Codes 

Explanation

M 

Method

MA 

Method with Accuracy

CA 

Consistent Accuracy

A 

Accuracy

C 

Conversion

D 

Define

J 

Justification/Reason/Explain

S 

Simplification

RD 

Reading from a table OR a graph OR a diagram OR a map OR a plan

F 

Choosing the correct formula

SF 

Substitution in a formula

O 

Opinion

P 

Penalty, e.g. for no units, incorrect rounding off, etc.

R 

Rounding Off

NPR 

No penalty for rounding OR omitting units

AO 

Answer only

Ques. 

Solution 

Explanation 

Level

1.1.1 

Emerald – R 19 089 +✔ 
Onyx R 23 551 + ✔

1A Emerald 
1A Onyx (2)

F 

L2

1.1.2 

Emerald = member + adult + child 
 = R2 477 + R1 761 + R914 ✔ 
 = R5 152 ✔ 

Onyx = member + adult + child 
 = R3 587 + R2 362 + R1 149 ✔ 
 = R7 098 ✔ 

Difference = 7 098 – 5 152 ✔ 
 = R1 946 ✔

1RT Correct values 
1CA Add values 
1RT Correct values 
1CA Add values 
1M Subtracting 
1CA Difference (6)

F 

L2

1.1.3 

Government subsidy = R5 152 – R2 530 ✔ 
 = R2 622 

Government % = 2 622   × 100 ✔ 
                            5 152 
 = 50,89%  
 = 50,9%✔

1MA Difference 
1M ×100 
1CA % Rounded to  1decimal place (3)

F 

L2

1.1.4 

It is important for people to be healthy. ✔✔ 

OR 

Accept any other relevant reason.

2O Importance  (2)

D 

L4

1.2.1 

Volume = π × r² × h ✔ 
 = 3,142 × 0,225 × 0,225 × 0,84 ✔✔ 
 = 0,1336 m³  

Volume of traditional beer = 70   × 0,1336 
                                            100
 = 0,09353 m³ ✔ 

OR 

Height of container = 0,7 × 0,84 
 = 0,588 m ✔ 

Volume = π × r² × h 
 = 3,142 × 0,225 m × 0,225 m × 0,588 m ✔✔✔ 
= 0,09353 m³

1M Calculating radius
1 Conversion 
1SF Substitution in  formula
1MA Finding 70% 
1MA Finding 70% 
1M Calculating radius
1A Conversion 
1SF Substitution in  formula (4)

M 

L3

1.2.2 

Length of store room = 2 m = 200 cm ✔ 
Number of the containers along the length 
= 200 / 45 ✔ 
= 4,444… 
= 4 containers ✔ 

Width of store room = 1,5 m = 150 cm 
Number of the containers along the width  
= 150 / 45  
= 3,333… 
= 3 containers ✔ 

Number of containers in total 
= 4 × 3 
= 12 containers ✔ 
Statement is invalid ✔

1M Conversion 
1M Dividing by 45 
1CA Number of  containers across the  length 
1CA Number of  containers across the  width 
1CA Total number of  containers 
1O Invalid (6)

M 

L4

Ques. 

Solution 

Explanation 

Level

1.3.1 

Amount before increase = 336000/ 106,5%✔✔ 
= R315 492,96✔

1M Dividing  
1M Using 106,5% 
1CA Amount (3)

F 

L2

1.3.2 

Bonus = 336 000 / 12 
 = 28 000 ✔ 

Year 1 = 105,8  × 28 000 ✔ 
                100
 = R 29 624 ✔ 

Year 2 = 106,5  × 29 624✔ 
                100
 = R 31 549,56 ✔

1M Monthly salary 
1M Using 5,8% 
1CA Amount 
1MA Finding 6,5%
1CA Amount (5)

F 

L3

Ques. 

Solution 

Explanation 

Level

2.1.1 

10 hours = 10 × 60  
 = 600 minutes ✔ 
 1 page = 26 minutes 
600 minutes = 600 / 26 ✓ 
 = 23 pages ✔ 
Supposed to develop 23 papers therefore 20 papers are  below norm time. ✓

1MA minutes in 10  hours 
1M Dividing by 26
1CA Number of pages 
1O Conclusion (4)

M 

L2

2.1.2 

% increase = 2015 rate - 2013 rate × 100   ✔ 
                                  2013 rate
 = 169,30−147,36   × 100 ✔ 
              147,36 
 = 14,89 % ✔ Accept 14,9%

1M Difference 
1M × 100 
1CA answer % (3)

D 

L3

2.1.3 

Amount of developing material: 
norm time  × rate for developing × number of pages
     60
= 26 × 169,30 × 161 ✓ 
   60 
= 11 811,50 ✓ 

For 10 employees = 11 811,50 × 10  
 = 118 114,9667 ✓ 

Km travelled = 35×2×2×7+2×25×3×7+12×2×5×7✓ 
= 980 + 1 050 + 840 
 = 2 870 km ✓ 

Transport = rate for transport × number of km 
Amount = 2 870 × 2,82 ✓ 
 = R8 093,40 ✓ 

Total amount = 118 114,9667 + 8093,40  
 = R 126 208,37 ✓ 

Balance = R130 000 - R 126 208,37 
 = R 3 791,63 ✓ 

Statement invalid; Balance less than R4 000 ✓

1SF Substituting  correct values 
1S Simplification 
1CA For 10 people 
1M Calculating  distance 
1CA Total distance 
1M Multiplying rate  per km 
1CA Amount 
1CA Total Amount
1CA Difference
1O Invalid (10)

M&F 

L4

Ques. 

Solution 

Explanation 

Level

2.2.1 

USA: 
46 075,25 + 33% of amount above 189 300 ✓ 
= 46 075,25 + 0,33 (350 500 – 189 300) ✓ 
= 46 075,25 + 0,33 × 161 200 
= 46 075,25 + 53 196 
= 99 271,25 ✓ 
        12 ✓ 
= 8 272,60 dollars ✓

1 correct tax bracket
1 SF 
1 simplification 
1M dividing by 12 
1CA (5)

F 

L3

2.2.2 

Income in South Africa  
= $350 500 × 14,11 
= R 4 945 555 ✓ 

Income Tax 
= 208 587 + 41% of amount above 701 300 ✓ 
= 208 587 + 0,41 × (4 945 555 - 701 300) 
= 208 587 + 0,41 × 4 244 255 
= 208 587 + 1 740 144,55 ✓ 
= 1 948 731,55 – 13 257 ✓ 
= 1 935 474,55 / 12 
= R 161 289,55 ✓ 
=R161 289,55/14,11 
= 11430,87 dollars✓ 

Statement is valid ✓

1 conversion 
1F Choosing correct  tax bracket 
1S Simplification 
1M Subtract rebate 
1CA Monthly Tax 
1C Answer in Dollars
1O Valid (7)

L4

2.2.3 

People in the higher tax brackets are feeding the  government bills. ✓✓ 

OR 

It is discouraging for people occupying higher positions  and earning higher salaries. ✓✓ 

OR 

People need to be treated equally✓✓ 
Accept any other valid reason

2R Reason (2)

D 

L4

2.3 

Speed = distance
                 time
Time taken = 08:55 – 06:00 
 = 2 hours 55 minutes ✔ 

Less time spent in Nanaga  
= 2 hrs 55 min – 0 h 30 min ✔ 
= 2 hrs 25 minutes  
= 2,416666…hrs ✔ 

Speed =    311       
            2,416666✔ 
 = 128,69 km/h✔ 

They travelled above the speed limit ✔

1M Travelling time 
1M Difference in time
1C Conversion 
1SF Substitution 
1CA Speed 
1O Opinion (6)

M 

L3

2.4 

School B ✔ has performed better.
The mean of School B is  higher, meaning 50% of the class were able to get 56. ✔✔
Minimum mark in School B is higher. ✔✔

1M Choosing school
2 O First reason 
2 O Second reason (5)

D 

L4

Ques. 

Solution 

Explanation 

Level

3.1.1

✓ ✓ ✓ 58 + 279 + 45 + 455 + 232 + 303 + 280 + 49 + 498 ✓  = 2 199 km ✓

3RT (1mark for every  three correct values)
1M Adding  

1CA Answer (5)

M&P 

L2

3.1.2 

N1 – South Africa ✓ 
N4 – South Africa ✓ 
A3 – Botswana ✓ 
A2 – Botswana✓ 
B6 – Namibia 

4 (1 mark for route  and country).  
If only roads  mentioned max 2 
If only countries  mentioned max 2 (4)

M&P 

L2

3.1.3 

OPTION 1: 
Accommodation = R 1 550 
Breakfast = R 95 × 4 
 = R 380 ✓ 
Total amount = R 1 550 + R 380  
 = R 1 930 ✓ 

OPTION 2: 
Accommodation with breakfast = R 550 × 4  
 = R 2 200 ✓ 
Difference = 2 200 – 1 930 
 = R 270 ✓ 
Not true, they would save R 270 ✓

1MA Cost of  breakfast 
1CA Total cost 
1MA Total cost 
1CA Difference 
1O Invalid (5)

F 

L4

3.1.4 

Probability of getting a self-catering unit at no extra cost
=  5✓  × 100  
    8✓
= 62,5% ✓ 
= 63% ✓

1A Numerator 
1A Denominator 
1CA % 
1 R Round to  nearest % (4)

P 

L2

3.1.5 

Distance travelled 
= 58+98+41+41+550+105+105+738✔ 
= 1 736 km✓ 

Difference = 2199 – 1736 ✔ 
 = 463 km✓ 
Statement is valid ✓

1M Adding distances
1 CA Total distance 
1MA Finding the  difference 
1CA distance  
1O Valid (5)

M&P 

L4

3.2.1 

Percentage achievement in Mathematical Literacy is  decreasing from 2013 to 2016✔✔

2 O Describing the  trend (2)

D 

L4

3.2.2 

Maths decreased from 2013 to 2015✔ 

Maths increased from 2015 to 2016✔

1O Description for  Mathematics for 2013  to 2015 
1O Description for  Mathematics for 2015  to 2016 (2)

D 

L4

3.2.3 

Mathematics = 265 810 – 263 903 
 = 1 907✔ 
Mathematical Literacy = 388 845 – 361 865 
 = 26 980✔ 
Ratio = 1 907 : 26 980✔

1A Increase in  Mathematics 
1A Decrease in  Mathematical Literacy
1CA Ratio (3)

D 

L3

Ques. 

Solution 

Explanation 

Level

4.1.1 

C = 2 × π × r  
157,1 = 2 × 3,142 × r ✔ 
157,1 = 6,284r 
r = 157,1 
6,284✔ 
 = 25 cm ✔ 
 = 0,25 m✔

1SF Substituting  correct formula 
1S Simplification 
1CA Calculate the  radius 
1C Convert to metres (4)

M 

L3

4.1.2 

D = 0,25 × 2 
 = 0,5 m ✔ 
Total height = 1 + 0,75 + 0,5 
 = 2,25 m ✔ 
Space without decoration = 4 – 2,25  
 = 1,75 m ✔ 
From top and from bottom = 1,75 
                                                2 
 = 0,875 m ✔

1CA finding diameter 1CA total height
1CA space without  decoration 
1CA (4)

M 

L3

4.1.3 

Area for red paint = area of rectangle + area of triangle  = ℓ × w + ½ b × h ✔
 = 1,5 m × 1 m + ½ × 0,75 m × 0,75 m ✔  = 1,5 m + 0,28125 m 
 = 1,78125 m² ✔ 
For 15 decorations = 1,78125 × 15 
 = 26,71875 ✔ 
Two coats = 26,71875 × 2
 = 53,4375 m² ✔ 
Litres of paint needed = 53,4375 
                                          8 
 = 6,6796875 ✔ 
5 litres = 6,6796873 
                      5 
 = 1,33 = 2 (5 litres of paint) ✔ 
Area for white paint 
= πr2 + ½ b × height  
= 3,142 × 0,25 × 0,25 + ½ × 0,75 × 0,75 
= 0,196375 + 0,28125 
= 0,477625 ✔ 
For 15 decorations = 0,477625 × 15 
 = 7,164375 
For 2 coats = 7,164375 × 2 
 = 14,32875 
Litres needed = 14,32875 
                                 8 
  = 1,79 
 = 1 (5 litre) ✔ 
White paint = R499 ✔ 
Red = 505 × 2 
 = R 1 010 ✔ 
Argument not valid, cost of red paint is not twice that of  white paint ✔

1M Using correct  formulae 
1SF Substituting 
1CA Area of shaded  parts 
1CA area for 15  decorations 
1CA for 2 coats 
1MA litres of paint  needed 
1MA number of 5 litre  tins 
1CA area for white  paint 
1CA number 5 litre  tins 
1MA cost for white  paint 
1MA cost for red  paint 
1O Not valid (12)

M&F 

L4

Ques. 

Solution 

Explanation 

Level

4.2.1 

North East 

2A Direction (2) 

M&P 

L2

4.2.2 

Mean = sum of values
                     21 
26,762 =   22+33+34+30+25+29+23+23+22 +30+29+30+28+24+25+25+58
                                                                 21✔ 
26,762 × 21 = 432 + 5B ✓ 
562 – 432 = 5B 
130 ✔=  5B 
130  = B  
  5
26 °C = B ✓

1SF Substitution 
1M Mean value ×21
1S Simplification 
1CA Value of B (4)

D 

L3

4.2.3 

Minimum temperatures: 
Lower quartile (Q₁) = 8  
Upper quartile (Q₃) = 11  
Interquartile range = 11 – 8 
 = 3 ✓ 

Maximum temperatures: 
22 22 23 23 24 25 25 25 26 26 26 26 26 28 29 29 30 30 30  33 34 ✔ 
Median = 26 ✔ 
Lower quartile = 24+25 
                               2 
 = 24,5 ✓ 
Upper quartile = 29+30 
                              2 
 = 29,5 ✓ 
Interquartile range = 29,5 – 24,5 
                             = 5 ✓ 
Difference = 5 – 3 
 = 2 ✓

CA from 4.2.2 
1MA Finding IQR for  min. temp.
1M Ascending order 1CA Median 
1MA Finding Q₁ 
1MA Finding Q₃ 
1CA Finding IQR 
1CA Difference (7)

D 

L3

Ques. 

Solution 

Explanation 

Level

4.2.4 

CA from 4.2.2 and  4.2.3 
5M (1 mark for every  set of bars plotted  correctly) 
1M Correct graph (6)

D 

L2

4.2.5 

Probability that a temperature is ≥ 28 °C
= 8 ✓ 
   21✓ 
= 0,380952381  
= 0,381 ✓

1A Numerator 
1A Denominator 
1R To 3 decimal  places (3)

P 

L2

4.2.6 

Measured distance = 6,6 cm✓ 
 = 6,6 cm : 1 045 km✓ 
 = 6,6 : 104 500 000✓ 
 =1 : 15 833 333,33✓ 
 = 1 : 16 000 000 ✓

1MA Measure on map (accept 6,4 - 6,8) 
1M Ratio 
1C Converting to cm
1S Simplification  
1R Round to nearest  million (5)

M&P 

L3