NIMCET 2015 Mathematics PYQ — A harbour lies in a direction South of West from a fort and at a … | Mathem Solvex | Mathem Solvex
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NIMCET 2015 — Mathematics PYQ
NIMCET | Mathematics | 2015
A harbour lies in a direction 60∘ South of West from a fort and at a distance 30 km from it. A ship sets out from the harbour at noon and sails due East at 10 km an hour. The time at which the ship will be 70 km from the fort is:
Choose the correct answer:
A.
7 PM
B.
8 PM
(Correct Answer)
C.
5 PM
D.
10 PM
Correct Answer:
8 PM
Explanation
Let the Fort be at the origin O(0,0).
The Harbour H is at 30 km in direction 60∘ South of West.
Coordinates of H:
xH=−30cos60∘=−30×21=−15
yH=−30sin60∘=−30×23=−153
The ship starts from H at noon and moves East at 10 km/h.
After t hours, the position of the ship S is:
xS=−15+10t
yS=−153
Distance of ship from fort (OS) is 70 km:
OS2=702
(−15+10t)2+(−153)2=4900
(10t−15)2+675=4900
(10t−15)2=4225
Taking square root:
10t−15=4225
10t−15=65
10t=80
t=8 hours
Since the ship started at noon (12:00 PM), 8 hours later the time will be 8:00 PM.
Correct Option: (b)
Explanation
Let the Fort be at the origin O(0,0).
The Harbour H is at 30 km in direction 60∘ South of West.
Coordinates of H:
xH=−30cos60∘=−30×21=−15
yH=−30sin60∘=−30×23=−153
The ship starts from H at noon and moves East at 10 km/h.
After t hours, the position of the ship S is:
xS=−15+10t
yS=−153
Distance of ship from fort (OS) is 70 km:
OS2=702
(−15+10t)2+(−153)2=4900
(10t−15)2+675=4900
(10t−15)2=4225
Taking square root:
10t−15=4225
10t−15=65
10t=80
t=8 hours
Since the ship started at noon (12:00 PM), 8 hours later the time will be 8:00 PM.
