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17 NATIONAL SENI CERTIFICATE GRADE 12 MATHEMATICAL LITERACY P1 NOVEMBER 2015 MEMANDUM MARKS: 150 Codes M MA CA A C D J S RD F SF O P R NP Explanation Method Method with Accuracy Consistent Accuracy Accuracy Conversion Define Justification/Reason/Explain Simplification Reading from a table a graph a diagram a map a plan Choosing the correct formula Substitution in a formula Opinion Penalty, e.g. for no units, incorrect rounding off, etc. Rounding Off No penalty for rounding omitting units This memorandum consists of 17 pages. Please turn over

18 Mathematical Literacy/P1 2 DBE/November 2015 NSC Memorandum KEY TO TOPIC SYMBOL: F = Finance; M = Measurement; MP = Maps, Plans and other representations DH = Data Handling; P = Probability QUESTION 1 [38] Ques Solution Explanation Level A = 150 1MA multiply by 2 and adding 16 1CA simplifying Answer only full marks L Cost = R225, = R Number of persons = R R225 = 152 (150 guests + bridal couple) 1M multiply by R225 1A for 152 1M divide by R225 1A number of persons L Cost per person = R = R225 R % Reception costs = 100% R M divide by 152 1A cost per person 1M correct fraction L1 = 53,16% 1CA percentage Answer only full marks NP rounding Flowers and decor = 1,8% R = R M percentage 1A amount Answer only full marks L1 Please turn over

19 Mathematical Literacy/P1 3 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level Rand value = GHS ,32253 R93 014,60 Shortfall = R R93 014,60 = R31 985,40 Cedi value = R ,32253 A = GHS 40316,25 Shortfall = GHS ,25 GHS = GHS ,25 Rand value = GHS ,25 0,32253 = R31 985,40 1M divide 1A correct rounding 1M subtraction 1CA amount 1MA multiply 1M subtraction 1A shortfall amount 1CA amount Answer only full marks NP rounding L2 (4) R1 349 = R188, Cost including VAT = R R188,86 = R1 537,86 Selling price in cedi = R1 537,86 0, A multiply by 14% 1M adding amount 1A amount with VAT 1M multiply by 0, CA value to nearest cedi L1 VAT inclusive cost = R ,14 = R1 537,86 Selling price in cedi = 1 537,86 0, A working with 14% 1M multiply by 1,14 1A amount with VAT 1M multiply by 0, CA value to nearest cedi Price in cedi = ,32253 = 435,09 1M multiply by 0, A cedi price Selling price including VAT in cedi = 435, , A working with 14% 1M multiply by 1,14 1CA value to nearest cedi Answer only full marks (5) Please turn over

20 Mathematical Literacy/P1 4 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level J Photographer (video) to create memories of the wedding day Wedding attire usually special wedding attire are required Wedding contract to pay for the lawyer s fees for drawing up the contract Gifts as a token for members who serve DJ to provide for the music at the reception (accept any valid wedding expense with an explanation ) Employee works and receives money for the work done D Employer is a person or institution that hires workers and pays wages/salary for work done D 1A wedding expense 1J explanation 1D employee 1D employer Unemployment Insurance Fund D 2D expanding R A amount No No amount allocated E Monthly tax credit = R A = R230 1A correct statement 1E reason 1MA divide correct value by 12 1CA monthly tax credit L1 L2 L1 L1 L1 L1 L1 Answer only full marks A = R R R8 640 = R M correct values 1CA total deductions L1 Answer only full marks Please turn over

21 Mathematical Literacy/P1 5 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level Gross non-retirement funding income = R R R8 640 = R Adding the amounts with source codes 3605, 3713 and 3810 Adding the annual payment other allowances and medical aid contributions Remaining monthly contributions = R R4 975,25 = R8 933,75 Average monthly contribution = R8 933,75 7 = R1 276,25 1M using the correct values/codes/words 1A addition 1A R CA subtracting R4 975,25 1M dividing the remaining amount 1A by 7 1CA pension per month (only if division by 4,5,6,7) Answer only full marks L1 L2 (5) [38] Please turn over

22 Mathematical Literacy/P1 6 DBE/November 2015 NSC Memorandum QUESTION 2 [31] Ques Solution Explanation Level SF Total area of a rectangular piece = 30 cm 12 cm = 360 cm² A Off-cut piece = 360 cm² 355,25 cm² = 4,75 cm² 1SF substitution 1A simplifying 1M subtraction 1CA area of off-cut L3 Total off-cut piece for both sides = 4,75 cm² 2 = 9,5 cm² SF Total area of 2 rectangular pieces = 2 30 cm 12 cm = 720 cm² A Area of both sides of stocking = 355,25 cm² 2 = 710,5 cm² Total off-cut piece = 720 cm² 710,5 cm² = 9,5 cm² Total off-cut area SF = (2 30 cm 12 cm) (355,25 cm² 2) A = 720 cm² 710,5 cm² = 9,5 cm² 1M multiply by 2 1CA area of off-cut 1SF substitution 1M multiply by 2 1A simplifying 1M multiply by 2 1M subtraction 1CA area of off-cut 1SF substitution 1M multiply by 2 1M multiply by 2 1A simplifying 1M subtraction 1CA area of off-cut Answer only full marks (6) Please turn over

23 Mathematical Literacy/P1 7 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level L SF 1 Area of a triangle = 3cm 5cm 1 SF substitution 2 = 7,5 cm² 1A simplifying Area of 6 triangles = 7,5 cm² 6 = 45 cm² SF 1 Area of triangles = 3cm 5cm 6 2 = 7,5 cm² 6 = 45 cm² 1M multiply by 6 1CA total area 1 SF substitution 1M multiply by 6 1A simplifying 1CA total area Answer only full marks Time taken = 9 18 minutes = 162 minutes A = 2 h 42 min 2,7 h C Finishing time = 08:25 + 2h42 = 11:07 NP -units (4) 1MA time in minutes 1C converting time 1M adding 1CA finishing time correct notation Answer only full marks L2 Two marks for 11: xx (4) Please turn over

24 Mathematical Literacy/P1 8 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level 2.2 Number of reels along length = 195 mm 23mm = 8, R Number of reels along breadth = 120 mm 23mm = 5, R Total = 5 8 = 40 1M dividing length by diameter 1A diameter 1R number rounded down 1R number rounded down 1CA total number Full marks for Total = 5 8 = 40 Max of 2 marks if divided by circle s area Max of 3 marks if divided by square area 1 mark for area of rectangle only L Painted surface area of the lid SF = 3,142 3,6 cm (3, ,9) cm 61 cm² Painted surface area of the lid SF = 3, mm ( ) mm = 6108,05 mm² 61 cm² C C 1A radius 1SF substitution 1C conversion 1CA surface area to nearest cm 2 1A radius 1SF substitution 1CA surface area to nearest cm 2 1C conversion (5) L2 Max of 3 marks if inner radius used Max of 2 marks if units are mixed (4) Please turn over

25 Mathematical Literacy/P1 9 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level Capacity = 75% 250 ml = 187,5 ml Volume = 187,5 cm³ Height of the water in the jar 3 Volume of the water (in cm ) = 2 π radius 3 187,5cm = SF 2 3,142 (3,25 cm) 3 187,5 cm 2 = 33, cm = 5,6497 cm 6 cm R 1M multiply by 75% 1CA capacity in ml 2SF substitution 1CA simplification 1R nearest cm L2 3 Volume of the water (in cm ) π radius 3 250cm SF 3,142 (3,25 cm) = 2 = cm 2 = 33, cm = 7,532 cm 2SF substitution 1CA simplification Height of the water in the jar = 75% 7,532...cm = 5,6497 cm 6 cm R 1M multiply by 75% 1CA height of water 1R nearest cm Answer only full marks = = M multiply by 2 1A fraction Accept 2 16 (6) L1 Answer only full marks [31] Please turn over

26 Mathematical Literacy/P1 10 DBE/November 2015 NSC Memorandum QUESTION 3 [24] Ques Solution Explanation Level Exit 3 RD 2RD reading from plan J No, there is no power outlet available in that seat 1A answer 1J reason RD C 109 RD 1RD correct row 1RD correct seat number Total seats = seats one side + seats in middle + seats other side = ( )+( ) + ( ) A A A = = 345 3MA adding correct number of seats in each section 1CA total seats Answer only full marks Max 2 marks if answer only 344 or 346 L1 L1 L2 L and 110 RD 2RD seat numbers Number of seats with access to a power supply = Probability = 345 (4) 1A counting seat 1CA numerator 1CA writing as a denominator from Max Answer only full marks L1 L times RD [Free State 15 times] (3) 2RD reading from map If 13 one mark L1 Please turn over

27 Mathematical Literacy/P1 11 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level Distance = 94,7 km 76 km = 18,7 km A 1MA subtracting from 94,7 1A distance Answer only full marks L Blue Hills RD 2RD reading from map RD RD WP 4, WP 5, WP 6 RD 3RD reading from map L1 L1 WP3 to WP4, WP 4 to WP5, WP5 to WP6 RD 3RD reading from map 2 marks for W4 to W6 (3) [24] Please turn over

28 Mathematical Literacy/P1 12 DBE/November 2015 NSC Memorandum QUESTION 4 [30] Ques Solution Explanation Level J L The data for the global regions is qualitative. 2J explanation The global regions cannot be expressed as numerical data J 2J explanation % RT and 8% RT 3RT Correct modal % L1 Two marks for first correct answer, one mark for second correct answer Median = % 2 (3) 2M for adding correct values and dividing by 2 L2 = 7,5% 1CA answer Answer only full marks RT Total usage = 3% + 8% + 11% = 22% 1RT correct values 1CA total (3) L % + 9% + 23% + 22% = 56% Note: Candidates that add the 4% of the Middle East is also correct. Answer only full marks 2M Adding all correct values. 1CA total Answer only full marks Answer only 60% full marks L (a) 16% RG 2RG correct value (3) L1 Please turn over

29 Mathematical Literacy/P1 13 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level (b) WLD POPULATION AND MEANS OF COMMUNICATION PERCENTAGES PER GLOBAL REGION 30 Percentage world population Percentage Internet communication Percentage cell phone communication 20 PERCENTAGES 10 0 A B C D E F G H I J K L 1A mark for every TWO points plotted correctly (Penalty of one mark if points are not joined) GLOBAL REGIONS (1 6) (6 ) L2 Please turn over

30 Mathematical Literacy/P1 14 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level South Asia I RD 2RD reading from graph or table A Rural Number = % 1MA multiplying with % 1A 48 % = A persons L1 L1 A Urban number = % = Rural = = MA multiplying with % 1A urban number 1A persons Social networking users Answer only full marks (3) L = 100% = 26,167 % SF 1SF dividing the correct value by CA answer in % Answer only full marks NP - rounding A for correct digits L1 [30] Please turn over

31 Mathematical Literacy/P1 15 DBE/November 2015 NSC Memorandum QUESTION 5[27] Ques Solution Explanation Level A M = = MA adding all values 1CA value of M Answer only full marks Full marks for Penalty of one if given as s F L Value for both N = ( R1 756) = R Each received = = R Sibiya: N = R1 970 R349 R1 = R1 620 Magome: N = R1 963 R342 R1 = R M subtracting from total 1CA cost for both 1M dividing by 2 1CA amount 1A for R M for subtracting R349 1M for subtracting R1 1CA total Sibiya 1A for R M for subtracting R342 1M for subtracting R1 1CA total Magome Answer only full marks F L2 Penalty of one if given as s Range = R R = R M concept of range 1CA range Answer only full marks Penalty of one if not given as s (4) D L Songelwa : Magome = 30 : 342 = 5 :57 = 1 : 11,4 1A correct values 1CA form NP - rounding F L1 Please turn over

32 Mathematical Literacy/P1 16 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level Sibiya: Increase = R R = R F L2 Phillips: Increase = R R = R M subtracting any two of Sibiya, Phillips, Mabilane Mabilane: Increase = R R = R Magome: Increase = R R = R Magome received the greatest increase 1A amount for Magome 2CA correct person Full marks if only Magome was calculated correctly with conclusion Mabunda MD 2A the correct person (5) D L1 Penalty one mark if an extra name is added % 2A correct % Accept 100 P L P = 18 7 = P = 1 = A numerator 1A denominator 1CA simplification 1M subtracting from 1 1A denominator 1CA simplification Answer only full marks (3) P L2 Please turn over

33 Mathematical Literacy/P1 17 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level 5.3 Growth 1 st year = % Total after the 1 st year = = A calculating 5% 1M adding 1CA first year total D L3 Growth 2 nd year = ,9% = Total after 2 nd year = = % + 5% = 105% Total after 1 st year = % = ,3 100% + 5,9% = 105,9% Total after 2 nd year = ,3 105,9% = , Total after 2 nd year = % 105,9% = , CA calculating 5,9% of total 1CA 2 nd year total 1A increasing with 5% 1M percentage calculation 1CA first year total 1CA increasing with 5,9% 1CA 2 nd year total, rounded 1M percentage calculation 1A increasing by 105% 1M percentage calculation 1A increasing by 105,9% 1CA 2 nd year total, rounded Answer only full marks (5) [27] TOTAL: 150

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54 NATIONAL SENI CERTIFICATE GRADE 12 MATHEMATICAL LITERACY P2 NOVEMBER 2015 MEMANDUM MARKS: 150 Symbol M MA CA A C S RD SF O P R NP Explanation Method Method with accuracy Consistent accuracy Accuracy Conversion Simplification Reading from a table/graph/diagram/map Correct substitution in a formula Opinion/Example Reason / Explanation /Deduction /Comment / Interpretation Penalty, e.g. for no units, incorrect rounding off, etc. Rounding off/reasoning No penalty for rounding off/units This memorandum consists of 20pages. Please turn over

55 Mathematical Literacy/P2 2 DBE/November 2015 NSC Memorandum QUESTION 1 [34 MARKS] Ques Solution Explanation Level L Gross monthly salary of one driver A 1A using the correct value = R734, = R3 182,96 1MA dividing by 12 and multiplying by 52 Weekly salary of one driver A = R3 182, = R734,53 1A using the correct value 1MA dividing by 52 and multiplying by Salary of one cleaner = 8 20 R18,66 = R2 985,60 Salary of one supervisor = R2 985,60 + R230,00 = R3 215,60 1M multiplying hours, days and rate 1CA salary of 1 cleaner 1CA salary of 1 supervisor L3 Salaries: Handymen = 11 R4 410,37 = R48 514,07 1A salaries Handymen Cleaners = 272 R2 985,60 = R ,20 1CA salaries Cleaners Supervisors = 12 R3 215,60 = R38 587,20 Drivers = 11 R3 182,96 = R35 012,56 1CA salaries supervisors 1CA salaries drivers Total salaries = R48 514,07 + R ,20 + R38 587,20 + R ,56 = R ,03 Total UIF payable = 2% R ,03 = R18 683,94 1CA Total salaries 1A 2% contribution 1CA total contribution Please turn over

56 Mathematical Literacy/P2 3 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level A Salary of one cleaner = 8 20 R18,66 1MA multiplying hours, days and rate = R2 985,60 1CA salary of 1 cleaner Salary of one supervisor = R2 985,60 + R230,00 = R3 215,60 Total UIF payable: For 11 handymen= 11 R4 410,37 2% = R970,28 For 272 cleaners= 272 R2 985,60 2% = R16 241,66 1CA salary of 1 supervisor 1A 2% contribution 1A UIF handymen 1CA UIF cleaners For 12 supervisors = 12 R3 215,60 2%= R771,74 For 11 drivers= R35 012,56 2%= R700,25 1CA UIF supervisors 1CA UIF drivers Total UIF payable = R970,28+ R16 241,66 + R771,74+ R700,25 = R18 683,93 1CA adding 1CA total contribution Total monthly salary A = 11 R4 410, R18, (8 20 R18,66 + R230,00) + 11 R3 182,96 = R48 514,07 + R ,20 + R38 587,20 + R35 012,56 = R ,03 Total UIF payable = 2% R ,03 = R18 683,94 1MA adding 1A multiplying numbers 1M multiplying hours, days and rate 1A salary of handymen 1CA salary of cleaners 1CA salary supervisors 1CA salary drivers 1CA total salary 1A 2% contribution 1CA total contribution NP rounding (10) Please turn over

57 Mathematical Literacy/P2 4 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level Mean salary = = 2, % R ,03 A 306 = R3 052,93 % difference = Mean salary cleaner's salary 100% cleaner's salary R3 052,93 R2 985,60 = 100% R2 985,60 1MA dividing total salary from Q1.1.2 by number of employees 1CA simplification 1M difference 1CA percentage calculation L4 2,3% 1CA percentage The statement is VALID. O R ,03 Mean salary = A 306 = R3 052,93 Mean as a percentage of the lowest salary R3 052,93 100% = 102,255...% R2 985,60 % difference = 102,255...% 100% 2,3% 1O conclusion 1MA dividing total salary from Q1.1.2 by number of employees 1CA simplification 1M percentage 1M subtracting 100% 1CA percentage The statement is VALID O 1O conclusion R18 683,93A Mean UIF payable = = 61, Mean UIF Cleaners UIF % difference = 100% Cleaners UIF 1MA dividing total UIF from Q1.1.2 by number of employees 1CA simplification 61, , = 59, = 2,255 % 2,3 % The statement is VALID. O 100% 1M subtracting 1M percentage 1CA simplification 1O conclusion Please turn over

58 Mathematical Literacy/P2 5 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level Mean salary = = 2,2054.% R ,03 A 306 = R3 052,93 % difference = Mean salary cleaner's salary 100% mean salary R3 052,93 R2 985,60 = 100% R3 052,93 1MA dividing total salary from Q1.1.2 by number of employees 1CA simplification 1M difference 1CA percentage calculation L4 2,2% The statement is VALID. O R ,03 Mean salary = A 306 = R3 052,93 Lowest salary as a percentage of the mean R2 985,60 100% = 97, % R3 052,93 % difference = 100% 97,7945% 2,2% The statement is VALID. O A R18 683,93 Mean UIF payable = = 61, Mean UIF Cleaners UIF % difference = 100% Mean UIF 1CA percentage 1O conclusion 1MA dividing total salary from Q1.1.2 by number of employees 1CA simplification 1M percentage 1M subtracting from 100% 1CA percentage 1O conclusion 1MA dividing total UIF from Q1.1.2 by number of employees 1CA simplification 61, , = 61, = 2,2054 % 2,2 % The statement is VALID. O 100% 1M subtracting 1M percentage 1CA simplification 1O conclusion (6) Please turn over

59 Mathematical Literacy/P2 6 DBE/November 2015 NSC Memorandum Number of additional employees is = 306 1A addition L2 3 Number of female cleaners = = 204 1A proportion Probability of selecting a female cleaner 204 = 306 1CA probability = 0, ,667 R Most unlikely, because the male supervisors are the smallest number of additional employees. O 1R rounding correctly Answer only full marks 2O explanation (4) L2 The fraction for the male supervisors is 3 smaller = 0, O RT R964,87 A= 100% M R RT reading from table 1M finding % L2 = 48,24 % B = R2 065,49 41,31% = R4 999,98 A M Last income 41,31% = R2 065,49 B = R2 065,49 41,31% = R4 999,98 A A M 1A value of A 1M dividing 1A value of B 1M dividing 1A value of B Accept R5 000 NP - rounding Answer only full marks (5) Please turn over

60 Mathematical Literacy/P2 7 DBE/November 2015 NSC Memorandum Ques Solution Explanation Level L2 A A CA CA B A 1Afor first 3 points plotted correctly 1CA for plotting points A and B 1A for plotting last 3 points 1CA joining the points up to R8 099 with a curve 1CA the line from R8 099 to R (5) [34] Please turn over

61 Mathematical Literacy/P2 8 DBE/November 2015 NSC Memorandum QUESTION 2 [30 MARKS] Ques Solution Explanation Level L2 8 1A numerator P (weight loss more than 20kg) = 100% 12 1A denominator 66,67% 1CA probability as % NP - rounding Answer only full marks pounds = 102 0, ,27 kg 55 pounds = 55 0, ,95 kg C 36 pounds = 36 0, ,33 kg Arranged weight loss for males: CA 13,2 ; 13,2 ; 16,33 ; 16,7 ; 18,8 ; 23,7 ; 24,95 ; 25,6 ; 31,6 ; 37,65 ; 43,36 ; 46,27. 23, ,95 Median weight loss of males = 2 1C converting one 1C converting other two (3) 1CA arranging weights 1CA identifying middle values 1M median concept L4 = 24,325 24,33kg Her statement is NOT correct. O 1CA simplification 1O conclusion Max 4 marks if using SA males only Max 3 marks if conversions are omitted IQR for males (in kg) = 34,63 16,52 = 18,11 IQR for females (in kg) = 64,87 27,97 = 36,9 1M IQR concept 1A males IQR 1A females IQR (7) L2 L4 The female IQR is more than the male IQR. R 2Rcomment relating to the IQR values (5) Please turn over

62 Mathematical Literacy/P2 9 DBE/November 2015 NSC Memorandum Ques Solution Working with 365days: Explanation L3 Mass in one can is 8,75 4g = 35g A 1MA mass in 1 can Mass for a year is =35g 365 = g For 2 cans = g = g A 1MA multiply by 365 1CA simplification 1CA annual mass intake Mass in one can = 8,75 4 g = 35 g Mass in TWO cans = 35g 2 = 70 g A A 1MA mass in 1 can 1MA mass for 2 cans Mass for a year = 70 g 365 = g 1M multiply by 365 1CA simplification In 1can 8,75 teaspoons 2 cans 17,5 teaspoons Mass per day = 17,5 4 = 70 g A Mass for the year = 70 g 365 = g Working with 366 days: Mass in one can : 8,75 4 g = 35g A Mass of sugar for 1 can for one year = 35g 366 = 12810g Mass of sugar in 2 cans for one year = g = g 1A number of teaspoons 1MA mass per day 1M multiplying by 365 1CA simplification 1MA mass of sugar in 1 can 1M multiply by 366 1CA simplification 1CA mass for two cans (4) Please turn over

63 Mathematical Literacy/P2 10 DBE/November 2015 NSC Memorandum Ques Solution Explanation Calories before = = 532 calories Calories after changing = = 248,67 calories Difference = 532 calories 248,673 calories = 283,33 calories Sugar intake before diet: 1A calculating calories 1M ratio 1M addition 1CA calculating calories 1CA difference NP - rounding (5) L3 L4 = 7, ,25+ 10,5 A = 33,25 tsp. 133grams Sugar intake after diet: 500 3,25 = = 2 6, ,00 = 15,54 tsp. 62,16 grams % Reduction of sugar (using teaspoons) (usings grams) 15,54 = 100% 33,25 62,16 = 100% ,74% A 46,74% A NOT VALID O 1MA adding correct values 1CA simplification 1A sugar in vitamin water 1CA simplification 1MA percentage 1O opinion Accept VALID as opinion only if an explanation provided Using Calories from Q ,67 % Calories = 100% = 46,7% 532 NOT VALID O 1CA total calories after 1M percentage 1M multiply by CA simplification 1A total calories before 1O opinion (6) [30] Please turn over

64 Mathematical Literacy/P2 11 DBE/November 2015 NSC Memorandum QUESTION 3 [31 MARKS] Ques Solution Explanation O 3.1 For easy access to save on costs no privacy required aesthetic value ease of movement between rooms ventilation purposes 2O explanation 3.2 Living room, bathroom and bedroom 2. 1A identified at O least two rooms No direct sunlight into the room. 2O reason L4 L2 L4 The sun's position is on the northern side of the house The living room floor side A C = 3,550 m (3,550 m 7,04%) = 3,3008 m 3,3m 3,3 m 3,3 m Area of 4 walls SF = 4 (3,3 m 2,650 m) = 34,98 m 2 A 100% 7,04% = 92,96% Side C = 3,550 m 92,96% = 3,3008 m 3,3 m 3,3 m Area of 2 door openings Area of opening to passage = 2 length width = length width = 2 2,032 m 0,750 m = 2, 082 m 0,75 m = 3,048 m 2 = 1,5615 m 2 Area of window = 1,511 m 0,949 m = 1,434 m 2 Area to cover with panelling = (34,98 3,048 1,5615 1,434) m 2 = 28,9365 m 2 29 m 2 R O (3) 1C conversion 1MA for subtracting 1M multiplication (3) 1SF area wall dimensions 1CA area of 4 walls 2M door opening dimensions 1CA area of opening to passage 1CA 2 door openings 1M window dimensions 1CA area of window 1M subtracting 1CA area 1R rounding L2 L3 Please turn over

65 Mathematical Literacy/P2 12 DBE/November 2015 NSC Memorandum Ques Solution Explanation Area of northern wall = Area of wall area of door 1M subtracting areas = (3,3 m 2,650 m) (2,082 m 0,750 m) = 8,745 m 2 1,5615 m 2 = 7,1835 m 2 1CA for calculating area of northern wall Area of eastern wall = Area of wall area of door = (3,3 m 2,650 m) (2,032 m 0,750 m) = 8,745 m 2 1,524 m 2 = 7,221 m 2 Area of southern wall = Area of wall area of door area of window = (3,3 m 2,650 m) (2,032 m 0,750 m) (1,511 m 0,949 m) = 8,745 m 2 1,524 m 2 1,434 m 2 = 5,787 m 2 1M subtracting areas 1CA for calculating area of eastern wall 1M subtracting areas 1A subtracting 1CA for calculating area of southern wall Area of western wall = (3,3 m 2,650 m) = 8,745 m 2 Area to cover = 7,1835 m 2 + 7,221 m 2 + 5,787 m 2 + 8,745 m 2 = 28,9365 m 2 29 m 2 R Area of wall including door and window openings = perimeter of floor height = 2 (width + width) height = 2 (3,3 m + 3,3 m) 2,650 m = 34,98 m 2 Area of window 1 opening = length breadth = 1,511 m 0,949 m =1, m 2 Area of 2 door openings Area of opening to passage = 2 length width = length width = 2 2,032 m 0,750 m = 2,082 m 0,75 m = 3,048 m 2 = 1,5615 m 2 Area to cover = 34,98 m 2 1, m 2 3,048 m 2 1,5615 m 2 = 28, m 2 29 m 2 R 1CA for calculating area of western wall 1M for adding 4 walls 1CA simplification 1R rounding 1M multiplying 1CA calculating total area of walls 1M area formula 1CA calculating area of window 2M area formula 2CA calculating area of door openings 1M for subtracting 1CA simplification 1R for rounding (11) Please turn over

66 Mathematical Literacy/P2 13 DBE/November 2015 NSC Memorandum Ques Solution Explanation 3.4 Surface area of one panel = 2 m 0,15 m = 0,3 m 2 1A area L m Number of panels needed = 2 0,3 m = 96, CA from Q3.3.2 simplification Total panels needed to be purchased = ,5% 97 4,5% = 4,365 = 101, = 102 R C Volume of 102 panels = 102 0,0125m 0,3 m 2 SF = 0,3825 m 3 1CA number of panels 1R rounding 1C convert to metre 1SF finding volume 1CA volume in m 3 Cost of panels excluding VAT = 0,3825 R5 000,00 = R1 912,50 Price of wood including VAT = R5 000 per m 3 114% = R5 700 per m 3 1CA cost excluding VAT Cost of the panels including VAT = 1,14 R1 912,50 = R2 180,25 Cost of the panels including VAT = R ,3825 = R2 180,25 1CA cost incl. VAT Labour cost = 29 R125,00 = R3625,00 Total cost = R2 180,25 + R3 625,00 = R5805,25 Budget is ENOUGH O 1CA labour cost (CA area from 3.3.2) 1CA total cost 1O conclusion Please turn over

67 Mathematical Literacy/P2 14 DBE/November 2015 NSC Memorandum Ques Solution Explanation Surface area of wood = 29 m 2 Volume of wood = 29m 2 0,0125 m 1CA from M calculating volume 1A correct thickness = 0,3625 m 3 Total volume of wood = 0, ,5% = 0, m 3 = 0,38 m 3 1CA simplification 1M % increase 1CA simplification 1CA rounding Cost of panels excluding VAT = 0,38 R5 000,00 = R1 900,00 Price of wood including VAT = R5 000 per m 3 114% = R5 700 per m 3 1CA cost excluding VAT Cost of the panels including VAT = 1,14 R1 900,00 = R2 166,00 Cost of the panels including VAT = R ,38 = R2 166,00 1CA cost incl. VAT Labour cost = 29 R125,00 = R3625,00 Total cost = R2 166,00+ R3 625,00 = R5 791,00 Budget is ENOUGH O 1CA labour cost (CA area from 3.3.2) 1CA total cost 1O conclusion NP - rounding (12) [31] Please turn over

68 Mathematical Literacy/P2 15 DBE/November 2015 NSC Memorandum QUESTION 4 [31 MARKS] Ques Solution Explanation Course modules have different costs O 2O relevant reason L4 - Course levels makes a difference. O Single rooms: O - Have more privacy and is more convenient; no disturbance. 2O relevant reason 2O relevant reason L4 - Better facilities. O Double rooms: - Are not private and not convenient. O O - Students share costs O Total fees for first year L2 = Tuition fees + hostel fees + non-sa citizen fee = R R R2 000 = R A all the values 1M adding fees 1CA total No penalty if deposit added Minimum payment on registration: Cost = appl. fee + 30% of tuition + non-sa additional + accommodation dep. + monthly residence fee = R0, % R R R1 220,00 + S S = R R R R1 720,73 = R13 481,73 R ,00 11 (3) 1A using correct amounts 1M adding amounts 1S tuition fee 1S accommodation fee 1CA minimum payment No penalty if deposit subtracted (5) L3 Please turn over

69 Mathematical Literacy/P2 16 DBE/November 2015 NSC Memorandum Ques Solution Explanation 4.2 Afrikaans Home Language is excluded because it is the lowest: L3 92 LO APS = A 2 = 46% rounded up to 50% LO will be allocated 4 APS R 1MA calculating % of LO 1R rounding up 1A LO APS Total APS based on final results: = = 42 She qualifies for 50% bursary Distance from Okahandja to Johannesburg = Windhoek to Pretoria + Okahandja to Windhoek + Pretoria to Johannesburg + 2 Gabarone A A = ( ) km = 1 602km Total distance Driving time = Average speed km = SF 108 km/h = 14,8333 hrs 14 hours 50minutes 1CA adding scores 1CA total 1CA identifying bursary % 1MA adding extra kilometres 1MA return on Gabarone 1CA total distance 1SF substitution 1CA Total time (6) L2 Distance from Okahandja to Johannesburg A A = [ (45) ]km = km Total distance Driving time = Average speed km = SF 108 km/h = 14,8333 hrs 14 hours 50 minutes 2MA for adding the distances to travel 1CA total distance 1SFsubstitution 1CA total time (5) Please turn over

70 Mathematical Literacy/P2 17 DBE/November 2015 NSC Memorandum Ques Solution Explanation O Strip charts are not drawn to scale. 2O for any valid explanation Total cost = P P50 + P50 + P20 = P A adding values 1CA total L4 L BWP = ,2454ZAR = 2 690,064ZAR 1M converting P to R 1CA amount 2 690, ,064ZAR = 0,998 =2 695,45491NAD 2 695,45NAD NAD Her estimation is NOT VALID. O 1CA amount 1O conclusion NAD 2160 = ,998 Rand = R2 155,68 Total cost in Pula = = P2 160 Total cost in Rand = ,2454 = 2 690,06 Her estimation is NOT VALID. O 1M converting NAD to Rand 1CA amount in Rand 1A adding values 1CA total 1CA cost amount 1O conclusion NP - rounding (6) [31] Please turn over

71 Mathematical Literacy/P2 18 DBE/November 2015 NSC Memorandum QUESTION 5 [24 MARKS] Ques Solution Explanation J More Chinese migrate to other countries. 2J interpretation L China's projected population A million 0,44% = 5,9664 million million + 5,9664 million = 1 361,966 4 million USA's projected population A 319 million 0,77% = 2,4563 million 319 million + 2,4563 million = 321,4563 million Difference = 1 361,966 4 million 321,4563 million = 1 040,5101 million China's projected population A = million 1,0044% = USA's projected population A = 319 million 1,0077% = Difference = MA calc. projected population growth 1A population in millions 1MA calc. projected population growth 1A USA population in million 1CA the difference (Accept 1041 million) 1MA calc. projected population 1A population in millions 1MA calc. projected population 1A USA population in million 1CA the difference L3 Max of 4 if rounded Max of 3 if millions omitted Middle East RD 2RD region (5) L2 Please turn over

72 Mathematical Literacy/P2 19 DBE/November 2015 NSC Memorandum Ques Solution North America sdifference million tons 410 million tons = 600 million tons Asia'sdifference million tons 380 million tons = 700 million tons Asia has a higherdifference of crude oil than North America J Asia consumes much more crude oil than North America They both have vibrant economies, therefore these regions need a lot more energy. O Both regions have more industries. O The regions have large populations. O They use large volumes of oil because they have outdated O technology. First world regions O O Developed regions Distance in km = 33 mm 25 mm km = km Distance in miles = km 1, = 4 101, miles 4 101,05 miles Accept measured distance from 27 to 29 mm and bar scale from 22 to 24 mm Explanation 1CA for calculating North American difference [Accept values in range of ±10 million tons.] 1CA for calculating Asia's difference 1J comment Penalise with one mark if millionsomitted 2O reason 1M for using the line scale 1CA for calculating distance 1CA for distance in miles (3) (3) L2 L4 L4 L3 Please turn over

73 Mathematical Literacy/P2 20 DBE/November 2015 NSC Memorandum Ques Solution Total amount of oil transported daily A RD 100% = 15 million barrels 30% = 50 million barrels per day 30 % ~ 15 million barrels 30 % ~ 15 million barrels RD 30 % ~ 15 million barrels 10 % ~ million barrels = 5 million barrels Therefore 100 % ~( ) million barrels = 50 million barrels It is not the shortest route It will take longer to transport the oil. It will cost more to transport the oil. O O O Explanation 1RD reading 15 million barrels 1MA dividing by 30% 1CA simplification 1RD reading 15 million barrels 1M calculating 10% 1CA simplification No penalty if millions omitted 2O relevant (time or distance related reason (3) 2O relevant cost related reason (4) [24] TOTAL:150 L2 L4