Refresher - MATHEMATICS Quiz 6
PROBLEM 1:
A passenger ferry makes trips from a town to an island community that is 7 km downshore from the town
and 3 km. off a straight shoreline. The ferry travels along the shoreline to some point and then proceeds
directly to the island. If the ferry travels 12 kph along the shoreline and 10 kph as it moves out to sea.
Determine the routes that have a travel time of 45 minutes.
Solution:
D2 = (7 - x) 2 + (3) 2
D = (7 - x) 2 + 9
D 3 km
D = 49 - 14 x + x 2 + 9 Town
Shoreline
D = x 2 - 14 x = 58
x 7-x
Distance
Time = 7 km
Velocity
x D
T= +
12 10
3
T = 45 minutes = hrs
4
3 x x 2 - 14 x + 58
- =
4 12 10
Multiply by 60 to both sides of the equation
45 - 5x = x 2 - 14x + 58(6)
5(9 - x) = 6 x 2 - 14x + 58
square both sides
25(9 - x) 2 = 36(x 2 - 14x + 58)
20250 - 450x + 25 x 2 = 36x 2 - 504x + 2088
(x - 3)(11x - 21) = 0
x = 3 km
21
x= = 1.9 km
11
, Refresher - MATHEMATICS Quiz 6
PROBLEM 2:
From boat sailing due North of 16.5 kph, a wreck ship k and an observation tower
T are observed in a line due East. One hour later the wrecked ship and the tower
have bearing S. 34˚40’ E. and S.65˚10’ E. Find the distance between the wrecked
ship and the tower.
Solution: N
CT Boat
tan 65˚10' =
16.5 65 10’
o
A
CT = 35.65 km
16.5
30 20’
o
CK 30 20’
o
tan 34˚40' =
16.5
CK = 11.41 km
C
T
KT = 35.65 – 11.41 K
Tower
Wrecked Ship
KT = 24.24 km
PROBLEM 1:
A passenger ferry makes trips from a town to an island community that is 7 km downshore from the town
and 3 km. off a straight shoreline. The ferry travels along the shoreline to some point and then proceeds
directly to the island. If the ferry travels 12 kph along the shoreline and 10 kph as it moves out to sea.
Determine the routes that have a travel time of 45 minutes.
Solution:
D2 = (7 - x) 2 + (3) 2
D = (7 - x) 2 + 9
D 3 km
D = 49 - 14 x + x 2 + 9 Town
Shoreline
D = x 2 - 14 x = 58
x 7-x
Distance
Time = 7 km
Velocity
x D
T= +
12 10
3
T = 45 minutes = hrs
4
3 x x 2 - 14 x + 58
- =
4 12 10
Multiply by 60 to both sides of the equation
45 - 5x = x 2 - 14x + 58(6)
5(9 - x) = 6 x 2 - 14x + 58
square both sides
25(9 - x) 2 = 36(x 2 - 14x + 58)
20250 - 450x + 25 x 2 = 36x 2 - 504x + 2088
(x - 3)(11x - 21) = 0
x = 3 km
21
x= = 1.9 km
11
, Refresher - MATHEMATICS Quiz 6
PROBLEM 2:
From boat sailing due North of 16.5 kph, a wreck ship k and an observation tower
T are observed in a line due East. One hour later the wrecked ship and the tower
have bearing S. 34˚40’ E. and S.65˚10’ E. Find the distance between the wrecked
ship and the tower.
Solution: N
CT Boat
tan 65˚10' =
16.5 65 10’
o
A
CT = 35.65 km
16.5
30 20’
o
CK 30 20’
o
tan 34˚40' =
16.5
CK = 11.41 km
C
T
KT = 35.65 – 11.41 K
Tower
Wrecked Ship
KT = 24.24 km
