Mathematics Grade 12 Paper 3 NSC November 2013 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 MATHEMATICS P3/WISKUNDE V3 NOVEMBER 2013 MEMORANDUM MARKS/PUNTE: 100 This memorandum consists of 14 pages. Hierdie memorandum bestaan uit 14 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P3/Wiskunde V3 2 DBE/November 2013 NSC/NSS – Memorandum NOTE: • If a candidate answered a question TWICE, mark only the first attempt. • 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 marking memorandum. LET WEL: • As 'n kandidaat 'n vraag TWEE keer beantwoord het, merk slegs die eerste poging. • As 'n kandidaat 'n antwoord deurgehaal en nie oorgedoen het nie, merk die deurgehaalde antwoord. • Volgehoue akkuraatheid is DEURGAANS in ALLE aspekte van die memorandum van toepassing. QUESTION/VRAAG 1 Calories 270 310 340 360 440 450 580 620 Total fat (in grams) 14 12 25 21 28 24 32 43 1.1 Scatter Plot showing number of calories versus total fat (in sandwiches 45 40 35 Total Fat (in grams) /Totale vetinhoud 30 25 (in gram) 20 15 10  all 8 points plotted correctly NOTE: • The English and Afrikaans diagrams are 2 marks for 5 – 7 points plotted different in size correctly • The point (300 ; 34) is from 1.4. Do not penalise for this point in question 1.1. 1 mark for 2 – 4 points plotted • The learner may have other points due to correctly trying to draw the least squares regression line No marks for 1 point or less plotted correctly. (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P3/Wiskunde V3 3 DBE/November 2013 NSC/NSS – Memorandum 1.2 a = – 5,62 (– 5,62181859...)  a or b NOTE: b = 0,07 (0,072396008...)  b or a Penalty 1 for rounding yˆ = −5,62 + 0,07 x  equation in question 1.2. (4) 1.3 Drawn on the scatter plot  graph passing close to (350 ; 19) / (300 ; 16) / (650 ; 40)  gradient (graph close to (450 ; 26)) (2) 1.4 The sandwich from Pinky’s restaurant has 30 grams of fat. In terms of the model, the sandwich with 16 grams of total  comparison with model (literal or fat should provide 300 calories. This suggests that the graphical (i.e. plotting the point)) sandwich from Pinky’s restaurant is far removed from expected model and is therefore an outlier.  outlier / explain this concept (2) Die toebroodjie van Pinky se restaurant het ‘n vetinhoud van 30 gram. In terme van die model, sal ‘n toebroodjie van  vergelyking met model 16 gram, 300 kalorieë verskaf. Dit veronderstel dat die toebroodjie van Pinky se restaurant ver van die verwagte  uitskieter model verwyder is en is dus ‘n uitskieter. (2) 1.5 r = 0,92 (0,92099548...)  answer (2) 1.6 There is a very strong, positive correlation between the  strong / reference to correlation number of calories and the total fat in the sandwiches. A  positive / increasing / explain the sandwich with higher number of calories will have a higher concept of increasing total fat content. (2) Daar is ‘n baie sterk positiewe korrelasie tussen die aantal kalorieë en die totale vetinhoud in die toebroodjies. ‘n Toebroodjie met ‘n hoë aantal kalorieë sal ‘n hoër totale vetinhoud hê. [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P3/Wiskunde V3 4 DBE/November 2013 NSC/NSS – Memorandum QUESTION/VRAAG 2 2.1 Random Sampling Method  answer (1) Ewekansige Steekproef Metode / Willekeurig / Onwillekeurig/ Lukraak 2.2 Although the sample is selected randomly, there exists a chance that this could be a homogenous sample.  homogenous sample / For example the sample could contain many learners from one the sample will be biased grade and few from other grades. towards one characteristic The sample could contain many more learners of a particular (1) gender and few from the other. Al is die steekproef ewekansig gedoen bestaan die kans wel dat die steekproef homogeen kan wees. Ter voorbeeld, die steekproef kan baie leerders van een graad insluit en min van  homogene steekproef ander grade. Die steekproef kan ook meer leerders van ‘n bepaalde geslag (1) insluit en min van die ander geslag. 2.3 The sampling should have equal representation across various  equal representation of criteria. In this case the criteria will possibly be grade and grade gender. There should possibly be a selection of 8 boys and 8  equal representation of girls randomly from each grade for the sample. gender Gelyke verteenwoordiging van verskillende kategorieë moet in any explanation that die steekproef teenwoordig wees. In die geval sal die indicates representing the kategorieë graad en geslag wees. Daar behoort moontlik 8 school for equally seuns en 8 meisies van elke graad ewekansig gekies te word Ideas to consider: vir die steekproef. gender; race; grade (2) [4] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P3/Wiskunde V3 5 DBE/November 2013 NSC/NSS – Memorandum QUESTION/VRAAG 3 3.1 Disagree. The events A and B have an intersection and are  disagree therefore not mutually exclusive.  intersection between events P(A and B) ≠ 0 exist (2) Verskil. Die gebeurtenisse A en B sny mekaar en is daarom nie onderling uitsluitend nie. P(A en B) ≠ 0 OR Disagree  disagree P(A) + P(B) + P(C) = 1,07 Since the sum of the probabilities of the events is greater than  intersection between events 1, these events must have an intersection. Hence A and B are exist not mutually exclusive events. (2) Verskil P(A) + P(B) + P(C) = 1,07 Siende dat die som van die waarskynlikhede van al die gebeurtenisse groter as 1 is, moet daar ‘n snyding wees. Vervolgens sal A en B nie onderling uitsluitend wees nie. 3.2.1 P(B or C) = P(B) + P(C)  substitution into correct = 0,3 + 0,32 NOTE: formula = 0,62 Answer only: full marks  answer OR (2) A C B 0,32 0,315 0,135 0,165 P(A and B) = P(A).P(B)  substitution into correct = 0,45 × 0,3 formula = 0,135  answer P(B or C) = 0,135 + 0,165 + 0,32 (2) = 0,62 3.2.2 P(A and B) = P(A).P(B) NOTE: = 0,45 × 0,3 Answer only: 1/2 = 0,135 marks  0,135 P(A or B) = P(A) + P(B) – P(A and B) = 0,45 + 0,3 – 0,135 = 0,615  answer = 0,62 (2) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P3/Wiskunde V3 6 DBE/November 2013 NSC/NSS – Memorandum QUESTION/VRAAG 4 4.1 The word “EQUATIONS” has 9 letters Number of different 5 letter codes  multiplication rule in five = 9×8× 7× 6×5 “slots” = 15120  answer (2) OR Number of different 5 letter codes 9!  multiplication rule in five = 4! NOTE: “slots” = 15120 Answer only: 2/2 marks  answer (2) ACCEPT: Number of different 5 letter codes (this is with repetition) = 95 = 59049 4.2 4 consonants 5 vowels Number of different codes OR Number of different codes 5 = 5 × 5 × 4 × 3 × 2 ×1 = 5 × 5!  5! = 600 = 600  answer (3) [5] QUESTION/VRAAG 5 5.1 The heights / frequency of the histogram for the interval  heights / frequency 500 ≤ x < 700 are incorrect. incorrect Die hoogtes / frekwensies van die histogram vir die wydte (2) 500 ≤ x < 700 is nie korrek nie. OR  width not taken into The histogram has not taken into account the width of the account classes. The first class is double the width of the other classes. The area of the rectangles for the class 500 ≤ x < 700 is double (2) what it should be. Die histogram het nie die wydte van die klasse in ag geneem nie. Die eerste klas is dubbel die wydte van die ander klasse. Die oppervlakte van die reghoeke vir die klas 500 ≤ x < 700, is dubbel wat dit behoort te wees. OR Area of each rectangle is not in the same ratio as the  area of rectangle not in frequencies. same ratio as frequencies Die area van elke reghoek is nie in dieselfde verhouding as (2) die frekwensies nie. 5.2 The interval 500 ≤ x < 700 should have a frequency of 5.  frequency should be 5  keep interval the same Die frekwensie behoort 5 vir die klas 500 ≤ x < 700 te wees. (2) [4] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P3/Wiskunde V3 7 DBE/November 2013 NSC/NSS – Memorandum QUESTION/VRAAG 6 6.1 A/M N/W    100 470 670  40  80 70 x P/O (6) 6.2 x + 80 + 40 + 70 + 470 + 100 + 670 = 2140  sum of values in Venn x + 1430 = 2140 diagram = 2140 x = 710  answer for x (must not be Number said lack of parental support negative) = 710 + 80 + 40 + 70 = 900  answer (3) 6.3 P(exactly 2 problems) 80 + 100 + 70  80 + 100 + 70 = 250 = 2140  2140 250 = 2140 25 = 214  answer = 0,12 (0,1168224299...) (3) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P3/Wiskunde V3 8 DBE/November 2013 NSC/NSS – Memorandum QUESTION/VRAAG 7 34% 34% 14% 14% 2% 2% 1 4 7 10 13 16 19 7.1.1 16 = 10 + 2(3)  2 std deviations from mean. P(More than 16 point) = 2%  2% (2) 7.1.2 84 NOTE:  84% × 200 = 168 learners  answer 100 If 16% is used: max 1 mark (2) 7.2 Mathematics test x + σ = 55 + 15 = 70  x + σ = 70 English test x + σ = 55 + (7,5) = 62,5  x + σ = 62,5 x + 2σ = 55 + 2(7,5) = 70 Matilda scored better relative to her classmates in the English  answer test as she scored between the 1st and the 2nd deviation from (3) the mean. Matilda het beter as die res van haar klasmaats in die Engelse toets gevaar aangesien haar punt tussen die 1ste en die 2de afwyking van die gemiddeld was. OR A mark of 67% in Maths lies within 1 standard deviation to the right of the mean. Hence it is approximately at the 76th  76th percentile percentile of distribution of the Maths marks. A mark of 67% in English lies between 1 and 2 standard deviations to the right of the mean. Hence it is approximately  95th percentile at the 95th percentile of distribution of the English marks. Matilda scored better relative to her classmates in the English test as she has outperformed more learners in English than in  answer Maths. (3) ‘n Punt van 67% in Wiskunde lê binne 1 standaardafwyking aan die regterkant van die gemiddeld. Vervolgens is dit ongeveer by die 76 ste persentiel van die verspreiding van die Wisk punte.’n Punt van 67% in Engels lê tussen die 1ste en 2de standaardafwyking regs van die gemiddeld. Vervolgens is die Engelse punt ongeveer by die 95 ste persentiel van die verspreiding van die Engelse punte. Matilda het relatief tot haar klasmaats beter in die Engelse toets gevaar. Sy het beter gevaar as meer van die leerders in Engels as in Wiskunde. [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P3/Wiskunde V3 9 DBE/November 2013 NSC/NSS – Memorandum QUESTION/VRAAG 8 8.1 T1 = −5 T2 = −5 + 3(2 ) − 4 = −3  T2 = −3 T3 = −3 + 3(3) − 4 = 2  T3 = 2 T4 = 2 + 3(4 ) − 4 = 10  T4 = 10 (3) 8.2 T30 = T1 + S 29 of the linear pattern  T1 + S 29 of the linear pattern 29 a=2 = −5 + [2(2) + 28(3)] d=3 2 = −5 + 1276 = 1271  answer (4) OR 2a = 3 3 3 a=  a= 2 2 3 3  + b = 2 2 5 5 b=−  b=− 2 2 3  5   +  −  + c = −5 2  2 c = −4  c = −4 3 5 Tn = n 2 − n − 4 2 2 3 5 T30 = (30) − (30) − 4 2 2 2 = 1271  answer OR (4) By expanding the formula: − 5 ; – 3 ; 2 ; 10 ; 21 ; 35 ; 52 ; 72 ; 95 ; 121 ; 150 ; 182 ; 217 ; 255 ; 296 ; 340 ; 387 ; 437 ; 490 ; 546 ; 605 ; 667 ; 732 ; 800 ; 871; 945 ; 1 022 ; 1 102 ; 1 185 ; 1 271  expansion th The 30 term = 1 271 NOTE:  answer Answer only: 4 marks (4) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P3/Wiskunde V3 10 DBE/November 2013 NSC/NSS – Memorandum QUESTION/VRAAG 9 A 2x B x C F E D 9.1 AF : FE =2:1 (Prop Th; FB || EC)  answer (Eweredigheid St; FB || EC)  reason (2) 9.2 AF 2 = FE 1 AF 8 FE = = = 4 cm  FE = 4 cm 2 2  AE = 12 cm AE = 12 cm ED 1 ED 1 = (BE || DC; Prop Th) / (BE || DC; Eweredigheid St)  = AE 2 AE 2 ED 1 = 12 2  answer ED = 6 cm (4) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P3/Wiskunde V3 11 DBE/November 2013 NSC/NSS – Memorandum QUESTION/VRAAG 10 B F 1 3 2 G •O C 4 56° 5 3 E 2 1 32° D H 10.1 Ê 2 = 90° – 56° (∠s in a semi circle)  Ê 2 = 34° Ê 2 = 34° (∠e in ‘n halwe sirkel)  ∠s in a semi circle (2) 10.2 CÊH = Ê1 + Ê 2 = 66°  CÊH = 66° ∴ EB̂C = B̂2 + B̂3 = 66° (tan ch th)  EB̂C = 66° (raaklyn koord) tan ch th OR (3) B̂ 2 = 32° (tan ch th)  tan ch th (raaklyn koord) B̂ 3 = 34° (∠s in the same segment)  B̂ 3 = 34° (∠e in selfde seg) EB̂C = 66°  answer (3) 10.3 Ê1 + Ê 2 + Ê 3 = 122°  F̂ = 122° F̂ = 122° (tan ch th) / (raaklyn koord)  reason OR Ĉ = 58° (sum of int ∠s of ∆) / (som van binne∠e ∆)  Ĉ = 58° F̂ = 122° (opp ∠s cyclic quad) / (oorst ∠ koordevierhoek)  ∠ sum ∆  F̂ = 122°  opp ∠s cyclic quad OR  D̂ = 58° D̂ = 58° (sum of int ∠s of ∆) / (som van binne∠e ∆)  ∠ sum ∆  F̂ = 122° F̂ = 122° (opp ∠s cyclic quad) / (oorst ∠ koordevierhoek)  opp ∠s cyclic quad (4) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P3/Wiskunde V3 12 DBE/November 2013 NSC/NSS – Memorandum QUESTION/VRAAG 11 O A B C Construct radii OA and OC.  construction In ∆OAB and ∆OCB i. OB is common  OB common ii. OA = OC (radii)  radii iii. OB̂A = OB̂C = 90° (given)  OB̂A = OB̂C = 90° ∆OAB ≡ ∆OCB (90°HS)  90°HS AB = BC (≡ ∆s) OR Construct radii OA and OC.  construction In ∆OAB and ∆OCB i. OB is common OR OA = OC (radii)  OB common / ii. Â = Ĉ (∠s opp = radii) OA = OC  ∠s opp = radii iii. OB̂A = OB̂C = 90° (given) ∆OAB ≡ ∆OCB (SAA)  OB̂A = OB̂C = 90°  SAA AB = BC (≡ ∆s) OR  construction Construct radii OA and OC. OB̂A = OB̂C = 90° (given) OA2 = OB2 + AB2 (Pythagoras)  OA2 = OB2 + AB2 OC2 = OB2 + BC2 (Pythagoras)  OC2 = OB2 + BC2 OC = OA (radii)  OA = OC / radii OB is common AB2 = BC2  AB2 = BC2 AB = BC [5] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P3/Wiskunde V3 13 DBE/November 2013 NSC/NSS – Memorandum QUESTION/VRAAG 12 K 1 2 3 x x 1 E 2 2 1 O3 4 x T 2 3 x 1 1 x 2 Y W 12.1 K̂ 3 = x (tan ch th)  K̂ 3 = x (raaklyn koord)  tan ch th Ŷ2 = x (∠s in same seg)  Ŷ2 = x (∠e in selfde seg)  ∠s in same seg K̂ 2 = x (∠s opp = radii) / (= chs subt = ∠s)  K̂ 2 = x (∠e oork = radiusse) / (= koorde = ∠e)  reason Ŵ2 = x (∠s in same seg) / (= chs subt = ∠s)  Ŵ2 = x (∠e in selfde seg) / (= koorde = ∠e)  reason (8) 12.2 Ô1 + Ô 4 = 180° − 2 x (sum of int ∠’s of ∆) / (opp∠ cyclic quad)  Ô1 + Ô 4 = 180° − 2 x (som van binne∠e ∆) / (oorst ∠ koordevierhoek)  reason T̂ = 90° − x (∠ at circ cent = 2∠ at circumference)  reason (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P3/Wiskunde V3 14 DBE/November 2013 NSC/NSS – Memorandum 12.3 Ê 2 = 90° (sum of int ∠’s of ∆) / (som van binne∠e ∆)  Ê 2 = 90° KE = ET (⊥ from centre to chord bisects chord)  sum of int ∠’s of ∆ (⊥ van middelpunt tot koord halveer koord)  ⊥ from centre to chord bisects chord OR (3) K̂1 + K̂ 2 + K̂ 3 = 90° − x (sum of int ∠’s of ∆)  K̂1 + K̂ 2 + K̂ 3 = 90° − x In ∆KWE and ∆TEW  sum of int ∠’s of ∆ 1. K̂1 + K̂ 2 + K̂ 3 = T̂ = 90° − x (proven above) 2. Ŵ1 = Ŵ2 = x (Proven in 12.1) 3. WE is common ∴ ∆KWE ≡ ∆TEW (∠∠S)  ∆KWE ≡ ∆TEW KE = ET (3) 12.4 In ∆KOE and ∆WTE  ∆KOE and ∆WTE i. K̂ 3 = Ŵ2 = x (proven)  K̂ 3 = Ŵ2 = x ii. Ê 2 = Ê 1 = 90° (∠s on str line / sum of int ∠’s of ∆)  Ê 2 = Ê 1 = 90° (∠e op reguit lyn / som van binne∠e ∆) iii. Ô 2 = T̂ = 90° − x (3rd ∠ of ∆)  Ô 2 = T̂ = 90° − x ∆KOE ||| ∆WTE (∠∠∠) KE OE KE OE = (||| ∆s)  = WE TE WE TE KE = TE (proven) KE.TE = OE.WE  KE = TE KE 2 = OE.WE (6) OR  ∆KOE and ∆WTE In ∆KOE and ∆KWE  K̂ 3 = Ŵ1 = x i. K̂ 3 = Ŵ1 = x (proven)  Ê1 is common ii. Ê1 is common iii. Ô 2 = K̂1 + K̂ 2 + K̂ 3 (3rd ∠ of ∆)  ∠∠∠ ∆KOE ||| ∆WKE (∠∠∠) KE OE KE OE = (||| ∆s)  = WE KE WE KE KE 2 = OE.WE (6) [20] TOTAL/TOTAAL: 100 Copyright reserved/Kopiereg voorbehou