Grade 12 NSC Maths Literacy P1 Preparatory 2018 Possible Answer hlayiso.com

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION 2018 MARKING GUIDELINES MATHEMATICAL LITERACY P1 (10601) Codes Explanation M Method MA Method with Accuracy CA Consistent Accuracy A Accuracy C Conversion D Define J Justification / Reason / Explain S Simplification Reading from a table OR a graph OR a diagram OR a map RT / RD / RG OR a plan F Choosing the correct formula SF Substitution in a formula O Opinion P Penalty, e.g. for no units, incorrect rounding-off, etc. R Rounding-off NP No penalty for rounding-off OR omitting units KEY TO TOPIC SYMBOLS: F = Finance; M = Measurement; MP = Maps; Plans and other representations; DH = Data Handling; P = Probability 13 pages

Downloaded from hlayiso.com 10601 / 18 GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION – 2018 MATHEMATICAL LITERACY (Paper 1) MARKING GUIDELINES QUESTION 1 Q ANSWER EXPLANATION LEVEL 1.1 1.1.1 VAT: value added taxA 2 A correct answer F1 (2) 1.1.2 R80,00 A 1 A Answer F1 (1) 1.1.3 R149,99 – R120,00MA 1 MA Subtraction = R29,99A 1A Answer F1 (2) 1.1.4 MA 1 MA multiplying (25 × R120, 00 = R3000) + 1 MA adding (25 × R40,00 = R1 000,00) + MA 1 A correct answer ( 3 × R60,00 = R180,00) = R4 180 A (3) F1 OR 25 × (R120,00 + R40) = R4 000 3 × R60 = R180 R4 000 + R180 = R4 180 1.1.5 A 1 MA multiply by 15% 1 A correct answer F1 MA (2) 1.1.6 MA 1 MA multiply by 100 R40,00 × 100 = 4000 cents A 1 A correct answer F1 Answer only: full marks (2) 1.1.7 RT 1 RT reading from table R120,00 : R60,00 1 A answer in simplest form 2: 1A F1 Answer only: full marks (2) 2

Downloaded from hlayiso.com 10601 / 18 Q ANSWER EXPLANATION LEVEL 1.2 1.2.1 126 cm ÷ 100MA 1 MA dividing by 100 M1 1 A correct answer A Answer only: full marks (2) 1.2.2 MA 111,6 cm + 69,1 cm + 111,6 cm + 69,1 cm 1 MA adding correct values 1 A correct answer = 361,4 cmA Answer only: full marks M1 OR (2) (2 × 111,6 cm) + (2 × 69,1 cm)MA = 361,4 cmA 1.3 1.3.1 DiscreetA 2 A for correct answer DH1 (2) 1.3.2 New Zealand RT 2 RT reading from table South Africa RT (2) DH1 1.3.3 MA 1 MA subtracting 1645 – 1 250 1 A correct answer = 395 A DH1 Answer only: full marks (2) 1.3.4 Australia RT 2 RT reading from table DH1 (2) 1.3.5 MA 1 MA adding all values 2303 + 1645 + 1250 + 1379 + 1487 1 A correct answer = 8064A DH1 Answer only: full marks (2) 1.3.6 EnglandRT 2 RT reading from table (2) DH1 [30] 3

Downloaded from hlayiso.com 10601 / 18 QUESTION 2 Q ANSWER EXPLANATION LEVEL 2.1 2.1.1 1¥ = = R0,124 A 2A correct answer (2) OR F1 12cA 2.1.2 R800,00 × 8,02 = ¥6 416,00A 2A correct answer F1 (2) MAMA 1MA adding 2.1.3 (R8 750,00 + R3 771 + (3 × R800,00) 1MA multiplying by 3 F1 = R14 921,00 A 1A answer (3) 2.1.4 MA 1 MA multiply by 6%, 106% or 1,06 1 A R10 600 1 CA final answer R10 000 + R600 = R10 600A (3) R10 600 + R636 = R11 236CA OR F2 OR R10 000 x 1,06 = R10 600 R10 600 x 1,06 = R11 236 OR 6 Year 1: R10 000 (1 + ) = R10 600 100 6 Year : R10 600 (1 + ) = R11 236 100 4

Downloaded from hlayiso.com 10601 / 18 Q ANSWER EXPLANATION LEVEL 2.2 2.2.1 Hire-purchase: a system by which one pays 2 J correct explanation F1 for item(s) in regular instalments while having the use of it. A (2) 2.2.2 R6 999,00 RT 2 RT correct answer F1 (2) 2.2.3 1 MA Method SF 1A Answer (2) = 14, 287… 14, 29% or 14% A OR F2 = 14, 287… 14, 29% or 14% MA 2.2.4 10 1 MA calculating 10% F1 × R8 999,00 = R899,90 A 100 1 A correct answer (2) 2.2.5 20,75% RT 2 RT correct answer (2) F1 2.2.6 MA A 1 MA Multiply by 20,75% or 0,2075 1 A R1 680,56 R1 680,56 x 3 = R5 041,69A 1 A R5 041,69 1 M Addition R8 099,10 + R5 041,69 M 1 CA Answer = R13 140,79CA OR MA A F2 1 MA Multiply by 20,75% or 0,2075 R8 099,10 + R1 680,56 = R9 779,66A 1 A R1 680,56 R9 779,66 + R1 680,56 = R11 460,22A 1 A R9 779,66 R11 460,22 + R1 680,56 = R13 140,78CA 1 A R11 460,22 1 CA Answer OR 5

Downloaded from hlayiso.com 10601 / 18 Q ANSWER EXPLANATION LEVEL Simple interest = F 1 F correct formula 1 SF substitution = SF 1 A correct interest amount = R5 041,69A 1 M addition 1 CA final answer Final Amount = R8 099,10 + R5 041,69M = R13 140,79CA (5) 2.2.7 MA 1 MA subtraction 1 A final answer 36 – 6 = 30 monthsA F1 Answer only: full marks (2) 2.3 2.3.1 TRANSACTION FEE (in Rand) Monthly account fee R51,00 A ATM cash withdrawals :  Commercial Bank R11,40 A ATM (5,70 x 2) 1 A for R51,00  Other bank R17,90 A 1 A for R11,40 (6,50 + 5,70 x 2) 1 A for R17,90 External debit order R8,80 A 1 A for R8,80 Internal debit order R3,30 A 1 A for R3,30 F2 External stop order R 17,60 A 1 A for R17,60 (2 x R 8,80) 1 A for R153,08 Online cheque deposit: R153,08 A 1 A for R29,40 (0,65 x R 23 5500 ÷100) 2 CA for final answer Cash deposit: R29,40 A (1,2% x R 2 425 ÷ 100) (10) Total for bank fees: R292,48 CA [37] 6

Downloaded from hlayiso.com 10601 / 18 QUESTION 3 Q ANSWER EXPLANATION LEVEL 3.1.1 Perimeter: the distance around a two- 2 J correct explanation dimensional shape. A (2) M1 3.1.2 70 m + (22 m + 100 m + 22 m) + 70 m + 1M Addition (22 m + 100 m + 22 m)  M 1A answer of 428 m M2 = 428 m  A (2) 3.1.3 1 A correct values 1 M addition A M 1 CA final answer 22 m + 100 m + 22 m = 144 m CA M2 OR 1 A correct values 1 M addition and times 100 m + (2 × 22 m) 1 CA final answer = 144 m (3) 3.1.4 Area of rugby field: 1 SF substitution of length 1 SF substitution of width A = length × width 1 CA answer M2 = 144 m × 70 m SF 1 U correct unit = 10 080 m2  CAU (4) 3.1.5 Lenght of sod of grass: 500 mm = 0,5 m 1 C conversion Width of sod of grass: 700 mm = 0,7 mC 1 SF substitution 1 CA answer Area of one sod of grass = length × width = 0,5 × 0,7 m SF = 0,35 m2 CA OR OR M2 1 SF substitution Area of 1 sod of grass 1 C conversion = length × width 1 CA answer = 500 mm × 700 mmSF (3) = 350 000 mm2 = C = 0,35 m2 CA 7

Downloaded from hlayiso.com 10601 / 18 Q ANSWER EXPLANATION LEVEL 3.1.6 Number of sods of grass needed to cover the 1 M division / rugby field: multiplication 1 CA answer = (2) = M = 28 800 sods of grassCA OR and Width : Length : M2 Total: 206 × 140 = 28 840 sods of grass OR and Width : 00 Length : Total: 100x 288 = 28 800 sods of grass 3.2 3.2.1 Radius of water tank = MA 1 MA division by 2 1 A correct answer M1 = 100 cm A Answer only: full marks (2) 8

Downloaded from hlayiso.com 10601 / 18 Q ANSWER EXPLANATION LEVEL 3.2.2 1C Conversion of height 2 Volume = π × r × h 1 SF substitution = 3,142 × (100 cm)2 × 350 cm 1 CA answer SFC 1 Mdivide / multiply by 1 = 10 997 000 cm3 CA 000 1 CA answer Note that: 1 000 cm3 = 1 litre 1 U litres as unit M (6) = 10 997 litresAU OR M3 C Volume = π × r2 × h = 3,142 × (1 m)2 × 3,5m SF = 10,997 m3 CA Note that: 1 m3 = 1 000 litres 10,997 m3 x 1 000M = 10 997 litresAU 3.2.3 Radius: 100 cm = 1 m C 1C conversion Height: 3,5 m 1 SF radius 1 SF height Surface area = 2 πr (r + h) 1 CA final answer SF SF M2 = 2 × 3,142 × 1 m (1 m + 3,5 m) NP No penalty for = 28,278 rounding 28,28 m2 CA (4) [28] 9

Downloaded from hlayiso.com 10601 / 18 QUESTION 4 Q ANSWER EXPLANATION LEVEL 4.1 4.1.1 N1RT 2 RT reading from map MP1 (2) 4.1.2 BloemfonteinRT 2 RT reading from map MP1 (2) 4.1.3 3 national roads RT 2 RT reading from map MP1 (2) 4.1.4 Bar scaleRT 2 RT reading from map MP1 (2) 4.1.5 N2 RT 2 RT reading from map MP1 (2) 4.1.6 Every 1 unit on the map represents 10 000 000 2 A correct answer units in reality. A (2) MP1 4.1.7 1 : 10 000 000 1 MA multiply by scale 13 cm : 130 000 000 cm MA 1 C conversion to km 1 A correct answer C MP2 = 1 300 km A (3) 4.1.8 Speed = 1 SF substitution 1 A answer 1 R correct rounding = SF = 117, 916 A = 118 km / h R (3) MP2 10

Downloaded from hlayiso.com 10601 / 18 Q ANSWER EXPLANATION LEVEL 4.1.9 1 M multiply by M MM consumption rate 1 415 km × × R14,70 1 M divide by 100 1 M times by fuel price / F3 litre = R1 622,44CA 1 CA answer (4) 4.2.1 A 1 A correct fraction S 1 S simplification P1 (2) 4.2.2 0,65 x 100 = 65 %A 2 A correct answer (2) OR P1 0,652 x 100 = 65,2 % A 4.2.3 No A 1 A correct answer There is still a 35% chance that it might not 1 J correct explanation P1 rain; it is not 100% certain. A (2) [28] 11

Downloaded from hlayiso.com 10601 / 18 QUESTION 5 Q ANSWER EXPLANATION LEVEL 5.1 5.1.1 53; 62; 63; 63; 65; 65; 66; 68; 69; 72; 72; 74; 2 MA arranging 80; 81MA Penalise with 1 mark if the values of the two teams are DH1 correct but separately done. (2) 5.1.2 Player 5 of team BRT 1 RT correct player 1,96 mRT 1 RT correct height DH1 (2) 5.1.3 63 kgRT 2 RT reading off correct answer DH1 (2) 5.1.4 15 yearsRT 2 RT reading from table (2) DH1 5.1.5 24,9RT 2 RT reading from table DH1 (2) 5.1.6 1 MA addition Mean height = 1 MA division by 7 1 A final answer MA (3) DH2 = MA = 1,66 mA 5.1.7 BMI = 1SF correct substitution 1 A correct answer = SF P Penalty for not DH2 writing unit (kg / m2 ) = 29,39 kg / m2 A(P) (2) 5.1.8 Continuous data – because of the decimal 2 J correct explanation numbers A (2) DH1 5.2 Height (in m) Tally Frequency 1,5 – 1,62 1111 5  1,63 – 1,75 1111 4 1,76 – 1,85 11 2 1,86 and taller 111 3 DH2 14 1 Mark for each correct row (tally and frequency) x 4 1 Mark for total of 14 (5) 12

Downloaded from hlayiso.com 10601 / 18 Q ANSWER EXPLANATION LEVEL 5.3 Masses of Players (in kg) 14 63 13 74 12 80 11 81 10 53 9 69 Players 8 63 7 65 6 66 5 68 4 72 3 DH2 65 2 72 1 62 0 10 20 30 40 50 60 70 80 90 100 Masses of players (in kg)  2 marks for correct bars for players 1 – 6  2 marks for correct bars for players 7 – 14  1 mark for correct bars with spaces. P : Penalise 1 mark if there are no spaces No marks if line graph is drawn (5) (27) TOTAL: 150 13