GRADE 12 TERM 2 3 SOLUTIONS

GRADE 12 TERM 2 CONTENT MANUAL MATHEMATICAL LITERACY SOLUTIONS TO ACTIVIIES May 2021 S

ACTIVITY 1 BANK LOAN AND INVESTMENT ACTIVITY 2 TIME TABLES ACTIVITY 3

580 000 _ Distance: ; = 580km 640 km Full tanks: Saag tas = 1,103 tanks For the return trip = 1,103 x 2 tanks = 2,206 tanks They will need more than 2 full tanks. Distance speed x time = 135 km/h x 2,5 hours 337.5 km Mr Moabi travelled 337,5 km. 1.1 1.2 1.3 2.1 2.2 2.3 2.4 2.5 2.6 3.1 3.2 3.3 1 yard = 36 inches 1 litre = 1,76 pints 3,784 litres =?? ? = 3,784 x 1,76 = 6,685984 3,784 litres = 6,686 pints lkilometre = 0,6284 miles 1190 miles = ?? kilometres ? = 1190 x 0,6284 = 747,796 1190 miles = 748 kilometres 44g = 0,044kg 330 x 0,07 = 231g 6 x 3 = 18 servings 3 x 230g = 690g 10:20=1:2 Cholesterol = 231g 13% + 5% + 7% + 30% + 16% + 7% + 6% + 20% + 10% + 4% = 118% For | pancake: 2 tablespoons sugar For 2 pancakes: 4 tablespoons; 4 x 15ml = 60ml sugar For 30 pancakes: 30 cups flour 30 x 240ml = 7 200m! flour For 30 pancakes: 30 tablespoons vegetable oil 30 x 15ml = 450ml vegetable oil

ACTIVITY 2 1 d=11,5em p= HEE =5,75em Area of 1 pancake = 3, 142 x (5, 75)2 = 103, 8882375 cm?* Area of 10 pancakes = 10 x 103, 8882375 cm? = 1038, 82 cm? 2, 2.1 Area of cake = 3,142 x 42 = 50,3 cm? 2.2 Area of tray = 3,142 x 802 = 20 108,8 mm2 2.3 Surface area of a cake = 2 x 3,142 x 42 + 2 x 3,142 x 4 x 8 = 301,63 cm? Cost of material = 301,63 + 100 = 3,0163 x R2,95 = R8,90 ACTIVITY 3 1. 1.1) 2 days 1.2 20x 60 = 1 200 minutes 13° 15 minutes 14 18:21 1s 0 1.6 1 hour 45 minutes 17 Speed = = 59.5mh ACTIVITY 4 1. LL 80 litres 12 = 255x899 = 2293 litres 13 92.65 8,99 = 58,44 litres 100 14 I Mitre = R13, 58,44 litres = RI? R13,90 x 58,44 = R812,32 1.5 False, the car will not be able to carry it because the maximum mass it can carry is 938kg.

ACTIVITY 5 Ll 12 13 14 21 22 23 31 32 33 34 35 4.1 4.2 43 ‘Total length: 45 cm = 450 mm Height of Athlete 2: 175 cm = 25 cm = 150cm + 100= 15m Height of the winner (Athlete 1): 1,5 m + 0,5 m = 2,0 m Volume of a cube: s* = (45) =91 125 em? Volume of a cylinder: me? h = 3,142 x (0,009)? x 1,850 = 0,004 m! = 4.000 litres ‘The first flush of water from the roof may contain bacteria from decomposed insects and skinks, and bird and animal droppings, Instead of flowing into the water tank, these pollutants are diverted with the initial flow of water into the chamber of the water diverter, Number of litres needed per day; 500 x 350 = 175 000 litres, Total daily supply by municipality: 100 000 litres 175.000 - 100 000 = 75 000 litres more will be pgeded. A perimeter is the distance around a two-dimensional shape or the length of the outline of a shape. Perimeter = (2)350 cm + (2)200 cm = 700 cm + 400 cm = 1 100 cm Width = diameter = 200 cm, but radius = 2m + 2=1m Surface area = 3,142 x 12 x 3,5 = 38,5m2 No, he will not be able to enter without bending because the height of the tunnel is 1,8m and he is 1,9m Total floor area of the house = Area of Room 1 + Area of Room 2 = (3,5m x 2,3 m) + (1,85 m x 1,9 m) = 8,05 m? + 3,515 m2 = 11,57m No. of tiles for Room 2 = Area of Room | + Area of a tile = 3,515 +0, 1225 = 28, 69388 = 29 tiles Costs of tiles for two rooms = (Total floor area of the house ~ Area of cabinet) = (11,57 m? - 0,6 m’ )x R159,84 = 10,97x R159, 84 = RI 753,45

MAP ACTIVITY 1 SOLUTIONS 1.1 Southerly direction / South direction 1.2 One way street 1.3 Commissioner street 1.4 North East/ NE 1.5 She has to turn to Commissioner street, go straight until she turn at Rissiki street then turn righto Bree street lastly turn right to King George MAP ACTIVITY 2 SOLUTIONS 2.1 Brits - Cradock Brits = 575 km Cradock = 284 km 575 – 284 = 291 km 2.2 Brits – East London Brits = 950 km East London = 375 km 950 – 375 = 575 km 2.3 T= D × SPEED = 284 × 93.3 km/h 2.4 JANSEENVILLE 2.5 3.7 cm : 95 km 3.7 95 𝑘𝑚 : 3.7 3.7 1 : 25.675 MAP ACTIVITY 3 SOLUTIONS 3.1 ENTRANCE 5 3.2 G 14 is DIS – CHEM 3.3 OPPOSITE 3.4 Enter in ENTRENCE 1,go straight passing blocks 23,24, 25,26and turn right at block 31 go straight passing CAN block 37,19 on the left then on the right hand side is G 40 which is bakery Fournos. 3.5 Impossible MAP ACTIVITY 4 SOLUTIONS 4.1 A 4.2.1 1 : 1.609 124 km : ? 1.609 × 124 km 199.52 km 4.2.2 100 km : 6.4 litres 124 km : ? = 7.936 litres 4.3 LESOTHO

4.4 South West 4.5 MPUMALANGA 4.6 MOOI RIVER,HOWICK, PETERMARITZBURG and PINETOWN MAP ACTIVITY 5 SOLUTIONS 5.1 From Melmoth to Durban Use R 34 –R 64 – R 102 5.2 R 34 – N2 5.3 From N 2 – R 102 Off ramp to left side to the Zinkwazini beach turn to the left and join N 2 - DURBAN 5.4 Scale: EXAM QUESTION

=21,8m 3.2 Surface area = 3,142 x 2,22+46,5 = 21,707 m? Amount of net = 21,707 x R71, 99 =R1 526,71 3.3.1 Volume of a cylinder = 3,142 x 2,752 x 1,2 m = 21,51 m} (a) Volume =12 x 0,6 x 0,1 =0,72 m3 072m )coa7m = 33.3 75:50 125: 2% x75 =125x 50 26,666 = 27 bags 75 83,3 kg of sugar will be needed. SCALE AND MAPS 1. yo YP A scale is a ratio of a map or plan distance to the actual distance. Bar scale South-west Approximately 107mm 20mm on the map represents 300km on the ground Distance (km) = etd =1605km 20mm 7.1 42mm 7.2. Distance (in km) = 42mm x 15 000 000 = 630 000 000mm = 630 km 798km x 20mm _ 53 kn mm Map distance (in mm) = (3) (4) (3) (2) (2) (3)

1, 7300mm 2, 7,1m 3. Area =7,3m x 7,1m = 51,83m2 4. Plan width = 71mm, actual width = 7 100mm therefore, scale is 1;100 9 Plan width (window) = a =9,49mm MAPS 1. 557km . N3 3. Distance from Harrismith to Pietermaritzburg = (289 — 69)km = 220 km OR (488 — 268)km = 220 km 4. 4.1 Take the N3 from Pretoria and drive past Heidelberg and Villiers to Harrismith. In Harrismith, turn left onto the N5 to Bethlehem. From Bethlehem take the N5 back to Harrismith. From Harrismith turn right onto the N3 and continue along this route. Turn left onto the N11 to Ladysmith, From Ladysmith take the N11 back to the N3 and left towards Escort. Continue on the N3 to Pietermaritzburg. 4.2 3,53 hours distance 4.3 Speed = ra 326 km “353h = 92,35 km/h 4.4 Given: speed = 120 km/h, time (for the journey) = 8 h, 35 min, time (stops) =2h, 15 min Time (driving) = 6,33 h Therefore, distance travelled = speed x time = 120 km/h x 6,33 h = 760 km Distance travelled — = 760 km, fuel consumption rate = 7,2 litres/100 km, fuel tank capacity = 50 litres Fuel used for 760 km = 760 km x 7,2 litres/100 km = 54,72 litres Mr Mtyanti did NOT travel on a single tank.

RESTAURANT ACTIVITY PLANS ACTIVITY SEATING

1. 9cm 2. Perimeter = 2(length + width) = 2(90 mm + 55 mm) =2x 145mm = 290 mm 3. Given: volume (1 litre) = 1 000cm? Volume = length x width x height eee volume «. Height = inet = 1 000cm3 / 49,5 cm? = 20,2 cm 4. Scale = picture length (in cm) : actual length (in cm) =15cm9 cm =16 OR : Scale = picture length (in mm) : actual length (in mm) = 15mm:90 mm =k6 5. 5.1 6-pack = 6 litres = 6 000m! ml +. 6-pack = pana” Ipint = 12,68 pints 5.2 Length (6-pack box) = 90 mm x 3 = 270 mm 5.3 Width (6-pack box) = 55 mm x 3 = 165 mm SEATING PLAN ACTIVITY 11.1.1 Row nearest to the court OR row with uninterrupted/best view. (2) 1.1.2 South-east (2) 1.13 Category 3 (2) 1.14 Walk straight east past the court, Turn right and walk behind the cameras, On-court seats will be on your lelt-hand side, (4)

ARENA ANSWERS 54 5.5 Packaging option 1: perimeter = 2(180 mm + 165 mm) =2x345 mm = 690 mm Packaging option 2: perimeter = 2(270 mm + 110 mm) =2 x 380mm = 760 mm . Packaging option 2 has the largest perimeter. Base length = 2,4 m = 2 400 mm Base width = 1 200 mm Box width along the base length Number of boxes (option 2) = ae 1200 mm 1100 mm Therefore, there are 8 x 10 = 80 boxes of option 2 (1 layer) Box width along the base length Number of boxes (option 2) = < — = 21 (along the length) = 8 (along the length) Number of boxes (option 2) = = 10 (along the width) Number of boxes (option 2) = ae Therefore, there are 21x 4 = 84 boxes Maximum number of boxes = 84 = 4 (along the width)

PACKAGING ANSWERS TO THE RDP FLOOR PLAN ACTIVITY

Floor area (bedroom 2) « 3550 mm x 3550 mm = 12.602 500 mm? Difference = (13 312.500 ~ 12.602 500) mm? = 710000 mm? (4) 3.1.4 No privacy required. (2) 3.2 3.211 South elevation (2) 3.2.2 Percentage discount = ec x 100% = 15% (3) 3.2.3 Length (in metres) = = 7,1 m Width (wall height) =2,7m Area (rectangular part) = 7,1 m x 2,7 m= 19,1 7m? Area (windows) = 2(1,511m x 0,949m) = 287m? Area (gable) = i x71 mx 24m) = 852m2 Area (to be painted) = total wall area ~ (area windows (W1) + area door opening) = (19,17m? + 8,52m?) - 287m? - 2,1m? = 22,72m? (12) 3.2.4 Total area to be painted = South area + North area + East area + West area South area = 22,72 from 3.2.3 above North area = 19,17 m? + 8,52 m?- 2,1 m? ~ (area W1) = 27,69 m? - 2,1 m?- 1,43 m? = 24,16 m? East area =7,3 mx 2,7 m=19,71 m? West area = 19,71 m2 = 2(1,022 m x 0,940 m) = 19,71 m?- 1,94 me? =17,77 mt? Therefore, total area to be painted = 22,72 m? + 24,16 m? + 19,71 m? + 17,77 m? = 84,36m2 2 So, number of litres of paint = = = 17 litres Two coats require 34 litres of paint Therefore, number of 5-litre buckets = =7 (11)

SOLUTIONS FREE STATE QUESTION 3 QUESTION 3 Marks Solutions Explanations Marks 3.1.1 Area of circle (hoop) = 3.142 ×( 22.5)2 1 A finding radius = 1590.6375 cm2 1 SF substitution Area of the ball = 𝜋 ×(radius)2 1 CA Simplification = 3.142 ×(12.4)2 1 A radius of the = 483.11392 cm2 ball 1590.6375 (7) The hoop = 483.1139 1 CA simplification = 3.3 Bigger than the ball 1 M Divide with curect values 1CA correct answer 3.1.2 12 12 Foot = m 3.28084 1 C convert to m = 3.657 76m 1 SF substitution (3) Perimeter of a circle = 3.142 ×3.657 76m 1 CA perimeter = 11.493m 3.1.3 Area = 4.9 × 5.8 ×2 1SF substitution = 56.84m2 1 A multiplying by (3) 2 1 CA area , Arranging scores in Ascending or descending 1 A arranging data1 order A identification (2) 27: 28: 30: 32: 34: 38: 41: 42: 43: 43: 44: 46; 53: 56: 62 Median point is 42

3.2.1(b) IQR – Upper Quartile – Lower Quartiles 2A identifying = Q3 – Q1 quartiles = 46 – 32 1 SF substitution (4) = 14 1CA simplification 3.2.2 B: G and M 2A if all three are (2) correct OR 1 A if two correct 3.2.3 2 1 A numerator of 2 Probability in % = 15 ×100% 1A denominator of = 13.33% 15 (3) 1 CA probability QUESTION MARKS Solutions Explanations Marks 4.2.1 4 Stops, stop 200 km (Leave Cape Town drive 200km Stop. 200km 2 A number of (2) Stop, 200km Stop. 200 km Stop 242km in stops EL) 4.2.2 Drive in SW direction on N1 to Collesburg 1 A N1 Proceed on the N9 to Middelburg 1 A Collesburg Continue in SE on the N 10 1 A Middelburg At the junction turn left or NORTH EAST 1A N10 (7) into N2 1A Left or NE Proceed on the N2pass Grahamstown to 1A Grahamstown EAST LONDON 4.2.3 Bloemfontein to East London = 575 km 1RT Distance from Bloemfontein to Collesburg = 219 km Bloem EL Collesburg to East London = 518 km 1RT Distances Total via Coleburg = 737 km 1CA total distance 1 RT KIM to EL (8) Kimberly to East London = 750 km 1RT KIM TO Kimberly to Collesburg = 284 km Colesburg Collesburg to East London = 518 km 1CA Total via KIM Total via Coleburg = 802 km 1 O choosing It will look better if Thaba picked her up. Thaba The detour that he would have to make is 1J reasoning with only 52 km. distances From Bloemfontein the detour is 162 km 4.2.4 Via Koffiefontein. Phillipolis 1 A using the Distance = SPEED ×TIME correct route 268 km = SPEED ×4 Hours 11 minutes 1SF substitution (4) 268 𝑘𝑚 1C converting time SPEED = 1CA speed 4.18333𝐻 = 64 .06 km/h