Mathematics Grade 12 Paper 1 NSC November 2016 Memo

Downloaded from hlayiso.com GRADE 12/GRAAD 12 NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT MATHEMATICS P1/WISKUNDE V1 NOVEMBER 2016 MEMORANDUM MARKS: 150 PUNTE: 150 This memorandum consists of 20 pages. Hierdie memorandum bestaan uit 20 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 2 DBE/November 2016 NSC – 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, sien slegs die EERSTE poging na. • Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing. QUESTION/VRAAG 1 1.1.1 x( x − 7 ) = 0 x = 0 or x = 7  x=0  x=7 (2) 1.1.2 2 x − 6x + 2 = 0 6 ± (−6) 2 − 4(1)(2) correct substitution into x= 2(1) correct formula 6 ± 28 x=  x = 0,35 2  x = 5,65 x = 0,35 or x = 5,65 (3) OR/OF x2 − 6x + 2 = 0 x 2 − 6 x + 9 = −2 + 9 (x − 3)2 = 7  ( x − 3) = 7 2 x−3= ± 7 x = 3± 7  x = 0,35 x = 0,35 or x = 5,65  x = 5,65 (3) 1.1.3 x −1 +1 = x x −1 = x −1  isolate x − 1 x − 1 = x 2 − 2x + 1  x2 − 2x +1 x 2 − 3x + 2 = 0 (x − 2)(x − 1) = 0  standard form  factors x = 2 or x = 1  both answers Both answers are valid (5) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 3 DBE/November 2016 NSC – Memorandum x −1 +1 = x  isolate x −1 x −1 = x −1 Let x − 1 = k k =k k ≥0 2 k=k  k2 2 k −k =0 k (k − 1) = 0  standard form (x − 1)(x − 2) = 0  factors x = 2 or x = 1 ; x ≥ 1  both answers Both answers are valid (5) OR/OF x −1 +1 = x  isolate x − 1 x −1 = x −1  x −1 = 0  x −1 = 1 By inspection : x − 1 = 0 or x − 1 = 1  x=2 x = 2 or x = 1  x =1 (5) 1.1.4 3 x +3 − 3 x + 2 = 486  expansion 3 x 33 − 3 x 3 2 = 486 ( 3 x 33 − 3 2 = 486)  common factor 3 x = 27 x 3  3 x = 27 3 =3 x=3  x=3 (4) OR/OF 3 x +3 − 3 x + 2 = 486  common factor ( ) 3 x + 2 31 − 1 = 486 (  31 − 1 ) 3 x + 2 = 243 3 x + 2 = 35  3 x + 2 = 243 x+2=5 x=3  x=3 (4) 1.2.1 f (x ) = x 2 + 3x − 4 0 = (x + 4)( x − 1)  factors x = − 4 or x = 1  both answers (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 4 DBE/November 2016 NSC – Memorandum 1.2.2 x 2 + 3x − 4 < 0 ( x + 4)( x − 1) < 0 + – + OR/ -4 1 –4 1 OF  − 4 < x <1 − 4 < x <1 OR/OF x ∈ (− 4 ; 1) (2) 1.2.3 2x + 3 ≥ 0  2x + 3 3 3 x≥−  x≥− 2 2 (2) f (x ) ≥ 0 when f is increasing / − 4 +1 − 4 +1 The turning point occurs at x =  x= 2 2 3 3 x≥−  x≥− 2 2 (2) 2 1.3 x = 2 y and x − 5 xy = −24 (2 y )2 − 5(2 y )( y ) = −24  substitution of 2y 2 2 4 y − 10 y = −24 − 6 y 2 = −24  − 6 y 2 = −24 y2 = 4 y = −2 or y = 2  both y – values x = −4 or x = 4  both x– values OR/OF (4) x = 2 y and x 2 − 5 xy = −24 x x y=  substitution of 2 2  x x 2 − 5(x )  = −24 2 5 x 2 − x 2 = −24 2 3 2 3  − x = −24 − x 2 = −24 2 2 x 2 = 16 x = −4 or x = 4  both x – values y = −2 or y = 2  both y – values (4) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 5 DBE/November 2016 NSC – Memorandum x x 2 + 24  equating = 2 5x x = 2 y and x 2 − 5 xy = −24 x y= 2 − x 2 − 24 y= − 5x 2 x x + 24 = 2 5x 5 x = 2 x 2 + 48 2  3 x 2 = 48 3x 2 = 48 x 2 = 16  both x – values  both y – values (4) x = −4 or x = 4 y = −2 or y = 2 [24] QUESTION/VRAAG 2 2.1 T4 = −7  –7 (1) 2.2 Tn = a + (n − 1)d − 87 = 5 + (n − 1)(−4)  a = 5 and d = –4 − 87 = 5 − 4n + 4  − 87 = 5 + (n − 1)(−4) 4n = 96 n = 24  n = 24 (3) OR/OF − 4n + 9 = − 87  –4n + 9  − 4n + 9 = − 87 − 4n = − 96 n = 24  n = 24 (3) 2.3 − 3 ; − 7 ; .......; − 87 n S n = [a + Tn ] 2  n = 22 22 S 22 = [− 3 − 87]  a = –3 2 = − 990  answer (3) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 6 DBE/November 2016 NSC – Memorandum − 3 ; − 7 ; .......; − 87 n S n = [2a + (n − 1)d ] 2 22 S 22 = [2(−3) + (22 − 1)(− 4)]  n = 22 2  a = –3 = − 990  answer (3) OR/OF All negative terms can be written down and added to get the answer of –990./Alle negatiewe terme kan neergeskryf word en dan bymekaar getel word om –990 te kry. a = –3  answer (3) OR/OF Sum = S 24 − (5 + 1) 24 24  [5 − 87] = [5 − 87] − 6 2 2  –6 = −990  answer (3) 2.4 5 ; − 15 ; − 35...... d = −20  d = –20 Tn = −20n + 25  Tn = −20n + 25 Last term in the sequence divisible by 5 is:/Laaste term in die ry deelbaar deur 5 is: − 4187 + 4(3) = − 4175 Tn = −20n + 25 − 4175 = −20n + 25  − 4175 = −20n + 25 20n = 4200 n = 210  n = 210 There will be 210 terms in the sequence that is divisible by 5./Daar is 210 terme in die ry deelbaar deur 5. (4) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 7 DBE/November 2016 NSC – Memorandum 5 ; 1 ; –3 ; ... ; –83 ; –87; ……; – 4187 Tn = −4n + 9  − 4n + 9 = −4187 − 4187 = −4n + 9 4n = 4196  n = 1049 n = 1049 There are 1049 terms in the sequence./Daar is 1049 terme in die ry. T1 ; T6 ; T11 ; T16 ... are divisible by 5./is deelbaar deur 5. The largest integer value of k such that  5k − 4 ≤ 1049 5k − 4 ≤ 1049 5k ≤ 1053 k ≤ 210,6  k = 210 k = 210 (4) OR/OF 5 ; 1 –3 ; –7 ; … ; –4175; –4179 ;–4183 ; –4187 Tn = a + (n − 1)d − 4175 = 5 + (n − 1)(− 4 )  d = –4 − 4180 = −4(n − 1) n = 1046  − 4175 = −4n + 9  1046 Number of terms divisible by 5 1046 − 1 = +1 5 = 210  n = 210 (4) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 8 DBE/November 2016 NSC – Memorandum QUESTION/VRAAG 3 3.1.1 – 1 ; x ; 3 ; x + 8 ; ... –1 x 3 x +8  x + 1 ; 3 − x and x + 5 x+1 3–x x+5 –2 x + 2 2x + 2  calculating second differences − 2x + 2 = 2x + 2  − 2x + 2 = 2x + 2 4x = 0 x =0 (4) x=0 3.1.2 First differences/Eerste verskille: 1 ; 3 ; 5 ; …… n S n = [2(1) + (n − 1)(2 )] 2  Sn = n2 = n2 250 < n 2  S n > 250 n > 250 ∴ n > 15,8  n > 15,8 The sum of the 16 first differences will be greater than 250. Therefore the 17th term of the quadratic number pattern is the first satisfying this condition./Die som van 16 eerste verskille  n = 17 sal groter as 250 wees. Gevolglik sal die 17de term van die (4) kwadratiese getalpatroon die eerste wees wat aan die voorwaarde voldoen. 3.2.1 21 + 21(0,85) + 21(0,85) + ......... 2 17 Tn = ar n −1  n = 10 ; r = 0,85 or 20 T10 = (21)(0,85) 9  substitution into correct = 4,86 cm formula  answer (3) 3.2.2 Sn = ( a 1− rn ) 1− r  n = 15 S15 = ( 21 1 − (0,85) 15 ) 1 − 0,85 = 127,77  127,77 Area of the page = 30 x 21 = 630 Percentage of paper covered in grey ink:  630 127,77 = × 100% 630  20,28 = 20,28% (4) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 9 DBE/November 2016 NSC – Memorandum QUESTION/VRAAG 4 4.1 y=0 y=0 (1) 4.2 R(0 ; 1)  answer (1) x 4.3 y=a 9 = a2  substitution ∴a = 3 a=3 (2) 4.4 DP = 2 – b y = 3x 1 = 3b 1 81  = 3b 81 3 − 4 = 3b  3−4 or use of logs b = −4  b = −4 DP = 2 − (−4) = 6 units  DP = 6 units (4) 4.5 h( x + 2) + k = 0 h( x + 2) = −k 1 1 1 0 < −k <  − k < or k > − 81 81 81 1 1 − <k <0  − <k <0 81 81 (3) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 10 DBE/November 2016 NSC – Memorandum QUESTION/VRAAG 5 5.1 f (x ) = − x 2 + 4 x − 3 4 f ′( x ) = 0 or x=−  − 2 x + 4 = 0 or 2(− 1) 4 − 2x + 4 = 0 x= 2 x=− 2(− 1) x=2 y = −(2 ) + 4(2 ) − 3  y = −(2 ) + 4(2 ) − 3 2 2 =1 (2) B(2 ; 1) OR/OF − x 2 + 4x − 3 = 0 x 2 − 4x + 3 = 0 (x − 3)(x − 1) = 0 x = 3 or x = 1 3 +1 3 +1 x=  x= 2 2 x=2 y = −(2 ) + 4(2 ) − 3 2  y = −(2 ) + 4(2 ) − 3 2 =1 (2) B(2 ; 1) 5.2 Range/Waardeversameling : y ≤ 1  y ≤1 OR/OF (1) Range/Waardeversameling : y ∈ (−∞ ; 1]  (−∞ ; 1] (1) 5.3 x ≤ −1 or x>2  critical values  x ≤ −1 or x > 2 OR/OF (2) (− ∞ ; − 1] ∪ (2 ; ∞ )  critical values  x ≤ −1 or x > 2 (2) 5.4 (x − p ) ( y + t ) = 3 Vertical asymptote of h(x) /vertikale asimptoot at x = 2 Translation 4 units to the left / Translasie 4 eenhede links  x = −2 x = 2 – 4 = –2 is the equation of the vertical asymptote of (1) h(x + 4) x = 2 – 4 = –2 is die vergelyking van die vertikale asimptoot OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 11 DBE/November 2016 NSC – Memorandum OR/OF 3 h( x ) = +1 x−2+4 3 = +1 x+2 x = −2 is the equation of the vertical asymptote / is die vergelyking  x = −2 van die vertikale asimptoot (1) 5.5 (x − p )( y + t ) = 3 3 (y + t) = (x − p ) 3 3  y= −t x− p x− p  −t B(2 ; 1) Point of intersection of the asymptotes Snypunt van die asimptote p=2  p=2 −t =1 t = −1  t = −1 (4) 5.6 x-intercepts of f / x-afsnitte van f : x 2 − 4x + 3 = 0 (x − 3)(x − 1) = 0  both critical values x = 1 or x = 3 g / ( x ) < 0 for x ∈ R ; x ≠ 2 Hence f ( x ) < 0  x ≤1 x ≤ 1 or x ≥ 3 OR/OF (− ∞ ; 1] ∪ [3 ; ∞ )  or  x≥3 (4) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 12 DBE/November 2016 NSC – Memorandum QUESTION/VRAAG 6 y 6.1 g: shape: increasing curve 3 g (1 ; 0): only on log graph x 0 1 3 f: f (3 ; 0) (0 ; 3) (4) 6.2 y = log 2 x  interchange x and y g −1 : x = log 2 y x y=2 y = 2x (2) 6.3 log 2 (3 − x ) = x 2x = 3 − x  2 x = − x + 3 2x = −x + 3 Reflect the graph of g about the line y = x to obtain g −1 and determine the point of intersection of f and g −1 ./ Reflekteer die  point of intersection of grafiek van g om die lyn y = x en bepaal die snypunt van f and f and g −1 g −1 (3) 6.4 x =1  answer (1) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 13 DBE/November 2016 NSC – Memorandum QUESTION/VRAAG 7 7.1 A = P( 1 + i ) n 2  0 ,15  substituting i and n = 250000 1 +   12  values in correct = R 256 289 ,06 formula  answer (2) 7.2 [ x 1 − (1 + i ) P= −n ] i 0,15   0 ,15  − 46  i= x 1 − 1 +   12   12    n = 46 256 289,06 =  substitution into 0 ,15 correct formula 12   0 ,15  − 46  3203,6133 = x 1 − 1 +     12    answer x = R 7 359 ,79 per month (4) OR/OF 0,15 −2   0,15  −46  i=  0,15  12 x1 +  1 − 1 +    12    12    n = 46 250 000 =  substitution into 0,15 correct formula 12 x = R 7 359,79  answer (4) 7.3   0,15  −n  9 0001 − 1 +   x = 9 000   12   256 289,06 = 0,15  substitute into correct 12 formula −n  0,15  1 +  = 0,6440429722  12   0,15   use of logs − n log1 +  = log 0,6440429722  12   n =35,42 n = 35,41872568 months/ maande ∴ 36 payments are required ∴36 paaiemente moet betaal word ∴ Thabiso will pay his loan off 10 months sooner./Thabiso los  10 months sy lening 10 maande vroeër af. (5) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 14 DBE/November 2016 NSC – Memorandum  0,15 n  90001 +  − 1  0,15  n  12    9 000 256289,061 +  =  12  0,15 12  substitute into correct n n formula  0,15   0,15  3203,613251 +  = 90001 +  − 9000  12   12  n  0,15  9000 = 5796,386751 +   12   use of logs n = log 0,15  1,5523691425  1+   12   n =35,42 n = 35,41872568 ∴ 36 payments are required ∴36 paaiemente moet betaal word ∴ Thabiso will pay his loan off 10 months sooner./Thabiso los  10 months sy lening 10 maande vroeër af. (5) th 7.4 The balance of his loan after the 35 payment was made: Die balans van sy lening nadat die 35ste paaiement betaal is:   0,15  35  9000 1 +  − 1 35  0,15  35   12    256289,061 + 0,15  Balance = 256289,061 +  −  12   12  0 ,15  0,15  35    12  9000 1 + 12  − 1 = R 3 735,45 0,15 12  0,15  Final instalment = 3 735,451 +   0 ,15   12   3 735,451 +   12  = R 3 782,14  answer OR/OF (4) P= [ x 1 − (1 + i ) −n ] i Final instalment  0,41872568   0,15  −0, 41872568    0,15  −0, 41872568  9 0001 − 1 +    9 0001 − 1 +  12     12    0,15     = 1 +  0,15 0,15  12  12 12  0,15   × 1 +  = R 3 782,14  12   answer (4) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 15 DBE/November 2016 NSC – Memorandum 36   0,15  36    0,15  9000 1 +  − 1 256289,061 +     12   0,15  36  12   Balance = 256289,061 +  −   0,15 36   1 +  12  0 ,15 9000   − 1   12   12 0,15 = R − 5 217,86 12  9 000 − 5217,86 Final payment = 9 000 − 5217,86  answer = R 3 782,14 (4) [15] QUESTION/VRAAG 8 8.1 f ( x + h) = 3( x + h) 2  3( x + h) 2 = 3( x 2 + 2 xh + h 2 ) = 3 x 2 + 6 xh + 3h 2 f ( x + h) − f ( x) = 3 x 2 + 6 xh + 3h 2 − 3 x 2  6 xh + 3h 2 = 6 xh + 3h 2 f ( x + h) − f ( x ) f ′( x) = lim h →0 h  f ′( x) = lim f ( x + h) − f ( x) h →0 h 6 xh + 3h 2 = lim h →0 h h(6 x + 3h)  lim(6 x + 3h ) = lim h →0 h →0 h = lim(6 x + 3h ) h →0 6x = 6x (5) OR/OF f ( x + h) − f ( x ) f ′( x) = lim h →0 h  f / ( x) = lim f ( x + h) − f ( x) h →0 h 3( x + h) 2 − 3x 2 = lim h →0 h  3( x + h) 2 − 3 x 2 3 x + 6 xh + 3h 2 − 3x 2 2 = lim h →0 h 6 xh + 3h 2  6 xh + 3h 2 = lim h →0 h h(6 x + 3h) = lim  lim(6 x + 3h ) h →0 h h →0 = lim(6 x + 3h ) h →0  6x = 6x (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 16 DBE/November 2016 NSC – Memorandum 8.2 4+h −2 lim h →0 h  answer g ( x) = x  answer a=4 (2) 8.3 3 5 3 y= x −  x 2 x3 3  − 5 x −3 y = x 2 − 5x −3 3 2 1  x dy 3 2 1 2 = x + 15 x − 4  15 x −4 dx 2 (4) 3 2 8.4 f ( x) = x + ax + bx + 18 f ( x ) = 3 x 2 + 2ax + b /  3 x 2 + 2ax + b At x = 1, mtan = −8 f / (1) = −8  f / (1) = −8 or 3(1) + 2a(1) + b = −8 2 3(1) + 2a(1) + b = −8 2 3 + 2a + b = −8 2a + b = −11 ..........(1) y = f (1) = g (1) = −8(1) + 20 = 12  1 + a + b + 18 = 12 1 + a + b + 18 = 12 a + b = −7 ............(2) a = −4  a = −4 b = −3  b = −3 (5) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 17 DBE/November 2016 NSC – Memorandum QUESTION/VRAAG 9 9.1 f ′( x ) = 3x 2 + 8 x − 3 = 0  equating derivative to zero (3x − 1)(x + 3) = 0  factors 1  x – values x= or x = −3 3 (3) // 9.2 f ( x) = 6 x + 8  6x + 8 6x + 8 < 0 4 4  x < − x<− 3 3 (3) OR 1 1 −3 −3 x= 3  3 2 2 4 = 3 4 ∴x < − 4 3  x < − 3 (3) 9.3 1  x ≤ −3 x ≤ −3 or x≥ 3 1  x≥ (2) OR/OF 3  [− ∞;−3] [− ∞;−3] ∪  1 ; ∞ 1    ; ∞ (2) 3  3  9.4 f (0) = −18 d = −18  d = −18 f ( x) = ax 3 + bx 2 + cx − 18  f ′(x ) = 3ax 2 + 2bx + c f ′(x ) = 3ax 2 + 2bx + c f ′( x ) = 3x 2 + 8 x − 3 3a = 3 2b = 8 a=1 b=4 a =1 b=4 c = −3  c = –3 f ( x ) = x 3 + 4 x 2 − 3x − 18 (5) OR/OF f ′( x ) = 3 x 2 + 8 x − 3 By integration/Deur integrasie  f (x ) = x 3 + 4 x 2 − 3x + d f (x ) = x 3 + 4 x 2 − 3x + d  d = −18 f (0) = d = −18 a=1 a =1 b=4 b=4  c = –3 c = −3 (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 18 DBE/November 2016 NSC – Memorandum QUESTION/VRAAG 10 10.1 M (t ) = −t 3 + 3t 2 + 72t  M(3) = −(3)3 + 3(3)2 + 72(3) M (3) = −(3) + 3(3) + 72(3) 3 2 = 216  216 (2) 216 molecules/molekules 10.2 M (t ) = −t 3 + 3t 2 + 72t  M / (t ) = −3t 2 + 6t + 72 M / (t ) = −3t 2 + 6t + 72 M / (2 ) = −3(2 ) + 6(2 ) + 72 2  M / (2 ) = 72  72 (3) 72 molecules per hour/molekules per uur 10.3 M (t ) = −t 3 + 3t 2 + 72t M / (t) = −3t 2 + 6t + 72  M // (t )  M // (t ) = 0 M // (t ) = 0 − 6t + 6 = 0  answer t =1 Maximum rate of change of the number of molecules of the drug in the bloodstream is after 1 hour./Maksimum tempo van verandering van die getal molekules in die bloedstroom is na 1 uur (3) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 19 DBE/November 2016 NSC – Memorandum QUESTION/VRAAG 11 11.1 Watches TV Do not watch Total during exams TV during exams Male 80 a 80+a Female 48 12 60 Total b 32 160 a + 12 = 32 a = 20 a = 20 b = 80 + 48 b = 128 = 128 (2) 11.2 No No 20 P(M and not watching TV) = ≠0 reason 160 (2) 128 4 128 11.3.1 P(watching TV ) = = = 0,8 = 80% 160 160 5 (2) 12 3 11.3.2 P(female and not watching TV ) = = = 0,075 = 7,5%  12 160 40 160 (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P1 20 DBE/November 2016 NSC – Memorandum QUESTION/VRAAG 12 12. We want to create codes that are even numbers greater than 5000. The digit 6 can be used in one of two places in these codes and therefore this presents two scenarios. Ons wil kodes kry wat ewe getalle groter as 5000 is. Die syfer 6 kan in twee posisies in die kode gebruik word en twee opsies is moontlik: CASE 1: The first digit is a 6./Die eerste syfer is’n 6. 2 6 4 1 × 5 × 4 × 2  1 × 5 × 4 ×2  40 Number of codes starting with 6./Getal kodes wat met 6 begin. = 1 × 5 × 4 ×2 = 40 CASE 2: The first digit is a 5 or 7./Die eerste syfer is’n 5 of 7. 2 5 4 7 6 2 × 5 × 4 × 3 Number of codes not starting with 6./Getal kodes wat nie met 6 begin = 2 × 5 × 4 ×3 = 120  2 × 5 × 4 ×3 Therefore total number of possible codes./Die totale getal  120 moontlike kodes = 40 + 120 = 160.  160 [5] OR/OF (3 × 5 × 4 × 1) + (3 × 5 × 4 × 1) + (2 × 5 × 4 × 1)  (3 × 5 × 4 ×1)  (3 × 5 × 4 ×1) = 60 + 60 + 40  (2 × 5 × 4 × 1) = 160  160 [5] OR/OF  (3 × 5 × 4 × 3) (3 × 5 × 4 × 3) − (1 × 5 × 4 × 1)  (1 × 5 × 4 × 1) = 180 − 20 = 160  160 [5] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou