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MATHEMATICAL LITERACY PAPER 1 GRADE 12 MEMORANDUM - NSC PAST PAPERS AND MEMOS MAY/JUNE 2019
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GRADE 12
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
MEMORANDUM
MAY/JUNE 2019
| Symbol | Explanation |
| M | Method |
| MA | Method with accuracy |
| CA | Consistent accuracy |
| A | Accuracy |
| C | Conversion |
| S | Simplification |
| RT | Reading from a table/a graph/document/diagram |
| SF | Correct substitution in a formula |
| O | Opinion/Explanation |
| P | Penalty, e.g. for no units, incorrect rounding off, etc. |
| R | Rounding off |
| NPR | No penalty for rounding |
| AO | Answer only |
| MCA | Method with constant accuracy |
NOTE:
- If a candidate answers a question TWICE, mark only 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.
- No CA mark follows after 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
QUESTION 1 [32 MARKS] ANSWER ONLY FULL MARKS
| Q/V | Solution | Explanation | T&L |
| 1.1.1 | Susan Visser / Susan / Visser | 2RT correct name Accept: Woolworths Financial Services (2) | F L1 |
| 1.1.2 | R548,37 | 2RT correct amount (2) | F L1 |
| 1.1.3 | 12 / twelve | 2A correct number of months (2) | F L1 |
| 1.1.4 | Debit order is a way for a third party, that you have given permission, to collect money from your bank account. It's typically used to collect monthly subscriptions, insurance premiums or loan repayments OR An instruction to the bank, authorising payment to the other person on a regular basiS OR It is an arrangement giving permission to a third party to withdraw money from a bank account on a regular basis OR A term used for bank references in order for them to deduct money owed to certain bank accounts on a regular basis OR When an individual has to pay a certain person on a regular basis, they set a date and how much should be taken from their account OR Pre-arranged monthly payment of a specific amount from your bank (on behalf of borrower) account to settle debt | 1A money taken out (deducted) of bank account 1A regular basis/monthly 1A money taken out of bank account/salary 1A regular basis/monthly (2) | F L1 |
| 1.1.5 | 26 days | 2A correct number of days Accept: 25 days(2) | F L1 |
| 1.1.6 | A = R6 859,99 + R144,04 + (– R221,89) =R6 782,14 OR A = (R6 859,99 + R144,04) – R221,89 = R7 004,03 – R221,89 = R6 782,14 OR R38+ R2 559,79 + A + R1 071,70 = R10 451,63 A = R10 451,63 – R3 669,49 = R6 782,14 | 1MA adding and subtracting the values 1A simplification OR 1MA adding and subtracting the values 1A simplification OR 1MA adding and subtracting the values 1A simplification NPU (2) | F L1 |
| 1.2.1 | 26°C | 2RT maximum temperature NPU (2) | M L1 |
| 1.2.2 | 8 June 2017 OR 08.06.2017 OR 08 /06 /2017 OR 8 June | 2RT correct date (2) | M L1 |
| 1.2.3 | 26°C ; 22°C ; 21°C ; 20°C ; 19°C ; 16°C ; 15°C ; 15°C ; 14°C | 1A correct values 1A correct order NPU (2) | D L1 |
| 1.2.4 | 6 June 2017 OR 06 / 06 / 2017 OR 6 June/ OR 6th/6de | 2RT correct date (2) | M L1 |
| 1.2.5 | 15°C - 3°C = 12°C | 1RT both correct values 1A simplification NPU (2) | D L1 |
| 1.3.1 | Age group: 20 – 29 | 2RT correct age group (2) | D L1 |
| 1.3.2 | Number of male voters under 40 = 109 224 + 2 443 115 + 3 095 538 = 5 647 877 OR/OF Number of male voters under 40 = 11 797 561 – 2 553 636 – 1 824 042 – 1 116 525 – 479 711 – 175 770 = 5 647 877 | 1M adding correct values 1CA answer 1M subtracting correct values 1CA answer (2) | D L1 |
| 1.3.3 | Two million eight hundred and fifty eight thousand nine hundred and ninety six | 2A correct number in words (2) | D L1 |
| 1.3.4 | Discrete | 2A discrete (2) | D L1 |
| 1.3.5 | 14 442 779 – 11 797 561 = 2 645 218 | 1MA subtracting correct values 1A correct answer (2) | D L1 |
| [32] |
QUESTION 2 [40 MARKS]
Related Items
| Q/V | Solution | Explanation | T&L |
| 2.1.1 | R1140,95 12 = R95,07916667 = R95,08 per kg | 1MA both values correct 1CA simplification 1R unit cost AO (3) | F L1 |
| 2.1.2 | R11,99 × 6 =R71,94 | 1MA multiply by 6 1CA total amount AO (2) | F L1 |
| 2.1.3 | Cost price of an item is the cost of making that item OR This is the amount that it costs per unit to either manufacture, purchase the item or to prepare for a service that will be delivered. This amount is pure cost, no markup or profit added yet OR Money spent when purchasing products/goods for resell OR Original price before profit is added | 2A explanation (2) | F L1 |
| 2.1.4 (a) | A – Cost of milo per cup koppie: R97,95 × 0,04 kg = R3,92 OR R97,95 ÷ 25 = R3,92 | 1MA multiply by 0,04 kg 1CA simplification OR 1MA divide by 25 1CA simplification AO (2) | F L1 |
| 2.1.4 (b) | B – amount of milk used R11,99R1,20 = 0,1 ℓ | 1MA dividing by R11,99 1A simplification AO NPU (2) | F L1 |
| 2.1.4 (c) | C – cost of 25 foam cups R1,78 × 25 = R44,50 | 1MA multiply by 25 1A simplification AO (2) | F L1 |
| 2.1.4 (d) | D – cost of one cup of milo: R3,92 + R1,20 + R0,13 + R1,78 + R0,26 = R7,29 | CA from Question 2.1.4(a) 1M adding 1A 5 correct values (2) | F L1 |
| 2.1.5 | Profit/Wins = R7,29 × 10025 Selling price = R1,8225 + R7,29 Selling price = R9,1125 = R9,11 OR R9,10 OR Selling price = R7,29 ×125 100 = R9,1125 = R9,11 OR/OF R9,10 OR Profit margin = Profit/wins ×100% selling price x - 7.29 × 100=25% x 1 100%x - 7.29 = 25% x 1 100%x – 7,29 = 25%x 100% - 25%x = 7,29 75%x = 7,29 =R9,72 OR 25% = SP/VP - CP/KP × 100% cost price 25% = SP/VP - 7.29 × 100% 7.29 Selling Price = (0,25 × 7,29) + 7,29 = R9,1125 = R9,11 | CA from question 2.1.4(d) 1M 25% of R7,29 only 1M adding 1CA simplification 1R rounding OR 1A 125% of R7,29 only 1M multiply 1CA simplification 1R rounding OR 1M creating formula 1M changing the subject of the formula 1M dividing by 75% 1CA simplification OR 1M 25% of R7,29 only 1M adding 1CA simplification 1R rounding Accept R9,15 and R9,20 (4) | F L2 |
| 2.2.1 (a) | P = 40 × R12,50 = R500,00 OR 80 cups = R1 000 ½ of 80 cups is 40 cups ∴ ½ of R1 000 is R500. ∴ P = R500 OR P = R375 + R125 = R500 | 1MA multiply by R12,50 1CA selling price OR 1MA trial and error method 1CA selling price 1MA adding 1CA selling price AO (2) | F L1 |
| 2.2.1 (b) | Income in rand = R12,50 × number of cups of milo/ n OR Income in rand = R12,50 × x = number of cups of milo/aantal koppies milo | 2A formula OR 1A Income in rand = R12,50 × x (in equation) 1A explaining variable (2) | F L2 |
| 2.2.1 (c) | Number of cups of milo | 2RT independent variable (2) | F L1 |
| 2.2.2 | R612,50 = R90,00 + (R9,50 × n) R612,50 − R90,00 = R9,50 × n n = 522.50 9.50 Q = 55 OR R90 + R 9,50 × 55 = R 612,50 Q = 55 | 1M changing subject of formula 1S simplification 1CA simplification OR 1M trial and error 1S simplification 1CA simplification AO (3) | F L2 |
| 2.2.3 | INCOME AND COST GRAPHS FOR MAKING AND SELLING OF CUPS OF MILO 1A start of graph – cost price (0;90) 1A end of graph – cost price (100;1 040) 1A joining the points in a straight line graph (3) | F L2 | |
| 2.2.4 (a) | The cost price for the number of cups of Milo sold and the selling price of that number is the same (equal). No profit or loss. OR Cost price = Selling price OR Income = Expenses OR The profit and loss are equal to 0 | 2A break-even (2) | F L1 |
| 2.2.4 (b) | 30 cups | CA from Question 2.2.3 (graph) 2RT number of cups (2) | F L2 |
| 2.3.1 | 1 200 ÷ 0,10976 = 10 932,94 Yen | 1RT correct values 1M dividing by exchange rate 1A simplification AO NPR (3) | F L2 |
| 2.3.2 | Yen is Weaker OR Rand is stronger | CA from Question 2.3.1 2A for stating weaker OR 2A for stating stronger (2) | F L1 |
| [40] | |||
QUESTION 3 [26 MARKS]
| Q/V | Solution | Explanation | T&L |
| 3.1.1(a) | Area of a face without a circular hole = side × side = 45cm × 45cm = 2 025 cm2 | 1SF substituting correct value 1A correct area 1A correct unit (3) | M L2 |
| 3.1.1(b) | Area of hole = π × radius2 = 3,142 × 9,5 cm × 9,5 cm = 283,5655cm2 Area of sides = 2 025 cm2 × 6 – 2 (283,5655 cm2 ) = 11 582,869 cm2 OR Area of hole = π × radius2 = 3,142 × (9,5cm)2 = 283,5655cm2 Area of faces without holes + area with faces with holes = (4 × 2 025cm2) + [ 2 × (2 025 – 283,5655)] = 8 100cm2 + 2 × 1 741,4345 = 8 100cm2 + 3 482,869cm2 = 11 582,869cm2 OR Area of hole/Oppervlakte van die gat = π × radius2 = 3,142 × (9,5cm)2 = 283,5655cm2 2 025 cm² × 6 = 12 150 cm² 3,142 × 9,5 cm² 283,5655 × 2 = 567,131 cm² 12 150 cm² − 567,131 cm² =11 582,869 cm² | CA from Question 3.1.1 (a) 1SF substituting correct value 1A correct area 1MA multiply by 6 1MA multiply by 2 1M subtracting the values OR 1SF substituting correct value 1A correct area 1CA total area without holes 1CA total area with holes 1M adding both values OR 1SF substituting correct value 1A correct area 1MA multiply by 6 1MA multiply by 2 1M subtracting the values (5) | M L3 |
| 3.1.1 (c) | Total surface area of 12 chairs: = 11 582, 869 cm2 × 12 = 138 994,428 cm2 Amount of paint = 138 994,428 mℓ ÷ 15 × 1,8 = (16 679,33136 ÷ 1 000) ℓ ≈ 17 ℓ OR Amount of paint per chair = 11 582,869cm² ÷ 15 × 1,8 = 1 389,94428 ml ÷ 1 000 = 1,38994428 ℓ Total surface area of 12 chairs: = 1,38994428 l × 12 = 16,67933136 ℓ ≈ 17 ℓ | 1M multiplying by 12 1MA ÷ 15 × 1,8 1C converting mℓ to ℓ 1R rounding up OR 1MA ÷ 15 × 1,8 1C converting mℓ to ℓ 1M multiply by 12 1R rounding up (4) | M L3 |
| 3.1.2 (a) | Diameter = 2 × r = 2 × 7 cm = 14 cm OR Diameter = 7 cm + 7 cm = 14 cm | 1MA multiplying by 2 1A simplifying OR 1MA adding correct values 1A simplifying AO NPU (2) | M L1 |
| 3.1.2 (b) | Volume of a cylinder = π × (radius)2 × height Volume silinder = π × (radius)2 × hoogte 5 000 cm3 = 3,142 × (7)2 × height Height = 5000 3142 x (7)2 = 32,476… cm ≈ 32,48 cm 1SF substitution – 5 000 cm3 and 7 | 1M changing the subject of the Formula 1CA correct height NPR (3) | M L2 |
| 3.2.1 | High Risk | 2RT correct answer (2) | M L1 |
| 3.2.2 | Waist-to-hip ratio = waist measurement hip measurement = 92105 = 1,141… | 1SF substituting correct values 1CA answer NPR (2) | M L2 |
| 3.2.3 (a) | 40 to 49 years of age OR 50 to 59 years of age OR 60 to 69 years of age | 2RT correct age (2) | M L1 |
| 3.2.3 (b) | Waist-to-hip ratio = waist measurement hip measurement 0,7826 = 72 hip measurement Hip measurement = 72 0,7826 = 91,5797507 ≈ 92 cm | 1SF substituting correct values in correct formula 1M changing the subject of the formula 1R rounding AO (3) | M L2 |
| [26] |
QUESTION 4 [21 MARKS]
| Q/V | Solution | Explanation | T&L |
| 4.1.1 | Number of passengers: = 60 / sixty/sestig | 2A number of passengers (2) | Map L1 |
| 4.1.2 | Row: K Number: 1 OR 6 | 1RT row K 1RT seat 1 OR 6 (2) | Map L1 |
| 4.1.3 | SE/South-east | 2A SE (2) | Map L2 |
| 4.1.4 | P = 9 × 100% 60 = 15% | CA from Question 4.1.1 1A numerator 1A denominator 1A probability as a percentage (3) | P L2 |
| 4.1.5 | A5 / 5A | 1A row A 1A seat number 5 (2) | Map L2 |
| 4.2.1 | 8/eight | 2RT correct number of airports (2) | Map L1 |
| 4.2.2 | 1 Unit on the map is equal to 10 000 000 units in real life OR 1 cm/mm on the map is equal to 10 000 000 cm/mm in real life OR The real one is 10 000 000 times bigger OR The drawing is 10 000 000 times smaller | 2A explanation (2) | Map L1 |
| 4.2.3 | Actual distance = 60 mm × 10 000 000 = 600 000 000 mm ÷ 1 000 000 = 600 km | 1M concept of scale 1M conversion 1CA distance AO (3) | Map L2 |
| 4.2.4 | Speed = Distance Time = 597 736/60 = 80,314 km/h | 1A substitution - 597 1A time calculation 1A correct average speed NPR (3) | Map L2 |
| [21] |
QUESTION 5 [31 MARKS]
| Q/V | Solution | Explanation | T&L |
| 5.1.1 | Survey / Questionnaire / Interviews | 2A correct instrument (2) | D L1 |
| 5.1.2 | KwaZulu-Natal/KZN | 2RT correct province (2) | D L1 |
| 5.1.3 | Mean number of voting stations = 22612 9 = 2 512,4444444 ≈ 2 512 OR 2 513 | 1A numerator 1MA dividing by 9 1R to the nearest whole number (3) | D L2 |
| 5.1.4 | 0 | 2RT for mode = 0 (2) | D L2 |
| 5.1.5 | Percentage = 1228 × 100 % 22612 = 5,43 % | 1RT correct values 1M percentage calculation 1A simplification NPR AO (3) | D L1 |
| 5.1.6 | P(Gauteng VD) = 0% OR 0 OR no chance OR impossible OR 0 2716 | 2A stating 0% or impossible (2) | P L2 |
| 5.1.7 | 3 111 – 2 966 = 145 OR 1 228 – (161+189+327+133+82+115+26+50) = 145 OR 1 228 – 1 083 = 145 | 1RT reading correct values 1M subtracting values in correct order OR 1M adding all the values 1M subtracting from 1 228 OR 1M adding all the values 1M subtracting from 1 228 (2) | D L1 |
| 5.1.8 |  1A for each correctly plotted bar × 6 (6) | D L2 | |
| 5.2.1 | Sport = 100% – (42,9 + 2,8 + 11 + 20,7 + 2,4 + 18,4 +0,7)% = 1,1% OR = 100% – 98,9% = 1,1% | 1MA subtract values 1CA correct percentage OR 1MA subtract values 1CA correct percentage AO (2) | D L1 |
| 5.2.2 | Car/Motorcar | 1RT correct modus (2) | D L1 |
| 5.2.3 | P(people travelling by bus) = 7,8% = 7.8 100 = 39 500 OR P(bus)= 7.8 x 10 100 10 = 7.8 1000 = 39 500 | 1RT correct percentage (7,8%) 1M out of 100 1A fraction form OR/OF 1RT correct percentage (7,8%) 1M out of 100 1A fraction form (3) | P L2 |
| 5.2.4 | Number of people/Aantal mense = 542 267 × 42,9% = 232 632,543 ≈ 232 632 OR/OF 232 633 | 1M multiplying correct values 1CA number of people NPR – whole number (2) | D L1 |
| [31] | |||
TOTAL:150
Published in 2019 May/June Grade 12 NSC Exam Papers and Memos