MATHEMATICAL LITERACY PAPER 1
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
NOVEMBER 2020

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.
• CA marks only apply if at least 1 correct value is used.
• If the candidate presents any extra solution when reading from a graph, table, layout plan and map, then penalise for every extra item presented.

QUESTION 1 [30 MARKS] ANSWER ONLY FULL MARKS
1.1.1 Vertical bar graph
Bar, Column graph
2A bar graph (2)
D
L1

1.1.2 A = R110 + R11
= R121
1MA adding correct values
1CA Simplification (2)
F
L1

1.1.3 B = R141 – R126
= R15
1MA subtracting correct values
1CA simplification (2)
F
L1

1.1.4 Difference
R126 – R110
= R16
1MA subtract lowest from highest
1A simplification (2)
F
L1

1.1.5 Increased Delivery fee
= R10,00 × 6,32%
= R0,632
= R0,63
OR
= R10,00 × 6,32
                   100
= R0,632
= R0,63
1MA calculating percentage
1A simplification

1.1.5 Increased delivery fee
= R10 × 1,0632
= R10,632
Increase in delivery fee
= R10,63 – R10,00
= R0,63
1MA calculating percentage
1A simplification (2)

1.2.1 2008
2RT reading correct year (2)
D
L1

1.2.2 Difference = R11,04 – R4,31
= R6,73
1MA subtracting correct values
1RT correct values
1CA simplification (3)
F
L1

1.2.3 5,56 : 12,48
1 : 2,24 OR 0,45 : 1
1MA concept of ratio in correct order
1RT correct values
1CA simplification (3)
F
L1

1.2.4 Total = 13,45 × R4,00
= R53,80
OR
R : ℓ
4 : 1
53,80 : 13,45
Total cost = R53,80
1MA multiplying correct values
1CA simplification (2)
F
L1

1.2.5 2007
2RT reading correct year (2)
D
L1

1.3.1 Strip Map (Chart)
2A strip map (chart) (2)
MP
L1

1.3.2 Distance in metre
= 779 × 1 000
= 779 000
1MA multiplying by 1 000
1A simplifying
NPU (2)
M
L1

1.3.3

1. Ladismith AND/EN Calitzdorp
   1A correct town
   1A correct town (2)
   MP
   L1
2. The distance from Riversdale to Oudtshoorn/
   Afstand vanaf Riversdal na Oudtshoorn
   = 82 km + 45 km + 53 km
   = 180 km
   1MA adding correct values
   1CA simplification (2)
   MP
   L1

[30]

QUESTION 2 [42 MARKS]
2.1.1 R4 656,71
2A correct balance (2)
F
L1

2.1.2 Full date
1 February 2019
01/02/19
01/02/2019
2A full date (2)
F
L1

2.1.3 R1 215,36
2A correct amount (2)
F
L1

2.1.4 R3 750,00
2A correct amount (2)
F
L1

2.1.5 FNB electroninc payments
R101,99 + R698,01
= R800,00
1RT 1st value correct
1RT 2nd value correct
1A simplification
AO (3)
F
L1

2.1.6 Price excluding VAT/Prys BTW uitgesluit
=R4000,00 x 100
                      115
= R3 478,26
OR
Price excluding VAT
R4000
 1,15
= R3 478,26
OR
VAT amount =  15  
                        115
R4000,00
= R521,74
Price excluding VAT
= R4 000 – R521,74
= R3 478,26
1RT price of item
1MA calculating VAT
1CA price excluding VAT
AO (3)
F
L2

2.2.1 South African Revenue Services/SARS
Revenue Services
2A name (2)
F
L1

2.2.2 2 / TWO
OR
7 / SEVEN
2A correct bracket (2)
F
L1

2.2.3 Annual tax before rebates
= R35 253 + 26% of taxable income above 195 850
= R35 253 + 26% × (R305 174,44 – R195 850)
= R35 253 + R28 424,35
= R63 677,35
Monthly tax before rebates
= R63 677,35 ÷ 12
= R5 306,45
OR
Annual tax before rebates
= R532 041 + 45% of taxable income above 1 500 000
= R532 041 + 45% × (R3 662 093,28 – R1 500 000)
= R532 041 + R972 941,98
= R1 504 982,98
Monthly tax before rebates
= R1 504 982,98 ÷ 12
= R125 415,25
CA from question 2.2.2
1SF correct substitution
1M adding correct amounts
1CA simplification
1MCA dividing by 12
1CA simplification
NPR (5)
F
L3

2.2.4

1. Primary rebate/Primêre korting OR/OF R14 067,00
   2RT reading from the table (2)
   F
   L1
2. 3/THREE
   2A correct number of rebates
   (2)
   F
   L1

2.3.1 Selling price of one photo
R500 OR R1000 OR R1600 OR R2 500 OR R3 000
  25            50               80              125             150
= R20
OR
R4 000 ÷ 200
= R20
1MA dividing
1A simplification
AO (2)
F
L1

2.3.2 Total income received
Income = R20,00 × n, where n = number of photos
Income = R20,00 × number of photos
CA from Question 2.3.1
1CA R20,00
1A multiply by unknown (2)
F
L2

2.3.3

Related Items

1. R5,00
   2A variable cost
   NPU (2)
   F
   L1
2. A : Expenses = R1 125 + number of photos × R5,00
   A = R1 125 + (80 × R5,00)
   A = R1 125 + R400
   = R1 525
   1SF substituting value
   1MCA adding values
   1CA simplification AO
   (3)
   F
   L2

2.3.4

1. Income and expenses of Ella's photography business
   2A correct heading (2)
   F
   L1
2. X
   2A correct graph
   (2)
   F
   L1
3. 75 photographs/foto's
   2A correct number of photographs
   (2)
   F
   L1

[42]

QUESTION 3 [31 MARKS]
3.1.1 Legs of ottomans
2 cubic ottomans × 4 legs
= 8 legs/pote
1 retangularottoman × 6 legs
= 6 legs
8 + 6
= 14 legs
1A number of legs
1MA adding 6 legs
1CA total number of legs
AO (3)
M
L1

3.1.2  75mm
             2
Radius
= 37,5 mm / 3,75 cm
1MA concept of radius
1A simplification
AO
NPR (2)
M
L1

3.1.3 Total height:
50 cm + 12 cm
= 62cm
OR
Total height:
= 120 mm + 500 mm
= 620 mm
= 62 cm
1C converting to cm
1A finding the height
AO (2)
M
L1
M
L2
Area
(50cm × 50cm) + (120cm × 50cm)
2 500 cm² + 6 000 cm²
Total Area
(10 × 2 500 cm²) + (2 × 6 000 cm²)
25 000 cm² + 12 000 cm²
37 000 cm²
1A area
1A area
1M multiplying correct values
1M adding the two areas
1CA simplification
OR
8 square sides × ( 50 × 50 )
= 20 000 cm
2 rectangular sides × ( 120 × 50 )
= 12000 cm
2 square sides × ( 50 × 50 )
= 5 000 cm
Total area to be painted:
= 20 000 cm²+ 12 000 cm² + 5 000 cm²
= 37 000 cm²
1A simplification
1A simplification
1A simplification
1M adding all values
1MA finding total area
OR
Total perimeter
= (50+50+50+50+50+50+50+50+120 +50+50+120) cm
= 740 cm
Total area to be painted:
= 740 cm × 50 cm
= 37 000 cm²
1A all correct values
1M adding correct values
1A simplification
1MA multiplying correct
values
1A simplification
(5)

3.1.5 2 2 37 000 cm ÷ 10 000 = 3,7 m²C
Total area to be painted/Totale area wat geverf moet word
= 3,7m² x 2 
= 7,4² m
Spread rate
7,4 m × 1 000
  8 m
= 925 millilitres
1C conversion
1M dividing by 8
1M area of 2 coats
1CA simplification
OR
Spread rate = 8 × 10 000 cm²/ℓ = 80 000 cm²/ℓ
Amount of paint in ℓ = 37 000
                                   80 000
= 0,4625
Amount of paint for 1 coat in mℓ
= 0,4625 × 1 000
= 462,5
Amount of paint for 2 coats
= 462,5 mℓ × 2
= 925 mℓ
CA Question 3.1.4
1C converting from cm² to m²
1M area for 2 coats
1M divide by spread rate
1CA answer in millilitres
OR
Total area to be painted:
=37 000 cm2 ÷ (100)2 = 3,7 m²
Amount of paint for 1 coat in ℓ
=3,7 × 1
   8
= 0,4625 ℓ
Total amount of paint
= 0,4625 × 1000 × 2
= 925 mℓ
1M multiplying by 8
1M dividing by 80 000
1C converting
1CA simplification
M
L2
8 m² : 1 ℓ
80 000 cm² : x
Amount of paint for 1 coat
x =1000 x 37000
         80000
= 462,5 mℓ
Amount of paint for 2 coats
= 462,5 mℓ × 2
= 925 mℓ
1C conversion
1M dividing by 80 000
1M area of 2 coats
1CA simplification
OR
Total area to be painted
= 37 000 ÷ 10 000
= 3,7 m² × 2
= 7,4 m²
Spread rate in mℓ/ ℓ
1 000 ÷ 8
= 125 mℓ/ ℓ
Amount of paint
125 × 7,4 m²
= 925 mℓ
1C conversion
1M area of 2 coats
1M dividing by 8
1CA simplification
(4)

3.1.6 Height =  Volume  
                      2π(radius)
=     1 000 cm  ³
 3,142 x (6,5 cm)
= 7,53298…. cm
1C conversion from litres to cm³
1SF substitution of radius
1CA simplification
NPR (3)
M
L2

3.2.1

1. W or White/Wit
2. SB or Synthetic Brown leather/Sintetiese bruin leer

2RT correct code
2RT correct code (4)
P
L1

3.2.2  P(not selecting red material) = ⁶/₉
= ²/₃
OR
P(not selecting red material) = 1 – ³/₉
= ⁶/₉
= ²/₃
1A numerator
1A denominator
1CA simplification (3)
P
L2

3.3.1 1 inch = 153,6 ÷ 60
= 2,56 cm
OR
Alternative solution method:
inch : cm
60 : 153,6   M
1 : 2,56    A
1 inch = 2,56 cm
1M dividing by 60
1A simplification (2)
M
L1

3.3.2 Perimeter = 2 × (5 m + 153,6 cm)
= 2 × (500 cm + 153,6 cm)
= 1 307,2 cm
1RT correct value – 153,6 cm
1C converting from 5 m to cm
1CA simplification
OR
Perimeter = 5 m + 5 m + 153,6 cm + 153,6 cm
= (500 + 500 + 153,6 + 153,6 ) cm
= 1 307,2 cm
OR
1RT correct value – 153,6 cm
1C converting from 5 m to cm
1CA simplification
(3)
M
L2
[31]

QUESTION 4 [17 MARKS]
4.1.1 R46
2A name of route (2)
MP
L1

4.1.2 Number scale OR Numeric scale OR Ratio scale
2A identifying the scale (2)
MP
L1

4.1.3 South West OR SW OR West of South West OR WSW
2A general direction (2)
MP
L1

4.1.4 A = 210 km – (62 km + 13 km + 82 km)
A = 53 km
1MA subtracting correct values
1CA simplification (2)
MP
L1

4.1.5 Ladismith
2A correct town (2)
MP
L2

4.2.1 Total length
= 20 cm + 229 cm + 20 cm + 20 cm + 229 cm + 20 cm
= 538 cm
1MA correct values (4×20)
1MA adding values (2×229)
1CA simplification
OR
Total length /Totale lengte
2 (20 cm + 229 cm + 20 cm)
2 × 269 cm
= 538 cm
1MA correct values (4×20)
1MA adding values (2×229)
1CA simplification
OR
Total length/Totale lengte
= (20 cm × 4) + (229 cm × 2)
= 80 cm + 458 cm
= 538 cm
1MA correct values (4×20)
1MA adding values (2×229)
1CA simplification (3)
MP
L2

4.2.2 D + 86 + 80 + 86 + D = 260
2D + 252 = 260
2D = 260 – 252
2D = 8
D = 8 ÷ 2
= 4 cm
1MA adding all values
1M subtracting from 260
1M dividing by 2
1CA simplification
OR
Length excluding D
= (86 cm × 2) + (20 cm × 4)
= 172 cm + 80 cm
= 252 cm
2D =260 cm – 252 cm
D = 8 cm
= 8 cm ÷ 2
= 4 cm
1MA calculating 252
1M subtracting from 260
1M dividing by 2
1CA simplification (4)
MP
L3
[17]

QUESTION 5 [30 MARKS]
5.1.1 TGA – team
2RT correct tea (2)
D
L1

5.1.2 Range = 9,625 – 9,100
= 0,525
1RT reading correct values
1CA concept of range (2)
D
L1

5.1.3 Mean
9,100 + 9,250 + 9,300 + 8,650 + 9,100 + 9,050 + 8,750 + 9,050 + 8,300 + 9,200
                                                                 10
= 8,975
D
L2
1RT correct values
1M concept of mean
1CA simplification
NPR (3)

5.1.4 A = 36,425 – (9,300 + 9,100 + 9,225)
= 8,800
1RT correct values
1M adding and subtracting
1A simplification (3)
D
L1

5.1.5 36,425
2A correct mode (2)
D
L1

5.1.6 ³/₅ x 100 %
= 60%
1A numerator
1A denominator
1CA percentage
NPR (3)
P
L2

5.1.7 Quartile 2
9,375 + 9,400
         2
= 9,3875
1RT arranging or correct values
1M dividing by 2
1A simplification NPR
(3)
D
L2

5.2.1 Fifty two million nine hunderd and eighty two thousand.
2A amount in words (2)
D
L1

5.2.2 Increase in population(2015-2016)
56 020 718 – 54 901 943
= 1 118 775
≈ 1 120 000
1RT correct values
1M subtracting
1R correct rounding (3)
D
L1

5.2.3   54 901 943 - 53 947 998 x 100%
                     53 947 998
Annual population growth(2015)
= 1,768%
≈ 1,8%
1SF substituting 54 901 943
1SF substituting 53 947 998
1CA simplification
NPR (3)
D
L2

5.2.4

1A – correctly plotted number of people
1CA – drawing of graph
1A – correctly plotted population growth
1CA – drawing of graph (4)
D
L2
[30]
TOTAL: 150