Mathematical Literacy Grade 12 P2 2019 Memo (Eng)

NATIONAL SENIOR CERTIFICATE GRADE 12 SEPTEMBER 2019 MATHEMATICAL LITERACY P2 MARKING GUIDELINE MARKS: 150 Symbol Explanation M Method M/A Method with accuracy MCA Method with consistent accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG/RM Reading from a table OR Reading from a graph OR Read from map F Choosing the correct formula SF Substitution in a formula J Justification P Penalty, e.g. for no units, incorrect rounding off etc. R Rounding Off OR Reason AO Answer only NPR No penalty for rounding This marking guideline consists of 10 pages.

2 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2019) 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 aflesings vanaf ʼn grafiek, tabel, uitlegplan en kaart geneem en ekstra antwoorde gee, penaliseer vir elke ekstra verkeerde item. Copyright reserved Please turn over

(EC/SEPTEMBER 2019) MATHEMATICAL LITERACY P2 3 QUESTION 1 [37] Ques. Solution Explanation Topic & Level M 1RT Correct opening L2 1.1.1 RT 0,8125% balance F Interest Amount = R749 299,39 × A 100 1M Multiply = R6 088,06 1A Monthly rate OR RT M A Interest Amount = R749 299,39 × 0,008125 = R6 088,06 (3) 1.1.2 750 000 1RT Home loan L2 Monthly repayment = × 8,59 RT 1 000 M 1M Divided by F = R6 442,50 1 000 1RT Correct factor OR RT RT 1RT Home loan Monthly repayment = 750 000 × 8,59 1RT Correct factor 6 442 500 = 1M Divided by 1 000 M 1 000 = R6 442,50 (3) 1.1.3 Closing balance for Month 3 1A Add and Subtract L2 = R749 299,39 + R6 088,06 – R6 442,50 A 1A Closing balance F = R748 944,95 A (2) 1.1.4 Statement not valid 1A Not valid L4 R R F The shorter the loan period, the higher the monthly repayment 1R Shorter period R OR R 1R Higher MRP The shorter the loan period, the higher the loan factor OR No A R R 1A No The loan factors are higher with shorter periods 1R Higher loan factor 1R Shorter periods Accept any other relevant response (3) 1.1.5 Amount at 119 months = R6 442,50 × 119 1MA Multiplying L4 = R766 657,50 CA correct values F 1CA Amount Difference = R766 657,50 – R750 000 M 1M Subtraction = R16 657,50 CA 1CA Difference Statement valid O 1O Valid (5) 1.1.6 It will reduce the interest amount. 2A Explanation L4 Accept any other relevant response (2) F 1.2.1 C 1SF Substitution M Volume = 2 m × 2 m × 0,3 m SF 1C Convert mm to m L2 = 1,2 m3 CA 1CA Answer in m3 (3) Copyright reserved Please turn over

4 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2019) 1.2.2 Bags of cement = 1,2 M CA from 1.2.1 M 0,26 1M Ratio concept L3 = 4,615384615 × 2 M = 9,2 bags S 1M Multiply by 2 ≈ 10 bags CA 1S Simplification OR 1R Number of bags 0,26 = 2 bags M M 2 1,2 1,2 = × 1M Ratio concept 1 0,26 = 9,2 bags S = 10 bags R 1M Multiply by 2 1S Simplification 1R Number of bags (4) 1.2.3 Litres of paint needed = 15 × 1,08 1MA Increased M = 16, 2 m2 MA surface area L3 16,2 1M Dividing by = M 5 spread rate = 3,24 litres = 3,24 litres × 2 6,48 CA 1CA Total number = 5‐𝑙𝑖𝑡𝑟𝑒 𝑡𝑖𝑛𝑠 of litres = 1,296 tins 1CA Number of 5- Number of 5-litre tins = 2 tins CA litre tins OR Litres of paint needed = 15 × 108% 1MA Increased = 16,2 m2 MA surface area Two coats = 16,2 × 2 1CA Two coats = 32,4 m2 CA 1M Dividing 32,4 Number of 5-litre tins = MM 1CA Number of 5- 5‐𝑙𝑖𝑡𝑟𝑒 𝑡𝑖𝑛𝑠 = 6,6 litres litre tins = 2 × 5 litres CA (4) 1.3.1 Price for 2015 = 1 251 158,39 + (1 251 158,39 × 0,05) MA 1MA Increase by 5% L3 = 1 251 158,39 + 62 557,9195 S 1S Simplification F = R1 313 716,31 CA 1CA Amount 2015 Price for 2016 = 1 313 716,31 + (1 313 716,31 × 0,04) = 1 313 716,31 + 52 548,65238 1CA Amount 2016 = R1 366 264,96 CA OR M M Price from 2015 – 2016 = 1 251 158,39 × 1,05 × 1,04 S = R1 366 264,96 CA (4) 1.3.2 RG 1RG Subtract correct L3 1 598 366,77 −1 029 331 values F Percentage change = × 100% 1 029 331 RG 1RG Correct 569 035,77 = × 100% denominator 1 029 331 1S Simplification 1R Nearest % = 0,5528… × 100% S = 55% R (4) [39] Copyright reserved Please turn over

(EC/SEPTEMBER 2019) MATHEMATICAL LITERACY P2 5 QUESTION 2 [36] Ques. Solution Explanation Topic and Level 2.1.1 Taxable Income = Gross Annual salary – Pension 1M Subtract 7,5% L2 = 401 137,75 – (0,075 × 401 137,75) M 1S Simplification F = 401 137,75 – 30 085,33 S 1CA Taxable Income = R371 052,42 CA (3) 2.1.2 Annual Tax A CA from 2.1.1 L4 = 63 853 + 31% of taxable income above 305 850 1A Correct Rates of F = 63 853 + 0,31 × (371 052,42 – 305 850) SF Tax = 63 853 + 0,31 × 65 202,42 1SF Substitution = 63 853 + 20 212,75 1S Simplification = 84 065,75 S – 13 635 M 1M Subtract rebate = 70 430,75 – 12 456 M 1M Subtract MAC = R57 974,75 CA 1CA Annual tax Annual School fees = 2 500 × 11 + 3 200 × 11 MA 1MA School fees × 11 = 27 500 + 35 200 1CA Annual School = R62 700 CA fees 1O Invalid Claim not valid O (9) 2.2.1 Probability other than African = 8,8% + 9,5% + 2,4% 1MA Adding correct L2 = 20,7% A values P 1A Probability OR Probability other than African = 100% – 79,3% M 1M Subtract from 100 = 20,7% CA 1CA Probability (2) 2.2.2 African % = 100% – 8,8% – 9,5% – 2,4% MA 1MA Subtract from L3 = 79,3% A 100 D 1A African % African Population in 2004 = 79,3% × 46,66 million MCA 1MCA Calculate % = 37, 001 380 million CA 1CA Total for 2004 in OR 37 001 380 millions African Population in 1911 = 67,3% × 5 972 757 = 401 9665,46 CA 1CA Total for 1911 Difference = 37 001 380 – 401 9665,46 M 1M Subtract = 32 981 714, 54 1CA Difference ≈ 32 981 714 CA NPR (7) 2.2.3 1911 = 0,08 × 5 972 757 1MA % for 1911 L2 = 477 820,56 CA 1CA Coloured D population 1911 2004 = 0,08 × 46 660 000 1CA Coloured = 3 732 800 CA population 2004 (3) 2.2.4 Percentage of the African race group increased 1A African increase L4 Percentage of the Indian race group decreased A 1A Indian decrease (2) Copyright reserved Please turn over

6 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2019) 2.3.1 Number of people = (25 × 2) + 17 1MA × 2 and addition L2 = 50 + 17 MA 1CA Number of D = 67 people CA people (2) 2.3.2 Cost for Option 1 = 1 500 + (250 × 67) CA from 2.3.1 L4 = 1 500 + 16 750 1MCA Adding and F = R 18 250 CA Multiplying Cost for Option 2 MA MCA 1CA Option 1 cost Cost for couples = 0,96 × 270 × 50 = R12 950 CA 1MA % decrease Cost for singles = 270 × 17 1MCA × 50 = R4 590 CA 1CA Couples cost Total for couples and singles = R12 950 + R4 590 1CA Singles cost = R 17 550 CA Statement invalid O 1CA Option 2 cost 1O Invalid (8) [36] Copyright reserved Please turn over

(EC/SEPTEMBER 2019) MATHEMATICAL LITERACY P2 7 QUESTION 3 [39] Ques. Solution Explanation Topic and Level 3.1.1 North West OR West of North A 2A Direction M&PL2 (2) 3.1.2 A 1A Measure scale M&PL3 Scale: 3,8 cm = 20 miles 1A Measured distance Distance = 8,6 cm A 1M Dividing and 20 multiplying Actual distance = × 8,6 M 3,8 1CA Distance to 3 dec places = 45,263 miles CA Scale Measured Accept 3,6 – 4 cm Measured distance Accept 8,4 – 8,8 cm (4) 3.1.3 3,8 cm = 20 miles CA from 3.1.2 M Convert miles to kilometres = 20 miles × 1,609 1MA Convert to miles to km L3 = 32,18 km MA 1MA Convert km to cm Convert km to cm = 32,18 × 100 000 1S Simplification = 3 218 000 cm MA 1R Nearest million Unit scale = 3,8 cm : 3 218 000 cm Penalise for units in unit = 1 cm : 848 842,1053 S ratio ≈ 1 : 1 000 000 R (4) 3.1.4 Noise pollution 2R Reason M&PL4 OR Air pollution R OR Danger R OR Length of runways R Accept any other relevant answer (2) 3.1.5 A 61 2A Road (2) M&PL2 3.1.6 Distance = Speed × Time 1SF Substitution M 78 miles = 40 miles per hour × Time SF 1M Changing subject of the L4 M 78 formula Time = 40 1S Time in hours = 1,95 hours S 1C Time in hours and min Time in hours and minutes = 1h 57 minutes C 1M Adding times Arrival time = 07:20 + 1:57 M 1CA Arrival time = 09:17 CA 1O Invalid Statement invalid O OR 1M Changing subject of the 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 formula Time = M 1SF Substitution 𝑆𝑝𝑒𝑒𝑑 78 1S Time in hours Time = SF 1C Time in hours and min 40 = 1,95 hours S 1M Travel time = 1h 57 minutes C 1O Not valid Travel time = 7:20 to 9:15 1O 2 min late = 1h 55 minutes M O (7) Not valid, she will be 2 minutes late O Copyright reserved Please turn over

8 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2019) 3.2.1 Range = Highest – Lowest 1M Concept of D 15°C = 13°C – Lowest M range L2 Lowest = 13°C – 15°C RT 1RT Correct = -2°C A values 1A Lowest value (3) 3.2.2 (a) Mistakes: D Data was not arranged A 1A Mistake 1 L3 Calculation was done without using the BODMAS-rule A 1A Mistake 2 (2) 3.2.2 (b) Correction: D 3 ; 4 ; 6 ; 7 ; 10 ; 13 ; 14 ; 19 ; 19 ; 22 ; 24 ; 24 M 1M Arrange L3 13 + 14 data ascending Median = 2 or descending 27 = 2 = 13,5 °C A 1A Median (2) 3.2.3 June, July and August A 1A Correct D Minimum temperatures are high R months L4 Maximum temperatures are high R 1R Min high 1R Max high (3) 3.2.4 A 𝟓 CA from 3.2.1 P Probability = 𝟏𝟐 A 1A Numerator L2 = 0,4166… 1A Denominator = 0,417 CA 1CA 3 dec places (3) 3.2.5 CA from 3.2.1 D Average Minimum and Maximum 1CA January L2 temperatures in Frankfurt 1A Feb – Apr 30 1A May – Jul 1A Aug – Nov 25 1A Dec 20 Temperature 15 A 10 A 5 A 0 JanAFeb Mar Apr May Jun Jul Aug Sep Oct Nov Dec A -5 Months (5) [39] Copyright reserved Please turn over

(EC/SEPTEMBER 2019) MATHEMATICAL LITERACY P2 9 QUESTION 4 [38] Ques. Solution Explanation Topic and Level 4.1.1 Area of unshaded part 1SF Correct values M = Area of rectangle – Area of ▲1 – Area of ▲2 for rectangle L3 1 1 = (Length × Width) – (2 × base × height) – (2 × base × height) 1SF Correct values ▲1 SF SF SF 1SF Correct values = (130 cm × 25 cm) – (0,5 × 50 cm × 25 cm) – (0,5 × 50 cm × 15 cm) = 3 250 cm2 – 625 cm2 – 375 cm2 M S for ▲2 = 2 250 cm2 CA × 5 M 1M Subtracting 2 Total area = 11 250 cm CA 1S Simplification OR 1CA Area of Area of rectangle = Length × Width unshaded part Area of 5 panels = 130 cm × 25 cm SF 1M Multiply by 5 = 32 500 cm2 × 5 M 1CA Total area = 16 250 cm2 CA OR 1 Area of ▲1 (5 panels) = × base × height 1SF Correct values 2 for rectangle = 0,5 × 50 cm × 25 cm × 5 SF 1M Multiply by 5 = 3 125 cm2 CA 1CA Area of rec 1 Area of ▲2 (5 panele) = × base × height 1SF Correct values 2 = 0,5 × 50 cm × 15 cm × 5 ▲1 = 1 875 cm2 CA 1CA Area of ▲1 Total Area = 16 250 cm2 – 3 125 cm2 – 1 875 cm2 M 1CA Area of ▲2 = 11 250 cm2 CA 1M Subtracting 1CA Area of unshaded part (8) 4.1.2 To glue the seams together 2R Reason M OR L4 To paste sides together R (2) 4.1.3 Diagram T – A ring should be attached to the wide side of the 1A Explain M&P hot air balloon A Diagram T L2 Diagram U – A paper clip to be attached to the ring A 1A Explain Diagram U (2) 4.1.4 To blow hot air in balloon 2R Reason M&P (2) L4 4.2.1 A 2A Temperature M&P When the temperature increases, the lift of the hot air balloon is increases L4 higher. A 2A Hot air balloon OR higher A A When the temperature decreases, the lift of the hot air balloon is lower. (4) Copyright reserved Please turn over

10 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2019) 4.2.2 0,972+0,946 1MA Concept of M Air Density of Hot air balloon B = 2 1,918 MA average L4 = 2 = 0,959 kg/m3 CA 1CA Air Density Lift = (Air density outside of the hot air balloon – Air density inside hot air balloon) × Volume of the hot air balloon RT RT 1RT Correct outside Lift = (1,204 kg/m3 – 0,959 kg/m3) × 2 400 m3 SF air density = 0,245 kg/m3 S × 2 400 m3 1RT Correct inside = 588 kg CA air density 1SF Substitution 1S Simplification 1CA Lift (7) 4.2.3 33+12 1A Numerator P Probability = 93 45 A 1A Denominator L2 = A 1CA Common 93 15 CA = fraction in the 31 simplest form (3) 4.3 Accommodation = 1 030 × 4 nights × 4 people A 1A 4 nights × 4 F = R16 480 CA 1CA Cost for L3 Hot air balloon rides = 750 – (750 × 0,15) MA accommodation = 750 – 112,50 1MA Less 15% S 1S Cost per person Cost per person = R637,50 × 4 1CA Cost for rides = R2 550 CA Total cost in Rand = R16 480 + R2 550 1CA Total cost = R19 030 CA 1C Dividing 19 030 1CA Cost in USD Cost in USD = C 13,63 1M Multiply = $1 396, 184886 CA 1CA Cost in Turkish Cost in Turkish Lira = $1 396,184886 × 5,25 M Lira = 7 329,97 Turkish Lira CA (10) [39] TOTAL: 150 Copyright reserved Please turn over