NATIONAL
SENIOR CERTIFICATE
GRADE 12
JUNE 2022
MATHEMATICAL LITERACY P2
MARKING GUIDELINE
MARKS: 100
| 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 [20] | |||
| Ques. | Solutions | Explanation | level |
| 1.1.1 | Mozambique ✓✓ | 2RT correct country (2) | L1 Maps |
| 1.1.2 | 3 ✓✓ | 2A no of tented camps (2) | L1 Maps |
| 1.1.3 | 9 ✓✓ | 2A correct number (2) | L1 Maps |
| 1.1.4 | Main camp ✓✓ | 2A type of camp (2 | L1 Maps |
| 1.2.1 | 260 ÷ 10 ✓ 26cm | 1C dividing by 1 000 1A answer (3) | L1 Meas. |
| 1.2.2 | 150 − 40 ✓ 110mm ÷ 1000 0.11m✓ | 1MA subtraction 1C dividing by 1000 1A answer (3) | L1 Meas. |
| 1.2.3 | 100:150 ✓✓ 2:3 ✓ | 1RT correct valu4es 1MA ration concept 1S simplification | L1 Meas. |
| 1.3.1 | 14 × 2 + 3✓ 31✓ | 1MA multiplication and addition 1A answer (2) | L1 Meas. |
| 1.3.2 | 6✓✓ | 2A (2) | L1 Probl |
| [20] |
| QUESTION 2 [32] | |||
| Ques. | Solutions | Explanations | Level |
| 2.1.1 | Distance = 1 029 km ✓ 1 029 x 1 000 ✓ 1 029 000 m ✓ | 1RT correct distance 1C conversion 1CA answer (3) | L2 Maps |
| 2.1.2 | Cape Town to Johannesburg = 1 402 km ✓ Johannesburg to Bloemfontein = 417 km ✓ Total = 1 402 + 417 = 1 819 km ✓ Cape Town to Nelspruit = 1 779 km ✓ Difference = 1 819 – 1 779 = 40 km ✓ Valid ✓ | 1RT distance CT to Johannesburg 1RT distance to Bloem 1CA total distance 1RT Nelspruit 1CA difference 1O valid | L4 Maps |
| 2.1.3 | Distance = 1 393 km ✓ Distance = Speed x Time 1 393 = 105 x T ✓ T = 1 393/105 ✓ T = 13,2666 hrs = 0,2666 x 60 ✓ = 16 min T = 13hrs 16 min + 2 hrs 30min ✓ = 15hrs 46min ✓ | 1RT correct distance 1A substitution 1S simplification 1C hours to minutes 1M adding times 1CA answer (6) | L3 Maps |
| 2.2.1 | Bar scale ✓ | 1A type of scale (2) | L1 Maps |
| 2.2.2 | Scale 1,2 cm = 5 km ✓ 4,3 cm = 5 x 4,3 ✓ = 21,5 ✓ 1,2 = 17,916666 x 1 000 = 1 7 917 m ✓ | 1A measuring the scale Accept 1,1 cm–1,3 cm 1A measuring the map Accept 4,1 cm–4,4 cm 1S simplification 1CA distance in metres (4) | L3 Maps |
| 2.2.3 | South East ✓✓ | 2A direction (2) | L2 Maps |
| 2.2.4 | 2 850 = 870 870 ✓ 2 850 0,30526 x 100 ✓ 30,526 ✓ = 30,53 cm ✓ | 1MA dividing by 2 850 1C to cm 1CA answer 1R rounding to two decimals (4) | L2 Maps |
| 2.2.5 | Fuel consumed: 10 km = 1litre 1 km = 1 litre 10 litre Therefore 929 km will require: 929 × 1/10 litre ✓M = 929 ÷ 10 = 92,9 litre ✓ A Cost of return journey = 2 (92,9 × R16,98) ✓ M = 2 (R1 577,442) = R3 154, 884 ✓ S = R3 154, 88 ✓ CA | 1M Determine litres 1A correct answer 1M using 929 km 1S simplifying CA correct answer | L2 Maps |
| [32] | L3 Maps |
| QUESTION 3 [29] | |||
| Ques. | Solutions | Explanations | Level |
| 3.1.1 | 274 +15,25 + 15,25 ✓✓ = 304,5 cm ✓ OR 274 + 2(15,25) ✓ 274 + 30,5 ✓ = 304,5 cm ✓ | 1RT all values correct 1MA adding overhang 1CA answer | L1 Meas. |
| 3.1.2 | 274 – 152,5 ✓ 121,5 x 10 ✓ 1 215 mm ✓ OR (274 x 10) – (152,5 x 10) ✓ = 2 740 – 1 525 ✓ = 1 215 mm ✓ | 1MA subtraction 1C to mm 1CA answer | L2 Meas. |
| 3.1.3 | 10:08 + 1:58 ✓ 11:66 ✓ 12:06 ✓ | 1MA adding time 1S simplification 1A correct time | L2 Meas. |
| 3.1.4 | 76 + 15,25 ✓ = 91,25 152,5 – 91,25 ✓ = 61,25 ✓ Not valid ✓ | 1MA addition 1MA subtraction 1A answer 1O not valid (4) | L4 Meas. |
| 3.2.1 | 100 +40 + 40 + 60 + 20 + 60 + 60 +120+20 +40 ✓✓ = 560 cm ✓ | 1A all values correct 1MA adding all values 1A answer (3) | L1 Meas. |
| 3.2.2 | Area = Length x Width FIGURE 1 = 100 x 40 = 4 000 cm2 ✓ FIGURE 2 = 20 x 60 = 1 200 cm2 ✓ FIGURE 3 = 120 x 40 = 4 800 cm2 ✓ Total area = 4 000 + 1 200 + 4 800 = 10 000 / 10 000 ✓ = 1 m2 ✓ | 1A area 1 1A area 2 1A area 3 1MA total area 1CA area in square metres | L2 Meas. |
| 3.2.3 | Area to paint = 1 x 2 x 2 ✓ = 4 m2 ✓ Litres needed = 4 ✓ 6,2 = 0,645 ✓ = 0,65 m2 ✓ | 1 MA multiplying by coats and no. of shapes 1 CA area to be painted 1 S simplification 1 CA no of litres 1 R rounding (5) | L3 Meas. |
| 3.2.4 | 0,65 x 1 000 ✓ = 650 mℓ ✓ Valid ✓ | 1 MCA multiplying by 1 000 1 CA simplification 1 O verification (3) | L4 Meas. |
| [29] |
| QUESTION 4 [19] | |||
| Ques | Solutions | Explanations | Level |
| 4.1.1 | 17 x 100 ✓✓ 255 = 6,67% ✓ | 1A numerator 1A denominator 1CA percentage NPR (3) | L2 Prob |
| 4.1.2 | D44 ✓ ✓ | 1A letter 1A number (2) | L1 Maps |
| 4.1.3 | D = 29 ✓ E = 32 ✓ H = 41 ✓ Total = 29 + 32 +41 = 102 ✓ | 3A all rows 1 mark for each row 1A correct total (4) | L2 Maps |
| 4.2.1 | 29,9 ✓✓ | 2RT correct amount (2) | L1 Meas |
| 4.2.2 | (height)2 x BMI = Weight 1,7 x 1,7 x BMI = 95 ✓ 2,89 x BMI = 95 BMI = 95/2,89 ✓ = 32,87 ✓ Obese / High health risk. Not valid ✓ | 1SF substitution 1S simplification 1CA 1O verification (4) | L4 Meas |
| 4.2.3 | Exercise ✓✓ Eat healthy food ✓✓ OR Any other relevant answer. | 2A 2A 2 for each suggestion (4) | L4 Meas |
| [19] | |||
| Total:100 |