and 2 down and the point in the fourth quadrant which is −9 the closest point on y = to the origin is (3 ; − 3) , answer/antwoord x (5) The closest point on f to A is (3 + 1 ; − 3 − 2 ) i.e. (4 ; − 5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 14 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum Onder die translasie 1 regs en 2 na onder, sal punte in die punte in 4de kwad bly vierde kwadrant steeds in die vierde kwadrant wees. in 4de kwad Die oorsprong word A onder die translasie 1 regs en 2 na oorsprong word A onder, en die punt in die vierde kwadrant wat die naaste naaste punt aan oorsprong −9 op moederfunksie is (3 ; − 3) punt aan y = tot die oorsprong is, is (3 ; − 3) . Die x naaste punt op f aan A is (3 + 1 ; − 3 − 2 ) d.i. (4 ; − 5) answer/antwoord (5) 5.1.5 9 9 y= +2  x −1 x −1  +2 (2) 5.2.1 For y-intercept/Vir y-afsnit substitution x = 0: x = 0 substitution into the y = 4.2 0 + 1 equation/in die vgl =5 y = 5 (2) H(0 ; 5) 5.2.2 For x-intercept/Vir y-afsnit y = 0 i.e./d.i. 4.2 − x + 1 = 0  4.2 − x + 1 = 0 4.2 − x = −1 1 1 2 − x = − , which is impossible, since 2 − x > 0 for x ∈ R  2− x = − and explanation/ 4 4 , wat onmoontlik is omdat 2 − x > 0 vir x ∈ R en verduideliking (2) Therefore/Dus: no solution/geen oplossing, which means there will be no x-intercept/wat beteken daar sal geen x-afsnit wees nie. OR/OF The graph lies above its asymptote y = 1 because the above/bokant coefficient of 2 − x is 4/Die grafiek lê bokant sy asimptoot y=1 (2) y = 1 want die koëffisiënt van 2 − x is 4. OR/OF The range is (1 ; ∞ ) or y > 1 correct range/korrekte Die waardeversameling is (1 ; ∞ ) of y > 1 waardeversameling (2) 5.2.3 y  shape/vorm y-intercept and other point/ h y-afsnit en ander punt  asymptote/asimptoot (3) y=1 5 . B(1 ; 3) O x Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 15 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum 5.2.4 ( g ( x) = 4 2 − x + 2 )  7 units/eenhede  upwards/opwaarts (2) = 4.2 − x + 8 [25] The graph of h is translated 7 units upwards to form g/ Die grafiek van h word 7 eenhede na bo getransleer om g te vorm. QUESTION/VRAAG 6 6.1 y = a(x − 2) + 9  substitution coordinates of 2 TP Substitution (0 ; 5) : 5 = a(0 − 2 ) + 9 2  substitution of/van (0; 5) 5 = 4a + 9 a = −1  value of/waarde van a y = −1( x − 2 ) + 9 2 ( = − x2 − 4x + 4 + 9 ) = −x2 + 4x + 5  simplification/vereenv (4) 6.2 9−5 5−9 9−5 5−9 Average Gradient = or  or 2−0 0−2 2−0 0−2 =2 answer/antwoord (2) 6.3 x-intercepts of/x-afsnitte van f: 1 2 1 2 x −8 = 0  x −8 = 0 2 2 x 2 = 16 x = 4 or − 4 At/By B: x = −4 –4 x-intercepts of/x-afsnitte van g: − x2 + 4x + 5 = 0 x2 − 4x − 5 = 0 (x − 5)(x + 1) = 0  factors/faktore x = −1 or 5 5 At/By D: x = 5 answer/antwoord (5) Length of/Lengte van BD: 4 + 5 = 9 6.4.1 x ≤ −4 or x ≥ 4  x ≤ −4 x≥4 (2) 6.4.2 0< x<2  endpoints/eindpunte  notation/notasie (2) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 16 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 7 7.1 CD = 2 x + 3 − (− 2 x 2 + 14 x + k )  f ( x) − g ( x) = 2 x + 3 + 2 x 2 − 14 x − k = 2 x 2 − 12 x + 3 − k  answer/antwoord (3) 7.2 Minimum value occurs at/Minimum waarde vind plaas by −b −b x= x= 2a 2a 12 = 2(2 ) =3 x-value for minimum Minimum value/Minimum waarde x-waarde vir minimum 5 = 2(3) − 12(3) + 3 − k 2  subst 5 5 = 18 − 36 + 3 − k  answer/antwoord (4) k = −20 OR/OF CD = 2 x 2 − 12 x + 3 − k ( ) = 2 x2 − 6x + 3 − k [ ] = 2 ( x − 3) − 9 + 3 − k 2  2( x − 3) 2 = 2(x − 3) − 18 + 3 − k 2  CD = 2( x − 3) − 15 − k 2 = 2(x − 3) − 15 − k 2 Hence the minimum value of CD is −15 − k The minimum value of CD is given to be 5 Vervolgens is die minimum waarde van CD −15 − k Die minimum waarde van CD is gegee as 5 5 = −15 − k  5 = −15 − k k = −20 answer/antwoord (4) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 17 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 8 8.1 A = P(1 + i ) n  140 000 = 140 000(1 + 0,061) 4  (1 + 0,061) 4 = R177 414,69 answer/antwoord (3) 8.2  0,07  2 1 + ieff = 1 +  0,07  2   2 1 + ieff = (1 + 0,035) 2  1 + ieff = (1 + 0.035) 2 ieff = (1 + 0,035) − 1 2 = 0,071225  answer/antwoord (3) The effective interest rate/Die effektiewe rentekoers is 7,12% p.a. 8.3  0,105  48  0,105  30 0,105 A = 240001+  − 70001 +    12   12  12  n=48 = R27 369,56  n=30 correct substitution into correct formula/korr subst in korr formule OR/OF answer/antwoord (5) 0,105  18  30   0,105  0,105  12 A = 240001+  − 70001 +    12   12   n=18 n=30 = R27 369,56 correct substitution into correct formula/korr subst in korr formule OR/OF answer/antwoord (5) 18  0,105  A after 18 months = 240001+  0,105  12   12 = R 28 074,70  n=18 R 28 074,70 − R7000 = R 21 074,70  21 074,70 30  0,105  A after 4 years = 21 074,701 +  n=30  12   answer/antwoord (5) = R27 369,56 8.4  r  3 102 755,34 = 198 0001 −  formula  100  102 755,34 r correct substitution into 3 =1− 198 000 100 correct formula/korr subst in korr formule r n =3 1− = 0,8036119818 100 r  3 102 755,34 = 1 − r − = − 0,1963880182 100 198 000 100 r = 19,64%  answer/antwoord (5) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 18 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 9 9.1 Given/Gegee: P(A) = 0,6 P(B) = 0,3 P(A or B) = 0,8 P(A and/en B) = 0,6 + 0,3 – 0,8 = 0,1 ≠ 0 P(A and/en B) = Therefore A and B are not mutually exclusive. 0,1 Dus is A en B nie onderling uitsluitend nie justification/regv Not mutually OR/OF exclusive/Nie onderling uitsluitend (4) A B Venn diagram 0,6 – x x 0,3 – x 0,6 − x + x + 0,3 − x = 0,8 x = 0,1 0,9 − x = 0,8 x = 0,1 There is an intersection between A and B/Daar is snyding tussen A en B justification/regv Therefore A and B are not mutually exclusive/Dus is A en B nie Not mutually onderling uitsluitend nie. exclusive/Nie onderling uitsluitend (4) 9.2.1a 1832 1832 = 69,24% 2646 (2) 2646 9.2.1b 460 230 answer/antwoord = = 17,38% (1) 2646 1323 9.2.1c 340 + 14 59 answer/antwoord = = 13,38% (2) 2646 441 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 19 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 19 DBE/November 2015 CAPS/KABV – Grade/Graad 11 – Memorandum 9.2.2 Let the event of a randomly selected person living in Area 1 be A. Let the event of a randomly selected person earning less than R3 200 be B. Laat die gebeurtenis van 'n persoon wat willekeurig gekies is wat in Gebied 1 woon, A wees. Laat die gebeurtenis van 'n persoon wat willekeurig gekies is wat minder as R3 200 verdien, B wees.  P(A and B) 500 P(A) P(A and B) = = 18,90% 2646 P(B) P(A) × P(B) = 1832 × 960 = 25,12%  P(A) × P(B) 2646 2646 Clearly/Duidelik, P(A and B) ≠ P(A) × P(B) conclusion with Hence A and B are not independent/Vervolgens is A en B nie onafhanklik justification/ nie. gevolgtrekking met motivering (5) OR/OF Let the event of a randomly selected person living in Area 2 be C. Let the event of a randomly selected person earning less than R3200 be D. Laat die gebeurtenis van 'n persoon wat willekeurig gekies is wat in Gebied 2 woon, C wees. Laat die gebeurtenis van 'n persoon wat willekeurig gekies is wat minder as R3 200 verdien, D wees. 460  P(C and D) P(C and D) = = 17,38% 2646 P(C) P(C) × P(D) = 814 × 960 = 11,16% P(D) 2646 2646  P(C) × P(D) Clearly, P(C and D) ≠ P(C) × P(D) Hence C and D are not independent/Vervolgens is C en D nie onafhanlik conclusion with nie. justification/ gevolgtrekking met motivering (5) 9.2.3 500  27,29% P(Area 1 person earns less than R3200) = = 27,29% 1832  56,51% 460 P(Area 2 person earns less than R3200) = = 56,51% 814 A person from Area 2 is more likely to earn less than R3200 conclusion/vgl (3) 500 P(Gebied 1 persoon verdien minder as R3200) = = 27 ,29% 1832 460 P(Gebied 2 persoon verdien minder as R3200) = = 56 ,51% 814 Dis meer waarskynlik dat 'n persoon uit Gebied 2 minder as R3 200 sal verdien. 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