
600 
Total cost  300  5 

300 

 300(5  2)
(iii)
MATHEMATICS
4016/02
 $2100
600
y  5
x
600
5.2  5 
x
600
0.2 
x
600
 3000 copies
x
0.2
1
Total amount  1299   24  40.30
3
 $1400.20
1400.20  1299
1027
 100%  7
%
1299
1299  7.79%
27 OCTOBER 2010
1 (iv)
(a) (i) MB  ( DB)2  ( DM )2
 72 2  432
 3335
 57.749
(ii)  57.7 m
AB  AM  MB
3
(a)
43
 57.749 
tan 62  80.613
(ii)
 80.6 m
(iii) 72
CD 
sin 23
 184.27  184 m
(b) ˆ  sin 1 43
MBD
72  36.671
Bearing of D from B  270  36.671
(b)
1299(1.06 3  1)  $248.13 (c)
759 
 233.329
 233.3
2 (a) x 50

8
x
4
(a)
4
( x  ( 5))
3
3 y  4 x  4(5)  3(4)
3 y  4 x  32
(b)
x  8  50  400 The equation of line AB is 3 y  4 x  32 . 2 x  9 y  68
2 x  68  9 y 4 x  136  18 y
x   400
100
 $660 115
y  y1  m( x  x1 )
y4 
2
 20
(b) t  p q

4
5
4q
tp
5
4q
t
p
5
4q  5 p
t 
5
(c) (i) 600
y  a
x
600
17  a 
50
 a  17  12  5
(ii) 600
y  5
x
600
 5
100
 $11
(i)
Substitute this equation into 3y  4 x  32 . 3 y  136  18 y  32
21y  168
y8
68  9(8)
 2
2
The coordinates of B is ( 2,8) . 
(i)
AE  6 2  12  37  6.08 E( 5  6, 4  1)  E(1, 5)
(ii)
(iii)     1   4   3 
DE  OE  OD           5  2  3 
    2   4   6 
DB  OB  OD           8  2  6 
(iv) 1. D, E and B are collinear since there is  1 
a common vector   . 1
2. E is exactly midway between B and 

D since DB  2 DE . x
(c)
5 (a) (i) (ii) ˆ  360  24 XCD
15
ˆ
CXD  180  2(24)  132 (d)
(i)
(ii)
(b) ˆ  24, triangle XCD is isosceles.
Since XDC
 XC  XD
Given BC  DE,
(c) XB  XC  BC  XD  DE  XE
ˆ  CDX
ˆ  24
BED
8
(a)
(i)
Let Y denote a point on FE extended.
ˆ  24 (exterior angle)
DEY
6 (a) (b) (ii)
(c) 42
x 1
2

(d) 2(2)
x  11.478 OR x  10.978
7 (e) (iii)
2
( 1)  ( 1)  4(2)( 252)
(b)
Since x is positive, let x  11.478 . 42
Time taken 
11.478
 3.659 hours
 3 h 39 min 33 s
(a) ˆ
170 2  952  102 2  2(95)(102) cos PQR
(b) 2
2
2 

ˆ  cos 1  170  95  102 
PQR
 2(95)(102) 


 119.255
 119.3
Angle of depression of Q from B
23
95
 13.6097
 13.6
1
(170)( RS) sin 52  5200
2
2(5200)
RS 
170 sin 52
 77.634
(c)  77.6 m
(i)
(ii)
1 2
r θ
2
1
 11.225  (8 2 )(3.5) 2
 11.225  112
Total area  11.225 
 123 m 2
Total volume
 Volume of pyramid
 Volume of cuboid
1
  10  10  12  30  10  10
3
 3400 cm 3
Let M denote the midpoint of AB. MN  10  2  5
VM  ( MN )2  (VN )2
 52  12 2
 13
1
 10  13
2
2
 65 cm
Total area  4(65)  4(30  10)
 1460 cm 2
See graph below (i)
m  5.15 (ii)
t  17
(iii) t  31
(i)
3.55  2.20
Gradient 
10  40
1.35

50
 0.027
Area of triangle VAB 
 tan 1
ˆ  2π  3 1 POQ
2
Area of triangle POQ
 11.2 m 2
2 x 2  x  252  0
x
1
2
1
ˆ
(8)(8) sin POQ
2
 11.225
6  42 x  42( x  1 )   x( x  1 ) 2 
2

x
252 x  252 x  126  x 2 
2
2 x 2  x  252  0
 26  1
 27
Total cost  26(28.50)  27(14.95)
 $1144.65
PRQ  44  2(8)  28
From s  r θ ,
28  8 θ

42 10

x 60
Number of posts required
θ 3
ˆ  180  24  24  132
BEF
42
Number of hours taken 
x
42
Number of hours taken 
x 1
2
77.634  3  25.878
 26 panels need to be bought
9
(a)
(b)
(c)
(ii) The gradient represents the rate of change of mass at the particular instant. When t  7 , the mass is decreasing at –0.027 kg/day instantaneously. (d) The relationship between m and t may be different for values of t beyond the given range. In this case, t  365  70 . a  28  4  7
10 (a) (i) b  60  (12  15  10  7  4  0  2  1)  9
c  12  0  0
d  3  9  27
e  0  15  20  27  28  20  0  14  8
 132
(ii) Σ fx 132
1
Mean 

2 Σf
60
5
Standard deviation 

Σ fx 2  Σ fx 
 

Σf
Σf 
510  1 
2 
60  5 
2
2
183
50
 1.91

(b) (c) Probability  0 Number of pupils who had read more than 4
books
 4 21  7
P(both had read more than 4 books)
7
6
7



60 59 590
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9 (a)
m
6
5
4
(–10, 3.55)
3
(40, 2.20)
2
0
10
20
30 40 50 60
70
t