Symbol

Explanation

M

Method

MA

Method with accuracy

CA

Consistent accuracy

A

Accuracy

C

Conversion

S

Simplification

RT/RG/RM

Reading from a table/Reading from a graph/Reading from a 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/Reason

AO

Answer only

NPR

No penalty for rounding

MARKING GUIDELINE

NOTE:

• If a candidate answers a question TWICE, only mark the FIRST attempt.
• If a candidate has crossed out (cancelled) an attempt to a question and NOT redone the solution, mark the crossed out (cancelled version)
• Consistent accuracy (CA) applies in ALL aspects of the marking guidelines, however it stops at the second calculation error.
• If the candidate presents any extra solution when reading from a graph, table, layout plan and map, then penalise for every extra incorrect item presented.

QUESTION 1 [20 MARKS] INTEGRATED QUESTION

Question

Solution

Explanation/Marks AO: FULL MARKS

Topic/L

1.1

1.1.1

VAT inclusive price means the price that has added VAT value J

2A justification              (2)

F
L1

1.1.2

                                  M      M
% Profit== 0,30 × 100 = 42,86%   CA
                 0,70

1M subtraction for profit
1M division and multiplication by100
1CA       (2)

F
L1

1.1.3

Joy’s profit= 2 × 50 × 0,30  RT

=R 30,00 üCA

1 RT correct values
1CA answer.   (2)

F
L1

1.2

1.2.1

Rate = R96,61      C
               100
= R0,9661 per kWh CA

1C Conversion
1CA rate  (2)

F
L1

1.2.2

Total amount = 96,61 × 50
= 4830,5 cents  M
= 4830,5
     100
=R48,30   CA

OR

Total amount = R 0,9661 × 50      M
= R48,30     CA 
                             NPR

1M multiply by 50
1CA amount
1M multiply by 50
1CA amount  (2)

F
L1

1.2.3

Maximum kWh = 400 - 50           M
= 350 kWh            CA

1M subtracting 50
1CA maximum number (2)

F
L1

1.3

1.3.1

% of energy produced by Others

= 100% ─ (85,7+5,2+3,2+1,7+0,9+0,9) üM

= 2,4% üCA

1M subtracting from 100% all values

1CA simplifying.         (2)

D
L1

1.3.2

                                         M
Natural Gas = 3,2 × 237,006
                      85,7
= 8,85 GWh          CA

1M dividing correct values
1CA simplification and answer
NPR                           (2)

D
L1

1.3.3

Nuclear: Diesel
5,2 : 1,7       RT    A

1RT correct values
1S simplification  (2)

D
L1

1.3.4

Production from coal
= 1 000 000 ×237,006    RT    M
= 237 006 000 KWh

1 RT production
1M multiplication by 1 000 000
1CA answer  (3)

D
L1

[21]

QUESTION 2 [27 MARKS] FINANCE

Question

Solution

Explanation/Marks

T/L

2.1

2.1.1

Number of copies 2A

2A number of copies

(2)

F L1

2.1.2

ANNEXURE C: GRAPH

F L2

1 mark for starting point (0;500)
1 mark any other correct point plotted correctly
1mark for the straight line    (3)

2.1.3

400 copies    RT

2RT                            (2)

F
L2

2.1.4

From graph for 600 copies:
on Option S cost= R1200 RT
on Option M cost= R800  RT
difference in cost = 1200-800= R400 A

OR

For S= 600 × 2 = R1 200 M
For M = 500 + 600 × 0,50 = R800 M
Difference= 1 200 – 800 = R400CA

1RT for value of S
1RT for value of M
1A the difference R400

1M for R1200
1M for R800
1A for R400             (3)

F
L3

2.2

2.2.1

= 7 880 970      
% difference in Energy
=8 145 975−7 880 970 × 100% SF
         7 880 970
= 265 005 × 100%
  7 880 970
= 3,36% S
= 3%R

1M adding 368182
1CA answer
1 SF substitution of correct values
1S simplification
1R    (5)

L3

2.2.2

=1 550 794,505 S
Projected income = 8 382 673 + 1 550 794,505        M
= R9 933 476,505
= R 9 933 476,51 CA

OR

Projected income =1,185 × 8 382 673 M
= R9 933 467,505 S
= R9 933 467,51    CA

1M multiplication by 0,185
1S simplification
1M addition
1CA answer
1M adding the increase
1M multiplication
1S simplification
1CA answer  (4)

F
L3

2.2.3

                       A
Probability = 4
                     6
= 0,6666 S

1A numerator
1A denominator.
1S simplified   (3)

F L2

2.3

Deposit at ATM = R4,80 + 1,2 × 5000 SF
                                         100
= ?64,80       S
Deposit at Branch = R8,00 + 1,5 × 5000
                                             100
= ?83,00    CA
Difference = 83,00-64,80 M
= R18,20
Statement valid.  A

1SF substitution
1S simplification
1CA answer for branch deposit
1M subtraction
1CA answer   (5)

F
L4

[27]

QUESTION 3 [15 marks]

Question

Solution

Explanation/Marks AO: FULL MARKS

Topic/L

3.1

43+21+149+72+34+20+32+11+83M
= 465CA

1M addition
1CA answer         (2)

D L1

3.2

149; 83; 72; 43; 34; 32; 21; 20; 11RT A

1RT reading all the values
1A correct order (2)

D L1

3.3

20:40   RT   RT
=1:2CA

1RT for 20
1RT for 40 1CA answer in simplified ratio    (3)

D L2

3.4

675 - (30 + 175 +19+17+140+182+12+40)M
=60 CA

1M addition
1CA answer        (2)

D L1

3.5

Total schools in NW = 32+182
= 214        M
Total schools in SA = 465+675
= 1140 CA
% age                  = ??? × ???%   M
                              ????
= ??, ??%CA

1M addition and total schools in NW
1CA total schools in SA adding the value from 3.1 and 675
1Mmultiplication; a fraction and 100%
1CA answer    (4)

D L3

3.6

Median = 34         RT
The province is Limpopo          CA

1RT median value
1CA province       (2)

D L2

[15]

QUESTION 4 [23 marks]

Ques.

Solution

Explanation

Lev el

4.1

Value of A = 222 – 121     MA 
=   101   A
Value of B = 123 – 59
= 64     A

OR

Value of B = 406 – 121 – 103 – 76 – 42
= 64     A
Value of C = 222 + 103 + 95 + 154 + 123 + 75 M
= 772     CA

OR

Value of C
= 121 + 101 + 103 + 95 + 76 + 78 + 64 + 59 + 42 + 33    MCA
= 772 CA

1MA Subtract correct values
1A Value of A
1A Value of B

1M Adding ALL values
1CA Value of C

1MCA Adding ALL values CA from A and B
1CA Value of C  (5)

D L2

4.2

Trend – From Grade 8 to Grade 12 the number of male learners decreases       A
Reason – Male learners drop out. A

OR

Male learners fail the grades A
Accept any other relevant reason

CA from 4.1.1
2A Trend
2A Reason (4)

L4 D

4.3

No of teachers = 772 A
                            35
= 22,057…S
≈ 22 CA
Statement not valid O

CA from 4.1
1M Dividing by 35
1S Simplification
1CA No of teachers
1O Not valid (4)

L4 D

4.4

Probability (Grade 8 or Grade 9 female)
= 121+103 ×100%
      772
= 224          A
  772         MCA
= 29% CA

CA from 4.1
1A Numerator
1MCA Denominator
1CA Percentage    (3)
NPR

L2 P

No of learners in 2019 = 772 × 1,03
= 796 CA

Term 1 = 796 × 3,18 × 51MCA
= R129 095,28 CA

Term 2 = 796 × 3,18 × 46
= R116 438,88 CA

Total Amount = R129 095,28 + R116 438,88 CA
= R245 534,16 CA
Statement is valid O

QUESTION 5 [15 marks] 

5.1

R43 500 ×12 M
= R522 000 CA

1M multiply by 12
1CA annual income    (2)

F L2

5.2

Pension Fund = R43 500 × 7,5%
= R3 262,50 CA
= R3 262,50 × 12
= R39 150 CA
Taxable Income = R522 000 – R39 150
= R482 850 CA
R482 850 = 110 739 + (482 850 – 467 500) × 36%
= 110 739 + 5526
= 116 265 – 15 714     CA
= 100 551
       12   
= 8379,25 – 888    M
= 7 491,25       CA

1CA pension value
1CA annual pension fund
1CA subtracting Pension Fund
1CA subtracting rebate
1M subtracting MTC
1CA monthly tax  (6)

F L3

5.3

                        A
95, 98 ,99 ,100 ,101,101,102,103,105,107,110,111,114,115,121
103 is the median CA

1A arrangement
1CA median   (2)

D L2

5.4

Q1= 100 CA
Q2 = 103 CA
Q3 = 111 CA
IQR = 111 – 100 M
= 11 CA

1CA Q1
1CA Q2
1CA Q3
1M subtraction
1CA IQR   (5)

D L2

[15]

TOTAL:

100