CE 2007 Math I Solution:
Section A1
1. 5 p − 7 = 3( p + q)
5 p − 7 = 3 p + 3q
2 p = 3q + 7
p=
2.
3q + 7
2
m6
m 6 n5 n5
=
= 3
m9 n −5
m9
m
3.
(a) r 2 + 10r + 25 = (r + 5) 2
(b) r 2 + 10r + 25 − s 2 = (r + 5) 2 − s 2
= (r + 5 − s)(r + 5 + s)
4.
Median = 67 kg
Range = 75 − 50 = 25 kg
S.D. = 7.65 kg
5.
Δ<0
142 − 4k < 0
196
k>
4
k > 49
6.
(a) Answer = 400 × 80% = $320
(b) Cost = 320 − 70 = 250
Percentage Profit =
7.
70
× 100% = 28%
250
Let x be the number of elderly patients consulted.
120 x + 160(67 − x) = 9000
120 x + 10720 − 160 x = 9000
40 x = 1720
x = 43
8.
x = 180 − 110 = 70
y = 110 − 90 = 20
Let ∠CBA = a
a = y = 20
∴ z = a = 20
1
9.
(a) Let ∠AOB = θ
2π ⋅ 40 ⋅
θ
360
= 16π
θ = 72
(b) Area = π (40) 2 ⋅
72
= 320π cm 2
360
Section A2
10. (a) The least possible length of the metal mire = 5 −
(b) (i)
Length of the wire ≤ 2.0 +
1
= 4.5 cm
2
0.1
= 2.05 m = 205 cm
2
∴ It is impossible that the actual length of this metal wire exceeds 206 m
1
= 4.5 cm
2
∵ 4.5 × 46 = 207 cm, ∴ the answer is impossible.
(ii) Maximum length of each shorter wire = 5 −
1
11. (a) Volume of the cone = π ⋅182 ⋅ 24 = 2592π cm 3
3
3
⎛ 8 ⎞
⎛1⎞
Volume of water = 2592π × ⎜ ⎟ = 2592π × ⎜ ⎟
⎝ 24 ⎠
⎝3⎠
(b) (i)
= 96π cm 3
Let the slant height of the cone be
3
,
= 182 + 242 = 30
Curved surface area of the cone = π ⋅18 ⋅ 30 = 540π cm 2
2
⎛1⎞
Wet area = 540π ⎜ ⎟ = 60π cm 2
⎝3⎠
(ii) Let the slant height of the bigger cone be s ,
s = 302 + 27 2 = 45
Curved surface area of the bigger cone = π ⋅ 27 ⋅ 45 = 1215π
1
Volume of the larger cone = π (27) 2 ⋅ 36 = 8748π
3
Let A be the required wet area
By similar solid,
1215π 3 8748π
=
= 4.5
A
96π
2
∴A=
12. (a)
1215π
= 60π cm 2
2
4.5
k
63
=
17 153
∴ k = 17 ×
63
=7
153
(b) Number of students = 17 ×
(c) Prob. =
360
= 40
153
4
1
=
40 10
(d) For the Bar Chart:
The "Number of students" need modification.
For the Pie Chart:
No modification is required
13. (a)
y −3
4
=−
3
x − 10
3 y − 9 = −4 x + 40
4 x + 3 y − 49 = 0
AB :
(b) 4 ⋅ 4 + 3h − 49 = 0
h = 11
(c) (i)
k + 10
=4
2
∴ k = −2
(ii) ΔABC =
1
(12)(11 − 3) = 48
2
AC = 62 + 82 = 10
10 ⋅ BD
= 48
2
∴ BD = 9.6
3
Section B
14. (a) (i)
f ( −3) = 0
4(−3)3 + k (−3) 2 − 243 = 0
−108 + 9k − 243 = 0
k = 39
(ii) f ( x) = 4 x 3 + 39 x 2 − 243
4
39
−3
0
−81
−12
−81
4
27
2
∴ f ( x) = ( x + 3)(4 x + 27 x − 81)
= ( x + 3)( x + 9)(4 x − 9)
(b) (i)
−243
243
0
C = k1 x 3 + k2 x 2 , where k1 and k2 are constants.
7381 = k1 (5.5)3 + k2 (5.5) 2
3
2
⎛ 11 ⎞
⎛ 11 ⎞
7381 = k1 ⎜ ⎟ + k2 ⎜ ⎟ , ∴ 488 = 11k1 + 2k2 .................(1)
⎝2⎠
⎝2⎠
9072 = k1 ⋅ 63 + k2 ⋅ 62 , ∴ 252 = 6k1 + k2 ...................(2)
2 × (2) − (1) :16 = k1
∴ k2 = 156
∴ C = 16 x 3 + 156 x 2
(ii) 972 = 16 x3 + 156 x 2
4 x 3 + 39 x 2 − 243 = 0
9
∴ x = cm (As x > 0 )
4
15. (a) (i)
Prob. =
48 3
=
80 5
(ii) Prob. =
12 3
=
80 20
(iii) Prob. =
48 + 4 52 13
=
=
80
80 20
(iv) Let L = Large size shirt , and B = Boy
3
P( L and B) 20 3 5 1
P( L | B) =
=
= × =
3 20 3 4
P( B)
5
4
(b) (i)
16 ⋅15
C216
3
Prob. = 80 = 2 =
80 ⋅ 79 79
C2
2
(ii) P(Same Size) =
C228 + C236 + C216 28 ⋅ 27 + 36 ⋅ 35 + 16 ⋅15
=
C280
80 ⋅ 79
2256 141
=
= 0.357
6320 395
∴ P(Different Sizes) = 1 − 0.357 = 0.643
=
∴ P(Different Sizes) > P(Same Sizes)
16. (a)
s=
5+6+9
= 10
2
ΔABC = 10(10 − 9)(10 − 5)(10 − 6) = 200 = 10 2
Volume of ABCDEF = 20 × 200 +
(b)
1
200 × 3 = 210 2 cm 3
3
DE = 62 + 32 = 45
FE = 9
DF = 32 + 52 = 34
Let ∠DEE = θ
34 + 81 − 45
cos θ =
= 0.6669389422
2 34 ⋅ 9
θ = 48.1688 = 48.2 (3 sig. fig.)
Let the shortest distance be L
L = DF sin θ = 34 sin 48.1688 = 4.344714
= 4.34 cm (3 sig. fig.)
(c) Area of the rectangle = 4 × 5 = 20
1
FE ⋅ L = 19.6 cm 2
2
As 20 > 19.6 , ∴ the answer is impossible.
Area of ΔFDE =
17. (a) (i)
∠ABG = ∠DBG ( I is the In-center)
BG is common
AB = BD (Given)
∴ΔABG ≅ ΔDBG (SAS)
(ii) ∠ABE = 90 ( ∠ in semi-circle)
∠AGB = 90 (Isos. Δ )
∴∠AGI = ∠ABE
5
Also ∠IAG = ∠BAE (In-center)
∴ΔAIG ~ ΔAEB (AAA)
∴
(b) (i)
GI BE
=
AG AB
O is the mid-point of AC
∴ A = (−25, 0)
⎛ 11 − 25 ⎞
, 0 ⎟ = (−7, 0)
Also G is the mid-point of AD , ∴ G = ⎜
⎝ 2
⎠
(ii)
BE 1
=
AD 2
By (a)(ii),
GI 1
=
AG 2
1
(18) = 9
2
∴ I = (−7,9)
∴ GI =
Radius of the required circle = GI = 9
∴ Equation of the inscribed circle:
( x + 7) 2 + ( y − 9) 2 = 92
6
CE 2007 Math II Key:
1.
11.
21.
31.
41.
51.
ADAAD
DDABC
DDBDC
DAACB
ACBCC
CCBD
6.
16.
26.
36.
46.
BBDAB
AABAD
BDBCC
CCDBA
DBAAC
7