Mathematical Literacy Grade 12 P2 2021 Memo (Eng)

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 SEPTEMBER 2021 MATHEMATICAL LITERACY P2 MARKING GUIDELINE MARKS: 150 Symbol Explanation M Method MA Method with accuracy CA Consistent accuracy RCA Rounding consistent accuracy A Accuracy C Conversion S Simplification SF Correct substitution in a formula J Justification O Opinion/Example/Definition/Explanation/Justification/Verification RT/RG/RM Reading from a table/graph/map P Penalty, e.g. for no units, incorrect rounding off etc. R Rounding off NPR No penalty rounding or omitting units AO Answer only, full marks This marking guideline consists of 10 pages.

Downloaded from hlayiso.com 2 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2021) MARKING GUIDELINES NOTE:  If a candidate answers a question TWICE, only mark the FIRST attempt.  If a candidate has crossed out (cancelled) an attempt to a question and NOT redone the solution, mark the crossed out (cancelled version)  Consistent Accuracy (CA) applies in ALL aspects of the marking guidelines; however, it stops at the second calculation error.  If the candidate presents any extra solution when reading from a graph, table, layout plan and map, then penalise for every extra incorrect item presented. LET WEL:  As ʼn kandidaat ʼn vraag TWEE keer beantwoord merk slegs die EERSTE poging.  As ʼn kandidaat ʼn antwoord van ʼn vraag doodtrek (kanselleer) en nie oordoen nie, merk die doodgetrekte (gekanselleerde) poging.  Volgehoue akkuraatheid (CA) word in ALLE aspekte van die nasienriglyn toegepas, maar dit hou by die tweede berekeningsfout op.  Wanneer ʼn kandidaat aflees van ʼn grafiek, tabel, uitlegplan en kaart en ekstra antwoorde gee, penaliseer vir elke ekstra item. Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICAL LITERACY P2 3 KEY TO TOPIC SYMBOL: F = Finance; M = Measurement; MP = Maps, plans and other representations; P= Probability QUESTION 1 [26 MARKS] Quest Solution Explanation Level 1.1.1 25 mg : 1 000 mg M (divide by 25) 1M ratio concept M = 1 : 40 MA 1MA Simplification L1 (2) 1.1.2 Tablets per day: M 25 × 2 MA 1MA multiplying by 2 L1 = 50 Tablets per week: 1MA multiplying by 7 50 × 7MA 1A answer = 350 mg A (3) 1.1.3 Number of days to use tablets: 1M finding the number of M 60 tablets days M days L1 Last day: 30 March 2021 M 1M Referring to last date Therefore, from the 31st March 2021 a refill is 1O Conclusion needed. O (3) 1.1.4 25 mg × 120 RT 1RT multiply correct values M = 3 000 mg ÷ 1 000 C 1C divide by 1 000 L1 = 3 g A 1A answer (3) 1.2.1 The “North Elevation” shows the side view of the 2A correct explanation. MP building from the northern direction. A L1 (2) 1.2.2 (a) A floor plan shows a top view of the inside of a 2A correct explanation MP building. A (2) L1 (b) An elevation plan shows the side view of the 2A correct explanation MP outside of the building. A (2) L1 1.2.3 Scale = 1 : 100 MP Measured length: 5 cm L1 Actual length: 5 cm M 1M multiply by scale = 500 cm MA 1MA answer (2) 1.3.1 90 km 1000 100 × 10 C M = 90 000 000 mm A 1C multiply by 1000 000 L1 OR C 1A correct answer 90 km × 1 000 000 = 90 000 000 mm A (2) 1.3.2 No of tiles per box M 1M dividing by 0,36 M 1A number of tiles L1 4 tiles per box A (2) 1.3.3 1,125 hours M 1 hour 1C finding minutes L1 0,125 hr. × 60 = 7,5 min C 1C finding seconds 0,5 min × 60 = 30 seconds C 1A correct answer 1,125 hours = 1 hr. 7 min 30 seconds A (3) [26] Copyright reserved Please turn over

Downloaded from hlayiso.com 4 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2021) QUESTION 2 [28 MARKS] Quest Solution Explanation Level 2.1.1 George; Knysna; Mossel Bay RT 2RT names of towns MP (Any two of the three) (2) L1 2.1.2 South East  A (accept East) 2A Direction MP North East A 2A Direction L1 (4) 2.1.3 National roads connect major cities. O 2O Reason 1 MP Save on fuel consumption. O 2O Reason 2 L4 Fewer traffic lights O Travel faster O [Any two or any acceptable explanation.] (4) 2.1.4 2,3 cm = 300 km A 1A measured bar scale MP Measured distance = 7 cm A 1A measured distance L2 range for bar 2.2cm to 2,4cm Actual distance = km M measured map range distance from 6,9 to 7,1 1M concept for ratio 913 km CA 1CA rounded answer (4) 2.1.5 Distance = Speed Time Accept CA from Q 2.1.4 MP 913 = 100 TimeSF 1SF substitute correct values L4 1M changing the subject of the = M formula 1CA time in hours 1 C hours to minutes = 9,13 hrs. CA 1M adding time 1CA arrival time = 0,13 60 1J justification = 7,8 minutes ≈ 8 minutes C Time of arrival = 9hrs 08 min 1hr 08:00 M = 18:08 CA They will arrive later than the planned time. J (7) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICAL LITERACY P2 5 Quest Solution Explanation Level 2.2.1 7,6 litres = 100 km Accept CA from MP Petrol Consumption = M Q 2.1.4 L2 1M concept of ratio = 9,13 = 69,388 litres S 1S simplification 1M multiply by 2 Return trip Consumption = 69,388 2 M 1CA Answer = 138,776 litres CA (4) 2.2.2 1 642 cents 100 R16,42 M Accept CA from MP Q 2.2.1 L2 Petrol cost R16,42 138,776 litres CA 1M Conversion R 2 278,701 1CA multiply by 138,78 litres R 2 278,70 CA 1CA Answer (3) [28] Copyright reserved Please turn over

Downloaded from hlayiso.com 6 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2021) QUESTION 3 [34 MARKS] Quest Solution Explanation Level 3.1.1 Radius: 18 cm 2 = 9 cm M 1M radius M Conversion to mm = 9 cm × 10 = 90 mm C 1C conversion L3 Area = 1SF correct substitution = 3,142 90 mm × 90 mm SF 1S simplification = 25 450,2 S 1R rounding = 25 450 mm2 R (5) 3.1.2 Radius: 250 mm ÷ 2 = 125 mm MA 1MA finding radius M Circumference = 2 × π × r L2 = 2 × 3,142 × 125 mm SF 1SF correct values = 785,5 + 50 mm M 1M adding 50 mm = 835,5 mm CA 1CA answer OR OR Radius: 250 mm ÷ 2 = 125 mm MA 1MA finding radius Circumference = π × D = 3,142 × 250 SF 1SF correct values = 785,5 + 50 M 1M adding 50 mm = 835,5 mm CA 1CA answer (4) 3.1.3 Conversion: length = 400 mm 10 = 40 cm C 1C conversion M Volume =l b h 1SF in formula L3 3 42 000 cm = 40 cm 30 cm h SF 1S changing the subject Height = 42 000 cm3 ÷ 1200 cm2 S of the formula = 35 cm CA 1CA answer (4) 3.1.4 25,4 mm = 1 inches M Diameter in mm = 250 mm RT 1RT using correct L4 Diameter in inches = 250  25,4 values 250 mm and 25,4 = 9,84 inches MA 1MA convert to inches Statement is not valid O 1O (3) 3.1.5 Radius value = 250 mm ÷ 2 1MA finding the radius M = 125 mm MA 1M percentage L2 Height = 0,64 × 125 mm M calculation = 80 mm CA 1A answer (3) 3.1.6 Flour : Sugar (3 cups) M 7 cups : 3 cups (× 2) L2 14 cups : 6 cups 1MA dividing by 3 = MA =2 1MA multiplying by 7 = 2 × 7 MA = 14 cups of flour A 1A answer OR 1MA dividing by 3 MA × 6 MA = 14 cups of flour A 1MA multiplying by 6 1A answer (3) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICAL LITERACY P2 7 Quest Solution Explanation Level 3.1.7 °C = (°F – 32°)  1,8 M  SF 1SF correct value L2 = (365 °F – 32°)  1,8 = 333 °F  1,8  S 1S simplify = 185°C 1R rounding to 10  190 °C  R degrees (3) 3.2.1 Tree diagram A 2A correct answer P (2) L1 3.2.2 Missing value P: Boy RT 1RT correct answer P Missing value Z: GBB RT 1RT correct answer L1 (2) 3.2.3 (a) Probability (2 girls at least) = RTRT 1RT Numerator P 1RT Denominator L2 = CA 1CA simplification (3) 3.2.3 (b) Probability (BGB) = % A 2A correct percentage P (2) L2 [34] QUESTION 4 [32 MARKS] Quest Solution Explanation Level 4.1.1 Basin / Wash basin RT 2RT correct feature MP (2) L1 4.1.2 1 Window RT 2RT correct number MP (2) L1 4.1.3 Width (bedroom 1 and bedroom 2) = 4 680 + 5 130 MA 1MA adding correct M = 9 810 A values L2 1A length Length of bathroom = 13 680 – 9810  M 1M finding length of = 3870 CA bathroom 1CA answer Wall (minus door opening) = (3 870 – 860) mm  M 1M subtracting door = 3 010 mm  CA width 1CA answer (6) Copyright reserved Please turn over

Downloaded from hlayiso.com 8 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2021) 4.2.1 Bathroom Area = Length × Width M = 3 870 mm × 2 250 mm SF 1SF correct values L3 A 1A finding area = 8 707 500 mm2 ÷ 1 000 000 C 1C ÷ by 1 000 000 = 8,7075 m2  CA 1CA for the area Kitchen Area = Length × Width SF 1SF correct values = 6 030 mm × 5 130 mm A 1A finding area = 30 933 900 mm2 ÷ 1 000 000  C 1C ÷ by 1 000 000 = 30,9339 m2 CA 1CA for the kitchen area 1CA total area Total Area = 8,7075 + 30,9339 (9) = 39,6414 m2 CA NPR 4.2.2 Area of 1 tile: 500 mm 500 mm CA from 4.2.1 M = 250 000 mm2 MA (1000 000) C 1M area of tile L4 = 0,25 m2 A 1C divide by 1 000 000 No. of tiles needed = 39,6414 m2 ÷ 0,25 m2 MCA 1MCA ÷ total area by =158,5656 × 1,05M 0,25 m2 =166,49388 tiles 1M increasing by 5% = 167 tiles CA 1CA rounded no. of tiles Statement not valid. O 1O conclusion (7) 4.2.3 No of boxes = 167  4 MCA CA tiles from 4.2.2 M = 41,75 1MCA dividing tiles by 4 L4 = 42 boxesCA 1 number of boxes 1M multiply by cost Amount for tiles = 42 × 249,90 M 1CA cost of tiles = R10 495,80 CA 1M adding labour cost Total amount = 10 495,80 + 8 186,09 = R18 681,89 M 1O conclusion Claim is valid. O (6) [32] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/SEPTEMBER 2021) MATHEMATICAL LITERACY P2 9 QUESTION 5 [30 MARKS] Quest Solution Explanation Leve l 5.1.1 Convert: 175 100 1,75 m C 1C conversion M ass in L3 BMI ei ht × m2 ass in 1SF correct values 25,1 SF ,7 m × ,7 m Mass BMI (height)2 1S changing subject of 25,1 (1,75)2 S formula Mass in (kg) 76,86875 kg 1CA answer 76,87 kg CA (4) 5.1.2 Overweight  J 1J conclusion M (2) L1 5.1.3 The young promising rugby player can eat 2O Explanation M healthier. O L4 OR He should exercise more regularly O (2) 5.2.1 3 laps = 5 000 m ÷ 1000 C 1C divide by 1 000 MP 1 lap = 5 km ÷ 3 MA 1MA divide by 3 L2 = 1,67 km CA 1CA distance per lap OR 1MA divide by 3 1 lap = 5000 m ÷ 3 MA 1C divide by 1000 = 1 666,666667 ÷ 1 000 C 1CA distance per lap ≈ ,67 mCA (3) 5.2.2 Incomplete runs: = 4 laps × 1,67 km M 1M multiply by lap MP = 6,68 km A distance L4 1A correct distance Complete runs = 3 × 5 km 1MA finding complete run = 15 km + 6,68 km MA distance = 21,68 km CA 1CA finding total distance His statement is not valid. O 1O correct conclusion OR 1M multiply 4 by lap M M distance (4 × 1,67) + (9 × 1,67) 1M multiply 9 by lap A A distance = 6,68 + 15,03 1A correct distance = 21,71 km CA 1A correct distance His statement is not valid. O 1CA finding total distance 1O correct conclusion (5) 5.2.3 Difference in Time = 9 min 30 sec − min 4 sec M 1M subtract time MP = 3 min 45 sec A 1A difference in time L3 Difference in Time per lap = 3 min 45 sec ÷ 3 M 1M divide by 3 = 1 min 15 sec A 1A total time Statement is valid  1O reason (5) Copyright reserved Please turn over

Downloaded from hlayiso.com 10 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2021) 5.3.1 Number of candles 1MA ÷ by correct MP Length : 24 cm 8 cm 3 MA values L3 Width : 16 cm 8 cm 2 MA 1MA for 2 correct Height : 24 cm ÷ (1 11 cm) 2 M values 3 and 2 S 1M adding 1cm 1S simplify no of Total number of candles: 3 2 2 12 candles CA candles 1CA conclusion OR OR 1MA × by correct Length: 8 cm × 3 = 24 therefore 3 will fit MA values Width: 8 cm × 2 = 16 therefore 2 will fit MA 1MA for 2 correct Height: (11 cm + 1 cm + 11 cm) = 23 cm, therefore 2 will values 3 and 2 fit MA 1M adding 1cm 1S simplify no of S candles Total number of candles: 3 2 2 12 candles CA 1CA conclusion (5) 5.3.2 Total area = 2 (H × L) + 2 (W × H) C 1C converting to m M = 2 (0,24 m × 0,24 m) + 2 (0,16 m × 0,24 m) SF 1SF correct values L2 = 0,1152 m2 + 0,0768 m2 S 1S simplification = 0,192 m2 CA 1CA answer (4) [30] TOTAL: 150 Copyright reserved Please turn over