and 11¼ minutes as 675 seconds.
Rate upstream = | ❨ | 750 | ❩m/sec | = | 10 | m/sec. |
675 | 9 |
Rate downstream = | ❨ | 750 | ❩m/sec | = | 5 | m/sec. |
450 | 3 |
∴ Rate in still water = | 1 | ❨ | 10 | + | 5 | ❩m/sec |
2 | 9 | 3 |
= | 25 | m/sec |
18 |
= | ❨ | 25 | x | 18 | ❩km/hr |
18 | 5 |
= 5 km/hr.
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
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 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
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 = 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.
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