# Mathematical Literacy Grade 12 P1 2023 Memo (Eng) > Grade 12 Mathematical Literacy Paper 1 memorandum from the Eastern Cape Mock Exam 2023 (English), 9 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/maths-lit-p1-gr12-memo-sept-2023-english/758e09c5-d36a-4034-8209-54a4c4a50c91 - Updated: 2026-09-27 - Subject: Mathematical Literacy - Grade: Grade 12 - Language: English - Document type: Memorandum - Year: 2023 - Exam period: Eastern Cape Mock Exam - Paper: 1 - Curriculum: CAPS - Pages: 9 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 P1 2023 Memo (Eng)](https://www.hlayiso.com/library/maths-lit-p1-gr12-memo-sept-2023-english/758e09c5-d36a-4034-8209-54a4c4a50c91) ## About this document Mathematical Literacy · Grade 12 · Eastern Cape Mock Exam · 2023 · English. Memorandum, 9 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 NATIONAL SENIOR CERTIFICATE GRADE 12 SEPTEMBER 2023 MATHEMATICAL LITERACY P1 MARKING GUIDELINE MARKS: 150 Symbol Explanation M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG/RM Reading from a table/a graph/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 9 pages. ### Page 2 2 MATHEMATICAL LITERACY P1 (EC/SEPTEMBER 2023) QUESTION 1 [30 MARKS] Ques. Solution Explanation AO: FULL MARKS T&L 1.1.1 Bank D RT 2RT identify correct F bank L1 (2) 1.1.2 ATM withdrawal RT 2RT F (2) L1 1.1.3 R1 600 – R1 250 = R350 M 1M for subtracting F 1MA multiplying L1 350 correct value ∴ R1,85 × ( 100 ) 1A answer = R1,85 × 4 MA = R7,40 A (3) 1.1.4 50% or B A 2A answer P (2) L1 M D 1.2.1 Total = 30 980 110 – (25 085 330 + 2 737 987 + 2 396 679) 1M subtracting correct L1 = 760 114 A values 1A answer (2) 1.2.2 Two million, six hundred and one thousand, nine hundred 1A correct words D and thirty-two A (2) L1 1.2.3 %= 25 085 330 × 100 RT M 1RT correct values D 30 980 110 1M multiplying by 100 L1 1A percentage = 80,97% A (3) 1.2.4 Numerical A 2A correct answer D (2) L1 1.3.1 Total = R249,95 + (50% × R135,95) MA 1 MA multiplying F correct value by 50% L1 = R249,95 + R67,98 M 1M adding correct values = R317,93 A 1A correct total (3) 1.4.1 Facebook RT 2RT correct social D media platform L1 (2) 1.4.2 Bar graph A 2A correct graph D (2) L1 RT 1RT correct values D 1.4.3 Difference = 1 309 million – 200 million M 1M subtraction L1 1CA difference = 1 109 million CA (3) 1.4.4 Instagram RT 2RT correct social D media platform L1 (2) [30] Copyright reserved Please turn over ### Page 3 (EC/SEPTEMBER/2023) MATHEMATICAL LITERACY P1 3 QUESTION 2: FINANCE [35 MARKS] Ques. Solution Explanation/Marks AO: FULL MARKS T&L RT F 2.1.1 500 % discount = 13 999 × 100 M 1RT correct values L2 1M multiplying by 100 = 3,57% CA 1CA simplification Accept 3,6% (3) F 2.1.2 18 L4 Deposit = 100 × 13 499 1M calculating = R2 429,82 M deposit Outstanding balance = R13 499 – R2 429,82 = R11 069,18 M 1M outstanding 16,25 MA balance Total due = ( 100 × R11 069,18 × 1 yr ) + R11 069,18 1MA multiplying = R1 798,74 + R11 069,18 correct value by percentage = R12 867,92 A 1A total due 𝑅12 867,92 Monthly payment = M 12 1M divide by 12 = R1 072,32 CA 1CA final answer ∴ No, it will not be enough. O 1O reason (7) MA F 2.1.3 3 1MA correct ratio L2 Mrs Rudolph = 8 × R13 499 M method 1M multiplying by = R5 062,13 CA cash price 1CA answer (3) Copyright reserved Please turn over ### Page 4 4 MATHEMATICAL LITERACY P1 (EC/SEPTEMBER 2023) Ques. Solution Explanation T&L 2.2.1 Difference = R21,40 – R20,75 MA 1MA subtracting correct values F = R0,65 A 1A correct answer L1 (2) 2.2.2 Cost = R21,40 + R2,06 MA 1MA adding correct values F 1A simplification L2 = R23,46 A (2) 2.2.3 Total fuel used: 1M fuel per session F = (2,2ℓ × 2,5 hrs) × 2 M 1A fuel per day L3 = 11 ℓ per day A 1CA amount of fuel per month 11ℓ × 31 = 341ℓ per month CA 1CA cost for the month Total cost = R20,45 × 341ℓ = R6 973,45 CA (4) 2.3.1 0,3 kℓ RT 2RT correct value F (2) L2 2.3.2 R28,96 ÷ 1,15 MA 1M dividing by 1,15 F L2 = R25,18 1M subtracting VAT = R28,96 – R25,18 M = R3,78 A 1A total VAT (3) 2.3.3 Cost: M 1M step 1 F Step 1: 0,3 kℓ × R20,81 = R6,24 A 1A step 1 L4 Step 2: 0,2 kℓ × R31,80 = R6,36 CA 1CA step 2 Step 3: 0,23 kℓ × R63,60 = R14,63 CA 1CA step 3 Total = R27,23 × 1,15 M 1M adding VAT = R31,31 CA 1CA total including VAT 1O statement ∴ Incorrect/Invalid O OR Cost: M 1M step 1 Step 1: 0,3 kℓ × R20,81 = R6,24 A 1A step 1 Step 2: 0,2 kℓ ×R31,80 = R6,36 CA 1CA step 2 F Step 3: 0,23 kℓ × R63,60 = R14,63 CA 1CA step 3 L4 VAT = R27,23 × 15% 1CA adding VAT = R4,0845 CA 1CA cost Total = R27,23 + R4,0845 = R31,31 CA 1O opinion ∴ Incorrect/Invalid O (7) 2.3.4 Holidays, therefore home more often. F Visitors during holidays. O 2O reason L4 Any valid reason (2) [35] Copyright reserved Please turn over ### Page 5 (EC/SEPTEMBER/2023) MATHEMATICAL LITERACY P1 5 QUESTION 3 [31 MARKS] Ques. Solution Explanation T&L 3.1.1 Total: M 1M subtracting correct D 101 447 – (48 319 + 8 542 + 7 987 + 14 135 + 4 374 + 9 114) values L2 = 101 447 – 92 471 1CA simplification = 8 976 million OR 8 976 000 000 CA (2) 3.1.2 General dealers RT 2RT correct retailer D (2) L1 3.1.3 92 747 + A 1SF substitution D 14 275,86 = SF 7 L3 1M multiplying 14 275,86 × 7 = 92 747 + A M 99 931,02 – 92 747 = A M 1M subtracting 7 184,02 = A CA 1CA simplification ∴ 𝐴 = 7 billion R 1R rounding off to the nearest billion (5) 3.1.4 Median: 1M correct order D M RT 1RT correct values in L2 8 298 ; 8 462 ; 8 542 ; 8 607 ; 8 978 ; 9 050 order 8 542 + 8 607 1M median concept = M 2 1CA simplification = 8 574,5 million OR 8 574 500 000 CA (4) 3.1.5 42 716 : 7 987 RT 1RT correct values D 1 : 0,19 A OR 5,35 : 1 1A simplification to L2 unit ratio (2) 3.1.6 RT RT 1RT correct value April D 101 113 – 98 502 1RT correct value Sept L2 % change = × 100% MA 98 502 1MA correct % calculation = 2,65% CA 1CA simplification (4) Copyright reserved Please turn over ### Page 6 6 MATHEMATICAL LITERACY P1 (EC/SEPTEMBER 2023) Ques. Solution Explanation T&L 3.1.7 5 RT RT M 1RT numerator P 42 1RT denominator L2 1R rounding off to 3 decimal = 0,119 R places (3) 3.2.1 50 + years RT 2RT correct age group D (2) L1 3.2.2 785 1RT correct percentage D RT 1 000 1S simplification L2 157 = 200 S (2) 3.2.3 1; 3 ; 3,45 ; 4 ; 4,2 ; 6,55 ; 20 ; 23 ; 36 M 1M values in correct order D and finding median L3 3 + 3,45 Q1 = 2 1M for Q1 = 3,225 M 1M for Q3 20 + 23 Q3 = 1M subtracting 2 = 21,5 M 1CA answer IQR = Q3 – Q1 = 21,5 – 3,225 M = 18,28% CA (5) [31] Copyright reserved Please turn over ### Page 7 (EC/SEPTEMBER/2023) MATHEMATICAL LITERACY P1 7 QUESTION 4 [28 MARKS] Ques. Solution Explanation T&L 4.1.1 B RT 2RT correct answer F (2) L1 A A F 4.1.2 Cost = 8000 + 80 × number of people 2A correct answer L2 (2) A 1A cost the same F 4.1.3 The point where both venues cost exactly the same for 1A same amount of people L1 the same number of people. A (2) 4.1.4 Venue 2: R180 × 180 guests M F = R32 400 A 1M multiply rate by guests L4 Venue 1: R8000 + R80 × 180 M 1A simplification = R22 400 A 1M correct multiplication Difference = R32 400 – R22 400 1A simplification = R10 000 CA 1CA difference Valid O 1O opinion (6) 4.2.1 Body Mass Index A 2A correct answer D (2) L1 4.2.2 Outlier A 2A correct answer D (2) L1 4.2.3 Scatterplot A 2A correct graph D (2) L1 4.2.4 50 kg RT 2RT weight D (2) L2 4.2.5 weight in kg D BMI = L3 height in m2 1RT weight 60 kg 1RT height in cm = RT RT (148 cm)2 1C height in m 60 kg = C (1,48 m)2 1CA correct BMI = 27,39 kg/m2 CA 1O weight status ∴ Overweight O (5) 4.2.6 5 RT RT 1RT numerator P 12 1RT denominator L2 1R correct percentage = 41,7% R rounded to 1 dec. place (3) [28] Copyright reserved Please turn over ### Page 8 8 MATHEMATICAL LITERACY P1 (EC/SEPTEMBER 2023) QUESTION 5: FINANCE, DATA HANDLING [26 MARKS] Ques. Solution Explanation T&L 5.1.1 South African Revenue Services A 2A correct name F (2) L1 F 5.1.2 Minimum amount that an individual must earn in order to 2O correct explanation L1 pay tax O OR up to that amount you do not pay tax (2) OR if you earn less than that amount you do not pay tax F 5.1.3 Annual Taxable Income = Gross – Pension – Charity L3 Pension = 7,5% × R46 523,16 = R3 489,24 M 1M calculating pension Tax Income = R46 523,16 – R3 489,24 – R1 600 M 1M subtracting values = R41 433,92 from gross Annual Tax Income = R41 433,92 × 12 M 1M multiplying by 12 = R497 207,04 CA 1CA simplification (4) F 5.1.4 Annual Tax L4 RT SF 1RT correct tax bracket = 115 762 + 36% (502 185,60 – 488 700) 1SF subtracting correct = 115 762 + 36% (13 485,60) values = 115 762 + 4 854,82 = 120 616,82 – 16 425 M 1M subtracting rebate = R104 191,82 CA 1CA simplification Quarter of taxable income = R502 185,60 × ¼ M = R125 546,40 A 1M multiplying by quarter 1A answer ∴ Invalid O 1O invalid OR 104 191,82 × 100 M 502 185,60 = 20,74% A (7) Copyright reserved Please turn over ### Page 9 (EC/SEPTEMBER/2023) MATHEMATICAL LITERACY P1 9 Ques Solution Explanation T&L 5.2.1 NZ$ : R 1M correct values F 0,088205 : 1 M 1MA dividing correct L3 values 1 ∴ MA 0,088205 1A simplification = NZ$ 1: R11,337 A (3) F 5.2.2 Total cost: L2 = 4 880 + 6 860 MA 1MA adding correct values = NZ$11 740 A 1A correct answer 11 740 × 11,337 M 1M multiplying by rate = R133 096,38 1R rounding to nearest 100 ≈ R133 100 R OR OR Total cost: = 4 880 + 6 860 MA 1MA adding correct values = NZ$11 740 A 1A correct answer 11 740 ÷ 0,088205 M 1M dividing by rate = R133 099,03 ≈ R133 100 R 1R rounding to nearest 100 (4) M MA 1M correct method F 5.2.3 500 000 × 1,0875 = NZ$ 543 750 A 1MA multiplying by L3 correct rate 543 750 × 1,0875 = NZ$ 591 328,13 CA 1A amount after 1 year 1CA final amount after 2 OR years MA Interest = 500 000 × 8,75% = 43 750 M Balance after 1st year = 500 000 + 43 750 = 543 750 A 2nd year interest = 543 750 × 8,75% = 47 578,13 (4) Total =543750 + 47 578,13 = NZ$ 591 328,13 CA [26] TOTAL: 150 Copyright reserved Please turn over ## Related documents - [MATHS LIT P1 GR12 MEMO SEPT2022 English (Approved)](https://www.hlayiso.com/library/maths-lit-p1-gr12-memo-sept2022-english-approved/5a64a94d-eb36-49e3-baaa-d90dae2dbde8) — Mathematical Literacy · Memo · 2022 · Eng - [MATHS LIT P1 GR12 MEMO SEPT 2021 English](https://www.hlayiso.com/library/maths-lit-p1-gr12-memo-sept-2021-english/9e0090b1-9cc7-4ae4-abe6-f4cf6ea5bce3) — Mathematical Literacy · Memo · 2021 · Eng - [MATHS LIT P2 QP GR12 SEPT 2023 English](https://www.hlayiso.com/library/maths-lit-p2-qp-gr12-sept-2023-english/f2531ae8-7751-4b26-9fdc-2c577eac4b74) — Mathematical Literacy · QP · 2023 · Eng - [MATHS LIT P2 ADD GR12 SEPT 2023 English](https://www.hlayiso.com/library/maths-lit-p2-add-gr12-sept-2023-english/6db29d67-251d-4ade-ab45-b41f079fc0a3) — Mathematical Literacy · QP · 2023 · Eng - [MATHS LIT P1 GR12 ADDENDUM SEPT 2021 English](https://www.hlayiso.com/library/maths-lit-p1-gr12-addendum-sept-2021-english/1141e20a-9cef-43c2-9fab-b805c645f238) — Mathematical Literacy · Addendum · 2021 · Eng - [MATHS LIT P1 GR12 QP SEPT2020 English (Printready 18 August 2020)](https://www.hlayiso.com/library/maths-lit-p1-gr12-qp-sept2020-english-printready-18-august-2020/f1819888-26f6-4ed6-8382-6909354c4789) — Mathematical Literacy · QP · 2020 · Eng ## More from Grade 12 Mathematical Literacy - [MATHS LIT P2 GR 12 SEPT 2025 ADDENDUM AFR watermark](https://www.hlayiso.com/library/maths-lit-p2-gr-12-sept-2025-addendum-afr-watermark/813c6e54-f134-474a-b01a-1260e38b4c8f) — Addendum · 2025 · Afr - [Maths LIT Grade 12 NSC P1 QP September 2025 Free State](https://www.hlayiso.com/library/maths-lit-grade-12-nsc-p1-qp-september-2025-free-state/45f0ab2a-afdd-46df-bf98-5505c4021e64) — QP · 2025 - [Maths LIT Grade 12 NSC P1 ANSWER BOOK September 2025 KZN](https://www.hlayiso.com/library/maths-lit-grade-12-nsc-p1-answer-book-september-2025-kzn/b0b358eb-c897-437a-8694-910fe61dfcbc) — Addendum · 2025 - [Maths LIT Grade 12 NSC P1 QP September 2025 Limpopo](https://www.hlayiso.com/library/maths-lit-grade-12-nsc-p1-qp-september-2025-limpopo/c5d7a167-2d27-4e33-80ef-fad9cde4ebd2) — QP · 2025 - [Maths LIT Grade 12 NSC P2 QP September 2025 Limpopo](https://www.hlayiso.com/library/maths-lit-grade-12-nsc-p2-qp-september-2025-limpopo/f22349c1-8dbf-420e-ae8d-13d53bcb300b) — QP · 2025 - [Maths LIT Grade 12 NSC P2 QP September 2025 North West](https://www.hlayiso.com/library/maths-lit-grade-12-nsc-p2-qp-september-2025-north-west/7ce5b7fe-941b-4148-80dc-d9cde7bdbfd7) — QP · 2025 ## Collection - [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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