Symbol

Explanation

M

Method

MA

Method with accuracy

CA

Consistent accuracy

A

Accuracy

C

Conversion

S

Simplification

RT

Reading from a table/graph/document/diagram

SF

Correct substitution in a formula

O

Opinion/Explanation

P

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

R

Rounding off/

NPR

No penalty for rounding

AO

Answer only

MCA

Method with consistent accuracy

RCA

Rounding consistent with accuracy

QUESTION 1 [30 MARKS] AO

Q

Solution

Explanation

T&L

1.1.1

Numerical data A

2A correct identification

(2)

D L1

1.1.2

Modal allowance
= R1 780 A

2A mode

(2)

D L1

1.1.3

R1 715; R1 715; R1 695; R1 695; R1 695; R960; R405 üüA

2A descending order Accept the names

(2)

D L1

1.1.4

Increase in rand
RT
R1 780 – R1 695
= R85,00 A

1RT correct 2 values 1A simplification

(2)

F L1

1.1.5

Pension allowances older than 75 A
War veteran allowances A

1A correct allowance
1A correct allowance

2)

D L1

1 kg = 1 000 g
?    = 400 g
∴ Quantity in kg =  400  g MA
                             1000
= 0,4 kg A

OR
400 g =  400  kg MA
            1 000
= 0,4 kg    A

OR
MA 400 g = 400 × 0,001kg
= 0,4 kg    A

1MA dividing by 1 000 1A amount in kg

OR
1MA dividing by 1 000 1A amount in kg

OR
1MA multiply by 0,001 1A amount in kg

NPU (2)

RT Profit = R14,30 – R10,99 M
= R3,31 CA

1RT correct values
1M subtracting values
1CA simplification
(3)

Number of packets
2,5 kg × 1000 MA
               250   M
= 10 packets CA

OR
2,5 kg    C
0,25 kg   M
= 10 packets üCA

OR
250g : 2,5kg MA
250g : 2500g C
1: 10
= 10 packets üCA

1MA multiply by 1 000 1M dividing by 250g 1CA simplification

OR
1C converting into kg 1M dividing by 0,25 kg 1CA simplification

OR
1MA ratio concept
1C conversion to same unit
1CA simplification  (3)

Selling price
R29,20 MA
    8     
 =R3,65 CA

OR
2 kg = 0,25kg
  8
∴ 2kg = R29,20
0,25 kg = 0,25 x R29,20
                         2              MA
= R3,65 CA

1MA dividing correct value by 8
1CA simplification (only if dividing by 8 or correct value used)

OR 

1MA dividing by 2 AND multiply by 0,25
1CA simplification (2)

1.3.1 (a)

2A correct value (2)

1.3.1 (b)

2A correct value (2)

Difference
RT 84% – 64%
= 20% üCA

1RT both correct values 1CA simplification (2)

2A correct value (2)

Probability
= 20% OR 0,2 OR 20 OR 2  OR 1 
                             100      10       5

OR
unlikely

OR
less likely  A

2A correct value/word  (2)

QUESTION 2 [42 MARKS]

Q

Solution

Explanation

T&L

2.1.1

Market value
= R944 630,00
Nine hundred and forty four thousand six hundred and thirty rand. A

2A correct value in words
NPU (2)

F L1

2.1.2

Amount of VAT
R836,02 × 15   MA
                 100

= R125,40 CA

OR
R836,02 × 1,15 MA
= R961,42
R961,42 – R836,02
= R125,40 CA

1MA correct value ×  15  
                                 100
1CA simplification

OR
1MA correct value × 1,15
1CA simplification

(2)

F

L1

2.1.3

Litres OR ℓ A

2A correct unit Accept dm ³
(2)

F L1

2.1.4

Monthly sewer charge
A = R378,95 A

2A correct charge (2)

F L1

2.1.5

Total water charge
MA            RT
B = (6 × R8,28) + (4 × R8,79) + (2 × R15,00)
= R49,68 + R35,16 + R30,00 M
= R114,84 CA

1MA identify 6, 4, 2
1RT identify R8,28; R8,79; R15,00
1M adding (at least 2 correct values)
1CA simplification
(4)

F L2

2.2.1

Inverse proportion A
OR
Indirect proportion

2A type of proportion  (2)

F L1

2A correct number (2)

Amount per person  RT
= R3 000,00
           7           MA
= R428,57 CA

1RT correct cost (R3 000)
1MA dividing by 7
1CA simplification
(3)

2.2.4

(a)

R17 000,00   MA
  R500,00
= 34 months CA

1MA dividing by R500,00
1CA simplification

AO (2)

2.2.4

(b)

Interest rate
= 8,30% A

2A correct interest rate (2)

2.2.4

(c)

Interest for 1 year
= R17 000,00 × 8,30  M
                          100
Interest for 3 years
= R1 411,00 × 3
= R4 233,00 CA
= R4 200,00 R

OR
Interest earned for 3 years
R17 000,00 × 8,30 x 3 M
                       100
= R4 233,00 CA
= R4 200,00 R

CA from Question 2.2.4 (b)
1M interest calculation

1CA simplification
1R rounding

OR
1M interest calculation
1CA simplification
1R rounding (3)

2.2.4

(d)

Percentage point difference
8,46% − 7,76% RT
= 0,7% CA

1RT correct values
1CA simplification AO (2)

2.2.4

(e)

RT
18 months
A            A
= 1 year and 6 months

1RT reading from table
1A number of years
1A number of months AO
(3)

RT
R242 700 million A

OR
RT
R242 700 000 000 A

1RT correct value (2 427)
1A number in millions NPU 

(2)

Total income received:
1 370 + 242,7 + 180,3 + 31,5 MA
A = 1 824,5 CA

1MA adding ALL correct values
1CA simplification
NPU (wrote billions or rands)
AO (2)

Other   RT
1 823,72 – (278,4+262,4+222,6+211,0 +209,2+208,5+ 202,2 +112,7)
M
B = 1 823,72 – 1 707 MA
= 116,72 CA

1RT reading correct values
1M adding all the values
1MA subtracting from total
1CA value of B
NPU (4)

Community development
RT
=    R208,5   × 100% M
   R1 823,72
= 11,43267607%   CA

ACCEPT ONLY FOR AFRIKAANS CANDIDATES:
Social developmentRT
=    R278,4     × 100% M
    R1 823,72
= 15,26550128%    CA

1RT both correct values
1M percentage calculation
1CA simplification
1RT both correct values
1M percentage calculation
1CA simplification
NPR
(3)

QUESTION 3 [26 MARKS]

Q

Solution

Explanation

T&L

3.1.1

Volume = It is the amount of solids or liquids an object can take/hold.  A
OR
Volume is the amount of space occupied by an object

2A explanation

(2)

M L1

3.1.2

Volume = side × side × height   C
= 0,5 m × 0,5 m × 0,08 m SF
= 0,02 m³ CA

OR
20 000 cm³    SF
1000 000
50 cm × 50 cm × 8 cm
= 0,02 m³  C   
CA

1SF correct substitution
1C conversion
1CA simplification

OR
1 SF correct substitution
1C conversion
1CA simplification

(3)

M L2

3.2.1

Area of one block = length × breadth
= 50 cm × 50 cm SF
= 2 500 cm²
Area of 12 blocks = 0,25 m² × 12  MA
= 3 m² CA

OR
Area of one block = length × breadth
= 0,5 m × 0,5 m SF
= 0,25 m²
Area of 12 blocks = 0,25 m² × 12 MA
= 3 m² CA

OR
Area of 12 blocks   = 12 × (side × side)
= 12 × (0,5 m × 0,5 m) SF
= 12 × 0,25 m² MA
= 3 m² CA

OR
Area of 12 blocks   = 12 × (side × side)
= 12 × (50 cm × 50 cm) SF
= 12 × 2 500 cm² MA
= 3 m² CA

CA from Question 3.1.2
1SF substituting correct values
1MA multiply by 12
1CA answer in m²

OR
1SF substituting correct values
1MA multiply by 12
1CA answer in m² 

OR

1SF substituting correct values
1MA multiply by 12
1CA answer in m²

OR
1SF substituting correct values
1MA multiply by 12
1CA answer in m²
(3)

M

L2

Area of walkway SF
4,05 m × 1,45 m
= 5,8725 m² A

Area to be covered with pebbles
= 5,8725 m² – 3 m² MCA
= 2,8725 m² CA

OR
Area to be covered with pebbles SF
(4,05 m × 1,45 m) – 3 m²  A
= 5,8725 m² – 3 m² MCA
= 2,8725 m² üCA

OR
Area of walkway SF
405 cm × 145 cm
= 58 725 cm² A

Area to be covered with pebbles
= 58 725 cm² – 30 000 cm²   MCA
= 28 725 cm² CA

OR
Area to be covered with pebblesSF
(405 cm × 145 cm) – 30 000 cm²  A
= 58 725 cm² – 30 000 cm²  MCA
= 28 725 cm² CA

CA from Question 3.2.1

1SF substitution
1A simplification
1MCA subtracting area of blocks
1CA simplification

OR
1SF substitution
1A simplification
1MCA subtracting area of blocks
1CA simplification

OR
1SF substitution 1A simplification
1MCA subtracting area of blocks
1CA simplification

OR
1SF substitution
1A simplification
1MCA subtracting area of blocks
1CA simplification 
NPR (4)

5,7 m²
0,36 m²  MA
= 15,833  CA
= 16 bags of pebbles  RCA

1MA dividing by 0,36 m²
1CA simplification
1RCA rounding
(3)

Length of large window frame
890 mm MA
    10
= 89 cm CA

1MA dividing by 10
1CA simplification 
AO
(2)

Perimeter  MA
= 18,5 cm + 18,5 cm + 18,5 cm  + 18,5 cm
= 74 cm CA 

OR
Perimeter
= 4 × 18,5 cm   MA
= 74 cm CA

AFRIKAANS ONLY OMIT SUB QUESTION 3.3.2 −
UPSCALE FROM 24 TO 26

1MA adding 4 sides
1CA simplification

OR
1MA side multiplied by four
1CA simplification
(2)

Diameter = 1,85 cm × 2
= 3,7 cm    A
18,5 cm
  3,7cm   M
= 5 beads    CA

1A diameter
1M dividing by diameter
1CA simplification
(3)

MA
2 × 18,5 cm = 3   of the width of the large window
                       4
A 
37 cm = 3 of the width of the large window
             4
Width of large window
=37 cm × 4 MA
                3
= 49,33 cm CA

1MA multiply 18,5 by 2
1A simplification
1MA multiply with inverse
1CA simplification
NPR
(4)

QUESTION 4 [24 MARKS]

Q

Solution

Explanation

T&L

4.1.1

Camping,swimming, dining(eating)and
checking-in (enquiries/registration/making payments).

4A 4 correct activities (4)

MP L1

4.1.2

Umngeni  RT

2RT reading from map (2)

MP L1

4.1.3

5 restaurants RT

2RT reading from map (2)

MP L1

4.1.4

Bar Scale A

2A correct scale
Accept: Line scale (2)

MP L1

4.1.5

A
4,2 cm  = 4 km
1 cm      =  0,9524 km M 
MA
∴10 cm = 9,524 km
≈ 10 km CA

OR
10 cm  x 4 km M
4,2 cm         MA
A 
= 9,524 km
≈ 10 km CA

OR
A
2,1 cm  = 2 km
1 cm      =  0,9524 km M
MA
∴10 cm = 9,524 km
≈ 10 km CA 

OR
10 cm x  2 km MA
2,1 cm
A
= 9,524 km
≈ 10 km CA

1A measure bar scale
1M concept of scale
1MA multiply by scale
1CA conversion

OR
1A measure bar scale
1M concept of scale
1MA multiply by scale
1CA conversion

OR
1A measure bar scale
1M concept of scale
1MA multiply by scale
1CA conversion 

OR

1A measure bar scale
1M concept of scale
1MA multiply by scale
1CA conversion 

Accept 4,1 cm – 4,3 cm
Accept 2 cm – 2,1 cm
(4)

MP L2

Total distance
= 10 km × 2
= 20 km MA
Time =  20 km    SF
           30 km/h
C Time = 0,6666666667 hours × 60
= 40 minutes CA

OR
Time =   10 km     SF
           30 km/h
= 0,3333   C
∴ In minutes = 0,3333 × 60
= 20 minutes MA
∴ Total time  = 20 × 2
= 40 minutes CA

1MA total distance (20 km)
1SF correct substitution
1C conversion
1CA simplification

OR
1SF correct substitution
1C conversion
1MA simplification
1CA simplification
(4)

2A number of doors
Accept 3 (2)

RT      RT
Bedroom 1, Bathroom and Bedroom 2
OR ONLY AFRIKAANS CANDIDATES:
RT          RT
Slaapkamer 1, Kombuis

1RT first room
1RT other 2 rooms

OR
1RT bedroom 1
1RT kitchen (2)

⁰/₂ OR 0 OR 0%
OR    
Impossible

2A probability (2)

QUESTION 5 [28 MARKS]

Q

Solution

Explanation

T&L

5.1.1

Questionnaires OR Interviews OR Survey OR
Document analysis OR Research OR Observation

2A means of collecting data
(2)

D L1

5.1.2

% Yard trimmings
MA
= 100% − (3,4% + 11,2% + 49,7% + 3,3% +9,0%)
= 100% − 76,6% M
= 23,4% CA

1MA adding all correct values
1M subtracting from 100%
1CA simplification
AO (3)

L2

5.1.3

% Textiles
= 11,2% − (1,6% + 2,3% + 2,9% + 1,7%)
= 11,2% − 8,5% MA
= 2,7% CA

1MA subtracting from 11,2%
1CA simplification
AO (2)

L2

Tons of plastic  RT
91 160 000 x 3,4  MA
                     100
= 3 099 440 tons CA 

OR
RT
91,16 x   3,4   MA
              100
= 3,09944 million tons CA

1RT correct total
1MA multiply by 3,4%
1CA simplification

OR
1RT correct total
1MA multiply by 3,4%
1CA simplification
NPR
(3)

D

5.1.5

Cans, pieces of a motor vehicles, household appliances;
scrap metal OR any other product that includes metal

2A metal products that are recyclable
(2)

D L1

2A type of graph (2)

Probability/Other= 11,2%
RT    MA
1,7% + 1,6% + 2,3% + 2,9% = 8,5%
 8,5 M
11,2
= 0,7589285 CA

OR
A RT
1 −  2,7  MA
     11,2
= 0,7589285 CA

1RT correct values
1MA adding all values
1M dividing
1CA simplification

OR
CA from Question 5.1.3
1RT correct values
1A for the number one
1MA subtracting
1CA simplification
NPR
(4)

2A correct number (2)

Number of seats   A
33 : 27 M
= 11 : 9 CA

1A correct values
1M ratio in correct order
1CA simplified ratio
Accept unit ratio
or fractional form
(3)

2RT reading from table (2)