# Mathematics Grade 10 Paper 2 KZN June 2024 Memo

> Grade 10 Mathematics Paper 2 memorandum from the KZN June Exam 2024, 11 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/kzn-maths-grade-10-june-2024-p2-and-memo/b45ec604-cb94-431c-a784-549d39bdeb95
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 10
- Document type: Memorandum
- Year: 2024
- Exam period: KZN June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 11

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## About this document

Mathematics · Grade 10 · KZN June Exam · 2024. Memorandum, 11 pages. Read online or download the PDF.

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## Document text

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### Page 1

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                               NATIONAL
                          SENIOR CERTIFICATES

                               GRADE 10

                            MATHEMATICS P2

                             COMMON TEST

                                JUNE 2024

 MARKS:       50

 TIME:1 Hour

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                                                       (2)

                                                       (2)

                                                        (4)

                                                       [9]

                                         TOTAL: [50]

### Page 6

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                                       FINAL

                                MATHEMATICS P2

                                  COMMON TEST

                                       JUNE 2024

                            MARKING GUIDELINES

                                  NATIONAL
                             SENIOR CERTIFICATE

                                       GRADE 10

 MARKS:       50

                         These marking guidelines consist of 6 pages.

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### Page 7

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Mathematics                          2         com         Common Test June 2024
                               GRADE 10
                             Marking Guideline

QUESTION 1

1.1      x + x y + y2 
       M 1 2 ; 1
         2          2                             A Formula
         6 + (−5) 6 + 3                            ✓A Substitution
       M         ;      
         2          2 
        1 9                                        CA Answer
       M ; 
        2 2                                                                       (3)
1.2            y2 − y1
        mWT =
               x2 − x1                               ✓A Formula
            6−0                                      A Substitution
        =
          6 − (−3)
          2                                          CA Answer
        =
          3                                                                         (3)

1.3           y2 − y1
        mRW =
              x2 − x1
                 3−0
            =
              −5 − (−3)                              A substitution
              −3
            =                                              −3
               2                                     A
                                                           2
                     2 −3
       mWT  mRW =       = −1
                     3 2                                   2 −3
                                                     A        = −1
       ∴ RW is perpendicular to WT                         3 2
                                                                                    (3)

1.4         x1 + x2                    y1 + y2
         x=                         y=                       1 a −3
               2                          2          CA       =
                                                             2   2
        1 a −3                     9 b+0
          =                          =               CA a = 4
        2    2                     2    2
          a=4                      b=9                          9 b+0
                                                     ✓CA          =
                                                                2   2
       OR
       Sliding method                                ✓CA        b=9

        a = 6−2          b = 6+3                     ANSWER ONLY
                                                     MAX: 2/4
        a=4              b=9

                                                                                    (4)

                                                                                   [13]
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### Page 8

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Mathematics                          3         com                          Common Test June 2024
                               GRADE 10
                             Marking Guideline

 QUESTION 2
2.1.1 x2 + y 2 = r 2
         x 2 + 52 = 132
                                                                    A substitution
         x = 12
         x = −12                                                    CA x = −12
                                                                    (CA provided x is negative)
                                                                                                       (2)
2.1.2   cos
          −12                                                       CA     answer
        =                                                                                              (1)
          13

2.1.3       sin  
        12        
            cos  
              5 
                    
        = 12  13                                                  CA substitution
              −12 
              13 
        = −5                                                        CA answer
                                                                                                       (2)

2.2.1   cos 2 ( 21 )
        = cos 42           NOTE: Penalise only here for rounding
                                                                    A    answer
        = 0, 74                                                                                        (1)

2.2.2   sin ( 20 + 21 )                                           A    (20 + 21)
        = sin 41
        = 0, 66                                                     A    answer
                            ANSWER ONLY: FULL MARKS                                                    (2)
2.2.3   sec ( 20 )
                1
        =                                                                 1
            cos ( 20 )                                             A
                                                                       cos20
        = 1, 06                                                     A answer
                            ANSWER ONLY: FULL MARKS                                                    (2)

                                                                                                      [10]

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Mathematics                          4         com                       Common Test June 2024
                               GRADE 10
                             Marking Guideline
QUESTION 3
              2                                                                                    2
         1  1               1                                          1 
                                                                                2
                                                                                          2
3.1         − 2 + tan10 tan10                                A             accept    
         2                                                             2               2 
          1 1                                                             1
        = − +1                                                   ✓A
          2 2                                                             2
        =1                                                                  1
                                                                 A
                                                                        tan10

                                                                 CA     answer
                                                                                                       (4)
        6sin x = 1                                               A 6sin x = 1
3.2                                                              A simplification
                1
        sin x =
                6
                                                                 CA answer
        x = 9, 6
                                                                                                       (3)
                            NOTE: do not penalise for rounding

                                                                                                       [7]

QUESTION 4

4.1     In MRC and XCR
        𝑅̂1      = 𝐶̂ 2            Alt  s. MR//CX.              A    S/R
        𝐶̂ 1     = 𝑅̂2             Alt  s. MC//RX.
                                                                 A    S/R
        CR       = CR              Common side.
         MRC  XCR               (AAS)                         A    MRC  XCR …(AAS)
        MR=XC                                                    A    conclusion
        MC=XR                                                                                          (4)
        RC=CR

        NOTE: Penalise ONCE if parallel lines are not stated

4.2.1   8𝑥 − 16° = 5𝑥 + 5°         opp  s of parm               A S/R
        3𝑥 = 21°                                                 A 3𝑥 = 21°
        𝑥 = 7°
                                                                 CA    x-value
                                                                 (CA provided x is positive)
                                                                                                       (3)
4.2.2
        Dˆ = Bˆ                    opp  s of parm               A    S/R

        Bˆ + Dˆ + Aˆ + Cˆ = 360    Sum  s of quad              A S/R

        2Bˆ + 80 = 360                                         CA S

        2 Bˆ = 280
        Bˆ = 140                                                CA answer
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### Page 10

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Mathematics                          5         com                   Common Test June 2024
                               GRADE 10
                             Marking Guideline

        OR

        Cˆ = 8 ( 7 ) − 16
                                                            CA
        Cˆ = 40
                                                                    40°

                                                            SR
        Bˆ + Cˆ = 180                  Co-int  s. AB DC
                                                            CA answer
        Bˆ = 180 − 40                                                                       (4)
        Bˆ = 140

                                                                                              [11]

QUESTION 5
5.1.1   Vˆ1 + 30 = 100             Ext  of              A S/R

        Vˆ1 = 70                                           A answer
                                                                                               (2)
        OR

        R̂1 = 80         s on a str. line                 A S/R
        V̂1 = 70         sum of 
                                                            A answer

5.1.2   Sˆ1 = Vˆ2            s opp = sides                 A S/R
        Sˆ + Vˆ + VRS
         1       2
                   ˆ = 180     Sum  s                    A S

        2 Sˆ1 + 100 = 180
        2 Sˆ = 80
             1

        Sˆ1 = 40
                                                            A    answer
        OR                                                                                     (3)
        Rˆ1 + VRS
               ˆ = 180               s on a str line

        Rˆ1 = 80                                           A    Rˆ1 = 80

        Sˆ1 = Vˆ2                 s opp = sides

        Sˆ1 + Vˆ2 = Rˆ1          Ext  of 
                                                            A S/R
        2 Sˆ = 80
             1

        Sˆ1 = 40                                           A    answer

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### Page 11

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Mathematics                          6         com                      Common Test June 2024
                               GRADE 10
                             Marking Guideline
5.2    ES = ER                      diagonals of rectangle   A   S/R
       R̂2 = t                       s opp =sides
                  ˆ = 180
       Eˆ + Rˆ + ESR               Sum  of 
          2       2

        Eˆ 2 + t + t = 180
        Eˆ = 180 − 2t
          2                                                  A   Ê2
        Eˆ 4 = Eˆ 2                Vert opp  s              A   S/R
        Eˆ = 180 − 2t
          4                                                  A   answer
                                                                                                (4)
                      OR

        Qˆ1 = t                   Alt ∠s PS∥QR               ✓A    S/R
                                  ∠s opp = sides             ✓A
        Pˆ = t
                                                                   S/R
         1

        Eˆ 4 + Qˆ1 + Pˆ1 = 180   Sum ∠s ∆                   ✓A    S
                                                             ✓A
        Eˆ = 180 − 2t
                                                                   answer
          4

                                                                                                [9]

                                                                            TOTAL:              50

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