# Mathematical Literacy Grade 12 P2 2021 Memo (Eng)

> Grade 12 Mathematical Literacy Paper 2 memorandum from the Eastern Cape Mock Exam 2021 (English), 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/maths-lit-p2-memo-gr12-sept2021-english-final/d8c48090-02c0-4e40-ab84-087617f9c91c
- Updated: 2026-09-27
- Subject: Mathematical Literacy
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2021
- Exam period: Eastern Cape Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 10

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## About this document

Mathematical Literacy · Grade 12 · Eastern Cape Mock Exam · 2021 · English. Memorandum, 10 pages. Read online or download the PDF.

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## Document text

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### Page 1

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                               NATIONAL
                          SENIOR CERTIFICATE

                                  GRADE 12

                             SEPTEMBER 2021

             MATHEMATICAL LITERACY P2
                MARKING GUIDELINE

MARKS:     150

  Symbol                                          Explanation
M            Method
MA           Method with accuracy
CA           Consistent accuracy
RCA          Rounding consistent accuracy
A            Accuracy
C            Conversion
S            Simplification
SF           Correct substitution in a formula
J            Justification
O            Opinion/Example/Definition/Explanation/Justification/Verification
RT/RG/RM     Reading from a table/graph/map
P            Penalty, e.g. for no units, incorrect rounding off etc.
R            Rounding off
NPR          No penalty rounding or omitting units
AO           Answer only, full marks

                  This marking guideline consists of 10 pages.

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2                            MATHEMATICAL LITERACY P2                            (EC/SEPTEMBER 2021)

MARKING GUIDELINES

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.

LET WEL:

    As ŉ kandidaat ŉ vraag TWEE keer beantwoord merk slegs die EERSTE poging.

    As ŉ kandidaat ŉ antwoord van ŉ vraag doodtrek (kanselleer) en nie oordoen nie, merk die
     doodgetrekte (gekanselleerde) poging.

    Volgehoue akkuraatheid (CA) word in ALLE aspekte van die nasienriglyn toegepas, maar dit
     hou by die tweede berekeningsfout op.

    Wanneer ŉ kandidaat aflees van ŉ grafiek, tabel, uitlegplan en kaart en ekstra antwoorde gee,
     penaliseer vir elke ekstra item.

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    (EC/SEPTEMBER 2021)               MATHEMATICAL LITERACY P2                                          3

KEY TO TOPIC SYMBOL:
F = Finance; M = Measurement; MP = Maps, plans and other representations; P= Probability

QUESTION 1 [26 MARKS]
Quest                    Solution                                       Explanation               Level
1.1.1 25 mg : 1 000 mg M (divide by 25)                         1M ratio concept                 M
      = 1 : 40 MA                                               1MA Simplification               L1
                                                                                            (2)
1.1.2   Tablets per day:                                                                          M
        25 × 2 MA                                               1MA multiplying by 2             L1
        = 50
        Tablets per week:                                        1MA multiplying by 7
        50 × 7MA                                                1A answer
        = 350 mg A                                                                        (3)
1.1.3   Number of days to use tablets:                           1M finding the number of         M
        60 tablets          days M                              days                             L1
        Last day: 30 March 2021 M                               1M Referring to last date
        Therefore, from the 31st March 2021 a refill is          1O Conclusion
        needed. O                                                                         (3)
1.1.4   25 mg × 120 RT                                          1RT multiply correct values      M
        = 3 000 mg ÷ 1 000 C                                    1C divide by 1 000               L1
        = 3 g A                                                 1A answer
                                                                                           (3)
1.2.1   The “North Elevation” shows the side view of the         2A correct explanation.          MP
        building from the northern direction. A                                                 L1
                                                                                            (2)
1.2.2   (a)   A floor plan shows a top view of the inside of a   2A correct explanation         MP
              building. A                                                                 (2) L1
        (b) An elevation plan shows the side view of the         2A correct explanation         MP
              outside of the building. A                                                  (2) L1
1.2.3   Scale = 1 : 100                                                                         MP
        Measured length: 5 cm                                                                   L1
        Actual length: 5        cm M                            1M multiply by scale
                        = 500 cm MA                             1MA answer
                                                                                            (2)
1.3.1   90 km 1000 100 × 10 C                                                                    M
        = 90 000 000 mm A                                       1C multiply by 1000 000          L1
                                OR
                 C                                              1A correct answer
        90 km × 1 000 000 = 90 000 000 mm A                                                (2)
1.3.2   No of tiles per box           M                         1M dividing by 0,36              M
                                                                 1A number of tiles               L1
                               4 tiles per box A                                           (2)
1.3.3   1,125 hours                                                                               M
        1 hour                                                   1C finding minutes               L1
        0,125 hr. × 60 = 7,5 min C                              1C finding seconds
        0,5 min × 60 = 30 seconds C                             1A correct answer
        1,125 hours = 1 hr. 7 min 30 seconds A                                             (3)
                                                                                           [26]

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    4                            MATHEMATICAL LITERACY P2                          (EC/SEPTEMBER 2021)

QUESTION 2 [28 MARKS]
Quest                    Solution                                  Explanation                Level
2.1.1 George; Knysna; Mossel Bay RT                       2RT names of towns                MP
      (Any two of the three)                                                              (2) L1
2.1.2 South East  A (accept East)                         2A Direction                      MP
      North East A                                        2A Direction                      L1
                                                                                          (4)
2.1.3    National roads connect major cities. O           2O Reason 1                       MP
         Save on fuel consumption. O                      2O Reason 2                       L4
         Fewer traffic lights O
         Travel faster O
         [Any two or any acceptable explanation.]                                         (4)
2.1.4    2,3 cm = 300 km A                                 1A measured bar scale               MP
         Measured distance = 7 cm A                        1A measured distance                L2
                                                            range for bar 2.2cm to 2,4cm
         Actual distance =               km    M           measured map range distance
                                                            from 6,9 to 7,1
                                                            1M concept for ratio
                             913 km CA                     1CA rounded answer
                                                                                          (4)
2.1.5    Distance = Speed Time                              Accept CA from Q 2.1.4              MP
          913 = 100 TimeSF                                 1SF substitute correct values       L4
                                                            1M changing the subject of the
               =          M                                formula
                                                            1CA time in hours
                                                            1 C hours to minutes
               = 9,13 hrs. CA                              1M adding time
                                                            1CA arrival time
               = 0,13 60                                    1J justification
               = 7,8 minutes ≈ 8 minutes C

         Time of arrival = 9hrs 08 min 1hr 08:00 M
                          = 18:08      CA
         They will arrive later than the planned time. J                                 (7)

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    (EC/SEPTEMBER 2021)                     MATHEMATICAL LITERACY P2                               5

Quest                                    Solution                         Explanation      Level
2.2.1    7,6 litres = 100 km                                           Accept CA from      MP
         Petrol Consumption =                              M          Q 2.1.4             L2
                                                                       1M concept of ratio
                                = 9,13

                                = 69,388 litres     S
                                                                       1S simplification
                                                                       1M multiply by 2
         Return trip Consumption = 69,388           2 M
                                                                       1CA Answer
                                      = 138,776 litres CA
                                                                                     (4)
2.2.2    1 642 cents      100      R16,42    M                        Accept CA from    MP
                                                                       Q 2.2.1           L2
         Petrol cost      R16,42     138,776 litres CA                1M Conversion

                          R 2 278,701                                  1CA multiply by
                                                                       138,78 litres
                          R 2 278,70 CA
                                                                       1CA Answer
                                                                                       (3)
                                                                                      [28]

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    6                             MATHEMATICAL LITERACY P2                (EC/SEPTEMBER 2021)

QUESTION 3 [34 MARKS]
Quest                          Solution                            Explanation           Level
3.1.1 Radius: 18 cm 2 = 9 cm M                              1M radius                   M
      Conversion to mm = 9 cm × 10 = 90 mm C                1C conversion               L3
      Area =                                                 1SF correct substitution
            = 3,142 90 mm × 90 mm SF                        1S simplification
            = 25 450,2 S                                    1R rounding
            = 25 450 mm2 R                                                        (5)
3.1.2 Radius: 250 mm ÷ 2 = 125 mm MA                        1MA finding radius          M
      Circumference = 2 × π × r                                                          L2
                    = 2 × 3,142 × 125 mm SF                 1SF correct values
                    = 785,5 + 50 mm M                       1M adding 50 mm
                    = 835,5 mm CA                           1CA answer
                                 OR                          OR
      Radius: 250 mm ÷ 2 = 125 mm MA                        1MA finding radius
      Circumference = π × D
                    = 3,142 × 250 SF                        1SF correct values
                    = 785,5 + 50 M                          1M adding 50 mm
                    = 835,5 mm CA                           1CA answer
                                                                                   (4)
3.1.3    Conversion: length = 400 mm 10 = 40 cm C           1C conversion            M
         Volume        =l b h                                1SF in formula           L3
                   3
         42 000 cm     = 40 cm 30 cm h SF                   1S changing the subject
         Height        = 42 000 cm3 ÷ 1200 cm2 S            of the formula
                       = 35 cm CA                           1CA answer
                                                                                  (4)
3.1.4    25,4 mm = 1 inches                                                           M
         Diameter in mm         = 250 mm RT                 1RT using correct        L4
         Diameter in inches     = 250  25,4                 values 250 mm and 25,4
                                = 9,84 inches MA            1MA convert to inches
         Statement is not valid O                           1O
                                                                                  (3)
3.1.5    Radius value = 250 mm ÷ 2                           1MA finding the radius M
                      = 125 mm MA                           1M percentage            L2
         Height       = 0,64 × 125 mm M                     calculation
                      = 80 mm CA                            1A answer
                                                                                  (3)
3.1.6    Flour : Sugar (3 cups)                                                       M
         7 cups : 3 cups (× 2)                                                        L2
         14 cups : 6 cups
                                                             1MA dividing by 3
                 =       MA
                 =2                                          1MA multiplying by 7
                 = 2 × 7 MA = 14 cups of flour A           1A answer
                                   OR
                                                             1MA dividing by 3
              MA × 6 MA = 14 cups of flour A              1MA multiplying by 6
                                                             1A answer
                                                                                 (3)

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    (EC/SEPTEMBER 2021)                  MATHEMATICAL LITERACY P2                                     7

Quest                                Solution                            Explanation               Level
3.1.7    °C      = (°F – 32°)  1,8                                                                M
                        SF                                         1SF correct value              L2
                 = (365 °F – 32°)  1,8
                 = 333 °F  1,8  S                                 1S simplify
                 = 185°C                                            1R rounding to 10
                  190 °C  R                                       degrees
                                                                                             (3)
3.2.1    Tree diagram A                                           2A correct answer        P
                                                                                         (2) L1
3.2.2    Missing value P: Boy RT                                   1RT correct answer       P
         Missing value Z: GBB RT                                   1RT correct answer       L1
                                                                                         (2)
3.2.3    (a)     Probability (2 girls at least)   =   RTRT        1RT Numerator            P
                                                                    1RT Denominator          L2
                                                  =   CA           1CA simplification
                                                                                         (3)
3.2.3    (b)     Probability (BGB) =      % A                     2A correct percentage    P
                                                                                         (2) L2
                                                                                        [34]

QUESTION 4 [32 MARKS]

Quest                          Solution                                  Explanation         Level
4.1.1    Basin / Wash basin RT                                    2RT correct feature      MP
                                                                                         (2) L1
4.1.2    1 Window RT                                              2RT correct number       MP
                                                                                         (2) L1
4.1.3    Width (bedroom 1 and bedroom 2) = 4 680 + 5 130 MA        1MA adding correct       M
                                         = 9 810 A                 values                   L2
                                                                    1A length
         Length of bathroom       = 13 680 – 9810  M               1M finding length of
                                  = 3870 CA                        bathroom
                                                                    1CA answer
         Wall (minus door opening)        = (3 870 – 860) mm  M    1M subtracting door
                                          = 3 010 mm  CA           width
                                                                    1CA answer
                                                                                         (6)

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    8                           MATHEMATICAL LITERACY P2                 (EC/SEPTEMBER 2021)

4.2.1    Bathroom Area = Length × Width                                                  M
                      = 3 870 mm × 2 250 mm SF             1SF correct values           L3
                                A                          1A finding area
                      = 8 707 500 mm2 ÷ 1 000 000 C        1C ÷ by 1 000 000
                      = 8,7075 m2  CA                      1CA for the area

         Kitchen Area = Length × Width      SF
                                                            1SF correct values
                         = 6 030 mm × 5 130 mm A           1A finding area
                         = 30 933 900 mm2 ÷ 1 000 000  C   1C ÷ by 1 000 000
                         = 30,9339 m2 CA                   1CA for the kitchen area
                                                            1CA total area
         Total Area      = 8,7075 + 30,9339                                      (9)
                         = 39,6414 m2 CA                                      NPR
4.2.2    Area of 1 tile: 500 mm 500 mm                      CA from 4.2.1                M
                         = 250 000 mm2 MA (1000 000) C   1M area of tile              L4
                         = 0,25 m2 A                       1C divide by 1 000 000
         No. of tiles needed = 39,6414 m2 ÷ 0,25 m2 MCA    1MCA ÷ total area by
                                =158,5656 × 1,05M          0,25 m2
                                =166,49388 tiles            1M increasing by 5%
                                = 167 tiles CA             1CA rounded no. of tiles
         Statement not valid. O                            1O conclusion
                                                                                   (7)
4.2.3    No of boxes     = 167  4 MCA                     CA tiles from 4.2.2          M
                         = 41,75                            1MCA dividing tiles by 4     L4
                         = 42 boxesCA                      1 number of boxes
                                                            1M multiply by cost
         Amount for tiles      = 42 × 249,90 M             1CA cost of tiles
                               = R10 495,80 CA
                                                            1M adding labour cost
         Total amount          = 10 495,80 + 8 186,09
                               = R18 681,89 M              1O conclusion

         Claim is valid. O                                                        (6)
                                                                                  [32]

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    (EC/SEPTEMBER 2021)                       MATHEMATICAL LITERACY P2                                 9

QUESTION 5 [30 MARKS]
Quest                                     Solution                         Explanation             Leve
                                                                                                   l
5.1.1    Convert:          175     100       1,75 m C              1C conversion                  M
                                 ass in                                                            L3
                    BMI
                             ei ht × m2
                                ass in                              1SF correct values
                    25,1                         SF
                             ,7 m × ,7 m
         Mass         BMI (height)2                                 1S changing subject of
                      25,1 (1,75)2 S                               formula
         Mass in (kg) 76,86875 kg                                   1CA answer
                      76,87 kg CA                                                           (4)
5.1.2    Overweight  J                                            1J conclusion                M
                                                                                             (2) L1
5.1.3    The young promising rugby player can eat                   2O Explanation               M
         healthier. O                                                                          L4
                                   OR
         He should exercise more regularly O                                               (2)
5.2.1    3 laps        = 5 000 m ÷ 1000 C                          1C divide by 1 000             MP
         1 lap         = 5 km ÷ 3 MA                               1MA divide by 3                L2
                       = 1,67 km CA                                1CA distance per lap

                                   OR                               1MA divide by 3
         1 lap        = 5000 m ÷ 3 MA                              1C divide by 1000
                      = 1 666,666667 ÷ 1 000 C                     1CA distance per lap
                      ≈ ,67 mCA                                                            (3)
5.2.2    Incomplete runs:     = 4 laps × 1,67 km M                 1M multiply by lap             MP
                              = 6,68 km A                          distance                       L4
                                                                    1A correct distance
         Complete runs          = 3 × 5 km                          1MA finding complete run
                                = 15 km + 6,68 km MA               distance
                                = 21,68 km CA                      1CA finding total distance
         His statement is not valid. O                             1O correct conclusion
                                     OR                             1M multiply 4 by lap
              M          M                                        distance
         (4 × 1,67) + (9 × 1,67)                                    1M multiply 9 by lap
              A A                                                 distance
         = 6,68 + 15,03                                             1A correct distance
         = 21,71 km CA                                             1A correct distance
         His statement is not valid. O                             1CA finding total distance
                                                                    1O correct conclusion
                                                                                            (5)
5.2.3    Difference in Time = 9 min 30 sec − min 4 sec M           1M subtract time               MP
                             = 3 min 45 sec A                      1A difference in time          L3
         Difference in Time per lap = 3 min 45 sec ÷ 3 M           1M divide by 3
                                      = 1 min 15 sec A             1A total time
         Statement is valid                                        1O reason
                                                                                            (5)

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10                          MATHEMATICAL LITERACY P2                          (EC/SEPTEMBER 2021)

5.3.1    Number of candles                                          1MA ÷ by correct       MP
         Length : 24 cm 8 cm 3 MA                                  values                 L3
         Width : 16 cm 8 cm      2 MA                              1MA for 2 correct
         Height : 24 cm ÷ (1 11 cm)   2          M                 values 3 and 2
                                   S                               1M adding 1cm
                                                                    1S simplify no of
         Total number of candles: 3    2     2   12 candles CA     candles
                                                                    1CA conclusion
                                      OR                                    OR
                                                                    1MA × by correct
         Length: 8 cm × 3 = 24 therefore 3 will fit MA             values
         Width: 8 cm × 2 = 16 therefore 2 will fit MA              1MA for 2 correct
         Height: (11 cm + 1 cm + 11 cm) = 23 cm, therefore 2 will   values 3 and 2
         fit MA                                                    1M adding 1cm
                                                                    1S simplify no of
                                           S                       candles
         Total number of candles: 3    2     2   12 candles CA     1CA conclusion
                                                                                     (5)
5.3.2    Total area = 2 (H × L) + 2 (W × H) C                      1C converting to m     M
         = 2 (0,24 m × 0,24 m) + 2 (0,16 m × 0,24 m) SF            1SF correct values     L2
         = 0,1152 m2 + 0,0768 m2 S                                 1S simplification
         = 0,192 m2 CA                                             1CA answer
                                                                                     (4)
                                                                                    [30]

                                                                         TOTAL: 150

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