Speed of Boy is B = 4.5 kmph

Let the speed of the stream is S = x kmph

Then speed in Down Stream = 4.5 + x

speed in Up Stream = 4.5 - x

As the distance is same,

=> 4.5 + x = (4.5 - x)2

=> 4.5 + x = 9 -2x

3x = 4.5

x = 1.5 kmph

Let the distance covered be D km. 

  D 10 + 4 + D 10 - 4 = 5

  D 14 + D 6 = 5

 10D = 42 x 5 = 210

=> D = 21 km

Speed of the boat downstream  s=a/t= 60/3 = 20 kmph

Speed of the boat upstream s= d/t = 30/3= 10 kmph

Therefore,  The speed of the stream =  s p e e d o f d o w n s t r e a m - s p e e d o f u p s t r e a m 2=5 kmph

Speed in upstream = Distance / Time = 3 x 60/20 = 9 km/hr.

Speed in downstream = 3 x 60/18 = 10 km/hr

Rate of current = (10-9)/2 = 1/2 km/hr.

Speed in still water = 6 kmph

Stream speed = 1.2 kmph

Down stream  = 7.2 kmph

Up Stream = 4.8 kmph

x/7.2 + x/4.8 = 1

x = 2.88

Total Distance  = 2.88 x 2 = 5.76 kms

Let the speed of the boat upstream be p kmph and that of downstream be q kmph

Time for upstream = 8 hrs 48 min =  8 4 5hrs

Time for downstream = 4 hrs

Distance in both the cases is same.

=> p x   8 4 5= q x 4

=> 44p/5 = 4q

=> q = 11p/5

 Now, the required ratio of Speed of boat : Speed of water current

=  q + p 2 : q - p 2

=> (11p/5 + p)/2  : (11p/5 - p)/2

=> 8 : 3

Let the speed of the stream = x kmph

From the given data,

  12 9 + x + 12 9 - x = 3 h r s

3 x 2 = 27

=> x = 3 kmph

Therefore, the speed of the stream = 3 kmph

Let the distance be d.

  d 8 - 2 - d 8 + 2 = 4

=> 2d = 120 

=> d = 60 kms.

As the distance travelled is constant, the time taken is inversely proportional to speed.

Let 'u' be the speed of the current and 'v' be the speed of the boat.

Speed of the boat downstream = v + u & upstream = v - u

=> v+u/v-u = 30/20 => v/u = 5 => v = 5u

=> v + u = 6u = 10/20/60 miles/hr

=> 6u = 30 m/h

=> u = 5 m/h

Speed in still water = Average of Speed in Upstream and speed in Downstream

= 1/2 (12 + 6) kmph = 9 kmph.