Latest Wipro Aptitude Question SOLUTION: A ship develops a leak 12 km from the shore. Despite the leak, the ship is able to move towards the shore at a speed of 8 km/hr. However, the ship can stay afloat only for 20 minut

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A ship develops a leak 12 km from the shore. Despite the leak, the ship is able to move towards the shore at a speed of 8 km/hr. However, the ship can stay afloat only for 20 minutes. If a rescue vessel were to leave from the shore towards the ship, and it takes 4 minutes to evacuate the crew and passengers of the ship, what should be the minimum speed of the rescue vessel in order to be able to successfully rescue the people aboard the ship?

(1) 53 km/hr
(2) 37 km/hr
(3) 28 km/hr
(4) 44 km/hr

Asked In Wipro (12 years ago)

Unsolved Read Solution (4)

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Read Solution (4) :

abhishek khowala  (12 years ago)

total distance 12km from the shore
4 miutes needed for the rescue
time left to sail the leaky ship=> (20-4)=16 mints
speed=8km/h
distance covered with this speed in 16/60 hr =(8*16)/60 =32/15 hr
distance left to sail the rescue ship=12-32/15=148/15
so the speed required= distance/time=(148/15)/(16/60)
=148/4
=37 km/h

-Submitted by Banke

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sravani  (9 years ago)

12km ------------- 20mins

12/20 * 60=36 km/hr + 8km/hr =44km/hr.........
4 mins for evacuvation...
8 ------------ 1hr
small value.. ------------ 4 min
so 44 km/hr.....

Is this solution Helpfull? Yes (2) | No   

Eswarlal B Krishnamoorthy  (9 years ago)

Relative speed = Distance/Time
S-8 = 12/(16/60) = 12*60/16 = 45
S = 45-8 = 37 kmph

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jothieswaran p  (8 years ago)

v1+v2 = (d/t)
v1+v2(+ sign since both sinking and rescue ship travel in opp direction) i.e. relative speed concept
now v1=8
v2=?
d=12km
t= 20-4=16 min.
t=(16/60)hrs
8+v2=(12*60)/(16)=45
v2=37km/hr

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