Mathematical Literacy Grade 12 P2 November 2024 NSC Memo

NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 MATHEMATICAL LITERACY P2/ WISKUNDIGE GELETTERDHEID V2 NOVEMBER 2024 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 Symbol/Kode Explanation/Verduideliking MA Method with accuracy/Metode met akkuraatheid MCA Method with consistent accuracy/Metode met volgehoue akkuraatheid CA Consistent accuracy/Volgehoue akkuraatheid A Accuracy/Akkuraatheid C Conversion/Herleiding S Simplification/Vereenvoudiging RT Reading from a table/graph/document/diagram/Lees vanaf tabel/grafiek/dokument/diagram SF Correct substitution in a formula/Korrekte vervanging in 'n formule O Opinion/Explanation/Opinie/Verduideliking P Penalty, e.g. for no units, incorrect rounding off, etc./Penalisering, bv. vir geen eenhede, verkeerde afronding, ens. NPR No penalty for correct rounding/Geen penalisering vir korrekte afronding nie NPU No penalty for omitting unit, but wrong unit is penalised/Geen penaliseringe indien die eenheid uitgelos is nie, maar wel indien 'n verkeerde eenheid gebruik word. AO Answer only/Slegs antwoord These marking guidelines consist of 17 pages. Hierdie nasienriglyne bestaan uit 17 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 2 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne 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 item presented. • Rounding is an independent mark. • General principle of marking, if the candidate makes one mistake one mark is deducted. 1 • A conclusion mark can only be given if 3 of the total marks for the sub-question have been awarded. • No penalty for rounding (NPR) if the first decimal is correct. LET WEL: • As 'n kandidaat 'n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na. • As 'n kandidaat 'n antwoord van 'n vraag doodtrek (kanselleer) en nie oordoen nie, sien die doodgetrekte (gekanselleerde) poging na. • Volgehoue akkuraatheid (CA) word in ALLE aspekte van die nasienriglyne toegepas; dit hou egter op by die tweede berekeningsfout. • Wanneer 'n kandidaat aflesings vanaf 'n grafiek,, tabel, uitlegplan en kaart aanbied en ekstra antwoorde gee, penaliseer vir elke ekstra item. • Afronding tel as 'n afsonderlike punt. • Die algemene beginsel van nasien is, as 'n leerder een fout maak, word een punt afgetrek. 1 • 'n Gevolgtrekkingspunt kan slegs gegee word indien 3 van die totale punte vir die subvraag toegeken is. • Geen penalisering vir afronding (NPR) nie as die eerste desimaal korrek is. QUESTION/VRAAG 1 [26 MARKS/26 PUNTE] ANSWER ONLY FULL MARKS Q/V Solution/Oplossing Explanation/Verduideliking T&L * M 1.1.1 D ✓✓ A 2A correct option L1 (2) E * ✓✓ A MP 1.1.2 E 2A correct option L1 (2) E * M ✓✓ A 1.1.3 C 2A correct option L1 (2) E * M 1.1.4 G ✓✓ A 2A correct option L1 (2) E ✓ MA ✓A M 1.2.1 220 mm ÷1 000 = 0,22 m 1MA ÷ 1 000 L1 1A conversion E (2) * M 1.2.2 A ✓✓ A 2A correct option. L1 (2) M * M 1.2.3 Number of bricks / Aantal stene 1RT correct values L1 ✓RT ✓MA 1MA dividing M = 2 860 mm ÷ 220 mm 1A number of bricks = 13 ✓A (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 3 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L * MP 1.3.1 18 ✓✓A 2A correct number L1 (2) E * MP 1.3.2 Number of cross pieces / Getal dwarsstutte L1 ✓RT 1RT 6 M = 6 × 3 ✓A 1A multiply by 3 = 18 ✓A 1A pieces (3) * MP 1.3.3 Chair support / Rugleuningbalk ✓✓RT 2RT correct option L1 (2) E MP 1.3.4 1,9 cm ✓✓A 2A correct dimension L1 NPU E (2) MP 1.3.5 Space between cross pieces:/Opening tussen dwarsstutte: L1 ✓ RT E Space/Opening = 1,27 × 10 mm 1RT correct value = 12,7 mm ✓ MCA 1MCA simplification NPU (2) [26] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 4 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 2 [31 MARKS/31 PUNTE] Q/V Solution/Oplossing Explanation/Verduideliking T&L MP 2.1.1 View from the top. L1 ✓✓ A Aansig van bo. E OR/OF View of the landscape from a certain height above ground like from a satellite or drone. ✓✓ A Aansig van die landskap vanaf 'n seker hoogte bo die grond soos vanaf ‘n satelliet of hommeltuig. 2A correct explanation OR/OF View from an elevated height. ✓✓ A Aansig vanaf 'n verhewe hoogte. OR/OF Birds-eye view. ✓✓ A Voëlperspektief. (2) MP 2.1.2 Number of campers/Getal kampeerders L1 ✓RT ✓MCA 1RT all correct values E = 6 +15 + 4 + 15 + 4 + 4 + 5 + 15 + 8 + 10 + 6 + 6 + 20 1MCA adding values = 118 ✓CA 1CA simplification AO OR/OF OR/OF Number of campers/Getal kampeerders 1RT all correct values ✓RT ✓MCA =20 + 15(3) + 10 + 8 + 6(3) + 5 + 4(3) 1MCA adding values = 118 ✓CA 1CA simplification AO (3) MP 2.1.3 South West/SW/Suidwes SW ✓✓A 2A compass direction L1 (2) E * MP 2.1.4 2 ✓✓O 2O identifying correct L2 site M (2) P 2.1.5 3 ✓A 1A correct numerator L2 × 100% 13 ✓A 1A correct denominator D = 23,076923% ✓CA 1CA simplification NPR (3) MP 2.1.6 9 showers/storte ✓✓A 2A correct number L1 (a) (2) E Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 5 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L * ✓A MP 2.1.6 90 mm = 8,2 m 1A measured distance L2 (b) ✓C 1C convert to mm M 90 mm = 8 200 mm (÷ 90) ✓M 1M divide by 90 1 : 91,11 1 : 91 ✓R 1R rounded answer OR/OF OR/OF ✓A 9 cm : 8,2 m 1A measured distance 9 cm : 820 cm ✓C 1C convert to cm = 1 : 91,11.. ✓M 1M divide by 9 = 1 : 91 ✓R 1R rounded answer Accept ± 1 mm deviation on measurement per (4) province Aanvaar ± 1 mm-afwyking op meting per provinsie MP 2.1.6 To allow people to sit while waiting for a toilet or shower L4 (c) to become available. ✓✓O M Vir mense om op te sit terwyl hulle wag dat 'n toilet of stort beskikbaar word. OR/OF To place your clothes or belongings on while you are 2O reason showering. Om jou klere of besittings neer te sit terwyl jy stort. OR/OF To sit while you change your outfit, or getting dressed or apply body lotion or for baby nappy change. Om op te sit terwyl jy jou uitrusting verander of terwyl jy aantrek of lyfroom aansmeer of babadoeke verander. (2) MP 2.2.1 Day 2 / Dag 2 ✓✓A 2A correct description L2 (2) E * MP 2.2.2 D ✓✓RT 2RT correct option L2 (2) M * MP 2.2.3 12,5 km ✓✓RT 2RT correct distance L2 (2) E Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 6 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L MP 2.2.4 The part shows a continuous downward slope, it is L4 downhill. ✓✓O E Die part het 'n aaneenlopende afwaartse helling getoon, dit is afdraand. 2O correct explanation OR/OF That part does not have many uphills. Daardie deel het nie baie opdraandes nie. (2) * MP 2.2.5 Difference in height/Verskil in hoogte L4 ✓RT ✓RT 1RT 1st correct value M Difference/Verskil = 1 050 m – 900 m 1RT 2nd correct value = 150 m ✓O He is CORRECT. / Hy is KORREK 1O conclusion (3) [31] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 7 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 [31 MARKS/PUNTE] Q/V Solution/Oplossing Explanation/Verduideliking T&L * ✓MA M 3.1.1 = 14:13 ̶ 12:55 1MA subtracting time L2 ✓A M = 1 hour 18 minutes/1 uur 18 minute 1A simplification [1hr18min] OR/OF 78 minutes OR/OF 1,3 hours/uur AO (2) * M 3.1.2 Total height of 4 pillows/Totale hoogte van 4 kussings L2 E = 11 cm × 4 ✓MA 1MA multiplying by 4 = 44 cm ✓CA 1CA simplification Difference/Verskil ✓RT = 48 cm ̶ 44 cm 1RT height = 4 cm ✓CA 1CA simplification OR/OF OR/OF ✓RT ✓MA ✓MA 1RT height Difference = 48 cm – 11 cm – 11cm – 11 cm – 11 cm 1MA subtracting 11 cm = 4 cm ✓CA 1MA subtracting all the 11’s 1CA simplification AO (4) M 3.1.3 Perimeter = 2 (length + width)/ L3 Omtrek = 2 (lengte + breedte) M Perimeter/Omtrek = 2 (46 cm + 30 cm) ✓SF 1SF correct substitution = 2 (76 cm) = 152 cm ✓CA 1CA simplification Total length for 4 bags/Totale lengte vir 4 sakke = 152 × 4 ✓MA 1MA multiply by 4 = 608 cm 608 cm = 100 = 6,08 m ✓C 1C simplification ✓R 1R correct rounding ∴ she must buy 6,5 m/ Sy moet 6,5 m koop OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 8 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L 30 cm ÷ 100 = 0,3 m 46 cm ÷ 100 = 0,46 m ✓C 1C metre Perimeter/Omtrek = 2(0,3 m + 0,46 m) ✓SF 1SF correct substitution = 1,52 m ✓CA 1CA simplification Total / Totaal = 1,52 m × 4 ✓MCA 1MCA multiply by 4 = 6,08 m ✓R ∴ she must buy 6,5 m/ Sy moet 6,5 m koop 1R correct rounding OR/OF OR/OF Using ½ metre lengths/ Gebruik ½ metre lengtes ✓SF Perimeter/Omtrek = 2 (46 cm + 30 cm) 1SF correct substitution = 152 cm ✓CA 1CA simplification Total length for 4 bags / Totale lengte vir 4 sakke = 152 × 4 ✓MA 1MA multiply by 4 = 608 cm ½ m = 50 cm ✓ C 1C to centimetre Number of half metre lengths / Getal half-meter lengtes = 608 cm ÷ 50 cm = 12,16 ≈ 13 ✓ R 1R correct rounding (5) * M 3.2.1 Circumference / Omtrek L2 ✓ SF 1SF substitute diameter E = 3,142 × 8 cm 1A simplification = 25,136 cm ✓ A NPR AO (2) M 3.2.2 Radius/Radius L1 E 8 cm 1MA concept of radius = ✓MA 2 = 4 cm ✓A 1A simplification NPU AO (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 9 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L CA from Question 3.2.2 M 3.2.3 Area of a circle/Area/Oppervlakte van 'n sirkel L4 ✓SF 1SF correct substitution M = 3,142 × (4 cm)2 = 3,142 × 16 cm2 ✓MCA 1MCA squaring = 50,272 cm2 ✓CA 1CA simplification Area to be painted / Oppervlakte wat geverf moet word Area = 50,272 – 0,3142 = 49,9578 cm2 ✓CA 1CA difference Total area / Totale opp.= 36 × 49,9578 cm2 = 1 798,4808 cm2. ✓MCA 1MCA multiply by 36 Total area in m2 / Totale opp.in m2 = 1 798,4808 ÷ 1002 1C dividing by 10 000 or = 0, 179848 m2. ✓C 1002 6 m2 = 1 ℓ = 1000 mℓ ✓ MA 1MA using ratio … m2 = 50 ml 0,3 m2 = 50 ml 1CA comparing areas. 0,3 m2 > 0,179848 m2 ✓ CA Therefore 50 mℓ will be more than sufficient. / ✓ O 1O verification Daarom sal 50 mℓ meer as genoeg wees. OR/OF OR/OF Area of a circle/Area/Oppervlakte van 'n sirkel ✓SF 1SF correct substitution = 3,142 × (4 cm)2 ✓MCA 1MCA squaring 1CA simplification = 50,272 cm2 ✓CA Area to be painted/Oppervlakte wat geverf moet word 1CA difference Area = 50,272 cm2 – 0,3142 cm2 = 49,9578 cm2 ✓CA Total area / Totale oppervlakte 1MCA multiply by 36 = 36 × 49,9578 cm2 = 1 798,4808 cm2. ✓MCA = 1 798,4808 ÷ 1002 1C dividing by 10 000 or = 0, 179848 m 2 ✓C 1002 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 10 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L Paint/ verf 6 m2 with/met 1 ℓ ∴ 0, 17984808 m2 with /met n ℓ 0,17984808 n= ℓ 6 ✓MCA 1MCA using ratio = 0,02997466 ℓ ≈ 30 mℓ ✓CA 1CA paint needed VALID / GELDIG ✓O 1O verification OR/OF OR/OF Area of ONE circle/Oppervlakte van EEN sirkel ✓SF 1SF correct substitution = 3,142 × (4 cm)2 1MCA squaring ✓MCA = 50,272 cm2 ✓CA 1CA simplification Area to be painted/Oppervlakte wat geverf moet word = 50,272 cm2 – 0,3142 cm2 = 49,9578 cm2 ✓CA 1CA difference 6m2 : 1ℓ 60 000 cm2 : 1 000 mℓ ✓C 1C converting 49,9578 : ? ' 49,9578 ×1 000 Paint needed/ Verf benodig ✓MCA 60 000 1MCA using ratio = 0,83263 mℓ Paint for 36/ Verf vir 36 = 0,83263mℓ × 36 ✓MCA 1MCA multiply by 36 = 29,97 mℓ ✓CA 1CA paint needed ∴ VALID / GELDIG ✓O 1O verification OR/OF 4 𝑐𝑚 OR/OF Radius = = 0,04 m 100 ✓SF 1SF correct substitution Area of circle/ Opp van sirkel = 3,142 × (0,04) ✓MCA 2 1MCA squaring = 0,0050272 m2 ✓CA 1CA simplification 0,3142 Area of circular hole/ Opp van gaatjie = 10 000 ✓C 1C dividing by 10 000 = 0,0000314 m2 Area to be painted = 0,0050272 m2 – 0,00003142 m2 1CA difference = 0,00499578 m2 ✓CA 2 ∴ 0,00499578 m × 36 1MCA multiply by 36 = 0,17984808 m2 ✓MCA Amount of paint/Hoeveelheid verf 0,17984808 1MCA using ratio = × 1 000 mℓ ✓MCA 6ℓ 1CA paint needed = 29,97468 mℓ ≈ 30 mℓ ✓CA ∴ 30 mℓ is less than 50mℓ 1O verification VALID / GELDIG ✓O Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 11 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L OR/OF OR/OF Area of a circle / Oppervlakte van ‘n sirkel = 3,142 × 42 ✓SF 1SF correct substitution = 3,142 × 16 ✓MCA 1MCA squaring = 50,272 cm2 ✓CA 1CA simplification Area to be painted / Oppervlakte wat geverf moet word Area / Opp = 50,272 – 0,3142 = 49,9578 cm2 ✓CA 1CA difference 6 m2 / ℓ = 60 000 cm2 / ℓ = 60 000 cm2 / 1 000 mℓ ✓C 1C conversion Amount of paint for one tag / Hoeveelheid van verf per houtplaatjie = 49,9578 ÷ 60 000 × 1 000 ✓MCA 1MCA using ratio = 0,83263 mℓ Paint for 36 tags/ Verf vir 36 houtplaatjies 0,83263 mℓ × 36 ✓MCA 1MCA multiply by 36 = 29,97468 mℓ ✓CA 1CA paint needed VALID / GELDIG ✓O 1O verification (9) * M 3.3.1 Volume of a cube = side × side × side/ L3 Volume van 'n kubus = sy × sy × sy M ✓ SF ✓ SF 1SF substitution number 2 744 cm³ = side × side × side 1SF cube unit (side)3 = 2 744 cm³ ✓MA 1 MA change subject of the formula 14 × 14 × 14 = 2 744 Side/Sy = 14 cm ✓CA 1CA simplification (4) P 3.3.2 8 + 7 = 15 L2 1A numerator 15 1A denominator E P = ✓A 35 ✓ A 1CA simplification = 0,42857… ≈ 0,43 ✓ CA OR/OF OR/OF 8 7 1A denominator P= + 35 35 ✓ A 1A writing as decimals = 0,22857… + 0,2 ✓ A = 0,42857… 1CA simplification ≈ 0,43 ✓ CA NPR (3) [31] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 12 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 4 [29 MARKS] Q/V Solution/Oplossing Explanation/Verduideliking T/L * MP 4.1.1 Kerosene or lamp oil 2RT correct product L1 /Keroseen of Lampolie ✓✓ RT (2) E MP 4.1.2 Gasoline or petrol/ ✓✓ RT 2RT correct product L1 Brandstof of Petrol (2) E * M 4.1.3 0C = ( F 0 - 32) L3 . 1RT correct value 300 M 1, 8 ✓ RT ( 0 F - 32) ✓ SF 1SF substituting information 3000C = . correctly 1, 8 ✓S 1S changing subject of the ⁰F = 1,8 × 300 + 32 formula = 572 ✓ CA 1CA simplification AO (4) * M 4.1.4 Surface area of an open cylinder/ L3 Buite-oppervlakte van 'n oop silinder D = 3,142 × diameter × height / = 3,142 × deursnee × hoogte ✓ SF = 3,142 × 6 m × 54 m 1SF substitution ✓ CA 1CA simplification = 1 018,008 m2 Area of pipes/Oppervlakte van pype 2,5 1 018,008 = × ✓ MCA 1MCA percentage calculation 100 1 = 25,4502 m2 ✓ CA 1CA simplification Total Surface Area/Totale buiteoppervlakte ✓ MCA ✓ MA 1MCA subtracting pipe area = 1 018,008 m – 25,4502 m + 150,816 m2 2 2 1MA adding A + C = 1 143,3738 m2. ✓ CA 1CA total surface area OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 13 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L Surface area of an open cylinder/Buite-oppervlakte van 'n oop silinder SA = 3,142 × diameter × height/ BO = 3,142 × deursnee × hoogte = 3,142 × 6 m × 54 m ✓ SF 1SF substitution = 1 018,008 m2 ✓CA 1CA simplification Excluding area of pipes/Oppervlakte van pype uitgesluit ✓ MA ✓ MCA 1MA less 2,5% 97,5 1 018,008 1MCA percentage calculation Area (excluding)/Opp(uitgesluit) = × 100 1 1CA simplification = 992,5578 ✓ CA ✓ MA 1MA adding A + C Total SA/Totale BO = 992,5578 m2 + 150,816 m2 = 1 143,3738 m2. ✓ CA 1CA total surface area NPR (7) * M 4.2.1 Number of bricks in 1 row of a double brick wall L2 Getal stene in een ry van ‘n dubbelsteenmuur M = 19 ✓✓ RT 2RT bricks in double row Number of bricks for 1 garage door Getal stene vir 1 motorhuis deur = 19 × 20 ✓ A 1A number of layers = 380 Total number of bricks needed /Totale getal stene nodig = 380 × 2 ✓MCA 1MCA doubling = 760 ✓R ∴ 2 Pallets of bricks /Stapelborde met stene 1R number of pallets OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 14 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L ✓ RT ✓ RT Single line wall/Enkelmuur = 20 × 10 1RT bricks (height) 1RT bricks (row) = 200 bricks/ stene ✓ CA Double line wall/Dubbelmuur = 2 × 200 = 400 bricks 1CA bricks on double walls To cover space of two garage doors:/ Om die spasie van twee motorhuisdeure te dek Number of bricks/Getal stene = 2 × 400 ✓ MCA 1MCA doubling = 800 Number of pallets needed/ Stapelborde met stene benodig =2 ✓R 1R number of pallets (5) CA pallets from 4.2.1 M/Fin 4.2.2 ✓ MA ✓ SF 1MA doubling L4 Area of 2 doors/Opp van 2 deure = 2 × 2,13 × 3 1SF correct values M = 12,78 m2 ✓A 1A simplification Labour cost/Arbeidskoste = 12,78 m2 × R500 = R 6 390 ✓ CA 1CA labour cost Brick cost/Steenkoste = 2 × 525 × R6,45 = R 6 772,50 ✓ CA 1CA brick cost COST = Other material + Labour + Bricks cost/ KOSTE = Ander materiaal + Arbeid + Steenkoste Total cost/Totale koste = R2 000 + R6 390 + 6 772,50 = R15 162,50 ✓ CA 1CA amount Not VALID/Nie GELDIG. ✓ O 1O verification OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 15 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L For one door/ Vir een deur Area/ oppervlakte = 2,13 m × 3 m ✓ SF 1SF correct values = 6,39 m2 ✓ A 1A simplification Labour cost/Arbeidskoste = 6,39 m2 × R500 = R3 195 ✓ MCA 1MCA labour cost ∴ for two / vir twee = 2 × R3 195 = R6 390 ✓ CA 1CA doubling Cost per pallet/ koste per stapelbord = R6,45 × 525 = R3 386,25 ∴ for two / vir twee = 2 × R3 386,25 = R6 772,50 ✓ CA 1CA brick cost Total cost/Totale koste = R2 000 + R6 390 + 6 772,50 = R15 162,50 ✓ CA 1CA amount Not VALID/Nie GELDIG. ✓ O 1O verification OR/OF OR/OF ✓ SF Area of 2 doors / Opp van 2 deure = 2 × 2,13 × 3 1SF substitution = 12,78 m2 ✓ MA 1MA doubling ≈ 13 m (cost per m2) 2 ✓ CA 1CA simplification Labour Cost / Arbeidskoste = 13 ×R500 = R6 500 ✓ CA 1CA labour cost Brick cost / Steen koste = 2 × 525 × R6,45 = R6 772,50 ✓ CA 1CA brick cost Total Cost / Totale Koste = R6 500 + R6 772,50 + R2 000 = R15 272,50 ✓ CA 1CA amount Not VALID / Nie GELDIG nie ✓O 1O verification (7) ✓✓ O M 4.2.3 To be easily lifted by a fork lift./Om maklik met 'n L4 vurkhyser op te lig word. E OR/OF Storage/ loading / transportation are made easier /Berging/ laai / vervoer word vergemaklik. OR/OF 2O reason It keeps them from breaking / Dit keer dat hulle breek OR/OF It makes counting easier/Dit is makliker om te tel To protect the bricks from damage/ Dit beskerm die stene van beskadiging To keep it tight and compact/ Om die stene stewig en (2) kompak te hou [29] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 16 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 5 [33 MARKS] Q/V Solution/Oplossing Explanation/Verduideliking T/L ✓RT ✓RT MP 5.1.1 Adelaide and/en Melbourne 1RT 1st city L1 1RT 2nd city E (2) * ✓RT ✓RT 1RT first correct month MP ✓RT 5.1.2 Aug, Sep, Oct, Nov 1RT second correct L1 Aug, Sep, Okt, Nov month M 1RT ALL correct months (3) ✓RT MP 5.1.3 14 655 – (738 + 1 062 + 922 + 1 705 + 2 850 + 1 871 + 2 811) 1RT all correct values L2 = 14 655 – 11 959 ✓MCA 1MCA subtracting M = 2 696 km ✓CA 1CA simplification AO (3) * MP 5.1.4 Tasmania./Tasmanië ✓✓RT 2RT correct state L1 (2) E M 5.1.5 Distance/Afstand = Speed × time L2 1 705 km = s × 20 h 40 min ✓SF 1SF substitution M 1 705 km = s × 20,67 h ✓C 1C converting to hours 1 705 km Speed/Spoed = ✓S 1S change subject of 20,67 h = 82,5 km/h ✓CA formula 1CA simplification (4) ✓RT M 5.2.1 1 142 feet/voet = 348 m ✓RT 1RT 1 142 L2 348 1RT 348 M 1 foot/voet = 1 142 = 0,304728546 1A 0,305 1 foot/voet ≈ 0,305 m ✓A (3) * M 5.2.2 Uluru : Eiffel Tower : Big Ben/ L2 Uluru : Eiffeltoring : Big Ben 1RT correct values E 1A correct order 348 : 324 : 96 ✓RT ✓A = 29 : 27 : 8 ✓✓CA 2CA simplified ratio (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 17 DBE/November 2024 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L * ✓✓A P 5.3.1 P = 22,5% L2 E OR/OF 2A correct probability 1 729 742 P= 7 688 220 ✓✓A = 0,224986… or 22,5% (2) MP 32 159 ✓RT 1RT area of islands L4 5.3.2 Area of island/Opp van eiland = 64 519 ✓RT 1RT area of Tasmania M = 0,498 ≈ 0,5 ✓CA 1CA simplification OR/OF OR/OF Half of Tasmania / Helfte van Tasmanië 1RT Tasmania area ✓RT 1 1MA simplification 64 519 × = 32 259,5 km² ✓MA 2 Islands / Eilande = 32 159 km² ✓RT 1RT island area OR/OF ✓RT ✓RT OR/OF 64 519 : 32 159 1RT area of islands ≈ 2 : 1 1RT area of Tasmania ✓CA 1CA simplification (3) * M 5.3.3 Population density/Bevolkingsdigtheid L3 (a) 1RT correct 454 499 D 454 499 ✓RT 1RT correct area 2 358 = 2 358 ✓RT = 192,74… ≈ 193 ✓R 1R simplification rounded up (3) * M ✓RT 5.3.3 2 358 1RT correct 2 358 L4 (b) Land % area/Land % opp = × 100% 1RT 7 688 220 D 7 688 220 ✓ RT = 0,0306..% ✓CA 1CA simplification Rounds off to zero/Rond af na nul. ✓O 1O opinion (4) [33] Copyright reserved/Kopiereg voorbehou