Symbol 

Explanation 

M 

Method 

A 

Accuracy

CA 

Consistent accuracy 

RT/RG/RM 

Reading from a table/Reading from a graph/Read from map

SF 

Substitution in a formula 

P 

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

S 

Simplification

R 

Rounding/Reason

NPR 

No penalty for rounding

QUESTION 1 [30 MARKS]

QUES

Solution 

Explanation/m 

T/L

1.1.1 

   750  × 100% ✔ 
 3 200
= 23,44% ✔

1M Multiply by 100
1CA (NPR) (2)

F 

LI

1.1.2 

Balance = 3 200 – 750 ✔ 
 = 2 450 ✔

1M subtraction 
1 CA (2)

F 

L1

1.1.3 

Total amount paid = 750 + 5 × 300 ✔ 
 = R2 250 ✔

1M Adding and  Multiplying 
1 CA (2)

F 

L1

1.2.1 

Rate in cents = 0,8865 × 100 ✔ 
 = 88,65 ✔

1M Multiplying  
1A (2)

F 

L1

1.2.2 

Amount charged = 50 × 88,65 ✔ 
 = 4 432,5 cent✔  
OR 50 × 0,8865✔ 
 = R44,33 ✔

1M Multiplying  
1A 
(NPR)  (2)

F 

L1

1.2.3 

Value Added Tax ✔✔ 

2 R   (2) 

F  

L1

1.3.1 

Time to clock off = 7:30 + 7 hr. = 14: 30 ✔ 
Then 15:00 + (15 min + 45 min) = 15:30 ✔

1M adding 7hr 
1A (2) 

M 

L1

1.3.2 

Income = 26 × 7 × 20 ✔ 

 =R3 640 ✔

1M multiplication 
1CA (2)

F 

L1

1.4.1 

Distance = 1 141 km ✔✔ 

2RT (2) 

M  

L1

1.4.2 

East London ✔ and Polokwane ✔ 

2RT (2) 

M  

L1

1.4.3 

Durban ✔and Cape Town ✔ 

2RT (2) 

M  

L1

1.5.1 

Parents in the survey = 24 + 32 + 16 + 13 + 5 ✔  = 90 ✔

1M Adding 
1A (2)

D 

L1

1.5.2 

Less than 20% = 24 + 32 ✔ 
 = 56 ✔

1M Adding 
1A (2)

D 

L1

1.5.3 

Modal range = 10–19 ✔✔ 

2RG (2) 

D  

L1

1.5.4 

Bar graph ✔✔ 

2RG (2) 

D  

L1

[30]

QUESTION 2 [42 MARKS]

QUES. 

Solution 

Explanation/m 

T/L

2.1.1 

Child support ✔✔ 

2RT (2) 

L1

2.1.2 

R1 525 + 2 x R350 ✔= R1 525 + R700 ✔  = R2 225 ✔

1MA amount  
1 S 
1CA amount (3)

L1

2.1.3 

R1 435 x 6,27% = 89,97 ✔ 
R1 435 + 89,97 = R1 524,97 ✔ 
R1 525 ✔ 

OR 

1435 x 1,0627 ✔= R1 524,97 ✔ 
 = R1 525 ✔

1MA multiply by  6,27% 
1CA addition 
1CA Rounding  
1MA multiply by  1,0627 
1CA  
1CA Rounding (3)

L1

2.2.1 

4 x R1 155,26 ✔ 
= R4 621,04 ✔ 

OR 

R5 620 – (115,80 + 192,98 + 690,18) 
R5 620 – R998,96 ✔ 
= R4 621,04 ✔ 

OR 

R4 929,82 – R115,80 – R192,98 
= R4 621,04

1M 
1CA 
1S 
1CA 
1M 
1A (2)

L1

2.2.2 

R0,00 ✔✔ 

2RT (2)

2.2.3 

R4 621,04 x 5,6%  
R258,78 ✔ 
R4 621,04 – R258,78 
R4 362,26 ✔ 
Price with VAT = 1,14 x 4362,26 ✔ 
 = R4 972,98 ✔ 

OR 

100% – 5,6%  
= 94,4% ✔ 
R4 621,04 x 94,4%  
R4 362,26 ✔ 
Price with VAT = 1,14 x 4362,26 ✔ 
 = R4 972,98 ✔

1MA Value of 5,6% 
1S Difference 
1M Subtraction 
1CA  
1M Multiply with 1,14
Price with VAT 
1M Difference in % 
1 CA Value of 94,4%
1M Multiply with 1.14
1CA Price with VAT  (4)

L3

2.3.1 

R8,3 billion ✔✔ 

OR 

8 300 000 000 ✔✔

2RT 
Penalise with 1 mark if  answer is written  without billion   (2)

L1

2.3.2 

R19,84 + R2,71 + R0,45 + R1,8 + R2,71 + R17,59 ✔
R45,1 billion ✔ 

OR 

R45 100 000 000 ✔✔

1M addition 
1CA 
2A 
Penalise with 1 mark if  answer is written  without billion (2)

L1

2.3.3 

1 Euro = R15,3728 
 ? = R1,8 billion 
? =    1.8    = € 0,1170899251 billion ✔ 
   15,3728✔ 
 = 0,1170899251 × 1 000 000 000 
 = €117 089 925,1

1M division 
1 CA  (2) 

L2

2.3.4 

96,0 – 36,8 ✔✔ 
= 59,2 billion ✔

1M correct values 
1M subtraction 
1A  
Penalise with 1 mark if  answer is written  without billion  (3)

L1

2.3.5 

17,59   x 100% ✔ 
45,1✔
 = 39% ✔

1M Division 
1M Multiply by 100 
1CA (3)

L2

2.3.6 

1,8 : 36,8 ✔✔ 
 1 : 20,44 ✔

1 Ratio  
1 Correct values 
1A (3)

L2

2.4.1 

R10,00 x 25 ✔ 
= R250,00 ✔

1M identifying R10
1CA (2)

L1

2.4.2 

260 x R1,50 ✔ 
= R390,00 ✔

1M 
1A (2) 

L1

2.4.3 

R11,00 ✔✔ 

2RT (2) 

L1

2.5.1 

Inflation is the increase in prices of goods and  services over time. ✔✔

2R   (2)

F 

L1

2.5.2 

Inflation rate = new prices - old prices × 100% 
                                old prices
 =55,95−52,95  × 100% ✔ 
         52,95
 = 5,665 ✔ 
 = 6% ✔

1M Substitution 
1M simplification 
1A Rounded in %  (3)

F 

L2

[42]

QUESTION 3 [23 MARKS]

QUES. 

Solution 

Explanation/m 

T/L

3.1.1 

28 + 2 × 10,6 + 41,2 + 28 ✔✔ 
= 118,4 g ✔

1M Addition 
1M 10,6 x 2 
1CA (One value missing)  (3)

L2

3.1.2 

Number of spoons = 2 7 , 5 ✔ 
                                  4 ,1 8 
= 6,6 teaspoons ✔ 
= 7 teaspoons ✔

1M Division 
1 CA when one of the  values is incorrect 
1R   (3)

L1

3.1.3 

50 × 500  = 25 000 
     250            250✔ 
= 100 cups ✔

1M  
1A (2)

L1

3.2.1 

Volume = length x width x height 
 = 28 cm x 15 cm x 8 cm ✔ 
 = 3 360 cm³✔

1SF  
1A (2)

L2

3.2.2 

Volume of chocolate =3 360 
                                     80✔ 
 = 42 cm³✔ 
Volume = area of base x thickness 
42 cm³ = 35 cm² x thickness ✔ 
Thickness of chocolate = 42
                                         35 
= 1,2 cm ✔

CA from 3.2.1 
1M 
1A Volume of one  chocolate 
1S Substitution 
1A Thickness of chocolate  (4)

L3

3.3.1 

Diameter 
130 cm + 25 cm × 2 + 1,8 cm x 2 ✔✔ 
= 183,6 cm ✔

1M x 2 and addition 
1CA Diameter (2)

L2

3.3.2 

Area = 2,3 m x 2,3 m ✔ 
 = 5,29 m² ✔

1SF  
1CA (2)

L2

3.3.3 

Area = 3,142 x (0,918 m)² ✔✔ 
 = 2,647 m² 
 = 2,65 m²✔

1M SF 
1CA radius from 3.3.1 value of diameter 
1CA (3)

L2

3.3.4 

Wasted material = Area of material – Area of  tablecloth to cut 
 = 5,29 m² ̶ 2,65 m²✔ 
 = 2,64 m²✔ 

CA from 3.3.2 and 3.3.3
1M subtraction 
1A   (2)

L2

[23]

QUESTION 4 [20 MARKS]

QUES 

Solution 

Explanation/m 

T/L

4.1.1 

Two ✔✔ 

2RM (2) 

L1

4.1.2 

Sample point E ✔✔ 

2 RM (2) 

L1

4.1.3 

Taaibuschspruit ✔ 
Leeuspruit ✔

1RM 1st side stream 1RM 2nd side stream  (2)

L1

4.1.4 

Scaled length = 3  x 100 000
                           25 000✔ 
 = 25 000 ✔ 
 = 12 cm ✔ OR 0,12 m

1M x 100 000 
1M Division 25 000
1A Distance in cm  (3)

L2

4.2.1 

(a) 

16 ✔✔ 

2RD (2) 

L2

(b) 

Fixed side ✔ and  Drop side ✔

1RD 1ste wood part 
1RD 2nd wood part (2)

L2

(c ) 

B ✔✔ 

2RD (2) 

L2

4.2.2 

16 ✔✔ 

2RM (2) 

L1

4.2.3 

 4 ✔ 
12 ✔ 
1 ✔
3

1M value of  numerator 
1M denominator 
1M simplified answer  (3)

L2

[20]

QUESTION 5 [35 MARKS]

QUES. 

Solution 

Explanation/m 

T/L

5.1.1 

Northern Cape ✔ 
Western Cape ✔

2RT  (2)

L1

5.1.2 

A  =11 705 + 5 105+4 568+10 070 + 13 022 + 5 091 + 1 963 + 3 543+3 589 ✔
= 58 656 ✔

1A adding 
1CA if one value is  missing (2)

L1

5.1.3 

613; 1 404; 2 122; 2 149; 2 290; 2 600; 2625; 3 492; 4 765 ✔✔

2M Ascending order  (2)

L1

5.1.4 

2 290 ✔✔ 

2CA from 5.1.3 (2) 

L2

5.1.5 

2 625 – 1 404 ✔✔ 
= 1 221✔

1M correct values 
1M subtraction 
1CA (3)

L1

5.1.6 

58 656 
     9 ✔ 
= 6 517,3 ✔ 
= 6 517 ✔

CA from 5.1.2  
1M dividing by 9 
1S 
1CA (using from  value from 5.1.2) (3)

L2

5.1.7 

Limpopo ✔✔ 

2RT (2) 

L1

5.2. 

A – Minimum value ✔ 
B – Lower Quartile ✔ 
C – Upper Quartile ✔ 
D – Maximum value ✔

1M Minimum 
1M Lower Quartile
1M Upper Quartile
1M Maximum (4)

L2

5.3.1 

Sample points 2 , 3 and 9 ✔✔ 

1M 1st sample point 1M 2nd and 3rd sample   points (2)

L1

5.3.2 

Significant risk of gastrointestinal disorders ✔✔ 

2RT (2) 

L1

5.3.3

1 Mark per two points correctly plotted (5 x 1) 
1 Mark for joining the points 

(6)

5.4.1 

A ---- (H;H) ✔ 
B----- (T) ✔ 
C------(T;H) ✔

3A 1 mark for each  outcome

P 

L2

5.4.2 

P(HH) =14✔✔ 

1M numerator 
1M denominator (2)

P 

L2

[35]

TOTAL: 

150