# Mathematical Literacy Grade 12 P2 2026 Memo (Eng)

> Grade 12 Mathematical Literacy Paper 2 memorandum from the NSC November Mock Exam 2026 (English), 7 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-mathematical-literacy-2026-nsc-november-mock-exam-memorandum-paper-2/8f7f2d4c-c15c-483b-9282-3ba807ec89b3
- Updated: 2026-10-29
- Subject: Mathematical Literacy
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Mock Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 7

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## About this document

Mathematical Literacy · Grade 12 · NSC November Mock Exam · 2026 · English. Memorandum, 7 pages. Read online or download the PDF.

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## Document text

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### Page 1

hlayiso mock exams
                           Practice Series:
                           Grade 12 NSC Preparation
                           WWW.HLAYISO.COM

                                NATIONAL
                           SENIOR CERTIFICATE

                                   GRADE 12

                       MATHEMATICAL LITERACY P2
                          MOCK EXAMINATION 2026
                            MARKING GUIDELINES

  MARKS: 150

                     These marking guidelines consist of 7 pages.

Copyright reserved                                                  Please turn over

### Page 2

Mathematical Literacy/P2                         2                                   Hlayiso/Mock 2026
                                       NSC – Marking Guidelines

SYMBOL AND EXPLANATION
 Symbol   Explanation                                         Symbol    Explanation
 M        Method                                              A         Accuracy
 MA       Method with accuracy                                CA        Consistent accuracy
 C        Conversion                                          D         Definition
 F        Choosing the correct formula                        SF        Substitution in a formula
 J        Justification                                       O         Opinion/Reason/Example
 P        Penalty (e.g. no units, incorrect rounding)         R         Rounding
 RT/RG/RM Reading from a table/graph/map                      S         Simplification
NOTE:

    • If a candidate answers a question TWICE, only mark the FIRST attempt.
    • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines.
    • For verification questions, the final conclusion (J) is only awarded if supported by calculations.
    • Measurements from maps/plans: accept a tolerance of ±2 mm.

QUESTION 1

 1.1.1       P = R2 400 + R18,50 ×100 = R4 250                          ✓M method
             Q = R2 400 + R18,50 ×200 = R6 100                          ✓A P
                                                                        ✓A Q                        (3)

 1.1.2                                                     Income       ✓A starts at (0 ; 0)
              8000                                                      ✓A point e.g. (100 ; 3 500)
                                                           Cost
                                                                        ✓A straight line to
                                                                        (250 ; 8 750)
                                                                        ✓ label                    (4)

              4000

                 0
                     0           100             200

 1.1.3       Break-even point ≈ 145 kotas and R5 100                    ✓RG number of kotas
             (Accept 140 – 150 kotas; R5 000 – R5 200)                  ✓RG amount
             At the break-even point income equals total cost:          ✓✓O explanation             (4)
             Sipho makes neither a profit nor a loss.

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### Page 3

Mathematical Literacy/P2                      3                               Hlayiso/Mock 2026
                                    NSC – Marking Guidelines

 1.1.4       Kotas sold = 6 × 22 = 132                           ✓M 132 kotas
             Income = 132× R35 = R4 620                          ✓A income
             Cost = R2 400 + R18,50 ×132 = R4 842                ✓A cost
             Profit = R4 620 − R4 842 = −R222 (a loss of R222)   ✓CA loss
             The statement is NOT VALID.                         ✓J conclusion              (5)

 1.1.5       New cost per kota = R18,50 ×1,12 = R20,72           ✓M 12% increase
             Contribution per kota = R35 − R20,72 = R14,28       ✓M difference
             Break-even = R2 400 ÷ R14,28 = 168,07               ✓M dividing
             ≈ 169 kotas                                         ✓R rounding up             (4)

 1.2.1       Block 1: 350× R2,45 = R857,50                       ✓A block 1
             Block 2: 150× R3,10 = R465,00                       ✓A block 2
             Block 3: (620 − 500)× R3,65 = 120× R3,65 =          ✓M 120 kWh
             R438,00                                             ✓A block 3
             Total = R1 760,50                                   ✓CA total                  (5)

 1.2.2       New use = 620 × 0,8 = 496 kWh                       ✓M 496 kWh
             Cost = R857,50 +(496 − 350)× R3,10                  ✓M block 2 units
             = R857,50 + R452,60 = R1 310,10                     ✓A new cost
             Saving = R1 760,50 − R1 310,10 = R450,40            ✓M difference
             % decrease =
                           450,40
                                   × 100% = 25,58%               ✓A percentage
                          1 760,50                               ✓J conclusion              (6)
             25,58% > 25%, so the claim is VALID.

 1.2.3       R2,45 ×1,127 = R2,76 per kWh                        ✓M increase
                                                                 ✓A answer                  (2)

                                                                                           [33]

QUESTION 2

 2.1.1       Radius = 1,55 m ÷ 2 = 0,775 m                       ✓M dividing by 2
                                                                 ✓A answer                  (2)

 2.1.2       Volume = 3,142 × (0,775)2 × 2,18                    ✓SF radius
             = 4,114 m3                                          ✓SF height
             Capacity = 4,114 × 1 000 = 4 114 ℓ                  ✓A volume
                                                                 ✓C ×1 000
                                                                 ✓CA litres                 (5)

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### Page 4

Mathematical Literacy/P2                       4                                Hlayiso/Mock 2026
                                     NSC – Marking Guidelines

 2.1.3       Roof area = 9 × 7 = 63 m2                              ✓M roof area
             Rainfall = 25 ÷ 1 000 = 0,025 m                        ✓C 0,025 m
             Volume = 63 × 0,025 = 1,575 m3                         ✓A 1,575 m3
             Water reaching tank = 85% × 1,575 = 1,33875 m3         ✓M 85%
             = 1 338,75 ℓ                                           ✓CA litres                 (5)

 2.1.4       Days = 4 114 ÷ 180 = 22,86                             ✓M dividing
             = 22 full days                                         ✓A 22,86
                                                                    ✓R rounding down           (3)

 2.1.5       Three events = 3 × 1 338,75 = 4 016,25 ℓ               ✓M ×3
             4 016,25 ℓ < 4 114 ℓ                                   ✓A 4 016,25 ℓ
             The tank will NOT be full. The statement is NOT        ✓M comparing
             VALID.                                                 ✓J conclusion              (4)

 2.2.1       Wall area = 2(8 × 3) + 2(7 × 3) = 48 + 42 = 90 m2      ✓M wall method
             Doors = 2 × 0,9 × 2,1 = 3,78 m2                        ✓A 90 m2
             Windows = 4 × 1,5 × 1,2 = 7,2 m2                       ✓A doors
             Area to paint = 90 − 3,78 − 7,2                        ✓A windows
             = 79,02 m2                                             ✓M subtracting
                                                                    ✓ answer shown             (6)

 2.2.2       Two coats = 2 × 79,02 = 158,04 m2                      ✓M ×2
             Paint = 158,04 ÷ 8 = 19,76 ℓ                           ✓M dividing by 8
             Tins = 19,76 ÷ 5 = 3,95                                ✓A 19,76 ℓ
             ≈ 4 tins                                               ✓M dividing by 5
                                                                    ✓R 4 tins                  (5)

 2.2.3       Paint = 4× R649,99 = R2 599,96                         ✓CA paint cost
             Hours = 158,04 ÷ 12 = 13,17 ≈ 14 hours                 ✓M hours
             Labour = 14× R180 = R2 520                             ✓R 14 hours
             Total = R2 599,96 + R2 520 = R5 119,96                 ✓A labour
             R5 119,96 > R5 000, so the budget is NOT sufficient.   ✓J conclusion with total (5)

                                                                                              [35]

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### Page 5

Mathematical Literacy/P2                       5                           Hlayiso/Mock 2026
                                     NSC – Marking Guidelines

QUESTION 3

 3.1.1       F2                                                 ✓✓RM                        (2)

 3.1.2       North-east                                         ✓✓A                         (2)

 3.1.3       From the taxi rank, travel east along Main Road.   ✓✓RM Main Road east
             Turn left (north) into Mandela Drive.              ✓RM left into Mandela
             The clinic is on the right-hand side.              Drive
                                                                ✓RM clinic on the right (4)

 3.1.4       Bar scale: 20 mm represents 500 m                  ✓M measuring bar scale
             Distance =
                         63
                            × 500 m = 1 575 m                   ✓M using scale
                         20                                     ✓A 1 575 m
             = 1,58 km                                          ✓C km                  (4)

 3.1.5       20 mm : 500 m                                      ✓C converting to mm
             = 20 mm : 500 000 mm                               ✓M dividing by 20
             = 1 : 25 000                                       ✓A answer                   (3)

                      1,8 km
 3.2         Time =           = 0,4 hours                       ✓SF substitution
                     4,5 km/h
             = 0,4 × 60 = 24 minutes                            ✓A 0,4 h
                                                                ✓C minutes                  (3)

 3.3.1       22 tables                                          ✓✓A                         (2)

 3.3.2       Seats = 22 × 8 = 176                               ✓M seats
             People = 172 + 10 = 182                            ✓A 182 people
             176 < 182: there are NOT enough seats (6 short).   ✓J conclusion               (3)

 3.3.3       Actual side = 30 mm × 200 = 6 000 mm = 6 m         ✓M using scale
             Area = 6 × 6 = 36 m2                               ✓C 6 m
             Dancers = 36 ÷ 0,5 = 72 people                     ✓A 36 m2
                                                                ✓CA 72 people               (4)

 3.3.4       Floor area = 24 × 15 = 360 m2                      ✓A area
             Maximum = 360 ÷ 1,2 = 300 people                   ✓M dividing
                                                                ✓A answer                   (3)

                                                                                           [30]

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### Page 6

Mathematical Literacy/P2                              6                            Hlayiso/Mock 2026
                                            NSC – Marking Guidelines

QUESTION 4

 4.1.1       Total                                                      ✓M adding
             = 72,5 + 78,1 + 80,4 + 84,0 + 86,2 + 88,9 = 490,1          ✓M dividing by 6
             Mean = 490,1 ÷ 6 = 81,68%                                  ✓A answer                  (3)

                                88,9 − 72,5
 4.1.2       % increase =                   × 100%                      ✓M difference
                                   72,5
             = 22,62%                                                   ✓M dividing by 72,5
                                                                        ✓A answer                  (3)

 4.1.3       Range P = 88,9 − 72,5 = 16,4                               ✓A range P
             Range Q = 87,0 − 79,6 = 7,4                                ✓A range Q
             School Q has the smaller range, so its results were        ✓M comparing
             more consistent. The claim is VALID.                       ✓J conclusion              (4)

 4.1.4       School P: pass rate increased every year (steady           ✓✓O trend P
             upward trend).                                             ✓O trend Q
             School Q: pass rate fluctuated but generally increased.    ✓O school P
             School P is likely to have the higher pass rate in 2026,   ✓O reason                  (5)
             because it improved faster and was already higher in
             2025.

 4.2.1       Minimum = 34                                               ✓A
             Q1 = 48                                                    ✓A
             Median = 60                                                ✓A
             Q3 = 71                                                    ✓A
             Maximum = 90                                               ✓A                         (5)

 4.2.2                                                                  ✓A box from 48 to 71
                                                                        ✓A median at 60
              30     40    50     60   70     80   90                   ✓A whisker to 34
                                                                        ✓A whisker to 90           (4)

 4.2.3       Learners with 50% or more: 11                              ✓RT 11 learners
             11                                                         ✓M percentage
                × 100% = 73,33%
             15                                                         ✓A 73,33%
             73,33% > 70%, so the claim is VALID.
                                                                        ✓J conclusion              (4)

 4.3         The vertical axis starts at 70% instead of 0%, which       ✓✓O explanation            (2)
             makes the increase in the pass rate look much bigger
             than it is.

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### Page 7

Mathematical Literacy/P2                         7                              Hlayiso/Mock 2026
                                       NSC – Marking Guidelines

                                                                                            [30]

QUESTION 5

 5.1.1       Cash price = R12 999 ×0,95 = R12 349,05                ✓M 5% discount
                                                                    ✓A answer                (2)

 5.1.2       Deposit = 10%× R12 999 = R1 299,90                     ✓A deposit
             Instalments = 24× R620 = R14 880                       ✓A instalments
             Total = R16 179,90                                     ✓CA total                (3)

 5.1.3       A = 12 999(1 + 0,18 × 2)                               ✓SF substitution
             = 12 999 × 1,36                                        ✓M simplifying
             = R17 678,64                                           ✓A answer                (3)

 5.1.4       Difference = R16 179,90 − R12 349,05 = R3 830,85       ✓M difference
                         3 830,85                                   ✓M percentage of cash price
             % more =             × 100% = 31,02%
                        12 349,05                                   ✓A 31,02%
             31,02% > 30%, so the statement is VALID.               ✓J conclusion            (4)

 5.2         Investment: A = 11 000(1 + 0,085)3 = R14 050,18        ✓SF substitution
             Laptop price: A = 12 349,05(1 + 0,045)3 = R14 092,32   ✓A R14 050,18
             R14 050,18 < R14 092,32 (short by R42,14)              ✓SF substitution
             The statement is NOT VALID.                            ✓A R14 092,32
                                                                    ✓M comparing
                                                                    ✓J conclusion            (6)

 5.3.1       3% × 2 400 = 72 laptops                                ✓M✓A                     (2)

               1
 5.3.2             = 33,33                                          ✓M dividing
             0,03
             1 in 33                                                ✓A answer                (2)

                                                                                            [22]

                                                                                  TOTAL:     150

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