# Mathematical Literacy Grade 11 P1 2023 Memo (Eng) > Grade 11 Mathematical Literacy Paper 1 memorandum from the Eastern Cape November Exam 2023 (English), 6 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/maths-lit-p1-memo-gr11-nov2023-english/96ab8acd-7e2b-492f-be89-dc29c2556bc4 - Updated: 2026-09-27 - Subject: Mathematical Literacy - Grade: Grade 11 - Language: English - Document type: Memorandum - Year: 2023 - Exam period: Eastern Cape November Exam - Paper: 1 - Curriculum: CAPS - Pages: 6 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 P1 2023 Memo (Eng)](https://www.hlayiso.com/library/maths-lit-p1-memo-gr11-nov2023-english/96ab8acd-7e2b-492f-be89-dc29c2556bc4) ## About this document Mathematical Literacy · Grade 11 · Eastern Cape November Exam · 2023 · English. Memorandum, 6 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 NATIONAL SENIOR CERTIFICATE GRADE 11 NOVEMBER 2023 MATHEMATICAL LITERACY P1 MARKING GUIDELINE MARKS: 100 Symbol Explanation M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG/RM Reading from a table/Reading from a graph/Reading from a 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/Reason AO Answer only NPR No penalty for correct rounding This marking guideline consists of 6 pages. ### Page 2 Downloaded from hlayiso.com 2 MATHEMATICAL LITERACY P1 (EC/NOVEMBER 2023) QUESTION 1 [23 MARKS] Ques. Solution Explanation T&L AO: FULL MARKS 1.1.1 A = R33 150 + R1 625 MA 1MA for adding F = R34 775 A correct values L1 1A answer (2) 1.1.2 R33 150 ÷ 26 MA 1MA for dividing F = R1 275,00 A correct values L1 1A answer (2) 1.1.3 B = R7 927,86 – (R375 + R850,25 + R1 500) MA 1MA for subtracting F = R7 927,86 – R2 725,25 correct values L1 = R5 202,61 A 1A answer (2) 1.1.4 RT F R850,25 1RT correct values L1 Percentage = × 100 M R7 927,86 1M multiplying by 100 1CA answer = 10,72% CA (3) 1.2.1 R22,50 RT 2RT reading correct F value L1 (2) 1.2.2 Vaal Plaza Main line RT 2RT correct tollgate F L1 (2) 1.2.3 140 : 225 RT 1RT correct values F 28 : 45 A 1A simplifying L1 correctly (2) 1.2.4 R271,00 × 2 MA 1MA multiplying F = R542,00 A correct rate by 2 L1 1A answer (2) 1.3.1 Pie Chart RT 2RT correct graph D (2) L1 1.3.2 MA D Percentage comedies = 100% ‒ (21% + 6% + 24% + 11%) 1MA subtracting L1 = 38% A 1A percentage (2) 1.3.3 Number of romance = 24% × 220 MA 1MA multiplying D = 52,8 correct L1 = 53 students A percentage 1A correct answer Accept 52 (2) [23] Copyright reserved Please turn over ### Page 3 Downloaded from hlayiso.com (EC/NOVEMBER/2023) MATHEMATICAL LITERACY P1 3 QUESTION 2 [25 MARKS] Ques Solution Explanation/Marks AO: FULL MARKS T/L 2.1.1 M RT 1RT correct rate F Jan 2023: R12,47 × 1,069 = R13,33 A 1M multiplying L3 MA 1A answer Feb 2023: R13,33 × 1,07 = R14,26 CA 1MA multiplying with correct rate 1CA answer (5) 2.1.2 Prices increased A 1A increased F Prices still went up, but by a lower percentages/ the 1O reason L4 percentages are not negative. O (2) 2.2.1 A A 1A 7th F 7 May 2023 1A May L1 (2) 2.2.2 Total = R349 × 3 MA 1MA multiplying by 3 F = R1 047 A 1A answer L2 ≈ R1 000 R 1R rounding to nearest 1 000 (3) 2.2.3 Labour per hour = R6 500 ÷ 7,5 hrs MA 1MA dividing by 7,5 F = R866,67 A 1A correct answer L2 (2) 2.2.4 Total = R5 499 + R1 047 + R1 699 + R6 500 M CA from 2.2.2 F = R14 745 × 1,15 MA 1M adding values L2 = R16 956,75 CA 1MA multiplying by 1,15 1CA answer (3) 2.2.5 RT F 5 795 – 5 499 1RT correct values L4 % change = × 100% MA 5 499 1MA correct % calculation = 5,38% CA 1CA simplification ∴ Valid O 1O statement (4) 2.3 Block 1: 600 kWh × 229,00c = 137 400c MA 1MA multiplying F Block 2: 112 kWh × 278,46c = 31 187,52c MA Block 1 L3 Total = 168 587,52c ÷ 100 C 1MA answer Block 2 = R1 685,88 CA 1C divide by 100 1CA answer (4) [25] Copyright reserved Please turn over ### Page 4 Downloaded from hlayiso.com 4 MATHEMATICAL LITERACY P1 (EC/NOVEMBER 2023) QUESTION 3 [19 MARKS] Ques Solution Explanation T&L 3.1 Forty-five million eight hundred thousand dollars RT 2RT correct value D (2) L1 3.2 Median: D RT 1RT correct values L3 5; 25; 47; 50; 52; 55; 55; 80; 90 MA 1MA method median 1A correct answer = $52 000 000 / $52 million A (3) 3.3 Federer total = $121,2 – $30,5 M 1M subtracting D = $90,7 million A 1A simplification L3 Total = $130 + $121,2 + $115 + $95 + $92,8 + $92,1 1M adding + $90,7 + $90 + $83,9 M 1CA simplification = $910,7 million / $910 000 000 CA (4) 3.4 M D $451,7 1M adding values L2 Mean = MA 9 1MA dividing by 9 1CA answer = $50,2 million / $50 200 000 CA (3) 3.5 55 1MA correct values D Percentage = × 100 MA 115 multiply by 100 L2 1A simplification = 47,83% A (2) 3.6 70 : 90 RT 1RT correct values D 7 : 9 CA 1CA simplification of L2 ratio (2) 3.7 4 RT 1RT numerator P Probability = RT 9 1RT denominator L2 1CA rounding to 3 = 0,444 CA decimal places (3) [19] Copyright reserved Please turn over ### Page 5 Downloaded from hlayiso.com (EC/NOVEMBER/2023) MATHEMATICAL LITERACY P1 5 QUESTION 4 [33 MARKS] Ques Solution Explanation/Marks T&L 4.1.1 Dependent: Amount in Rands RT 1RT dependent F Independent: Number of curry bunnies RT 1RT independent L1 (2) 4.1.2 R250 ÷ 10 MA 1MA dividing by 10 F = R25,00 A 1A simplification L1 (2) RT RT F 4.1.3 Total Expenses = 450 + (10 × number of curry bunnies) 1RT fixed cost L2 1RT cost per curry bunny (2) 4.1.4 F Income and Expenses of Curry L2 Bunny Sales 5400 4950 4500 Amount in Rands 4050 3600 3150 2700 2250 1800 1350 900 450 0 1A starting point 0 50 100 150 200 250 1A 2 correct points Number of Curry Bunnies 1A end point (3) 4.1.5 Profit ONE month = R5 000 – R2 450 F = R2 550 A 1A profit one month L4 Profit FOUR months = R2 550 × 4 MA 1MA profit 4 months = R10 200 CA 1CA answer ∴ Valid O 1O statement (4) 4.2.1 A = 20 + 13 RT 1RT adding correct D = 33 A values L2 1A answer (2) 4.2.2 15:00 – 17:59 A 2A correct answer D (2) L2 4.2.3 18:00 – 20:59 RT 2RT correct time interval D (2) L2 Copyright reserved Please turn over ### Page 6 Downloaded from hlayiso.com 6 MATHEMATICAL LITERACY P1 (EC/NOVEMBER 2023) Ques Solution Explanation/Marks T&L 4.2.4 6,5 ℓ × 77 = 500,5 ℓ MA 1MA calculating litres D 1M multiplying L3 Cost = 500,5 ℓ × R21,92 M 1CA answer = R10 970,96 CA (3) Probabilty = 77 RT 4.2.5 32 1RT numerator P RT 1RT denominator L2 (2) 4.3.1 BANK B RT 1RT correct bank F Interest amount is the same for each year. O 1O statement/opinion L2 (2) 4.3.2 BANK A: R1 277,29 × 1,085 MA 1MA multiplying by rate F = R1 385,86 A 1A simplification L4 1M difference Difference = R1 385,86 – R1 340 M = R45,86 ∴ Valid O 1O statement (4) 4.3.3 68,49 × 19,83 MA 1MA multiplying by F = R1 358,16 A exchange rate L4 1A simplification ∴ Yes, she will have enough. O 1O statement (3) [33] TOTAL: 100 Copyright reserved Please turn over ## Related documents - [MATHS LIT P2 QP GR11 NOV 2023 English](https://www.hlayiso.com/library/maths-lit-p2-qp-gr11-nov-2023-english/84506bee-de17-47de-8b37-7901b6b5c5e8) — Mathematical Literacy · QP · 2023 · Eng - [MATHS LIT P1 GR11 MEMO NOV2020 English](https://www.hlayiso.com/library/maths-lit-p1-gr11-memo-nov2020-english/08e52e8a-ef68-4338-9e8e-a0babd449184) — Mathematical Literacy · Memo · 2020 · Eng - [MATHS LIT P1 GR11 MEMO NOV2018 English 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