# Mathematical Literacy Grade 12 P1 EC June 2019 Memo (Eng)

> Grade 12 Mathematical Literacy Paper 1 memorandum from the EC June 2019 (English), 8 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/maths-lit-p1-memo-june-2019-eng/a8fdfb2a-fad2-4fd3-9db8-09771365ffde
- Updated: 2026-09-27
- Subject: Mathematical Literacy
- Grade: Grade 12
- Document type: Memorandum
- Year: 2019
- Exam period: EC June
- Paper: 1
- Curriculum: CAPS
- Pages: 8

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Mathematical Literacy · Grade 12 · EC June · 2019. Memorandum, 8 pages. Read online or download the PDF.

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### Page 1

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                           NATIONAL
                      SENIOR CERTIFICATE

                              GRADE 12

                                JUNE 2019

              MATHEMATICAL LITERACY P1
                 MARKING GUIDELINE
MARKS:   100

   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

                       This marking guideline consists of 8 pages.

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2                                MATHEMATICAL LITERACY P1                              (EC/JUNE 2019)

 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.

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 (EC/JUNE/2019)                          MATHEMATICAL LITERACY P1                                         3

QUESTION 1 [22 MARKS] INTEGRATED QUESTION
 Question                Solution                                   Explanation/Marks
                                                                    AO: FULL MARKS               Topic/L
1.1   1.1.1       DM Row Travel Agency  RT                     2 RT correct agency                F
                                                                                      (2)          L1

      1.1.2       Reference Number: 0674582  RT                2 RT correct Voucher no.          F
                                                                                       (2)         L1

      1.1.3       3 nights  RT                                 2RT correct no of nights          F
                                                                                         (2)       L1

      1.1.4       3 × R1 440,00  M                              1M×1 440,00                       F
                  = R4 320,00  CA                               1CA simplification        (2)     L1

1.2   1.2.1       250 + 330 + 450 + 500 + 1 000 + 1500 + 2 000   1M addition all values            M
                  = 5 580 mℓ M                                                                    L1
                                                                 1CA dividing and
                       = 5,58 ℓ  CA                             simplifying
                                        OR
                  0,25 + 0,33 + 0,45 + 0,5 + 1 + 1,5 + 2 l MA
                  = 5,58 l CA                                                             (2)

      1.2.2                              M                      1M division correct values        M
                  No. of A containers=        = 6CA             1CA No of A containers            L1
                                                                                         (2)

1.3   1.3.1       42,2 km  RT                                  2 RT distance                    M&P
                                                                                           (2)     L1

      1.3.2       Difference in height above sea level           1RT correct values               M&P
                  = 46 – 35  RT                                                                   L1
                  =11 m  CA                                     1CA difference            (2)

1.4   1.4.1       Mode = 61  RT                                 2RT mode                          D
                       = 47                                                               (2)     L1

      1.4.2       Range = 86 – 18  MA                           1MA subtracting correct
                  = 68 CA                                       values                            D
                                                                 1CA range               (2)       L1

      1.4.3       Raw mark =         150  MA                    1MA % of 150                      D
                                                                                                   L1
                          = 129  CA                             1CA raw mark
                                      OR
                  Raw mark = 0,86 x 150 MA
                          = 129  CA                                                       (2)
                                                                                          [22]

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 4                               MATHEMATICAL LITERACY P1                              (EC/JUNE 2019)

QUESTION 2 [29 MARKS] FINANCE
Question                 Solution                                 Explanation/Marks     Topic
                                                                  AO: FULL MARKS        /L
2.1   2.1.1 Total amount = 1 200 × 60  M                      1M multiplication           F
                         = R72 000  CA                        1CA amount           (2)   L1

      2.1.2           20 000  R6 000 M  CA                   1M multiplication of correct
                                                               values
                                     OR
                                                               1CA simplification        (2)
              School’s contribution = 0,3 x 20 000  M                                                  F
                                   = R6 000  CA                                                        L1

      2.1.3 Ratio 2 000 : 20 000 MA A                        1MA for ratio and 2000                   F
            1 : 10                                             1A for 20 000                (2)         L1

      2.1.4 Balance = 2 000 – (10 50                           1M balance after sale of 10 T
                    = R1 500 M                                shirts
                 = 50 M                                       1M division
                                                               1CA total number of T-shirts
            Total T-Shirts = 50 + 10                                                                    F
                          = 60 CA                                                          (3)         L2

2.2   2.2.1 D = 24 901 – 23 901 M                             1M subtracting values                    F
             = 1000 kwh A                                     1A simplifying               (2)         L1

      2.2.2 Costs = 100 132,70+50         155,30 M            1M multiplication and
                  = 13 270 + 7 765                             addition
                  = 21 035 M                                  1M simplification
                           CA                                 1CA dividing by 100
                                                                                                        F
                  = R210,35                                                                 (3)         L2

      2.2.3 VAT amount=      210,35 M                         1M multiplication of 15% by
                                                               210,35                                   F
                      = R31,55 A
                                                               1A simplification        (2)             L1

2.3   2.3.1 Inflation is the increase in price of goods and    2 Explanation
                                                                                                        F
            services e.g. the price of petrol increases with
                                                                                                        L1
            time.  O                                                                      (2)

      2.3.2 Price increase = 16,21 – 16,02 M                  1M subtraction of values
                           = 19c OR R0,19 A                   1A difference (with units cents          F
                                                               or Rands)                  (2)           L1

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 (EC/JUNE/2019)                       MATHEMATICAL LITERACY P1                                          5

      2.3.3                  M                                       1CA from 2.3.2
              Distance = 5 50 2  500 km M                            1M distance for a day (50     2)
              Fuel used =   M 40      CA                            1M distance for 5 days
                            ,
                                                                      1M division by 12,5
              Increase in fuel cost= 40    0,19 = R7,60 CA           1CA simplification
                                                                      1CA increase
                                      OR
              Distance = 50 x 2  = 100 km for a day
              Fuel used =     =8ℓ
                            ,
              Increase in cost of fuel for 1 day = 8 x 0,19
                                                 = R1,52 
              Increase in cost of fuel for 5 days = R1,52 x 5 
                                                  = R7,60 

                                          OR
              2 x 50 x 5 = 500 km 
              Petrol used =     = 40 ℓ 
                                ,
              Cost of fuel after increase = 40 x 16,21 = R648,40 
              Cost of fuel before increase = 40 x 16,02 = R640,80                                           F
              Increase = R648,40 – R640,80 = R7,60                                               (5)       L3

2.4   2.4.1 Balance = R13 502,64 RT                                 2 RT                                  F
                                                                                                  (2)       L1

      2.4.2 Fee = 42,37+17,47+100,88 = R160,72  RTCA                1 RT adding correct values
                                                                      1 CA simplification                   F
                                                                                                (2)         L1
                                                                                               [29]

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  6                                         MATHEMATICAL LITERACY P1                    (EC/JUNE 2019)

QUESTION 3 [20 MARKS] MEASUREMENT
Ques.                   Solution                                        Explanation/Marks
                                                                        AO: FULL MARKS             Level
3.1.1                         M                                       1M Addition of correct
        Total distance = 6,3 + 13,02 + 5,04                            values                        M
                      = 24,36 km CA                                   1CA Simplification (2)        L1

3.1.2                       M                                         1M distance around
        Perimeter = 120 × 2 + 90 × 2                                   1A simplification             M
                 = 420 m (or 0,42 km) A                                                    (2)      L1

3.1.3                              ,                                   1 RT value of 1,6 km
        No of rounds =                  RT M
                                                                       1 M Division
                            C                                         1 C Conversion
                    = 3,81 CA                                         1 CA simplification
        Completed rounds =3 R                                         1 Rounding to completed
                                                                       rounds                        M
                                                                                           (5)       L2

3.1.4   :              %                           %                   1M substitution
                                                                       1CA simplification
                                                                       (NPR)
                               =        100%M
                                                                                                     M
                               = 97,65%CA                                                  (2)      L2

3.1.5                              M                                  1M conversion to m
                           ,                                           1M division
        Distance =          M
                                                                       1CA simplification
                 = 0,6665 CA                                                                        M
                                                                       1CA rounding       (4)
                 = 0,67m CA                                                                         L2

3.2.1        V    πr h                                                 1M substituting radius
                         M                                            1M substituting height
            V = 3,142 2.5   12,75                                      1CA simplification            M
              = 250,38 cm CA                                          (NPR)                (2)      L2

3.2.2   100 mℓ : 211 KJ                M                              1Mconcept of ratio
        250,38 : ?                                                     1M method
                                                                       multiplication of correct
                       ,                                               values and dividing by
        ?=                     M = 528,30 KJCA
                                                                       100                           M
                                                                       1CA simplification (3)        L2
                                                                                            [20]

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      (EC/JUNE/2019)                        MATHEMATICAL LITERACY P1                                   7

QUESTION 4 [11 MARKS] MAPS AND REPRESENTATIONS
Ques.                    Solution                                       Explanation/Marks
                                                                       AO: FULL MARKS             Level
4.1        South RT                                                  2RT correct direction      M&p
                                                                                           (2)      L1

4.2        RT                                                         1RT measuring scale
           2,5 cm : 20 km                                              1M conversion from
           1.6    : 10 1000            100 M                          km to cm
                                                                       1CA expressing in
                   1 :                                                 unit ratio
                               ,
                   1 : 800 000 CA
                                            OR
           2,5 cm : 20 km
           1 cm : 8 km x 1000 000                                                                 M &p
           1 : 800 000                                                                      (3)     L2

4.3        2,5 cm : 20 km                                              1CA from 4.2
           6,5 cm : ? (distance from Queenstown CBD and Whittesea)M   1M measuring
                          M                                           distance on map
                             ,
           distance =          M
                           ,                                           1M multiplication of
                       = 52 km CA                                     values
                                                                       1M division by 2,5
                                     OR                                1CA simplification (4)
           1 : 800 000 (CA from 4.2)
           6,5 : 800 000 6,5cm MM                                    1M measuring
                                   ,
                                                                       distance on map
           Distance =                  M                              1M multiplication
                                                                       1M conversion to km
                   = 52 km CA                                         1CA simplification
           (Accept 6,2 – 6,7 cm)                                                                  M &p
           (Accept 49,6 – 53,6 km)                                                          (4)     L2

4.4        Numeric scale  A                                          2A Correct scale
           Representative scale / Ratio scale A                                              M &p
           Verbal scale / Word scale A                                                   (2)   L1
                                                                                          [11]

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8                              MATHEMATICAL LITERACY P1                               (EC/JUNE 2019)

 QUESTION 5 [18 MARKS] DATA HANDLING
 Ques.                  Solution                                    Explanation/Marks
                                                                    AO: FULL MARKS Level
 5.1       Gauteng RT                                             2RT correct province   D
                                                                                       (2) L1

 5.2       1 145 861; 2 745 590; 3 509 953; 4 039 939; 5 404 868;   2RT arranging all            D
                                                                    values starting with         L1
           5 822 734; 6 562 053; 10 267 300; 12 272 263 RT        smallest.
                                                                                        (2)

 5.3       Limpopo RT                                             2RT middle position          D
                                                                    CA from 5.2 and              L1
                                                                    check province    (2)

 5.4       5.4.1 Free State and North West CA                     CA from 5.2                  D
                                                                    1 mark for each              L1
                                                                    province             (2)
           5.4.2 Eastern Cape and KwaZulu-Natal CA                CA from 5.2                  D
                                                                    1 mark for each              L1
                                                                    province             (2)

 5.5       IQR represents 50% A                                   2A                           D
                                                                                         (2)     L1

 5.6       No. of listeners in NW =      3 509 953M                1M 90 % of the               D
                                                                    population                   L2
                                  = 3 158 957,71S
                                                                    1S simplification
                                  = 3 158 958CA
                                                                    1CA (Rounded to the
                                                                    nearest whole number
                                                                    people cannot be
                                                                    decimals)
                                                                                      (3)

 5.7       % Mean =                               M                1M adding the                D
                                                                    percentages                  L2
                                                                    1M dividing by 9
                      =     M                                      1CA simplification
                                                                     (NPR)
                      = 89,78% CA                                                      (3)
                                                                                       [18]

                                                                                 TOTAL:         100

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