# Mathematical Literacy Grade 12 P2 2019 Memo (Eng)

> Grade 12 Mathematical Literacy Paper 2 memorandum from the Eastern Cape Mock Exam 2019 (English), 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/maths-lit-p2-memo-gr12-sept2019-english/8e666634-65ff-4adf-997e-9e006994bf46
- Updated: 2026-09-27
- Subject: Mathematical Literacy
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2019
- 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 · 2019 · English. Memorandum, 10 pages. Read online or download the PDF.

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## Document text

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### Page 1

NATIONAL
                           SENIOR CERTIFICATE

                                    GRADE 12

                             SEPTEMBER 2019

                 MATHEMATICAL LITERACY P2
                    MARKING GUIDELINE

MARKS:     150

Symbol           Explanation
M                Method
M/A              Method with accuracy
MCA              Method with consistent accuracy
CA               Consistent accuracy
A                Accuracy
C                Conversion
S                Simplification
RT/RG/RM         Reading from a table OR Reading from a graph OR Read from 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 OR Reason
AO               Answer only
NPR              No penalty for rounding

                        This marking guideline consists of 10 pages.

### Page 2

2                             MATHEMATICAL LITERACY P2                            (EC/SEPTEMBER 2019)

    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 aflesings vanaf ŉ grafiek, tabel, uitlegplan en kaart geneem
        en ekstra antwoorde gee, penaliseer vir elke ekstra verkeerde item.

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### Page 3

(EC/SEPTEMBER 2019)                  MATHEMATICAL LITERACY P2                                       3

QUESTION 1 [37]
Ques.                                 Solution                               Explanation                Topic
                                                                                                         &
                                                                                                        Level
                                           M                              1RT Correct opening            L2
1.1.1                             RT      0,8125%                         balance                         F
          Interest Amount = R749 299,39 ×               A
                                                  100                     1M Multiply
                          = R6 088,06                                     1A Monthly rate
                                            OR
                                 RT       M      A
          Interest Amount = R749 299,39 × 0,008125
                          = R6 088,06                                                         (3)
1.1.2                             750 000                                 1RT Home loan                  L2
          Monthly repayment =          × 8,59      RT
                                   1 000
                                       M                                  1M Divided by                  F
                              = R6 442,50                                 1 000
                                                                          1RT Correct factor
                                     OR
                                   RT      RT                             1RT Home loan
          Monthly repayment = 750 000 × 8,59                              1RT Correct factor
                                  6 442 500
                              =                                           1M Divided by
                                    1 000     M                           1 000
                             = R6 442,50
                                                                                           (3)
1.1.3     Closing balance for Month 3                                     1A Add and Subtract            L2
          = R749 299,39 + R6 088,06 – R6 442,50 A                         1A Closing balance             F
          = R748 944,95 A                                                                  (2)
1.1.4     Statement not valid                                             1A Not valid                   L4
                     R                                        R                                          F
          The shorter the loan period, the higher the monthly repayment   1R Shorter period
                        R                OR        R                      1R Higher MRP
          The shorter the loan period, the higher the loan factor
                                         OR
          No A                 R                    R                     1A No
          The loan factors are higher with shorter periods                1R Higher loan
                                                                          factor
                                                                          1R Shorter periods
          Accept any other relevant response                                               (3)
1.1.5     Amount at 119 months = R6 442,50 × 119                          1MA Multiplying                L4
                                = R766 657,50 CA                          correct values                 F
                                                                          1CA Amount
          Difference = R766 657,50 – R750 000       M                     1M Subtraction
                      = R16 657,50 CA                                     1CA Difference
          Statement valid        O                                        1O Valid         (5)
1.1.6     It will reduce the interest amount.                             2A Explanation                 L4
          Accept any other relevant response                                               (2)           F
1.2.1                                 C                                   1SF Substitution               M
          Volume = 2 m × 2 m × 0,3 m SF                                   1C Convert mm to m             L2
                    = 1,2 m3 CA                                           1CA Answer in m3
                                                                                           (3)

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### Page 4

4                                MATHEMATICAL LITERACY P2                    (EC/SEPTEMBER 2019)

1.2.2 Bags of cement = 1,2              M                           CA from 1.2.1             M
                             0,26                                   1M Ratio concept          L3
                            = 4,615384615 × 2           M
                            = 9,2 bags S                            1M Multiply by 2
                            ≈ 10 bags CA                            1S Simplification
                                       OR                           1R Number of bags
         0,26 = 2 bags
                  M         M
                2   1,2
          1,2 = ×                                                   1M Ratio concept
                1    0,26
              = 9,2 bags S
              = 10 bags R                                           1M Multiply by 2
                                                                    1S Simplification
                                                                    1R Number of bags
                                                                                      (4)
1.2.3 Litres of paint needed = 15 × 1,08                            1MA Increased             M
                             = 16, 2 m2 MA                          surface area              L3
                                        16,2                        1M Dividing by
                                    =               M
                                         5                          spread rate
                                    = 3,24 litres
                                    = 3,24 litres × 2
                                         6,48 CA                    1CA Total number
                                    =
                                        5‐𝑙𝑖𝑡𝑟𝑒 𝑡𝑖𝑛𝑠                of litres
                                = 1,296 tins                        1CA Number of 5-
         Number of 5-litre tins = 2 tins CA                         litre tins
                                      OR
         Litres of paint needed = 15 × 108%                         1MA Increased
                                = 16,2 m2 MA                        surface area
                      Two coats = 16,2 × 2                          1CA Two coats
                                = 32,4 m2 CA                        1M Dividing
                                             32,4
         Number of 5-litre tins =                       MM          1CA Number of 5-
                                        5‐𝑙𝑖𝑡𝑟𝑒 𝑡𝑖𝑛𝑠
                             = 6,6 litres                           litre tins
                             = 2 × 5 litres CA                                     (4)
1.3.1 Price for 2015 = 1 251 158,39 + (1 251 158,39 × 0,05)     MA 1MA Increase by 5%         L3
                     = 1 251 158,39 + 62 557,9195 S                1S Simplification          F
                     = R1 313 716,31 CA                            1CA Amount 2015

      Price for 2016 = 1 313 716,31 + (1 313 716,31 × 0,04)
                     = 1 313 716,31 + 52 548,65238                  1CA Amount 2016
                     = R1 366 264,96 CA
                                   OR             M      M
      Price from 2015 – 2016 = 1 251 158,39 × 1,05 × 1,04 S
                              = R1 366 264,96 CA                                      (4)
1.3.2                                    RG                         1RG Subtract correct      L3
                           1 598 366,77 −1 029 331                  values                    F
      Percentage change =                          × 100%
                                  1 029 331 RG                      1RG Correct
                           569 035,77
                         =             × 100%                       denominator
                                 1 029 331                          1S Simplification
                                                                    1R Nearest %
                                = 0,5528… × 100%            S
                                = 55% R
                                                                                      (4)
                                                                                     [39]

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### Page 5

(EC/SEPTEMBER 2019)               MATHEMATICAL LITERACY P2                                  5

QUESTION 2 [36]
Ques.                              Solution                        Explanation                  Topic
                                                                                                 and
                                                                                                Level
2.1.1     Taxable Income = Gross Annual salary – Pension           1M Subtract 7,5%               L2
                          = 401 137,75 – (0,075 × 401 137,75) M    1S Simplification              F
                          = 401 137,75 – 30 085,33 S               1CA Taxable Income
                          = R371 052,42 CA                                            (3)
2.1.2     Annual Tax                      A                        CA from 2.1.1                 L4
          = 63 853 + 31% of taxable income above 305 850           1A Correct Rates of           F
          = 63 853 + 0,31 × (371 052,42 – 305 850) SF              Tax
          = 63 853 + 0,31 × 65 202,42                              1SF Substitution
          = 63 853 + 20 212,75                                     1S Simplification
          = 84 065,75 S – 13 635 M                                 1M Subtract rebate
          = 70 430,75 – 12 456 M                                   1M Subtract MAC
          = R57 974,75 CA                                          1CA Annual tax

          Annual School fees = 2 500 × 11 + 3 200 × 11     MA      1MA School fees × 11
                             = 27 500 + 35 200                     1CA Annual School
                             = R62 700 CA                          fees
                                                                   1O Invalid
          Claim not valid O                                                         (9)
2.2.1     Probability other than African = 8,8% + 9,5% + 2,4%      1MA Adding correct            L2
                                         = 20,7% A                 values                        P
                                                                   1A Probability
                                        OR
          Probability other than African = 100% – 79,3% M          1M Subtract from 100
                                         = 20,7% CA                1CA Probability (2)
2.2.2     African % = 100% – 8,8% – 9,5% – 2,4% MA                 1MA Subtract from             L3
                     = 79,3% A                                     100                           D
                                                                   1A African %
          African Population in 2004 = 79,3% × 46,66 million MCA   1MCA Calculate %
                                    = 37, 001 380 million   CA     1CA Total for 2004 in
                                   OR 37 001 380                   millions

          African Population in 1911 = 67,3% × 5 972 757
                                     = 401 9665,46 CA              1CA Total for 1911
          Difference = 37 001 380 – 401 9665,46 M                  1M Subtract
                     = 32 981 714, 54                              1CA Difference
                     ≈ 32 981 714 CA                               NPR                (7)
2.2.3     1911 = 0,08 × 5 972 757                                  1MA % for 1911                L2
               = 477 820,56 CA                                     1CA Coloured                  D
                                                                   population 1911
          2004 = 0,08 × 46 660 000                                 1CA Coloured
               = 3 732 800 CA                                      population 2004 (3)
2.2.4     Percentage of the African race group increased           1A African increase           L4
          Percentage of the Indian race group decreased A          1A Indian decrease
                                                                                      (2)

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### Page 6

6                           MATHEMATICAL LITERACY P2              (EC/SEPTEMBER 2019)

2.3.1     Number of people = (25 × 2) + 17                   1MA × 2 and addition         L2
                             = 50 + 17 MA                    1CA Number of                D
                             = 67 people CA                  people            (2)
2.3.2     Cost for Option 1 = 1 500 + (250 × 67)             CA from 2.3.1                L4
                             = 1 500 + 16 750                1MCA Adding and              F
                             = R 18 250 CA                   Multiplying
          Cost for Option 2     MA           MCA             1CA Option 1 cost
          Cost for couples = 0,96 × 270 × 50
                           = R12 950 CA                      1MA % decrease
          Cost for singles = 270 × 17                        1MCA × 50
                           = R4 590 CA                       1CA Couples cost
          Total for couples and singles = R12 950 + R4 590   1CA Singles cost
                                        = R 17 550 CA
          Statement invalid O                                1CA Option 2 cost
                                                             1O Invalid        (8)
                                                                              [36]

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### Page 7

(EC/SEPTEMBER 2019)                       MATHEMATICAL LITERACY P2                                7

QUESTION 3 [39]
Ques.                                 Solution                Explanation                       Topic
                                                                                                 and
                                                                                                Level
3.1.1     North West OR West of North              A          2A Direction                     M&PL2
                                                                                         (2)
3.1.2         A                                               1A Measure scale             M&PL3
          Scale: 3,8 cm = 20 miles                            1A Measured distance
          Distance = 8,6 cm A                                 1M Dividing and
                                20                            multiplying
          Actual distance =           × 8,6   M
                                3,8                           1CA Distance to 3 dec places
                              = 45,263 miles CA
                                                              Scale Measured Accept
                                                              3,6 – 4 cm
                                                              Measured distance Accept
                                                              8,4 – 8,8 cm
                                                                                        (4)
3.1.3     3,8 cm = 20 miles                                   CA from 3.1.2                      M
          Convert miles to kilometres = 20 miles × 1,609      1MA Convert to miles to km         L3
                                       = 32,18 km MA          1MA Convert km to cm
          Convert km to cm = 32,18 × 100 000                  1S Simplification
                              = 3 218 000 cm MA               1R Nearest million
          Unit scale = 3,8 cm : 3 218 000 cm                  Penalise for units in unit
                     = 1 cm : 848 842,1053 S                  ratio
                     ≈ 1 : 1 000 000 R                                                  (4)
3.1.4     Noise pollution                                     2R Reason                        M&PL4
                                   OR
          Air pollution     R
                                   OR
          Danger      R
                                   OR
          Length of runways       R
          Accept any other relevant answer                                               (2)
3.1.5     A 61                                                2A Road                    (2)   M&PL2
3.1.6     Distance = Speed × Time                             1SF Substitution                  M
          78 miles = 40 miles per hour × Time SF              1M Changing subject of the        L4
               M 78                                           formula
          Time =
                     40                                       1S Time in hours
                 = 1,95 hours S                               1C Time in hours and min
          Time in hours and minutes = 1h 57 minutes C         1M Adding times
          Arrival time = 07:20 + 1:57 M                       1CA Arrival time
                       = 09:17 CA                             1O Invalid
          Statement invalid O
                                 OR                           1M Changing subject of the
                     𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒                                 formula
          Time =                 M                            1SF Substitution
                      𝑆𝑝𝑒𝑒𝑑
                     78                                       1S Time in hours
          Time =       SF                                     1C Time in hours and min
                     40
               = 1,95 hours S                                 1M Travel time
               = 1h 57 minutes C                              1O Not valid
          Travel time = 7:20 to 9:15                          1O 2 min late
                       = 1h 55 minutes M
               O                                                                         (7)
          Not valid, she will be 2 minutes late    O

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### Page 8

8                                                 MATHEMATICAL LITERACY P2                     (EC/SEPTEMBER 2019)

3.2.1      Range = Highest – Lowest                                                      1M Concept of         D
            15°C = 13°C – Lowest M                                                       range                 L2
           Lowest = 13°C – 15°C RT                                                       1RT Correct
                  = -2°C A                                                               values
                                                                                         1A Lowest
                                                                                         value       (3)
3.2.2 (a) Mistakes:                                                                                            D
          Data was not arranged A                                                        1A Mistake 1          L3
          Calculation was done without using the BODMAS-rule                       A     1A Mistake 2
                                                                                                        (2)
3.2.2 (b) Correction:                                                                                          D
          3 ; 4 ; 6 ; 7 ; 10 ; 13 ; 14 ; 19 ; 19 ; 22 ; 24 ; 24          M               1M Arrange            L3
                                     13 + 14                                             data ascending
           Median =
                                          2                                              or descending
                                     27
                                 =
                                     2
                   = 13,5 °C A                                                           1A Median (2)
3.2.3      June, July and August  A                                                      1A Correct            D
           Minimum temperatures are high R                                               months                L4
           Maximum temperatures are high R                                               1R Min high
                                                                                         1R Max high
                                                                                                      (3)
3.2.4                        A                𝟓                                          CA from 3.2.1         P
           Probability =
                         𝟏𝟐 A                                                            1A Numerator          L2
                       = 0,4166…
                                                                                         1A Denominator
                       = 0,417 CA
                                                                                         1CA 3 dec places
                                                                                                      (3)
3.2.5                                                                                    CA from 3.2.1         D
                                      Average Minimum and Maximum                        1CA January           L2
                                         temperatures in Frankfurt                       1A Feb – Apr
                            30                                                           1A May – Jul
                                                                                         1A Aug – Nov
                            25
                                                                                         1A Dec
                            20

              Temperature
                            15
                                                                A
                            10
                                                      A
                             5
                                                                               A
                             0
                                     JanAFeb Mar Apr May Jun Jul Aug Sep Oct Nov Dec A
                            -5
                                                          Months
                                                                                                      (5)
                                                                                                     [39]

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### Page 9

(EC/SEPTEMBER 2019)                    MATHEMATICAL LITERACY P2                                       9

QUESTION 4 [38]
Ques.                                     Solution                                Explanation          Topic
                                                                                                        and
                                                                                                       Level
4.1.1     Area of unshaded part                                                1SF Correct values        M
          = Area of rectangle – Area of ▲1 – Area of ▲2                        for rectangle             L3
                                1                     1
          = (Length × Width) – (2 × base × height) – (2 × base × height)       1SF Correct values
                                                                               ▲1
                      SF                   SF                    SF            1SF Correct values
          = (130 cm × 25 cm) – (0,5 × 50 cm × 25 cm) – (0,5 × 50 cm × 15 cm)
          = 3 250 cm2 – 625 cm2 – 375 cm2 M S                                  for ▲2
          = 2 250 cm2 CA × 5         M                                         1M Subtracting
                                 2
          Total area = 11 250 cm CA                                            1S Simplification
                                        OR                                     1CA Area of
          Area of rectangle = Length × Width                                   unshaded part
          Area of 5 panels = 130 cm × 25 cm SF                                 1M Multiply by 5
                           = 32 500 cm2 × 5 M                                  1CA Total area
                           = 16 250 cm2 CA                                     OR
                                   1
          Area of ▲1 (5 panels) = × base × height                              1SF Correct values
                                   2                                           for rectangle
                                = 0,5 × 50 cm × 25 cm × 5             SF       1M Multiply by 5
                                = 3 125 cm2 CA                                 1CA Area of rec
                                      1
          Area of ▲2 (5 panele) = × base × height                              1SF Correct values
                                  2
                                = 0,5 × 50 cm × 15 cm × 5                      ▲1
                                = 1 875 cm2 CA                                 1CA Area of ▲1
          Total Area = 16 250 cm2 – 3 125 cm2 – 1 875 cm2 M                    1CA Area of ▲2
                     = 11 250 cm2 CA                                           1M Subtracting
                                                                               1CA Area of
                                                                               unshaded part (8)
4.1.2     To glue the seams together                                           2R Reason                   M
                                            OR                                                             L4
          To paste sides together   R                                                           (2)
4.1.3     Diagram T – A ring should be attached to the wide side of the        1A Explain              M&P
          hot air balloon A                                                    Diagram T                L2
          Diagram U – A paper clip to be attached to the ring A                1A Explain
                                                                               Diagram U        (2)
4.1.4     To blow hot air in balloon                                           2R Reason               M&P
                                                                                                (2)     L4
4.2.1                            A                                             2A Temperature          M&P
          When the temperature increases, the lift of the hot air balloon is   increases                L4
          higher.     A                                                        2A Hot air balloon
                                      OR                                       higher
                             A                                    A
          When the temperature decreases, the lift of the hot air balloon is
          lower.                                                                                (4)

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### Page 10

10                                MATHEMATICAL LITERACY P2                         (EC/SEPTEMBER 2019)

4.2.2                                          0,972+0,946                   1MA Concept of                M
          Air Density of Hot air balloon B =
                                                   2
                                              1,918   MA                     average                       L4
                                            =
                                                2
                                            = 0,959 kg/m3 CA                 1CA Air Density

          Lift = (Air density outside of the hot air balloon – Air density
          inside hot air balloon) × Volume of the hot air balloon
                      RT             RT                                      1RT Correct outside
          Lift = (1,204 kg/m3 – 0,959 kg/m3) × 2 400 m3 SF                   air density
               = 0,245 kg/m3 S × 2 400 m3                                    1RT Correct inside
               = 588 kg CA                                                   air density
                                                                             1SF Substitution
                                                                             1S Simplification
                                                                             1CA Lift          (7)

4.2.3                     33+12                                              1A Numerator                  P
          Probability =
                            93
                          45      A                                          1A Denominator                L2
                      =           A                                          1CA Common
                          93
                          15      CA
                      =                                                      fraction in the
                          31
                                                                             simplest form   (3)

4.3       Accommodation = 1 030 × 4 nights × 4 people A                      1A 4 nights × 4               F
                            = R16 480 CA                                     1CA Cost for                  L3
          Hot air balloon rides = 750 – (750 × 0,15) MA                      accommodation
                                = 750 – 112,50                               1MA Less 15%
                                       S                                     1S Cost per person
          Cost per person        = R637,50 × 4                               1CA Cost for rides
                                 = R2 550 CA
          Total cost in Rand = R16 480 + R2 550                              1CA Total cost
                              = R19 030 CA                                   1C Dividing
                          19 030                                             1CA Cost in USD
          Cost in USD =            C
                           13,63
                                                                             1M Multiply
                        = $1 396, 184886 CA
                                                                             1CA Cost in Turkish
          Cost in Turkish Lira = $1 396,184886 × 5,25 M
                                                                             Lira
                                 = 7 329,97 Turkish Lira CA
                                                                                            (10)
                                                                                            [39]

                                                                                        TOTAL:         150

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## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

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Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.
