Mathematical Literacy P2 Nov 2013 Memo Eng hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICAL LITERACY P2 NOVEMBER 2013 MEMORANDUM MARKS: 150 SYMBOL EXPLANATION A Accuracy CA Consistent accuracy C Conversion J Justification (Reason/Opinion) M Method MA Method with accuracy P Penalty, e.g. for no units, incorrect rounding off, etc. R Rounding off RT/RG Reading from a table/Reading from a graph S Simplification SF Correct substitution in a formula O Own opinion/Example NPR No penalty for rounding This memorandum consists of 22 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 2 DBE/November 2013 NSC– Memorandum QUESTION 1 [24 MARKS] Ques Solution Explanation AS 12.1.2 1.1 Amount of juice (in litres) L2 400 kg M OR 2,5 kg makes 1  = 400 kg M 2,5 kg 400 kg makes 1M dividing by 2,5 A 2,5 kg /  = 160 = 160  A 1A simplification Number of 5  bottles Number of 5  bottles OR 160  160  = = 5 5 = 32 CA = 32 CA OR 1 : 2,5 = x : 400 2,5x = 400 400 M x= 1M using proportion 2,5 x = 160 A 1A simplification 160  Number of 5  bottles = 5 = 32 CA 1CA simplification OR OR 5  juice is made from 5 × 2,5 kg = 12,5 kg fruit A 1A mass of fruit 400 kg ∴ Number of 5  bottles = M 1M dividing by 12,5 12,5 kg = 32 CA 1CA simplification OR OR 400 kg 1A using proportion = 80 kg /  A 5 M 1M dividing by 2,5 80 kg /  Number of 5  bottles = = 32 CA 1CA simplification 2,5 kg /  Correct answer only: full marks (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 3 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS 12.3.1 1.2.1 90 1A value of radius L2 Radius (in mm) = = 45 A 2 Surface area (in mm2) = 4 × 3,14 × 452 SF 1SF substitution = 25 434 CA 1CA simplification Accept 25 446,90 using π Using diameter max 2 marks NPR Correct answer only: full marks (3) CA from 1.2.1 12.3.1 4 L2 1.2.2 Volume (in mm3) = × 3,14 × 45 3 SF 1SF substitution 3 = 381 510 CA 1CA simplification Accept 381 703,51 using π NPR Correct answer only: full marks (2) 12.3.1 30 A 1A radius of basket 1.3 Radius of basket = = 15 cm 12.1.2 2 SF 1SF substitution L3(6) Volume of basket = 3,14 × (15 cm)2 × 25 cm = 3,14 × (150 mm)2 × 250 mm C 1C converting to mm L4(1) 3 = 17 662 500 mm CA 1CA volume of basket Accept 17 671 458,68 using π M/A 1M/A subtracting 17 662 500 mm3 − 113 040 mm3 space The number of oranges = 381510 mm3 M/CA 1 M dividing by = 46 volume of an orange CA from 1.2.2 ∴ Franz’s statement is not correct CA 1CA conclusion OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 4 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS OR 30 1A value of radius Radius of basket = = 15 cm A 2 SF Volume of basket = 3,14 × (15 cm)2 × 25 cm 1SF substitution = 17 662,5 cm3 CA 1CA volume of basket Accept 17 671,46 using π 17 662,5 cm 3 − 113 040 mm 3 1M dividing by The number of oranges = 381 510 mm 3 M volume of an orange 1M subtracting space 17 662,5 cm 3 − 113, 040 cm 3 M = 381,51 cm 3 C 1C converting to cm = 46 (46 > 44) ∴ Franz’s statement is not correct CA 1CA conclusion OR OR 30 A 1A radius of basket Radius of basket = = 15 cm 2 1SF substitution SF Volume of basket = 3,14 × (15 cm)2 × 25 cm 1C converting to mm = 3,14 × (150 mm)2 × 250 mm C 1CA volume of = 17 662 500 mm3 CA basket Space in the basket for oranges (in mm3) = 17 662 500 – 113 040 = 17 549 460 M 1M subtracting space 3 Space occupied by oranges (in mm ) A 1A calculating the = 381 510 mm2 × 44 = 16 786 440 mm2 space occupied by the oranges (∴ there is space for more oranges) ∴ Franz’s statement is not correct CA 1CA conclusion Correct conclusion only: 1 mark (7) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 5 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS 12.1.1 1.4 Trailer length C 1C conversion 12.3.2 = 394 × 2,54 cm = 1 000,76 cm OR 10,0076 m 12.3.1 1C conversion Trailer breadth C L2(1) = 119 × 2,54 cm = 302,26 cm OR 3,0226 m L3(3) Option 1: L4(4) Maximum number of boxes packed lengthwise along the breadth of the trailer: 302,26 OR 3,0226 = M = M 1M dividing 30 0,3 = 10,075… = 10,075… ≈ 10 ≈ 10 Maximum number of boxes packed breadthwise along the length of the trailer: 1 000,76 OR 10,0076 = = 21,5 0,215 = 46,54… = 46,54… 1R rounding down ≈ 46 R ≈ 46 R Maximum number of boxes of oranges = 10 × 46 = 460 CA 1CA maximum number of boxes Option 2: Maximum number of boxes packed breadthwise along the breadth of the trailer: 302,26 OR 3,0226 1M dividing = M = M 21,5 0,215 = 14,05… = 14,05… ≈ 14 ≈ 14 Maximum number of boxes packed lengthwise along the length of the trailer: 1 000,76 OR 10,0076 = = 30 0,3 = 33,35… = 33,35… ≈ 33 R R ≈ 33 1R rounding down Maximum number of boxes = 33 × 14 = 462 CA 1CA maximum number of boxes ∴ OPTION 2 is the best CA 1CA conclusion OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 6 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS OR OR Trailer length C = 394 × 2,54 cm = 1 000,76 cm OR 10,0076 m 1C conversion Trailer breadth C 1C conversion = 119 × 2,54 cm = 302,26 cm OR 3,0226 m Height = 94,6 × 2,54 cm = 24 003 cm OR 240,03 m 240,03 Number of layers of boxes = = 10,214... ≈ 10 0,235 Option 1: Maximum number of boxes packed lengthwise along the breadth of the trailer: M 3,0226 1M dividing = = 10,075… ≈ 10 0,3 Maximum number of boxes packed breadthwise along the length of the trailer: 10,0076 = = 46,54… ≈ 46 R 0,215 1R rounding down Number of boxes to be packed in this option = 10 × 10 × 46 = 4 600 CA 1CA total number of boxes Option 2: Maximum number of boxes packed breadthwise along the breadth of the trailer: M 3,0226 1M dividing = = 14,05… ≈ 14 0,215 Maximum number of boxes packed lengthwise along the length of the trailer: 10,0076 = = 33,35… ≈ 33 R 1R rounding down 0,3 Number of boxes to be packed in this option = 14 × 33 × 10 1CA total number of = 4 620 CA boxes 1CA conclusion ∴ OPTION 2 is the best. CA Correct conclusion only: 1 mark (9) [24] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 7 DBE/November 2013 NSC– Memorandum QUESTION 2 [26 MARKS] Ques Solution Explanation AS NOTE: No variable 12.2.1 2.1.1 (symbol or words), L3(2) NO marks Amount claimed (in rand) A A 1A correct fuel tariff × = 4,67 number of kilometres travelled 1A multiplying tariff in rand by number of OR kilometres travelled A A = 467 cents × number of kilometres travelled OR A A × = 467 number of kilometres travelled ÷ 100 OR A Amount claimed (in rand) = 4,67 × n where n = number of kilometres travelled A OR A Amount claimed (in rand) = 467 cents × n where n = number of kilometres travelled A (2) 12.2.1 2.1.2 Amount claimed (in rand) = 4,67 × 1 960 SF 1SF substitution in formula from Q 2.1.1 L4(3) = 9 153,20 CA 1CA simplification ∴ The amount claimed by Rodney was incorrect. CA 1CA conclusion OR OR M A 1M concept 9 430 The rate of claim used = = 4,8112... 1A calculated rate 1960 (4,8112... is more than the correct rate of 4,67) ∴ The amount claimed by Rodney was incorrect.CA 1CA conclusion OR OR M A 9 430 1M concept Number of kilometres claimed = = 2019,27... 4,67 1A number of km (2019,27... is more than the 1960 km travelled.) ∴ The amount claimed by Rodney was incorrect. CA 1CA conclusion Correct conclusion only: 1 mark (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 8 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS M/A 12.1.1 2.2.1 Petrol cost (in rand) = 1960 × 1,013 = 1 985,48 1M/A petrol cost L2 M/A Maintenance cost (in rand) = 450 + 125 + 500 + 200 = 1 275 1M/A maintenance Monthly cost (in rand) = 1 985,48 + 1 275 = 3 260,48 CA 1CA monthly cost OR OR Monthly cost (in rand) M/A = (450 + 125 + 500 + 200) + 1 960 × 1,013 M/A 1M/A maintenance = 1 275 + 1 985,48 1M/A petrol cost = 3 260,48 CA 1CA monthly cost Correct answer only: full marks (3) 12.2.1 2.2.2 Finding remaining amount using the 1,5  vehicle: October 12.1.1 Claim amount 1M multiplying the M M L2(3) = 2994 cents × 1 960 km OR = R2,994 × 1 960 km tariff with distance = 586 824 cent = R5 868,24 CA L3(3) = R5 868,24 CA 1CA claim amount L4(3) 1M subtracting the Remaining amount = R5 868,24 – R3 260,48 M monthly cost = R2 607,76 CA (Q2.2.1) from a calculated claim amount 1CA remaining amount Finding remaining amount using the 2,3  vehicle: November Petrol cost (in rand) = 1960 × 1,317 = 2 581,31 M/A 1M/A Petrol cost M/A Maintenance cost (in rand) = 700 + 210 + 800 + 450 = 2 160 1M/A maintenance Monthly cost (in rand) = 2 581,31 + 2 160 = 4 741,32 CA 1CA monthly cost Using CORRECT claim Using RODNEY's amount: claim amount: Remaining amount Remaining amount = R9 153,20 – R4 741,32 OR = R9 430 – R4 741,32 1CA remaining = R4 411,88 CA = R4 688,68 CA amount (Q2.1.2) ∴ Difference in ∴ Difference in remaining amounts remaining amounts = R4 411,88 – R2 607,76 = R4 688,68 – R2 607,76 1CA difference = R1 804,12 CA = R2 080,92 CA NPR except if R2,99 is used then max 8 marks (9) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 9 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS 12.1.3 2.3 i = 9% pa n = 24 months A = R104 753,89 L3 1A interest rate per 9% A month R104 753,89 × [Note: do not penalise 12 SF x= if % sign is omitted but  9%  24  A calculation is done 1 +  − 1  12   correctly] 1SF substitution 1A number of = R4 000 CA months 1CA simplification OR OR A 0,09 1A interest rate per R104 753,89 × 12 SF month x=  0,09  24  A 1SF substitution 1 +  − 1 1A number of  12   months = R4 000 CA 1CA simplification OR OR A 1A interest rate per R104 753,89 × 0,0075 month x= SF 1SF substitution  0,09  24  A 1 +  − 1 1A number of  12   months 1CA simplification x = R4 000 CA OR OR 1A interest rate per R104 753,89 × 0,01 A x= month (NPR) [ ] (1 + 0,01) 24 − 1 SF A 1SF substitution x = R3 883,59 CA 1A number of months 1CA simplification NPR Correct answer only: full marks (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 10 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS 12.1.3 2.4 Tax(before rebate) M/A L2(3) A 1A identifying = R51 300 + 30% × (R315 054 – R250 000) correct tax interval L3(2) 1M/A finding 30 amount above = R51 300 + × R65 054 R250 000 100 = R51 300 + R19 516,20 = R70 816,20 CA 1CA tax amount Tax payable (after rebate) = R70 816,20 – R11 440,00 – R6 390 M 1M subtracting = R52 986,20 CA both rebates from the tax amount. 1CA simplification If rebates are subtracted before calculating the tax max 3 marks [If incorrect tax bracket used max 3 marks] Correct answer only: full marks (5) [26] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 11 DBE/November 2013 NSC– Memorandum QUESTION 3 [38 MARKS] Ques Solution Explanation AS 12.4.4 3.1.1 Total number of persons 20 years and older in 1996 is 1M total L4 21 251 533 A M 1A population in Total number of persons 20 years and older in 2001 is 1996 25 472 770 A 1A total number in 2001 ∴ The increase in the total population from 1996 to 2001 is greater than the increase in the number of persons with no 2O explanation schooling.  O OR explanation with calculation OR Total number of persons 20 years and older in 1996 is 21 251 533 A M 1M total Total number of persons 20 years and older in 2001 is 1A population in 25 472 770 A 1996 1A total number in 2001 Percentage growth of persons with no schooling in 2001 4 567 498 − 4 055 646 = × 100% = 12,6207…% 4 055 646 1CA percentage CA growth Percentage growth of persons 20 years and older in 2001 25 472 770 − 21 251 533 = × 100% = 19,8632…% 21 251 533 Percentage growth of persons 20 years and older was more than the percentage growth of people with no schooling. O 1O explanation (5) 12.4.1 3.1.2 Total number 20 years and older in 2011 = 30 915 706 A 1A total 20 years and 12.1.1 older L3 59,7% of population = 30 915 706 30 915 706 Total population = M 59 ,7% 1M dividing by 30 915 706 59,7% = 0,597 = 51 785 102,18 ≈ 51 785 102 CA 1CA population Total younger than 20 years = 51 785 102 – 30 915 706 OR = 40,3% of 51 785 102 = 20 869 396 CA =20 869 396,11 ≈ 20 869 396 CA 1CA solution OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 12 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS OR OR Total number 20 years and older in 2011 = 30 915 706 A 1A total 20 years and older Total younger than 20 years 30 915 706 M 1M dividing by = × 40,3% 59,7% 59,7% M 1M multiplying by = 20 869 396 CA 40,3% 1CA solution (4) 12.4.4 3.1.3 Number of persons with Gr 12 in 2001 = 5 200 602 L3 P(Grade 12) 1A number with 5 200 602 A Gr 12 = 1A denominator 44 819 778 A 2 600 301 866 767 = OR OR 22 409 889 7 469 963 1 11,6% OR ≈ 0,12 OR CA 1CA simplifying 8,6 Correct answer only: full marks (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 13 DBE/November 2013 NSC– Memorandum Ques Solution and Explanation AS 12.4.2 3.2.1 PERCENTAGE HIGHEST EDUCATION LEVEL L2 35 30 A 25 A A 20 1996 Percentage A 2001 15 A CA 10 2011 5 0 Grade 12 Completed primary Tertiary Education Some secondary No schooling Some primary Highest Education Level 1 or 2 points plotted incorrectly max 5 marks 3 points plotted incorrectly max 4 marks 4 points plotted incorrectly max 3 marks 5 points plotted incorrectly max 2 marks 1CA joining all the points by means of a line Penalty of one mark if graph is moved either left or right (6) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 14 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS 12.4.4 3.2.2 ANY TWO possible trends: L4 * From 1996 to 2011 there was an increase in the number of persons with Grade 12. CA 2CA per trend * From 1996 to 2011 there was an increase in the number of persons with Tertiary education. CA 2CA per trend * The percentage increase of persons with Grade 12 is higher than that of persons with Tertiary education. CA * There are always more persons in Grade 12 than persons with Tertiary education. CA (4) 12.4.4 3.3.1 The percentages given represent the number of people with 2O acceptable L4 Grade 12 as a percentage of the number of people 20 years explanation and older in each province and not nationally. O OR Data is per province O (2) 12.4.3 3.3.2 The ascending order is M/A 1M/A arranging in L3(2) 19,8 ; 22,4 ; 22,7 ; 25,2 ; 26,8 ; 28,2 ; 29,0 ; 30,9 ; 34,4 ascending order L4(1) ∴ Free State has the median percentage CA 1CA province OR OR The ascending order is EC; LP; NC; NW; FS; WC; MP; KZN; GP M/A 1M/A ascending order ∴ Free State has the median percentage CA 1CA province Correct answer only: full marks (2) 1A EC 12.4.3 A A 3.3.3 Eastern Cape and Limpopo 1A LP L4 (2) 12.4.2 3.3.4(a) The percentages do not add up to 100% J 2J explanation L4 OR The degrees to not add up to 3600 J OR There are too many sectors J (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 15 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS 12.4.2 J 3.3.4(b) The histogram cannot be used since the data is qualitative 2J explanation L4 OR J The data is not continuous OR Data is not given in class intervals J (2) A A 12.3.3 3.4.1 Northern Cape; Gauteng 1A Northern Cape L4 1A Gauteng Limpopo can also be included (2) 12.3.3 3.4.2 TS ≈ 7 mm A 1A measurement L4 [accept answers from 5 mm to 8 mm] Actual distance ≈ 7 mm × 10 000 000 M 1M using scale = 70 000 000 mm CA 1CA simplifying = 70 km C 1C converting to km [accept answers from 50 km to 80 km] OR OR Scale is 1 mm : 10 000 000 mm ∴ 1 mm : 10 km C 1C converting scale to km TS ≈ 7 mm A 1A measurement [accept answers from 5 mm to 8 mm] Actual distance ≈ 7 mm × 10 km/mm M 1M using scale = 70 km CA 1CA simplifying [accept answers from 50 km to 80 km] Correct answer only: full marks (4) [38] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 16 DBE/November 2013 NSC– Memorandum QUESTION 4 [34 MARKS] Ques Solution Explanation AS 12.3.1 1M/A multiplying 4.1.1 Perimeter = 5 × 270 mm M/A L2 side by 5 only = 1 350 mm A 1A simplification OR OR Perimeter = (270 + 270 + 270 + 270 + 270) mm M/A 1M/A adding 5 sides = 1 350 mm A 1A simplification Correct answer only: full marks (2) 12.3.1 4.1.2 Area of rectangle = length × breadth 12.3.2 = 360 mm × 270 mm SF 1SF substituting into L3 = 0,36 m × 0,27 m C area formula = 0,0972 m2 1C converting Surface area of front pentagon (in m2) = 0,13 – 0,017 – 0,013 = 0,1 M 1M subtracting the Surface area of rear pentagon (in m2) = 0,13 – 0,013 openings = 0,117 M 1M five rectangles Total surface area (in m2) = 5 × 0,0972 + 0,1 + 0,117 = 0,703 CA 1CA simplification using all faces OR OR Total surface area = 2 × pentagons + 5 × rectangles – (letter opening + 2 × newspaper openings) 1M five rectangles M SF M 1SF substituting area = 2 × 0,13 m2 + 5 × 360 mm × 270 mm – (0,017 m2 + 1M subtracting the 2 × 0,013 m2) C openings = 0,26 m2 + 5 × 0,36 m × 0,27 m – 0,043 m2 1C converting = 0,26 m2 + 0,486 m2 – 0,043 m2 = 0,703 m2 CA 1CA simplification using all the faces Correct answer only: full marks (5) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 17 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS 12.3.1 4.1.3 Area of a newspaper opening = π × r2 L3 (3) 0,013 m2 = 3,14 × r2 SF 1SF substitution L4 (2) 0,00414... m2 = r2 41,401... cm2 = r2 C 1C conversion r ≈ 6,434... cmCA 1CA value of r The radius of the newspaper is 6 cm A 1A radius of newspaper ∴ The newspaper will fit. CA 1CA conclusion OR OR 12 Newspaper radius (in cm) = =6 A 1A radius 2 Area of a circle = π × r2 = 3,14 × (6 cm)2 SF 1SF substitution = 3,14 × (0,06 m) 2 C 1C converting ≈ 0,0113 m2 CA 1CA simplification 1CA conclusion ∴ The newspaper will fit. CA Answer only 1 mark (5) NOTE No variable in 12.2.1 A M M second term (symbol L3(3) 4.2.1 Cost = R30,50 + R4,50 × mass of parcel greater than 1kg or words), max 1 mark 1A basic rate R30,50 OR 1M the rate for more A M Cost = R30,50 + R4,50 × a M than 1 kg where a is the mass of a parcel greater than 1 kg 1M multiplied with the mass greater than 1 kg OR A M M Cost = R30,50 + R4,50 × (mass of parcel – 1) (3) SF 12.2.1 1SF substitution (CA 4.2.2 A = R30,50 + R4,50 × (2,5 – 1) = R37,25 CA L2 from question 4.2.1) M 1CA value of A R 70,55 − R 30,50 Additional mass in kg = 1M subtracting R30,50 R 4,50 M 1M dividing R4,50 = 8,9 CA 1CA additional mass ∴ B = 1 + 8,9 = 9,9 CA 1CA value of B OR SF OR A = R30,50 + R4,50 × (2,5 – 1) = R37,25 CA 1SF substitution (CA from question 4.2.1) 1CA value of A R70,55 = R30,50 + R4,50 × a SF 1SF substitution R40,05 = R4,50 × a S 1S simplification 8,9 = a CA 1CA value of a ∴ B = 1 + 8,9 = 9,9 CA 1CA value of B Answer only: full marks (6) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 18 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS 12.2.2 4.2.3 L3 THE COST OF AN ORDINARY PARCEL PER MASS 50 45 A 40 A Cost (in rand) CA 35 A A 30 A 0 25 0 1 2 3 4 5 6 Mass (in kilogram) 1A plotting points (0,5; 30,50) and (1; 30,5) 1A plotting point (3; 39,50) 1A drawing horizontal line with open circle between 0 and 0,5 1A drawing horizontal line between 0,5 to 1 1CA drawing the line from 1 to 3 1A continue line beyond (3; 39,50) with correct slope (6) 12.3.4 4.3.1 Walmer Health Centre A 2A correct place L3 across Main Road 1A place on left If DIY Store 2 marks (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 19 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS 12.3.4 4.3.2 The length of the vacant land on the map ≈ 16 mm 1A measurements L3 (1) A The width of the land on the map ≈ 13 mm (accept lengths from L4 (3) 15 mm to 19 mm; Accept widths from 12 mm to 14 mm) Area of vacant land on the map = 1,6 cm × 1,3 cm = 2,08 cm2 CA 1CA area of vacant land 2,08 cm 2 Number of sites = 0,15 cm 2 = 13,866 1CA number of sites ≈ 13 CA She can only get 13 sites on the vacant land ∴ Her claim is not valid CA 1CA verification OR OR The length of the vacant land on the map ≈ 16 mm 1A measurements A (accept lengths from The width of the land on the map ≈ 13 mm 15 mm to 19 mm; Accept widths from 12 mm to 14 mm) Area of vacant land on the map = 1,6 cm × 1,3 cm = 2,08 cm2 CA 1CA area of vacant land Area covered by the sites = 14 × 0,15 cm2 = 2,1 cm2 CA 1CA area of the sites This area is more than the area on the map ∴ Her claim is not valid CA 1CA verification Answer only: NO marks (4) [34] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 20 DBE/November 2013 NSC– Memorandum QUESTION 5 [28 MARKS] Ques Solution Explanation AS 12.4.4 5.1.1 Schools and industries are closed therefore more people 2O explanation L4 book their drivers test in December O OR With schools etc. closed there are less cars on the road during holidays, so less chance to make mistakes and fail the test. O Any other valid explanation (2) 12.4.3 5.1.2 Minimum = 16 and maximum = 60 M 1M identifying min L2 Range = 44 CA and max values (accept minimum values of 14 to 18) 1CA range (accept values from 42 to 46) Correct answer only: full marks (2) 12.4.6 5.1.3 Toni did not arrange the bars in calendar/chronological L4 order, hence creating the impression that there was an increase. J 2J explanation Example: CA January the number of learners was 52 and February was 24 1CA example OR any other suitable example (3) 12.2.3 5.2.1 No change in the cost after 15 hours. J 2J correct description L4 OR Constant cost from 15 hours onwards. J OR For 15 hours or more of driving lessons there is a fixed rate of R1 500. J (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 21 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS 12.2.3 5.2.2 No payment for zero lessons. J 2J correct description L4 (a) OR Payment will only be made once the driving lessons start. J (2) A 12.2.3 5.2.2 • A learner driver pays a basic amount of R600 1A R600 L4 (b) for the first two hours A 1A time period • Then R50 per hour for every additional hour. A 1A rate in rand (3) O O 12.2.1 5.2.3 At point Q, both Options cost the same at the same time. 1O same cost L4 1O same time OR OR O O There were 10 hours of driving that cost R1 000 for both 1O time Options. 1O cost Accept " breakeven point " ONLY 1 mark (2) 12.2.3 5.2.4 A J 1A correct option L4 (a) With Option B Zaheera will get 14 hours of driving lessons. 1J justification OR A J Zaheera must choose Option B to get 2 more hours of driving lessons than in Option A. (2) A 12.2.3 5.2.4 Toni would benefit more from Option A. She still gets 1A correct option L4 (b) R1 200 but in a shorter time than Option B J 1J justification OR A J Option A, she will have 2 hours to train someone else. (2) J 12.2.3 A 5.2.5 Option A is cheaper for Zaheera. 1A correct option L4 2J justification OR A J She must choose Option A she will pay R600 for the driving lessons. (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 22 DBE/November 2013 NSC– Memorandum Ques Solution Explanation AS 12.2.3 5.2.6 Option A: L3(3) A Cost for 30 hours = R1 500 1A cost option A L4(2) Option B: A A 1A basic rate Cost for 30 hours = R600 + (R50 per hour × 28 hours) 1A rate multiplied by = R600 + R1 400 hours = R2 000 CA 1CA cost ∴ Difference in cost = R2 000 – R1 500 = R500 CA 1CA difference in cost OR OR Option A: A Cost for 30 hours = R1 500 1A cost option A Option B: Cost for 30 hours A A 1A basic rate = R600 + (R100 per two hours × 14 two hour periods) 1A rate multiplied by = R600 + R1 400 period = R2 000 CA 1CA cost ∴ Difference in cost = R2 000 – R1 500 = R500 CA 1CA difference in cost OR OR Option B: For 22 hours it costs R1 600 It is increasing with R100 every 2 hours A 1A rate ∴ Extra cost = 4 × R100 = R400 A 1A extra cost Cost for 30 hours = R1 600 + R400 = R2 000 CA 1CA cost Option A: A Cost for 30 hours = R1 500 1A cost option A ∴ Difference in cost = R2 000 – R1 500 = R500 CA 1CA difference in cost Correct answer only: full marks (5) [28] Total: 150 Copyright reserved