Maths Lit SOLUTIONS(SECOND CHANCE)

Grade 12 – JIT 2020 – Term 2 SECOND CHANCE MATRIC PROGRAMME KZN DEPARTMENT OF EDUCATION MATHEMATICAL LITERACY JUST IN TIME MATERIAL GRADE 12 PROGRESSED LEARNERS 2020 SOLUTIONS SYMBOL EXPLANATION M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG/RD/RM Reading from a table/ graph/ diagram/Map SF Correct substitution in a formula O Opinion/ reason/deduction/example/Explanation J Justification R Rounding off F deriving a formula AO Answer only full marks P Penalty e.g. for units, incorrect rounding off etc. NPR No penalty for rounding / units E Explanation R Rounding 1

Grade 12 – JIT 2020 – Term 2 TABLE OF CONTENTS No. TOPIC SECTION PAGE NO. QUESTION 1 1 QUESTION 2 1 1 MEASUREMENT QUESTION 3 2 QUESTION 4 2 QUESTION 1 4 QUESTION 2 4 2 FINANCE QUESTION 3 5 QUESTION 4 5 QUESTION 5 6 QUESTION 1 7 QUESTION 2 7 3 DATA HANDLING QUESTION 3 8 QUESTION 4 8 QUESTION 5 9 QUESTION 1 10 QUESTION 2 10 QUESTION 3 11 QUESTION 4 11 QUESTION 5 11 QUESTION 6 12 4 MAPS AND PLANS QUESTION 7 13 QUESTION 8 13 QUESTION 9 14 QUESTION 10 14 QUESTION 11 14 QUESTION 12 15 QUESTION 13 15 This document has been compiled by the FET Mathematical Literacy Subject Advisors together with Lead Teachers. It seeks to unpack the contents and to give more guidance to teachers

Grade 12 – JIT 2020 – Term 2 1. TOPIC: Measurement QUESTION 1 [15] Que Solution Explanation T/L 1.1.1 2 layers 2 A answer M L1 1.1.2 Radius is the line from the centre to the 2A definition M circumference. L1 1.1.3 Diameter =8.5cm × 2 1M multiply by 2 M =17cm 1A L1 AO 1.1.4 Total height = 17cm + 19cm 1MA adding M = 36cm × 10 correctvalues L1 = 360mm 1C multiply by 10 1CA answer 1.1.5 Volume of a cylinder = π × (radius)2× height 1 SF M = 3.142 × (8.5cm)2 × 19 cm L2 = 4 313.18cm3 1.2.1 Mass of butter = .   1M divide by2.25 M 1A answer L2 . = 1.11kg 1.2.2 F = (1.8 × C) + 32ᵒ ᵒ ᵒ 1 SF M = (1.8 × 180ᵒ) + 32ᵒ 1A L2 = 324ᵒ + 32ᵒ = 356ᵒ QUESTION 2 [15] Que Solution Explanation T/L 2.1.1 Volume =  ×  ×  1C converting cm to m M = 8m × 0.5m ×   1SF correct substitution L2 1CA answer = 8 ×0.5 × 0.5 = 2m3 2.1.2 Total cost of the concrete = R180 × 2  1 CA from 1.1.1. M = R360 1M multiplication L1 1CA answer 2.1.3 Cost per m3 =   1M dividing by 2 M 1A answer L1  = R160 2.2.1 Leaving time = 6:00 + 1h45min  1M adding M = 7:45pm 1A answer L1 2.2.2 24-hour format = 19:45 2A 24-hour format M (2) L1 1

Grade 12 – JIT 2020 – Term 2 2.3 Average speed =  1SF correct substitution M 1C conversion L2  =      1CA answer  =  . = 121,94km/h  QUESTION 3 [15] Que Solution Explanation T/L 3.1 Length = 160cm RT 2 RT length M L1 3.2 Breadth = 156.25cm RT 2 RT breadth M L1 3.3 Height = 100cm RT 2 RT height M L1 3.4 Surface Area = 95 250cm2 A 2 RT surface area M L1 3.5 Volume = 2 000 000cm3 ÷ 1000 C 1 C divide by 1000 M = 2000ℓ A 1 A answer M L1 3.6 Area = Length × Breath M = 160cm × 100cm SF 1 SF correct values L1 = 16 000cm2 A 1 A answer 3.7 Surface Area = 2(L×B) + 2(L×H) + 2(B×H) M SF 1 SF correct substitution L1 = 2(200cm×125cm) + 2(200cm×80cm) 1CA answer +(125cm×80cm) 1 U unit = 102 000CAcm2U QUESTION 4 [21] Que Solution Explanation T/L 4.1 Mass (kg) = 12 pounds ÷ 2,205 lb/kg 1 C - divide by 2,205 M = 5,44217… kg 1 A - answer L2 = 5,44 kg NPR 4.2 Chestnuts(lb) : Servings 1 MA - correct values M 2 : 8 and order L2 1 : 4 1 S - simplification 4.3 Amount (ml) = 2 cups x 250 ml/cupM 1 MA - multiply by 250 M = 500 ml ml L2 1 A - correct answer 4.4 Number of grams = 2 cups x 250 g/cup 1 M A- multiply by 250 M = 500 g 1 A - correct answer L2 2

Grade 12 – JIT 2020 – Term 2 4.5 Preparation time (h) = 45 min ÷ 60 min/h 1 C - divide by 60 M = 0,75 hours 1 A - correct answer L2 4.6 Time (min) = (4 h x 60min/h) + 30 min 1 C - multiply by 60 M = 240 min + 30 min 1 S - simplification L2 = 270 min 1 CA - answer 0 4.7 C = ( F – 320) ÷ 4,8 0 1 SF M = ( 3470 - 320) ÷4,8 1S L2 = 3450÷4,8 1 CA = 4750 4.8 Earliest finishing time = 08:49 + 45 min+ 4hours 30 2M - add prep and M min cooking time L2 = 44:04 1 CA - finishing time 4.9 ‘’Simple but seriously yummy’’. 2 A - correct answer M L4 3

Grade 12 – JIT 2020 – Term 2 2. TOPIC: Finance QUESTION 1 [15] Que Solution Explanation T/L 1.1.1 Expenses = R2 000 + R2 300 + R500 + R1500 + 1M F R275 + R350 + R250 + R460  1A L1 = R7 635,00 AO 1.1.2 Annual salary = R12 000 × 12  1M F = R144 000,00 1A L1 AO 1.1.3 Savings = R12 000,00 – R7 635,00  1M F = R4 365,00 1A L1 AO 1.2.1 Taxable income is the amount of income used to 2E F calculate how much tax an individual owes to the L2 government in a given year. 1.2.2 Pension contribution = 7,5% × R12000 1M F = R9000× 12 1A L2 = R108 000  AO 1.2.3 Taxable income = R144 000 – R10800 1M F = R133 200,00  1A L1 AO 1.2.4 Tax bracket 2 will be used.  2RT F L1 1.2.5 R7 740,00  2RT F L1 QUESTION 2 [15] Que Solution Explanation T/L 2.1 = R2 375.50 + (R2 375.50 × 0.09)  1 MA, multiplying by F = R 2 589.295  9%. L2 New price= R 2 589.30  1CA. 1R OR R2 375.50 × 0.09 = R 213.795  1 MA, multiplying by New price = R 2 375.50 + R213. 795 9%. New price = R2 589.295 = R 2 589.30 1M, adding 1CA, NPR OR = R 2 375.50 x  2 MA, multiplying by 109% = R 2 589.295.  1CA New price = R 2 589. 30. 4

Grade 12 – JIT 2020 – Term 2 2.2 Inflation is the increase in prices 2 E Accept Any valid F answer L1 2.3 Yes, the supply demand may decrease due to the 1A, Answer. F rise of prices. 1J, Justification L2 2.4 Inflation rate=  .    ×100 1 SF F 1 S L2   = 18.775% 1R Inflation rate = 18.78% 2.5 Deflation 2A F L1 QUESTION 3 [15] Que Solution Explanation T/L 3.1 29 February 2020 2 A, Answer with day, F month and year L3 3.2 Tax rebate is a tax concession or a type of discount. 2 E Accept any F  Reasonable Answer L3 3.3 Primary rebate - R34 220 1A F Secondary rebate - R7 794 1A L3 3.4 Annual salary = R22 370 x 32 2 MA, multiply by 32 B = R266 040 1 CA L3 3.5 Monthly Pension Fund Contribution: 1 M, multiply F = . x R22 370 1 SF L3 1A  = R3 662,75 3.6 Annual Pension Fund Contribution: 2 MA, multiply by 32 B = R3 662,75 x 32 1 CA L3 = R39 953 3.7 Taxable Income =Gross Salary - Pension Fund Contribution F = R266 040 -  R39 953 1 SF, CA from 3.4 & 3.5 L2 R246 087 1 M, subtraction 1 CA 3.8 Tax Bracket 2 2 CA, use answer from F Q3.7 to determine tax L2 bracket QUESTION 4 [20] Que Solution Explanation T/L 4.1 Mrs K. Ndlovu 2A F L4 4.2 490 kWh 2 RT F L4 5

Grade 12 – JIT 2020 – Term 2 4.3 24/04/2020 to 23/02/2020 2 RT F L4 4.4 05/03/2020 2 RT F L4 4.5 The account holder will be charged interest. 2A F L2 4.6 The amount has been deducted from the opening 2A F balance. L2 4.7.1 A = 4 349 kl - 4 342 klMA 1 MA subtract correct F = 37 kl values L2 1A 4.7.2 B = 45/400 x R64,04M 1 M - multiply by 45% F = R 9,456 1A L2 = R 9,46 1R 4.7.3 C = R752,69 - (R392,73 + R237,59 + R64,04) 1 MA subtract correct F = R 752,69 - R 694,36 values L2 = R 64,33 1S 1CA QUESTION 5 [19] Que Solution Explanation T/L 5.1 Cost/g = R 9,99 ÷ 400 g 1 MA divide by 400g F = R 0,02475 L2 = R 0,02 1R 5.2 Amount of avalable oil (ml) = 5 l x 1 000 ml/l 1C M = 5 000 mlA 1A L2 Number of bottles = 5 000 ml ÷ 750 ml/bottle 1 M divide by 750 = 6,6666…. bottles = 6 bottles 1 R round down 5.3 Selling price = Cost price + Profit 1 MA add correct F = R 119,99 + R 48,27MA values L2 = R 168,26A 1 A correct answer 5.4 Number of tins = 2packs/week x 12 tins/pack x 2 2 M multiply by 12 and M weeks by 2 weeks L2 = 96 tins 1 CA Number of grams = 96 tins x 400 g/tin 1 M multiply by 400g = 38 400 g 1 CA 5.5 VAT excusive price = R 119,99 ÷ 1,15 1 MA divide by 1,15 F = R 104,3391304 1A L2 = R 104,34 1R 5.6 Total cost = R89,99 + R 119,99 + (2 x 12R 9,99) 2M add all items , F = R 209,98 + R 239,76 multiply by 2 and 12 for L2 = R 449,74 beans 1 CA 6

Grade 12 – JIT 2020 – Term 2 3. TOPIC: Data Handling QUESTION 1 [17] Que Solution Explanation T/L 1.1 18 , 25 ,33 , 35 , 39 , 42 , 42 ,42 , 44 , 43 , 49 , 50 , 52 2 A for arranging DH , 65 , 74 , 78 , 80 , 81 , 86 , 92 , 98 L1 1.2 5 2A DH L1 1.3 Median= 49A 2A DH L1 1.4 3rd Quartile =     1 M adding scores DH  1 M diving by 2 L2 = 79 1 CA mark 1.5 Range = 98 – 18 1 M for range DH = 80 1A L2 1.6 Discrete data – a data that can be counted.E 2E DH Continuous data – data that is measured.E 2E L1 1.7 Mode = 42A 2A DH L2 QUESTION 2 [20] Que Solution Explanation T/L 2.1 Bar graph 2A DH L2 2.2 Minimum - R2 400 2 RG DH Maximum - R6 000 2 RG L2 2.3 Variable expense 2A F L2 2.4 Water and electricity : Rent B 3000 : 6000 2 RG L2 2 : 2 2S 2.5 R2 400 ; R2 500 ; R2 500 ; R2 500 ; 2A, correct ordering of DH R3 000 ; R3 000 ; R4 000 ; R6 000 numbers L2 2.6 Median = R2 500 2A, correct value DH L2 2.7 Mean = sum of data values 2MA, divide correct DH no. of data values values L2 = R2 400 +( R2 500 x 3) + (R3 000x 2) + 2 CA, correct answer R4 000 + R6 000 2.8 Mode = R2 5000 2A, correct answer DH L2 2.9 House maintenance, insurance policies, gifts, 2O, Accept any F entertainment, etc reasonable expense not L4 listed 7

Grade 12 – JIT 2020 – Term 2 QUESTION 3 [16] Que Solution Explanation T/L 3.1 Observation 2 A correct method DH L3 3.2 50 plants 2RT correct answer from DH table L3 3.3 20cm - 24cm 2RT correct answer from DH table L3 3.4 35cm - 39cm 2RT correct answer from DH table L3 3.5 36 plants 2RT correct answer from DH table L3 3.6 6 MA one mark for DH correct tally in each L2 category of plant height QUESTION 4 [18] Que Solution Explanation T/L 4.1 Number who wrote = 20A 2A DH L1 4.2 97 62 48 48 47 44 44 37 33 32 32 24 18 18 18 18 14 2 A - descending order DH 11 9 A L1 4.3 Mean = 694 %M ÷ 20 learnersM 1 M add all values DH = 34,7 1 M divide by 20 L2 = 34 %CA 1 CA NPR 4.4 Median = (32 % + 32%) ÷ 2M 1 M median concept DH = 64% ÷ 2 with correct values L2 = 32%CA 1 CA -answer 4.5 Range = 97% - 9%M 1 M range concept DH = 81 %CA 1 CA- answer L2 8

Grade 12 – JIT 2020 – Term 2 4.6 Outlier A 1 A - correct answer DH L1 4.7 Mark interval Tally Frequency 1 A - correct tallies DH (%) 2A - correct frequencies L2 0 - 29 IIII III 8 30 - 39 IIII 4 40 - 49 III A 4 A 40 - 49 I 1 60 - 69 I 1 70 - 79 0 80 - 100 I 1 4.8 Modal interval = 0 - 29 %A 1 A - correct answer DH L2 QUESTION 5 [10] Que Solution Explanation T/L 5.1 A = Tails A answer P B = Heads A answer L1 C = H, H, H A answer 5.2 8 2 RD reading from the P diagram L1 (2) 5.3 Probability (2 heads+1 tails) =   P   1 A numerator L1 =   1 A denominator =   1 CA answer  (3) 5.4 Probability(6) =   P   1 A numerator L1 =   1 A denominator (2) 9

Grade 12 – JIT 2020 – Term 2 4. TOPIC: Maps and Plans QUESTION 1 [15] Que Solution Explanation T/L 1.1.1 One unit measurement on the map represents four 2E MP million and five hundred thousand units on actual L 1 ground/reality. 1.1.2 Mkuze game reserve 2 RT MP Tembe Elephant game reserve  L1 1.1.3 1 : 4 500 000 1 M measured length MP 55mm : ? (allow 54 -56 mm) L 3 ? = 247 500 000mm÷ 1 000 000 1 M dividing by ? = 247,5 km 1 000 000 1 CA 1.1.4 East 2 A: Direction MP L1 1.1.5 Time = MP 1M distance L3 = 1 M Speed =3,031860226 hours x 60 = 181,9116136 minutes  - 180 minutes 1S = 3 hour 2 minutes  1CA hours & minutes QUESTION 2 [17] Que Solution Explanation T/L 2.1.1 4 water points  2 RM MP L1 2.1.2 Difference in km= 25km – 10km 1 M : values MP = 15km 1 MA L1 1 M :answer in km 2.1.3 This map is an elevation because it shows information 2 E MP about profile of a route as seen from the side. OR L1 Because it shows the height of the physical part of the earth relative to sea level 2.1.4 25 km × 1000 1M multiply by 1000 MP = 25000m 1 CA L1 2.1.5 To help participants who need medical attention. 2E MP (any valid explanation) L1 2.1.6 Time 1M: distance MP 1M :time L3 = 25km ÷ 3,6491 hours 1M:Conversion = 6,85 km/h 1CA: answer QUESTION 3 [13] 10

Grade 12 – JIT 2020 – Term 2 Que Solution Explanation T/L 3.1.1 C 2 2 M Grid MP L1 3.1.2 Bar : 42 mm : 100km 1M : A MP 55 mm : 131 km L3 Number scale : 55mm × 2 500 000 1M : dividing by = 1 375 000 ÷ 1 000 000 1 000 000 = 137,5 km 1M:CA 3.1.3 Bar scale is more reliable 2E MP L1 3.1.4 From Nongoma take R 66 in the northerly to Piet 2 M correct direction MP Retief 2 M correct direction L1 From Piet Retief take same R 66 to reach Ermelo QUESTION 4 [13] Que Solution Explanation T/L 4.1.1 Port Shepstone 2 RT MP Umtata L1 4.1.2 Bar scale 2 A Type of scale MP L1 4.1.3 It stays accurate when maps are reduced or 2E MP enlarged L1 4.1.4 20mm = 300 km 1 M Concept of scale MP 20mm =300 x 1000 000  1 M measured distance L2 20mm = 300 000 000 1A 1 : 15 000 000 4.1.5 20mm = 300 km 1 M measured length MP 32mm= ? 1 M multiplication by 300 L2 1A = 480 km  QUESTION 5 [10] Que Solution Explanation T/L 5.1 Four 2 RM MP L1 5.2 Distance = 964 – 504 1 M Diff in distance MP = 460km 1A L2 5.3 N2 2 RM MP L1 5.4.1 Port Alfred 2 RM MP L1 5.4.2 South East 2 M Direction MP L1 QUESTION 6 [43] 11

Grade 12 – JIT 2020 – Term 2 Que Solution Explanation T/L 6.1. Four members  2A correct answer MP L1 6.2 964 km  2A correct reading from MP map L1 6.3 R61 2RM MP L1 6.4.1 9 NATIONAL ROADS 2 RM MP L1 6.4.2 N1 2RM MP L1 6.5 Addo Elephant national park 2 RM MP L1 6.6 MP 1 M dividing correct L2  values  1A  1R 6.7 N2 coffee bay  1 RM MP 1 A correct place L1 6.8 Distance = 193km + 904km 1 M addition MP =1097km  1A L1 6.9 Time taken = 15:45 – 04:30 1Msubtraction of times MP =11hours 15minutes 1 A(number of hours) L2  1 M correct formula speed  1S =85.69km/hour 1A 6.10 Arrival= 3:45pm departure= 4:30am 1 RT correct time MP 1 RT correct time L1 6.11 It is because the strip map only shows the actual 2E MP distances not how they have been scaled  L1 6.12 8litres = 100km 1 M proportion MP 60litres =km  1M L2  1A  6.13 8litres = 100km MP Litre s = 964km  1 M proportion L3  M 1M  1A Cost of petrol= 77.12 x R15,80 1 M multiplication =R1218.496 1A =R 1218.50 1R 12

Grade 12 – JIT 2020 – Term 2 QUESTION 7 [16] Que Solution Explanation T/L 7.1 Three 2 RM MP L1 7.2. Bontebok and Tsitsikamma National Parks  1RM MP 1RM L1 7.3. BEAUFORT WEST  2 RM correct answer MP L2 7.4. DIRECTIONS MP • From Ladysmith take R323 to the N2 1 E road R323 L2 • Turn left on the N2 1 E correct instruction • Go straight with N2 to port Elizabeth 1 E road R335 • From port Elizabeth take R335 1A • Go straight until you see Addo Elephant park  7.5 Route one MP Distance = 46km + 50km +29km +146km 1 M addition of values L2 = 271km from Map 1A Route two Distance = 51km + 304km + 82km  1 M addition of values = 437km from Map Route one is the shortest 1 CA QUESTION 8 [16] Que Solution Explanation T/L 8.1.1 Bar scale 2 A type MP L1 8.1.2 4mm = 20 km 1 M Concept of scale MP 4mm = 20 000 000 mm 1 M Conversion L1 1 : 5 000 000 1 M Ratio 8.1.3 Number sale is represented in a ratio form while a bar 2E MP scale has segments. L1 Bar scale is more reliable because even when the 2E map is resized the measurements remain the same. 8.2.1 70mm 2 M measured length MP L1 8.2.2 Five pockets  2 M Number of pockets MP Accept: 5 pockets L1 8.2.3 1 : 20 1 M concept of scale MP 70mm ? L1 ? = 1400 ÷ 1000 1 M dividing by 1000 ? = 1,4m 1 M Answer QUESTION 9 [13] 13

Grade 12 – JIT 2020 – Term 2 Que Solution Explanation T/L 9.1.1 Floor plan: plan which shows designs and dimensions MP of the inside of a building from side view. 2E L1 9.1.2 North West 2 MA direction MP L1 9.1.3 1 : 100 2M MP L1 9.1.4 One unit measurement on the floor plan represent 2M MP ONE hundred units on actual map  L1 9.1.5 1 : 100 MP 140mm : ? 1 M 130 mm L2 1 M dividing by 1000 1 M answer 9.1.1 Floor plan : plan which shows designs and MP dimensions of the inside of a building from side 2E L1 view. QUESTION 10 [10] Que Solution Explanation T/L 10.1 Number or ratio scale 2 M correct scale MP L1 10.2 1 unit plan represent 20 unit in reality  2M correct explanation MP L1 10.3 12 rooms  2 RP MP L1 10.4 SE or south east 2M correct direction MP L2 10.5 Probability (RS) = 0 2M correct answer MP L2 QUESTION 11 [17] Que Solution Explanation T/L 11.1 Bar scale  1 correct scale MP L1 Advantage: if the plan is enlarged or reduced, the 1E scale also reduce or enlarge with the same proportion  11.2 3m 2 RP MP L1 11.3 2 baths 1 RP correct baths MP L1 2 bedroom 1 RP correct bedrooms 11.4 Windows= 10  1 RP correct windows MP L1 Doors = 5 1 RPcorrect doors 11.5 Left-handed  1 A correct choice MP L1 14

Grade 12 – JIT 2020 – Term 2 Because the circular part which shows the path of the 2E door is on the right, to open that door one must use left hand 11.6 Bathrooms : bedrooms 2 A correct ratio MP L1 2  : 3 11.7 Measured length of the scale = 2,8cm  1 A length MP L1 2,8cm = 3m 1 A proportion 8cm = m  1M  1 CA =8,6 m QUESTION 12 [10] Que Solution Explanation T/L 12.1 10 seats  2 RM MP L1 12.2 195 seats  2RM MP L1 12.3 A10, A9 2RM any two correct MP L1 A8,A7 seats A6 ,A5 12.4  1M probability fraction MP L2 1M multiplying by 100  1answer  12.5 B16 2M correct seat MP L2 QUESTION 13 [8] Que Solution Explanation T/L 13.1 9 rooms 2 A correct value MP L1 13.2 11 windows  2 A correct number of MP windows L1 13.3 9 doors  2Acorrect number of MP doors L1 13.4 South East 2A correct compass MP or direction L1 SE 15