MATHEMATICAL LITERACY
GRADE 12
NATIONAL SENIOR CERTIFICATE EXAMINATION
MEMORANDUM
MAY/JUNE 2021

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 item presented.
• General principle of marking, if the candidate makes one mistake he loses one mark.

QUESTION 1 [32 MARKS] 
1.1.1 Radius = 300 mm ÷ 2
= 150 mm OR 15 cm
1MA dividing by 2
1A simplify
(2)

1.1.2 R330,00; R275,00; R220,00; R110,00
1RT reading all the values
1CA correct order
(2)

1.1.3 VAT/BTW = R275,00 × 15%
= R41,25

OR
Price including VAT
= R275 1,15
= R316,25 
VAT = R316,25 – R275
= R41,25 
1MA multiplying by 15%
1A simplify

OR
1MA calculating VAT
1A simplify
(2)

1.1.4 150 ÷ 60
= 2,5 OR 2 hours
1A divide by 60
1A 2,5 hours
(2)

1.1.5 Total cost
= R330,00 + R275,00 + R220,00 + R220,00 + R165,00
+ R110,00
= R1 320
1MA adding all correct values
1MCA simplify (at least 5 correct values)
(2)

1.1.6 Discount/ = R330,00 × 7,5%
= R24,75
1MA multiplying by 7,5%
1A simplification
(2)

1.2.1
1 250 g
 1 000
= 1,25 kg
1MA dividing by 1 000
1A simplification
(2)

1.2.2 Cost Price
= R55,00 – R30,30
= R24,70
1MA subtracting correct values in the correct order
1A simplification
(2)

1.2.3 1 250 : 500
5 : 2 OR 2,5 : 1 OR 1: 0,4
1MA values in correct order
1CA simplified form
(2)

1.2.4 Number of packets
4 000 g = 8
 500 g
Mass
100g
  8
= 12,5 g
1MA number of 500g packs
1A dividing 4 000 g
1CA simplification

OR
Mass
500g x 100g = 12,5g
    4 000 g
1A number of 500g packs
1MA dividing 100 g by 8
1CA simplification

OR
4000 g : 100 g
500 g : mass of raisins/massa van rosyntjies
4 000
50 000g
Mass of raisins 
Mass of raisins/Massa van rosyntjies
= 12,5 g
1MA correct ratio concept
1A dividing 4 000 g
1CA simplification
(3)

1.2.5 Number of cups/aantal koppies
= 2 × 5
= 10

OR
4 000 : 5
8 000 : 10
The number of cups = 10
1MA multiply by 2 and 5
1A simplification
OR
1MA correct ratio
1A simplification
(2)

1.3.1 Money earned on an investment
2A definition
(2)

1.3.2 25 months
2A correct number of months
(2)

1.3.3 Bank A
2A correct bank
(2)

1.3.4 Difference
7,50% 6,7%
= 0,8%
1RT correct value from tables
1RT correct value from tables
1CA simplification (one value must be correct)
(3)
[32]

QUESTION 2 [37 MARKS]
2.1.1 Dr. JJ Ndlovu
2A correct name
(2)

2.1.2 Year of birth
1982 / '82
2RT reading from table
(2)

2.1.3 R0,00/nothing
2A correct amount
(2)

2.1.4 Amount excluding VAT
R1 744,75 ÷ 115 OR × 100
                    100          115
R1 744,75 ÷ 1,15
= R1 517,17

OR
VAT amount/BTW bedrag
R1744,75 x 15 
                  115
= R227,58
Amount excluding VAT
= R1 744,75 – R227,58
= R 1 517,17
1CA simplification
OR
1A amount VAT
1M subtracting VAT
1CA simplification
(3)

2.1.5 One infection control /Een infeksiebeheer
=R40,55 ÷ 2
= R20,28
1MA divide by 2
1A simplification
NPR
AO
(2)

2.2.1 Total fixed cost/Totale vaste koste
= R140,00 + R60,00
= R200,00
1RT correct values
1CA simplification (one value must be correct)
(2)

2.2.2 Expenses
Expenses (A) = R200,00 + R12,50 × number of packets
A = R200,00 + R12,50 × 10
A = R200,00 + R125,00
= R325,00
B = 400 ÷ 25 OR B = (400 – 200 ) ÷ 12,5
           = 16                      = 16
1SF correct substitution
1A simplification
1SF correct substitution
1A simplification
AO
(4)

2.2.3 

1. 
   
   1A starting point (0; 0)
   1A endpoint (50; 1 250)
   1A straight line (must be joining at least 3 points stated in the table; CA for using B)
   (3)
2. Where the cost price of mixed vegetable packs equals the selling price of the packs
   2A explanation
   (2)
3. 16 packs
   CA from Q2.2.2 / 2.2.3 (a)
   2A correct number of packs
   (2)

2.3.1

1. Deposit
   R1 799,00 × 20  
                       100
   = R359,80
   1MA calculating 20%
   1A simplification
   (2)
2. Total amount/Totale bedrag
   = R359,80 + (24 × R95,00)
   = R359,80 + R2 280,00
   = R2 639,80
   = R2 640,00
   CA from Question 2.3.1(a)
   1MA multiplying by 24
   1MCA adding the deposit
   1CA simplification
   1R to the nearest rand
   (4)

2.3.2

1. The value of one currency relative to the value of another currency
   1A value of one currency
   1A relative to the value of another currency
   (2)
2. yen / Ұ OR Japanese yen / Japanese jen
   2A correct currency
   (2)
3. 1 ZAR = 0,067251 dollar ($)
        $130     x R1
   $0,06725      1
   
   = R1 933,056758
   = R1 933,00
   1RT exchange rate
   1C conversion
   1R correct rounding
   OR
   Dollar ($) = ZAR14,86966737
   $130 x R14,86966737
     $1
   = R1 933,056758
   = R1 933,00
   1RT exchange rate
   1C conversion
   1R correct rounding
   (3)

[37]

QUESTION 3 [22 MARKS]
3.1.1 Width = 3 × 10,4 cm
= 31,2 cm
Length = 4 × 10,4 cm
= 41,6 cm
1MA for multiplying diameter by 3
1A simplification
1MA for multiplying diameter by 4
1A simplification
(4)

3.1.2 Ribbon needed for one candle (cm)
= 2 × 3,142 × radius + 3 cm
= 2 × 3,142 × 5,2 cm + 3 cm
= 35,6768 cm
20 × 100
= 2 000 cm
Number of candles
2 000 cm ÷ 35,6768 cm
= 56, 05883936
= 56 candles
1SF correct substitution (radius)
1A length for 1 candle
1C conversion
1MCA dividing by length of ribbon
1R correct number of candles
(5)

3.1.3 Volume = 3,142 × (5,2cm)² × 11,4cm
= 968,54 cm³
Volume of horsehead
=²/₃ x 968,54 cm³
              3
= 645,69 cm³ 
CA from Question 3.1.2
1SF substituting correct values
1CA answer in cm³
1MCA multiply by 2 and dividing by 3
1CA simplification

OR
968,54 = 322,84666 × 2
   3
= 645,69 cm³
2CA answer in cm³
1MCA multiply by 2 and dividing by 3
1CA simplification

OR
Volume = 3,142 × (5,2)² × 11,4 cm
= 968,54 cm³
Volume of horsehead = 968,54 cm³ - ¹/₃(968,54 cm³)
= 968,54 – 322,85
= 645,69cm³
1SF substituting correct values
1CA answer in cm³
1MCA subtracting
1CA simplification
(4)

3.2.1

1. Ribbon/Lint OR R/L
   2A ribbon
   (2)
2. HBN /PSG
   2A HBN/PSG
   (2)

3.2.2

1. P_{[candle with ribbon]} = ½ x100 % OR  4  x 100 %
                                           1               8      1
                                    = 50%                 = 50%
   1A fraction
   1M concept of percentage
   1CA for percentage
   AO
   (3)
2. P/W P[Gold horsehead candle] = 0
   OR
   Impossible/⁰/₈/0%/0,0
   2A correct probability (2)
   [22]
   

QUESTION 4 [21 MARKS]
4.1.1 North West / NW
2RT reading from map
(2)

4.1.2 N8
(2)

4.1.3 Campbell
2RT town
(2)

4.1.4 04:00 – 09:30
= 5 hours 30 min / 5,5 hours
Average Speed 
= 496,9 
    5,5
= 90,3454545 km/h
= 90 km/h
1A calculating 5,5 hours
1MCA dividing correct values in correct order
1CA simplification
1R rounding
(4)

4.2.1 1 unit on the plan represents 380 units in real life
2A explanation
(2)

4.2.2 Lifts
OR
Ground Floor
OR
Stairs
2A lifts (2)

4.2.3 27mm
2A correct value
(2)

4.2.4 Bloed street entrance
OR
South entrance
2RT correct entrance
(2)

4.2.5 27 mm
2A for correct measurement
1A correct wall
(Accept 26 – 28 mm)
(3)
[21]

QUESTION 5 [38 MARKS]
5.1.1 Range is the difference between the highest/maximum value and the lowest/minimum value in a data set.
2A correct definition (2)

5.1.2 Line graph
OR
Broken line graph
2A correct graph (2)

5.1.3 Discrete data
2A discrete (2)

5.1.4 1 749 + 2 239 + 1 618 + 903 + 429 +150 + 16
= 7 104
1RT correct values
1M adding ALL values
1CA simplification (at least 6 values correct)
Accept 7 136 = full marks
AO
(3)

5.1.5 L2
3RT correct level
(3)

5.1.6 Median level descriptor
= 62; 223; 551; 935 1 231; 1 357; 1 990
CA from Question 5.1.4
1MCA arranging in order
1CA correct median
1CA level descriptor

OR
L2 : L3 ; L1 ; L4 ; L5 ; L6 , L7
Median level
= L4
1MCA arranging
1CA correct order
1CA level descriptor
AO
(3)

5.2.1 IIII I
2A correct tally (2)

5.2.2 6
CA from Question 5.2.1
2CA correct frequency
(2)

5.2.3
43 + 17 = 60
OR
0 + 3 + 6 + 12 + 7 + 15 + 17 = 60
1RT correct values
1MA simplification
OR
1RT correct values
1MA simplification (2)

5.3.1 Stacked bar graph/Stapel staafgrafiek
1A stacked
1A bar graph
(2)

5.3.2 Two hundered and ninety four thousand two hundred and two/
1A first part of wording
1A second part of wording
(2)

5.3.3 298 607 – 222 034 9 670
= 66 903
1RT correct values
1M subtracting
1CA simplification (two values must be correct)
AO
(3)

5.3.4 Mean
225458 + 263903+ 265810 + 245103+ 233858 + 222034
                                          6
= 242 694,33
1RT correct values
1M concept of mean
1CA simplification
NPR
(3)

5.3.5 Range
388 845 – 294 202
= 94 643
1MA concept of range
1CA simplification (one value must be correct)
(2)

5.3.6 % for Mathematics
=222034 x 100
  530311      1
41,8686%
% for Mathematical Literacy
=298607 x 100
  530311      1
56,3079%
56,3079% 41,8686%
= 14,4 %
1RT correct values
1MA percentage calculation
1CA simplification
1CA simplification
1CA simplification with correct rounding

OR
298607 - 222 034× 100
       530311
= 14,4%OR
1RT correct values
1M subtracting values
1CA correct denominator
1MA percentage calculation
1CA simplification with correct rounding
(5)
[38]
TOTAL: 150