Mathematical Literacy Grade 12 P2 NSC November 2025 Memo

NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 MATHEMATICAL LITERACY P2/ WISKUNDIGE GELETTERDHEID V2 NOVEMBER 2025 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 Correct answer only/Slegs korrekte antwoord These marking guidelines consist of 15 pages. Hierdie nasienriglyne bestaan uit 15 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 2 DBE/November 2025 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. • NOTE: consistent accuracy (CA) does not apply in cases of a 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. • Rounding is an independent mark. • A conclusion mark can only be given if relevant calculations precede it. • 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 op by die tweede berekeningsfout. • Let wel: volgehoue akkuraatheid (CA) geld nie in die geval van 'n afbreuk 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 • Afronding tel as 'n onafhanklike punt • 'n Gevolgtrekkingspunt kan slegs gegee word indien relevante berekeninge dit voorgaan. • Geen penalisering vir ronding (NPR) as die eerste desimaal korrek is nie. . Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 3 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 [28 MARKS/PUNTE] ANSWER ONLY FULL MARKS Q/V Solution/Oplossing Explanation/Verduideliking T&L MP 1.1.1 B ✓✓ A 2A correct option L1 (2) E MP 1.1.2 G ✓✓ A 2A correct option L1 (2) E M 1.1.3 A ✓✓ A 2A correct option L1 (2) E MP 1.1.4 E ✓✓ A 2A correct option L1 (2) E M 1.1.5 F ✓✓ A 2A correct option L1 (2) E ✓A M ✓ RT 1.2.1 Quarter past two in the afternoon. 1RT correct time in words L1 Kwart oor twee in die middag. 1A correct time of day E OR/OF OR/OF ✓ RT ✓A Fifteen minutes after two in the afternoon. 1RT correct time in words Vyftien minute na twee namiddag. 1A correct time of day (2) ✓ MA M 1.2.2 1 1 L1 Number of passengers/Aantal passasiers = 333 % × 189 1MA multiply with 333 % M = 63 ✓ A 1A simplification OR/OF OR/OF 1 1 Number of passengers/Aantal passasiers = × 189 ✓ MA 1MA multiply with 3 3 = 63 1A simplification ✓A (2) M 1.2.3 B ✓✓ A 2A correct letter L1 (2) M M 1.2.4 Thursday/Donderdag ✓✓ A 2A correct day L1 (2) E Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 4 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T/L MP 1.3.1 (a) R ✓A 1A correct letter L1 (b) S ✓A 1A correct letter M (c) U ✓A 1A correct letter (d) P ✓A 1A correct letter (e) Q ✓A 1A correct letter (f) T ✓A 1A correct letter (6) M 1.3.2 C ✓✓ A 2A correct formula L1 (2) E M 1.3.3 Length of geyser/Lengte van waterverwarmer L1 = 1,2 m × 1 000 ✓ MA 1MA multiply by 1 000 E = 1 200 mm ✓ A 1A conversion (2) [28] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 5 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 [34 MARKS/PUNTE] Q/V Solution/Oplossing Explanation/Verduideliking T&L MP 2.1.1 Number of seats/ Getal sitplekke 1MA addition and subtraction L1 = 59 + 22 – 4 ✓MA of correct values E = 77 seats/sitplekke ✓A 1A simplification AO (2) ✓A MP 2.1.2 F2 ✓A 1A correct row L2 1A correct seat M (2) MP 2.1.3 a) Right/Regs ✓A 1A correct choice L2 b) Left/Links ✓A 1A correct choice E c) Aisle/Gang ✓A 1A correct choice d) Second/Tweede ✓A 1A correct choice (4) ✓✓RT MP 2.2.1 Camps Bay/Kampsbaai 2RT correct stop L1 (2) E MP 2.2.2 Bar scale / Line scale /Linear scale / Graphic scale L1 Staafskaal/Lynskaal /Grafiese skaal ✓✓RT 2RT correct scale E (2) MP 2.2.3 Clockwise/Kloksgewys ✓✓A 2A correct direction L1 (2) E ✓A MP 2.2.4 13 mm = 500 m ✓RT 1A measured distance L3 1RT correct ratio M Real distance in meter/Regte afstand in meter 19,2 km = 19 200 m ✓C 1C convert to m Map distance/Kaart afstand = 19 200 ÷ 500 × 13 ✓MCA 1MCA divide and multiply = 499,2 mm ✓CA 1CA simplification OR/OF OR/OF ✓A 1A measured distance Bar/Staaf: 1,3 cm = 500 m ✓RT 1RT correct ratio 500 m = 0,5 km ✓C 1C convert to km Map distance / kaart afstand 19,2 km × 1,3 cm 1MCA divide and multiply = ✓MCA 0,5 km = 49,92 cm ✓CA 1CA simplification OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 6 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L Measure distance / gemete afstand ✓A 1A measured distance 1,3 cm : 500 ✓RT 1RT correct ratio ✓C 0,013 m : 500 m 1C convert to m 1 : 38 461,53846 ?: 19,2 19,2 ?= ✓MCA 38 461,53846 1MCA divide by scale factor = 0,0004992 km = 49,92 cm OR 499,2 mm ✓CA 1CA simplification Accept 12 mm – 13 mm Accept 460,8 for 12 mm (5) ✓✓A MP 2.3.1 Right / Regs 2A correct direction L1 (2) E MP 2.3.2 Longer dimension/Langer afmeting = 12’8” ✓RT 1RT 12’8” L3 Length in cm/Lengte in cm M 12’ × 30,48 ✓MCA 1MCA multiply by 30,48 = 365,76 cm ✓C 1C convert feet to cm 8” × 2,54 = 20,32 cm ✓C 1C convert inches to cm Total/Totaal = 365,76 + 20,32 = 386,08 cm ✓CA 1CA simplification OR/OF OR/OF Longer dimension/Langer afmeting = 12’8” ✓RT 1RT 12’8” ✓ ✓MCA Length/ Lengte = (12 × 30,48 cm) + (8 × 2,54 cm) 1MCA multiply by 30,48 ✓C ✓C 1C convert feet to cm = 365,6 cm + 20,32 cm 1C convert inches to cm = 386,08 cm ✓CA 1CA simplification (5) ✓✓RT MP 2.3.3 One has a living room and the other has a bedroom/ 2RT identify first difference L1 Een het 'n woonkamer en die ander een 'n slaapkamer M ✓✓RT 2RT identify second The balcony is on the first floor/ Die balkon is op die eerste verdieping difference (4) MP ✓✓O 2.3.4 It is on the ground floor. /Dit is op die grondverdieping. 2O explanation L4 (2) M MP 2.3.5 There must be more flats adjacent to it. ✓✓O 2O explanation L4 Daar moet meer woonstelle langsaan wees. (2) M [34] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 7 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 [32 MARKS/PUNTE] Q/V Solution/Oplossing Explanation/Verduideliking T/L ✓ MA M 3.1.1 13:12 – 3 hours and 27 min/3 ure en 27 min 1MA subtracting time L2 = 09:45 ✓A 1A simplification E OR/OF OR/OF Time = 13,2 – 3,45 = 9,75 hours ✓ MA 1MA subtracting the hours = 09:45 ✓A 1A simplification AO (2) M ✓RT ✓ MA 1RT interpreting pattern L2 3.1.2 Number/Getal = 1 + 2 + 3 + 4 + 5 + 6 + 7 1MA adding values E 1CA simplification = 28 ✓CA OR/OF OR/OF ✓RT 1RT interpreting pattern 7(1+7) Number/Getal = 2 1MA multiplication of = 7 × 4 ✓ MA values = 28 ✓CA 1CA simplification AO (3) M 3.1.3 Height in cm/Hoogte in cm = 240 mm L2 M = 24 cm ✓C 1C conversion 1RT 28 1 ✓ RT Area/Oppervlakte = 2 (28 cm × 24 cm) ✓ SF 1SF substitution in formula = 336 cm2 ✓CA 1CA simplification OR/OF OR/OF ✓ RT ✓ SF 1SF substitution in formula Area/Oppervlakte = ½ × 280 mm × 240 mm 1 RT 280 = 33 600 mm2 ✓CA 1CA simplification = 336 cm2 ✓C 1C conversion (4) ✓ MA CA from 3.1.3 M 3.1.4 Area of C/Oppervlakte van C = 30 240 mm2 × 1,6 1MA multiply by 1,6 L3 ✓ CA 1CA simplification M = 48 384 mm2 Total Area/Totale oppervlakte ✓C 1C convert to mm2 = 48 384 mm2 + 30 240 mm2 + 33 600 mm2 ✓ MCA 1MCA adding values = 112 224 mm2 ✓ CA 1CA simplification OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 8 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L Part/Deel B: 30 240 mm2 ÷102 = 302,4 cm2 ✓ MA 1MA multiply by 1,6 Part /Deel C: 302,4 cm2 ×1,6 = 483,84 cm2 ✓ CA 1CA simplification ✓ MCA ✓ CA 1MCA adding values Total/Totaal: (336 + 483,84 + 302,4) cm = 1 122,24 cm2 2 1CA simplification  (1 122,24 ×100) mm2 = 112 224 mm2 ✓ C 1C convert to mm2 (5) P 3.1.5 9 + 7 + 28 + 8 + 22 = 74 ✓ A 1A correct number of circles L2 M 16 ✓ MA 1MA correct numerator P= × 100 % ✓ MA 1MA concept percentage 74 ≈ 21,62% ✓CA 1CA simplification OR/OF OR/OF 9 + 7 + 28 + 8 + 22 = 74 ✓ A 1A correct number of circles 9 P(red) = 74 × 100% ✓ MA 1MA P(red) = 12,16216216% 7 P(green) = 74 × 100% ✓MA 1MA P(green) = 9,459459459% P(red or green) ≈ 12,16 % + 9,46% = 21,62% ✓CA 1CA simplification OR/OF OR/OF 9 + 7 + 28 + 8 + 22 = 74 ✓ A 1A correct number of circles ✓ MA 1MA P(red) P(red or green) = 74 + 74 ✓ MA 9 7 1MA P(green) = 0,1216216… + 0,945945… = 0,216216… × 100% = 21,62% ✓CA 1CA simplification (4) M 3.2.1 Radius of small circle/ Radius van klein sirkel L1 4 1MA multiplication by 14 E = 7 × 14 mm ✓MA 1A correct length = 8 mm ✓A AO (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 9 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L M 3.2.2 Circumference/Omtrek = 3,142 × 2 × 14 mm ✓SF 1SF substitution in formula L2 = 87,976 mm ✓CA 1CA simplification M ≈ 88 mm ✓R 1R rounding (3) M 3.2.3 Area of material/Oppervlakte van materiaal L4 ✓C 1C converting to cm D = 30 cm × 137 cm ✓SF 1SF substitution in formula = 4 110 cm² ✓CA 1CA correct area Total area of all circles/Totale oppervlakte van alle sirkels 1SF substitution in formula ✓SF ✓MA ✓MA 1MA multiplying by 48 = 3,142 × (1,4 cm)² × 48 × 2 × 2 1MA multiplying by 2 × 2 = 1 182,40 cm² ✓CA 1CA simplification Area of off-cuts/Oppervlakte van oorblyfsels = 4 110 cm² – 1 182,40 cm² 1CA area of off cuts = 2 927,60 cm² ✓CA ✓O 1O opinion Her statement is CORRECT/Haar stelling is KORREK OR/OF OR/OF Area of material/Oppervlakte van materiaal ✓C 1C converting to cm = 30 cm × 137 cm ✓SF 1SF substitution in formula = 4 110 cm² ✓CA 1CA correct area Area of circle /Oppervlakte van sirkel = 3,142 × radius² = 3,142 × (1,4 cm)² ✓SF 1SF substitution in formula =6,15832 cm2 ✓MA 1MA multiplying by 48 Number of circles / Getal sirkels = 48 × 2 = 96 For both sides/ Vir beide kante = 96 × 2 = 192 ✓MA 1MA multiplying 96 by 2 Total area/ Totale oppervlakte = 6,15832 cm2 × 192 = 1 182,39744 cm2 ✓CA 1CA simplification Area of off-cuts/Oppervlakte van oorblyfsels = 4 110 – 1 182,439744 ≈ 2 927,60 cm² ✓CA 1CA area of off cuts ✓O 1O opinion Her statement is CORRECT/Haar stelling is KORREK (9) [32] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 10 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 [30 MARKS/PUNTE] Q/V Solution/Oplossing Explanation/Verduideliking T&L MP 4.1.1 De Aar ✓✓ A 2A correct town L1 (2) E MP 4.1.2 8 ✓✓ A 2A correct number L2 (a) (2) E ✓✓ A ✓A MP 4.1.2 Kimberley and/en Britstown 2A 1st correct town L2 (b) 1A 2nd correct town E (3) MP 4.1.3 Provincial borders are indicated by a broken line / dotted L4 line/ dashed line ✓✓ O 2O correct description M Provinsiale grense word deur 'n gebroke lyn/ stippellyn voorgestel (2) M 4.2.1 17 months/maande ✓✓ A 2A number of months L2 (2) E M 4.2.2 Time of arrival/Aankomstyd L3 ✓ MA 1MA adding the flight time M = 19:00 + 14 hours 25 min/14 ure 25 min = 09:25 1A 16 June ✓A ✓A 1A arrival time 16 June 2023 at 09:25 / 16 Junie 2023 teen 09:25 OR/OF OR/OF From/Van 19:00 to 24:00 = 5 hours/ure ✓ A 1A 5 hours 14 hours/ ure 25 min – 5 hours/ure = 09:25 ✓A ✓A 1A 16 June 16 June 2023 at 09:25 / 16 Junie 2023 teen 09:25 1A arrival time AO (3) P 4.2.3 0 ; 0% , Impossible /Onmoontlik ✓✓ A 2A correct probability L2 (2) E M 4.2.4 Distance = Speed × time /Afstand = Spoed × tyd L2 770 km = speed/spoed × 14 hours 25 min/14 ure 25 min 1SF substitute into formula D ✓ SF 770 km ✓S Speed/Spoed = 1S change subject of formula 14 hours 25 min 770 km = ✓C 1C convert minutes to hours 14, 4166... hours ≈ 53,41 km/h ✓CA 1CA simplification (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 11 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L M 4.2.5 1 mile/myl = 1,60934 km L4 311,72 miles/myle = ? km 1MA multiply with conversion M ? = 311,72 × 1,60934 km ✓MA factor = 501,6634… km ✓A 1A simplification Extra distance/Ekstra afstand = 770 km – 501,6634… km ✓MCA 1MCA subtracting = 268,336… km 1CA simplification ✓CA INVALID/NIE GELDIG NIE ✓O 1O verification OR/OF OR/OF Distance/ Afstand = 311,72 × 1,60934 km ✓MA 1MA multiply with 1,60934 = 501,6634… km ✓A 1A simplification Total distance/ Totale afstand 1MCA adding 268,13 = 501,6634… km + 268,13 km ✓MCA = 769,7934… km ✓CA 1CA simplification INVALID/ NIE GELDIG NIE ✓O 1O verification (5) M 4.2.6 Number of adults/Aantal volwassenes L3 = 78% × 333 million/miljoen ✓MA 1MA calculating 78% M = 259,74 million/miljoen Number of adult women/Aantal volwasse vroue = 50% × 259,74 million/miljoen ✓MA 1MA calculating 50% of = 129,87 million/miljoen previous value Number shorter than flamingo/Aantal korter as flamink =10% × 129,87 million/miljoen ✓MA 1MA calculating 10% of = 12,987 million/miljoen ✓CA previous value ≈ 13 million/miljoen ✓R 1CA simplification 1R correctly rounded OR/OF OR/OF ✓MA 1MA calculating 50% Number of women/ vroue = 50% × 333 000 000 = 166 500 000 ✓MA 1MA calculating 78% of Number of adults/ volwasse = 78% × 166 500 000 = 129 870 000 previous value Number shorter / korter = 10% × 129 870 000 ✓MA 1MA calculating 10% of = 12 987 000 ✓CA previous value ≈ 13 000 000 ✓R 1CA simplification 1R correctly rounded (5) [30] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 12 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 [26 MARKS/PUNTE] Q/V Solution/Oplossing Explanation/Verduideliking T&L MP 5.1.1 F OR/OF G ✓✓ RT 2RT for correct letter L2 (a) (2) M MP 5.1.1 C ✓✓ A 2RT correct letter L2 (b) (2) M ✓A P 5.1.2 1 1A numerator L2 P= OR/OF 0,333 OR/OF 33,3% 3 1A denominator E ✓A NPR (2) M 5.1.3 The length of the bookcase is 60 cm and the height of L4 (a) the bookcase is 90 cm. ✓✓ O 2O explanation M Die lengte van die boekrak is 60 cm en die hoogte van die boekrak is 90 cm. OR/OF OR/OF It covers the entire back of the bookcase ✓✓ O 2O explanation Dit bedek die hele agterkant van die boekrak (2) M 5.1.3 Dimensions/Afmetings: 1MA using scale L3 (b) Width/Breedte = 60 cm ÷ 20 ✓ MA D = 3 cm ✓ A 1A simplification Height/Hoogte = 90 cm ÷ 20 = 4,5 cm ✓CA 1CA simplification 3 cm ✓ CA 1CA width drawn 1CA height drawn ✓ CA 4,5 cm ✓ CA 1CA shape (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 13 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L 5.2 M Area of complete piece of wood/Oppervlakte van volledige L3 stuk hout: 1SF substitute into M = 42 cm × 430 cm  SF formula = 18 060 cm2  A 1A simplification Area of pieces of one bookcase/Oppervlakte van dele van een boekrak:  MA 1MA total length and = (90 + 60 + 56) cm × (20 + 20) cm width = 8 240 cm2  CA 1CA simplification Area of unused wood/Oppervlakte van ongebruikte hout: = 18 060 – 2(8 240)  MCA 1MCA multiplying by 2 = 1 580 cm2  CA 1CA simplification OR/OF OR/OF Area complete piece/Oppervlakte van volledige stuk 1SF substitute into  SF formula = 42 cm × 430 cm = 18 060 cm²  A 1A simplification For 1 bookcase:  MA 1MA total length and Area /oppervlakte E & D = 60 cm × 20 cm × 2 = 2 400 cm² width Area /oppervlakte F & G = 56 cm × 20 cm × 2 = 2 240 cm² Area /oppervlakte A & B = 90 cm × 20 cm × 2 = 3 600 cm² Total/Totaal = (2 400 + 2 240 + 3 600) cm² = 8 240 cm²  CA 1CA simplification For 2 bookcases/ Vir 2 boekrakke = 8 240 cm² × 2 = 16 480 cm²  MCA 1MCA multiplying by 2 Remaining part/ Oorblywende deel = 18 060 cm² – 16 480 cm² = 1 580 cm²  CA 1CA simplification OR/OF OR/OF For 2 bookcases/Vir 2 boekrakke  MA E&D Area /Oppervlakte= 60 cm × 20 cm × 4 1MA total length and = 1200 cm2 × 4  MCA width = 4800 cm2 1MCA multiplying by 4 A&B Area /Oppervlakte = 90 cm × 20 cm × 4 = 1 800 cm2 × 4 = 7 200 cm2 F&G Area /Oppervlakte = 56cm × 20 cm × 4 = 1 120 cm 2 × 4 = 4 480 cm2 Area complete piece/Oppervlakte van volledige stuk 1SF substitute into = 430 cm × 42 cm  SF formula 2A 1A simplification = 18 060 cm Remaining part/ Oorblywende deel =  MCA 1MCA subtraction 18 060 cm²– (4480 + 7200 + 4800) cm2 = 1 580 cm²  CA 1CA simplification (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 14 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L OR/OF OR/OF Total length of one bookcase/Totale lengte van een boekrak: 1MA adding values and = 2(90) + 2(60) + 2(56)  MA multiplied by 2 = 412 cm A 1A simplification Wood left over /Hout wat oorbly = 430 cm – 412 cm  MA 1MA subtraction = 18 cm  CA 1CA simplification Area of unused wood/Oppervlakte van ongebruikte hout: = 18 cm × 40 cm + (2 × 430)  MCA 1MCA adding unused areas = 720 cm + 860 cm = 1 580 cm2  CA 1CA simplification OR/OF OR/OF Total length of one bookcase/Totale lengte van een 1MA adding values and boekrak:  MA multiplied by 2 = 2(90cm) + 2(60cm) + 2(56cm) = 412 cm  A 1A simplification Total width of 2 bookcases /Totale breedte van 2 boekrakke  MA 1MA subtraction 42 cm – 20 cm – 20 cm = 2 cm  CA 1CA simplification Wood left /Oorblywende hout = 430 cm – 412 cm = 18 cm Area of unused wood/Oppervlakte van ongebruikte hout: = (412 cm × 2cm) + (18cm × 42 cm)  MCA 1MCA adding unused areas = 1580 cm2  CA 1CA simplification (6) M 5.3 Volume of wood per month/Volume hout per maand L4 = 0,4 m3 × 100 3 = 400 000 cm3  C 1C converting m3 to cm3 D Mass per month/ Massa per maand = 0,75 g/ × 400 000 cm3  MA 1MA substitution  CA = 300 000 g ÷ 1000 1CA mass in gram = 300 kg 1 ton = 1000 kg Number of months to accumulate one ton/ 1C tonnes to kg Getal maande om ‘n ton mekaar te maak C 1CA number of months = 1000 kg ÷ 300 kg = 3,33  CA 1O verification VALID/GELDIG O (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematical Literacy/P2/Wiskundige Geletterdheid/V2 15 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne Q/V Solution/Oplossing Explanation/Verduideliking T&L OR/OF OR/OF Unused wood produced per month Ongebruikte hout per maand  MA C 1MA substitution = (0,4 × 1 000 000) cm3 × 0,75 g/ cm3 1C conversion = 300 000 g  CA 1CA mass in grams = 300 000 ÷ 1 000 = 300 kg = 300 ÷ 1 000 = 0,3 ton  C 1C convert to tonnes Number of tons in 3 months / Aantal ton in 3 maande 1CA tons in 3 months = 0,3 × 3 = 0,9  CA VALID/ GELDIG O 1O verification OR/OF OR/OF 0,4 m3 × 3 = 1,2 m3  A 1A volume in 3 months Mass of used wood/Massa van ongebruikte hout 1MA substitution  MA 1C converting m3 to cm3 = (1,2 × 100 × 100 × 100 × 0,75) g C 1CA mass in g = 900 000 g  CA = 900 kg C 1C converting to kg in 3 months < 1000 kg His statement is correct/Sy stelling is korrek O 1O verification OR/OF OR/OF  C 1C converting to gram 1 ton = 1 000 000 g C 1C converting m3 to cm3 Volume = 0,4 m3 × 1003 = 400 000 cm3 / month/maand  MA 1MA substitution Mass/Massa = (0,75 × 400 000)g = 300 000 g Number of months/ Getal maande  CA 1CA mass in g 1 000 000 g = 300 000 g = 3,3 months  CA 1CA number of months O Statement is valid/Sy stelling is korrek 1O verification (6) [26] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou