Question: A fery carries passengers to Rock of Vivekananda and back from Kanyakumari. The distance of Rock of Vivekananda from Kanyakumari is 100 km. One day, the ferry started for Rock of Vivekananda with passengers on board, at a speed of 20 km per hour. After 90 minutes, the crew realized that there is a hole in the ferry and 15 gallons of sea water had already entered the ferry. Sea water is entering the ferry at the rate of 10 gallons per hour. It requires 60 gallons of water to sink the ferry. At what speed should the driver now drive the ferry so that it can reach the Rock of Vivekananda and return back to Kanyakumari just in time before the ferry sinks?
(Current of the sea water from Rock of Vivekananda to Kanyakumari is 2 km per hour.)
Explanation:
Since the current is from the Rock towards the Kanyakumari, the boat travels the first part against the stream and the second part with the stream.
Thus, in 90 minutes the boat has travelled 27 kms.
Since 15 gallons of water is already filled, there are 45 more gallons required to sink the ship which will take 4.5 hours at 10 gallons/hr.
Thus, the boat needs to cover 73 kms of first part and 100 kms of the second part within 4.5 hrs.
Let the speed of the boat be SB
Since the first part is upstream, the time required will be = 73/(SB − 2)
The time required for the second part will be = 100/(SB + 2)
Thus,
Time taken by the ship ≤ 4.5 i.e.
73 S B - 2 100 S B + 2
We now put the values of SB from the options and calculate the time taken by the ship.
Thus, the calculated times from options (1), (2), (3), (4) would be 4.36, 4.37, 4.455, 4.65 hrs respectively.
Please note that in the question, the ferry needs to complete the journey just in time and hence the answer would be the option closest to 4.5 hrs i.e. 4.455 hrs.
Hence, option (c).