# Mathematical Literacy Grade 11 P2 2012 Memo (Afr) > Grade 11 Mathematical Literacy Paper 2 memorandum from the Eastern Cape November 2012 (Afrikaans), 8 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/mlit-p2-gr11-2012-memo-afr/b784093e-7b98-4716-81a2-3983fd3ffb42 - Updated: 2026-09-27 - Subject: Mathematical Literacy - Grade: Grade 11 - Document type: Memorandum - Year: 2012 - Exam period: Eastern Cape November - Paper: 2 - Curriculum: CAPS - Pages: 8 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Mathematical Literacy](https://www.hlayiso.com/library/grade-11/mathematical-literacy) › [Mathematical Literacy Grade 11 P2 2012 Memo (Afr)](https://www.hlayiso.com/library/mlit-p2-gr11-2012-memo-afr/b784093e-7b98-4716-81a2-3983fd3ffb42) ## About this document Mathematical Literacy · Grade 11 · Eastern Cape November · 2012. Memorandum, 8 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 Downloaded from hlayiso.com Province of the EASTERN CAPE EDUCATION NASIONALE SENIOR SERTIFIKAAT GRAAD 11 NOVEMBER 2012 WISKUNDIGE GELETTERDHEID V2 MEMORANDUM PUNTE: 100 Simbool Verduideliking M Metode MA Metode met akkuraatheid CA Voortgesette akkuraatheid A Akkuraatheid (Antwoord) C Omskakeling S Vereenvoudiging RT/RG/RM Lees van tabel/Lees van grafiek/Lees van kaart F Kies van korrekte formule SF Substitusie in formule J/O Mening P Penalisering vir geen eenhede, verkeerde afronding, ens. R (Afronding / Rede) Hierdie memorandum bestaan uit 8 bladsye. ### Page 2 Downloaded from hlayiso.com 2 WISKUNDIGE GELETTERDHEID V2 (NOVEMBER 2012) VRAAG 1 1.1 LU3 AS 1.1.1 1 tlp = 5 mℓ OF tsp = 15/5  1:C (mℓ 11.3.2 1 etlp = 15 mℓ = 3 tlp x 2 na tlp) 15 mℓ = 3 tlp  = 6 tlp  1:M (x2) Daarom is 30 ml = 3 tlp x 2  1:A = 6 tlp  (3) LU3 AS 1.1.2 1 blikkie = 4 mense 11.3.2 20 mense  4 mense 1:M =5 (20/4) Om 20 mense te bedien sal 5 blikkies nartjies benodig 1:A word. (2) LU3 AS 1.1.3 ºF = ºC x 1,8 + 32 11.3.2 = 220º x 1,8 + 32  1:SF = 396º  + 32 1:S = 428º  1:A 400º ≠ 430 º 1:R (10º) Nee, Gretchen het die oond op die verkeerde temperatuur ºC gestel.  (4) 1.2 1 lb (pond) 2 oz (onse) Volstruisfilet LU 3 AS 1 lb = 0,45359 kg 11.3.2 1 oz = 0,0625 lb 1:C (lb na 0,0625 lb x 0,45359 = 0,028349375 kg x 2  kg) = 0,05669875 kg 1: M (x2) Kilogram volstruisfilet = 1 lb + 2 oz = 0,45359 kg + 0,05669875 kg  1:M = 0,51028875 kg  1:A = 0,5 kg  (5) 1:R 1.3 Volume = r2h 1:A LU3 AS = 3,14 x 11 cm x 11 cm x 9 cm  (radius) 11.3.1 = 3419,46 cm3  1:SF 1:A As 1 000 cm3 = 1l dan is 3419,46 cm3/1 000 = 3,4 ℓ 1:C (cm3 Ja, die kastrol is groot genoeg vir die gekookte dis.  na l) (5) 1:J 1.4 LU1 AS 1.4.1 500 g = ½ kg 11.1.1 ½ x R67 = R 33,50  (1) 1:A ### Page 3 Downloaded from hlayiso.com (NOVEMBER 2012) WISKUNDIGE GELETTERDHEID V2 3 LU1 AS 1.4.2 BTW bedrag = (R33,50 + R12,59)  x 0,14 1:M 11.1.2 = R 46,09 x 0,14  (korrekte = R 6,45  waardes) 1:M (x 14%) (3) 1:A LU1 AS 1.4.3 Alhoewel nartjies vrugte is, is dit in hierdie geval 11.1.3 geprosesseer (geblik), terwyl die lemoene ʼn vars produk is.  OF Geen BTW is betaalbaar op vars produkte soos lemoene nie, maar wanneer dit geblik word, is belasbaar.  (Enige ander relevante verduideliking.) (2) 2:A LU1 AS 1.4.4 Omdat 1 en 2 sent muntstukke nie meer in sirkulasie is 11.1.3 nie, word die finale bedrag afgerond tot die naaste 5 sent.  (Enige ander relevante verduideliking.) (2) 2:A LU1 AS 1.4.5 Geen kleingeld is aan Gretchen verskuldig nie, 11.1.3 omdat sy slegs die verskuldigde bedrag betaal.  (2) 2:A LU1 AS 1.4.6 Middag  16:42  11.1.1 (2) 2:A [31] ### Page 4 Downloaded from hlayiso.com 4 WISKUNDIGE GELETTERDHEID V2 (NOVEMBER 2012) VRAAG 2 2.1 LU1 AS 2.1.1 Waarde van deposito = R250 000 x 0,16  1:M 11.1.1 = R40 000  (2) 1:A LU1 AS 2.1.2 P = R250 000 – R40 000 1:A (P- 11.1.1 = R210 000  waarde) n = 72 / 12 = 6 jare  1:A (n- i = 9,5 / 100 waarde) = 0,095 = 210 000 (1 + 6 x 0,095)  1:SF = 210 000 (1+ 0,57) = 210 000 (1,57)  1:S = R329 700  (5) 1:A LU1 AS 2.1.3 I=A–P 1:M 11.1.1 = R329 700 – R210 000  1:CA = R119 700  (2) 2.2 LU2 AS 2.2.1 1:SF 11.2.1 = 250 000 (1 – 0,2)2  1:S = 250 00 (0,8)2 1:A = 250 000 (0,64)  = R160 000  OF = 200 000 (1 – 0,2)1  = 200 000 (0,8)1 = R160 000  (3) LU1 AS 2.2.2 11.2.2 1 punt vir elk (0,250 000) (1,200 000) (2,160 000) (3,128 000) (4,102 400) (5) ### Page 5 Downloaded from hlayiso.com (NOVEMBER 2012) WISKUNDIGE GELETTERDHEID V2 5 LU1 AS 2.2.3 Indirekte of Omgekeerde eweredigheid  11.2.3 Soos die jare toeneem, so neem die waardes van 1:A die motor af.  (2) 1:R 2.3 P (silwer) = 3/14  OF 0,214  OF 21,4%  LU4 1:A (teller) AS 11.4.5 1:A (2) (noemer) [21] ### Page 6 Downloaded from hlayiso.com 6 WISKUNDIGE GELETTERDHEID V2 (NOVEMBER 2012) VRAAG 3 3.1 LU4 AS 3.1.1 5  1:A 11.4.5 Spanne kan nie teen hulself speel nie.  (2) 1:R LU4 AS 3.1.2 10  11.4.5 (2) 2:A LU4 AS 3.1.3 4  11.4.5 (2) 2:A LU4 AS 3.1.4 10  11.4.5 (2) 2:A LU4 AS 3.1.5 ¼  OF 0,25  OF 25%  1:A (teller) 11.4.5 (2) 1:A (noemer) 1 LU4 AS 3.1.6 /16  OF 0,063  OF 6,3%  1:A (teller) 11.4.5 (2) 1:A (noemer) 3.2 LU2 AS 3.2.1 (a) s = 5t + 2c + 3p  11.2.1 (3) 3:F LU2 AS (b) s = 5t + 2c + 3p 1:SF 11.2.1 = (5 x 6) + (2 x 5) + (3 x 3)  (korrekte = 30 + 10 + 9 waardes) = 49  (2) 1:CA LU2 AS 3.2.2 Vir 1 strafskop  11.2.1 (1) 1:A 3.3 LU1 AS 3.3.1 1 ZAR (Suid-Afrikaanse Rand) = 0,15761 NZD 11.1.1 Kategorie B = 123 NZD  1:RT (123) ZAR = 123 NZD/0,15761 NZD  1:M = 780,4073346 = 780,41  (3) 1:A LU1 AS 3.3.2 1 NZD = R6,3450 ZAR 11.1.1 200 NZD ZAR = 200 x 6,3450  1:M = R1 269  (2) 1A [23] ### Page 7 Downloaded from hlayiso.com (NOVEMBER 2012) WISKUNDIGE GELETTERDHEID V2 7 VRAAG 4 4.1 LU2 AS 4.1.1 Kosprys vir 1 CD = (80 – 30)/10 OF Kosprys vir 1 CD = (130 – 30)/20  11.2.3 = 50/10 = 100/20 = R5  = R5  (180 – 30) Kosprys vir 1 CD = /30 OF Kosprys vir 1 CD = (230 – 30)/10  = 150/30 = 200/40 = R5  = R5  (280 – 30) Kosprys vir 1 CD = /50OF Kosprys vir 1 CD = (330 – 30)/60  250 = /50 = 300/60 = R5  = R5  (380 – 30) Kosprys vir 1 CD = /70OF Kosprys vir 1 CD = (430 – 30)/80  350 = /70 = 400/80 = R5  = R5  Kosprys vir 1CD = (480 – 30)/90  = 450/90 1:M =R5  (2) 1:A LU2 AS 4.1.2 Verkoopprys van 1 CD = 60/10 OF Verkoopprys van = 120/20  11.2.3 = R6  = R6  180 Verkoopprys van 1 CD = /30 OF Verkoopprys van 1 CD = 240/40  = R6  = R6  300 Verkoopprys van 1 CD = /50  OF Verkoopprys van 1 CD = 360/60  = R6  = R6  420 /70  OF Verkoopprys van 1CD = 480/80  Verkoopprys van 1 CD = =R6 =R6 OF Verkoopprys van 1CD = 540/90  1:M = R6  (2) 1:A LU2 AS 4.1.3 % Wins = (Inkomste – Uitgawes) x 100 11.2.3 Uitgawes = 6 – 5 x 100 5 = 1/5 x 100  1:M = 20%  (2) 1:A LU2 AS 4.1.4 30  ; 180  1:A 11.2.1 (30) 1:A (2) (180) ### Page 8 Downloaded from hlayiso.com 8 WISKUNDIGE GELETTERDHEID V2 (NOVEMBER 2012) LU2 AS 4.1.5 Gelykbreekpunt  11.2.1 Vir 30 CD’s is die inkomste en uitgawes presies dieselfde 1:A (R180).  (2) 1:R LU2 AS 4.1.6 Voor die gelykbreekpunt is die inkomste minder as die 11.2.3 uitgawes.  OF Voor die gelykbreekpunt is die uitgawes meer as die inkomste.  (2) 2:A LU2 AS 4.1.7 Daar is ʼn aanvangskoste van R30 (vervoerkoste).  11.2.3 (2) 2:A 4.2 LU4 AS 4.2.1 25% van die verkope was 15 en minder CD’s verkoop vir die 11.4.3 maand.  (2) 2:A LU4 AS 4.2.2 75% van die verkope was 37 en meer CD’s verkoop vir die 11.4.3 maand.  (2) 2:A LU4 AS 4.2.3 Ja  11.4.3 Meeste van die CD’s wat hy verkoop het, is meer as 15 1:A (75%).  (2) 1:R 4.3 Deursnit van buitesirkel = 118 mm = 11,8 cm  LU3 AS Radius van buitesirkel = 5,9 cm  11.3.1 11.3.2 Radius van binnesirkel = 0,75 cm 1:C Opp. van CD = Opp. van buitesirkel – Opp. van binnesirkel (mm = r2 – r2 na cm) = 3,14 x 5,92 – 3,14 x 0,752  1:A = 109,30 cm2 – 1,77 cm2  (vind r) = 107,53 cm2  OF Opp van CD = Opp. van buitesirkel – Opp. van binnesirkel = r2 – r2 = 3,14 x 5,9 cm x 5,9 cm – 3,14 x 0,75 cm x 0,75 cm  1:SF = 109,3034 cm2 – 1,76625 cm2  1:S = 107,53715 cm2 1:CA = 107,54 cm2  (5) [25] TOTAAL: 100 ## Related documents - [MLIT-P2-GR11 2012-QP-Eng](https://www.hlayiso.com/library/mlit-p2-gr11-2012-qp-eng/80083a22-c6c0-4e01-9f24-a2dd6d54615f) — Mathematical Literacy · QP · 2012 · Eng - [MLIT-P2-GR11 2012-QP-AFR](https://www.hlayiso.com/library/mlit-p2-gr11-2012-qp-afr/0c37794d-4008-4951-b606-b8d92770d008) — Mathematical Literacy · QP · 2012 · Afr - [MLIT P1 N12 MEMO Afr](https://www.hlayiso.com/library/mlit-p1-n12-memo-afr-1/afc99f01-178a-42e9-b5fc-bab13e80a480) — Mathematical Literacy · Memo · 2012 · Afr - [MLIT P1 N12 MEMO Eng](https://www.hlayiso.com/library/mlit-p1-n12-memo-eng/9e15aa07-6686-4bc4-9db4-9ff15ac2e675) — Mathematical Literacy · Memo · 2012 · Eng - [MLIT P1 N12 MEMO Afr](https://www.hlayiso.com/library/mlit-p1-n12-memo-afr/21590ca5-de1b-4d88-b1ac-408fcdd06420) — Mathematical Literacy · Memo · 2012 · Afr - [MATHS LIT GR11 P2 MEMO NOV2022 Afrikaans](https://www.hlayiso.com/library/maths-lit-gr11-p2-memo-nov2022-afrikaans/8c7033a8-caf2-4285-a7e0-5e3c22a2f68a) — Mathematical Literacy · Memo · 2022 · Afr ## More from Grade 11 Mathematical Literacy - [Grade 11 Mathematical Literacy P2 November 2026 Mock 2](https://www.hlayiso.com/library/grade-11-mathematical-literacy-2026-november-mock-exam-2-question-paper-paper-2/9e28596c-b065-4744-84e1-a7995223ca68) — QP · 2026 · Eng - [Gr 11 Maths Lit ATP 2023-24](https://www.hlayiso.com/library/gr-11-maths-lit-atp-2023-24/729b7b1d-095d-46f4-8082-c0c019b4c6c7) — ATP - [MATHS LIT AFR GRADE 11 NOV 2025 ADDENDUM (3) watermark](https://www.hlayiso.com/library/maths-lit-afr-grade-11-nov-2025-addendum-3-watermark/e7865fd3-d528-472d-a551-0585ea50cee1) — Addendum · 2025 · Afr - [MATHS LIT P2 GR11 NOV25 ADDENDUM ENGLISH FINAL (1) watermark](https://www.hlayiso.com/library/maths-lit-p2-gr11-nov25-addendum-english-final-1-watermark/6d3d1a99-ee4e-4b35-9dfc-765595836d35) — Addendum · 2025 · Eng - [LP Maths Literacy Grade 11 June 2025 P1 only](https://www.hlayiso.com/library/lp-maths-literacy-grade-11-june-2025-p1-only/010db24d-7d07-4b57-b916-5b4856296406) — QP · 2025 - [FS Maths Literacy Grade 11 June 2025 P1 and Memo](https://www.hlayiso.com/library/fs-maths-literacy-grade-11-june-2025-p1-and-memo/913740d3-d3f7-470f-ad0e-289f4f74f68b) — QP + Memo · 2025 ## Collection - [Grade 11 past papers](https://www.hlayiso.com/grade-11/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.