Download 8 hours = 8 hrs × 60 mins × 60 secs = 28800 seconds 1) Radio

1)
Radio signals travel at a speed of 3 × 108 metres per second.
A radio signal from Earth to a space probe takes 8 hours.
What is the distance from Earth to the probe?
Give your answer in scientific notation.
Soln
d
s
d
t
8 hours
= 8 hrs × 60 mins × 60 secs
= 28800 seconds
= s × t
= ( 3 × 108 ) metres/sec × ( 8 × 60 × 60 ) secs
= 3 × 108 × 8 × 60 × 60
= 3 × 108 × 28800
= 3 × 108 × 2.88 × 104
= 3 × 2.88 × 108 × 104
= 8.64 × 10(8 +4)
= 8.64 × 1012 metres
Change to km
÷ by 1000 = 103
= 8.64 × 1012
103
= 8.64 × 10(12 − 3)
= 8.64 × 109 kilometres
2)
A tank which holds 100 litres of water has a leak.
After 150 minutes, there is no water left in the tank.
v
100
litres
0
minutes
t
150
The above graph represents the volume of water ( v litres ) against
time ( t minutes ).
(a) Find the equation of the line in terms of v and t.
(b) How many minutes does it take for the container to lose 30 litres of
water?
Soln
(a) The equation of a line is y = mx + c
gradient
m
(0, 100)
(150, 0)
÷ by 50
( top and bottom)
y = mx + c
v = mt + c
=
y2 − y1
x2 − x1
=
0 − 100
150 − 0
=
− 100
150
=
−2
3
y intercept (0, 100)
c = 100
( from graph above y = v and x = t )
v = mt + c
v = −2 t + 100
3
(b)
÷ 100
× 30
litres
minutes
100
150
1
1.5
30
45
÷ 100
× 30
It takes 45 minutes for the container to lose 30 litres of water
3)
Bottles of juice should contain 50 millilitres.
The contents of 7 bottles are checked in a random sample.
The actual volumes in millilitres are as shown below.
52,
50,
51,
49,
52,
53,
50
Calculate the mean and standard deviation of the sample.
Soln
Mean
Σx
n
Mean
=
52 + 50 + 51 + 49 + 52 + 53 + 50
7
=
357
7
=
51
Standard Deviation
√
x2 − ( x )2 / n
(n−1)
x2 = 522 + 502 + 512 + 492 + 522 + 532 + 502
x2 = 18219
n = 7
(n−1) = 6
=
√
√
=
√2
=
18219 − ( 127449 / 7 )
6
12
6
Standard Deviation
=
1.41
(2 d.p.)
x = 357
( x )2 = (357)2
( x )2 = 127449
4)
250 milligrams of a drug are given to a patient at 12 noon.
The amount of the drug in the bloodstream decreases by 20% every hour.
How many milligrams of the drug are in the bloodstream at 3pm.
Soln
12 noon
250
losing 20% each hour
1pm
250 × 0.8
2pm
250 × 0.8 × 0.8
left with 80%
of the previous hour
2
= 250 × (0.8)
3pm
250 × 0.8 × 0.8 × 0.8
= 250 × (0.8)3
= 128 milligrams
80% = 80 = 0.8
100
5)
A helicopter, at point H, hovers between two aircraft carriers at points A
and B which are 1500 metres apart.
H
500
A
550
1500 m
B
From carrier A, the angle of elevation of the helicopter is 500.
From carrier B, the angle of elevation of the helicopter is 550.
Calculate the distance from the helicopter to the nearest carrier.
Soln
H
Shortest side is opposite the smallest angle
750
xm
A
500
550
1500 m
B
Sine Rule
x
sin 50
=
1500
sin 75
( x )( sin 75 ) = ( 1500 )( sin 50 )
( ÷ sin 75 both sides )
x = ( 1500 )( sin 50 )
( sin 75 )
x = 1189.6 m
( 1 d.p. )
The distance to the nearest carrier is 1189.6 metres.
6)
The diagram below shows a spotlight at point S, mounted 10 metres
directly above a point P at the front edge of a stage.
The spotlight swings 450 from the vertical to illuminate another point Q,
also at the front point of the stage.
S
450
10 m
P
Q
B
Through how many more degrees must the spotlight swing to illuminate
a point B, where Q is the mid-point of PB?
Soln
Diag. 1
S
(a)
0
10 m
P
y0
(a)
10 m
45
x
(o)
Q
Diag. 2
S
P
20 m ( 2x )
(o)
tan 450 = o = x
a
10
0
tan 45 = x
10
( × 10 both sides )
tan y0 = o = 20
a
10
0
tan y = 20
10
0
tan y = 2
( 10 )( tan 450 ) = x
x = 10
y0 = 63.40
Spotlight must swing through 63.40 − 450 = 18.40 more
to illuminate a point at B.
B
7)
A square trapdoor of side 80 centimetres is held open by a rod as shown.
Trapdoor
Rod
80 cm
760
40 cm
40 cm
The rod is attached to a corner of the trapdoor and placed 40 centimetres
along the edge of the opening.
The angle between the trapdoor and the opening is 760.
Calculate the length of the rod to 2 significant figures.
Soln
Rod
x cm
80 cm
760
40 cm
Cosine Rule
x2 = 802 + 402 − (2)(80)(40)(cos 760)
x2 = 8000 −(6400)(cos 760)
x2 = 6451.699……
x = √ 6451.699……
x = 80.322……
x = 80
( 2 s.f. )
8)
The curved part of a doorway is an artic circle with radius
500 millimetres and centre C.
The height of the doorway to the top of the arc is 2000 millimetres.
The vertical edge of the doorway is 1800 millimetres.
C
500 mm
1800 mm
2000 mm
Calculate the width of the doorway.
Soln
200 mm
x mm
300 mm
C
1800 mm
x mm (a)
300 mm
(b)
500 mm
(h)
500 mm
Pythagoras
2000 mm
a2
x2
x2
x2
x
x
=
=
=
=
=
=
h2 − b2
5002 − 3002
250000 − 90000
160000
√ 160000
400 mm
The width of the doorway is 2x mm = (2)(400) mm = 800 mm
9)
A gift box, 8 centimetres high, is prism shaped.
A
B
O
8 cm
10 cm
The uniform cross-section is a regular pentagon.
Each vertex of the pentagon is 10 centimetres from the centre O.
Calculate the volume of the box.
Soln
A
A
10 cm
B
0
72
O
O
10 cm
Area of
3600 = 720
5
720
B
10 cm
AOB
= 1 × 10 × 10 × sin 720
2
= 47.55 cm2 ( 2 d.p. )
Area of Pentagon = 5 × 47.55
= 237.75 cm2
Volume of the box
= Area of Pentagon × Height of Box
= 237.75 cm2 × 8 cm
= 1902 cm3
10)
Solve algebraically the equation
4 sin xo + 1 = −2
0 ≤ x ≤ 360.
Soln
4 sin xo + 1 = −2
( − 1 both sides )
4 sin xo = −3
( ÷ 4 both sides )
sin xo = −3
4
xo = (180 + 48.6)0 or xo = (360 − 48.6)0
xo = 228.60
or
xo = 311.40
180
0
S
A
T
C
sin yo = 3
4
o
y = 48.60
0/3600
11) A rectangular lawn has a path, 1 metre wide, on 3 sides as shown.
xm
1m
The breadth of the lawn is x metres.
The length of the lawn is three times its breadth.
The area of the lawn equals the area of the path.
(a)
Show that 3x2 − 5x − 2 = 0.
(b)
Hence find the length of the lawn.
Soln
3x m
(a)
1m
(x + 1) m
(x + 1) m
xm
1m
Area of Lawn
3x m
=
=
(3x)(x)
3x2
Area of lawn = Area of path
3x2
= 5x + 2
3x2 − 5x − 2 = 0
1m
Area of Path
=
=
=
=
2(1)(x + 1) + (3x)(1)
2(x + 1) + 3x
2x + 2 + 3x
5x + 2
(b)
3x2 − 5x − 2 = 0
(3x + 1)(x − 2) = 0
(3x + 1) = 0 or
(x − 2) = 0
3x = −1
x=2
or
x = −1
3
( NOT possible
x is a length )
The length of the garden is 3x = (3)(2)
= 6 metres