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Dear Student,

L e t t h e d i s \tan c e b e t w e e n t h e 2 p o r t s b e d L e t t h e s p e e d o f t h e s t r e a m e r i n s t i l l w a t e r b e x T h e s p e e d o f t h e s t r e a m i s g i v e n t o b e 1 k m / h r . W h e n t h e s t r e a m e r m o v e s d o w n s t r e a m, i t i s m o v i n g w i t h t h e c u r r e n t, s o t h e e f f e c t i v e s p e e d i s x + 1 W h e n t h e s t r e a m e r m o v e s u p s t r e a m, i t i s m o v i n g a g a i n s t t h e c u r r e n t, s o t h e e f f e c t i v e s p e e d i s x - 1 N o w, T i m e = \frac{D i s \tan c e}{S p e e d} A c c o r d i n g t o t h e g i v e n c o n d i t i o n s, \frac{d}{x + 1} = 5, i . e d = 5 (x + 1) \frac{d}{x - 1} = 6, i . d d = 6 (x - 1) E q u a t i n g b o t h t h e v a l u e s o f d, w e g e t, 5 (x + 1) = 6 (x - 1) i . e 5 x + 5 = 6 x - 6 i . e x = 11 R e s u b s t i t u t i n g v a l u e o f x, w e g e t, d = 5 (x + 1) = 5 (11 + 1) = 5 \times 12 = 60 H e n c e d i s \tan c e b e t w e e n p o r t s = d = 60 k m S p e e d o f s t r e a m e r i n s t i l l w a t e r = x = 11 k m / h r

Regards.

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