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 [37]

Ques. 

Solution 

Explanation 

Topic  &  Level

1.1.1 

Interest Amount = R749 299,39 × 0,8125%  ✔ A  ✔RT✔ M 
                                                       100 
 = R6 088,06 

OR 

Interest Amount = R749 299,39 × 0,008125 ✔A  ✔RT  ✔ M
 = R6 088,06 

1RT Correct opening  balance 
1M Multiply 
1A Monthly rate   (3)

L2 
F

1.1.2 

Monthly repayment = 750 000 × 8,59  ✔RT 
                                     1 000   ✔M 
 = R6 442,50 

OR 

Monthly repayment = 750 000 × 8,59 ✔RT ✔RT  
       = 6 442 500  ✔M 
             1 000 
 = R6 442,50 

1RT Home loan 
1M Divided by  1 000 
1RT Correct factor   
1RT Home loan 
1RT Correct factor
1M Divided by   1 000    (3)

L2 
F

1.1.3 

Closing balance for Month 3  

✔A 

= R749 299,39 + R6 088,06 – R6 442,50 

✔A

= R748 944,95 

1A Add and Subtract
1A Closing balance  (2)

L2 

F

1.1.4 

Statement not valid  ✔R✔R 
The shorter the loan period, the higher the monthly repayment
OR 

The shorter the loan period, the higher the loan factor ✔R ✔R
OR 

No ✔A ✔R ✔R 
The loan factors are higher with shorter periods 
Accept any other relevant response  

1A Not valid 
1R Shorter period 
1R Higher MRP 
1A No 
1R Higher loan factor 
1R Shorter periods  (3)

L4 

F

1.1.5 

Amount at 119 months = R6 442,50 × 119  ✔CA 
 = R766 657,50  ✔M 
Difference = R766 657,50 – R750 000  
 = R16 657,50  ✔CA 
Statement valid  ✔O

1MA Multiplying  correct values 
1CA Amount 
1M Subtraction 
1CA Difference 
1O Valid (5)

L4 

F

1.1.6 

It will reduce the interest amount. 
Accept any other relevant response 

2A Explanation   (2)

L4 

F

1.2.1

Volume = 2 m × 2 m × 0,3 m   ✔C
 = 1,2 m3✔CA   ✔SF 

1SF Substitution 
1C Convert mm to m
1CA Answer in m3  (3)

M 

L2

1.2.2 

Bags of cement = 1,2  ✔M 
                             0,26  ✔M 
 = 4,615384615 × 2 
 = 9,2 bags   ✔S 
 ≈ 10 bags ✔CA 

OR 

0,26 = 2 bags ✔M ✔M 
 1,2 = 2× 1,2 
          1  0,26 
 = 9,2 bags ✔S 
 = 10 bags  ✔R

CA from 1.2.1 
1M Ratio concept 

1M Multiply by 2 
1S Simplification 
1R Number of bags   

1M Ratio concept 

1M Multiply by 2 
1S Simplification 
1R Number of bags  (4)

M 

L3

1.2.3 

Litres of paint needed = 15 × 1,08 
 = 16, 2 m²   ✔MA 
 = 16,2   ✔M 
      5
 = 3,24 litres 
 = 3,24 litres × 2  ✔CA 
 =      6,48      
     5‐litre tins
 = 1,296 tins 
Number of 5-litre tins = 2 tins ✔CA 

OR 

Litres of paint needed = 15 × 108% 
 = 16,2 m²   ✔MA 
 Two coats = 16,2 × 2 
 = 32,4 m²   ✔CA 
Number of 5-litre tins = 32,4  ✔MM 
                                    5‐litre tins
= 6,6 litres 
 = 2 × 5 litres   ✔CA

1MA Increased  surface area 
1M Dividing by  spread rate 

1CA Total number  of litres 
1CA Number of 5- litre tins  

1MA Increased  surface area 
1CA Two coats 
1M Dividing 
1CA Number of 5- litre tins   (4)

M 

L3

1.3.1 

Price for 2015 = 1 251 158,39 + (1 251 158,39 × 0,05) ✔MA 
 = 1 251 158,39 + 62 557,9195 ✔S
 = R1 313 716,31 ✔CA 

Price for 2016 = 1 313 716,31 + (1 313 716,31 × 0,04) 
= 1 313 716,31 + 52 548,65238 
 = R1 366 264,96  ✔CA 

OR 

Price from 2015 – 2016 = 1 251 158,39 × 1,05 × 1,04 ✔M ✔M✔S 
 = R1 366 264,96 ✔CA

1MA Increase by 5%
1S Simplification 
1CA Amount 2015 
1CA Amount 2016   (4)

L3 

F

1.3.2

                                                            ✔RG 
Percentage change = 1 598 366,77 −1 029 331  × 100% 
                                              1 029 331 
 = 569 035,77  × 100%  ✔RG
   1 029 331
 = 0,5528… × 100% ✔S 
 = 55%  ✔R 

1RG Subtract correct  values 
1RG Correct  denominator 
1S Simplification 
1R Nearest %  (4)

L3 

F

[39]

QUESTION 2 [36]

Ques. 

Solution 

Explanation 

Topic and Level

2.1.1 

Taxable Income = Gross Annual salary – Pension  
= 401 137,75 – (0,075 × 401 137,75) ✔M 
 = 401 137,75 – 30 085,33  ✔S 
 = R371 052,42  ✔CA

1M Subtract 7,5% 
1S Simplification 
1CA Taxable Income  (3)

L2 

F

2.1.2 

Annual Tax   ✔A 
= 63 853 + 31% of taxable income above 305 850 ✔SF 
= 63 853 + 0,31 × (371 052,42 – 305 850) 
= 63 853 + 0,31 × 65 202,42 
= 63 853 + 20 212,75 
= 84 065,75 – 13 635 ✔S ✔M
= 70 430,75 – 12 456   ✔M 
= R57 974,75  ✔CA Annual School fees = 2 500 × 11 + 3 200 × 11  ✔MA
 = 27 500 + 35 200 
 = R62 700 ✔CA 
Claim not valid ✔O

CA from 2.1.1 
1A Correct Rates of  Tax  
1SF Substitution 
1S Simplification 
1M Subtract rebate
1M Subtract MAC 
1CA Annual tax 
1MA School fees × 11
1CA Annual School  fees  
1O Invalid   (9)

L4 

F

2.2.1 

Probability other than African = 8,8% + 9,5% + 2,4% 
= 20,7% ✔A 

OR 

Probability other than African = 100% – 79,3% ✔M 
 = 20,7% ✔CA

1MA Adding correct  values 
1A Probability  
1M Subtract from 100
1CA Probability (2)

L2 

P

2.2.2 

African % = 100% – 8,8% – 9,5% – 2,4%  ✔MA
 = 79,3%  ✔A 
African Population in 2004 = 79,3% × 46,66 million  ✔MCA 
 = 37, 001 380 million ✔CA 

 OR 37 001 380 

African Population in 1911 = 67,3% × 5 972 757 ✔CA 
 = 401 9665,46  
Difference = 37 001 380 – 401 9665,46  ✔M 
 = 32 981 714, 54 
 ≈ 32 981 714 ✔CA

1MA Subtract from  100 
1A African % 
1MCA Calculate %
1CA Total for 2004 in  millions  
1CA Total for 1911
1M Subtract 
1CA Difference  
NPR (7)

L3 

D

2.2.3 

1911 = 0,08 × 5 972 757 ✔CA 
 = 477 820,56 
2004 = 0,08 × 46 660 000 
 = 3 732 800 ✔CA

1MA % for 1911 
1CA Coloured  population 1911 
1CA Coloured  population 2004 (3)

L2 

D

2.2.4 

Percentage of the African race group increased 
Percentage of the Indian race group decreased✔A

1A African increase
1A Indian decrease   (2)

L4

2.3.1 

Number of people = (25 × 2) + 17 
 = 50 + 17 ✔MA 
 = 67 people ✔CA

1MA × 2 and addition
1CA Number of  people (2)

L2 

D

2.3.2 

Cost for Option 1 = 1 500 + (250 × 67) 
 = 1 500 + 16 750 
 = R 18 250 ✔CA 
Cost for Option 2 
Cost for couples = 0,96 × 270 × 50 ✔MA ✔MCA 
 = R12 950  ✔CA 
Cost for singles = 270 × 17  
 = R4 590 ✔CA 
Total for couples and singles = R12 950 + R4 590 
= R 17 550  ✔CA 
Statement invalid  ✔O

CA from 2.3.1 
1MCA Adding and  Multiplying
1CA Option 1 cost 
1MA % decrease 
1MCA × 50 
1CA Couples cost 
1CA Singles cost 
1CA Option 2 cost 
1O Invalid (8)

L4 

F

[36]

QUESTION 3 [39]

Ques. 

Solution 

Explanation 

Topic  and  Level

3.1.1 

North West OR West of North ✔✔A

2A Direction    (2)

M&PL2

3.1.2

Scale: 3,8 cm = 20 miles  ✔A 
Distance = 8,6 cm  ✔A 
Actual distance = 20 
3,8× 8,6 ✔M 
 = 45,263 miles ✔CA

1A Measure scale 
1A Measured distance 
1M Dividing and  multiplying 
1CA Distance to 3 dec places   
Scale Measured Accept  3,6 – 4 cm 
Measured distance Accept  8,4 – 8,8 cm  (4) 

M&PL3

3.1.3 

3,8 cm = 20 miles  
Convert miles to kilometres = 20 miles × 1,609 
= 32,18 km ✔MA 
Convert km to cm = 32,18 × 100 000 
 = 3 218 000 cm ✔MA 
Unit scale = 3,8 cm : 3 218 000 cm ✔S 
 = 1 cm : 848 842,1053 ✔R
 ≈ 1 : 1 000 000  

CA from 3.1.2 
1MA Convert to miles to km
1MA Convert km to cm
1S Simplification 
1R Nearest million 
Penalise for units in unit  ratio  (4)

M 

L3

3.1.4 

Noise pollution 

OR 

Air pollution  ✔✔R 

OR 

Danger  ✔✔R 

OR 

Length of runways  ✔✔R
Accept any other relevant answer 

2R Reason   (2)

M&PL4

3.1.5 

A 61 

2A Road (2) 

M&PL2

3.1.6 

Distance = Speed × Time 
78 miles = 40 miles per hour × Time ✔SF 
Time ✔M = 78 
                   40 
 = 1,95 hours  ✔S 
Time in hours and minutes = 1h 57 minutes ✔C
Arrival time = 07:20 + 1:57 ✔M  
 = 09:17  ✔CA 
Statement invalid  ✔O 

OR 

Time = Distance   ✔M 
             speed
Time = 78  ✔SF 
            40 
 = 1,95 hours   ✔S 
 = 1h 57 minutes   ✔C 
Travel time = 7:20 to 9:15 
 = 1h 55 minutes   ✔M 
Not valid, she will be 2 minutes late   ✔O✔O

1SF Substitution 
1M Changing subject of the  formula 
1S Time in hours 
1C Time in hours and min
1M Adding times 
1CA Arrival time 
1O Invalid  
1M Changing subject of the  formula  
1SF Substitution 
1S Time in hours 
1C Time in hours and min
1M Travel time 
1O Not valid  
1O 2 min late   (7)

M 

L4

3.2.1 

Range = Highest – Lowest 
 15°C = 13°C – Lowest  ✔M 
Lowest = 13°C – 15°C  ✔RT  ✔A
 = -2°C  

1M Concept of  range 
1RT Correct  values 
1A Lowest  value (3)

D 

L2

3.2.2 (a) Mistakes: 
Data was not arranged   ✔A
Calculation was done without using the BODMAS-rule ✔A

1A Mistake 1 
1A Mistake 2  (2)

D 

L3

3.2.2 (b) Correction: ✔M 
3 ; 4 ; 6 ; 7 ; 10 ; 13 ; 14 ; 19 ; 19 ; 22 ; 24 ; 24 
Median = 13 + 14 
                     2 
 = 272 
 = 13,5 °C  ✔A

1M Arrange  data ascending  or descending 
1A Median (2)

D 

L3

3.2.3 

June, July and August ✔A
Minimum temperatures are high  ✔R 
Maximum temperatures are high  ✔R 

1A Correct  months 
1R Min high 
1R Max high    (3)

D 

L4

3.2.4 

Probability = 5 ✔A 
                    12 ✔A
 = 0,4166… 
 = 0,417  ✔CA 

CA from 3.2.1
1A Numerator 
1A Denominator
1CA 3 dec places (3)

P 

L2

3.2.5 

CA from 3.2.1
1CA January 
1A Feb – Apr 
1A May – Jul 
1A Aug – Nov
1A Dec   (5)

D 

L2

[39]

QUESTION 4 [38]

Ques. 

Solution 

Explanation 

Topic  and  Level

4.1.1 

Area of unshaded part 
= Area of rectangle – Area of ▲1 – Area of ▲2 
= (Length × Width) – (½× base × height) – (½× base × height)
                    ✔SF                      ✔SF                                ✔SF 
= (130 cm × 25 cm) – (0,5 × 50 cm × 25 cm) – (0,5 × 50 cm × 15 cm)
= 3 250 cm² – 625 cm² – 375 cm²  ✔M ✔S 
= 2 250 cm² × 5  ✔CA ✔M 
Total area = 11 250 cm²   ✔CA 

OR 

Area of rectangle = Length × Width 
Area of 5 panels = 130 cm × 25 cm  ✔SF 
 = 32 500 cm² × 5  ✔M 
 = 16 250 cm²  ✔CA 
Area of ▲1 (5 panels) = ½× base × height  ✔SF 
 = 0,5 × 50 cm × 25 cm × 5  ✔CA 
 = 3 125 cm² 
Area of ▲2 (5 panele) = ½ × base × height  
 = 0,5 × 50 cm × 15 cm × 5  = 1 875 cm²  ✔CA 
Total Area = 16 250 cm² – 3 125 cm² – 1 875 cm²  ✔M 
= 11 250 cm²   ✔CA

1SF Correct values  for rectangle 
1SF Correct values  ▲1 
1SF Correct values  for ▲2 
1M Subtracting 
1S Simplification
1CA Area of  unshaded part 
1M Multiply by 5
1CA Total area 

OR 

1SF Correct values  for rectangle 
1M Multiply by 5 1CA Area of rec 
1SF Correct values  ▲1 
1CA Area of ▲1
1CA Area of ▲2
1M Subtracting 
1CA Area of  unshaded part (8)

M 

L3

4.1.2 

To glue the seams together 

OR 

To paste sides together✔✔R

2R Reason    (2)

M 

L4

4.1.3 

Diagram T – A ring should be attached to the wide side of the  ✔A  hot air balloon
Diagram U – A paper clip to be attached to the ring  ✔A

1A Explain  
Diagram T 
1A Explain  
Diagram U (2)

M&P 

L2

4.1.4 

To blow hot air in balloon 

2R Reason   (2)

M&P 

L4

4.2.1

When the temperature increases, the lift of the hot air balloon is  higher.  ✔✔A ✔✔A 

OR 

When the temperature decreases, the lift of the hot air balloon is  lower. ✔✔A ✔✔A 

2A Temperature  increases 
2A Hot air balloon  higher   (4)

M&P 

L4

4.2.2 

Air Density of Hot air balloon B = 0,972+0,946 
                                                             2 
= 1,918   ✔MA 
      2 
 = 0,959 kg/m³   ✔CA 
Lift = (Air density outside of the hot air balloon – Air density  inside hot air balloon) × Volume of the hot air balloon
                ✔RT ✔RT ✔SF 
Lift = (1,204 kg/m³ – 0,959 kg/m³) × 2 400 m³   ✔S
 = 0,245 kg/m³ × 2 400 m³   ✔CA 
 = 588 kg 

1MA Concept of  average 
1CA Air Density 
1RT Correct outside  air density 
1RT Correct inside  air density 
1SF Substitution 
1S Simplification 
1CA Lift (7)

M 

L4

4.2.3 

Probability = 33+12  ✔A 
                        93 
= 45  ✔A 
   93 
 = 15   ✔CA
    31 

1A Numerator 
1A Denominator 
1CA Common  fraction in the  simplest form (3)

P 

L2

4.3 

Accommodation = 1 030 × 4 nights × 4 people  ✔A
 = R16 480  ✔CA 
Hot air balloon rides = 750 – (750 × 0,15) ✔MA 
 = 750 – 112,50 ✔S 
Cost per person = R637,50 × 4 ✔CA 
 = R2 550 
Total cost in Rand = R16 480 + R2 550  ✔CA 
 = R19 030 
Cost in USD = 19 030  ✔C 
                         13,63 
 = $1 396, 184886   ✔CA 
Cost in Turkish Lira = $1 396,184886 × 5,25  ✔M
 = 7 329,97 Turkish Lira ✔CA 

1A 4 nights × 4 
1CA Cost for  accommodation 
1MA Less 15% 
1S Cost per person
1CA Cost for rides  
1CA Total cost 
1C Dividing 
1CA Cost in USD
1M Multiply 
1CA Cost in Turkish  Lira   (10)

F 

L3

[39]

TOTAL: 

150