# Mathematics Grade 12 Paper 2 Free State Mock 2018 Memo

> Grade 12 Mathematics Paper 2 memorandum from the Free State Mock Exam 2018, 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2018-grade-12-trial-memo-p2-fs/007a3161-10b4-423a-81bf-8f74f4a2101b
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: Free State Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 14

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Mathematics · Grade 12 · Free State Mock Exam · 2018. Memorandum, 14 pages. Read online or download the PDF.

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### Page 1

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                 PREPARATORY EXAMINATION

                                 GRADE 12

                     MATHEMATICS P2

                          SEPTEMBER 2018

                               MARKS: 150

                     MARKING GUIDELINE

                     This marking guideline consists of 14 pages.

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Mathematics P2                                   2                    FS/September 2018
Grade 12                                 Marking Guidelines

QUESTION 1

 1.1    Lower quartile : Q1 = 89                                ✓      89
        Upper quartile : Q3 = 113                               ✓      113

                                                                                  (2)
 1.2                                                            ✓      Min = 58 and
                                                                       max = 145
                                                                ✓      Q1 = 89 and
                                                                       Q3 = 113
                                                                ✓      Q2 = 100

                                                                                    (3)
 1.3    Symmetrical                               accept both   ✓      Answer
        Marginally skewed to the right                                              (1)
 1.4                  1 522                                        1522
           Mean : x =                                           ✓✓
                        15                                          15
                     x = 101,47                                                     (2)

 1.5    Standard deviation :  = 19,07                              ✓  = 19,07
                                                                                    (1)
 1.6    (x −  ; x +  )                                        ✓       82,4
        = (82,4 ; 120,54)                                       ✓       120,54
                                                                ✓       2 days
        2 days
                                                                                    (3)

                                                                                  [12]

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Mathematics P2                                    3                     FS/September 2018
Grade 12                                  Marking Guidelines

QUESTION 2

 2.1     𝑟 = −0,95                                             ✓ ✓ value of r
                                                                                       (2)
 2.2     There is a strong negative relationship.              ✓strong
                                                                     negative
                                                                        (1)
 2.3     𝑎 = 11,71323529                                       ✓ 𝑎 = 11,71
         𝑏 = −1,11764                                          ✓ 𝑏 = −1,12
         𝑦̂ = 11,71 − 1,12𝑥                                    ✓ correct equation
                                                                                       (3)
 2.4     𝑦̂ = 11,71 − 1,12(6)                                  ✓ Substitution
         𝑦̂ = 4,99                                             ✓ 𝑦 ≈ 5 times
         𝑦̂ ≈ 5 times

         OR

         𝑦̂ = 5,007                                            ✓ calculator
            = 5 times                                          ✓ 𝑦̂ ≈ 5 times
                                                                                       (2)
                                                                                       [8]

QUESTION 3

 3.1.1    M EF = M FG                                          ✓ M EF = M FG
          − 1 − 3 1 − (−1)
                 =                                             ✓ Substitution into the
          0−4         t
                                                                  correct formula
          − 4 1+1
               =
          −4      t                                                  2
                  2                                            ✓   1=
              1=                                                     t
                  t                                            ✓ Answer
              t=2                                                                      (4)

 3.1.2   M EF  M FG = −1
                                                                   2
            2
         1  = −1                                              ✓ 1  = −1
                                                                   t
             t
                                                               ✓ Answer
         t = −2                                                                        (2)

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Mathematics P2                                 4                              FS/September 2018
Grade 12                               Marking Guidelines

 3.2.1 𝑦 = 1 𝑥 + 8                                                   ✓ rewrite equation of
           2
                                                                     PR
                             1                                                  1
          M QR = M PS =                QR//PS                        ✓ M PS =
                             2                                                  2
           y −5 =
                     1
                       (x − 4)                                       ✓ Substitute (4; 5)
                     2
              1                                                         correctly
           y=   x−2+5
              2
                1                                                    ✓ Equation
          y = x+3
                2                                                                            (4)
 3.2.2       1
          0 = x+3
             2                                                       ✓ 𝑦=0
          x = −6
         ∴ S(−6; 0)                                                  ✓ 𝑆(−6; 0)
                                                                                             (2)
 3.2.3       2 + ( −6) 6 + 0                                         2 + (−6)
         M            ;                                            ✓
             2           2                                               2
                                                                       6+0
         = M (−2; 3)                                                 ✓
                                                                         2
                                                                                            (2)
 3.2.4 Midpoint of QS = midpoint of PR          (diagonals bisect)              x+4
                                                                     ✓ −2=
              x+4                                                                2
        −2=
                2
         𝑥 = −8                                                           y+5
                                                                     ✓ 3=
             y+5                                                            2
         3=
               2                                                     ✓ R(−8 ; 1)
              y =1
              ∴ R(−8 ; 1)                                                                    (3)
         OR
         P → Q : (x − 2; y + 1)
         S → R : (x − 2 : y + 1)                                     ✓ → R : (x − 2 : y + 1)
             R(− 6 − 2 : 0 + 1)                                     ✓✓  R(− 8 : 1)
             R(− 8 : 1)
                                                                                           (3)
 3.2.5               1
          𝑚𝑃𝑆 = 2               PS // QR
                              1                                                               1
           tan O 𝑆̂ 𝑃 = 𝑚𝑃𝑆 =                                        ✓ tan O 𝑆̂ 𝑃 = 𝑚𝑃𝑆 = 2
                                   2
         ∴ 𝑂𝑆̂𝑃 = 26,57°                                             ✓ 𝑂𝑆̂𝑃 = 26,57°
                         1                                                        1
           𝑚𝑅𝑆 = − 2                                                 ✓ 𝑚𝑅𝑆 = − 2

           𝑅𝑆̂O = 180° − 26,57°
                                                                     ✓ 𝑅𝑆̂ O = 153,43°
                = 153,43°
              ̂
          ∴ 𝑅𝑆𝑃 = 153,43° − 26,57°
                = 126,86°                                            ✓ 126,86°
                                                                                             (5)

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Mathematics P2                                      5                    FS/September 2018
Grade 12                                    Marking Guidelines

                                                                                      [22]
QUESTION 4

 4.1         − 2 + 4 3 + (−7)                                  ✓ Subst. into the
          M         ;                                             correct formula
             2           2 
          = M (1;−2)                                             ✓ Centre of the circle

          ∴ centre of the circle is (1; −2)
          (x − 1)2 + ( y − (−2))2 = r 2                          ✓ Subst. 𝑁(−2; 3)
          (− 2 − 1)2 + (3 + 2)2 = r 2
                     9 + 25 = 𝑟 2
                                                                 ✓ 34 = 𝑟 2
                       ∴ 34 = 𝑟 2
           (x − 1)2 + ( y + 2)2 = 34                             ✓ Equa of the circle
                                                                                     (5)

 4.2                  −2−3                                       ✓ Subst. into the
          M rad =                                                  correct formula
                     1 − ( −2 )
                                                                             −5
                  −5                                             ✓ M rad =
                 =                                                            3
                   3
                   3
           M tan =                     radius ⊥ tangent          ✓ M tan =
                                                                             3
                   5                                                         5
           y − 3 = (x − (−2) )
                   3
                   5                                             ✓ Subst. 𝑁(−2; 3)
                   3   6                                           and gradient
            y = x + +3
                   5   5
                   3    21
                 = x+
                   5     5                                       ✓ Equation
                                                                                        (5)

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Mathematics P2                                     6                           FS/September 2018
Grade 12                                   Marking Guidelines

 4.3     Let the angle of inclination of line NP be α and the angle                   5
                                                                      ✓ tan  = −
         adjacent to it be θ                                                          3
                  −5
         tan  =
                   3
                          5
           = 180 − tan                                            ✓   𝛼 = 120,96°
                           3
             = 180 − 59,036243...                                 ✓  = 59,04
               = 120,96°                     OR
                  = 59,04                  𝑁𝑄̂ 𝑃 = 120,96° − 90°         ^

             ^                                                        ✓ NPQ = 30,96
          NPQ = 180 − (59,04 + 90)               = 30,96°                                 (4)
              = 30,96°                              (ext.< 𝑜𝑓 ∆)

         OR
                                                                      ✓ 𝑁𝑄̂ 𝑃 = 90°
           𝑁𝑄̂ 𝑃 = 90°        diameter subtends a right angle
                                                                      ✓NQ = 6 and PQ =
             𝑁𝑄 = 6 and 𝑃𝑄 = 10                                       10
                          6                                                          6
           ∴ tanN 𝑃̂ 𝑄 =                                              ✓ tanN 𝑃̂ 𝑄 = 10
                         10                                           ✓ 𝑁𝑄̂ 𝑃 = 30,96°
           ∴ 𝑁𝑄̂ 𝑃 = 30,96°
                                                                                             (4)
 4.4     QP = 10 units and NQ = 6 units
                          1
         Area ∆NPQ = 2 (6)(10) ✓
                       = 30 square units
            Area ∆NPQ        30✓
         ∴ Area 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒 = 34𝜋 ✓
                          = 0,28 ✓                                                          (4)
                                                                                          [𝟏𝟕]

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Mathematics P2                                            7                         FS/September 2018
Grade 12                                          Marking Guidelines

QUESTION 5
 5.1.1    Reflex BÔP = 𝜃
                                              y

                                      𝜃

                                          ෡
                                          O                            x
                                                              B

                                                              P൫𝑐; −√21 ൯

           c2 +      ( 21) = 5
                          2       2                                         ✓ subst into pyth

           c 2 = 25 − 21
          c2 = 4
                                                                            ✓   Answer
          c = 2                                                                                      (2)
 5.1.2                        2
          a)      cos =
                              5                                             ✓ Answer
                                                                                                      (1)
                                          2                                                      2
                      − 21  − 21                                              21  21 
          b)      =       +     
                                                                           ✓       +      
                                                                                             
                       2      5                                               2       5 
                      − 25 21 + 42                                            for corr subst
                  =                                                         ✓ Answer
                           50                                                                         (2)
          c)      2 sin  cos                                               ✓ Identity
                       − 21  2                                           ✓ Answer
                  = 2       
                                                                                                     (2)
                       5  5 
                      − 4 21
                  =
                        25
 5.2       (− sin x). tan x. cos(360 − 30)
                        (sin x) 2                                           ✓ − sin 𝑥
                                                                            ✓ (sin 𝑥)2
               − sin x. tan x(cos30)                                       ✓ cos 30°
           =
                       sin 2 x.
               −√3
           =          tan 𝑥 ÷ sin 𝑥
                 2
                                                                               sin x
                                                                            ✓
            −√3sin 𝑥     1                                                     cos x
          =          X
             2 cos 𝑥   sin𝑥                                                 ✓ answer
            −√3
          =                                                                                           (5)
            2cos𝑥
                                                                                                     [𝟏𝟐]

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Mathematics P2                                       8                         FS/September 2018
Grade 12                                     Marking Guidelines

QUESTION 6

 6.1                      2 sin 2 x                               ✓
                                                                      sin x
           LHS =
                   sin x                                              cos x
                     2    − 2 sin x cos x
                   cos x
                                                                  ✓ 2𝑠𝑖𝑛 𝑥𝑐𝑜𝑠 𝑥
                         2 sin 2 x
              =
                        1                                                      1
                2 sin x         − cos x                         ✓ 2 sin 𝑥 (cos 𝑥 − cos 𝑥)
                        cos x           
                   sin x
              =                                                        sin x
                1 − cos2 x
                                                                  ✓ 1 − cos x
                                                                            2

                   cos x                                               cos x
                         cos x
              = sin x  2
                        sin x                                     ✓ 𝑠𝑖𝑛2 𝑥
                cos x
              =
                sin x
          ∴ LHS = RHS
                                                                                             (5)

 6.2      This identity will be undefined when
           2 + 2 cos 2 x = 0
           cos 2 x = −1
                                                                  ✓ cos 2 x = −1
           ∴ 2𝑥 = 180° + 𝑘. 360° 𝑘𝜖ℤ
             𝑥 = 90° + 𝑘. 180°                                    ✓ 2𝑥 = 180° +
                                                                  𝑘. 360°; 𝑘𝜖ℤ
          NB: Penalise 1 mark for leaving out 𝑘𝜖ℤ                 ✓ 𝑥 = 90° +
                                                                  ✓𝑘. 180°, ; 𝑘𝜖ℤ
                                                                                             (4)
                                      OR
          2 + 2 cos 2𝑥 = 0
          1 + cos 2𝑥 = 0                                          ✓ Identity
          1 + 2𝑐𝑜𝑠 2 𝑥 − 1 = 0
          cos 𝑥 = 0                                               ✓ cos 𝑥 = 0
          ∴ 𝑥 = 90°
          𝑥 = ±90° + 360°𝑘; 𝑘 ∈ ℤ                                 ✓ ± 90° +
                                                                  ✓k 360°; 𝑘 ∈ ℤ
                                                                                             (4)
                                                                                             [9]

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Mathematics P2                                    9                       FS/September 2018
Grade 12                                  Marking Guidelines

QUESTION 7

 7.1 𝐷𝐸෡ 𝐹 = 180° − 𝜃            opp angle of cyclic quad      ✓ Answer
                                                               (1)
 7.2                                                            ✓ correct sub into
          𝐷𝐹 = √𝑟 2 + 𝑟 2 − 2𝑟 2 cos(180° − 𝜃)                     cosine 𝑟𝑢𝑙𝑒

               = √2𝑟 2 + 2𝑟 2 cos 𝜃                            ✓ simplification
                                                                                          (2)
                 = 𝑟√2 + 2 cos 𝜃            given

 7.3 𝐷𝐹̂ 𝐺 = 90° angl in semi circle
                       𝐷𝐹
     In ∆𝐷𝐹𝐺: sin 𝜃 = 𝐷𝐺
                                                                          𝑟√2 + 2 cos 𝜃
                                                               ✓sin 𝜃 =
                 𝑟√2 + 2 cos 𝜃                                                 2𝑟
       sin 𝜃 =
                      2𝑟
                                                               ✓
       (2 sin 𝜃)2 = (√2 + 2 cos 𝜃)2                            4 sin2 𝜃 = 2 + 2 cos 𝜃
       4 sin2 𝜃 = 2 + 2 cos 𝜃                                  ✓2 sin2 𝜃 = 1 + cos 𝜃
                                                                                          (3)
       2 sin2 𝜃 = 1 + cos 𝜃

 7.4                                                           ✓ 2(1 − cos 2 𝜃) = 1 +
                           2
                  2(1 − cos 𝜃) = 1 + cos 𝜃                       cos 𝜃

                   2 − 2 cos2 𝜃 = 1 + cos 𝜃
        −2 cos2 𝜃 − cos 𝜃 + 1 = 0
          2 cos 2 𝜃 + cos 𝜃 − 1 = 0                            ✓ standard form
       (2 cos 𝜃 − 1)(cos 𝜃 + 1) = 0
                                                               ✓ factors
       cos 𝜃 = 2
                 1
                       or cos 𝜃 ≠ −1                           ✓ both equations

                                                               ✓   𝜃 = 60°
       ∴ 𝜃 = 60°               θ ≠ 180°
                                                                                          (5)
                                                                                        [11]

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Mathematics P2                                     10                              FS/September 2018
Grade 12                                    Marking Guidelines

QUESTION 8

 8.1      𝑎=1                                                    ✓ 𝑎=1
          𝑏=2                                                    ✓𝑏=2
          𝑐=2                                                    ✓𝑐=2
          𝑑=1                                                    ✓𝑑=1
                                                                 (4)

 8.2       𝑃(21,44°; 0,73)                                       ✓ correct substitution
                                                                 (2)

 8.3.1 𝑥 = 90°                                                   ✓ 90°                        (1)

 8.3.2    x  [45;135]                                         ✓ 45° 𝑎𝑛𝑑 135°
                 OR                                              ✓ Notation
          45  x  135                                           (2)
                                                                                                 [9]

QUESTION 9

 9.1.1    𝐶̂ = 𝐴̂1 = 78∘ alt <'s AP // CB                                    ✓ R
                                                                            (1)
 9.1.2    𝐴𝐵̂ 𝐶 = 𝐴̂1 = 78∘ tan chord theorem                               ✓ S✓R
                                                                            (2)
 9.1.3    𝐴̂3 = 𝐶̂ = 78∘ tan chord theorem                                  ✓ S                 (1)

 9.1.4    𝐴̂1 + 𝐴̂2 + 𝐴̂3 = 180∘ straight line                              ✓ S
          78∘ + 𝐴̂2 + 18∘ = 180∘
          𝐴̂2 = 24∘                                                                              (1)

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Mathematics P2                                  11                            FS/September 2018
Grade 12                                 Marking Guidelines

 9.2.1       𝑌̂1 = 𝑎                 tan chord theorem               ✓S✓R✓
          𝑌̂1 = 𝑌𝑋෡ 𝑃 = 𝑎                 alt angle XY // QT         ✓S
          𝑌𝑋෡ 𝑃 = 𝑄̂ = 𝑎            = angles opp = 𝑠𝑖𝑑𝑒𝑠             ✓S
          𝑌̂2 = 𝑄̂ = 𝑎              angles in same circle segment    ✓S✓R✓
                                                                                            (6)
 9.2.2    ෡2 = XY
          T      ෡T              alt angles; XY // QT                ✓    S/R
             =෡Y1 + ෡
                    Y2
             =෡ T1 + ෡
                     T1                                              ✓
                 ෡
             = 2T1                                                                          (2)

                                           1
 9.2.3    𝑇̂2 + 𝑇̂1 = 90°         angle in 2 ⨀                       S✓R✓
          𝑇̂2 = 90° − 𝑎
                                                                                            (2)
 9.2.4    𝑇̂2 = 90° − 𝑎          proven
          But YO෡ Q = 2𝑇̂2      angle at centre = 2 angl on circle   ✓ S
                     = 2(90° − 𝑎)
                     = 180° − 2𝑎
                QO෡ R= 180° − 180° + 2𝑎
                     = 2𝑎
          ∴ SORT is a cyclic quad ext angl = opp int angle           ✓ R

          OR

           ෡O = 90°
          XY                          tan ⊥ radius
           ෡
          XOY = 90°−∝                 sum of < 𝑠 of ∆                ✓S
           ෡    ̂
          XOY = 𝑇2
          ∴ SORT is a cyclic quad/converse: ext. <s of cyclic quad   ✓R

          OR

           ෡ Y = 2𝑇̂1
          XO                          < at centre = 2 < at circum. ✓S
           ෡     ̂
          XOY = 𝑇2
          ∴ SORT is a cyclic quad/converse: ext. < of cyclic quad  ✓R
                                                                                            (2)
 9.2.5    𝑇̂1 + 𝑇̂2 = 90°                                            ✓ S
          𝑎 + 2𝑎 = 90°
               3𝑎 = 90°
               ∴ 𝑎 = 30°                                             ✓ answer
                                                                                            (2)
 9.2.6    𝑋𝑌̂𝑂 = 90°               rad ⊥ tangent                     ✓    S
          𝑅̂2 = 90°                alt.angles; XY // TQ              ✓    S
          ∴ 𝑇𝑅 = 𝑅𝑄                 line from centre of circle ⊥     ✓    R
          chord                                                                            (3)
                                                                                          [22]

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Mathematics P2                                       12                           FS/September 2018
Grade 12                                      Marking Guidelines

QUESTION 10

 10.1.1      Const: Join KZ & LY & draw ℎ1 from K ⊥ XL & ℎ2                ✓ const
                    from L ⊥ XK

             Proof
                                                                               𝐴𝑟𝑒𝑎 ∆𝑋𝐾𝐿 𝑋𝐾
                             1                                             ✓            =
             𝐴𝑟𝑒𝑎 ∆𝑋𝐾𝐿       2
                               𝑋𝐾×ℎ1     𝑋𝐾                                    𝐴𝑟𝑒𝑎 ∆𝐿𝑌𝐾 𝐾𝑌
                         =   1         = 𝐾𝑌
             𝐴𝑟𝑒𝑎 ∆𝐿𝑌𝐾         𝐾𝑌×ℎ1
                             2

                             1                                                 𝐴𝑟𝑒𝑎 ∆𝑋𝐾𝐿 𝑋𝐿
             Area ∆XKL
                         =   2
                               XL×ℎ2     XL
                                       = LZ                                ✓            =
             Area ∆KLZ       1
                               LZ×ℎ2                                           𝐴𝑟𝑒𝑎 ∆𝐾𝐿𝑍 𝐿𝑍
                             2

                                                                           ✓ R
             Area of ∆𝑋𝐾𝐿 = 𝐴𝑟𝑒𝑎 ∆ 𝑋𝐾𝐿                    common

             But Area ∆ 𝐿𝐾𝑌 = 𝐴𝑟𝑒𝑎 ∆ 𝐾𝐿𝑍             same base & height;
                                                     LK// YZ               ✓S
                                                                                                (5)
             𝐴𝑟𝑒𝑎 ∆ 𝑋𝐾𝐿 𝐴𝑟𝑒𝑎 ∆𝑋𝐾𝐿
                       =
             𝐴𝑟𝑒𝑎 ∆𝐿𝑌𝐾 𝐴𝑟𝑒𝑎 ∆𝐾𝐿𝑍
               𝑋𝐾 𝑋𝐿
             ∴    =
               𝐾𝑌 𝐿𝑍

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Mathematics P2                                           13                       FS/September 2018
Grade 12                                          Marking Guidelines

 10.2.1      𝐸𝐷       𝐴𝑇        2                                        ✓S✓R
                = 𝐴𝐶 = 3                     line // to one side of 
             𝐷𝐶
                      2
             DE = 3 × 9
                =6                                                      ✓DE = 6
             ∴ 𝐷 is the midpoint of BE                                                           (3)

 10.2.2      BF= FT line from midpt drawn // to 2nd side bisct          ✓       S/R
             3rd
                  1
             FD= 2TE midpoint theorem                                   ✓R
                                                                        ✓ 𝑇𝐸 = 4
             ∴ 𝑇𝐸 = 4
                                                                                                 (3)
 10.2.3(a             𝐴𝑟𝑒𝑎 𝑜𝑓 ADC

 )                    𝐴𝑟𝑒𝑎 𝑜𝑓ABD

                           𝐷𝐶
                      = 𝐵𝐷          same height                         ✓ same height

                           3𝑦
                      = 2𝑦

                           3
                      =2
                                                                            3
                                                                        ✓2
                                                                                                 (2)
 10.2.3(b        1
 )               2 × 𝑇𝐶 × 𝐸𝐶 × sin 𝐶                                     1
                 1                                                         × 𝑇𝐶 × 𝐸𝐶 × sin 𝐶
                 2 × 𝐴𝐶 × 𝐵𝐶 × sin 𝐶                                    ✓2
                                                                         1
                                                                         2 × 𝐴𝐶 × 𝐵𝐶 × sin 𝐶
                      (𝑥)(𝑦)
                 =                                                          (𝑥)(𝑦)
                     (3𝑥)(5𝑦)                                           ✓ (3𝑥)(5𝑦)

                      1
                 =
                     15                                                      1
                                                                        ✓
                                                                            15
                                                                                                (3)
                                                                                               [16]

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### Page 14

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Mathematics P2                               14                    FS/September 2018
Grade 12                              Marking Guidelines

QUESTION 11

 11.1    In TPS and  QSR

         𝑃𝑆 1,5 3                                                     𝑃𝑆 1,5
            =   =                                                 ✓      =
         𝑄𝑆   4   8                                                   𝑄𝑆   4

         𝑇𝑃 4,5 3                                                     𝑇𝑃 4,5
           =   =                                                  ✓     =
         𝑆𝑅 12 8                                                      𝑆𝑅 12
                                                                      𝑇𝑆 3,6
         𝑇𝑆 3,6 3                                                 ✓     =
                                                                      𝑄𝑅 9,6
           =   =
         𝑄𝑅 9,6 8

         ∴ TPS/// QSR        sides of triangles in proportion   ✓✓ S/R

         ∴ 𝑃𝑇̂S= 𝑅̂             s are equiangular                ✓ 𝑃𝑇̂S= 𝑅̂

         ∴ 𝑇𝑃 is a tangent   converse of tan chord theorem        ✓R

                                                                                  (7)
 11.2    𝑃̂ = 𝑄𝑆̂𝑅           s are equiangular                   ✓ S/R

         𝑄𝑆|| TP             corres. <s are =                     ✓ S/R

         𝑇𝑄
             = 12
                 1,5
                             proportion theorem; PT||SQ           ✓✓ S/R
         9,6

                                                                  ✓ 𝑇𝑄 = 1,2
         ∴ 𝑇𝑄 = 1,2                                                              (5)
                                                                                [12]

                                                                       TOTAL: 150

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