A man can row a boat in still water at the rate of 6 km per hour. If the stream flows at the rate of 2 km/hour, he takes half the time going downstream than going upstream the same distance. His average speed for upstream and down stream trip is -

6 km/hour

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16/3 km/hour

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Insufficient data to arrive at the answer

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none of the above

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Solution

The correct option is B 16/3 km/hour
$L e t t h e d i s t a n c e t r a v e l l e d = d k m G i v e n t h a t t h e s p e e d o f t h e b o a t i n s t i l l w a t e r = 6 k m / h r a n d t h e s p e e d o f t h e w a t e r c u r r e n t = 2 k m / h r . ∴ T h e r e l a t i v e s p e e d d o w n s t r e a m = (6 + 2) k m / h r = 8 k m / h r . S o t h e t i m e t a k e n d o w n s t r e a m = \frac{d i s t a n c e}{r e l a t i v e s p e e d} = \frac{d}{8} h r . ∴ A g a i n t h e r e l a t i v e s p e e d u p s t r e a m = (6 - 2) k m / h r = 4 k m / h r . t h e t i m e t a k e n u p s t r e a m = \frac{d i s t a n c e}{r e l a t i v e s p e e d} = \frac{d}{4} h r . ∴ T h e t o t a l t i m e f o r g o i n g d o w n a n d c o m i n g u p = (\frac{d}{8} + \frac{d}{4}) = \frac{3 d}{8} h r A l s o t h e t o t a l d i s t a n c e f o r g o i n g d o w n a n d c o m i n g u p = d + d = 2 d . ∴ A v e r a g e s p e e d = \frac{t o t a l d i s t a n c e}{t o t a l t i m e t a k e n} = \frac{2 d}{\frac{3 d}{8}} = \frac{2 d \times 8}{3 d} k m / h r = \frac{16}{3} k m / h r . A n s - O p t i o n B .$

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