Symbol | Explanation |
M | Method |
MA | Method with accuracy |
CA | Consistent accuracy |
A | Accuracy |
C | Conversion |
S | Simplification |
RT/RG/RM | Reading from a table/Reading from a graph/Reading from a 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/Reason |
AO | Answer only : Full Marks |
NPR | No penalty for rounding |
QUESTION 1 [32] | |||
Quest. | Solution | Explanation/Marks | Lev. |
1.1.1 | R8 750 + R2 530 ✔ M OR -R2530 + Net Salary= R8 750 ✔ M | 1M adding correct values 1A net salary | F L1 |
1.1.2 | Gross Salary =175× R8 750 ✔ M OR R8 750 × 1,75 ✔ M | 1M multiplication by the % and 8 750 | F L1 |
1.2.1 | Withdrawal fee = R2 ✔✔ A | 2A fee amount (2) | F L1 |
1.2.2 | Discount price paid =80× R300 ✔ M OR Discount price = R300 – (R300 × 20%) } ✔ M | 1M multiplication by the % and R300 OR 1M subtracting R60 | F L1 |
1.2.3 | Elapsed time = 16:21–14:35 ✔ A | 1A Subtraction | M L1 |
1.3.1 | Number of slices = 800 ✔ M OR | 1M division | M L1 |
Quest. | Solution | Explanation/Marks AO | Lev. |
1.3.2 | Amount of sugar 3,20 g × 8 ✔ M OR Amount of sugar 1,6 × 16 ✔M | 1M Multiplication | M L1 |
1.3.3 | 1 003,90 × 1 000 J | 1M multiplication by 1000 1A simplification (2) | M L1 |
F | |||
1.3.4 | Cost of two slices = 100 × 14,99 ✔M OR 16 Slices = R14,99 | 1M 100/800 and multiplying by 14,99 OR 1M fraction using 162and multiplying by 14,99 | L1 |
1.4.1 | Ratio scale or Number scale | 2A type of scale (2) | MP L1 |
1.4.2 | 1 cm = 500 000 ✔M | 1M division | MP L1 |
1.5.1 | Nelson Mandela Metro ✔✔ A | 2A correct district (2) | D L1 |
1.5.2 | 50,2%; 53,3%; 56,0%; 61,0%; 61,1%; 61,5%; 61,7%; 64,5%; 69,2%, 70,6%,72,1%,75,8%✔✔ A | 2A ascending order of pass percentages (2) | D L1 |
1.5.3 | OR Tambo Inland; Sarah Baartman; Buffalo City; Nelson Mandela Metro ✔✔ A | 1A correct two districts 1A correct two other districts (2) | D L1 |
Quest. | Solution | Explanation/Marks AO | Lev. | ||
1.5.4 | % failed learners = 100 – 69,2 ✔M | 1M subtraction of correct values | D L1 | ||
1.5.5 | 64,4% ✔✔ A | 2RT correct probability (2) | P L1 | ||
[32] | |||||
QUESTION 2 [41 MARKS] FINANCE | |||||
Quest. | Solution | Explanation/Marks AO | Lev. | ||
2.1 | 2.1.1 | 500 + 450 + 2 250 +100 + 300 +1 200 + 100 + 350 + 1600 ✔M | 1M adding correct values | L1 | |
2.1.2 | 063 656 2015 ✔✔ RT | 2RT correct cellphone number (2) | L1 | ||
2.1.3 | R105 + R70+ R70 ✔RT✔M = R245 ✔A | 1RT correct values 1M adding | L1 | ||
2.2 | 2.2.1 | B = R500 ✔✔ A | 2A Income (2) | L1 | |
2.2.2 | Income = R500 + R750 × number of events | 1SF substitution | L2 | ||
2.3 | 2.3.1 | Income = R500 + R750 × number of events ✔M | 1M subtracting correct values | L2 |
Quest. | Solution | Explanation/Marks AO | Lev. | |
2.3.2 |
1 Mark – straight line drawn | L2 | ||
2.3.3 | 5 Events ✔✔RG | 2 RG reading from the graph (2) | L1 | |
2.3.4 | From Graph Income = R6 500 | 1M subtracting both correct values | L3 |
Quest. | Solution | Explanation/Marks AO | Lev. | |
2.4 | 2.4.1 | Deposit = 10 × 49 999,99 ✔M OR Deposit= 76 353,12 – 71 353,13 ✔M OR Deposit = (49 999,99 + 1 000 + 500 + 375) – 46 875 ✔M | 1M multiplying by 10% 1M subtracting the correct values (2) | L1 |
2.4.2 | Credit is obtaining goods and services before payment and payment will be done later on agreement and including interest✔A | 1A goods before payment | L1 | |
2.4.3 | Interest = 10 078,13 ✔A OR Interest = 10,75% x 46 875 ✔A | 1M division | L1 | |
2.4.4 | 1/8/2021 ✔✔A | 2A correct date (2) | L1 | |
2.4.5 | VAT =14 400 - 100 × R14 400 ✔M OR VAT amount = 14 400 OR VAT amount=14 400× 15% ✔M | 1M amount exclusive of VAT OR 1M divide by 1,15 OR 1M working with ratio % 2A answer (3) | L2 |
Quest. | Solution | Explanation/Marks | Lev. | |
2.5 | 2.5.1 | GBP £1 = R17,268 ✔✔RT | 2A answer (2) | L1 |
2.5.2 | GBP £1 = R17,268 | CA from 2.5.1 | L3 | |
[41] | ||||
QUESTION 3 [23 MARKS] MEASUREMENT | |||
Ques. | Solution | Explanation/Marks AO | Lev. |
3.1.1 | 29 3 ✔RT | 1RT correct values | L1 |
3.1.2 | ✔RT | 1RT correct value | L2 |
3.1.3 | 5' 5'' ✔✔RT | 2A answer (2) | L1 |
3.1.4 | BMI = 62 kg ✔SF | 1SF substitution | L2 |
3.2.1 | 530 ÷ 10 ✔M | 1C dividing by 10 | L1 |
3.2.2 | Volume = length ×width × height | CA from 3.2.1 | L2 |
3.2.3 | Number of litres = 66 250 ✔ C | 1C dividing by 1000 | L1 |
Ques. | Solution | Explanation/Marks AO | Lev. | |
3.3.1 | Perimeter is the total distance around the outside of the shape.✔✔A | 2A explanation (2) | L1 | |
3.3.2 | Perimeter = 2(17,68 m +3,66 m) ✔SF | 1SF substitution | L1 | |
3.3.3 | Area = 22,56 m ×3,66 m ✔SF | 1SF substitution | L2 | |
[23] | ||||
QUESTION 4 [16 MARKS] MAPS AND PLANS | ||||
Ques. | Solution | Explanation/Marks AO | Lev. | |
4.1. | Guanting Reservoir ✔✔RM | 2 RM Correct Reservoir (2) | L1 | |
4.2 | North ✔✔RM | 2 RM North | L1 | |
4.3 (a) | 45 km + 40 km + 5 km ✔M | 1M adding correct values | L1 | |
(b) | 46,8 km + 30 km +35 km +15 km = 126,8 km ✔M | 1M adding correct values | L2 | |
4.4 | Tongzhou ✔✔RM | 2 RM Correct town (2) | L2 | |
4.5 | Time = distance | 1M substitution | L2 | |
4.6 | 121,7 – (46,8+ 43,7) ✔M | 1M subtracting from121,7 | L1 | |
[16] |
QUESTION 5 [38 MARKS] DATA HANDLING | |||
Ques. | Solution | Explanation/Marks AO | Lev. |
5.1.1 | 742 + 753 ✔M | 1M addition | L1 |
5.1.2 | Kwazulu-Natal ✔✔ RT | 2RTcorrect province (2) | L1 |
5.1.3 | 247 739 – (22 784 + 23 415) ✔RT✔ M | 1RT correct values | L2 |
5.1.4 | P(selecting a passenger vehicle from Western Cape) = 37 848 ✔A | 1A numerator | L2 |
5.1.5 | No. of cars sold = 23 415 – 22 189 ✔A ✔M | 1A correct values | L1 |
5.1.6 | Range is the difference between the maximum value and minimum value.✔✔ A | 2 A Explanation (2) | L1 |
5.1.7 | Range = Maximum value – Minimum value ✔RT | 1RT correct values | L2 |
5.1.8 | Mean = 247 739✔ RT OR Mean= 12 017 + 7 002 +114822 + 41912 +9 447 +13071+8 681+ 2 939 +37 848 ✔M
Mean= 247 739 ✔ | 1RT total, IM divide by 9 | L2 |
Ques. | Solution | Explanation/Marks | Lev. |
5.1.9 | IQR = 39 880 – 7 842 ✔RD ✔M | 1RD correct values | L2 |
5.2.1 | 2013 ✔✔RT | 2RT (2) | L2 |
5.2.2 | 1,30 ×1000 000 000 ✔RT | 1RT | L2 |
5.2.3 | Mode (net migrants) = -4,9 millions ✔✔RT | 2RT correct value | L2 |
5.2.4 | 2015 ✔✔RT | 2RT greatest change (2) | L1 |
5.2.5 | P(a year with urban population less than 4,2)= 3 ✔A | 1A numerator | L2 |
5.2.6 | 1 140 000 000 – 2 880 000 ✔M ✔RT | 1RT correct values | L2 |
[38] | |||
TOTAL: | 150 |