# Mathematical Literacy Grade 12 P1 2018 Memo (Eng)

> Grade 12 Mathematical Literacy Paper 1 memorandum from the Eastern Cape Mock Exam 2018 (English), 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-mathematical-literacy-2018-eastern-cape-mock-exam-memorandum-paper-1-english/a18cfa95-33e7-4b2c-8855-ccec00a4ae5e
- Updated: 2026-09-28
- Subject: Mathematical Literacy
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2018
- Exam period: Eastern Cape Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 10

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## About this document

Mathematical Literacy · Grade 12 · Eastern Cape Mock Exam · 2018 · 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 2018

              MATHEMATICAL LITERACY P1
                 MARKING GUIDELINE
MARKS:   150

   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 10 pages.

### Page 2

2                                      MATHEMATICAL LITERACY P1                      (EC/SEPTEMBER 2018)

QUESTION 1 [30 MARKS] INTEGRATED QUESTION
Question       Solution                                              Explanation/Marks
                                                                     AO: FULL MARKS                    Lev.
1.1   1.1.1    Inflation is the increase in the price of goods       2A Explanation                     F
               and services over time.A                                                               L1
                                       OR
               The decrease in the purchasing value of the
               money over time. ✓✓A                                                              (2)

      1.1.2    Difference = R125 – R120 = R5 M                      1M Difference                      F
                          = 5 × 100 = 500 cents A                   1CA In cents                (2)    L1

      1.1.3               112 . 5                                    1M Multiply112,5 by 125 and        F
                                     125
               Price =     100              M                       dividing by 100                    L1
                                                                     1CA Price
                     = R140,60 CA                                                OR
                                    OR                               1M For profit
                        12,5                                         1CA Price
               Profit = 100 × 125 = R15,625M     OR
                                                                     NPR
               Price = 125 × 1,12                                    (Concept of money)
               Price = 125 + 15,625 = R140,625 = 140, 625
                                   = R140,60 CA                                                 (2)
               = R140,60

1.2   1.2.1                                 90                       1MA Divide by 60 and              F
                Cost of a dozen =                 12    MA
                                            60                       multiply by 12                     L1
                                      = R18,00  CA                  1CA Cost
                                           OR
                             60
               Dozens =
                             12
                          =5                                        1MA Number of dozens
                                            90
               Cost of a dozen =
                                            5
                                                                     1CA Cost                    (2)
                                      = R18 

      1.2.2                     (( 60  75 )  90 )                  1M Subtraction for profit         FL1
                                                         100
               % Profit =               90                      M
                                                                     1M Fraction 45/90 multiply
                             45
                                                                     by 100
                          = 90 x 100% M                             1CA Percentage
                           = 50% CA                                                           (3)

1.3   Time = 24:00 – 16:25 = 7h 35 min M                            1M For 7h 35min
        From midnight to 3:30 am = 3h30 min M                       1M For 3h 30min
       ∴ Total time = 7 hours 35 min + 3h30min                       1CA Time                           M
                    = 11h 05 min CA                                                             (3)    L1

1.4                     2  1000                                     1M Correct numerator and
      Plastic cups =                M
                          275                                        denominator
                      = 7,2727                                       1A Rounding down
                      = 7 A                                                                     (2) ML1

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### Page 3

(EC/SEPTEMBER/2018)                    MATHEMATICAL LITERACY P1                                           3

1.5   1.5.1                2𝑐𝑚 × 75 𝑚                           1M for Numerator
                Length =                2M
                            100 𝑐𝑚                              1M for Denominator
                                                                                                 ML1
                                                                1CA With correct unit
                       = 1,5 m                                                           (3)

      1.5.2     Scale = 100 : 7 500 2M                        1M Ratio form
                      = 1 : 75  1CA                            1C Conversion
                                                                1CA Answer                        MP
                                                                                          (3)      L1

1.6   1.6.1     72,67% = Nelson Mandela District  A           2A District
                                                                                          (2)    DL1

      1.6.2     Districts are: Chris Hani East, OR Tambo        1A First 2 districts
                Coastal, A                                     1A Second 2 districts
                Amathole East and Amathole West A                                        (2)    DL 1

      1.6.3     Nelson Mandela, Sarah Baartman OR Tambo         1M Identifying 7 best
                Inland, Chris Hani West, Alfred Nzo West, Joe   performing districts in
                Gqabi, Buffalo City M                          order
                                                                1CA for middle district
                ∴ 𝑀𝑖𝑑𝑑𝑙𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑐𝑡 = CHRIS HANI WEST CA                                   (2)    DL1

      1.6.4                           26 ,1                     1M Correct values
                Candidates failed =           × 313 030 M
                                      100                       1CA Candidates
                                 = 81 701 CA
                                     OR
                         73,9
                Passed = 100 × 313 030
                                                                1MA Candidates passed
                       = 231 329 M                             1CA Candidates failed
                Number failed = 313 030 – 231 329               Accept 81 700         (2)
                              = 81 701CA                                                        DL1
                                                                                        [30]

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### Page 4

4                             MATHEMATICAL LITERACY P1                    (EC/SEPTEMBER 2018)

QUESTION 2 [46 MARKS] FINANCE
Question      Solution                                          Explanation/Marks
                                                                AO: FULL MARKS               Lev.
2.1   2.1.1 Value of A = R3 250 + R4 500 + R1 200  M           1M Addition
                       = R8 950  A                             1A Value of A
                                   OR
            Value of A = R7 950 + R1 000  M                    1M Addition
                      = R8 950 A                               1A Value of A         (2)     L2

      2.1.2 Value of B = R1 440 – R1 500 M                     1M Subtracting correct
                      = -R60  A                                values
            Value of B = R1 440 – R1 500 M                     1A Negative answer
                      = (R60)  A                               1M Subtracting correct
                                                                values
                                                                1A Answer in brackets
                                                                                     (2)

      2.1.3 Value of C = R2 600 – R200 M                       1M Subtraction
                       = R2 400  CA                            1CA Value of C
                                    OR
            Value of C = R5 780 – (R880 + R1 000 + R1 500) M   1M Subtraction
                        = R2 400 CA                            1CA Value of C        (2)     L1

      2.1.4 Delivery A                                        2A Correct answer     (2)     L1

2.2   2.2.1 Annual gross salary = R65 000 ×12 M                1M Multiply by 12
                                = R780 000 CA                  1CA Annual salary
                                                                                      (2)     L1

      2.2.2 Mrs John’s annual pension fund contribution:
                  7 ,5                                          1M Multiply by 12
              = ( 100 × 65 000) ×12 M
                                                                CA Pension contr.
              = R58 500 CA                                                           (2)     L1

      2.2.3 Medical Aid contribution = R1 050 ×12 M            1M Multiply by 12
                                     = R12 600 A               1A Contribution       (2)     L1

      2.2.4                                  75                 1M Correct values
              Mrs John’s performance bonus = 100 ×65 000 M     1M Multiply by 75%            L1
                                           = R48 750 A                               (2)

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### Page 5

(EC/SEPTEMBER/2018)                    MATHEMATICAL LITERACY P1                                       5

      2.2.5 Annual taxable income                                  CA from 2.2.1, 2.2.2
            = R780 000 + R48 750 – (58 500 + R12 600) M          and 2.2.3
            = R828 750 – R71 100 M                                1M Total income
            = R757 650 CA                                         1M Total contributions
                                                                   1M Subtraction
                                                                   1CA Taxable income (4)          L2

2.3   2.3.1 R701 301 and above RT                                2RT Correct group
                                                                                             (2)   L1

      2.3.2 Rebate Mrs John will receive = R13 500 + R7 407 M 1M Adding correct
                                         = R20 907 A          rebates
                                                               1A Total rebate      (2)            L1
      2.3.3 Actual tax = Income tax calculated on taxable      CA from 2.2.5, 2.3.1
            income Rebate                                      and 2.3.2
            = 206 964 + 41 % of the amount above 701 300 –     1SF Substitution
              R20 907                                          1S Simplification
            = 206 964 + 0,41 × (757 650 – 701 300) –20 907 SF 1S Simplification
            = 206 964 + 0,41 × 56 350 – 20 907 S              1CA Nearest rand
            = 206 964 + 23103,50 – 20 907 S                                        (4)
            = R209 160,50 CA                                                                      L3

      2.3.4 Net annual salary = Annual taxable income –            CA from 2.2.5 and 2.3.3
            Actual annual tax                                      1SF Substitution
                                                                                                   L1
                        = R757 650 – R209 160,50 SFM             1M Subtraction
                        = R548 489,50 CA                          1CA Net salary      (3)

2.4   2.4.1                       15 , 5                                                           L2
                       st
              Interest 1 year = 100 × 400 000
                               = R62 000 A                        1A Interest
              2nd year amount = R400 000 + R62 000
                               = R462 000 S                       1S New amount 2nd year
                                  15 , 5
                                                                   1S Interest in 2nd year
                       nd
              Interest 2 year   = 100  × R462 000                  1A Total interest
                                = R71 610S                                                  (4)
              Total interest    = R62 000 + R71 610
                                = R133 610  A

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### Page 6

6                             MATHEMATICAL LITERACY P1               (EC/SEPTEMBER 2018)

    2.4.2 Phone D:
                                 100                       1M Cost without VAT
            Cost excluding VAT = 115 × 1 750              1M Subtraction
                               = R1 521,74 M             1A VAT value
                           VAT = R1 750 – R1 521, 74 M
                               = R228,26 A
                                   OR                     1M Fraction
                    15                                    1M Multiplication
            VAT = 115M × 1 750 M                        1A VAT value          (3)
                 = R228,26 A                                                           L2

    2.4.3 Phone D : Phone E
                 3:2
                60 : E M                                 1M Ratio form
                     60  2

            E= 3                                          1CA Number of phones
             = 40 CA                                                        (2)

    2.4.4 Total cost = 60 × 1 750 + 40 × 2 000            CA from 2.4.3
                     = 105 000 M + 80 000 M             1M Cost for phones D
                                                                                        L1
                     = R185 000  CA                      1M Cost for phones E
                                                          1CA Total cost       (3)

    2.4.5            1: 0,52709                           1M Conversion
            185 000 : Total cost in CYN                   1A Total cost
                                                                                        L2
            Total cost = 185 000 × 0,52709 M             1A Answer in Yuan
                        = CYN 97511,65 AA                                     (3)
                                                                               [46]

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### Page 7

(EC/SEPTEMBER/2018)              MATHEMATICAL LITERACY P1                                       7

QUESTION 3 [25 MARKS] MEASUREMENT
Ques. Solution                                                  Explanation/Marks
                                                                AO: FULL MARKS              Level
3.1.1   Length = 4 880 mm ÷ 1 000 C                            1C Divide by 1000
               = 4,88 m A                                      1A Metres
                                                                                     (2)      L1

3.1.2   Distance A = 3 (150 mm) M                              1M Multiplication by 3
                   = 450 mm A                                  1A Distance
                                                                                    (2)       L1

3.1.3   Height of the wall = 2,1 m + 450 mm ÷ 1 000 C          CA from 3.1.2
                           = 2,1 m + 0,45 m S                  1C Conversion
                           = 2,55 m CA                         1S Simplification
                                                                1CA Height          (3)       L1

3.1.4   Area = Length × Width                                   CA from 3.1.1
                4,88                                            1M Dividing by 2
             = 2  × 2,1 m  M
                                                                1M Multiplication
             = 5,124 m2 CA Unit
                                                                1CA Area
                                                                1A Unit              (4)      L2

3.1.5   Area covered by bricks                                  CA from 3.1.1 and
        = Area of garage – Area of double door                  3.1.3
        = (2,55 m × 5,18 m)  – (4,88 m × 2,1 m) MM           1M Area of garage
        = 13,209 m2 – 10,248 m2 S                              1M Area of double door
        = 2,961 m2 CA                                          1M Subtraction
                                                                1S Simplification
                                                                1CA Area covered by
                                                                bricks             (5)        L2

3.2.1   Height of the bricks and cement = (12 x 2) + 76 mm M   1M Multiplication by 2
                                       = 100 mm CA             and addition
                                                                1CA Height
                                                                                    (2)       L1

3.2.2                             2 500                         CA from 3.2.1
        Number of rows of bricks = 100 M M
                                                                1M Conversion
                                                                1M Division
                               = 25 CA
                                                                1CA Number of rows
                                                                of bricks         (3)         L1

3.2.3   Volume = 23 cm ×11 cm × 7,6 cm SF C                   1SF Substitution
               = 1 922,8  A cm3 A                             1C Conversion
                                                                1A Volume
                                                                1A Units
                                                                                     (4)      L2
                                                                                    [25]

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### Page 8

8                            MATHEMATICAL LITERACY P1                (EC/SEPTEMBER 2018)

QUESTION 4 [31 MARKS] DATA HANDLING
Ques. Solution                                                         Explanation/Marks
                                                                       AO: FULL MARKS            Level
4.1.       Range = 3,316 kg – 0,182 kg RT M                          1RT Correct values
                                                                       1M Subtraction
                   = 3,134 kg CA                                      1CA Range
                                                                                        (3)           L2

4.2        1,668 kg A                                                2A Median
                                                                                          (2)         L2

4.3        Average =
           1,26 ×2 +1,371 ×9 +1,668 ×8 + 1,746 ×4 + 1,849 ×8 + 2,163   1M Adding
           +2,333 + 3,128 ×2  M                                       1M Concept of mean
               60,731
             = 35 M
                                                                       1R Average to
             = 1,735
                                                                       nearest kg
             = 2 kg R
                                                                                          (3)         L2

4.4        22  A                                                     2A Explanation
                                                                                          (2)         L1

4.5                         8                                          1M Fraction value
           Probability = 35 M × 100% M                               1M Multiply by 100
                       = 22,9 % CA                                    1CA Percentage
                                                                                         (3)          L1

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### Page 9

(EC/SEPTEMBER/2018)                  MATHEMATICAL LITERACY P1                                              9

4.6                                                                                                               L2

         4M For the first 4 bars plotted correctly
         1M For the next 2 bars plotted correctly
         1M For the last 2 bars plotted correctly                                                      (6)

4.7      0,182; 0,182; 0,182; 0,309; 0,729; 0,729; 0,729; 0,856; 0,856; 0,856; 0,856   1M Correct values          L1
         0,936; 2,448; 2,448; 2,449; 3,038; 3,316; 3,316; 3,316; 3,316; 3,316; 3,316   1M Value or
         Q2                                                                            position of median
                                              Q3                                       1CA Q3
                   0,856+0,936
         Q2=Median=     2
                               = 0,896  MM
         Q3=3,316  CA                                                                                  (3)

4.8      3,316 kg                                                                    2A RT Modal                L1
                                                                                       value
                                                                                                        (2)

4.9                           8                                                        1M Fraction                L1
         % of total fishes = 57 × 100 M                                              1M Multiply by 100
                           = 14,04 %  CA                                              1CA Percentage
                                                                                                      (3)

4.10     3rd hour                                                                    2RT Correct hour           L1
                                                                                                      (2)

4.11     1,1,1,1,2,3,3,4,6                                                           2M Arrangement             L1
                                                                                                    (2)
                                                                                                   [31]

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### Page 10

10                                    MATHEMATICAL LITERACY P1               (EC/SEPTEMBER 2018)

 QUESTION 5 [18 MARKS] MAPS, PLANS and OTHER REPRESENTATIONS
 Ques Solution                                   Explanation/Marks
                                                 AO: FULL MARKS Level
 5.1  There is no seat for Lundi here A        2A No seat          MP
                                                                 (2) L1

 5.2       South A                                              2A Direction                 L1
                                                                                        (2)

 5.3       A8  RP                                                1RP First seat               L1
           A11  RP                                               1RP Second seat
           A15  RP                                               1RP Third seat       (3)

 5.4       35 RP                                                2RP Available seats          L1
                                                                                     (2)

 5.5       B14  RP                                              2RP Asi’s seat no.           L1
                                                                                        (2)

 5.6       Row J  RP                                            2RP Furthest row             L1
                                                                                        (2)

 5.7       Side BB  RP                                          2RP Correct side             L1
                                                                                        (2)

 5.8                                20                            1A Numerator                 P
           P(seat from side AA) =         A A
                                    194                           1A Denominator               L2
                                                                  1CA Answer
                            = 0,103 CA                                                 (3)

                                                                                       [18]
                                                         TOTAL:                        150

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