Upstream speed = B-S
Downstream speed = B+s
B-S = 15/5 = 3 km/h
Again B= 4S
Therefore B-S = 3= 3S
=> S = 1 and B= 4 km/h
Therefore B+S = 5km/h
Therefore, Time during downstream = 15/5 = 3h
B.G. =(T.D.)^2/P.W.
= Rs. (160*160)/1600 = Rs. 16.
Sum =(B.D. x T.D.)/(B.D. - T.D.)
= Rs. (72 x 60) / (72 - 60)
= Rs. (72 x 60) /12
= Rs. 360.
F = Rs. 8100
R = 5%
T = 3 months = 1/4 years
Therefore BD - TD = 101.25-100 = Rs.1.25
Let the speed of the boat = p kmph
Let the speed of the river flow = q kmph
From the given data,
=> 56p - 56q -28p - 28q = 0
=> 28p = 84q
=> p = 3q.
Now, given that if
Hence, the speed of the boat = p kmph = 9 kmph and the speed of the river flow = q kmph = 3 kmph.
Speed in downstream = (14 + 4) km/hr = 18 km/hr;
Speed in upstream = (14 ? 4) km/hr = 10 km/hr.
Let the distance between A and B be x km. Then,
x/18 + (x/2)/10 = 19 ? x/18 + x/20 = 19 ? x = 180 km.
If t1 and t2 are the upstream and down stream times. Then time taken in still water is given by
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
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