Let the distance he covered each way = d kms
According to the question,
d/45 - d/50 = 1
=> d = 450 kms.
Hence, the total distance he covered in his way = d + d = 2 d = 2 x 450 = 900 kms.
Rate of her upstream = 12/2.5 = 4.8 km/hr
Then, ATQ
Rate of downstream = 4.8 x 3 = 14.4 km/hr
Hence, the distance she covers downstream in 5 hrs = 14.4 x 5 = 72 kms.
Let the place be at a distance of 'd' kms
From the given data,
5d/12 = 5/4 => d = 3 kms.
Hence, the place is 3 kms far.
Let the speed of the boat in still water is 'w'
Speed of the current is 'c'
Let the distance between two places is 'd'
According to the question, motorboat takes half time to cover a certain distance downstream than upstream.
=> 2w - 2c = w + c
=> w = 3c
=> c : w = 1 : 3
Hence, the ratio between rate of current(c) and rate of boat in still water(w) = 1 : 3
Speed of the current = 24-16/2
= 8/2
= 4 km/hr.
Let the speed of current = 'C' km/hr
Given the speed of boat in still water = 6 kmph
Let 'd' kms be the distance it covers.
According to the given data,
Boat takes thrice as much time in going the same distance against the current than going with the current
i.e,
Hence, the speed of the current C = 4 kmph.
Speed of boat in still water = 1/2 (12 + 6) = 9 kmph.
Given that, upstream distance = 7 kms
Upstream speed = 7/42 x 60 = 10 kms
Speed of the stream = 3 kmph
Let speed in still water = M kmph, then
Upstream speed = M - 3 = 10
=> M = 13 kmph.
let speed of boat= X, speed of stream= Y
Upstream speed= X-Y
Downstream speed= X+Y
Sum of upstream & downstream= (X-Y) +(X+Y)= 2X
So, 2X= 40
X= 20 km/hr
Speed of boat : speed of stream= 600+100 :100= 7:1
So speed of Stream= 20/7 km/hr
ATQ, D/( X-Y) + D/( X+Y) = 5
D/(120/7) + D/(160/7)= 5
D= 480×5/49= 48.97 km= 50 Km(approx)
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