Sample Individual Event (1986 Final Individual Event 2) SI.1 Given that 3x

  

Individual Events

  4

  5 3 1000

  5

  17 SI h I1 a I2 p I3 a I4 a I5 a

  32

  4

  36

  8

  12

  5 k b q b b b p

  3 c 10 k 12 c 16 c 4 c

  23

  16 34 150

  1

  12

  9 q d m d d d

  

Group Events

SG a

  2 G6 a 150 G7 C

  47 G8 A

  2 G9 S 1000 G10 A 1584

• –3

  10

  2

  3

  98

  14 b b K B K k

  60

  37.5

  1

  7 20 160 p k A C t x

  136

  6

  5

  9

  5

  15 q d B k d n

  Sample Individual Event (1986 Final Individual Event 2) h

2 Given that 3x – 4x + = 0 has equal roots, find h.

  3

  h

  2

  = (–4) – 4(3) = 0

  ∆ ⋅

  3

  h = 4

  If the height of a cylinder is doubled and the new radius is h times the original, then the new

  SI.2 volume is k times the original. Find k.

  2 Let the old height be x, old radius be r, then the old volume is . r x

  π

  The new height is 2x, the new radius is 4r,

  2

  2

  then the new volume is (4r) (2x) = 32

  r x π π k = 32 If log 210 + log – log 56 + log 40 – log 120 + log 25 = p, find p.

  SI.3

  10 10 k

  10

  10

  10

  10

   210

  32

  40 25 

  × × ×

   

  p = log

  10

   56 120 

  ×

  = log 1000 = 3

  10 p cos q A

  SI.4 If sin A = and , find q.

  =

  5 tan A

  15

  3 sin A = 5 cos

  A q =

  tan A

  15

  2

  cos

  A q =

  sin A

  15

  2

  2

  3

  1 sin

  1

  16 A − q

  − ( )

  5

  =

  = =

  3

  sin A

  15

  15

  5

  = 16

  q

  Individual Event 1

  2 Find a if 2t + 1 is a factor of 4t + 12t + a.

  I1.1

  2 Let f(t) = 4t + 12t + a

  2

   1   1   1       

  f =

  4 12 a = 0

  − − − + +

   2   2   2 

  a = 5

  I1.2 K denotes the nonnegative square root of K, where K

  0. If b is the root of the equation

  ≥ a x = x – 3, find b. −

  2

  2

  5 x = (x – 3)

  − ( )

  2 ⇒

  5 – x = x – 6x + 9

  2

  x – 5x + 4 = 0

  ⇒

  x = 1 or 4 When x = 1, LHS = 2 –1 = RHS

  ≠ When x = 4, LHS = 1 = RHS.

  x = b = 4

  ∴

  20 If c is the greatest value of , find c.

  I1.3

  2 cos

• b

  θ

  20

  20

  10 = = 2 cos

  4 2 cos 2 cos

  b θ θ θ + + +

  10

  c = the greatest value = = 10

  2

  1

  −

  A man drives a car at 3c km/h for 3 hours and then 4c km/h for 2 hours. If his average speed

  I1.4 for the whole journey is d km/h, find d.

  Total distance travelled = (30 3 + 40 2) km = 170 km

  × ×

  170

  d = = 34

  2

• 3

  Individual Event 2

  1

  3 cos 4 has roots. Find .

  I2.1 p p

  ° ≤ θ ° θ θ =

• +

  If 0 < 360 , the equation in :

  cos

  θ

  2

  3 cos + 1 = 4 cos

  θ θ

  2 ⇒

  3 cos – 4 cos + 1 = 0

  θ θ

  1

  ⇒

  cos = or 1

  θ

  3 = 3

  p

  1

  1

3 I2.2 If x p and x q , find q.

  − = − =

  3 x x

  Reference: 2009 FI2.3

  2

  1  1   

  x

  3 ; x

  9

  − = − =

   

  x x

  1

  ⇒

  x

  11

• 2

  =

  2 x

  1

  3 q = x

  −

  3 x

   1   1 

  2

      = x x

  1

  − + +

  2

     

  x x

  = 3(11 + 1) = 36 A circle is inscribed in an equilateral triangle of perimeter

  I2.3 A

  2 cm. If the area of the circle is k cm , find k. q

  π Reference: 1984 FG9.4

  12 cm

  Let the equilateral triangle be ABC, the centre of the

  12 cm E

  inscribed circle is O, which touches the triangle at D and

  O E , with radius r cm r cm

  Perimeter = 36 cm ⇒ Each side = 12 cm

  B C D 12 cm

  = 60 ( s of an equilateral )

  ACB ∠ ° ∠ ∆

  = 90 (tangent radius)

  ODC ∠ ° ⊥

  = 30 (tangent from ext. pt.)

  OCD ∠ ° CD = 6 cm (tangent from ext. pt.) r = 6 tan 30 =

  2

  3

  °

  2

  2

2 Area of circle =

  2 3 cm = 12 cm

  π π ( ) k = 12 Each interior angle of a regular polygon of k sides is m . Find m.

  I2.4 °

  Angle sum of 12-sides polygon = 180 (12 – 2) = 1800

  ° °

  Each interior angle = m = 1800 12 = 150

  ° ° ÷ ° m = 150

  Individual Event 3

2 If 998a + 1 = 999 , find a.

  I3.1

  2 998a = 999 – 1 = (999 – 1)(999 + 1)

  = 998 1000 ×

  = 1000 a If log = log , find b.

  I3.2 a b

  10

  2 log 1000 = log b

  10

  2 log b = 3

  2

  3 = 8 ⇒ b = 2

  The area of the triangle formed by the x-axis, the y-axis and the line 2x + y = b is c sq. units.

  I3.3 Find c. Reference: 1994 FI5.3 2x + y = 8; x-intercept = 4, y-intercept = 8

  1

  = area =

  4 8 = 16

  c ⋅ ×

  2

2 If 64 + is a perfect square, find + .

  I3.4 t ct d d

  2

  64 has a double root + + 16

  t t d

  2

  = 16 – 4 64 = 0

  d ∆ ×

  = 1

  d

  Individual Event 4 a a a

• 1 –1 Solve the equation 2 + 2 + 2 = 112 in a.

  I4.1

  1

  a 2 (2 + 1 + ) = 112

  ⋅

  2

  a

  2 = 32

  a = 5 Method 2

  6

  5

  4

  112 = 64 + 32 + 16 = 2 + 2 + 2 = 5

  a

2 If is one root of the equation + 35 = 0, find . –

  I4.2 a x bx b

  2

• – One root of x bx + 35 = 0 is 5

  2 ⇒

  5 – 5 b + 35 = 0

  ⇒

  = 12

  b b c

  − I4.3 If sin = , where 180 < < 270 , and tan = , find c.

  θ ° θ ° θ

  15

  3

  12

  4 sin =

  θ − = −

  15

  5

  4

  ⇒

  tan =

  θ

  3

  ⇒

  c = 4

  1 The probability of getting a sum of c in throwing two dice is . Find d.

  I4.4 d

  P(sum = 4) = P((1,3), (2, 2), (3, 1))

  3

  1

  1 =

  = =

  36

  12

  d ⇒

  d = 12

  Individual Event 5 I5.1 In the figure, find a.

  12

  2

  2

  2

  2

• – 8 = 12 + 9 (Pythagoras’ Theorem)

  a

  9

  8 = 17 a a If the lines ax + by = 1 and 10x – 34y = 3 are perpendicular to each other, find b.

  I5.2 17x + by = 1 is perpendicular to 10x – 34y = 3 ⇒ product of slopes = –1

  17

  10

  − × = − b

  34

  ⇒

  b = 5

  th th

  If the b day of May in a year is Friday and the c day of May in the same year is Tuesday,

  I5.3 where 16 < c < 24, find c. th

  5 May is a Friday

  th ⇒

  9 May is Tuesday

  th ⇒

  16 May is Tuesday

  rd ⇒

  23 May is Tuesday = 23

  c th c is the d prime number. Find d.

  I5.4 Reference: 1985 FSG.2, 1989 FSG.3

  The first few prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23

  th

  23 is the 9 prime number = 9

  d

  Sample Group Event (1986 Sample Group Event) The sum of two numbers is 50, and their product is 25. SG.1 If the sum of their reciprocals is a, find a.

  Let the 2 numbers be x, y.

• y = 50, xy = 25

  x

  1

  1

• ⇒ a =

  x y

• x y

  = xy

  50

  25 + If the lines + 2 + 1 = 0 and 3 + 5 = 0 are perpendicular, find .

  SG.2 ax y x by b

  2 + 2 + 1 = 0 is to 3 + 5 = 0 +

  x y x by ⊥

  ⇒

  product of slopes = –1

  2

  3

  −

  1

  − × = −

  2 b

  ⇒

  b = –3

  2 SG.3 The area of an equilateral triangle is 100 3 cm . If its perimeter is p cm, find p.

  Let the length of one side be x cm.

  1

  2 o

  sin 60 100

  3

  x =

  2

  ⇒

  x = 20

  ⇒

  p = 60

  3

  2 If x – 2x + px + q is divisible by x + 2, find q.

  SG.4

  3

  2 Let f(x) = x – 2x + 60x + q

  (–2) = –8 – 8 – 120 + q = 0

  f

  = 136

  q

  Group Event 6 ( ) ( )

  3

  3

  3

  68

  65

  32

  18 − ⋅

• 3

  If a = , find a.

  G6.1

  2

  2

  2

  2

  32

  32

  18

  18

  68

  68

  65

  65 − × ⋅ × + + +

  ( ) ( )

  2

  2

  

2

  2

  32

  18

  32

  32

  18

  18

  68

  65

  68

  68

  65

  65 − × ⋅ − × + + + +

  ( ) ( ) ( ) ( ) a =

  2

  2

  2

  2

  32

  32

  18

  18

  68

  68

  

65

  65 − × ⋅ × + + +

  ( ) ( ) = 50 3 = 150

  × If the 3 points (a, b), (10, –4) and (20, –3) are collinear, find b.

  G6.2 b

  4

  3

  4

  − + + The slopes are equal:

  =

  150

  10

  20

  10

  − − ⇒

  b = 10 If the acute angle formed by the hands of a clock at 4:15 is k , find k.

  G6.3 °

  Reference 1984 FG7.1, 1985 FI3.1, 1987 FG7.1, 1989 FI1.1, 2007 HI1

  1 = 30 + 30 = 37.5

  k ×

  4 R In the figure, PQ = 10, RS = 15, QS = 20. If XY = d, find d.

  G6.4 Reference: 1985 FI2.4, 1989 HG8 P

  1

  1

  1

  25

  1

  15 = + = =

  X d

  10 15 150

  6

  10 d d = 6

  Q S Y

  20

  Group Event 7 2 apples and 3 oranges cost 6 dollars. G7.1 4 apples and 7 oranges cost 13 dollars.

  16 apples and 23 oranges cost C dollars. Find C. Let the cost of one apple be $x and one orange be $y. 2x + 3y = 6 (1)

  …… 4x + 7y = 13 (2)

  …… (2) – 2(1): y = 1, x = 1.5 = 16x + 23y = 24 + 23 = 47 C

  5 sin

• 6 cos

  θ θ G7.2

  

K

θ

  3 sin

• 2 cos

  θ θ Reference: 1986 FG10.3, 1987 FG8.1, 1989 FSG.4, 1989 FG10.3

  6

  5 cos θ cos θ

• cos θ sin θ

  =

  K θ

  2

  3 cos θ cos θ

• cos sin θ

  5 tan

• 6

  θ

  =

  3 tan

• 2

  θ

  5

  2

• 6

  ×

  = = 2

  3

  2

• 2

  ×

  and G7.4 , are positive integers less than 10 such that 21 104 11 = 2 8016 9.

  G7.3 A B A B

  × × Similar Questions 1985 FG8.1-2, 1988 FG8.3-4 Find . G7.3 A 11 and 9 are relatively prime, 21 104 is divisible by 9.

  A

  2 + 1 + + 1 + 0 + 4 = 9

  A m ⇒

  8 + = 9

  A m ⇒

  = 1

  A Find .

  G7.4 B 2 8016 is divisible by 11.

  B

  2 + 8 + 1 – ( + 0 + 6) = 11

  B n ⇒

  11 – ( + 6) = 11

  B n ⇒

  = 5

  B

  Group Event 8 In the multiplication shown, the letters A, B, C and K (A < B) represent different integers from 1 to 9.

  2 = 64, 9

  C = 7 G8.4 Find K.

  B = 3 G8.3 Find C.

  ∴ A = 2 G8.2 Find B.

  ÷ 37 = 27

  10B + C = 37 or 74 When B = 3, C = 7, K = 9 999

  37K, 37 is a prime number Either 10A + C or 10B + C is divisible by 37

  100K + 10K + K = 111K = 3 ×

  2 = 81 Possible K = 1, 4, 5, 6, 9

  2 = 49, 8

  (Hint: KKK = K ×

  2

= 36, 7

  2 = 25, 6

  2 = 16, 5

  2 = 9, 4

  2 = 4, 3

  2 = 1, 2

  1

  ) × G8.1 Find A.

  111.) K K K C B C A

  K = 9

  Group Event 9 If S = ab – 1 + a – b and a = 101, b = 9, find S. G9.1 Reference: 1985 FG8.4, 1986 FG9.3, 1988 FG6.3 = (a – 1)(b + 1) = 100 10 = 1000

  S ×

  K

  &

  If x = 1 .

  9

  8 9 – 1 = , find K.

  G9.2 & and x

  99

  89 = 1.9 +

  x

  990

  K

  9

  89

  x

• – 1 = =

  99 10 990

  99 89 980

  98

  ×

• 9

  =

  = =

  990 990

  99 = 98

  K The average of p, q and r is 18. The average of p + 1, q – 2, r + 3 and t is 19. Find t.

  G9.3 p q r

  18

  =

  3

  ⇒

  p + q + r = 54

  p

  1 q 2 r 3 t

  − + + + + +

  19

  =

  4

  ⇒

  p + q + r + 2 + t = 76

  ⇒

  54 + 2 + t = 76 = 20

  t

  Z

  ) ) )

  In the figure, QR , RP , PQ are 3 arcs, centres at X, Y and Z

  G9.4

  respectively, touching one another at P, Q and R. If ZQ = d,

  d

  P

  XR = 3, YP = 12, X = 90 , find d.

  ∠ ° Reference: 1986 FG7.1

  12 Q

  XZ = 3 + d, XY = 3 + 12 = 15, YZ = 12 + d

  2

  2

  2 XZ + XY = YZ (Pythagoras’ theorem)

  Y

  2

  2

  2 X

  3 R (3 + d) + 15 = (12 + d)

  2

  2

  9 + 6d + d + 225 = 144 + 24d + d 18d = 90

  ⇒

  d = 5

  Group Event 10 G10.1 If A = 1 + 2 – 3 + 4 + 5 – 6 + 7 + 8 – 9 + + 97 + 98 – 99, find A.

  … Reference: 1985 FG7.4, 1988 FG6.4, 1991 FSI.1, 1992 FI1.4

  = (1 + 2 – 3) + (4 + 5 – 6) + (7 + 8 – 9) + + (97 + 98 – 99) A

  

…

  96

• 3

  = 0 + 3 + 6 + + 96 =

  32 = 99 16 = 1584

  A … × ×

  2

2 G10.2 If log ( – 1) – log ( – 5 + 4) + 1 = 0, find .

  k k k k

  10

  10

  2

  10( – 1) = – 5 + 4

  k k k

  2

• – 15 + 14 = 0

  k k

  = 1 or 14

  k

  When = 1, LHS is undefined rejected

  k ∴

  When = 14, LHS = log 13 – log (14 – 1)(14 – 4) + 1 = RHS

  k

  10

  10

  = 14

  k ∴

  and One interior angle of a convex -sided polygon is . The sum of the remaining

  G10.3 G10.4 n x ° interior angles is 2180 .

  ° Reference: 1989 HG2, 1992 HG3, 2002 FI3.4, 2013 HI6 G10.3 Find . x

  2180 + = 180( – 2) ( s sum of polygon)

  x n ∠

  2160 + 20 + = 180 12 + 20 + = 180( – 2)

  x x n ×

  < 180

  x Q

  20 + = 180

  x ∴

  = 160

  x G10.4 Find . n

• – 2 = 12 + 1

  n

  = 15

  n