HKCEE 2008 Math Paper I Suggested Solution:
Section A(1)
1. [3 marks]
(ab)3 a 3b3
= 2 = ab3
2
a
a
2.
[3 marks]
(a)
14 x
≥ 2x + 7
5
14 x ≥ 10 x + 35
4 x ≥ 35
x≥
35
4
(b) Ans = 9
3.
[3 marks]
(a) m + 2n = 5
m = 5 − 2n
When n = 1 , m = 3 .
When n = 2 , m = 1
(b) (2 x + m)( x + n) ≡ 2 x 2 + (m + 2n) x + mn
∴ m + 2n = 5 and k = m ⋅ n
By (a): k = 3 ⋅1 or 1 ⋅ 2
i.e. k = 3 or 2
4.
[3 marks]
⎛9⎞
The true bearing of Q from P = 180D + sin −1 ⎜ ⎟ ≈ 220D (3 sig. fig.)
⎝ 14 ⎠
5.
[3 marks]
1 1 1 1 1
The required probability = ⋅ + ⋅ =
3 2 3 2 3
6.
[4 marks]
(a)
2s + t 3
=
s + 2t 4
8s + 4t = 3s + 6t
2t = 5s
t=
5s
2
(b) s + t = 959
s+
5s
= 959
2
7 s = 959 × 2
s = 274 , t = 685
7.
[4 marks]
(a) Ans. = 9 × 4 + 17 × 3 + 5 × 2 = 97
(b) Since 97 < 100 , he has enough money to buy all things.
8.
[5 marks]
(a) Number of girls = 625 × (1 − 28%) = 450
(b) (i)
Ans. =
860
× 100% = 80%
450 + 625
(ii) 625 ⋅ 80% + 450 ⋅ x% = 860
∴x =
9.
86000 − 625 ⋅ 80
= 80
450
[5 marks]
x = 33D
y = 43D + 33D = 76D
2 y + z = 180D
∴ z = 180D − 76D = 28D
Section A(2)
10. (a) [4 marks]
a = 4 , k = 12 + 4 = 16
b = 37 − 16 = 21
A = 37 + 10 = 47
c = 50 − 47 = 3
(b) [3 marks]
Mean = 3.276 kg
Standard Deviation ≈ 0.299 (3 sig. fig.)
11. (a) [3 marks]
42 + b ⋅ 4 − 15 = 9
4b = 8
b=2
∴ f ( x) = x 2 + 2 x − 15
Solve f ( x) = 0
( x − 3)( x + 5) = 0
x = −5 or 3
∴ The x -intercepts of y = f ( x) are given by −5 and 3 .
(b) [4 marks]
f ( x) − k = 0
x 2 + 2 x − (15 + k ) = 0
As Δ > 0 , ∴ 4 + 4(15 + k ) > 0
k > −16
(c) [1 mark]
y = −16
12. (a) [2 marks]
B = (−3, 4) , C = (4, −3)
(b) [3 marks]
4
−3
, but slope of OC =
4
−3
hence O , B and C is not collinear.
Slope of OB =
(c) [4 marks]
4+3
= −1
−3 − 4
∴ Slope of CD = 1
Slope of BC =
Equation of CD :
y+3
=1
x−4
i.e. x − y − 7 = 0
Let D = (a,3) , where a is a real number
a −3−7 = 0
∴ a = 10
∴ D = (10,3)
13. (a) [4 marks]
Let the radius and the height of the cone X be r and h respectively.
2π r = 2 × 20π ×
216
360
∴ r = 12 cm
r 2 + h 2 = 202
∴ h = 400 − 122 = 16 cm
(b) [2 marks]
1
1
Volume of the cone X = π r 2 h = π ⋅ 122 ⋅ 16 = 768π cm 3
3
3
(c) [3 marks]
Let the radius and the height of the cone Y be r ' and h ' respectively.
2π r ' = 2π ⋅10 ⋅
108
360
∴r ' = 3
r '2 + h '2 = 102
h ' = 100 − 9 = 91
h' h
As
, ∴ X and Y are not similar.
≠
r' r
Section B
14. (a) [6 marks]
9
P( x < 5500 and "Good") P(4500 < x < 5500) 36 3
=
=
=
P ( x < 5500 | "Good") =
15 5
P("Good")
P("Good")
36
(i)
⎛ 8 ⎞ ⎛ 15 ⎞ 4
(ii) (1) Required Probability = 2 × ⎜ ⎟ ⎜ ⎟ =
⎝ 36 ⎠ ⎝ 35 ⎠ 21
⎡⎛ 8 ⎞ ⎛ 7 ⎞ ⎛ 15 ⎞ ⎛ 14 ⎞ ⎛ 13 ⎞ ⎛ 12 ⎞ ⎤
(2) Required Probability = 1 − ⎢⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ ⎥
⎣⎝ 36 ⎠ ⎝ 35 ⎠ ⎝ 36 ⎠ ⎝ 35 ⎠ ⎝ 36 ⎠ ⎝ 35 ⎠ ⎦
=
838
419
=
36 × 35 630
(b) [5 marks]
(i) Median = 5000
I.R. = 6400 − 4300 = 2100
(ii) Increase all the bonus by 5000 × 20% = 1000 .
15. (a) [2 marks]
Let BH = A
By Sine Formula,
50
A
=
D
sin15 sin130D
50sin130D
≈ 147.9884223 m
sin15D
≈ 148 m(3 sig. fig.)
A=
(b) [9 marks]
(i) Let ∠CBH = θ
cosθ =
2102 + A 2 − 1302
≈ 0.789968057
2 ⋅ A ⋅ 210
θ ≈ 37.81747348D ≈ 37.8D (3 sig. fig.)
(ii) Let N be the foot of perpendicular from H to BC
HN = A sin θ ≈ 90.73880496
HA = A sin 35D ≈ 84.88267191
HA
sin ∠HNA =
HN
∠HNA ≈ 69.30285559 ≈ 69.3D (3 sig. fig.)
(iii) HN is the shortest distance between the point H and any point on line BC , thus
the maximum angle of elevation of H from any point on BC is 69.3D .
Hence, it is impossible for Christine to find a point K on BC such that the angle of
elevation of H from K is 75D .
16. (a) [3 marks]
10 + 24 = 2b
∴ b = 17
a + b = 2 ⋅ 10
a=3
(b) [4 marks]
(i) Ans. = P × 20% = 0.2 P
(ii) In order to pay standard tax rate :
30000 × (3% + 10% + 17%) + ( P − 172000 − 90000) × 24% ≥ 0.2 P
9000 + 0.24 P − 62880 ≥ 0.2 P
0.04 P ≥ 53880
P ≥ 1347000
Hence P = 1347000
(c) [4 marks]
Tax = 1400000 × 20% = 280000
⎡⎛ 3% ⎞12 ⎛ 3% ⎞11 ⎛ 3% ⎞10
⎛ 3% ⎞ ⎤
Saving = 23000 × ⎢⎜ 1 +
⎟ + ⎜1 +
⎟ + ⎜1 +
⎟ + ⋅ ⋅ ⋅ + ⎜1 +
⎟⎥
12 ⎠
12 ⎠
12 ⎠
12 ⎠ ⎦⎥
⎝
⎝
⎝
⎣⎢⎝
12
⎛ 3% ⎞ ⎡1.0025 − 1 ⎤
= 23000 × ⎜1 +
≈ 280526.3706
⎟
12 ⎠ ⎢⎣ 1.0025 − 1 ⎥⎦
⎝
∴ Peter have enough money to pay the tax.
17. (a) [3 marks]
Let ∠BAP = a , we have ∠PAC = a
∴ BP = PC
( I is the in-centre)
(eq. ∠ eq. chord)
Let ∠ACI = b , we have ∠ICB = b
( I is the in-centre)
∠BCP = a
∴∠ICP = a + b
( ∠ in the same segment)
∠PIC = a + b
∴∠ICP = ∠PIC
(ext. ∠ of Δ )
∴ IP = PC
(base ∠ isos. Δ )
Hence BP = CP = IP .
(b) [8 marks]
(i) Let P = (a, b)
By (a):
(a + 80) 2 + b 2 = a 2 + (b − 32) 2 = (a − 64) 2 + b 2
(a − 64) 2 + b 2 = (a + 80) 2 + b 2
a 2 − 128a + 4096 = a 2 + 160a + 6400
a = −8
Also, a 2 + b 2 − 64b + 1024 = a 2 + 160a + 6400 + b 2
−64b + 1024 = 160a + 6400
∴ b = −64
PB = (80 − 8) 2 + 642 = 9280
Hence equation of the circle is given by: ( x + 8) 2 + ( y + 64) 2 = 9280
(ii) As PGQ is the perpendicular bisector of BC , we may let Q = (−8, b)
As PQ is a diameter of the circle A, B, P, C , we have QB ⊥ BP
∴ (slope of QB) × (slope of BP) = −1
⎛ q ⎞ ⎛ −64 ⎞
⎜
⎟×⎜
⎟ = −1
⎝ 80 − 8 ⎠ ⎝ 80 − 8 ⎠
722
= 81
64
∴ Q = (−8,81)
q=
(iii) Slope of BQ =
Slope of QI =
81
81
=
80 − 8 72
81 − 32 −49
=
−8
8
⎛ 81 ⎞⎛ −49 ⎞
As ⎜ ⎟⎜
⎟ ≠ −1 ,
⎝ 72 ⎠⎝ 8 ⎠
∴∠BQI ≠ 90D
∴ B , Q , I and R are not concyclic.