# Physical Sciences Grade 12 Paper 1 Mpumalanga Mock 2022 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Mpumalanga Mock Exam 2022, 37 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/mp-physical-science-grade-12-sept-2022-p1-and-memo/739f41a8-d1d9-4b3d-b784-ecc1267fa209
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2022
- Exam period: Mpumalanga Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 37

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## About this document

Physical Sciences · Grade 12 · Mpumalanga Mock Exam · 2022. Memorandum, 37 pages. Read online or download the PDF.

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Sept 2022 September Exam - Mpumalanga Errata

Fout by kortvraag 1.10. Moet A wees. En tekenfout by 2.3; moet +f wees. (Antw: F = 56,17N)

Error: 1.10 A is correct and sign error in 2.3; should be +f (Anwer: F = 56,17 N)

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                                   NATIONAL
                              SENIOR CERTIFICATE
                                  NASIONALE
                              SENIOR SERTIFIKAAT

                                 GRADE/GRAAD 12

                      PHYSICAL SCIENCES: PHYSICS (P1)
                      FISIESE WETENSKAPPE: FISIKA (V1)

                                   SEPTEMBER 2022

                   MARKING GUIDELINES/NASIENRIGLYNE

MARKS/PUNTE: 150

                    These marking guidelines consists of 15 pages

                      Hierdie nasienriglyne bestaan uit 15 bladsye

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Physical Sciences P1/Fisiese Wetenskappe V1    2            MDE/MDO/September 2022
                                              NSC/
                                              NSS

QUESTION 1 / VRAAG 1

1.1    D                                                                          (2)

1.2    B                                                                          (2)

1.3    B                                                                          (2)

1.4    D                                                                          (2)

1.5    A                                                                          (2)

1.6    B                                                                          (2)

1.7    C                                                                          (2)

1.8    C                                                                          (2)

1.9    A                                                                          (2)

1.10   B                                                                          (2)
                                                                                   [20]

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 Physical Sciences P1/Fisiese Wetenskappe V1     3                           MDE/MDO/September 2022
                                                NSC/
                                                NSS

QUESTION 2 / VRAAG 2
 2.1 Marking criteria/Nasienkriteria
     If any one of the underlined key words/phrases in the correct context is omitted,
     deduct 1 mark. /Indien enige van die onderstreepte sleutelwoorde/frases in die
     korrekte konteks uitgelaat is, trek 1 punt af.
     A body will remain in its state of rest or motion at constant velocity unless a non-
     zero resultant/net/unbalanced force acts on it. 
     ‘n Liggaam sal in ‘n toestand van rus of beweging teen ‘n konstante snelheid
     volhard, tensy ‘n nie-nul resulterende/netto/ongebalanseerde krag daarop
     inwerk.
     OR/OF
     A body will remain in its state of rest or uniform motion in a straight line unless a
     non-zero resultant/net/unbalanced force acts on it. 
     ‘n Liggaam sal in ‘n toestand van rus of uniforme beweging in ‘n reguit lyn
     volhard, tensy ‘n nie-nul resulterende/netto/ongebalanseerde krag daarop
     inwerk.
                                                                                                     (2)

 2. 2
                         N                                     Accept/Aanvaar
                                           F                    N
                                                                                      F

                 T
                                                        T                         w┴ 
                               w                             w// 
                                                                                                     (4)
        Accepted labels / Aanvaarde benoemings
        w     Fg / Fw / force of earth on block / weight / mg / gravitational force / 29,4 N
        F     Fapplied / FA / Applied force
        T     Tension in rope / FT
        N     Normal force / FN / 25,46 N

        Notes/Aantekeninge:
            Any additional forces: deduct 1 mark: max 3⁄4
            No labels: deduct 1 mark: max 3⁄4
            No arrows: 0⁄4
              Force(s) not touching object: deduct 1 mark: max 3⁄4
              Ignore relative sizes of the vectors

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Physical Sciences P1/Fisiese Wetenskappe V1           4                          MDE/MDO/September 2022
                                                     NSC/
                                                     NSS

2.3    F = (Fg//)5kg + (Fg//)3kg - fs
           = 5(9,8)Sin30° + 3(9,8)Sin30° - 16,97 = 22,23 N
                                                                                                         (3)

2.4    Marking criteria/Nasienkriteria:
            Formula for block P or block Q 
            Substitution of Fnet for block P 
            Substitusion of Fnet for block Q 
            5a OR 3a 
            Answer: a=5,16 m∙s-2 
       For Block P:                             For Block Q:
       Fnet = ma                                Fnet = ma
                                Any one 
       T + (-Fg//) + (-f) = ma                  F + (-Fg//) + (-T) = ma
       T – (5×9,8)Sin30° - 4,5  = 5a          85 – (3×9,8)Sin30° - T = 3a
       T = 5a +29                               T = 70,3 – 3a

                                            ∴ 5a + 29 = 70,3 – 3a
                                                  8a = 41,3
                                               a = 5,16 m∙s-2                                           (5)

2.5
                                                       Marking criteria/Nasienkriteria
                                        Q
                                                           Both straight lines

        velocity                                          For Q: positive gradient and larger
                                                            than initially
                                                P         For P: negative gradient
                                                time                                                     (3)
                                                                                                        [17]

QUESTION 3 / VRAAG 3
 3.1 9,8 m·s-2  downwards / afwaarts                                                                   (2)

3.2.1 Note: do not penalize for the omission of zero(s).
                                         OPTION 1
      Upwards positive:                          Downwards positive:
      ∆y = -50-(-1,5) = -48,5 m                 ∆y = 50-1,5 = 48,5 m

       vf2 = vi2 + 2a∆y                                    vf2 = vi2 + 2a∆y 
          = 0 + 2(-9,8)(-48,5)                                 = 0 + 2(9,8)(48,5) 
       vf = 30,83 m·s-1                                    vf = 30,83 m·s-1 

                                       OPTION 2
       (mgh+½mv2)top = (mgh+½mv2)P         (mgh+½mv2)top = (mgh+½mv2)P 
       m(9,8)(50)+0 = m(9,8)(1,5)+½(m)v2 m(9,8)(48,5) + 0 = 0 + ½(m)v2                               (4)

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Physical Sciences P1/Fisiese Wetenskappe V1    5                              MDE/MDO/September 2022
                                              NSC/
                                              NSS

                    v = 30,83 m·s-1                           v = 30,83 m·s-1 

3.2.2 OPTION 1
      Upwards Positive                               Downwards Positive
      Δy = viΔt + ½aΔt2                             Δy = viΔt + ½aΔt2 
      -48,5 = 0 + ½ (-9,8) Δt2                      48,5 = 0 + ½ (9,8) Δt2 
      Δt = 3,15 s                                   Δt = 3,15 s 
      OPTION 2
      POSITIVE MARKING FROM Q3.2.1
      Upwards Positive                               Downwards Positive
      vf = vi +a Δt                                 vf = vi +a Δt 
      -30,83 = 0 + (-9,8) Δt                        30,83 = 0 + (9,8) Δt 
      Δt = 3,15 s                                   Δt = 3,15 s 
      OPTION 3
      Upwards Positive                               Downwards Positive
      Fnet∙Δt = mvf – mvi                           Fnet∙Δt = mvf – mvi
      m(-9,8) Δt = m(-30,83) – 0                    (m×9,8) Δt = m(30,83) – 0
      Δt = 3,15 s                                   Δt = 3,15 s                                  (3)

3.3    Marking criteria/Nasienkriteria:
           Formula for motion at constant velocity
           Substitution for motion at constant velocity
           Formula to calculate vi for vertical motion
           Substitution to calculate vi
           Final answer: 13,28 m∙s-1
       Motion at constant velocity:
       Δx = viΔt + ½aΔt2 OR/            (vf +vi )         OR/ Distance = speed×time
       11 = 2,3 Δt                 ∆x=            ∆t     OF 11 = 2,3 Δt 
                             OF              2
       Δt = 4,78 s                       (2,3+2,3)             Δt = 4,78 s
                                    11=               ∆t 
                                               2
                                    Δt = 4,78 s

       Vertical motion:
       Upwards positive                              Downwards positive
       Δy = viΔt + ½aΔt2                            Δy = viΔt + ½aΔt2 
       -48,5 = vi(4,78) + ½(-9,8)(4,78)2            48,5 = vi(4,78) + ½(9,8)(4,78)2
       vi = 13,28 m∙s-1                             vi = -13,28 m∙s-1
                                                     vi = 13,28 m∙s-1
                                                                                                   (5)
                                                                                                  [14]

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Physical Sciences P1/Fisiese Wetenskappe V1     6                         MDE/MDO/September 2022
                                               NSC/
                                               NSS

QUESTION 4 / VRAAG 4

       Marking criteria/Nasienkriteria
       If any one of the underlined key words/phrases in the correct context is
       omitted, deduct 1 mark. /Indien enige van die onderstreepte
       sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af.
4.1    A system on which the resultant/net external force is zero
       ‘n Sisteem waarin die resulterende/netto eksterne krag nul is.                             (2)

4.2.1 p = mv 
      8,4 = m(560) 
      m = 0,015 kg                                                                               (3)

4.2.2 POSITIVE MARKING FROM 4.2.1

                  Σpi = Σpf
                                           Any one
       (mvi)1 + (mvi)2 = (mvf)1 + (mvf)2
       (0,015)(560) + (3)(-2,5)  = (0,015)(80) + (3)(v) 
       v = -0,1
       v = 0,1 m∙s-1                                                                             (4)

4.2.3 POSITIVE MARKING FROM QUESTION 4.2.1
      OPTION 1
      Fnet∙Δt = mvf – mvi
      Fnet(0,02) = 0,015(80-560) 
      Fnet = -360
      Fnet = 360 N East/Oos      Accept: Right/Regs
      POSITIVE MARKING FROM QUESTION 4.2.2
      OPTION 2
      Fnet∙Δt = mvf – mvi
      Fnet(0,02) = 3(0,1 – 2,5) 
      (Fnet)block = -360

       (Fnet)bullet = 360 N East/Oos        Accept: Right/Regs                                  (4)
                                                                                                 [13]

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Physical Sciences P1/Fisiese Wetenskappe V1    7                       MDE/MDO/September 2022
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QUESTION 5 / VRAAG 5

5.1    OPTION 1
                  (Em)A = (Em)B
       (mgh + ½mv2)A = (mgh +½ mv2)B       Any one 
        m(9,8)(0,7) + ½ m(2)2 = 0 + ½mv2 
              v = 4,21 m·s-1 
       OPTION 2
       Wnc = ΔEk + ΔEp                         Any one 
       0 = [½mvf2 – ½mvi2] + [mghf – mghi]
       0 = [½mv2 – ½m(2)2] + [0 – m(9.8)(0,7)] 
        v = 4,21 m·s-1                                                                        (3)

5.2    Marking criteria/Nasien kriteria
           Formula for Ek = 11,17
           Substitution of mv=5,79
           Final answer: v = 3,86 m·s-1
       OPTION 1
       p = mv = 5,79            ½mv∙v = 11,17
                2
       Ek = ½mv = 11,17         ½(5,79)v = 11,17
                                v = 3,86 m·s-1 
       OPTION 2
       p = mv = 5,79            Ek = ½mv2 = 11,17
           5,79                 1 5,79 2
       m=                         (     ) v =11,17
             v                  2 v
                                v = 3,86 m·s-1                                                (3)

5.3    Marking criteria/Nasienkriteria
       If any one of the underlined key words/phrases in the correct context is
       omitted, deduct 1 mark. /Indien enige van die onderstreepte
       sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af.
       The net work done on an object is equal to the change in kinetic energy of the
       object. 
       Die netto arbeid op ‘n voorwerp verrig is gelyk aan die verandering in die
       kinetiese energie van die voorwerp.                                                     (2)

5.4    POSITIVE MARKING FROM Q5.2
       OPTION 1               OR/OF
       mv = 5,79                                     ½mv2 = 11,17
       m(3,86)=5,79                                 ½ m(3,86)2 = 11,17
       m = 1,5 kg                                    m = 1,5 kg

       Wnet = ΔEk                  Any one 
       fΔxcosΘ = ½mvf2 – ½mvi2
       (3)(X)cos180°= 0 – ½(1,5)(3,86)2
       X = 3,72 m 

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Physical Sciences P1/Fisiese Wetenskappe V1    8                       MDE/MDO/September 2022
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       OPTION 2
       mv = 5,79                                 ½mv2 = 11,17
       m(3,86)=5,79                             ½ m(3,86)2 = 11,17
       m = 1,5 kg                                m = 1,5 kg

       Wnc = ΔEk + ΔEp
       fΔx cosΘ= [½mvf2 – ½mvi2 ]+ [mghf – mghi]      Any one 
       (3)(X)cos180°= [0 – ½(1,5)(3,86)2] + 0
       X = 3,72 m                                                                            (5)
                                                                                             [13]

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Physical Sciences P1/Fisiese Wetenskappe V1       9                         MDE/MDO/September 2022
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QUESTION 6 / VRAAG 6

6.1     Marking criteria/Nasienkriteria
        If any one of the underlined key words/phrases in the correct context is
        omitted, deduct 1 mark. /Indien enige van die onderstreepte
        sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af.
        The change in frequency/pitch/wavelength of the sound detected by a listener
        because the sound source and the listerner have different velocities relative to
        the medium of sound propagation. 
        Die verandering van frekwensie/toonhoogte/golflengte van die klank
        waargeneem deur ‘n luisteraar omdat die klankbron en die luisteraar
        verskillende snelhede relatief tot die medium van klankvoorplanting het.

        OR/OF
        An (apparent) change in observed frequency/pitch/wavelength as the result of
        the relative motion between a sounce and an observer/listener. 
        ‘n (Skynbare) verandering in waargenome frekwensie/toonhoogte/golflengte as
        gevolg van die relatiewe beweging tussen die bron en ‘n waarnemer/luisteraar.               (2)

6.2     Waves in front of source are more compact/wavelength decreases 
        More waves per second reaches the detector/listener 
        Golwe voor die bron kompakteer / golflengte neem af 
        Meer golwe per sekonde bereik die detector                                                 (2)

6.3.1                 Moving towards                              Moving away
             v±vL                 v                           v±vL                v
        fL =      fS  OR fL =       fS                  fL =      fS   OR fL =     f
             v±vS               v-vS                          v±vS              v+vS S
                     330                                        330
        950 =             fs                    750 =              f 
                   330 –vs                                    330 +vs s
              950(330 − 𝑣𝑠 )                           750(330 + 𝑣𝑠 )
        𝑓𝑠 =                                      𝑓𝑠 =
                    330                                        330
                                    ∴ 950(330-vs) = 750(330+vs)
                                        200(330) = 1700vs
                                         vs = 38,82 m∙s-1                                          (6)

6.3.2 POSITIVE MARKING FROM Q6.3.1
      OPTION 1            OPTION 2                              OPTION 3
      Δt = 12,5-8= 4,5 s Δt = 12,5-8= 4,5 s                   Δt = 12,5-8= 4,5 s

        Δx = viΔt + ½aΔt2               (vf +vi )              Distance = speed×time
           = (38,82)(4,5)        ∆x=              ∆t                   = (38,82)(4,5) 
                                             2
           = 174,69 m                    (38,82+38,82)                  = 174,69 m
                                        =               (4,5) 
                                                   2
                                        = 174,69 m                                                 (4)
                                                                                                   [14]

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Physical Sciences P1/Fisiese Wetenskappe V1    10                        MDE/MDO/September 2022
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QUESTION 7 / VRAAG 7

7.1.1 Marking criteria/Nasienkriteria
      If any one of the underlined key words/phrases in the correct context is
      omitted, deduct 1 mark. /Indien enige van die onderstreepte
      sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af.

          Note: If masses used ( 0⁄2 )

        The magnitude of the electrostatic force exerted by one point charge (Q 1) on
        another point charge (Q2) is directly proportional to the product of the
        magnitudes of the charges  and inversely proportional to the square of the
        distance (r) between them 
        Die grootte van die elektrostatiese krag wat een puntlading (Q1) op 'n ander
        puntlading (Q2) uitoefen, is direk eweredig aan die produk van die groottes van
        die ladings en omgekeerd eweredig aan die kwadraat van die afstand (r)
        tussen hulle.                                                                           (2)

7.1.2 Negative/Negatief                                                                         (1)

7.1.3       kQ1 Q2
        F=          
               r2
                         2
               (9×109 )Q
        57,6 =             
                  (0,1)2
         Q = 8×10-6 C       Accept: Q = -8×10-6 C                                               (3)

7.2.1 Marking criteria/Nasienriglyne:
      -1 mark for each of the 5 key words omitted in the correct context.
      -1 punt vir elk van die 5 sleutelwoorde weggelaat in die korrekte konteks.
      The electric field at a point is the (electrostatic) force experienced per unit
      positive charge placed at that point.
      Die elektriese veld by 'n punt is die (elektrostatiese) krag wat per positiewe
      eenheids-lading wat by daardie punt geplaas is, ondervind word.                            (2)

7.2.2       kQ
        E=      
             r2
              (9×109 )(8×10-9 )
           =                    
                  (0,2)2
           = 1 800 N∙C-1 to the left                                                           (4)

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Physical Sciences P1/Fisiese Wetenskappe V1    11                        MDE/MDO/September 2022
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7.2.3 Enet = Ex - Ey
             (9×109 )Q (9×109 )(4×10-9 )
      120 =           -                 
                 0,22       0,082
      Q = 2,55×10-8 C

              Q
       n=        
              e-
             (2,55×10-8 -8×10-9 )                    -2,55×10-8 -(-8×10-9 )
         =                             OR/OF
              1,6×10-19                                   -1,6×10-19
         = 1,1×1011                                                                             (5)
                                                                                                [17]

QUESTION 8 / VRAAG 8

8.1    7,2 (V)                                                                                  (1)

      NOTE: Penalize once in Q8.2 for incorrect use of ×10-2
8.2.1                V2 -V1
      Gradient= -r =                 NOTE: Can use any coordinates on the line, eg
                      I2 -I1
                                     (100×10-2 ; 2,4)
                          7,2-0
                   =            -2
                                    (67,5×10-2 ; 4,0)
                     0-150×10        (17,5×10-2 ; 6,4)
                 r = 4,8 Ω                                                                      (2)

8.2.2 POSITIVE MARKING FROM Q8.1 AND Q8.2.1
      OPTION 1                        OPTION 2
      ɛ = I(R+r)                     V = IR
      7,2 = (117,5×10 )[R+4,8] 
                     -2               1,6 = (117,5×10-2)R
      R = 1,33 Ω                     R = 1,36 Ω                                                (3)
                                                                                                 [6]

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QUESTION 9 / VRAAG 9

9.1    The potential difference across a conductor is directly proportional to the current
       in the conductor  at constant temperature. 
       Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die geleier
       by konstante temperatuur.                                                                  (2)

9.2.1 OPTION 1                                       OPTION 2
      V6Ω = IR                                       R6Ω : R12Ω
         = 0,5(6)                                      6 : 12
         =3V                                         I6Ω : I12Ω
                                                       12 : 6
       I12Ω = V⁄R = 3⁄12 = 0,25 A                   0,5 : 0,25
                                                     I1 = 0,5+0,25=0,75 A
       I1 = 0,5+0,25 = 0,75 A                                                                    (3)

9.2.2 POSITIVE MARKING FROM Q9.2.1
      V20Ω=IR
          = 0,75(20)
          = 15 V

       P=VI
       16 = (15+3)I
       I2 = 0,89 A                                                                               (3)

9.2.3 POSITIVE MARKING FROM Q9.2.1 & Q9.2.2
      OPTION 1
      IT=0,75+0,89 = 1,64 A

       ɛ = I(R+r) 
         = (1,64)(10,98+1) 
         = 19,65 V
       OPTION 2
       IT=0,75+0,89 = 1,64 A

       ɛ = Ve + Ir 
         = 18 + (1,64)(1) 
         = 19,64 V                                                                               (4)
                                                                                                 [12]

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QUESTION 10/VRAAG 10

10.1.1 Split ring/Split ring commutator/commutator
       Splitring/splitring kommutator/kommutator                                                  (1)

10.1.2 Anticlockwise/Antikloksgewys                                                             (2)

10.1.3 Electrical (energy) TO mechanical/kinetic (energy)
       Elektriese (energie) NA meganiese/kinetiese (energie)                                      (1)

10.2.1 Marking criteria/Nasienkriteria
       If any one of the underlined key words/phrases in the correct context is
       omitted, deduct 1 mark. /Indien enige van die onderstreepte
       sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af.
       The AC current which dissipates the same amount of energy as an equivalent
       DC current. 
       Die WS-stroomsterkte wat dieselfde hoeveelheid energie verbruik as die
       ekwivalente GS-stroomsterkte.

         ACCEPT/AANVAAR
         The DC current which dissipates the same amount of energy as an equivalent
         AC current.
         Die GS-stroomsterkte wat dieselfde hoeveelheid energie verbruik as die
         ekwivalente WS-stroomsterkte.                                                            (2)

10.2.2 OPTION 1                         OPTION 2
              Imax                             Imax
       Irms =                          Irms =      
               √2                               √2
                8                                8
          =                               =       
               √2                               √2
           = 5,66 A                         = 5,66 A
                                        Vrms = Irms R
         Pave = I2rms R                      = (5,66)(40)
              = (5,66)2 (40)                = 226,4 V
              = 1280 W                                                          2
                                                                               Vrms
                                        Pavg = VrmsIrms      OR/OF   Pave         
                                                                                R
                                             = (226,4)(5,66)
                                             = 1281,42 W                      (226,4)2
                                                                             =          
                                                                                  40
                                                                           = 1281,42 W 

                                                                                                  (5)
                                                                                                 [11]

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QUESTION 11 / VRAAG 11

11.1     Marking criteria/Nasienkriteria
         If any one of the underlined key words/phrases in the correct context is
         omitted, deduct 1 mark. /Indien enige van die onderstreepte
         sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af.
         The minimum energy of light needed to eject electrons from a metal/surface.
         
         Die minimum energie benodig om 'n elektron uit die oppervlak van 'n metaal
         vry te stel.
          Note: If referred to frequency: 0⁄2
                                                                                          (2)

11.2     Zinc
         Photons that have shorter wavelengths will have higher frequencies and
         thus a higher energy. 
         OR/OF
         photons that have longer wavelengths will have lower frequencies, and
         thus a lower energy 
         OR/OF
                1
         𝑊0 ∝ , ℎ&𝑐 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
               𝜆0
         OR/OF
              1
         𝑓0 ∝ , 𝑐 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
              𝜆0
         𝑊0 ∝ 𝑓0 , ℎ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
         OR/OF
         W0 is inversely proportional to λ0  (with h and c staying constant).           (3)

11.3     Frequency of photon is less than the threshold frequency of the metal
         Die frekwensie van die foton is minder as die drumpelfrekwensie van die
         metaal.
         OR/OF
         Energy of the photon is less than the work function of the metal. 
         Die energie van die foton is minder as die arbeidsfunksie/werksfunksie van die
         metaal.                                                                          (2)

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11.4     OPTION 1
         E = W 0 + Ek(max)
                      1 2
         hf = hf0 +    mv
                      2 max     Any one 
                hc
         hf =      + Ek(max)
                λ0
                                        (6,63×10-34 )(3×108 )
         (6,63×10-34 )(5×1014 ) =                               + Ek(max)
                                              10×10-7
         Ek(max) = 1,33 × 10-19 J 

         OPTION 2
         c = f0λ0
         3×108 = f0(10×10-7)
         f0 = 3×1014 Hz

         hf = hf0 + Ek(max)
         (6,63×10-34 )(5×1014 ) = (6,63×10-34 )(3×1014 ) + Ek(max)
         Ek(max) = 1,33 × 10-19 J 
                                                                                                  (4)

11.5.1 Remains the same/Bly dieselfde                                                            (1)

11.5.2 Increases/Toeneem                                                                         (1)
                                                                                                  [13]

                                                                              TOTAL/TOTAAL: 150

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