Symbol 

Explanation

M 

Method

MA 

Method with accuracy

CA 

Consistent accuracy

A 

Accuracy

C 

Conversion

S 

Simplification

RT/RG/RD 

Reading from a table/graph/map/diagram

SF 

Correct substitution in a formula

O 

Opinion/reason/deduction/example

P 

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

R 

Rounding off

NP 

No penalty for rounding

AO 

Answer only full marks

J 

Justification

QUESTION 1 [36 MARKS]

Ques 

Solution 

Explanation 

T&L

1.1.1 

P(_{even number date}) = 11   ✔✔A 
                                 22   ✔A
 = ½ or 0,5 or 50% 

2A numerator 
1A denominator 
 AO (3)

P 

L2 

1.1.2 

• Quality of bank services / security / perks. ✔✔O 

 OR 

• Proximity or accessibility of the bank. ✔✔O 

 OR 

• Marketing/advertising appeal  ✔✔O   

 OR 

•  Loyalty to bank  ✔✔O

OR 

•  Religious reasons / Economical reasons  ✔✔O
  Any other suitable reason 

2O reason   (2)

F  

L4

1.1.3 

2014 Fee = R3,50 + 1,1% × R1 000  ✔SF 
 = R14,50 ✔CA 
% change =[ R15,50 −1] × 100%    ✔SF 
                   [  R14, 50  ] 
 =    R1,00   × 100% 
      R14,50 
 = 6, 8965517…  ✔CA
 A ≈ 6,9%  ✔R 

OR 

% change = [           R15,50  - 1      ]  × 100%    ✔SF
                    [   R3,50 0,011 R1000 ]     ✔SF 

=  [R15,50  −  1 ]  × 100%
         R14,50     ✔CA 
 = 6,8965517…  ✔CA 
 A ≈ 6,9%  ✔R

1SF substituting R1000
1CA 2014 fee  
1SF correct values 
1CA simplification 
1R rounding 

OR 

1SF correct values 
1SF substituting R1000
1CA 2014 fee  
1CA simplification 
1R rounding  (5)

F 

L2

Ques 

Solution 

Explanation 

T&L

1.1.4 

Withdrawal fee R15 000 at Bank X 
                    ✔SF 
= R3,95 + 0,013 × R15 000 
= R198,95 ✔CA 
Fees for 4 withdrawals 
 = R198,95 × 4  
                            ✔CA 
 = R795,80 

Withdrawal fee for R15 000 at Bank Y 
 = R4,00 + R15 000 × 1,15% 
 = R176,50  ✔CA  
 Fees for 4 withdrawals = 4 × R176,50 
                         ✔CA 
 = R706,00 
Difference in fees = R795,80 – R706,00  
               ✔CA 
 = R89,80 
                       ✔O 
It is NOT VALID. 

OR 

Withdrawal fee R15 000 at Bank X 
                      ✔MA 
= R3,95 + 0,013 × R15 000 
= R198,95   ✔CA

Withdrawal fee for R15 000 at Bank Y 
= R4,00 + R15 000 × 1,15% 
= R176,50 ✔CA 
                                                                          ✔CA 
Difference in fees = R198,95 – R176,50 = R22,45
                                                       ✔M 
Saving on 4 withdrawals = R22,45× 4 = R89,80  ✔CA
                         ✔O 
It is NOT VALID 

OR 

1SF substituting 
1CA weekly charges
1CA fees for 4 withdrawals
1CA charges 
1CA fees for 4  withdrawals 
1CA difference 
1O conclusion  

OR 

1MA substituting 
1CA weekly charges 
1CA charges 
1CA difference 
1M fees for 4  withdrawals 
1CA October charges 
1O conclusion  

OR 

F 

L4

Ques 

Solution 

Explanation 

T&L

Bank X: 
Fee per R1 000 = R3,95 + R1,30 ÷ 100 × 1 000
 = R16,95 ✔
Withdrawal fee for R15 000 = R16,95 × 15 
 = R254,25 ✔
For 4 withdrawals : R254,25 × 4 
 = R1 017 ✔
Bank Y: 
Withdrawal fee for 4 times R15 000 
= R15,50 × 4 × 15 
= R930 ✔
Difference in fees = R1 017 – R930 = R87 
It is NOT VALID ✔

1MA substituting 
1CA weekly charges 
1M fees for 4  withdrawals 
1CA charges 
1CA October charges 
1CA difference 
1O conclusion  
(Max of 6 marks for a  total withdrawal of  R60 000 .) 

 (7)

1.1.5 

Wage for 4 full weeks = R2 142,85 × 4 
 = R8 571,40 
Wage for 2 days = R2142,85  × 2 
                                     5 
 = R857,14 
Total wage = R8 571,40 + R857,14 
 = R9 428,54 ✔CA 

OR 

Average day wage = R2142,85     OR    R2142,85 × 4 
                                       5                                   20 
 = R428,57 
Total wage for October = 22 × R428,57 
 = R9 428,54 ✔

OR 

2 days of a five day week = ²/₅  of a week ✔
Total number of weeks = 4²/₅  OR 4,4 ✔
Total wage for October =4²/₅  × R2142,85 52 ✔
 = R9 428,54 ✔

OR 

1A 4 weeks wage 
1M divide by 5 
1M multiply by2 
1CA total wage 

OR 

1M divide by 5 
1A daily wage 
1M multiply by 22 
1CA total wage 

OR 

1M divide by 5 
1A number of weeks 
1M multiply by weekly  wage 
1CA total wage 

OR

F 

L2

Ques 

Solution 

Explanation 

T&L

                                                   ✓M
Monthly wage =  R2142,85 × 52     ✓A 
                                               12      ✓MA 
 = R9 285,68  ✓CA 

1M multiplying 
1A 52 weeks in year 1MA dividing by 12 
1CA total wage  

 (4)

1.2.1 

• More small/local companies may have entered the  market ✓✓O
• The increased use of smartphones, laptops and tablets ✓✓O 
• Locally produced no need to import. ✓✓O 
• Cost of transport increased ✓✓O 
• Economical reasons / factors ✓✓O
• Maritime piracy / security ✓✓O 
• Other means of transport used ✓✓O
• Durability - demand for new computers became less Or any other valid factors with reasons ✓✓O  

2O factor with reason
2O factor with reason 

 (4)

D 

L4

1.2.2 

Q1 of 2012: 
                         ✔MA
(15,7 + 11,7 + 10,1 + 9 + 5,4 ) million  
                    ✔CA 
= 51,9 million or 51 900 000 
Q1 of 2013: 
= ( 12 + 11,7 + 9 + 6,2 + 4,4 ) million 
                 ✔MA 
= 43,3 million or 43 300 000 
Difference between 2013 and 2012  
                                            ✔CA
= 51,9 mil – 43,3 mil = 8,6 million or 8 600 000 OR 

1MA adding correct  values 
1CA total shipment in  2012 
1MA total shipment in  2013 
1CA difference in  million 

OR

D 

L2

Ques 

Solution 

Explanation 

T&L

Differences (in millions) for 
A = 15,7 – 12,0 = 3,7 ✔A 
B = 11,7 – 11,7 = 0 
C = 10,1 – 9,0 = 1,1 ✔A 
D = 9,0 – 6,2 = 2,8 
E = 5,4 – 4,4 = 1 
                                                         ✔M 
Total difference = (3,7 + 1,1 + 2,8 + 1) million 
 = 8,6 million ✔CA

2A differences in  millions 
1M adding all  differences 
1CA total difference  in million 
Penalty if million  omitted 

(4)

1.2.3

                                                           ✔RT     ✔M 
% change A = 12 000 000 -15 700 000 × 100 %  
                                 15 700 000 
 = – 23,56687898%  ✔CA
                                                        ✔RT 
% change D = 6 200 000 -  9 000 000 × 100 %   ✔M 
                                    9 000 000 
 = – 31,11111111%  ✔CA 
The statement is NOT VALID.  ✔O

OR 

Percentage of 2012 shipped in 2013: 
               ✔RT 
By A: 12,0 ×  100% 
          15,7 
 = 76,43%  ✔A 
∴ Percentage decrease = 100% – 76,43% = 23,57% ✔M
✔RT 
By D: 6,2  × 100% 
           9 
 = 68,89% ✔A                                           ✔M 
∴ Percentage decrease = 100% – 68,89% = 31,11%
                                                                                     ✔O
D shows the greatest decrease, the statement is NOT VALID 

1RT correct values 
1M calculating %  change 
1CA % change 
1RT correct values 
1M calculating %  change 
1CA % change 
1O conclusion 

OR 

1RT correct values 
1A percentage  
1M % change 
1RT correct values 
1A percentage  
1M % change 
1O conclusion  

D 

L4

NP

(7)

[36]

QUESTION 2 [47 MARKS]

Ques 

Solution 

Explanation 

T&L

2.1.1 

(a)

                       ✔A 
Amount × 109,7% = R218,9 billion 
Total amount spent =  R218,9 billion    
                                        109,7%        ✔M
 = R199 544 211 500   ✔CA 

OR

R199,54 billion or R1,9954 × 10¹¹ 

1A correct value and % 

1M dividing by  

109,7% 

1CA total amount

F 

L2

NP 

(3)

2.1.1 

(b)

                                      ✔A 
It is more appropriate to round to one decimal place. 
If a rand value in billions is rounded off to a whole number, the amount that is added or lost is hundreds of millions of  rands.  ✔✔O

OR 

                         ✔A 
It is not appropriate to round to off to a whole number since it  has a big financial implication✔✔O 

1A statement 
2O explanation 
(Note: More  appropriate can be  implied in the  statement) 

(3)

F 

L4

2.1.2 

                          ✔A                                 ✔A 
International: 43% of R 218,9 billion = R94,127 billion
Number of visitors = 14,3 million or 14 300 000  

Average spent per visitor =  R94 127 000 000  ✔C
                                                 14 300 000 ✔MA 
 = R6 582,31  ✔CA 
This is NOT correct.  ✔O 

OR 

                        ✔A                                              ✔A 
International: 43% × R 218,9 billion = R94,127 billion 
Average spent per visitor = R94,127 × 1000 million   ✔C 
                                                 14,3million ✔MA 
 = R6 582,31 ✔CA 
This is NOT correct.  ✔O

OR 

1A percentage 
1A amount 
1C conversion 
1MA average 
1CA value 
1O conclusion 

OR 

1A percentage  
1A amount 
1C conversion 
1MA average 
1CA value 
1O conclusion 

OR

F 

L3

Ques 

Solution 

Explanation 

T&L

Amount spent by the International visitors  
                           ✔MA 
= R6 580 × 14,3 million 
                     ✔A                        ✔C
= R94 094 million = R94,094 billion  
But spent by international tourists is 
 ✔A                                                   ✔A 
43% × R 218,9 billion = R94,127 billion 
The amount was NOT CORRECT✔O 

1MA multiplying 
1A amount  
1C conversion 
1A percentage 
1A amount 
1O conclusion  (6)

2.1.3 

Air transport and road transport ✔A✔A

1A for each item   (2)

F 

L2

2.1.4 

 Payment of tourism levy  ✔✔O
OR 
Purchase of souvenirs   ✔✔O 
OR 
Entrance fees to tourist attractions  ✔✔O 
OR 
Any other suitable example  ✔✔O 

2O example (2)

F 

L4

2.1.5 

Growth in 2014 = 2,9% × R103,6 billion ✔M 
 = R3,0044 billion 
                                                                              ✔M 
GDP contribution (2014) = (R3,0044 + R103,6) billion 
= R106,6044 billion ✔CA 
Growth in 2015 = 2,9% × R106,6044 billion 
 = R3,0915276 billion  ✔CA 
GDP contribution (2015) = (R3,0915276 + R106,6044) billion 
 = R109,6959276 billion 
Growth in 2016 = 2,9% × R109,6959276 billion 
 = R3,1811819 billion 
GDP contribution (2016) = (R3,1811819 + R109,6959276) bil.
 = R112,8771095 billion   ✔CA 
 = R112 877 million  ✔R  or R112 877 000 000 or R112,877 billion 

OR 

1M multiplying 
1M adding 
1CA amount in 2014 
1CA amount in 2015 
1CA amount in 2016 
1R correct rounding

OR

Ques 

Solution 

Explanation 

T&L

2.1.5 

                                                    ✔A                     ✔M 
GDP contribution (2014) = 102,9% × R103,6 billion
= 106,6044 billion ✔CA 
GDP contribution 2015 = 102,9% × R106,6044 billion 
= 109,6959276 billion ✔CA 
GDP contribution 2016 = 102,9% × R109,6959276 billion 
= R112,8771095 billion. ✔CA
 = R112 877 million  or R112 877 000 000  ✔R

OR 

GDP contribution 2016  
                       ✔M             ✔A                   ✔A 
= R103,6 billion × 102,9% × 102,9% × 102,9% 
= R112,8771095 billion. ✔CA 
                                                         ✔C 
= R112,877 billion or R112 877 million  or R112 877 000 000  ✔R 

1M multiplying 
1A 102,9% 
1CA amount in 2014 
1CA amount in 2015 
1CA amount in 2016 
1R correct rounding 
1M multiplying 
2A 102,9% 
CA amount in 2016 
1C conversion 
1R correct rounding (6)

F 

L3

2.2.1 

(a)

                                                         ✔✔✔RT 
Stopover times = 5 + 20 + 5 + 2 + 8 + 2 + 2 + 2 + 23 + 26 + 3 + 17 + 3 + 14 + 3 + 3  ✔M 
                                                                                         
                     ✔CA
 = 138 minutes or 2 hrs and 18 minutes  or 2,3 hours 

3RT correct stopover times 
1M adding  stopover times 
1CA total stopover time   
Stopover times: 
One or two errors only 1 mark penalty,  
Three or four errors 2 mark penalty 

D 

L2

AO

(5)

2.2.1 

(b) 

2 and 3 minutes ✔✔CA

CA From Q2.2.1 (a) 
2CA modal time   (2)

D 

L2

Ques 

Solution 

Explanation 

T&L

2.2.1 

(c)

Actual train travel time:  
✔RT 
13:24 (day2) to 17:30 (day1) – stopover time 
✔CA 
= 19 hr 54 min – 2 hr 18 min  ✔M 
= 17 hr 36 min = 17, 6 hr  ✔C 

D = S × T 
992 km = S × 17hr 36 min   ✔SF 
S =     992 km                ✔S 
        17,6 hour 
= 56,36 km/h    ✔CA 

OR 

                                                             ✔RT         ✔CA
Total time = 24 hours – 17h30 + 13h24 = 19hr 54 min
                                ✔M 
19hr 54 min – 2 hrs 18 min = 17 hrs 36 min = 17,6 hr  
                                                                               ✔C
 D = S × T     
992 km = S × 17,6 hr      ✔SF 
S =   992 km      ✔S
     17,6 hour 
 ≈ 56 km/h  ✔CA 

OR  

From 17:30 to 00:00 = 6 hrs 30 min ✔RT 
From 00:00 to 13:24 = 13hrs 24 min ✔RT
Time of journey = 19 hrs and 54 minutes  ✔CA 
Travel time = 19 hr 54 min – 2 hr 18 min   ✔M 
 = 17 hr 36 min  

D = S × T 
                               ✔SF 
992 km = S × 17,6 hr 
Average Speed =  992 km   ✔S 
                            17,6 hour   ✔C 
 = 56,36 km/h  ✔CA 

CA From Q2.2.1(a) 
1RT start and end time 
1CA 19 hours 54 min 1M subtracting  stopover time 
1C conversion 
1SF substitution 
1S changing subject of  formula 
1CA simplification 

OR 

1RT start and end time 1CA 19 hours 54 min 1M subtracting  stopover time
1C conversion 
1SF substitution 
1S changing subject of  formula 
1CA simplification 

OR 

1RT start and end times 
1CA trip time 
1M subtracting  stopover time 
1SF substitution 
1S changing subject of  formula 
1C conversion 
1CA simplification

M 

L3

NP

(7)

Ques 

Solution 

Explanation 

T&L

2.2.2 

Forward trip in January: 
Parents = 2 × R560 = R1 120  ✔MA 
Father = R560 – R560 × 25% OR R560 × 75% ✔MA 
= R420 ✔CA 
Children's fare = R560 × 80% = R448 ✔MA 
Two children = 2 × R448 = R896   ✔CA
Total fare for family: R1 120 + R420 + R896 = R2 436  ✔CA  

Return trip in February: 
Parents fare = 2 × R490 = R980  ✔A 
Father = R490 minus R490 × 25% or R490 × 75% 
 = R367,50  ✔A 
Two children = 2 × (R490 – R490 × 50%) 
 = R490   ✔A 
Total fare for return trip = R980 + R490 + R367,50   
 = R1 837,50  ✔CA 
Total cost for both trips = R2 436 + R1 837,50  
 = R4 273,50  ✔CA

OR 

1MA two adult price 
1MA discounted price  for over 55 yrs 
1CA father's fare 
1MA children fare 
1CA total children's  fare 
1CA Jan total  fares 
1A adults Feb fare 
1A senior citizen fare 
1A children Feb fare  
1CA total Feb trip's fare 
1CA total trip fare 
(Note: Max of 6 marks  if only one trip is calculated ; Max of 9  marks for using the same fare for both trip)

OR

Fin 

L3

Ques 

Solution 

Explanation 

T&L

                                  ✔MA               ✔MA 
Father's fare = (R560 + R490) × 75%   ✔M 
= R787,50  ✔CA 
Parents' fare = 2 ×( R560 + 490)   ✔MA 
 = R2 100  ✔CA 
                                        ✔MA                   ✔MA ✔A 
Children's fare = (R560 × 80% + R490 × 50%) × 2  
 = R1 386   ✔CA 
Total fare for both trips = R787,50 + R2 100 + R1 386  
= R4 273,50 ✔CA

1MA adding correct  values 
1MA 75 % 
1M % calculation 
1CA simplification 
1MA adding and  multiplying
1CA simplification 
1MA 80%  
1MA 50% 
1A correct values 
1CA simplification 
1CA total return trip  fare (11)

[47]

QUESTION 3 [31 MARKS]

Ques 

Solution 

Explanation 

T&L

3.1.1 

Capacity of section C = 5 m × 1, 2 m × 15 m  ✓SF 
 = 90 m³   ✓CA 
Capacity of section A = 2 m × 12,5 m × 15 m    ✓SF 
= 375 m³ ✓CA 
Maximum capacity = 90 m³ + 375 m³ + 300 m³   ✓MA 
= 765 m³  

OR 

Maximum capacity = Capacity of section (A + B + C)
          ✓SF                                                     ✓SF 
= 2 m × 12,5 m × 15 m + 300 m³ + 5 m × 1, 2 m × 15 m
          ✓CA                        ✓CA 
= 375 m³ + 300 m³ + 90 m³ ✓MA 
= 765 m³ 

OR 

Volume = 30 m × 15 m × 2 m  ✓SF
 = 900 m³  ✓CA 
Volume beneath C = 5 m × 15 m × 0,8 m 
 = 60 m³  
Volume beneath B = 21 × 12,5 m × 15 m × 0,8 m ✓SF  
 = 75 m³  ✓CA
Maximum capacity = 900 m³ – 60 m³ – 75 m³  
= 765 m³     ✓MA 

1SF correct values 
1CA capacity section C 
1SF correct values 
1CA capacity section A 
1MA adding capacities in  m³ 

OR 

1SF Correct values for A
1SF correct values for C
1CA capacity section A
1CA capacity section C 
1MA adding capacities in  m³ 

OR 

1SF volume 
1CA volume section A 
1SF volume beneath B
1CA volume beneath B 
1MA subtracting volume  in m³(5)

M 

L3

3.1.2

                                                ✓M 
Volume of water = 94% × 765 m³ = 719,1 m³  ✓C 
 = 719 100 ℓ  
 = 719 100 × 1 gallons   ✓C 
            3,785 
 ≈ 189 986,79 gallons ✓CA

OR 

1M calculating % 
1C convert to litres 
1C convert to gal. 
1CA simplification 

OR

M 

L3

Ques 

Solution 

Explanation 

T&L

Capacity (in litres) = 765 m³ × 1 000 = 765 000 ℓ ✓C 
Capacity( in gallons) = 765 000   ✓C
                                      3,785 
 = 202 113,6063 
Volume of water = 94% × 202 113,6063   ✓M
 = 189 986,79 gallons  ✓CA

1C convert to litres 
1C convert to gal. 
1M calculating % 
1CA simplification

NP

(4)

3.1.3 

In 1 hour 2 350 litres of water will flow. 
In 1 day: 24 ×2 350 litres ✓MA 
= 56 400 litres will flow ✓CA
                                                                             ✓M 
In 2½ days amount of water flowing = 2½ × 56 400 litres  
 = 141 000 litres ✓CA
∴ Statement is NOT VALID.  ✓O 

OR 

Time to fill swimming pool = 135 000 ✓MA 
                                               2 350 /h 
 ≈ 57,4468 hours   ✓CA 
57,4468 hrs = 2 days and 9 h 27 min ✓M 
Two and a half days = 2 days 12 hours ✓C
∴ Statement is NOT VALID ✓O 

OR 

Time to fill swimming pool = 135 000 L          ✓MA 
                                              2 350L /h 
 ≈ 57,4468 hours  ✓CA 
                                                     ✓MA 
. Two and a half days = (2 ×24 + 12) hours = 60 hours ✓A
∴ Statement is NOT VALID ✓O 

OR 

1MA using flow rate 
1CA water in 1 day 
1M multiplying 
1CA simplification 
1O conclusion 

OR 

1MA finding time taken 1CA time 
1M splitting calc. hrs 
1C converting two and a  half days
1O conclusion 

OR 

1MA finding time taken 
1CA time 
1MA multiply with 24  and add 12 
1A hours 
1O conclusion 

OR

Ques 

Solution 

Explanation 

T&L

3.1.3 

Time to fill swimming pool = 135 000 L  ✓MA
                                              2 350 L/h 
 ≈ 57,4468 hours   ✓CA
                                            ✓MA       ✓CA 
57,4468 hours ÷ 24 hours/day = 2,3936  
NOT VALID  ✓O 

OR 

                          ✓MA         ✓A 
2½ days × 24 h/d = 60 hours 
✓MA 
Volume of water = 60 hours × 2 350 ℓ/hour 
 = 141 000 ℓ  ✓CA 
This is more than the 135 000 ℓ to be topped up 
The statement is NOT VALID ✓O

1MA finding time taken 1CA time 
1MA dividing by 24 h/d 1CA days 
1O conclusion 

OR 

1MA multiplying with 24  h/d 
1A number of hours 
1MA multiplying hours  with flow rate
1CA simplification 
1O conclusion   (5)

M 

L3

3.2.1 

Total = 18 × 15 = 270  ✓MA 
                             ✓M 
Difference = 270 – 236 = 34 
x = 34 ÷ 2  ✓M
 = 17  ✓CA 

OR 

Mean = 2x + 236 = 15  ✓MA 
                 18 
 2x = 270 – 236 ✓M 
 = 34 
 x = ³⁴/₂ ✓M   
 = 17  ✓CA 

OR 

1MA multiplying 
1M subtracting totals 
1M dividing by 2 
1CA value of x 

OR 

1MA adding correct  values 
1M subtracting totals 
1M dividing by 2 
1CA value of x 

OR

Data 

L3

Ques 

Solution 

Explanation 

T&L

✓M 
Mean =  2x + 236  =   2x  +  13,1111 
                 18              18  ✓M 
15 – 13,1111 = 1,8888... 
2x= 1,8888...  ✓CA 
18 
x = 1,888... × 18 ÷ 2   
 = 17  ✓CA

1M adding correct values
1M mean concept 
1CA manipulating  formula
1CA value of x

AO

(4)

3.2.2

Q₁ = 15 ✓RG     and Q₃ = 20 ✓RG 
IQR = 20 – 15 ✓M 
 = 5 ✓CA 

1RG finding Q1 
1RG finding Q3 
1M subtracting 
1CA IQR value

Data 

L3

AO

(4)

3.2.3 

It is more convenient for them to go in the evening  ✓✓O 

 OR 

During daytime other distractions keep people away. ✓✓O 

OR 

Small groups receive individual attention  ✓✓O 

OR 

Any other sensible reason   ✓✓O 

2O reason  (2)

D 

L4

3.2.4 

✓A 
P(_{Day Group full attendance}) =   6   × 100% 
                                           18✓A
 ≈ 33%✓R 

1A numerator 
1A denominator 
1R whole %

P 

L2

AO

(3)

3.2.5 

The range of the afternoon group was smaller.  ✓✓O 
The afternoon group has a higher median.  ✓✓O 
The afternoon group has smaller inter-quartile range. ✓✓O
Minimum of the afternoon group is higher.  ✓✓O 
(Any TWO acceptable reasons) 

2O reason 
2O reason (4)

D 

L4

[31]

QUESTION 4 [36 marks]

Ques 

Solution 

Explanation 

T&L

4.1.1 

✓MA 
0,21875 miles = 385 yards 
Hence, 1 mile =     385    yards   ✓MA 
                         0,21875 
 = 1 760 yards 

OR 

       1         = 4,571428571 ✓MA 
0,21875 
                  ✓MA
385 × 4,571428571 = 1760 yards 

1MA recognising equal  parts 
1MA correct fraction   

OR 

1MA conversion factor 
1MA multiplying 385  with conversion factor (2)

M  

L2

4.1.2 

Approximately 4,5 miles ✓✓RG
(Accept distances in the range 4,3 miles to 4,7 miles) 

2RG correct distance. (2)

MP 

L2

4.1.3

          ✓RG                  ✓C        ✓CA
700 ft = 700 × 0,3038 m = 212,66 m 
(Accept heights in the range 700 ft to 710 ft) 

1RG correct distance
1C converting to m 
1CA max height

MP 

L2

NP

(3)

4.1.4 

It is uphill. (steep) ✓✓O 

OR 

This runner found it difficult to run uphill.     ✓✓O 

OR 

It is easier to run downhill. ✓✓O

2O reason    (2)

MP 

L4

4.2.1 

✓A     ✓A 
6 + 3 or 9 
[Due to the annexure of Limpopo full marks can be awarded  if only 6 is given as the number of venues] 

2A number of venues (2)

MP  

L2

4.2.2 

Hippo ✓✓A

2A correct enclosure (2)

MP 

L2 

Ques 

Solution 

Explanation 

T&L

4.2.3 

✓✓A 
Zoo is 6 times bigger than the elephant exhibit.
                  ✓M                 ✓CA
∴ 6 × 4 = 24 football fields 
Also accept 5 or 7 as a correct estimation. 
ANSWER ONLY full marks if 20 to 28 football fields. 

2 A estimation 
1M multiplying 
1CA solution 
(Max 2 marks for  number of football fields  for estimated areas of 3,4  ,8 or 9.) (4)

MP  

L4

4.2.4 

                                                    ✓A 
The distance on the map = 85 mm 
                        ✓A         ✓M 
Bar scale 20 mm is 200 m 

Real distance using the bar scale =   85 mm × 200m  ✓M  
                                                           20mm 
          = 850 m  ✓CA
1,6 km = 1 600 m    ✓C 
∴ The scale is NOT correct.   ✓O 

OR 

                      ✓A               ✓M 
Bar scale 20 mm is 200 m 
1,6 km = 1 600 m  ✓C 

Calculated map distance =  1600 m ×   20mm  ✓M 
                                             200m 
 = 160 mm  ✓CA 
Measured distance = 85 mm  ✓A 
∴ The scale is NOT correct. ✓O 
(Accept a range from 82 mm to 87 mm for the distance  between streets and 18 mm to 22 mm for the bar scale.) 

1A measured  distance  
1A measured bar 
1M relating to bar to  measurement 
1M using the given scale
1CA simplification 
1C conversion 
1O conclusion 

OR 

1A measured bar 
1M relating to bar to  measurement 
1C conversion 
1M using the given scale 1CA simplification 
1A measured distance  1O conclusion (7)

MP 

L4

4.3.1 

Saturday ✓✓A

2A correct day (2)

D 

L2

4.3.2 

Monday is NOT reflected on the given graph. ✓✓O

2O reasoning   (2)

P  

L4

Ques 

Solution 

Explanation 

T&L

4.3.3 

The number of visitors increase to about 12:00. on weekdays and then decrease again till 16:00. ✓✓O 

 OR  

The number of visitors on weekends is more than the  visitors on weekdays. ✓✓O 

 OR 

The number of visitors increase to about 13:00 on  weekends and then decrease again till 16:00. ✓✓O
Any TWO trends relating time and number of visitors. 

2O trend 
2O trend  (4)

D 

L4

4.3.4 

The number indicated by the height of the column on  Saturday is a little more than double the height of the  mean number for a Tuesday  ✓✓O 

OR 

People work during the week  ✓✓O 

OR 

Saturdays they go with their families to the zoo.  ✓✓O 

OR 

Cheaper to go during the weekends  ✓✓O 

OR 

More activities at the zoo on Saturday.  ✓✓O 

2O reason
2O reason   (4)

D  

L4

[36]