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 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 = =5 kmph
Let the distance covered be D km.
10D = 42 x 5 = 210
=> D = 21 km
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
Speed of the stream = 1
Motor boat speed in still water be = x kmph
Down Stream = x + 1 kmph
Up Stream = x - 1 kmph
[35/(x + 1)] + [35/(x - 1)] = 12
x = 6 kmph
If t1 and t2 are the upstream and down stream times. Then time taken in still water is given by
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 = hrs
Time for downstream = 4 hrs
Distance in both the cases is same.
=> p x = q x 4
=> 44p/5 = 4q
=> q = 11p/5
Now, the required ratio of Speed of boat : Speed of water current
=
=> (11p/5 + p)/2 : (11p/5 - p)/2
=> 8 : 3
Let the speed of the stream = x kmph
From the given data,
=> x = 3 kmph
Therefore, the speed of the stream = 3 kmph
Let the distance be d.
=> 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
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