Individual Event 1 (1998 Sample Individual Event) I1.1

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Individual Event 1 (1998 Sample Individual Event)

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Individual Event 2

I2.1 By considering: 1 1

The given is equivalent to:

3

I2.2 A triangular pyramid is cut from a corner of a cube with

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I3.1 The average of a, bc and d is 8. If the average of a, b, c, d and P is P, find P.

8

4 

  b c d a

a + b + c + d = 32

P P d c b

a    _

5

 32 + P = 5P P = 8

I3.2 If the lines 2x + 3y + 2 = 0 and Px + Qy + 3 = 0 are parallel, find Q.

Their slopes are equal:

Q

8 3 2

  

Q = 12

I3.3 The perimeter and the area of an equilateral triangle are Q cm and 3 cmR 2 respectively. Find R.

Perimeter = 12 cm, side = 4 cm

Area = 4 sin60 2

1 2

=4 3

R = 4

I3.4 If (1 + 2 +  + R)2_{= 1}2_{+ 2}2₊₊_R2₊_S_{, find}_S_.

(1 + 2 + 3 + 4)2 = 12 + 22 + 32 + 42 +S

100 = 30 + S

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I4.1 If each interior angle of a n-sided regular polygon is 140, find n.

Reference: 1987 FG6.3

Each exterior angle is 40 (adj. s on st. line)

 

40 360



n n = 9

I4.2 If the solution of the inequality 2x2 – nx + 9 < 0 is k < x < b, find b.

2x2 – 9x + 9 < 0

(2x – 3)(x – 3) < 0

2 3

< x < 3

b = 3

I4.3 If cx3 – bx + x – 1 is divided by x + 1, the remainder is –7, find c.

f (x) = cx3 – 3x + x – 1

f (–1) = –c + 3 – 1 – 1 = –7

c = 8

I4.4 If c x

x1  and d x

x2 1₂  , find d.

8 1

 

x

 1 64

2

       

x x

64 2 1

2 2  

x x

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I5.1 The volume of a hemisphere with diameter a cm is 18 cm3_{, find}_a_.

        

 18

2 4 2

1 a 2

a = 6

I5.2 If sin 10a = cos(360 – b) and 0 < b < 90, find b.

sin 60 = cos(360 – b)

360 – b = 330

b = 30

I5.3 The triangle is formed by the x-axis and y-axis and the line bx + 2by = 120. If the bounded

area of the triangle is c, find c.

30x + 60y = 120 x + 2y = 4

x-intercept = 4, y-intercept = 2

c = 4 2 2 1_ _

= 4

I5.4 If the difference of the two roots of the equation x2 – (c + 2)x + (c + 1) = 0 is d, find d.

x2 – 6x + 5 = 0  (x – 1)(x – 5) = 0

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Group Event 1

G1.1 In the given diagram, A + B + C + D + E = a, find a.

Reference: 1989 FI5.1, 2005 FI2.3

In APQ, B + D = AQP(1) (ext.  of ) C + E = APQ(2) (ext.  of )

A+B + C + D +E = A + AQP +APQ(by (1) and (2))

= 180 (s sum of ) a = 180

P Q

A

B

C D

E

G1.2 There are x terms in the algebraic expression x6 + x6 + x6 +  + x6 and its sum is xb_{. Find}_b_.

xx6 = xb x7₌_xb

b = 7

G1.3 If 1 + 3 + 32_{+ 3}3₊₊₃8₌

2 1 3c

, find c.

2 1 3 2

1 39  _ c

c = 9

G1.4 16 cards are marked from 1 to 16 and one is drawn at random. If the chance of it being a

perfect square number is

d

1

, find d.

Reference: 1995 HI4

Perfect square numbers are 1, 4, 9, 16.

Probability =

d

1 16

4 _

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Group Event 2

G2.1 If the sequence 1, 6 + 2a, 10 + 5a,  forms an A.P., find a.

6 + 2a = 2

5 10 1  a

12 + 4a = 11 + 5a a = 1

G2.2 If (0.002540)b₌

100 1

, find b.

100 1 40

400

1 _

   



 _ b

 ₂

10 1 10

1 _

b

b = 2

G2.3 If c is an integer and 33 3 1₃ 8 c c c

c , find c.

8 1 3 _

      

c c

 12 c c

c2 – 2c + 1 = 0 c = 1

G2.4 There are d different ways for arranging 5 girls in a row. Find d.

First position has 5 choices; 2nd_{position has 4 choices,}_{, the last position has 1 choice.}

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Group Event 3

G3.1 Let m be an integer satisfying the inequality 14x – 7(3x – 8) < 4(25 + x). Find the least value of m.

14x – 21x + 56 < 100 + 4x

–44 < 11x  –4 < x m = –3

G3.2 It is given that f

 

x x x x 3x 5x 7 4x 3

2 2 3

1 3 2  3 2   

 . If f(–2) = b, find b.

f (x) = x3 + x2 + x + 7

b = f(–2) = –8 + 4 – 2 + 7 = 1

G3.3 It is given that 0.5 2

logx and 0.1 5

logy  . If log xy = c, find c.

1 . 0 5 . 0 5 log 2

logx y  

log xy – 1 = 0.6 c = log xy = 1.6

G3.4 Three prime numbers d, e and f which are all less than 10, satisfy the two conditions d + e = f

and d < e. Find d.

Possible prime numbers are 2, 3, 5, 7. 2 + 3 = 5 or 2 + 5 = 7

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Reference: 2014 HI7

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Group Event 5

G5.1 If a is a positive multiple of 5, which gives remainder 1 when divided by 3, find the smallest

possible value of a. (Reference: 1998 FSG.1)

a = 5k = 3m + 1

52 = 33 + 1

The smallest possible a = 10.

G5.2 If x3 + 6x2 + 12x + 17 = (x + 2)3 + b, find b.

Reference: 1998 FG1.4

(x + 2)3 + b = x3 + 6x2 + 12x + 8 + b

b = 9

G5.3 If c is a 2 digit positive integer such that sum of its digits is 10 and product of its digit is 25,

find c. (Reference: 1998 FSG.3)

c = 10x + y, where 0 < x < 10, 0 y < 10.

x + y = 10

xy = 25

Solving these two equations gives x = y = 5; c = 55

G5.4 Let S1, S2,…, S10 be the first ten terms of an A.P., which consists of positive integers.

If S1 + S2 +…+ S10 = 55 and (S10 – S8) + (S9 – S7) +…+ (S3 – S1) = d, find d.

Reference: 1998 FSG.4

Let the general term be Sn = a + (n – 1)t







2 10 1



55 2

10 _ _ _

t a

 2a + 9t = 11

a, t are positive integers, a = 1, t = 1

d = (S10 – S8) + (S9 – S7) + … + (S3 – S1)

= [a + 9t – (a + 7t)] + [a + 8t – (a + 6t)] + …+ (a + 2t – a)

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Group Spare

GS.1 ABCD is a parallelogram and E is the midpoint of CD. If the ratio of the area of the triangle

ADE to the area of the parallelogram ABCD is 1 : a, find a.

1 : a = 1 : 4

a = 4

GS.2 ABCD is a parallelogram and E is the midpoint of CD.

AE and BD meet at M. If DM : MB = 1 : k, find k.

It is easy to show that ABM ~ EDM (equiangular)

DM : MB = DE : AB = 1 : 2

k = 2

1 1

M

E A

D C

B

GS.3 If the square root of 5 is approximately 2.236, the square root of 80 with the same precision is

d. Find d.

5 4 5 16

80   = 42.236 = 8.944

GS.4 A square is changed into a rectangle by increasing its length by 20% and decreasing its width

by 20%. If the ratio of the area of the rectangle to the area of the square is 1 : r, find r.

Let the side of the square be x

Ratio of areas = 1.2x0.8x : x2

= 0.96 : 1 = 1 : 24 25