Mathematics Form 3 Chapter 8 Loci in Two Dimensions                       (c)
 (c) P                                                                            Q
        SQ
4. (a)                                   R                                             P
                                                        P                         Q
                                                                          5. (a)
                                                                                                                                               S
(b)                                      Q                                               P
                                                                   P
              Q                                                                                                           R
                                                                          (b)
                                                                                                          R
                                                                                         P
                                                                                                                                            S
                                                                                                                            Q
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(c)                                                     Mathematics Form 3 Chapter 8 Loci in Two Dimensions
               R                                        (c) S B R
                               S
                                                                                                  P                             Q
                                                   Q
P
                                                                       C DA
6. (a) D P C
                                                             7. (a)                                                A
                                                                                                     QQ
                                                                               R
RS
                   A        Q  B                                                                     B DC
                        Q                                                                                                    P
    (b)
                                                             (b)                                                                   B
                     A
P                                                                   A
                     D         B                                                Z                                                  C
                                                    R                                          YX
                               C                             DE
                        S                                                                            © Penerbitan Pelangi Sdn. Bhd.
                                                         43                                                                 A2
Mathematics Form 3 Chapter 8 Loci in Two Dimensions                  PT3 Standard Practice 8
 (c)                                                        Section A
                                          A                  1. Answer: C
BQ                                                   E       2. Locus of X is a circle of radius 6 cm with
                                         P                        centre N. From the rectangle KLMN, locus
                          R                                       X is arc KQR.
           C                             D                        Answer: D
                                                             3. Answer: D
                                                             4. Answer: C
                                                             5. Locus X is arc STU and locus Y is line OB.
                                                                  Hence, the point of intersection is at A.
                                                                  Answer: A
                                                             6. Locus X is line BD and locus Y is arc ADE at
                                                                  centre B. Hence the point of intersection is
                                                                  at D.
                                                                  Answer: D
                                                            Section B
                                                             1. (a) An ant moves on a
                                                                         straight horizontal
                                                                         wire.
                                                                            Seekor semut bergerak
                                                                            pada wayar mendatar
                                                                            yang lurus.
                                                                   (b) The tip of a fan blade
                                                                         that is spinning.
                                                                            Hujung bilah kipas yang
                                                                            sedang berpusing.
                                                                   (c) The tip of hour hand
                                                                         in a clock from 4
                                                                         o’clock to 8 o’clock.
                                                                            Hujung jarum jam dari
                                                                            pukul 4 ke pukul 8.
                                                                   (d) A lift moving in tall
                                                                         building.
                                                                            Sebuah lif bergerak
                                                                            dalam bangunan tinggi.
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                                                        Mathematics Form 3 Chapter 8 Loci in Two Dimensions
2.                                                      (b) P           Q
       A moving point that its
       distance from the line KL and         Line            UR
       NM is always constant.                POR                                                                Locus L
        Suatu titik bergerak yang jaraknya   Arc             T Locus M  S
        sentiasa sama dari garis KL dan      KVM
        NM.                                             (c)
                                             Line
       A moving point that its               LON             S 5 cm        T
       distance is the same as LM
       from point L.                         Line
                                             QOS
        Suatu titik bergerak yang jaraknya
        adalah sama dengan LM dari titik L.
       A moving point that its
       distance from the line KN and
       NM is always constant.
        Suatu garis bergerak yang jaraknya
        sentiasa sama dari garis KN dan
        NM.
       A moving point that is
       equidistant from the points P
       and R.
        Suatu titik bergerak yang berjarak
        sama dari titik P dan R.
Section C
 3. (a) (i) The locus of the screw is a parallel
                   line to the horizontal line.
            (ii) The locus of the apple is a vertical
                   line going down to the ground.
            (iii) The locus of the cow is a circle and
                   the pole as its centre.
            (iv) The locus of the ship is a
                   perpendicular line between Johor
                   and Singapore.
                                                        45 © Penerbitan Pelangi Sdn. Bhd.
                                                                                                            A2
Mathematics Form 3 Chapter 8 Loci in Two Dimensions
    HOTS Challenge
                                                         13 m                                                                                                  Pool
Tree                                                                                                                                                             Kolam
Pokok                                                                                                                                             7.6 m
                                                                                                    X
                                                                                                                       Stool
                                                                                                                                          Bangku
                                                                                                                       Y
The possible locations of the treasure are marked X or Y.
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Top One (MATHS F3) Penerbitan Pelangi Sdn BhdCHAP
         TER
     9 Straight Lines
                           Garis Lurus
1. (a) m = undefined                             (c)     Gradient           m      =     –7  5  =  –7
    (b) m < 0                                                                          11 –        6
    (c) m = 0                                                     –7
                                                         y  =     6   x    +   c,
                                                         substitute (5, 0)
2. (a) y = –5x + 2                                       0     =  –7    (5)    +    c
                                                                  6
      (b) y = 5
                                                                  35
      (c) y = 3                                          c     =  6
      (d)  y  =  1  x                                    y     =  –  7   x  +      35
                 2                                                   6             6
      (e)  y  =  3  x  +  2
                 4        5
                                                 5. (a) 3y = –5x + 2
3. (a) m = –5, c = –6                                       y  =  –5     x  +      2
                                                                  3                3
                 1
      (b)  m  =  2  ,  c  =   7
      (c) m = 0, c = 8                           (b) 6y = 2x +3
                                                            y  =     2   x  +      3
                                                                     6             6
4.    (a)  Gradient       m   =  3–5       =  1             y  =     1   x  +      1
                          c,     –4 – 0       2                      3             2
                 1
           y  =  2  x  +
           substitute (0, 5)                     (c) 4y = –x – 1
           5= 0 + c                                         y  =  –     x   –    1
                                                                        4        4
           c=5                                   (d) –8y = –4x +12
           y  =  1  x  +  5                                           –4            12
                 2                                                    –8            8
                                                               y  =         x  –
      (b)  Gradient       m   =  –2  –  1                      y  =   1  x    –    3
                                 10  –  9                             2            2
                              =  –3                      y            x
                                 1                       8            4
                                                 6. (a)        =  –         +    1
                              = –3
           y = –3x + c,                                     y  =  –   8  x    +    8
                                                                      4
           substitute (9, 1)
                                                            y = –2x + 8
           1 = –3(9) + c
           c = 28
           y = –3x + 28
                                                 47 © Penerbitan Pelangi Sdn. Bhd.
                                                                                                     A2
Mathematics Form 3 Chapter 9 Straight Lines
   (b)     y       =     –   x    +    1                    9. (a) LHS: 3(2) = 6
           5                10                                       RHS: 2(0) + 6 = 6
                                                                     Conclusion: Yes
                y  =     –  5x    +    5
                            10                                  (b) LHS: 4(–2) = –8
                                                                       RHS: 3(–1) + 1 = –2
                y  =     –  1  x   +   5                               Conclusion: No
                            2
                                                                (c) LHS: 0
   (c)     –    y     =     –  x   +     1                             RHS: 5(–3) – 1 = –16
                4              3                                       Conclusion: No
                   y  =     4x    –   4
                            3
   (d)     –    y     =     4x    +    1
                9           9
                                                            10. (a) L1: 3y = –6x + 7
                   y = –4x – 9
                                                                                          7
                                                                      y   =  –2x      +   3  ,  gradient m1 = –2
   7. (a)                          y  =     1  x  –  9           L2: –4y = 8x – 5
                                            6
                                                                      y   =  –2x      +   5  ,  gradient  m2    =   –2
                                6y = x – 54                                               4
           x – 6y – 54 = 0                                       m1 = m2, therefore L1 is parallel to L2.
   (b)                x  +     y   =   1                    (b)  L1:  2y  +  1=–      1   x  +  6
                      2        8                                                      2
   	       	         x  +     y   =   1    ×    8                          y  =  –  1  x   +  5  ,  gradient  m1 =   –  1
                      2        8                                                      4         2                         4
                      4x + y = 8                                 L2: 3y = 8x – 3
           4x + y – 8 = 0                                                    8                                  8
                                                                             3                                  3
                                                                      y=        x  –  1,  gradient    m2  =
   8. (a)                      5y = 2x + 10                      m1 ≠ m2, therefore L1 is not parallel to
                                                                 L2.
                –2x + 5y = 10
           –    2x       +     5y  =      10
                10             10         10
                      x
                –     5     +  y   =      1                 (c) L1: 3y + 2x = –6
                               2
                                                                                      2                                   2
   (b) 6y – 2x – 8 = 0                                                       y  =  –  3   x  –  2,    gradient  m1 =   –  3
                (–2x + 6y = 8) ÷ 8                               L2: 6 – 5y = 8x – 1
                   –x    +     3y     =   1                                  y  =  –  8   x  +  7  ,  gradient  m2  =  –  8
                   4           4                                                      5         5                         5
   (c)                      y  =   1   x  –    6                 m1 ≠ m2, therefore L1 is not parallel to
                                   2                             L2.
	          		  1  x  –     y  =   6    ÷   6
                2
           112x       –  y     =   1
                         6
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11. (a) 2y = 6 – 3x                                        Mathematics Form 3 Chapter 9 Straight Lines
   y  =     –3    x  +3                      12. (a) y = 2x – 4
            2                                                xy
                                                             0 –4
m     =     –  3                                             20
               2
                                                         y = 6 – 2x
new equation                                                 xy
                                                             06
y  =     –  3     x  +     c                                 30
            2
                                                   From the graph,
pada titik (5, 0)                                  Point of intersection
                                                   = (2.5, 1)
0  =     –  3     (5)      +  c
            2                                                              y = 6 – 2x y
c  =     15                                                                             6
         2
                                                                                        4 y = 2x – 4
Maka,          y     =  –     3  x  +  15
                              2        2
(b) 6x – 3y = 8
               y  =     6  x  –     8
                        3           3
                  =  2x –        8
                                 3
            m=2
new equation                                                            2       (2.5, 1)  x
y = 2x + c                                                                 24
                                             –4 –2 0
at the point (8, –1)                                                   –2
–1 = 2(8) + c
 c = –17
Therefore, y = 2x – 17
(c) 2y = 3x + 10                             –4
   y  =     3  x     +  5
            2
m=          3
            2
new equation
y     =  3     x  +     c
         2
at the point(–2, –1)
–1 =        3  (–2)        +  c
            2
   c=2
Therefore,                 y  =  3  x  +  2
                                 2
                                             49 © Penerbitan Pelangi Sdn. Bhd.
                                                                                                 A2
Mathematics Form 3 Chapter 9 Straight Lines                From the graph,
                                                           Point of intersection = (2, 0)
 (b) y = –x – 2
                                                                                    y
           xy
           0 –2                                                  y = –x + 2 4
          –2 0
                                                                                   2             y = –21 x – 1
        y = 2x – 2
                                                          –4 –2 0                           24                  x
           xy                                                                    –2
          0 –2
          10
     From the graph,
     Point of intersection
     = (0, –2)
                              y
                        4 y = 2x – 2                13. (a) y = 2x – 1 ................(1)
                        2                                       y = 3 – x ..................(2)
                                                           (1) – (2)
y = –x – 2                                                 3x – 4 = 0
                                                                    x     =  1  1
                                                                                3
     –4 –2 0                             24  x      Substitute         x  =  1  1     into  (2)
                            –2                                                  3
                                                    y  =  3   –  1  1     =  1  2
                                                                    3           3
                                                    Therefore the point of intersection is
                                                    =  1  1  ,  1  2  
                                                           3        3
            1
(c)  y  =   2  x  –  1
        xy                                          (b) 2y = 3x – 2 ................(1)
                                                            y = 2x + 1 ................(2)
        0 –1
                                                          (2) × 2
        20                                                2y = 4x + 2 ................(3)
     y = –x + 2                                           (1) – (3)
                                                          0 = –x – 4
        xy                                                x = –4
       02
       20                                                 Substitute x = –4 into (2)
                                                          y = 2(–4) + 1
                                                          y = –7
                                                          Therefore the point of intersection is
                                                          = (–4, –7)
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                                                          Mathematics Form 3 Chapter 9 Straight Lines
(c)  y  =  1  x  –  3  ..................(1)              substitute           x=       7   into  (3)
           2                                                                            2
     4y = 6 – x ...................(2)                    y  =  8    7    –  26
                                                                      2
     (1) × 4
     4y = 2x – 12 ...............(3)                      y=2
     (2) – (3)                                            Therefore, the point of intersection is
     0 = –3x + 18
     x =6                                                 =    7  ,  2
                                                                2
     Substitute x = 6 into (1)                       (c)  y  =  1     x  –  16       ....................(1)
                                                                5           5
     y  =  1  (6)   –  3  =  0
           2
                                                          –3x + 7y = –8 ..................(2)
     Therefore, the point of intersection is
     = (6 , 0)                                            Substitute (1) into (2)
14. (a) 7x – 4y = –7 ..........(1)                         –3x+     7     1  x  –     16    = –8
            5x + y = 22 ............(2)                                     5           5
            From (2)                                         –3x      +  7  x     –  112      =  –8
            y = 22 – 5x ............(3)                                  5            5
            Substitute (3) into (1)                                                  –  8  x  =  72
             7x – 4(22 – 5x) = –7                                                       5        5
               7x – 88 + 20x = –7                                                          x = –9
                             27x = 81
                                x=3                       substitute x = –9 into (2)
                                                          –3(–9) + 7y = –8
            Substitute x = 3 into (3)
            y = 22 – 5(3)                                               y = –5
            y=7
                                                          Therefore the point of intersection is
            Therefore, the point of intersection is       = (–9, –5)
            = (3, 7)
                                                     15. (a) y = –2x + 5,
(b) 4x – 7y = 0 .......................(1)                       Gradient PQ = –2
      8x – y – 26 = 0 .................(2)                       Equation PQ, y = –2x + c
                                                                 Substitute x = 2 and y = –1,
     From (2)                                                    –1 = –2(2) + c
     y = 8x – 26 ........................(3)                       c=3
     substitute (3) into (1)                                     Therefore the equation is y = –2x + 3
     4x – 7(8x – 26) = 0
     4x – 56x +182 = 0
                       52x = 182
                          x=    7
                                2
                                                     51 © Penerbitan Pelangi Sdn. Bhd.
                                                                                                         A2
Mathematics Form 3 Chapter 9 Straight Lines
(b)  Gradient CD =                          2–  (–2)     =     4            Substitute x = 0, y = 4
                                            4–  (–1)           5
                                                                            4   =  2  (0)   +  c
     Gradient        AB =                     y–0        =     4                   3
                                            3 – (–3)           5            c =4
                                                      y  =     4            y  =   2  x  +  4
                                                      6        5                   3
                                                                                         4           2
                                                      y  =  4     4         mQR =     –  6     =  –  3
                                                                  5
            4
     y   =  5  x  +  c                                                      Equation of QR is :
     At the point (–3, 0 )                                                  y  =   –  2  x  +  c
                                                                                      3
     substitute x = –3 and y = 0                                                                                                2
                                                                            at  the   point       (0,   4),  c  =  4,  y  =  –  3  x  +  4
            4
     0   =  5  (–3)     +                c
     c   =  12                                                              (e) R(–3, 3)
            5                                                                     mRP = 0
                                                                                  Equation of RP is y = 3
     Therefore equation of AB is
     y   =  4  x  +  12                                                     (f) Q(4, 0)
            5        5
                                                                                         8
     and the equation of AB in general form                                 mPQ =     0  –  4     =  –2
     is                                                                     mRS = mPQ = –2
		                       y             =  4   x  +  12       ×5           Equaton of RS:
                                            5         5
                        5y = 4x + 12                                        y = –2x + c
     4x – 5y + 12 = 0                                                       Substitute x = –3, y = 8
                                                                            8 = –2(–3) + c
(c)  Gradient mPQ                        =  6–        1  =  5               c =8–6=2
                                            7–        6                     y = –2x + 2
     mOR = mPQ = 5
     Equation of OR: y = 5x + c
     Substitute x = 0, y = 0
     c=0
     y = 5x
(d) mPR = undefined	
      Equation of PR is x = 6
     Q (0,4)
     mPQ =        8–4      =                2
                    6                       3
     Equation of PQ                         is     y  =  2  x  +     c
                                                         3
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                                            Mathematics Form 3 Chapter 9 Straight Lines
          PT3 Standard Practice 9           Section B
Section A                                   1.                 Diagram                  Equation
                                                 (a)
                                6  –  (–4)                        Rajah                 Persamaan
                                8  –  (–2)
1.  Gradient     of    KL    =                                    y
                                                            K
                             =1                                                         y  =  –  2  +  2
                                                                2                                5
    Equation of KL                                                             5 Lx
    y=x+c                                                       O
    at L(8, 6)
    6=8+c                                   (b) y                              K(8, 4)     y=    1  x
    c = –2                                                                       x               2
                                                      LO
    ∴y=x–2
    Answer: D
2. 4x + 3y = 15                             2. Equation
          3y = –4x + 15                                     Persamaan             Point
           y  =  –     4  x  +  5                (a) y = 3x – 7                     Titik
                       3
                                                                                  (6, 11)           ✓
                 4
    ∴  m  =  –   3  ,  c  =  5
    Answer: B                               (b) –5x + y = 2                       (–3, –13) ✓
                                                                                   (2, –3) ✗
3. x-intercept, y = 0                       (c)       3y    =  1  x  +  8         (–10, 6) ✗
       2x – 3y = 6                                             2
    2x – 3(0) = 6                           (d)       1  y  +  1  x  =  1
             2x = 6                                   3        5
               x=3
                                            Section C
    ∴ x-intercept = 3
                                            1. (a) (i)         Rhombus has equal length of sides.
    Answer: B                                                  OK = KL = LM = MO
                                                               OK = (–5)2 + 122
4. mKL = –1, c = –5
                                                                   = 13 units
    ∴ equation of the straight line KL is                      ∴ OM = 13 units
        y = –x – 5
                                                                  M(–13, 0)
    Answer: C
5.            –  2        =  5                        (ii)     mKO   =   12 – 0
                 3                                                      –5 – (0)
          y         x
                                                                           12
          3y – 2x = 15                                               =  –  5
    2x – 3y + 15 = 0                                           y = mx + c
    Answer: C                                                  y = – 152x + c
                                                               Substitute x = 0, y = 0,
                                                               c=0
                                                               Equation of LM
                                                               y = – 152x
                                                      (iii) y = 12
                                            53 © Penerbitan Pelangi Sdn. Bhd.
                                                                                                          A2
Mathematics Form 3 Chapter 9 Straight Lines
(b) Equation of line PQ                                         2. (a) (i)  k located at line y = 3
                                                                            Therefore k(a, 3)
y  =  –     5  x   +     5               ..............(1)                  y = 2x – 4
            3                                                               3 = 2(a) – 4
                                                                            a = 3.5
Equation of line RS
y = x + 3 ....................(2)
(1) = (2),                                                                  ∴ k(3.5, 3)
 –  5  x  +   5    =   x               +   3  ×3              (ii) x = –4
      3
–5x + 15= 3x + 9
                                                                (iii) mML = mKN
            x=        3                                                      =2
                      4
                                                                       y = 2x + c
Substitute x =                           3   into (2)                  Substitute x = –4, y = 0,
                                         4                             0 = 2(–4) + c
         3                                                             c =8
y=       4  +3                                                         ∴ y = 2x + 8
y  =  15
      4
∴  k    3  ,   15                                             (iv) Coordinates P = (p, 0)
         4      4                                                     Substitute x = p, y = 0 into y = 2x – 4,
                                                                      0 = 2p – 4
(c) Substitute x = 2, y = 8,                                          p=2
                                                                      Length of MP = 2 – (–4)
   y  =     k  x  +   5m                                                                  = 6 units
            3                                                         Area of KLMP = (3 × 6)
                                                                                          = 18 units2
   8  =     2k    +   5m
            3
24 = 2k + 15m ...................(1)
Substitute x = –5, y = –13
   y     =  k   x    +   5m                                     (b) 2x + 3y – 4 = 0
            3
                                                                                   3y = –2x + 4
–13 =       –5k       +                  5m
             3                                                                     y   =  –  2  x  +  4
                                                                                             3        3
–39 = –5k + 15m ...............(2)
                                                                                             2
(1) – (2)                                                       gradient,          m   =  –  3
24 – (–39)= 2k – (–5k)
               63 = 7k                                          (c) 5x + 2y = 8
                  k=9                                           Divide both sides by 8,
Substitute k = 9 into (1)                                       5x          +  2y  =   8
                                                                8              8       8
   24 = 2(9) + 15m                                                             y
                                                                x           +  4   =1
15m = 6                                                         8
   m     =  2                                                   5
            5
                                                                Therefore,            x-intercept     =  8
∴m       =     2  ,  k   =9                                                                              5
               5                                                y-intercept = 4
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                                        Mathematics Form 3 Chapter 9 Straight Lines
HOTS Challenge
            Types   AB
Packaging           10 15
Packet (x)            8 20
Boxes (y)
                   740 1 350
10x + 8y = 740 ....................(1)
15x + 20y =1 350 ................(2)
From (1)
10x + 8y = 740
           x = (740 – 8y) ÷ 10
           x = 74 – 0.8y
Substitute (3) into (2)
15(74 – 0.8y) + 20y = 1 350
   1 110 – 12y + 20y = 1 350
                        8y = 240
                          y = 30
x = 74 – 0.8 × 30
x = 50
                                        55 © Penerbitan Pelangi Sdn. Bhd.
                                                                                            A2
PT3 Model Paper
Section A                                         5. 4.3 × 10–5 + 0.00174
 1.                                                   = 4.3 × 10–5 + 1.74 × 10–3
                                                      = (0.043 + 1.74) × 10–3
      1 +2 2 +2                                       = 1.783 × 10–3
      3 +2 4 +2                                       Answer: C
             =5
                                                      6.
      ∴ 5n + 2, n = 1, 2, 3, 4, …                       10 –3 =      17 3
      Answer: A
                                                        17 10
 2.
       8 8, 16, 64                                                 17  3  1
            1, 2, 8                                                  10
      Answer: B                                         =                     2
 3. Scale                                         	 		                  3
      = length of drawing : length of object
      = 0.16 cm : 5 m                                     17 2
      = 0.16 : 500
      = 1 : 3 125                                       = 10
      Answer: C
                                                  Answer: C
 4. Answer: D
                                                  7. x + 90° + 0.5x + 69° + 2x + 68° = 360°
                                                                                 3.5x + 227° = 360°
                                                                                           3.5x = 133°
                                                                                                x = 38°
                                                      Answer: A
                                                  8. Answer: B
                                                                     x1  +  x2      y1 + y2
                                                                         2             2
                                                   9. Midpoint =               ,
                                                                     –6  + (–2)        –9 +  3
                                                                          2               2
                                                         S =                      ,
                                                        = (–4, –3)
                                                  Answer: A
                                                  10. Coordinates of image
                                                        = (–1 + 7, –5 + –5)
                                                        = (6, –10)
                                                        Answer: C
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11. Total time = 2345 – 2015                                    16. Mean
                    = 3.5 hours
                                                                         (12 × 5) + (15 × 3) + (18 × 7) + (21 × 2)
     Average speed =         250                                      =                    + (24 × 6)
                             3.5                                                       5+3+7+2+6
                          = 71.43 km/h                                =  417
                                                                         23
     Answer: B
                                                                      = 18 .13
12. (Cos 60° + sin 45°)(Cos 60° – sin 45°)                            Answer: A
       1+  2  1  –   2                                  17. Percentage of students weight more than
               2    2      2                                          60 kg.
      2
     =  –  1                                                                           25 + 5
           4                                                             15 + 30 + 10 + 15 + 25 + 5
                                                                      =                                × 100%
     Answer: D
                                                                      = 30%
13. Price after discount                                              Answer: D
     P  =  RM60     ×  80     =    RM48                         18. The interest Vadivoo has to pay
                       100
                                                                                          5.85
     Q  =  RM78     ×   75    =    RM58.50                            =  RM55   000    ×  100    ×  2
                        100
                                                                      = RM6 435
     R  =  RM53     ×  95     =    RM50.35                            Answer: B
                       100
     S  =  RM94     ×  60     =    RM56.40                      19. Answer: A
                       100
     ∴ The cheapest type of diaper is diaper P                  20. K(0, y) J(–2, 6)
     Answer: A
                                                                      Gradient JK = 3
14. ∠UQP = ∠TSQ = 65°                                                    0  y–6        =  3
      ∠QUP = 180° – 65° – 72°                                               – (–2)
               = 43°
                                                                              y–6=6
      Answer: C
                                                                                 y = 12
                                                                      ∴ OK = 12 units
15.  Probability    of    green    dresses   =  1  –  2   –  1        OL = 132 – 122
                                                      5      5            = 5 units
                                                2                     ∴ L(5, 0)
                                                5
                                             =                        x-intercept of KL is 5
                                                                      Answer: C
     2   × Total    dresses     =  10
     5              dresses     =  10 × 
                                          5
           Total                          2
    = 25 pieces
     Answer: A
                                                                57 © Penerbitan Pelangi Sdn. Bhd.
                                                                                                                    A3
Mathematics Form 3 PT3 Model Paper                                   0.5 cm
                                                                  0.5 cm
Section B
21. (i)  K  =  –2.4   =  –2              2
                                         5
                                         4
         L  =  –1.8   =  –1              5
(ii) KL – 1.203
      = –2.4 (–1.8) – 1.203
      = 4.32 – 1.203
      = 3.117
22. (a) (i) Line KM / Garis KM                                    25. (i) ✓
            (ii) Line AC / Garis AC                                     (ii) ✗
      (b) 6 – 3y  4y – 8
            6 + 8  4y + 3y
            14  7y
               y2
23. (a) Centre                                                    (b) (i) FALSE
      (b) Chord                                                         (ii) TRUE
      (c) Major sector
      (d) Circumference                                           Section C
24. (a) Scale = 2 cm : 10 cm                                      26. (a) (i)     1  =    1
                    =1:5                                                        3 64        4
(b)      AB    =  36  ×  1               =  6  cm                    (ii)   2  –   9  2      2  –      2
                         6                                                      3                  3
                                                                                             =           3
         BC    =  EF  =  21              ×  1     =  3.5  cm                                =   –    7 2
                                            6
                                                                                                      3
         DE = 9       ×  1               = 1.5 cm                                               =  49
                         6                                                                         9
         AH    =  42  ×  1               =  7  cm
                         6
                                                                                x2              x –3          1
         HE    =  2  AB  ×               1                        (b) (i)       8y2   × x –5 =  8y2    =    8x3y2
                  3                      6
               =  2  (36)   ×            1     =  4  cm                                                         1
                  3                      6
                                                                  (ii)                 93k – 1  =  729 × 81 2
                                                                                                        3–2
                                                                                36k – 2 × 3–2 = 36 × 32
                                                                                       36k – 4 = 38
                                                                                ∴ 6k – 4 = 8
                                                                                        6k = 12
                                                                                          k=2
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                                                                        Mathematics Form 3 PT3 Model Paper
(c) Amount to be paid by instalment                            c  =        7
      = RM84.62 18                                                         2
      = RM1 523.16
                                                               ∴ Equation of the straight line is
      Amount of interest to be paid                                               1        7
      = RM1 523.16 – RM1 360                                            y  =  –   2  x  +  2
      = RM163.16
                                                               28. (a) x + y = 12 ......................(1)
             k + 5 + 6 + 6 + 8 + 12 + 12                                   4y – x = 18 .....................(2)
27. (a) (i)                     + 12                       =8  From (1)
                                  8               k + 61 = 64  y = –x + 12 ...................(3)
                                                        k=3    From (2)
     (ii) mode = 12                                            y  =        x  + 18   ......................(4)
                                                                               4
     (iii) 3, 5, 6, 6, 8, 12, 12, 12
                                   6     +  8                  (3) = (4)
             median,         m  =  7     2
                                =                                                       x  + 18
                                                                  –x       +  12  =         4
             k–m =3–7                                          –4x + 48 = x + 18
                     = –4
                                                                              –5x = –30
                                                                              x=6
(b) x2 – 2x – 3 = x(3 – x)                                     Substitute x = 6 into (1)
           x2 – 2x – 3 = 3x – x2                               6 + y = 12
          2x2 – 5x – 3 = 0                                          y= 6
        2x 1 x                                                 ∴ coordinates T is (6, 6)
        x –3 –6x
     2x2 –3 –5x                                                (b) (i)     P(Green         pen)  =1  –  3  =     4
                                                                                                        7        7
     (2x + 1)(x – 3) = 0
                                      1
                             x  =  –  2  ,  x  =  3                        Total number of pens:
                                                                           4  × Total      number of pens           =  24
                                                                           7
(c)  1  x    +  1   y  =  5                                                Total     number      of  pens  =     24 ×  7
     4          2                                                                                                      4
                1   y  =  –  1  x  +  5                                                                    = 42
                2            4
                                                                           Number of pink pens = 42 – 24
                    y  =  –  1  x  +  10                                                                 = 18
                             2
                                                                           New probability of pink pen:
     ∴       Gradient     of    the   line,    m  =  –  1                       18 + 2
                                                        2                  =  42 + 2 + 6
     Equation of straight line                                             =  2
             –1                                                               5
     y  =    2   x  +  c
     Substitute x = 3, y = 2
     2  =    –   1  (3)   +  c
                 2
                                                               59 © Penerbitan Pelangi Sdn. Bhd.
                                                                                                                   A3
Mathematics Form 3 PT3 Model Paper
         (ii) New probability of green pen:                  n() = 60
                          2                 3               17 – x + x + 8 + x + 2 + 23 – x + 6 + y = 60
                 =  1  –  5              =  5              17 – 3 + 3 + 8 + 3 + 2 + 23 – 3 + 6 + y = 60
(c) 2 – p  2p – 4                                                                                      56 + y = 60
         2 + 4  2p + p                                                                                        y=4
             3p  6
               p2                                         30. (a) (i) (a) Transformation A is reflection
        2p – 4  26 – 3p                                                            at line y = 1.
      2p + 3p  26 + 4                                                        (b) Transformation B is an
             5p  30                                                                enlargement with scale factor
               p6                                                                  of 2 at centre (– 4, 0).
              pϽ6                                 pϾ2                  (ii) ∆STU = 22 × ∆JKL
                                                                                      = 4 × 25 cm2
                 2                             6                                      = 100 cm2
         2ϽpϽ6
         Therefore, the possible values of p are           (b) (i)     z  =  1     x–2
         3, 4 and 5.                                                           3       y
29. (a)  sin x =    8                                                  3z =     x–2
                    17                                                              y
         AG      =  8                                                  9z2  =  x  –  2
         BG         17                                                            y
         16         8                                                  9yz2 = x – 2
         BG         17
                 =                                                     x = 9yz2 + 2
         BG = 34 cm                                        (ii) x = 9yz2 + 2
         DG = BG – BD
                                                                        x          1  2
              = 34 – 29                                                              3
              = 5 cm                                                   x
                                                                       =    9(–1)          +  2
                                                                       =    1
         cos  y=    5
                    13
                           5                               31. (a) Volume of small cubes
              y  =  cos–1  13
                                                                    Volume of sphere
              y = 67.4°                                    =                4 851
                                                            =     4  22   (21)3
                                                                    3  7
(b) Price of juice Kavita has to pay
      = (150 × RM0.10) + (200 × RM0.08)                                4 851
         + [(1 000 – 150 – 200) × RM0.03]
      = RM15 + RM16 + RM19.50                              = 8 cm3
      = RM50.50                                            Length of the side of small cube = 38
                                                                                                       = 2 cm
(c) n(P  R  Q) = 16                                      Area of one of the faces of the small
      x + 8 + x + 2 = 16                                   cube
                     2x = 6
                       x=3                                 =2×2
                                                           = 4 cm2
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(b)           Average     speed         =  Total distance       Mathematics Form 3 PT3 Model Paper
                                                Time
                                                           (ii) ∠FBC = 180° – 150°
[1.5  ×  (10 × 60)  +  5 000     +  x]  =  7  000                         = 30°
         10 + 32.5  +  17.5                   60
                                                                 ∴ x + 130° + 30° = 180°
                                    x = 1 100m                                         x = 20°
∴ Ken walked for 1 100 m.                                        ∠DEB + x = 180°
                                                                      ∠DEB = 180° – 20°
(c)      (i)  tan   ∠KJL  =   10.24                                            = 160°
                                8
                                                                 y = 360 – 160°
                                 10.24                             = 200°
                                   8
               ∠KJL
                       =  tan–1
              ∠KJL = 52°
                                                           61 © Penerbitan Pelangi Sdn. Bhd.
                                                                                                               A3
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