# Mathematical Literacy Grade 11 P2 2026 Memo (Eng)

> Grade 11 Mathematical Literacy Paper 2 memorandum from the November Mock Exam 2 2026 (English), 7 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-11-mathematical-literacy-2026-november-mock-exam-2-memorandum-paper-2/573f0a85-a015-40a2-866d-ceda7977c700
- Updated: 2026-10-29
- Subject: Mathematical Literacy
- Grade: Grade 11
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam 2
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 7

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## About this document

Mathematical Literacy · Grade 11 · November Mock Exam 2 · 2026 · English. Memorandum, 7 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Matching paper and memo

- [Grade 11 Mathematical Literacy P2 November 2026 Mock 2](https://www.hlayiso.com/library/grade-11-mathematical-literacy-2026-november-mock-exam-2-question-paper-paper-2/9e28596c-b065-4744-84e1-a7995223ca68) — QP · 2026 · Eng

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

Mock

                             hlayiso mock exams
                             Practice Series:
                             Grade 11 Exam Preparation
                             WWW.HLAYISO.COM

                                  NATIONAL
                             SENIOR CERTIFICATE

                                     GRADE 11

                         MATHEMATICAL LITERACY P2
                     NOVEMBER EXAMINATION 2026 (MOCK)
                              MARKING GUIDELINES

MARKS: 100

                       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/RM Reading from a table/map                         S         Simplification
 NPR    No penalty for rounding                         AO        Answer only, full marks
KEY TO TOPIC AND LEVEL: M = Measurement; MP = Maps, plans and other representations; P
= Probability; L1–L4 = cognitive levels 1 to 4
NOTE:

    • If a candidate answers a question TWICE, only mark the FIRST attempt.
    • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines; however, it stops at
      the second calculation error.
    • Answers only, without working, are awarded full marks unless stated otherwise. In ”verify”
      questions the conclusion (J) is only awarded if it is supported by a calculation.

QUESTION 1

 1.1.1       Sugar = 3 × 15 mℓ                                        ✓C multiplying by 15
             = 45 mℓ                                                  ✓A answer
                                                                      M   L1                     (2)

 1.1.2       (Cooking/deep-fry/oil) thermometer                       ✓✓A instrument
                                                                      M   L1                     (2)

 1.1.3       ◦ C = (350 − 32) ÷ 1,8                                   ✓SF substitution
             = 318 ÷ 1,8                                              ✓S simplification
             = 176,666 . . . ≈ 176,7 ◦ C                              ✓R answer to 1 decimal
                                                                      place
                                                                      M   L2                     (3)

 1.1.4       Number of recipes = 150 ÷ 24 = 6,25                      ✓M dividing by 24
             Flour = 6,25 × 1 kg = 6,25 kg                            ✓CA 6,25 kg flour
             Bags = 6,25 ÷ 2,5 = 2,5                                  ✓M dividing by 2,5
             ⇒ 3 bags                                                 ✓R rounding up to 3 bags
             (Alternative: flour per vetkoek = 1 000 g ÷ 24 = 41,67   M   L3                     (4)
             g; 150 × 41,67 g = 6 250 g)

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### Page 3

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

 1.1.5       05:50 + 20 min = 06:10                                ✓M adding kneading time
             06:10 + 1 h 15 min = 07:25                            ✓M adding rising time
             07:25 + 1 h 40 min = 09:05                            ✓CA finishing time 09:05
             09:05 is 5 minutes AFTER the market opens.            ✓J conclusion
             ∴ The claim is NOT valid.                             M   L3                     (4)
             (OR total time = 20 + 75 + 100 = 195 min = 3 h 15
             min; 05:50 + 3 h 15 min = 09:05)

 1.2.1       650 mℓ                                                ✓✓A reading
             (Accept 640 mℓ – 660 mℓ)                              M   L1                     (2)

 1.2.2       650 mℓ − 625 mℓ                                       ✓M subtracting 625 mℓ
             = 25 mℓ                                               ✓CA answer
             (CA from 1.2.1)                                       M   L2                     (2)

 1.3.1       Perimeter is the total distance (length) around the   ✓✓D definition
             outside of a shape / figure.                          M   L1                     (2)

 1.3.2       Diameter = 1,2 × 100 = 120 cm                         ✓C converting to cm
             C = 3,142 × 120 cm                                    ✓CA answer
             = 377,04 cm                                           (Answer in m, 3,77 m: max 1
             (Accept 377 cm; NPR)                                  mark)
                                                                   M L1                    (2)

                                                                                           [23]

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### Page 4

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

QUESTION 2

 2.1.1       N16                                                  ✓✓RM
                                                                  MP    L1                   (2)

 2.1.2       South-east (SE)                                      ✓✓A direction
                                                                  MP    L2                   (2)

 2.1.3       Map distance = 10,8 cm                               ✓A measuring map distance
             (Accept 10,6 cm – 11,0 cm)                           ✓A measuring bar
             Bar scale: 4 cm represents 40 km                     ✓M 1 cm = 10 km
             ∴ 1 cm represents 10 km                              ✓M multiplying
             Actual distance = 10,8 × 10 km                       ✓CA answer in km
             = 108 km                                             MP L3                  (5)
             (Accept 106 km – 110 km; otherwise CA from the
             candidate’s own map and bar measurements, i.e. map
             cm × 40÷ bar cm)

 2.1.4       4 cm : 40 km                                         ✓M bar length and distance
             = 4 cm : 4 000 000 cm                                ✓C converting to cm
             = 1 : 1 000 000                                      ✓CA simplified scale
             (Accept 1 : 975 610 – 1 : 1 025 641 (CA) from a      MP L3                    (3)
             measured bar of 3,9 cm – 4,1 cm)

 2.1.5       Towns on a national road: Lindela, Siyakhula,        ✓RM 5 towns
             Khanyisa, Emthonjeni, Nqabeni = 5                    ✓M dividing by 8
                 5
             P = × 100%                                           ✓CA percentage
                 8                                                P    L2                    (3)
             = 62,5%

                             150 km
 2.2.1       Driving time =          = 1,666 . . . h              ✓SF substitution
                             90 km/h
             = 1 h 40 min                                         ✓C 1 h 40 min
             Total time = 1 h 40 min + 20 min = 2 h               ✓M adding the stop
             Arrival = 07:50 + 2 h = 09:50                        ✓CA arrival time
             10:00 − 09:50 = 10 minutes before the start          ✓J conclusion
             10 min < 15 min ∴ The claim is NOT valid.            M    L3                    (5)

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### Page 5

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

 2.2.2       Return distance = 150 × 2 = 300 km                   ✓M return distance
             Diesel = 300 ÷ 9,6 = 31,25 ℓ                         ✓M dividing by 9,6
             Cost = 31,25× R21,35 = R667,19                       ✓M multiplying by R21,35
             Per player = R667,19 ÷15 = R44,48                    ✓CA amount per player
                                                                  (Accept R44,47 – R44,48)
                                                                  M L3                     (4)

                                                                                           [24]

QUESTION 3

 3.1.1       Volume is the amount of space that a 3-dimensional   ✓✓D definition
             object takes up (occupies).                          M   L1                    (2)

 3.1.2       Radius = 7 ÷ 2 = 3,5 cm                              ✓A radius
             V = 3,142 × (3,5)2 × 12                              ✓SF substitution
             = 461,874 ≈ 461,87 cm3                               ✓CA answer
             (Accept 461,9 cm3 or 462 cm3 ; NPR)                  M   L2                    (3)

 3.1.3       Mass of 1 candle = 461,874 × 0,9 g = 415,69 g        ✓M multiplying by 0,9
             Mass for 120 candles = 415,69 × 120                  ✓M multiplying by 120
             = 49 882,39 g = 49,88 kg                             ✓C converting to kg
             Blocks = 49,88 ÷ 5 = 9,98                            ✓M dividing by 5
             ⇒ 10 blocks                                          ✓R rounding up to 10
                                                                  M   L3                    (5)

 3.1.4       Length = 3,142 × 7 + 1 = 21,994 + 1 = 22,994 cm      ✓M circumference + 1 cm
             Height = 80% × 12 = 9,6 cm                           ✓A 9,6 cm
             Area = 22,994 × 9,6                                  ✓SF length × width
             = 220,74 cm2                                         ✓CA area
                                                                  M L3                   (4)

 3.1.5       Along 45 cm: 45 ÷ 7 = 6,43 ⇒ 6 candles               ✓M✓R along the length
             Along 30 cm: 30 ÷ 7 = 4,29 ⇒ 4 candles               ✓R along the width
             Candles per layer = 6 × 4 = 24                       ✓CA 24 per layer
             Layers: 26 ÷ 12 = 2,17 ⇒ 2 layers                    ✓M 2 layers
             Total = 24 × 2 = 48 candles                          ✓J 48 and conclusion
             48 > 40 ∴ The claim is VALID.                        (Using volume ratio only,
                                                                  35 100 ÷ 461,87 = 76: max 2
                                                                  marks)
                                                                  M L4                      (6)

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### Page 6

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

 3.1.6       TSA = 2(45 × 30) + 2(45 × 26) + 2(30 × 26)               ✓SF substitution
             = 2 700 + 2 340 + 1 560                                  ✓S simplification
             = 6 600 cm2                                              ✓CA answer
                                                                      M    L2                   (3)

 3.1.7       Diameter = 3,5 × 2,54 = 8,89 cm                          ✓C converting to cm
             Radius = 8,89 ÷ 2 = 4,445 cm                             ✓A radius
             1 000 = 3,142 × (4,445)2 × h                             ✓SF substitution
             h=
                   1 000
                             = 16,108 . . .                           ✓CA height (1 decimal)
                 62,08 . . .                                          M    L3                   (4)
             ≈ 16,1 cm

                                                                                               [27]

QUESTION 4

 4.1.1       Right-hand side                                          ✓✓RM
                                                                      MP    L1                  (2)

 4.1.2       Rows 1–13: 13 × 4 = 52 seats                             ✓M 52 seats
             Minus 2 (middle door) and 2 (toilet): 52 − 4 = 48        ✓M subtracting 4
             Plus back row: 48 + 5 = 53 seats                         ✓CA total 53
                                                                      (Counting AO: full marks)
                                                                      MP L2                    (3)

 4.1.3       Window seats: A-column 12 + D-column 12 + 14A            ✓RM 26 window seats
             and 14E = 26                                             ✓M dividing by 53
                  26                                                  ✓CA percentage
             P=       × 100%
                  53                                                  P    L2                   (3)
             = 49,06%
             (CA from 4.1.2)
             (Accept 24
                      49 = 48,98% if row 1 is excluded with a
             stated reason)

 4.1.4       Any ONE valid reason, e.g.                               ✓✓O valid reason
             - They are close to the driver and the front door.       MP    L4                  (2)
             - They can see/supervise the learners and talk to the
             driver.
             - They can get off first to control learners at stops.

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### Page 7

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

 4.1.5       53 seats = learners + teachers                        ✓M learners + teachers = 53
             If 48 learners: teachers needed = 48 ÷ 10 = 4,8 ⇒ 5   ✓M 1 teacher per 10 learners
             48 + 5 = 53 (fits)                                    ✓R rounding up teachers
             If 49 learners: teachers needed = 4,9 ⇒ 5;            ✓CA 48 learners
             49 + 5 = 54 > 53 (does not fit)                       MP L4                    (4)
             ∴ Maximum = 48 learners

 4.1.6       Cost = 386× R14,50 + R1 200                           ✓M multiplying by 14,50
             = R5 597 + R1 200 = R6 797                            ✓M adding R1 200
             Per learner = R6 797 ÷48 = R141,60                    ✓CA cost per learner
             R141,60 > R140 ∴ The claim is NOT valid.              ✓J conclusion
                                                                   M    L3                    (4)

 4.2.1       Washers = 12 × 15                                     ✓RT 12 per chair
             = 180 washers                                         ✓CA answer
                                                                   MP    L2                   (2)

 4.2.2       4 mm Allen key and 10 mm spanner                      ✓A Allen key
                                                                   ✓A spanner
                                                                   MP    L1                   (2)

                  15
 4.3.1       P=      × 100%                                        ✓M dividing by 400
                 400
             = 3,75%                                               ✓A percentage
                                                                   P    L1                    (2)

                               15   385
 4.3.2       P(not win) = 1 −     =                                ✓M 1− P(win) or 385 tickets
                              400 400
             = 0,9625 (accept 0,96)                                ✓CA decimal
                                                                   P L2                    (2)

                                                                                             [26]

                                                                                  TOTAL:      100

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