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MATHEMATICAL LITERACY PAPER 1 GRADE 12 MEMORANDUM - NSC PAST PAPERS AND MEMOS SEPTEMBER 2020 PREPARATORY EXAMINATIONS
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GRADE 12
SEPTREMBER 2020
MATHEMATICAL LITERACY P1
MARKING GUIDELINE
| 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 |
MARKING GUIDELINES
NOTE:
- If a candidate answers a question TWICE, only mark 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.
- If the candidate presents any extra solution when reading from a graph, table, layout plan and map, then penalise for every extra incorrect item presented.
| QUESTION 1 ( 32 MARKS) | |||
| Ques. | Solution | Explanation | T & L |
| 1.1.1 | Stop time = 8 × 30 ✓M = 240 minutes ✓CA | 1M Multiply by 30 1CA Answer in minutes (2) | M L1 |
| 1.1.2 | Arrive = 10:00 am ✓✓A | 2A Correct time (2) | M L1 |
| 1.2.1 | Profit is the amount of money gained after sale above the cost price. ✓✓A OR Profit is the sale price minus the cost price. ✓✓A | 2A Explanation (2) | F L1 |
| 1.2.2 | 125% : R1 200 25% : Profit Profit = 25%✓M × ?1 200 ✓M 125% = R240 ✓CA OR 125%: 1 200 100%: Cost Price Cost price = 100% × ?1 200 ✓M 125% = R960 ✓CA Profit = 1 200 – 960 = R240 ✓CA | 1M Division 25%
1M Cost price 1CA Cost price 1CA Profit (3) | F L1 |
| 1.3.1 | 28,239 litres = R434,61 ✓M 28,239 litres 28,239 ✓CA 1 litre = R15,39 | 1M Dividing by 28,239 1CA Cost per litre NPR (2) | M L1 |
| 1.3.2 | 383,5 km : 28,239 litres ✓M 13,58 km :1 litre ✓CA | 1M Method 1CA Number of litres NPR (2) | M L1 |
| 1.3.3 | 13,58 km : R15,39 ✓RT 1 km : R1,13328242 ✓CA | CA from 1.3.1 and 1.3.2 1RT Correct values used 1CA Answer NPR (2) | F L1 |
| 1.3.4 | 175 km ✓✓M 13,58 = 12,89 litres ✓CA OR 383,5 km : 28,239 litres 175 km : ? (Fuel required)✓ M Fuel required = 175 × 28,239 ✓ M 383,5 =12,89 litres ✓CA | CA from 1.3.2 2M Dividing 175 km by 13,58 1CA Number of litres OR 1M Concept of ratio 1M Fraction multiplied by 28,239 1CA Answer (3) | M L1 |
| 1.3.5 | 383,5 km : R434,61 Distance : R675,55 Distance = 25 9073,425 ✓M 434,61 = 596,11 km ✓CA | 1M Division: numerator (383,5 × 675,55) by 434,6 1CA Distance travelled NPR (2) | M L1 |
| 1.4.1 | It means 50 cm on the map represents 100 km on the ground. | 2A Scale concept (2) | M &:P L1 |
| 1.4.2 | 50 cm : 10 000 000 cm 50 cm : 10000000 ✓C 50 50 ✓M 1: 200 000 ✓CA | 1 C Conversion 1M Dividing by 50 1CA Unit ratio (3) | M &:P L1 |
| 1.5.1 | 2017 ✓✓RT | 2RT Correct year (2) | D L1 |
| 1.5.2 | 38 086 769 + 38 820 239 + 39 550 889 ✓M = 116 457 897 ✓CA | 1M Adding correct values 1CA Total urban population (2) | D L1 |
| 1.5.3 | ✓RT Difference =7 794 798 739 – 7 547 858 925 ✓M = 246 939 814 ✓CA | 1M Subtracting correct values 1RT Correct values 1CA Difference (3) | D L1 |
| [32] | |||
| QUESTION 2 (40 MARKS) | |||
| Ques | Solution | Explanation | T/L |
| 2.1.1 | R25 000 – R10 000 ✓M = R15 000 ✓ CA | 1M Subtract correct values 1CA Answer (2) | F L1 |
| 2.1.2 (a) | ✓RT 1 207,50 × 100% ✓M 10 000 = 12,08 % ✓CA | 1M Method 1RT Correct value 1CA % NPR (3) | F L2 |
| (b) | R355,95 + R69 = R424,95 ✓CA ✓M | 1M Adding correct values 1CA Answer (2) | F L1 |
| (c) | R424,95 × 48 = R20 397,60 + R1 207,50 ✓M = R21 605,10 ✓S = R21 605,10 ─ R10 000 ✓MA = R11 605,10 ✓CA | CA from 2.1.2(b) 1M Adding R20 397,60 and 1 207,50 1S Simplification 1MA Subtracting R10 000 1CA Difference (4) | F L2 |
| 2.1.3 | February 2024 ✓A ✓A | 1A Month 1A Year (2) | F L1 |
| 2.1.4 | 250 CAD = ? 1CAD = R11,0555 250 × R11,0555 ✓M = R2 763,875 ✓S =R2 763,88 ✓CA | 1M Multiplying by rate 1S Simplification 1CA Answer NPR (3) | F L2 |
| 2.2.1 | Inflation is the increase in prices over the period of time resulting in the fall of the purchasing value of money. ✓✓A | 2A Explanation (2) | F L1 |
| 2.2.2 | 2018 = R12,24 × (100% +4,62%) ✓ M = R12,81 ✓ S 2019 = R12,81× (100% + 4,38%) ✓ M = R13,37 ✓ CA | 1M Calculating percentage 1S Simplification 1M 2019 price 1CA Answer (4) | F L2 |
| 2.3.1 | 4 Tour packages ✓✓RG | 2RG Break-even-point (2) | F L2 |
| 2.3.2 | Income = R1 000 ×8 ✓RG ✓SF = R8 000 ✓ S OR R8 000 ✓✓✓ RG | 1RG Correct value 1SF Substituting 1S Simplification 3RG Correct value (3) | F L1 |
| 2.3.3 | ✓RG R6 000 ×15% ✓ M = R900 ✓CA | 1M VAT 1RG Correct value 1CA Answer (3) | F L1 |
| 2.3.4 | ✓RG Profit = R6 000 – R5 000 – 900 ✓SF = R100 ✓CA | CA from 2.3.3 VAT value 1SF Substitution 1RG Correct values 1CA Answer (3) | F L2 |
| 2.3.5 | Tour package ✓✓RG | 2RT Correct value (2) | F L1 |
| 2.4.1 | Unemployment Insurance Fund ✓✓A | Correct answer (2) | F L1 |
| 2.4.2 | R12 500 × 2% × 12 ✓✓M = R3 000 ✓A | 1M Using 2% 1M Multiplication by 12 1A Simplification and answer (3) | F L1 |
| [40] | |||
| QUESTION 3 [23 MARKS] | |||
| Ques | Solution | Explanation | T/L |
| 3.1.1 | 3 × 12 ✓M = 36 ✓CA 1/8 × 5 × 6 ✓C = 3,75 mℓ of salt ✓CA OR 1 pinch : 6 people ? : 36 (3 ×12) ✓M 1 pinch = 36/6 = 6 pinches ✓CA 1 pinch : 1/8 teaspoons 6 pinches : ? Teaspoon = 1/8 × 6 = 6/8 ✓C 1 teaspoon : 5 mℓ 6/8 : ? 6/8 × 5 = 3,75 mℓ ✓CA | 1M Multiply by 3 1M Calculating dozen 1CA Number of pinches
1C Conversion
1CA Number of millilitres (4) | M L2 |
| 3.1.2 | 55 + 20 = 75 minutes 75 mins × 12 ✓M = 900 minutes = 900 ✓C 60 = 15 hours ✓CA | 1M Adding 20 and 55 1M Multiply by 12 1C Conversion 1CA Answer in hours Accept 16,25 hours if 13 Sundays have been used (4) | M L2 |
| 3.1.3 | 6 : 250 g 66 : ? 64 × 250 ✓M = 16 500 ✓S = 16 500 ✓M 6 = 2 750 g ✓S = 2,75 kg ✓C | 1A Use 66 1M Multiplying by 250 1S Simplification 1M Dividing by 6 1S Simplification 1C Conversion (5) | M L3 |
| 3.1.4 | ℉ = (9/5 × 180) + 32 ✓SF = 324 + 32 ✓S = 356 ✓CA | 1SF Substitution 1S Simplification 1CA Correct degrees (3) | M L2 |
| 3.2.1 | Radius = 125 ✓M 2 = 62,5 mm ✓CA | 1M Dividing by 2 1CA Correct radius (2) | M L1 |
| 3.2.2 | Volume is the amount of space that an object occupies. ✓✓A | 2A Explanation (2) | M L1 |
| 3.2.3 | ✓C ✓SF Volume = 3,142 × 6,25 cm × 6,25 cm × 19 cm = 2 331,95 cm3 ✓CA | 1C Conversion 1SF Substitution 1CA Answer NPR (3) | M L2 |
| [23] | |||
| QUESTION 4 [22 MARKS] | |||
| Ques | Solution | Explanation | T/L |
| 4.1.1 | 4 ✓✓RP | 2RP Number of entrances (2) | M&P L1 |
| 4.1.2 | 39 ✓✓ RP | 2RP Number of shops (2) | M&P L1 |
| 4.1.3 | ✓ RP 20/39 × 100% ✓M = 51,28% = 51 % ✓R ✓S | CA from 4.1.2 1RP Correct values 1M Multiplying by 100 1S Simplification 1R Rounding (4) | P L2 |
| 4.1.4 | 159 ✓✓A | 2A Correct shop no (2) | MP L1 |
| 4.1.5 | South West ✓✓A OR SW ✓✓A | 2A Correct direction (2) | MP L1 |
| 4.1.6 | Woolworths ✓✓RP | 2RP Correct shop (2) | MP L1 |
| 4.1.7 | Entrance 2 ✓✓ RP | 2RP Correct entrance (2) | MP L1 |
| 4.2.1 | 12 parts ✓✓ RP | 2RP Number of parts (2) | MP L1 |
| 4.2.2 | ✓ ✓ ✓ ✓ A B, D, C, A | 1A B 1A D 1A C 1A A (4) | MP L2 |
| [22] | |||
| QUESTION 5 [33 MARKS] | |||
| Ques | Solution | Explanation | T & L |
| 5.1.1 | 5 524 ✓M 6 000 ✓CA | 1M Adding correct values 1CA Rounding (2) | D L1 |
| 5.1.2 | 1838 307 ✓RT 12 ✓M = 153,25 ✓ CA | 1RT Correct values 1M Dividing by 12 1CA Mean value NPR (3) | D L2 |
| 5.1.3 | Joe Gqabi 94 876 ✓RT 2 876 = 32,988 ✓M ≈ 33 ✓R | 1RT Correct values 1M Dividing by 2 876 | D L2 |
| 5.1.4 | Amathole East ✓✓A | 2A Correct district (2) | D L1 |
| 5.1.5 | ✓RT 7520 – 2498 ✓M =5022 ✓CA | 1RT Correct values 1M Subtracting values 1CA Range (3) | D L2 |
| 5.1.6 | 733, 647, 619, 599, 489, 459, 411, 398, 363, 327 ,254, 225 ✓✓A | 2A Arranged in descending order (2) | D L1 |
| 5.1.7 | Total male educators = 53 241 × (100 – 71,9)% ✓M = 53241 × 28,1% ✓S = 14 960,72 ✓S = 14 961 ✓CA | 1M Method 1S Simplification 1S Using 28,1% 1CA Female educators (4) | D L2 |
| 5.1.8 | 647 + 619 ✓RT = 1 266 ✓M 5 524 = 633 2 762 ✓CA | 1RT Adding correct values 1M Numerator and denominator 1CA Answer (3) | P L2 |
| 5.2.1 | ✓RT P = 213 225 × 100% ✓M 294 204 = 72,48 % ✓CA | 1RT Correct values 1M % Concept 1CA P-value as % NPR (3) | D L2 |
| 5.2.2 | ✓RT Difference(%) = 54,5% ─ 45,0% ✓M = 9,5 % ✓CA | 1RT Correct values 1M Subtraction 1CA Difference (3) | D L1 |
| 5.3.1 | 52 185, 80 369, 333 251, 550 684, 738 340 ✓A Median = 333 251 ✓A | AO 1A Arrangement (ascending or descending) 1A Median value (2) | D L1 |
| 5.3.2 |  | ||
| 1A For the first 2 bars correctly plotted 1A For Q3 bar only 1A For the last 2 bars correctly plotted (3) | L2 | ||
| [33] | |||
| TOTAL: 150 | |||
Published in Grade 12 September 2020 Preparatory Examinations