Let the speed of boat in still water is ?x? km/hr & that of stream is ?y? km/hr.
Then, ATQ
(x+y)/y = 9/1 9y = x + y
x = 8y
y = 3 km/hr
So, x = 24 km/hr
Upstream speed = 24-3 = 21 km/hr
Hence, distance travelled upstream in 5 hours = 21*5 = 105 km.
Speed of streamer = 4.5 km/hrSpeed of water = 1.5 km/hr
Downstream speed = 4.5+1.5 = 6 km/hr
Upstream speed = 4.5 -1.5 = 3 km/hr
Average Speed = (6 X 3) / 4.5 = 4km/hr
Speed of boat in still water = 9 km/hr
Speed of current = 3km/hr
Downstream speed = 9+3 = 12 km/hr
Upstream speed = 9-3 =6 km/hr
Let the distance between A to B be x km.
x/6 + x/12 = 3x + 2x = 36
3x = 36
x = 12 km
Let distance between A & B = d km
Let speed in still water = x kmph
Let speed of current = y kmph
from the given data,
d/x = 2
From A) we get d
From B) we get d/x+y
From C) we get y
So, Any one pair of A and B, B and C or C and A is sufficient to give the answer i.e, the speed of upstream.
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)
If the speed downstream is a kmph and the speedup stream is b kmph then
Speed of the stream = 1/2 x (a?b) kmph
Speed downstream a = 12kmph
Speed upstream b = 8 kmph
Speed of the stream = 1/2 x (a?b) = 1/2 x (12?8)= 4/2 = 2 kmph
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