The ratio of the speed of a motorboat to that of the current of water is 31 ∶ 6. The motorboat starts from a point and covers a certain distance along the current in 4 h 10 min. Find the time taken by the motorboat to come back to its initial point.

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SSC CGL 2023 Tier-I Official Paper (Held On: 20 Jul 2023 Shift 4)
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  1. 4 h 10 min
  2. 5 h 10 min
  3. 5 h 50 min
  4. 6 h 10 min

Answer (Detailed Solution Below)

Option 4 : 6 h 10 min
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Detailed Solution

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Given

Total time = 4 hr 10 min.

The ratio of the speed of a motorboat to that of the current of water is 31 ∶ 6.

Formula used

Speed = Distance/Time

Detailed solution:-

Let the common factor of speeds of a motorboat to that of the current of water be x.

Then, the speed of the motorboat = 31x km/h, 

The speed of the current = 6x km/h.

Motorboat goes along with the current in 4 h 10 m = 4 h +10/ 60 h = 25/6 h 

Speed of downstream = 31x + 6x = 37x km/h

Speed of upstream = 31x – 6x = 25x km/h

Now, the distance covered by the downstream

⇒ speed of downstream × Time taken = 37x × (25 /6) km

Similarly, the distance covered upstream is also 37x × (25 /6)km
 
Time is taken by upstream = Distance covered by upstream/ Speed of upstream

⇒ [37x × (25 /6) ]/25x

⇒ 37/ 6 = 6h + 1/ 6 = 6h 10 min

∴ The time the motorboat requires to return is equal to 6 hours 10 minutes. 

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