# Mathematical Literacy Grade 12 P2 NSC June 2024 Memo (Afr) > Grade 12 Mathematical Literacy Paper 2 memorandum from the NSC June Exam 2024 (Afrikaans), 20 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/grade-12-mathematical-literacy-2024-nsc-june-exam-memorandum-paper-2-afrikaans/6f2dc5d1-63b0-4017-86c2-cebd3188b1c1 - Updated: 2026-09-28 - Subject: Mathematical Literacy - Grade: Grade 12 - Language: Afrikaans - Document type: Memorandum - Year: 2024 - Exam period: NSC June Exam - Paper: 2 - Curriculum: CAPS - Pages: 20 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematical Literacy](https://www.hlayiso.com/library/grade-12/mathematical-literacy) › [Mathematical Literacy Grade 12 P2 NSC June 2024 Memo (Afr)](https://www.hlayiso.com/library/grade-12-mathematical-literacy-2024-nsc-june-exam-memorandum-paper-2-afrikaans/6f2dc5d1-63b0-4017-86c2-cebd3188b1c1) ## About this document Mathematical Literacy · Grade 12 · NSC June Exam · 2024 · Afrikaans. Memorandum, 20 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 SENIOR CERTIFICATE EXAMINATIONS/ NATIONAL SENIOR CERTIFICATE EXAMINATIONS SENIORSERTIFIKAAT-EKSAMEN/ NASIONALE SENIORSERTIFIKAAT-EKSAMEN MATHEMATICAL LITERACY P2/ WISKUNDIGE GELETTERDHEID V2 MAY/JUNE/MEI/JUNIE 2024 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 Symbol/Kode Explanation/Verduideliking MA Method with accuracy/Metode met akkuraatheid CA Consistent accuracy/Volgehoue akkuraatheid A Accuracy/Akkuraatheid C Conversion/Herleiding S Simplification/Vereenvoudiging RT Reading from a table/a graph/document/diagram/Lees vanaf tabel/grafiek/diagram SF Correct substitution in a formula/Korrekte vervanging in formule O Opinion/Explanation/Reasoning /Opinie/Verduideliking/Redenasie P Penalty, e.g. for no units, incorrect rounding off, etc./Penalisasie, bv. vir geen eenhede/verkeerde afronding, ens. R Rounding off/Afronding NPR No penalty for correct rounding/Geen penalisasie vir korrekte afronding nie NPU No penalty for omitting correct unit/Geen penalisasie vir die uitlos van die korrekte eenheid nie AO Answer only/Slegs antwoord MCA Method with constant accuracy/Metode met volgehoue akkuraatheid RCA Rounding consistent with accuracy/Afronding met volgehoue akkuraatheid These marking guidelines consist of 20 pages. Hierdie nasienriglyne bestaan uit 20 bladsye.. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 2 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/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 did NOT redo 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 or breakdown. • If the candidate presents any extra solution when reading from a graph, table, layout plan and map, then penalise for every extra item presented. • As a general marking principle, if a candidate has incurred one mistake and there is evidence of sound mathematics thereafter, then that candidate should lose one mark only. • A conclusion mark can only be given if relevant calculations precede it (at least 1 mark before conclusion). • Rounding is an independent mark. • No penalty for rounding (NPR) if the first decimal is correct, except questions involving money. 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 of afbreuk ’break down’ nie • Wanneer 'n kandidaat aflesings vanaf 'n grafiek, tabel, uitlegplan en kaart geneem en ekstra antwoorde gee, penaliseer vir elke ekstra item. • 'n Algemene nasienbeginsel is dat indien 'n kandidaat een fout maak en daarna voortgaan met korrekte wiskunde, dat die kandidaat slegs een punt verloor • 'n Gevolgtrekkingspunt kan slegs gegee word indien relevante berekeninge dit voorgaan (ten minste een punt voor die gevolgtrekking). • Afronding tel as 'n onafhanklike punt. • Geen penalisering vir ronding (NPR) as die eerste desimaal korrek is nie, behalwe as vrae geld insluit. QUESTION/VRAAG 1 [26 MARKS/PUNTE] Answer Only AO - full marks Q/V Solution/Oplossing Explanation/Verduideliking T/L 1.1* 1.1.1 E ✓✓A 2A correct option M L1 1.1.2 G ✓✓A 2A correct option P L1 1.1.3 F ✓✓A 2A correct option M L1 1.1.4 B ✓✓A 2A correct option M (8) L1 ✓✓ A MP 1.2.1 Numerical /Number/ ratio scale. 2A type of scale L1 Numeriese- / Nommer- /Getalle- /syfer-/verhouding- (2) E skaal. ✓✓A MP 1.2.2 1 unit on the map is equivalent to 50 000 units in real life. 2A relationship L1 1 eenheid op die kaart is gelykstaande aan 50 000 M eenhede in werklikheid OR/OF The map is 50 000 times smaller than real life. Die kaart is 50 000 keer kleiner as werklikheid (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 3 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L MP 1.2.3* 1: 25 000 ✓✓ A 2A correct scale L1 (Accept B) E (2) ✓A ✓A M 1.3.1* Rectangle and a circle. 1A rectangle L1 Reghoek en ‘n sirkel 1A circle E (2) ✓✓A MP 1.3.2 144 km 2A correct answer L1 Accept 144 E (2) MP 1.3.3* It is the maximum speed a motorist can travel on the L1 road. ✓✓A M Dit is die maksimum spoed wat ‘n motoris mag ry op die pad. 2A correct explanation. OR/OF The motorist can cover a distance of 120 km in 1 hour. Die motoris kan 120 km aflê in 1 uur (2) M 1.3.4* Distance/Afstand (Jhb – Trompsburg) L1 M = 534 – 27 ✓ RT 1RT both correct values ✓A = 507 km 1A distance NPU (2) MP 1.3.5 North /N/Noord / N ✓✓A 2A correct direction L1 (2) E M 1.3.6 90 𝑐𝑚 1MA dividing by 100 L1 ✓MA 100 E 1A simplification = 0,9 m ✓A (2) [26] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 4 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 [29 MARKS/PUNTE] Q/V Solution/Oplossing Explanation/Verduideliking T/L MP 2.1.1 2 ✓✓A 2A correct number L1 (2) E MP 2.1.2 6 ✓✓A 2A correct road L1 (2) M MP 2.1.3* 7 ✓✓A 2A correct number L1 (2) E ✓✓A P 2.1.4* C 2A correct choice L2 (2) M ✓✓A MP 2.1.5 South East (SE) / Suidoos (SO) 2A correct direction L2 (2) M MP 2.1.6 Length / Lengte = 65 m L3 = 65 000 mm ✓C 1C conversion M Scale/ Skaal: 1 : 8 000 n : 65 000 65 000 ✓ MA 1MA dividing n= 8 000 ✓CA 1CA simplification = 8,125 mm ✓R 1R rounding ≈ 8 mm OR/OF OR/OF Scale/Skaal: 1 : 8 000 n : 65 65 ✓ MA 1MA dividing n= 8 000 = 0,008125 m ✓CA 1CA simplification = 8,125 mm ✓C 1C conversion ≈ 8 mm ✓R 1R rounding (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 5 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L MP 2.2* 2.2.1 C ✓A L2 2.2.2 E ✓A M 2.2.3 D ✓A 5A correct order 2.2.4 B ✓A 2.2.5 A ✓A (5) ✓✓O MP 2.3* Only use the go cart on level ground. / smooth, flat, L4 hard, tarred, road surface E Gebruik die knortjor slegs op gelyke grond/ gladde , 2O Explanation for 1st picture plat, harde, pad, geteerde oppervlakte or for 2nd picture OR/OF Do not use the vehicle on a long grassy surface. Moet nie in lang gras ry nie. (2) ✓MA P 2.4.1 X = 2 840 – 1 476 – 1 024 = 340 ✓CA L1 1MA subtracting from total E OR/OF 1CA simplification ✓MA X = 565 – 163 – 62 = 340 ✓CA AO (2) P 2.4.2 2 840 + 796 ✓RT 1RT numerator L2 P(not a horse) = 1RT denominator M 4 996 ✓RT 3 636 = 4 996 909 = ✓A 1A simplification 1 249 OR/OF OR/OF 1 360 P(horse) = 1RT both values 4 996 ✓RT 1 360 P(not a horse) = 1 – ✓MCA 1MCA subtracting from 1 4 996 909 ✓A = 1A simplification 1 249 (3) ✓RT P st 2.4.3 363 1RT 1 value L3 P= × 100% 796 1RT 2nd value correctly place M ✓RT = 45,60301508 % ✓CA 1CA simplification NPR (3) [29] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 6 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 [33 MARKS/PUNTE]] Q/V Solution/Oplossing Explanation/Verduideliking T/L M 3.1.1 2 000 kW ✓✓RT 2RT correct kW L1 NPU E (2) M ✓✓O 3.1.2 To allow rotor blade to produce more energy. L4 Om met die rotorlem meer energie op te wek. E OR/OF Advanced technology to have material that can allow a big structure to stand firm on the ground. Verbeterde tegnologie om materiaal te hê wat toelaat dat so ‘n groot struktuur standvastig kan staan OR/OF 2O reason (more electricity) Larger rotor diameters allow wind turbines to sweep more area, capture more wind and produce more electricity Groter rotormiddellyne laat die windturbines ‘n groter area dek, meer wind vang en so meer krag opwek OR/OF Demand for electricity increased/ demand for cleaner electricity Verhoogde aanvraag vir elektrisiteit/ aanvraag vir skoner elektisiteit (2) M 3.1.3 Max. height (in m) = Poles height + radius of L2 rotor M Maks. Hoogte (in m) = Paal hoogte + radius van rotor ✓ RT 124 1RT both correct values = 114 + 2 ✓MA 1MA divide by 2 to determine the radius = 114 + 62 = 176 ✓CA 1CA simplification OR/OF OR/OF Pole + rotor/ Paal + rotor =124 + 114 ✓ RT 1RT both correct values = 238 Maximum height /Maksimum hoogte in m = 238 – (124÷2) ✓MA 1MA divide by 2 to determine = 238 – 62 the radius = 176 ✓CA 1CA simplification AO (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 7 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L M 124 3.1.4* Radius = 2 = 62 ✓ A 1A radius L4 M Area /Oppervlakte = 3,142 × (62)2 ✓ SF 1SF substitution squared = 12 077,848 m2 ✓ CA 1CA simplification ✓O Not valid. / Nie geldig nie 1O invalid (4) ✓ RT 3.1.5 5 000 − 800 1RT 1st correct value M % increase/verhoging = × 100% ✓ MA 1A denominator 800 L2 ✓A 1MA percentage M = 525 % ✓CA 1CA simplification OR/OF OR/OF Current percentage / Huidige persentasie ✓RT 1RT 1st correct value 5 000 = 800 × 100% 1A denominator ✓A = 625% % increase/verhoging 625% – 100% ✓ MA 1MA percentage difference = 525 % ✓CA 1CA simplification (4) ✓✓ A M 3.1.6* Generators OR solar power OR hydro-power OR L1 nuclear power 2A source E Kragopwekkers OF sonkrag OF hidro-elektrisiteit OF kernkrag (2) ✓ RT 1RT correct values M ✓ SF 3.2.1 Perimeter/Omtrek = 2 × (2,3 + 2,3 + 2,3 + 3) m 1SF substitution L2 ✓ MA E = 2 × (6,9 + 3) m 1MA 6,9 = 19,8 m ✓ CA 1CA answer OR/OF OR/OF Perimeter/Omtrek 1RT correct values ✓ RT ✓ SF 1SF substitution = 3 + 2,3 + 2,3 + 2,3 + 3 + 2,3 + 2,3 + 2,3 m ✓MA 1MA 6,9 = 19,8 m ✓CA 1CA answer (4) ✓RT ✓ A M 1RT correct numbers 5 and 7 3.2.2* 5 + 7 + 4 = 16 boards /planke ✓CA L3 1A on 4 M 1CA simplification adding AO (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 8 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L M 5 3.2.3 Height /hoogte B= × 0,5 m L3 7 D = 0,3571428571 m✓ A 1A height box B Vol B = ( 2,3 × 3 × 0,3571428571)𝑚3 = 2,464285714𝑚3 ✓ CA 1CA volume of B box Vol A = (2,3 × 3 × 0,5) 𝑚3 ✓ SF 1SF volume of A box = 3,45𝑚3 ✓A 1A simplification 3,45 CA total volume Total/Totaal = 2,4642514𝑚3 + 3,45𝑚3 = 5,914285714𝑚3 ✓ CA Capacity / Kapasiteit = 1 000 × 5,914285714 1CA answer in litres = 5 914,285714 litres. ✓ CA OR/OF OR/OF Height of section B/ Hoogte van boks B 5 = × 0,5 m 1A height box B 7 = 0,3571428571 m ✓ A Vol = (length×width×height) + (length×width×height) ✓ SF 1SF volume of A box = (2,3 m × 3 m × 0,357… m) + (2,3 × 3 m × 0,5 m) ✓ CA ✓A 1CA volume of B box = (2,464285714 + 3,45) m3. 1A 3,45m3 ✓ CA = 5,914285714 m3 1CA total volume Capacity / Kapasiteit = 1 000 × 5,914285714 = 5 914,285714 litres. ✓ CA 1CA answer in litres OR/OF OR/OF Volume A = length × width × height/ lengte × breedte ×hoogte = (3 m × 2,3 m× 0,5 m) ✓SF 1SF volume of A box = 3,45 m3 ✓A 1A simplification 3,45 ∴ 3 450 litres ✓C 1C conversion 3 450 ℓ 1A ratio Volume B = ×5 ✓A 7 = 2 464,285714 litres ✓CA 1CA volume box B ∴ 𝑇𝑜𝑡𝑎𝑙 = 3 450 + 2 464,285714 1CA answer in litres = 5 914,285714ℓ ✓ CA NPR (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 9 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L M 3.3* o C = 5 × (oF – 32o) L2 9 M ✓ SF 70° = 5 × (oF – 32o) 1SF substituting in formula 9 9 70° × = °F – 32 ✓ MA 5 1MA changing subject 126° = °F – 32 °F = 158 ✓ CA 1CA answer (3) [33] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 10 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 [33 MARKS/PUNTE] Q/V Solution/Oplossing Explanation/Verduideliking T/L ✓RT 1RT 30 MP 4.1.1 30 : 6 ✓RT 1RT 6 L2 = 5 : 1 ✓A 1A simplification E (3) ✓✓A ✓A 2A 1st room MP 4.1.2 Reading room and computer lab 1A second room L1 Leeskamer en rekenaarlokaal (3) E ✓✓A MP 4.1.3 Stairs / trappe 2A stairs L2 (2) E ✓✓A ✓A MP 4.1.4 Multi-media room 1 / Multi-mediakamer 1 2A correct room L3 1A correct number M (3) M 4.1.5 Area/ Opp A = length × width /lengte × breedte L4 = 11 m × 3,5 m ✓SF 1SF substitution D = 38,5 m2 ✓ MCA 1MCA simplification Area/ Opp B = length × width /lengte × breedte = 14 m × 3,5 m = 49 m2 ✓ MA 1MA simplification Floor area/Vloer opp. = 38,5 m2 + 49 m2 = 87,5 m2 ✓ MCA 1MCA simplification total area Area of tile = length × width Opp van teël = lengte × breedte = 600 mm × 600 mm = 360 000 mm2 ✓ A 1A area tile 360 000 ∴ = 0,36 m2 ✓C 1C conversion 1000 000 87,5 Number of tiles/Getal teëls = ✓ MCA 0,36 1MCA dividing ≈ 243,056 tiles✓ CA 1CA number of tiles 244 Number of boxes/ Getal bokse = 5 = 48,8 ✓ CA = 49 1CA number of boxes INVALID/ ONGELDIG. ✓ O 1O opinion OR/OF OR/OF 1SF substitution ✓ SF ✓ MA ✓A 1MA adding areas Floor Area/vloeropp = 11m ×7 m + 3,5 m × 3 m = 77 m2 + 10,5 m2 1A 3 = 87,5 m 2 ✓ CA 1CA area Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 11 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L 600 mm Tiles /Teëls = = 0,6 m ✓ C 1C conversion 1000 Area of a tile / Opp van teël 1MCA area of tile = 0,6 m × 0,6 m = 0,36 m2 ✓ MCA 87,5 1MCA dividing areas Number of tiles/Getal teëls = ✓ MCA 0,36 1CA number of tiles ≈ 243,056 tiles✓ CA 244 Number of boxes/ Getal bokse = 5 = 48,8 = 49 ✓ CA 1CA number of boxes INVALID/ ONGELDIG. ✓ O 1O conclusion OR/OF OR/OF ✓ SF ✓ A 1A 3 Floor Area/ Vloer opp = 11m ×7 m + 3,5 m × 3 m 1SF substitution = 77 m2 + 10,5 m2 ✓ MA 1MA adding areas = 87,5 m2 ✓ CA 1CA area 600 mm Tiles / Teëls = = 0,6 m ✓ C 1C conversion 1000 Area of a tile / Opp van ‘n teël = 0,6 m × 0,6 m = 0,36 m2 ✓ MCA 1MCA area of tile 87,5 Number of tiles / Getal teëls = ✓ MCA 1MCA dividing areas 0,36 ✓ CA ≈ 243,056 tiles 1CA number of tiles tiles in 40 boxes / teels in 40 bokse = 40 × 5 = 200 40 boxes is not enough or 200 < 244 ✓ CA 1CA less than 40 bokse is nie genoeg nie of 200 < 244 INVALID./ ONGELDIG ✓O 1O conclusion OR/OF OR/OF ✓ SF 1A 3 ✓A Floor Area/vloeropp = 14m ×7 m – 3,5 m × 3 m 1SF substitution = 98 m2 – 10,5 m2 ✓ MA 1MA subtracting areas = 87,5 m2 ✓ CA 1CA area 600 mm Tiles /Teëls = 1 000 = 0,6 m ✓C 1C conversion Area of a tile / Opp van teël = 0,6 m × 0,6 m = 0,36 m2 ✓ MCA 1MCA area of tile 87,5 Number of tiles/Getal teëls = ✓ MCA 1MCA dividing areas 0,36 ≈ 243,056 tiles ✓ CA 1CA number of tiles 244 Number of boxes/ Getal bokse = 5 = 48,8 ✓ CA 1CA number of boxes = 49 INVALID/ ONGELDIG. ✓ O 1O conclusion Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 12 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L OR/OF OR/OF Area A = length × width/ lengte × breedte = 11 m × 3,5 m ✓SF 1SF substitution = 38,5 m2 ✓ MCA 1MCA simplification Area B = length × width/ lengte × breedte = 14 m × 3,5 m = 49 m2 ✓ MCA 1MCA simplification Floor area/Vloer opp. = 38,5 m2 + 49 m2 1MCA simplification total = 87,5 m2 ✓ MCA area Area of tile = length × width / Opp van teël= l × b = 600 mm × 600 mm = 360 000 mm2 ✓A 1A area tile 360 000 1C conversion ∴ 1000 000 = 0,36 m2 ✓C 87,5 Number of tiles/Getal teëls = ✓ MCA 1MCA dividing 0,36 1CA number of tiles ≈ 243,056 tiles ✓ CA 244 Number of boxes/ Getal bokse = 5 = 48,8 ✓ CA 1CA number of boxes = 49 INVALID/ ONGELDIG. ✓ O 1O opinion OR/OF OR/OF Area of tile / Opp van teël = 600 mm × 600 mm = 360 000 mm2 ✓A 1A area tile L = 14 m × 1 000 = 14 000 mm ✓C 1C conversion B = 7 m × 1 000 = 7 000 mm ∴ 𝐴𝑟𝑒𝑎/𝑂𝑝𝑝 = 14 000 𝑚𝑚 × 7 000 𝑚𝑚 ✓SF 1SF substitution = 98 000 000 mm2 ✓ MCA 1MCA simplification ∴ 𝐴𝑟𝑒𝑎/𝑂𝑝𝑝 = 3 500 𝑚𝑚 × 3 000 𝑚𝑚 = 10 500 000 mm2 ✓ MCA 1MCA simplification Total area/ Totale opp = 98 000 000 mm2 – 10 500 000 mm2 1MCA simplification total = 87 500 000 mm2 ✓MCA area Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 13 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L Number of tiles/Aantal teëls 87 500 000 𝑚𝑚2 1MCA dividing = ✓MCA = 243,0555556 tiles ✓CA 360 000 𝑚𝑚2 1CA number of tiles Number of boxes/Getal bokse 243,0555556 = 5 = 48,61 ≈ 49 ✓CA 1CA number of boxes ∴ INVALID/𝑂𝑁𝐺𝐸𝐿𝐷𝐼𝐺 ✓O 1O opinion OR/OF OR/OF ✓SF ✓ A 1A 3 Area (Lab) = (7 × 14 – 3 × 3.5) 𝑚2 1SF substitution = (98 – 10,5) 𝑚2 ✓MA 1MA subtracting = 87,5 𝑚2 ✓ MCA 1MCA simplification total area Tile side / Teël sy = 600 ÷ 1 000 = 0,6 m ✓ C 1C conversion Area covered by a box of tiles Oppervlakte bedek deur ‘n boks teëls = (0,6 × 0,6) × 5 ✓MCA 1MCA area of 1 tile = 1,8 m2 ✓ CA 1CA area box of tiles Number of boxes / Getal bokse 87,5 1MCA dividing = 1,8 ✓ MCA = 48,6 ≈ 49 ✓ CA 1CA number of boxes of tiles INVALID / ONGELDIG ✓ O 1O opinion OR/OF OR/OF Calculating 3 areas/Berekening 3 opp. A1 = 3,5 ×11 ✓SF = 38,5𝑚2 1SF substitution A2 = 3 × 3,5 ✓ A 1A 3 =10.5𝑚2 A3 = 3,5 × 11 = 38.5𝑚2 1MA adding TOTAL =38,5𝑚2 +10,5𝑚2 +38,5𝑚2 ✓MA =8 7,5𝑚2 ✓MCA 1MCA simplification total area 87,5 Number of tiles/Getal teëls = ✓ MCA 1MCA dividing 0,36 ≈ 243,056 tiles✓ CA 1CA number of tiles 244 Number of boxes/ Getal bokse = 5 = 48,8 = 49 ✓ CA 1CA number of boxes INVALID/ ONGELDIG ✓O 1O opinion (10) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 14 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L MP 4.2.1* 4 ✓✓ A 2A number of countries L1 (2) E MP 4.2.2 Harare ✓✓ A 2A correct town L1 (2) E M 4.2.3* ✓ SF ✓A 1A 179 L3 (a) 97 km/h = 179 km ÷ tyd 1SF substitution 97 D Time =distance ÷ speed Tyd = afstand ÷ spoed 179 1MCA change formula = ✓ MCA 97 = 1,845 hours ✓ CA 1CA time in hours Time duration / tydsduur = 1,845 hours./uur = 1 hour/uur + 0,845 × 60 min = 1 hour/uur 51 min Arrival time / Aankomstyd: = 09:55 + 1 h 51 min = 11: 46 ✓ CA 1CA answer (5) ✓MA M 4.2.3 Distance/Afstand = (713 – 263) + 2(18) 1MA subtracting L2 (b) correct values M = 450 + 36 ✓MA 1MA getting 36 = 486 km ✓ CA 1CA total distance OR/OF OR/OF Distance /Afstand 1MA subtracting ✓ MA ✓ MA correct values = (713 – 552) + 18 + 18 + (552 – 263) 1MA adding distances = 161 + 18 + 18 +289 = 486 km ✓ CA 1CA answer (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 15 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L OR/OF OR/OF ✓MA Distance/Afstand = (455 – 5) + 2(18) 1MA subtracting correct values = 450 + 36 ✓MA 1MA getting 36 = 486 km ✓ CA 1CA total distance OR/OF OR/OF Distance /Afstand ✓ MA ✓ MA 1MA subtracting correct values = (166 – 5) + 18 + 18 + (455 – 166) 1MA adding values = 161 + 18 + 18 +289 = 486 km ✓ CA 1CA answer OR/OF OR/OF Distance/Afstand ✓MA ✓MA 1MA subtracting correct values = 179 + 18 + (552 – 263) km 1MA adding values = 486 km ✓CA 1CA answer (3) [33] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 16 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 [29 MARKS/PUNTE] Q/V Solution/Oplossing Explanation/Verduideliking T/L MP 5.1.1* 8 ✓✓A 2A correct number L1 (2) E MP 5.1.2 Front entrance portal /Voorste ingangsportaal 1MA subtracting from 58 L3 ✓MA ✓RT ✓RT 1RT room dimensions M = 58 – (11 × 4 + 2 × 4) ✓A 1RT wall thickness 1A multiplying with 4 = 6 feet/ voet ✓CA 1CA simplification (5) MP 5.1.3 There are no walls separating the kitchen, dining room L4 and living room. ✓✓O 2O reason E Daar is geen mure wat die kombuis, eetkamer en (2) woonvertrek skei nie MP ✓✓A 5.1.4* Toilet OR bath OR basin or sink 2A correct feature L1 Toilet OF bad OF wasbak (2) E ✓O ✓O MP rd 5.1.5 3 floor and B that it is the second apartment L4 3de vloer en B is die tweede woonstel M OR/OF ✓O Block B, Number 3 ✓O Blok B, nommer 3 OR/OF 1O numbering of the floors ✓O 1O numbering of the 3 Floor, unit on the left/right ✓O rd apartments de 3 vloer, die eenheid links/ regs OR/OF ✓O 3rd Floor, B-wing ✓O 3de vloer, B -vleuel (2) M 5.1.6 17,6784 m = 58 feet/voet L2 (a) Conversion factor/ Herleidings faktor: M ✓RT 1RT 58 = 3,28083989… ✓MA 58 1m= 1MA simplification 17, 6784 ≈ 3,281 feet ✓R 1R rounded answer (3) CA from 5.1.6 (a) M ✓RT 5.1.6 40 1RT correct width L2 Width / Breedte = (b) 3, 281 ✓MCA 1MCA dividing M = 12,191405 m ✓CA 1CA simplification Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 17 DBE/May/June/Mei/Junie 2024 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L OR/OF OR/OF 58 feet/voet = 17,6784 m 40 feet/voet = n ✓RT 1RT correct width 40 1MA working with ratio n = 58 × 17,6784 ✓MA 1CA simplification = 12,191405 m ✓CA NPR (3) M 5.2.1 Area /Oppervlakte = length × width / lengte × breedte L2 = 0,614 m × 0,474 m ✓SF 1SF substitution E = 0,291036 m2 = 0,3 m2 ✓R 1R simplification NPU (2) CA from Q 5.2.1 M 5.2.2* Area for 6 panels /Opp van 6 panele = 0,3 m2 × 6 L4 = 1,8 m2 ✓MCA 1MCA simplification D Cost for 6 panels /Koste van 6 panele = 1,8 m2 × R490/m2 = R882 ✓MCA 1MCA simplification cost Mass of the 6 panels / Massa van 6 panele = 1,8 m2 × 15 kg/m2 = 27 kg ✓MCA 1MCA simplification: mass Delivery mass / Aflewerings massa= 20 kg + 7 kg Cost of delivery / Afleweringskoste ✓MA 1MA cost of 1st 20kg = R820 + R53,50 × 7 kg ✓MCA 1MCA add and multiply = R1 194,50 ✓CA 1CA simplification Total cost / Totale koste = R882,00 + R1 194,50 = R2 076,50 ✓CA 1CA simplification INVALID/ ONGELDIG ✓O 1O verification OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Text truncated here. 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