Question: 26 -
A steamer moves with a speed of 4.5 km/h in still water to a certain upstream point and comes back to the starting point in a river which flows at 1.5 km/h. The average speed of steamer for the total journey is
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10 km/h
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6 km/h
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12 km/h
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4 km/h
Answer:
4 km/h
Solution:
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 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
Question: 27 -
Equal distance is covered by a boat in upstream and in downstream in total 5 hours. Sum of speed of a boat in upstream and downstream is 40 km/hr. Speed of boat in still water is 600% more than the speed of stream. Find theapproximate distance covered by boat in downstream (in km).
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45
-
55
-
60
-
50
Answer:
50
Solution:
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)
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)
Question: 28 -
The ratio of the speed of a boat downstream and speed of the stream is 9:1. If the speed of the current is 3 km per hr, find the distance travelled by the boat upstream in 5 hours.
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100 km
-
98 km
-
105 km
-
109 km
Answer:
105 km
Solution:
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.
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.
Question: 29 -
A boatman can row 3 km against the stream in 20 minutes and return in 18 minutes. Find the rate of current ?
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1/3 kmph
-
2/3 kmph
-
1/2 kmph
-
1/4 kmph
Answer:
1/2 kmph
Solution:
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 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.
Question: 30 -
Find the speed of stream if a boat covers 36 km in downstream in 6 hours which is 3 hours less in covering the same distance in upstream?
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0.5 kmph
-
0.75 kmph
-
1.5 kmph
-
1 kmph
Answer:
1 kmph
Solution:
Speed of the boat upstream = 36/9 = 4 kmph
Speed of the boat in downstream = 36/6 = 6 kmph
Speed of stream = 6-4/2 = 1 kmph
Speed of the boat upstream = 36/9 = 4 kmph
Speed of the boat in downstream = 36/6 = 6 kmph
Speed of stream = 6-4/2 = 1 kmph