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### Introduction:

Mathematics Literacy is a subject that focuses on practical applications of mathematics in everyday life. Grade 12 Mathematics Literacy is an important subject as it prepares students for real-world situations where mathematical concepts and skills are applied. Paper 2 of the Grade 12 Mathematics Literacy exam consists of a memorandum that provides solutions to the exam questions.

### How to access the question papers:

The Grade 12 Mathematics Literacy exam papers and memoranda are available on various educational websites and online resources. Students can also access these papers through their school or teacher.

### Subtopics and possible questions:

Paper 2 of the Grade 12 Mathematics Literacy exam covers various subtopics, including:

- Financial maths
- Measurement
- Data handling

### Possible questions in the exam could include:

- Calculating interest and repayment amounts on loans
- Converting between different units of measurement
- Interpreting and drawing graphs
- Analyzing data sets and calculating measures of central tendency

### How to prepare for exams:

To prepare for the Grade 12 Mathematics Literacy exam, students should:

- Practice past exam papers: Use past exam papers to familiarize yourself with the exam format and the types of questions that may be asked. This will also help you identify areas where you need to improve.
- Study regularly: Allocate regular study time for Mathematics Literacy and stick to your study schedule. Review concepts and practice calculations regularly.
- Seek help: If you are struggling with any concepts or calculations, seek help from your teacher, tutor, or classmates.
- Memorize formulas and definitions: Memorize formulas and definitions as these are essential for solving mathematical problems.
- Stay focused during the exam: During the exam, read the instructions carefully, manage your time, and stay focused on the task at hand.

**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] |
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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 cm ^{2} ✓FIGURE 2 = 20 x 60 = 1 200 cm ^{2} ✓FIGURE 3 = 120 x 40 = 4 800 cm ^{2} ✓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 m ^{2} ✓Litres needed = 4 ✓ 6,2 = 0,645 ✓ = 0,65 m ^{2} ✓ |
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] |
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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 = Weight1,7 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] |
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Total:100 |

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