Sample Individual Event SI.1 Given A = (b

  Individual Events SI A

  20 I1 n

  10 I2 a

  48 I3 a

  2 I4 A

  40 I5 a

  45 B 4 a 25 b 144 b –3 B 6 b

  15 C 5 z 205 c 4 c

  12 C 198 c

  12

  5 D S 1 d 572 d 140 D 7 d

  2

  2 Group Events

  SG 2550 G6 a

  1 G7 a –8 G8 A

  2 G9 x

  6 G10 c

  3 2452 b 52 b 10 b 171 y 6 a –2

  P 2501 c

  13 area 116 c

  3 T

  10 200 b

  5 Q 10001 d 3 tan 2 d 27 n 19 d

  5 

  Sample Individual Event m n m +n

  SI.1 Given A = (b ) + b . Find the value of A when b  4, m  n  1.

  1 1 1+1 A = (4 ) + 4 = 4 + 16 = 20

  A

10 SI.2 If 2 = B and B > 0, find B.

  20

  10

  2 = 4  B = 4

20 B  45 SI.3 Solve for C in the following equation:  C .

  C 20  4 

  45  C C

  3

  125 = C  C = 5 SI.4 Find D in the figure.

  5 D = C sin 30  =

  C

  2 D 30 

  Individual Event 1 I1.1 If the sum of the interior angles of an n-sided polygon is 1440 , find n.

  180 (n – 2) = 1440  n = 10

2 I1.2 If x – nx + a = 0 has 2 equal roots, find a.

  2

  (–10) – 4a = 0  a = 25 I1.3 In the figure, if z = p + q, find z.

  A

  Reference: 1989 HI19

  ACB = 180 – p (opp. s cyclic quad.) a 

  ABC = 180 – q (opp. s cyclic quad.) P p 

  180 – p + 180 – q + a = 180 ( s sum of )

  z = p + q = 180 + a = 205

  Q q  B C I1.4 If S = 1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 +  + z, find S.

  Reference: 1985 FG7.4, 1988 FG6.4, 1990 FG10.1, 1991 FSI.1 S = 1 + (2 – 3 – 4 + 5) + (6 – 7 – 8 + 9) +  + (202 – 203 – 204 + 205) = 1

  Individual Event 2

4 I2.1 If ar = 24 and ar = 3, find a.

  ₄

  3

  1

  3 ar r = = = ar

  24

  8

  1  r =

  2

  ar = 24

  1  a = 24

  2  a = 48 ₂

  a a

   ²   I2.2 If x  x   x  b , find b.

   

  4

  2  

  2

  2

  (x + 12) = x + 24x + 144  b = 144

  b

  I2.3 If c = log 2 , find c.

  9 144

  c = log 2 = log 2 16 = 4

  9

  c

2 I2.4 If d = 12 – 142 , find d.

  4

  2 d = 12 – 142

  2

  2

  = 144 – 142 = (144 + 142)(144 – 142) = 2(286) = 572

  Individual Event 3

  sin

  15

  1  ² I3.1 If a =   tan ₂ 15  , find a. cos

  75  sin 75  sin

  15  ^{2  2}

  a =  sec

  15  tan 15  sin 15 

  = 1 + 1 = 2 I3.2 If the lines ax + 2y + 1 = 0 and 3x + by + 5 = 0 are perpendicular to each other, find b.

  a

  3 

  

  1    

    2 b  

   b = –3

  c I3.3 The three points (2, b), (4, b) and (5, ) are collinear. Find c.

  2

  c The three points are (2, –3), (4, 3) and (5, ), so their slopes are equal. _c

  2

  3

  3

  3      ₂

   4  2 5 

  4

  c

   – 3 = 3

  2  c = 12

  1

  1

  1

  1 I3.4 If : : 1  3 : 4 : 5 and :  9c : d, find d.

  x y z x  y y  z

  1

  1

  1

  x : y : z = : :

  3

  4

  5

  20

  15

  12 = : :

  60

  60

  60 = 20 : 15 : 12

  x = 20k, y = 15k, z = 12k

  1

  1

  1

  1 :  :

  x  y y  z

  20 k  15 k 15 k  12 k = 27 : 35 = 108: 140 = 9c : d

   d = 140

  Individual Event 4

  2 I4.1 In the figure, the area of PQRS is 80 cm .

  2 If the area of QRT is A cm , find A.

  QRT has the same base and same height as the parallelogram PQRS.

  1  = 40

  A =

  80

  2

  8 A   I4.2 If = log , find B.

  B _{2  }

  5  

  8

  40   

  = log

  B _{2  }

  5  

  = log

  2

  64

  6

  = log

  2

  2 = 6

  1 ₃

  1 I4.3 Given x  = B. If C = x  , find C. ₃

  x x

  1

  x  = 6 x ₂

  1

  1

  2 x  = ( x  ) – 2 ₂ x x

  2

  = 6 – 2 = 34 ₃

  1 C = x  ₃ x

  1 ²

  1   

  =   x x

  1  

      ₂

  x x

      = 6(34 – 1) = 198 I4.4 Let (p, q)  qD + p. If (C, 2)  212, find D.

  2D + C = 212  2D = 212 – 198 = 14  D = 7

  Individual Event 5

  2

  

3

  3 I5.1 Let p , q be the roots of the quadratic equation x – 3x – 2 = 0 and a = p + q . Find a. p + q = 3, pq = –2

  2

  2 a = (p + q)(p – pq + q )

  2

  = 3[(p + q) – 3pq]

  2

  = 3[3 – 3(–2)] = 45

  A I5.2 If AH  a, CK  36, BK  12 and BH  b, find b.

  ABH ~ CBK (equiangular)

  45

  b

   (ratio of sides, ~ s)

  K

  12

  36 P

  b = 15 C B H L I5.3 Find c.

  Reference: 1985 FG6.4 b

  20

  2

  2

  2

  15 + 20 = 25

  c

   ML  LN (converse, Pythagoras’ theorem)

  M N

  25

  1 1520 = 1 25c Area of MNL =

  2

  2

  c = 12

  I5.4 Let

  2 x

  23 2 x 1 c and d 2 x

  23 2 x 1 . Find d.        

  Reference: 2014 HG1 cd 

  2 x  23  2 x 

  1 2 x  23  2 x 

  1

    

  12d = (2x + 23) – (2x – 1) = 24  d = 2

  Sample Group Event Reference HKCEE Mathematics 1990 Paper 1 Q14

  Consider the following groups of numbers: (2) (4, 6) (8, 10, 12) (14, 16, 18, 20) (22, 24, 26, 28, 30) ……………………………

  th SG.1 Find the last number of the 50 group.

  2 = 2 1 6 = 2(1 + 2) 12 = 2(1 + 2 + 3) 20 = 2(1 + 2 + 3 + 4) 30 = 2(1 + 2 + 3 + 4 + 5)

  th

  The last number of the 50 group = 2(1 + 2 +  + 50)

  1 = 2   50   1  50  = 2550

  2

  th SG.2 Find the first number of the 50 group. th There are 50 numbers in the 50 group. th

  The first number of the 50 group = 2550 – 2(50 – 1) = 2452

  th SG.3 Find P if the sum of the numbers in the 50 group is 50P.

  2452 + 2454 +  + 2550 = 50P

  1  50   2452  2550  

  50 P

  2 P = 2501

  th SG.4 Find Q if the sum of the numbers in the 100 group is 100Q.

  1

  th

  The last number in the 100 group = 2(1 + 2 +  + 100) = 2   100   1  100  = 10100

  2

  th

  The first number of the 100 group = 10100 – 2(100 – 1) = 9902 9902 + 9904 +  + 10100 = 100P

  1  100   9902  10100   100 P

  2 P = 10001

  

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  5

  3

  5

  12 sin PCX =

  25

  PCX =

  2

  PCX = 1 – cos

  2

  5 2 d , find d. sin

  PCX =

  G6.4 If sin

  13  c = 13

  5

  2 52  =

  cos PCX =

  7 Group Event 6 As shown in the figure, ABC and XYZ are equilateral triangles and are ends of a right prism. P is the mid-point of BY and BP  3 cm,

  a = 1 G6.2 If CX = b cm, find b. CX = ^{2 2}

  XY  4 cm.

  G6.1 If a = PX CP , find a.

  CP = ^{2 2}

  4

  3 

  cm = 5 cm = PX (Pythagoras’ theorem)

  4

  c , find c.

  6 

  cm = 52 cm (Pythagoras’ theorem) A Y

  X C B P Z

  b = 52 G6.3 If cos

  PCX =

  5

  2  d = 3

  Group Event 7 Given that OABC is a parallelogram.

  B(4, b) A(a, 9) G7.1 Find a. a – 0 = 4 – 12

   a = –8 G7.2 Find b.

  b – 1 = 9 – 0 C(12, 1) 

   b = 10

  O(0, 0) G7.3 Find the area of OABC.

  1

  12

  1 Area = 2 = 116 

  4

  10

2 G7.4 Find tan .

  OC = 145 _{5 2} OB = 116 _{2 2} 

  BC =  12  4    ₂ 1  10  = 145 _{2 2 1}

  145  116  145

  1 cos  = = 2 145 116

  5

    

  tan  = 2

  Method 2

  1 

  1

  m OC = 

  12 

  12 10 

  5

  m OB = 

  4  _{5 1}

  2 ₂  ₁₂ tan  = = 2 _{5 1}

  1   _{2 12}

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  G8.3 The roots of the equation x

  

  1 ₂    

  3

  1

  2

  6

  27 60 sin

   

  3 , find d. d =

  If the height of the pyramid is 27 cm, and its volume is d cm

  1 + 171 + c = sum of roots = 173  c = 3 G8.4 The base of a triangular pyramid is an equilateral triangle of side 2c cm.

  2 + 339x + 513 = 0 are 1, b and c. Find c.

   173x

  3

  = 171

  9 Group Event 8

  19 

  3

    ^{5 2 5}

  =

  b

    ⁵ 243 19 , find b.

  G8.2 If b ^A

   A = 2

   A

  1 ₂  

  2

  3 60 sin

  2 . Find A.

  G8.1 The area of an equilateral triangle of side A cm is 3 cm

  = 27

• x
• – 2x

  10

  cm

  2

  3 y =

  3

  6  y = 6

  G9.3 - G9.4 (Reference: 1991 FG8.1-2)

  Consider the following number pattern:

  T 1 = 2 T 2 = 8 T 3 = 18 T 4 = 32 G9.3 Find T 10 .

  8 – 2 = 6, 18 – 8 = 10, 32 – 18 = 14  T

  1 = 2, T 2 = 2 + 6, T 3 = 2 + 6 + 10, T 4 = 2 + 6 + 10 +14 T 10 =    

   

  4

  1

  2

  = 3 6

  2

  2

  10     = 200 G9.4 If T n = 722, find n.

       

  722

  4

  1

  2

  2

  2      n

  n n

  2

  = 361

  2

  2

  n = 19

  AC

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  10 Group Event 9 If the area of a regular hexagon ABCDEF is

  3 54 cm

  2

  and AB  x cm, AC  3 y cm, G9.1 find x. The hexagon can be cut into 6 identical equilateral triangles

   

  3

  54 60 sin

  2

  1

  6 ²    

   x

   x = 6 G9.2 find y. ABC = 120

  2

  1 ] cm

  = (x

  2

  2

  2

  cos 120 ) cm

  2

  = [6

  2

  2

  2

• 6
• – 2(6)

     

    

  2

  Group Event 10

  2

y

The following shows the graph of y = ax + bx + c.

  G10.1 Find c. x = 0, y = c = 3

  3 G10.2 Find a.

  x

  1

  1 O 

  3

  y = a(x + )(x – 3)

  2

  2 Sub. x = 0, y = 3

  3  – a = 3

  2

  a = –2 G10.3 Find b.

  1

  b

  3 – = sum of roots =  2   

  2

  b = 5

  2 G10.4 If y = x + d is tangent to y = ax + bx + c, find d.

  2 Sub. y = x + d into y = ax + bx + c

  2

• –2x + 5x + 3 = x + d

  2

  2x – 4x + d – 3 = 0

  2

   = (–4) – 4(2)(d – 3) = 0 4 – 2d + 6 = 0

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