Mathematics Grade 12 P3 NSC November 2012 Memo (English)

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P3 NOVEMBER 2012 MEMORANDUM MARKS: 100 This memorandum consists of 16 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 2 DBE/November 2012 NSC – Memorandum NOTE: • If a candidate answered a question TWICE, mark the FIRST attempt ONLY. • If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed out question. • Consistent accuracy applies in ALL aspects of the memorandum. QUESTION 1 1.1 The number of times the CD was played.  answer (1) Afrikaans: Getalkerewat die CD gespeel is. 1.2  all 10 points Scatter plot showing the number of times a CD was played vs the CD sales in the following week plotted correctly 5,000 2 marks if 5–9 4,500 points are plotted correctly 4,000 3,500 1 mark if 1–4 points are plotted 3,000 correctly. 2,500 (3) 2,000 1,500 1,000 500 0 0 5 10 15 20 25 30 35 40 45 50 55 1.3 a = 293,06 (293,057554...) calculating a b = 74,28 (74,28057554...) and b yˆ = 293,06 + 74,28 x  equation (4) 1.4 r = 0,95 (0,9458185...)  answer (2) 1.5 yˆ ≈ 293,06 + 74,28(45)  substitution ≈ 3635,66 Note: Penalise 1 mark for incorrect ≈ 3635 rounding off.  answer ≈ 3650 (to the nearest 50) (2) 1.6 There is a very strong positive relationship between the number of times strong that a CD was played and the sales of that CD in the following week. (1) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 3 DBE/November 2012 NSC – Memorandum QUESTION 2 2.1 Yes. The events Pass and Fail are mutually exclusive. Yes It is not possible for pass and fail to take place at the same  P(Pass and Fail) = 0 / no time. There is no intersection between the two sets. intersection between the sets. P(Pass and Fail) = 0 (2) Note: OR If a candidate answers ‘No’ P(Pass) = 0,59 then award 0 marks P(Fail) = 0,41  Yes P(Pass) + P(Fail) P(Pass and Fail) = 0 / = 0,59 + 0,41 No intersection between =1 the sets P(Pass and Fail) = 0 / No intersection of the sets The events Pass and Fail are mutually exclusive. (2) Afrikaans Ja. Die gebeurtenisse Slaag en Druip is onderling Ja uitsluitend.  P(Slaag en Druip) = 0 / geen Dit is nie moontlik dat slaag en druip gelyktydig plaasvind snyding nie. P(Slaag en Druip) = 0 (2) 2.2 PASS FAIL TOTAL Males 46 32 78 Females 72 50 122 Total 118 82 200 78 P(Male) = 78 = 0,39  P(Male) = = 0,39 or Note: 200 200 If a candidate answers ‘No’ 118 P(Pass) = 118 = 0,59 then award 0 marks P(Pass) = = 0,59 200 200 46 P(Male and Pass) = = 0,23  P(Male and Pass) = 0,23 200 P (Male) × P (Pass) = 0,39 × 0,59  P (Male) × P(Pass) = 0,23 = 0,23 (0,2301) ∴ P(Male) × P(Pass) = P(Male and Pass) conclusion ∴ Passing the competency test is independent of gender. (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 4 DBE/November 2012 NSC – Memorandum OR 120 122  P (Female) = = 0,61 or P(Female) = = 0,61 200 200 118 118 P(Pass) = = 0,59 P(Pass) = = 0,59 200 200 72  P(Female and Pass) = 0,36 P(Female and Pass) = = 0,36 200 P (Female) × P (Pass) = 0,61 × 0,59 = 0,36 (0,3599)  P (Female) × P (Pass) = 0,36 ∴ P(Female) × P(Pass) = P(Female and Pass) conclusion ∴ Passing the competency test is independent of gender. (4) [6] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 5 DBE/November 2012 NSC – Memorandum QUESTION 3 3.1 Histogram showing the frequency of the lifespan of a television (years) 20  intervals 18  3 bars correct 17  6 bars correct (3) 15 Note If the candidate 10 draws a bar graph, award max 6 2 marks 5 5 2 2 0 0 0 4,95 5,65 6,35 7,05 7,75 8,45 9,15 3.2 Lifespan (in years) Frequency Midpoint 4,95 ≤ x< 5,65 2 5,3 5,65 ≤ x< 6,35 6 6 Note: If candidate works out average (x ) of 6,35 ≤ x< 7,05 18 6,7 midpoints, answer is 7,05 then 0 marks 7,05 ≤ x< 7,75 17 7,4 7,75 ≤ x< 8,45 5 8,1 8,45 ≤ x< 9,15 2 8,8 2 × 5,3 + 6 × 6 + 18 × 6,7 + 17 × 7,4 + 5 × 8,1 + 2 × 8,8 x= frequencies × 50 midpoints 351,1 = 50  50 = 7,02 years  answer (x = 7,022) (3) 3.3 The required area is 98% to the right of some value. This value is at  x − 2σ 2 standard deviations on the left of the mean. x − 2σ  7,02 − 2(0,76) = 7,02 − 2(0,76)  answer = 5,5 years (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 6 DBE/November 2012 NSC – Memorandum 3.4 They can issue a 5-year guarantee.  Issue the 5-year The average lifespan of a set is 7,02 years - which is in excess of 5 guarantee years. 98% of the sets lasted for more than 5,5 years.  reason Very few sets have lasted less than 5 years. The number of sets of this (2) brand that will be returnedshould be minimal if a 5-year guarantee is issued. Afrikaans Hullekan ‘n 5 jaar-waarborguitreik. kan ‘n 5 jaar- Die gemiddelde lewens duur van 'n televisiestel is 7,02 jaar –wat waarborg uitreik 5 jaar oorskry. 98% van die stelle het langer as 5,5jaargehou. rede 'n Klein aantal stelle het vir minder as 5 jaar gehou. Die aantal stele (2) wat terug geneem sal moet word sal minimal wees indien 'n 5 jaar- waarborg uitgereik word. [11] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 7 DBE/November 2012 NSC – Memorandum QUESTION 4 4.1 Outcome(Sunny, cycle) Sunny branch Cycle 7/ Rainybranch 10  cycle, drive, train Drive 1/5 branches on both Outcome(Sunny, drive) weather types Sunny 1/  probabilities listed 4 10 /7 Train  outcomes listed Outcome(Sunny, train) (5) Outcome(Rainy, cycle) 1 Cycle /9 Rainy 3 5 /7 Drive /9 Outcome(Rainy, drive) Train 1 /3 Outcome(Rainy, train) OR Outcome(Sunny, cycle) Cycle 0,7 Drive 0,2 Outcome(Sunny, drive) Sunny 0,1 0,571428... Train Outcome(Sunny, train) Outcome(Rainy, cycle) 0,111111... Cycle Rainy 0,5555... 0,42857.. Drive . Outcome(Rainy, drive) Train 0,33333... Outcome(Rainy, train) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 8 DBE/November 2012 NSC – Memorandum 4.2.1 P(Rainy, Cycle) 3 1 Note: 3 1 = ×  × 7 9 3 1 7 9 If + then 0 marks  answerin any 1 7 9 = form 21 (must be from OR multiplication) P(Rainy, Cycle) (2) = 0,428... × 0,1111... = 0,04761904762 ≈ 0,05 or 4,76% 4.2.2 P(Train) P(Train) 4 3 1 4 3 1 = × 0,1 + ×  × 0,1 and × 4 3 1 7 7 3 7 7 3 = × 0,1 + × 7 7 3 = 0,05714... + 0,1428...  addition 1 1  answer = = (in any form) 5 5 (3) = 0,2 OR = 0,2 = 20% = 20% 4.3 4 3 5 4 3 5 P(Drive) = × 0,2 + ×  × 0,2 and × 7 7 9 7 7 9 37  addition = 105 37  = 0,35238... 105 37  answer Vusi drives for × 245 = 87 days (86,333...) (4) 105 Accept: 86 days 4 3 5 OR  × 0,2 and × 4 3 5 7 7 9 P(Drive) = × 0,2 × 245 + × × 245  addition 7 7 9  28 + 58,333 = 28 + 58,333  answer = 87 days (86,333...) (4) Accept: 86 days [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 9 DBE/November 2012 NSC – Memorandum QUESTION 5 5.1.1 Number of PIN codes = 10 × 10 × 10 × 10 × 10 10 5 = 10  answer = 100 000 (2) 5.1.2 Number of PIN codes = 10 × 9 × 8 × 7 × 6  multiplication = 30 240  answer (2) OR Number of PIN codes 10! 10! =  5! 5! = 30 240  answer (2) 5.2 Number of PINs that DO NOT contain 9s = 9×9×9×9×9 9 = 59 049  59 049 P(at least one 9) = 1 − P(no 9s) 59049 1 − 59049 100000 = 1−  answer 100000 (4) = 0,41 OR Number of PINs that DO NOT contain 9s 9 = 9×9×9×9×9  59 049 = 59 049 Number of PINs that contain AT LEAST one 9  40951 = 100 000 – 59 049 = 40 951 P(at least one 9) 40951 = 100000 = 0,41  answer (4) [8] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 10 DBE/November 2012 NSC – Memorandum QUESTION 6 6.1 Tk +1 = 2Tk + 3 where T1 = 1 , k ≥ 1  Tk +1 = 2Tk + 3  T1 = 1  k ≥1 (4) OR Tk +1 = Tk + 2 k +1 where T1 = 1 , k ≥ 1  Tk +1 = Tk + 2 k +1  T1 = 1  k ≥1 (4) OR  Tk +2 = 2(Tk +1 − Tk ) + Tk +1 where T1 = 1 , T2 = 5 , k ≥ 1 Tk +2 = 2(Tk +1 − Tk ) + Tk +1  T1 = 1 T2 = 5  k ≥1 (4) 6.2 4 7 13 24 44 2+1 22 + 2 23 + 3 24 + 4 The next term of the sequence is 44 + 2 5 + 5 answer = 81 OR 4 7 13 24 44 ? 4 7 13 24 44 79 3 6 11 20 35 3 5 9 15 2 4 6 The next term of the sequence is 79. answer Note: This sequence can be represented by the following recursive formula: 1 11 (2) Tn +1 = Tn + n3 − n 2 + n = where T1 4 and n ≥ 1 [6] 3 3 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 11 DBE/November 2012 NSC – Memorandum QUESTION 7 F L P Q M N G H 7.1 Draw a point P on FG such that FP = LM and a point Q on FH  construction such that FQ = LN. Note: No construction In ∆FPQ and ∆LMN constitutes a breakdown, hence no marks All three statements 1. F̂ = L̂ (given) 2. FP = LM (construction) must be given 3. FQ = LN (construction) ∴∆FPQ ≡∆LMN (SAS) ∆FPQ ≡∆LMN (SAS) FP̂Q = LM̂N (≡∆s)  FP̂Q = LM̂N But FĜH = LM̂N (given) FP̂Q = FĜH  FP̂Q = FĜH PQ || GH (corresponding angles =) FP FQ  PQ || GH = (PQ || GH ; Prop Th) FP FQ FG FH  = FG FH LM LN = (7) FG FH Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 12 DBE/November 2012 NSC – Memorandum V 2x – 10 4 2x – 10 T P 66 99 9 K R 7.2 VP VT VP VT = (PT || RK;Prop Th)  = PR TK PR TK 2 x − 10 4  (PT || RK; Prop Th) =  substitution 9 6 2 x − 10 = 6 2 x = 16 x=8  answer (4) OR VP VT VP VT = (PT || RK; Prop Th)  = VR VK VR VK 2 x − 10 4  (PT || RK; Prop Th) = 2 x − 1 10  substitution 20 x − 100 = 8 x − 4 12 x = 96  answer x=8 (4) [11] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 13 DBE/November 2012 NSC – Memorandum QUESTION 8 8.1 ... equal to the angle subtended by the chord in the alternate  answer segment. (1) 8.2 R Q 1 1 76° 41° 2 3 29° 29° M 117° 117° 63° 75° 34° 105° c 34° 41° 1 d b 76° U P a T 29°  a = 29°  tan ch. thm W  QP̂R = 34° ∠s in same seg a = 29° (tan ch.thm)  c = 41°  b = 76° QP̂R = 34° (∠s in same seg)  Q̂1 = 76°  d = 105° c = 41° ext∠ cyclic quad b = 76° (adj∠s on str. line) (9) Q̂1 = 76° (∠s in same seg) d = 105° (ext∠ cyclic quad)  a = 29°  tan ch. thm OR  T̂1 = c a = 29° (tan ch. thm)  tan ch. thm  c + 34° = 75° T̂1 = c (tan ch. thm)  tan ch. thm c + 34° = 75° (tan ch. thm)  c = 41° c = 41°  b = 76°  d = 105° b = 76° (adj∠s on str. line) (9) d = 105° (adj∠s on str. line) OR An alternative solution for calculating d: Q̂1 = RP̂T = 76° (∠s in same seg) [10] d + RT̂Q = PQ̂T + QP̂R + RP̂T (ext∠∆) d + 34° = 29° + 34° + 76° d = 105° Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 14 DBE/November 2012 NSC – Memorandum QUESTION 9 C x B 1 4x T 2 3 360° – 2x x 1 O 180° – x2 2x K 1 2 3 A 9.1 AÔB = 2 x (∠circ centre = 2 ∠ circumference)  AÔB = 2 x T̂ = 180° − 2 x (opp∠ cyclic quad suppl) ∠circ centre = 2 ∠ circumference opp∠ cyclic quad suppl (3) 9.2 CÂT = x (∠ sum ∆)  CÂT = x K̂ 1 = x (ext∠ cyclic quad) ∠ sum ∆ CÂT = K̂ 1  K̂1 = x BK || AC (corresponding ∠s =) ext∠ cyclic quad  corresponding ∠s = OR (5) K̂1 = Ĉ = x (ext∠ cyclic quad)  K̂1 = Ĉ = x B̂ 4 = x (∠ sum ∆) ext∠ cyclic quad  B̂ 4 = x B̂ 4 = Ĉ = x ∠ sum ∆ BK || CA (corresponding ∠s =)  corresponding ∠s = (5) OR  CÂT = x CÂT = x (∠ sum ∆) ∠ sum ∆ BK̂A = 180° − x (opp∠ cyclic quad)  BK̂A = 180° − x CÂT + BK̂A = 180° opp∠ cyclic quad BK || AC (coint∠s supp) co-int∠s supp (5) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 15 DBE/November 2012 NSC – Memorandum 9.3 In ∆BKT and ∆CAT 1. CÂT = K̂ 1 (= x)  CÂT = K̂ 1 2. T̂ is common  T̂ is common 3. AĈT = B̂4 (∠ sum ∆) ∆BKT ||| ∆CAT (∠∠∠) ∠∠∠ (3) 9.4 AC AT AC AT = (||| ∆s)  = KB KT KB KT AC 7  ||| ∆s =  answer KB 2 (3) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P3 16 DBE/November 2012 NSC – Memorandum QUESTION 10 B D 6,5x O 2,5x 4x M C A 10.1 DC = 13x  CD = 13 x (1) 10.2 13 13 OD = x  OD = x 2 2 5 OM = x  answer 2 (2) 10.3 BO = OD (radii) AM = MB = 12 units (line from circ cent ⊥ch) MB = 12 2 2 13   12 2 +  x  =  x  2 2  5   13  5 12 +  x  =  x  2 (Pythagoras) 2   2  2   2  or 12 + 6,25 x 2 = 42,25 x 2 2 25 x 2 169 x 2 25 2 169 2 144 + = or 12 2 + x = x 4 4 4 4 144 x 2 144 = 4 x =4 2 x = ±2 x=2 ( x > 0)  answer The radius = 13 (2) 2 = 13 units.  answer (4) [7] Copyright reserved