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MATHEMATICAL LITERACY PAPER 1 GRADE 12 MEMORANDUM - 2018 SEPTEMBER PREPARATORY EXAM PAPERS AND MEMOS
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GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
SEPTEMBER 2018
| Symbol | Explanation |
| M | Method |
| M/A | Method with Accuracy |
| MCA | Method with Consistent Accuracy |
| CA | Consistent Accuracy |
| A | Accuracy |
| C | Conversion |
| S | Simplification |
| RT/RG/RM | Reading from a table OR Reading from a graph OR Read from 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 OR Reason |
| AO | Answer only |
| NPR | No penalty for rounding |
QUESTION 1 [30 MARKS] INTEGRATED QUESTION
| Question | Solution | Explanation/Marks AO: FULL MARKS | Lev. |
| 1.1 1.1.1 | Inflation is the increase in the price of goods and services over time.✓✓A OR The decrease in the purchasing value of the money over time. ✓✓A | 2A Explanation (2) | F L1 |
| 1.1.2 | Difference = R125 – R120 = R5 ✓M = 5 × 100 = 500 cents ✓A | 1M Difference 1CA In cents (2) | F L1 |
| 1.1.3 | Price = 112.5 x 125 100 ✓M = R140,60 ✓CA OR Profit = 12,5 × 125 = R15,625✓M OR 100 Price = 125 × 1,12 Price = 125 + 15,625 = R140,625 = 140, 625 = R140,60 ✓CA = R140,60 | 1M Multiply112,5 by 125 and dividing by 100 1CA Price OR 1M For profit 1CA Price NPR (Concept of money) (2) | F L1 |
| 1.2 1.2.1 | Cost of a dozen = 90 x 12 ✓ MA 60 = R18,00 ✓ CA OR Dozens = 60 12 = 5 ✓ Cost of a dozen =90 5 = R18 ✓ | 1MA Divide by 60 and multiply by 12 1CA Cost 1MA Number of dozens 1CA Cost (2) | F L1 |
| 1.2.2 | % Profit = (( 60 + 75 ) 90 ) x 100 90 = 45 x 100% ✓M 90 = 50% ✓CA | 1M Subtraction for profit 1M Fraction 45/90 multiply by 100 1CA Percentage (3) | FL1 |
| 1.3 | Time = 24:00 – 16:25 = 7h 35 min ✓M From midnight to 3:30 am = 3h30 min ✓M ∴ Total time = 7 hours 35 min + 3h30min = 11h 05 min ✓CA | 1M For 7h 35min 1M For 3h 30min 1CA Time (3) | M L1 |
| 1.4 | Plastic cups = 2 x 1000 ✓M 275 = 7,2727 = 7 ✓A | 1M Correct numerator and denominator 1A Rounding down (2) | ML1 |
| 1.5 1.5.1 | Length = 2𝑐𝑚 × 75 𝑚 ✓✓2M 100 𝑐𝑚 = 1,5 m ✓ | 1M for Numerator 1M for Denominator 1CA With correct unit (3) | ML1 |
| 1.5.2 | Scale = 100 : 7 500 ✓✓2M = 1 : 75 ✓ 1CA | 1M Ratio form 1C Conversion 1CA Answer (3) | MP L1 |
| 1.6 1.6.1 | 72,67% = Nelson Mandela District ✓✓ A | 2A District (2) | DL1 |
| 1.6.2 | Districts are: Chris Hani East, OR Tambo Coastal, ✓A Amathole East and Amathole West ✓A | 1A First 2 districts 1A Second 2 districts (2) | DL 1 |
| 1.6.3 | Nelson Mandela, Sarah Baartman OR Tambo Inland, Chris Hani West, Alfred Nzo West, Joe Gqabi, Buffalo City ✓M ∴ 𝑀𝑖𝑑𝑑𝑙𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑐𝑡 = CHRIS HANI WEST ✓CA | 1M Identifying 7 best performing districts in order 1CA for middle district (2) | DL1 |
| 1.6.4 | Candidates failed = 26.1 × 313 030 ✓M 100 = 81 701 ✓CA OR Passed = 73,9 × 313 030 100 = 231 329 ✓M Number failed = 313 030 – 231 329 = 81 701✓CA | 1M Correct values 1CA Candidates 1MA Candidates passed 1CA Candidates failed Accept 81 700 (2) | DL1 |
| [30] |
QUESTION 2 [46 MARKS] FINANCE
| Question | Solution | Explanation/Marks AO: FULL MARKS | Lev. |
| 2.1 2.1.1 | Value of A = R3 250 + R4 500 + R1 200 ✓ M = R8 950 ✓ A OR Value of A = R7 950 + R1 000 ✓ M = R8 950✓ A | 1M Addition 1A Value of A 1M Addition 1A Value of A (2) | L2 |
| 2.1.2 | Value of B = R1 440 – R1 500 ✓M = -R60 ✓ A Value of B = R1 440 – R1 500 ✓M = (R60) ✓ A | 1M Subtracting correct values 1A Negative answer 1M Subtracting correct values 1A Answer in brackets (2) | |
| 2.1.3 | Value of C = R2 600 – R200 ✓ M = R2 400 ✓ CA OR Value of C = R5 780 – (R880 + R1 000 + R1 500) ✓M = R2 400 ✓CA | 1M Subtraction 1CA Value of C 1M Subtraction 1CA Value of C (2) | L1 |
| 2.1.4 | Delivery ✓✓A | 2A Correct answer (2) | L1 |
| 2.2 2.2.1 | Annual gross salary = R65 000 ×12 ✓M = R780 000 ✓CA | 1M Multiply by 12 1CA Annual salary (2) | L1 |
| 2.2.2 | Mrs John’s annual pension fund contribution: = 7.5 (100 × 65 000) ×12 ✓M = R58 500 ✓CA | 1M Multiply by 12 CA Pension contr. (2) | L1 |
| 2.2.3 | Medical Aid contribution = R1 050 ×12 ✓M = R12 600 ✓A | 1M Multiply by 12 1A Contribution (2) | L1 |
| 2.2.4 | Mrs John’s performance bonus = 75 100 ×65 000 ✓M = R48 750 ✓A | 1M Correct values 1M Multiply by 75% (2) | L1 |
| 2.2.5 | Annual taxable income = R780 000 + R48 750 – (58 500 + R12 600) ✓✓M = R828 750 – R71 100 ✓M = R757 650 ✓CA CA from 2.2.1, 2.2.2 and 2.2.3 | 1M Total income 1M Total contributions 1M Subtraction 1CA Taxable income (4) | L2 |
| 2.3 2.3.1 | R701 301 and above ✓✓RT | 2RT Correct group (2) | L1 |
| 2.3.2 | Rebate Mrs John will receive = R13 500 + R7 407 ✓M = R20 907 ✓A 1M Adding correct rebates | 1A Total rebate (2) | L1 |
| 2.3.3 | Actual tax = Income tax calculated on taxable income Rebate = 206 964 + 41 % of the amount above 701 300 – R20 907 = 206 964 + 0,41 × (757 650 – 701 300) –20 907 ✓SF = 206 964 + 0,41 × 56 350 – 20 907 ✓S = 206 964 + 23103,50 – 20 907 ✓S = R209 160,50 ✓CA | CA from 2.2.5, 2.3.1 and 2.3.2 1SF Substitution 1S Simplification 1S Simplification 1CA Nearest rand (4) | L3 |
| 2.3.4 | Net annual salary = Annual taxable income – Actual annual tax = R757 650 – R209 160,50 ✓SF✓M = R548 489,50 ✓CA | CA from 2.2.5 and 2.3.3 1SF Substitution 1M Subtraction 1CA Net salary (3) | L1 |
| 2.4 2.4.1 | Interest 1st year = 15.5 100 × 400 000 = R62 000 ✓A 2nd year amount = R400 000 + R62 000 = R462 000 S✓ Interest 2nd year = 15.5 100 × R462 000 = R71 610✓S Total interest = R62 000 + R71 610 = R133 610 ✓A | 1A Interest 1S New amount 2nd year 1S Interest in 2nd year 1A Total interest (4) | L2 |
| 2.4.2 | Phone D: Cost excluding VAT = 100 115 × 1 750 = R1 521,74 ✓M VAT = R1 750 – R1 521, 74 ✓M = R228,26 ✓A OR VAT = 15 ✓M × 1 750 ✓M 115 = R228,26 ✓A | 1M Cost without VAT 1M Subtraction 1A VAT value 1M Fraction 1M Multiplication 1A VAT value (3) | L2 |
| 2.4.3 | Phone D : Phone E 3 : 2 60 : E ✓M E = 60 x 2 3 = 40 ✓CA | 1M Ratio form 1CA Number of phones (2) | |
| 2.4.4 | Total cost = 60 × 1 750 + 40 × 2 000 = 105 000 ✓M + 80 000 ✓M = R185 000 ✓CA | CA from 2.4.3 1M Cost for phones D 1M Cost for phones E 1CA Total cost (3) | L1 |
| 2.4.5 | 1: 0,52709 185 000 : Total cost in CYN Total cost = 185 000 × 0,52709 ✓M = CYN 97511,65 ✓A✓A | 1M Conversion 1A Total cost 1A Answer in Yuan (3) | L2 |
| [46] |
Related Items
QUESTION 3 [25 MARKS] MEASUREMENT
| Question | Solution | Explanation/Marks AO: FULL MARKS | Lev. |
| 3.1.1 | Length = 4 880 mm ÷ 1 000 ✓C = 4,88 m ✓A | 1C Divide by 1000 1A Metres (2) | L1 |
| 3.1.2 | Distance A = 3 (150 mm) ✓M = 450 mm ✓A | 1M Multiplication by 3 1A Distance (2) | L1 |
| 3.1.3 | Height of the wall = 2,1 m + 450 mm ÷ 1 000 ✓C = 2,1 m + 0,45 m ✓S = 2,55 m ✓CA | CA from 3.1.2 1C Conversion 1S Simplification 1CA Height (3) | L1 |
| 3.1.4 | Area = Length × Width = 4,88 ✓ × 2,1 m ✓M 2 = 5,124 m2 ✓CA ✓Unit | CA from 3.1.1 1M Dividing by 2 1M Multiplication 1CA Area 1A Unit (4) | L2 |
| 3.1.5 | Area covered by bricks = Area of garage – Area of double door = (2,55 m × 5,18 m) ✓ – (4,88 m × 2,1 m) ✓M✓M = 13,209 m2 – 10,248 m2 ✓S = 2,961 m2 ✓CA | CA from 3.1.1 and 3.1.3 1M Area of garage 1M Area of double door 1M Subtraction 1S Simplification 1CA Area covered by bricks (5) | L2 |
| 3.2.1 | Height of the bricks and cement = (12 x 2) + 76 mm ✓M = 100 mm✓CA | 1M Multiplication by 2 and addition 1CA Height (2) | L1 |
| 3.2.2 | Number of rows of bricks = 2 500 ✓M ✓M 100 = 25 ✓CA | CA from 3.2.1 1M Conversion 1M Division 1CA Number of rows of bricks (3) | L1 |
| 3.2.3 | Volume = 23 cm ×11 cm × 7,6 cm ✓SF ✓C = 1 922,8 ✓ A cm3 ✓A | 1SF Substitution 1C Conversion 1A Volume 1A Units (4) | L2 |
| [25] |
QUESTION 4 [31 MARKS] DATA HANDLING
| Question | Solution | Explanation/Marks AO: FULL MARKS | Lev. |
| 4.1. | Range = 3,316 kg – 0,182 kg ✓RT ✓M = 3,134 kg ✓CA | 1RT Correct values 1M Subtraction 1CA Range (3) | L2 |
| 4.2 | 1,668 kg ✓✓A | 2A Median (2) | L2 |
| 4.3 | Average = 1,26 ×2 +1,371 ×9 +1,668 ×8 + 1,746 ×4 + 1,849 ×8 + 2,163 +2,333 + 3,128 ×2 ✓ M = 60,731 ✓M 35 = 1,735 = 2 kg ✓R | 1M Adding 1M Concept of mean 1R Average to nearest kg (3) | L2 |
| 4.4 | 22 ✓✓ A | 2A Explanation (2) | L1 |
| 4.5 | Probability = 8 ✓M × 100%✓M 35 = 22,9 % ✓CA | 1M Fraction value 1M Multiply by 100 1CA Percentage (3) | L1 |
| 4.6 |  4M For the first 4 bars plotted correctly 1M For the next 2 bars plotted correctly 1M For the last 2 bars plotted correctly (6) | L2 | |
| 4.7 | 0,182; 0,182; 0,182; 0,309; 0,729; 0,729; 0,729; 0,856; 0,856; 0,856; 0,856 0,936; 2,448; 2,448; 2,449; 3,038; 3,316; 3,316; 3,316; 3,316; 3,316; 3,316 Q2 Q2=Median= 0,856+0,936 = 0,896 ✓✓ MM 2 Q3=3,316 ✓ CA | 1M Correct values 1M Value or position of median 1CA Q3 (3) | L1 |
| 4.8 | 3,316 kg ✓✓ | 2A RT Modal value (2) | L1 |
| 4.9 | % of total fishes = 8 ✓× 100 ✓M 57 = 14,04 % ✓ | 1M Fraction 1M Multiply by 100 1CA Percentage (3) | L1 |
| 4.10 | 3rd hour ✓✓ | 2RT Correct hour (2) | L1 |
| 4.11 | 1,1,1,1,2,3,3,4,6 ✓✓ | 2M Arrangement (2) | L1 |
| [31] | |||
QUESTION 5 [18 MARKS] MAPS, PLANS and OTHER REPRESENTATIONS
| Question | Solution | Explanation/Marks AO: FULL MARKS | Lev. |
| 5.1 | There is no seat for Lundi here ✓✓A | 2A No seat (2) | MP L1 |
| 5.2 | South ✓✓A | 2A Direction (2) | L1 |
| 5.3 | A8 ✓ RP A11 ✓ RP A15 ✓ RP | 1RP First seat 1RP Second seat 1RP Third seat (3) | L1 |
| 5.4 | 35 ✓✓RP | 2RP Available seats (2) | L1 |
| 5.5 | B14 ✓✓ RP | 2RP Asi’s seat no. (2) | L1 |
| 5.6 | Row J ✓✓ RP | 2RP Furthest row (2) | L1 |
| 5.7 | Side BB ✓✓RP | 2RP Correct side (2) | L1 |
| 5.8 | P(seat from side AA) = 20 ✓A ✓A 19 4 = 0,103 ✓CA | 1A Numerator 1A Denominator 1CA Answer (3) | P L2 |
| [18] | |||
| TOTAL: 150 | |||
Published in Grade 12 September 2018 Preparatory Examinations