MATHEMATICAL LITERACY PAPER 2
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM

NOVEMBER 2019 

Symbol

Explanation

M

Method

MA

Method with accuracy

CA

Consistent accuracy

A

Accuracy

C

Conversion

S

Simplification

RT

Reading from a table/a 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

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 guideline; however it stops at the second calculation
  • If the candidate presents any extra solution when reading from a graph, table, layout plan and map, then penalise for each extra item
  • As a general marking principle, if a candidate has incurred one mistake and there is evidence of sound mathematics thereafter, then that candidate should lose one mark

QUESTION 1 [39 MARKS]

Q

Solution

Explanation

T&L

1.1.1

A      A
Bothaville and Viljoenskroon.

1A Bothaville
1A Viljoenskroon (2)

MP L2

1.1.2

A         A 
South West and South.

2A SW
2A S (any order)
(4)

MP L2

1.1.3

Bloemfontein Welkom NAMPO
= 152 km + 75 km = 227 km  A 
Bloemfontein Bultfontein NAMPO
= 100 km + 120 km = 220 km  A
via Bultfontein. O

OR
Bloemfontein – Welkom – NAMPO
220 km – 75 km = 145 km A
Bloemfontein – Bultfontein – NAMPO
220 km – 120 km = 100 km A
via BultfonteinüO

OR
Bultfontein to NAMPO = 120 km A
Bloemfontein to Bultfontein = 100 km A
120 km + 100 km = 220 km A

OR
Bloemfontein to/tot NAMPO= 220 km A
220 km – 100km to Bultfontein = 120 km A
120 km is the distance to NAMPO A 

OR
Bloemfontein to NAMPO = 220 km A
Bultfontein to/tot NAMPO = 120 km A
Bloemfontein to Bultfontein = 220 km – 120 km
= 100 km    A

OR
Nampo Park to Bothaville = 15 km
Bothaville to/tot Bultfontein = 105 km A
Nampo Park to Bloemfontein  A
= 15 km + 105 km + 100 km
= 220 km A

1A correct value
1A correct value
1O conclusion

OR
1A correct value
1A correct value
1O conclusion

OR
1A correct value
1A correct value
1A conclusion

OR
1A correct value
1A correct value
1A conclusion 

OR
1A correct value
1A correct value
1A conclusion

OR
1A correct value
1A correct value
1A conclusion

MP L4

1.1.4

Distance = speed x time
150 km = 88 km/h x time SF 
Time = 150 h  S
             88
= 1,7045…
= 1h 42 min   C
M
Arrival time = 18:45 + 1h42 min
= 20:27   CA
NOT CORRECT   O

OR
From 18:45 to 20:00 is 1 hour 15 min = 1,25 hour
Distance/ = speed x time 
= 88 km/h × 1,25h   SF
= 110 km    S
His timing is not correct, he is not yet in Sasolburg

OR
From 18:45 to 20:00 is 1 hour 15 min = 1,25 hour
Distance    = speed x time
150 km = speed × 1,25h    SF
Speed = 150  km/h   S
               1,25
O = 120 km/h
He is wrong, he will have to drive faster to get to Sasolburg on time.

OR
Distance = speed x time 
150 km = 88 km/h x time  SF
Time = 150 h        S
            88
= 1,7045…
= 1h 42 min     C  MA
From 18:45 to 20:00 is 1 hour 15 min = 1,25 hour
INCORRECT

1SF correct values into formula
1S changing subject of formula
1C conversion
1M adding
1CA arrival time
1O verification

OR

1M subtracting time
1A elapsed time
1C conversion
1SF into correct formula
1S simplification
1O verification

OR
1M subtracting time
1A elapsed time
1C conversion
1SF into correct formula
1S changing subject of formula
1O verification

OR
1SF into correct formula
1S changing the subject of the formula
1C conversion
1MA subtracting
1A elapsed time
1O verification
(6)

M L4
1.2.1

Volume of a rectangular prism
= length x width x height
= 300cm x 68,5 x 40cm    SF
= 822 000 cm3 or  822 ℓ
Capacity = 485 ℓ = 485 000 cm3 C
Volume of the concrete (in cm3)
= 822 000 – 485 000  MA
= 337 000 üCA

1C m to cm
1C mm to cm
1SF substitution
1A volume
1C conversion
1MA subtracting capacity
1CA concrete volume
(7)

M L3
1.2.2

Number of cows = 485  MA
                               56
= 8,66   A
CORRECT  O

OR
Volume = 56ℓ x 8 MA
= 448 ℓ A
CORRECT  O

OR
Volume per cows  = 485l MA
                                  8
= 60,625 ℓ   A
CORRECT  O

OR
56 × 8 ×1000 cm3  MA
= 448 000 cm3 A CORRECT

1MA dividing by 56
1A simplification
1O conclusion

OR
1MA multiplying by 8
1A simplification
1O verification

OR
1MA division by 8
1A simplification
1O verification

OR
1MA multiplying by 8; 1 000
1A simplification
1O verification
(3)

M L4
1.2.3

Volume = 485 = 242,5 ℓ  MA
                  2
Time =  242,5𝑙  MA
          14,5𝑙/min
= 16,724…
≈ 17 min  R

OR
Time to fill
= 485 ℓ ÷ 14,5 ℓ/min  MA
= 33,44827586 min
Time for half empty
= 33,44827586 min ÷ 2  MA
= 16,72413793
≈ 17 R

1MA dividing by 2
1MA dividing by rate
1R time

OR
1 MA dividing by rate
1MA dividing by 2 1
1R time (3)

M L2
1.3.1 9,2 m    A

2A estimated distance [accept answers in the range
9,0 m to 9,5m (2)

MP L2
1.3.2

Measured distance= 174 mm A
Distance from stand 10 to 17 = 4,5 x 7 + 5 = 36,5 A
Scale
174 mm : 36,5 m M
= 174 mm : 36 500 mm
≈  : 209,8  CA

OR
Measured distance= 174 mm A
Distance from stand 10 to 17 = 4,5 x 7 + 5 = 36,5 A
Scale
17,4 cm = 36,5 m
1 cm  = 2,0977011…m M
1 cm = 2,1 m    CA

1A measurement (as per province)
1A distance
1M concept of scale
1CA simplified scale

OR
1A measurement (as per province)
1A distance
1M concept of scale
1CA simplified scale
[accept measured answers in the range
± 2 mm from province measurement]
(4)

MP L3
1.3.3

4 m x 4 m = 16 m2 is R22 942.
1m2 = 22 942 = R1 433,88
             16
Area stand 26
= 4 m x 4,5 m = 18 m2
Cost = R1 433,88 x 18 m2  M
= R25 809,84   CA
NOT VALID   O

OR
Area stand 26
= 4 m x 4,5 m = 18 m2   RT
Cost = 22 942 x 18  M
              16        MA
= R25 809,75   CA
NOT VALID   O

OR
4 m × 4 m = 16 m2 is R22 942
Stand  26 = 4 m × 4,5 m RT
Cost of stand 26
= R22 942 ÷ 4 × 4,5
= R25 809,75    CA
NOT VALID 

OR
4 m × 4 m = 16 m2 is R22 942
1m2 = 22 942
             16
= R1 433,88  MA
4 m × 4,5 m = 18 m2  RT is R25 000
1m2 = 25 000
              18
= R1 388, 89  CA
R1 433,88 ≠ R1 388,89
NOT VALID

1MA unit price
1RT dimensions of stand 26
1M multiply by 18
1CA simplification
1O conclusion

OR
1RT dimensions of stand 26
1MA divide by 16
1M multiply by 18
1CA simplification
1O conclusion

OR
1RT dimensions of stand 26
1MA divide by 4
1M multiply by 4,5
1CA simplification
1O conclusion

OR
1MA unit price
1RT dimensions of stand 26
1M divide by 18
1CA simplification
1O conclusion NPR
(5)

F L4
    [39]  

   

QUESTION 2 [38 MARKS]

Q

Solution

Explanation

T/L

2.1.1

Mean = R287 240 000 000  C
                    148 266                     MA
= R1 937 328,855 per year
Monthly mean = R1 937 328,855 ÷ 12  MA 
= R161 444,07 CA
INCORRECT   O

OR
 Mean = 287 240 000 000
                     148 266        MA
= R1 937 328,855 per year
Then: R161 000 × 12 = R 1 932 000 per year
INCORRECT

OR
Total monthly income of millionaires
= 161 000 × 148 266    MA
= R23 870 826 000
Total annual income
= R23 870 826 000 × 12  MA
= R286 449 912 000 CA    C
Total taxable annual income is R287,24 billion
INCORRECT 

OR
Income per year per person
= R161 000 × 12   MA
Total income per year
= R1 932 000 × 148 266  MA    CA
= R286 449 912 000 = R286,449912 billion
≠ R287,24 billion
INCORRECT

OR
Income per year per person
= R0,161 million × 12    MA 
Total income
= R1,932 mil × 148 266    MA
= R286 449,912 mil  CA
= R286,449912 billion C
≠ R287,24 billion
INCORRECT   

OR
Income per year per person
= R0,000161 billi × 12 MA
Total income
= R0,001932 billion × 148 266 MA
= R286,449912 billion CA
≠ R287,24 billion 
INCORRECT  O

1C billion to rand
1MA dividing by 148 266
1MA dividing by 12
1CA monthly income
1O conclusion

OR
1C billion to rand
1MA dividing by 148 266
1MA multiply by 12
1CA yearly income
1O conclusion

OR
1MA multiply by 148 266
1MA multiply by 12
1CA yearly income
1C billion to rand
1O conclusion

OR
1MA multiply by 12
1MA multiply by 148 266
1CA yearly income
1C billion to rand
1O conclusion

OR
1MA multiply by 12
1MA multiply by 148 266
1CA yearly income
1C billion to rand
1O conclusion

OR

1C billion to rand
1MA multiply by 12
1MA multiply by 148 266
1CA yearly income
1O conclusion
(5)

D L4

2.1.2

Number = 148 266 ×    100      × 148 266 
                                105, 0065   1,050065 A
= 141 196,97
≈ 141 196 or 141 197  CA

1MA dividing
1A 105,0065%
1CA simplification
(3)

D L3
2.2.1

Medical scheme tax rebate
= R310 × 2 × 12 MA
= R7 440  CA

1RT correct value
1MA multiplying
1CA simplification AO (3)

F L2
2.2.2

Tax payable
= R532 041 + 45% (R2 045 364 – R1 500 000)
= R777 454,80 S
Tax after rebate
= R777 454,80 – R14 067 – R7 MA
= R755 674,80
Tax payable
= R755 674,80 – R7 440    MCA
= R748 234,80      CA

CA from Q2.2.1
1A correct tax bracket
1A for 2 045 364
1SF correct substitution
1S simplification
1M subtracting rebates
1MA both correct values
1MCA subtracting MST rebate
1CA tax
(8)

F L3
2.3.1

Earning in Euro = 600 000 MA
                                7,47
= 80 321,28514 A
MCA Earning in rand = 80 321,28514 × 15,64   CA
= R1 256 224,90

OR
Conversion ratio
15,64 = 2,093708166
7,47
Earning = Kr600 000 × 2,093708166
= R1 256 224,90      CA

OR
R15,64 = Kr7,47   M
R2,0937… = Kr1 A
Kr600 000 × R2,0937… M
= R1 256 224,90  CA

1MA dividing by euro
1A simplification
1MCA multiplying
1CA value

OR
1MA dividing by euro
1A simplification
1M multiplying
1CA simplification

OR
1M equation the rates
1A unit ratio
1M multiplying
1 CA simplification (4)

F L3
2.3.2

Total deductions
= Kr229 760 + Kr48 000 + Kr37 200r
= Kr314 960   A
Percentage = Kr314960 M × 100%
                      Kr600 000
≈ 52,49%    CA 
VALID

OR
Total deductions
= Kr48 000 + Kr37 200 + Kr229 760
= Kr314 960  A

Amount = Kr600 000 × 52%  M
= Kr312 000     CA
VALID     O

OR
220 760 + 48 000 + 37 200 = 314 960
To Euro = 314 960 ÷ 7,47 = €42 163,32
To rand = €42 163,32 × R15,64
= R659 434,32      A
Percentage = 𝑅659 434,32 × 100 % M
                     𝑅1 256 224,98
= 52,493%
= 52%    CA
VALID   O

1A total deductions
1M percentage calculation
1CA simplification
1O conclusion

OR
1A total deductions
1M percentage calculation
1CA simplification
1O conclusion

OR
1A total deductions
1M percentage calculation
1CA simplification
1O conclusion
(4)

 F L4
2.4.1 United States of America 2A correct country  D L2
2.4.2

P = 2/23
= 0,08695652174
≈ 0,087   R

1A numerator
1A denominator
1R correct form

P L2
2.4.3(a) Q2 = 40 2A median D L2
2.4.3(b)

Q1 = 33 A
Q3 = 45 A
IQR = 45 – 33 MCA
= 12
CORRECT

1A quartile 1
1A quartile 3
1MCA IQR with at least one correct value
1O verification
(4)

D L4

 

QUESTION 3 [35 MARKS]

Q

Solution

Explanation

T&L

3.1.1 

MA
Rate per h = R31 050 = R1 725/h
                        18
Rate per min = R1 725 = R 28,75/min
                            60

OR
Rate per 18 hours
= R31 050  = R517,50 /18 h    MA
       60
Rate per min = R517,50   M
                             18
= R28,75/min  CA

OR
18 hours × 60 = 1 080 minutes  MAM               
CA
Solo rate = 31 050 = R28,75/min
                   1 080

1MA dividing by 18
1M dividing by 60 1CA rate

OR
1MA dividing by 60
1M dividing by 18
1CA rate

OR
1MA conversion to minutes
1M dividing by 1 080
1CA rate
AO
(3)

F L2

3.1.2

Cost
= 28 × R2 050 + R31 050 + 15 × R1 242 + R700 + R6 544 + 7 × R190
                                             3
= R57 400 + R31 050 + R6 210 + R700 + R6 544 + R1 330
= R103 234   
CA

1MA multiplying cost by hours
1MA theory lesson cost
1MA number of exams by cost
1M adding ALL values
1CA simplification
(5)

F L3

3.2

Interest 1st year = R90 000 × 8,5%  MA
= R7 650    A
Balance year 1 = R 90 000 + R7 650
= R97 650    CA
Interest 2nd year = R97 650 × 8,5%
= R8 300,25      CA
Balance at end of 2nd year
= R97 650 + R8 300,25
= R105 950,25     CA
The amount is ENOUGH

OR
The amount is increasing by 108,5%  MA
Balance at the end of the second year
= R90 000 × 108,5% × 108,5%
=R105 950,25    CA
The amount is ENOUGH

1MA multiplying by the %
1A 1st year interest
1CA 1st year balance
1CA 2nd year interest
1CA 2nd year balance
1O conclusion CA from 3.1.2

OR
2MA percentage increase
1MA multiplying for 1st year
1MA multiplying for 2nd year
1CA simplification
1O conclusion CA from 3.1.2
(6)

F L4

3.3.1

Students study more after failing/ more serious about their work.

OR
They have seen what the tests look like and prepare better for
following tests/ gained experience. 

OR
They have more time to prepare/ more practice/ attended extra classes. 

2O reason D L4
3.3.2

24 is 20%
A is 80%  MA
A = 24 × 4 = 96       A
20% of B = 24
Or
24 ÷ 20% = 120
A = 120 – 24 = 96
B =  24 = 120 or B = 96 + 24 = 120
      20%
C = A = 96  CA
D = 96 – 67 = CA  or D = 30% × 96 = 28,8
= 29      
Total that passed
= 24 + 29 = 53      CA
Or
67 ÷ 70% = 95,7≈ 96
D = 96 – 67 = 29

OR                  
A = 80% × 24 MA      
      20%
= 96 
Or
A: 20% = 24
1% =  24  = 1,2   MA
         20%
80 ×1,2 = 96 A
B = 100% × 24
       20%                      
= 120   CA
C = 100% × 67  = 95,71 ≈ 96     CA
       70%
D = 30% × 67  = 28,71 ≈ 29     CA
      70%
Total that passed
= 24 + 29 = 53   CA

1MA multiplying by 4
1A value of A
1CA value of B
1CA value of C [accept 95]
1CA value of D [accept 28]
1CA total [accept 52]

OR
1MA multiplying by 4
1A value of A
1CA value of B
1CA value of C [accept 95]
1CA value of D [accept 28]
1CA total
NPR [accept 52]
(6)

D L3
3.4

Number of Days
= 26 000 ÷ 24 = 1083,333…CA
Number of hours = 0,333…× 24 = 8
Number of weeks M
= 1083 ÷ 7 = 154,7142857…CA
Number of days = 0,71428… × 7 = 5
154 weeks 5 days 8 hours
VALID  O

OR
Hours per week = 24 × 7 = 168
Weeks= 26 000 = 154,7619047619
                168
Days = 0,7619047619 weeks × 7
= 5,333... days = 5   CA
Hours = 0, 333...days × 24 = 8
⸫154 weeks 5 days 8 hours CA

OR
Days  = 154 × 7 = 1 078    M CA
Total days =1 078 + 5 = 1 083M
Hours  1 083 × 24 = 25 992
Total hours = 25 992 + 8 = 26 000 CA
VALID

OR M
1 week = 7 days  = 7 × 24 h= 168 hours
Hours = 154 × 168 = 25 872 CA
Hours = 5 × 24 M = 120   CA
Total hours = 25 872 + 120 + 8 = 26 000
VALID

1M dividing by 24
1CA hours
1M dividing by 7
1CA simplification
1O verification

OR
1M multiply by 7
1CA days
1M multiply by 24
1CA hours
1O verification

OR
1M multiply by 7
1CA simplification
1M multiply by 24
1CA simplification
1O verification

OR
1M multiply by 7
1CA simplification
1M multiply by 24
1CA simplification
1O verification
(5)

 M L4
3.5.1 33 2A value(2) MP L2
3.5.2

Place seat face down.    A
Attach the bench leg/s to the bench seat.
Attach the long panel to bench leg/s.       A

OR
Lift the bench leg, align dowels with hole on the bench seat and insert them.
Insert the long panel. A

OR
Insert the dowels of the bench leg into the seat,
Connect the long panel with the bench leg. A

2A first instruction
2A second instruction
[Any correct two]
(4)

MP L4
3.5.3

It stabilises the bench  O
Keeps the bench sturdy/ steady/ strong/safe to sit on
It prevents the bench from collapsing
It supports the bench legs

2O explanation (2)

MP L4
    [35]  

 

QUESTION 4 [38 MARKS]

Q

Solution

Explanation

T&L

4.1.1

Percentage increase
14,5 million -10.8 million x 100%
         10,8 million     A
≈ 34,26%   CA

OR
Percentage increase
= 14,5 million x 100% -100%  M
    10,8 million    A
≈ 34,26%  CA

1M subtracting values
1A denominator
1CA simplification

OR
1A denominator
1M subtracting values
1CA simplification NPU (million and %)

D L2

4.1.2 (a)

Two or 2

2A correct size

D L2

4.1.2 (b)

Three or 3

2A correct size

D L2

4.1.3

2001:
Number of households
= 33% × 10,8 millionc    MA
= 3,564 million  CA

2011:
Number of households
= 25% × 14,5 million  MA
= 3,625 million  CA
Increase = 3,625 mil – 3,564 mil
= 0,061 million

INCORRECT,  O
OR the number of households increased

1MA percentage calculation
1CA simplification
1MA percentage calculation
1CA simplification
1O conclusion
(5)

D L4

4.1.4

Rounding factor or effect of rounding.
Rounded-off the decimals.

2O reason

D L4

4.1.5

P(less than four)
= 27% + 19% + 15%
= 61%    üCA

1RT correct values
1MA adding correct values
1CA simplification(3)

P L2
4.2.1 R20 to R79

2RT correct class (2)

D L2
4.2.2 5,4 mil + 3,2 mil = 8,6 mil   CA

1 MA adding correct values
1CA number of households AO (2)

F L2
4.2.3

Total income = R817 500    A
Wong’s household annual per capita
=R817 500
        3,5      
= R233 571,43   CA
Wong’s household daily per capita
R233571,4285      MCA
             365
= R639,92 CA

OR
Total annual income
= R276 000 + R541 500 = R817 500 
Wong's household daily income
= R817 500       MCA   
       365
» R2 239,73     CA     
or 
R 276 000R 541 500
     365               365          
= R756,16 + R1 483,56
=R2 239,72
Family size = 1 + 1 + 1 + 0,5 = 3,5 
Wong’s household daily per capita
= R2 239,73
        3,5
= R639,92  CA

OR
Total income= R817 500 
Family size = 1 + 1 + 1 + 0,5 = 3,5
Wong's household daily per capita
= R817 500    SF
    365 x 3,5 
= R639,92 CA

1A total income
1A family size
1SF substitution
1CA annual per capita
1MCA dividing annual per capita by 365
1CA daily per capita

OR
1A total household income
1MCA dividing by 365
1CA daily income
1A family size
1SF correct substitution
1CA daily per capita

OR
1A total household income
1A family size
1A denominator
1MCA dividing by 365
1SF Substitution
1CA daily per capita
(6)

F L2
4.2.4

Total per day
= 4% × R280 = R11,20   A

Total per year  A
= R11,20 × 365 = R4 088   CA

OR
MCA
Rate per year = R280 ×365 = R102 200
Amount spent on cellphones
= R102 200 × 4%  A
= R4 088    ü CA

1A daily value
1A multiply by 365
1CA simplification

OR
1MCA multiply by year consistent with Q4.2.3
1A calculation 4%
1CA simplification AO
(3)

F L3
4.3.1 Neo.

2O correct name (2)

D L4
4.3.2

Elec  = R125 × 12,2 mil = R1 525mil MA
Water =R98 × 10,6 mil = R1 038,8 mil ü MA Monthly total in million
Total spent on electricity and tap water in millions:
= R2 563,8 × 12 = R30 765,6     CA

OR
Elec = R125 × 12,2 mil = R1 525 mil
Total for the year 
= R1 525 million × 12
= R18 300 million   MA 

Water = R98 × 10,6 mil = R1 038,8 mil
Total for the year
= R1 038,8 million × 12
= R12 465,6 million  MA
Total spent on electricity and tap water in millions:
M
= R18 300 + R12 465,6 = R30 765,6    CA

OR
Annual cost for electricity 
= R125 × 12 = R1 500
Total electricity
= R1 500 × 12,2 million  = R18 300 million
MA Annual cost for tap water
= R98 × 12 = R1 176
Total :water= R1 176 × 10,6 million
= R12 465,6 million   MA 

Total spent on electricity and tap water
= R18 300 million + R12 465,6 million   M
= R30 765,6 million = R30 765 600 000     CA

1MA electricity amount
1MA water amount
1M adding amounts
1CA simplification

OR
1MA electricity amount
1MA water amount
1M adding amounts
1CA simplification

OR
1MA electricity amount
1MA water amount
1M adding amounts
1CA simplification
(4)

F L3
4.3.3

The scale on the axis (vertical / y axis) of the two graphs differs.
The intervals on Graph A is 10% while Graph B is 40%

2O reason
(2)

D L4
    [38]  

 TOTAL : 150 

Last modified on Wednesday, 01 December 2021 12:59