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Basic Concepts of Questions on Boats and Streams
- A boat is said to go downstream, if the boat goes in the direction of stream.
- A boat is said to go upstream, if the boat goes opposite to the direction of stream.
Basic Formulas
- If speed of boat in still water is b km/hr and speed of stream is s km/hr,
- Speed of boat in downstream = (b + s) km/hr , since the boat goes with the stream of water.
- Speed of boat in upstream = (b - s) km/hr. The boat goes against the stream of water and hence its speed gets reduced.
Shortcuts With Explanation
Scenario 1: Given a boat travels downstream with speed d km/hr and it travels with speed ukm/hr upstream. Find the speed of stream and speed of boat in still water.
Let speed of boat in still water be bkm/hr and speed of stream be skm/hr.
Then b + s = d and b – s = u.
Solving the 2 equations we get,
b = (d + u)/2
s = (d – u)/2
Then b + s = d and b – s = u.
Solving the 2 equations we get,
b = (d + u)/2
s = (d – u)/2
Scenario 2: A man can row a boat, certain distance downstream in td hours and returns the same distance upstream in tu hours. If the speed of stream is s km/h, then the speed of boat in still water is given by
We know distance = speed * time
Let the speed of boat be b km/hr
Case downstream:
d = (b + s) * td
Case upstream:
d = (b - s) * tu
=> (b + s) / (b - s) = tu / td
b = [(tu + td) / (tu - td)] * s
Let the speed of boat be b km/hr
Case downstream:
d = (b + s) * td
Case upstream:
d = (b - s) * tu
=> (b + s) / (b - s) = tu / td
b = [(tu + td) / (tu - td)] * s
Scenario 3: A man can row in still water at bkm/h. In a stream flowing at s km/h, if it takes him t hours to row to a place and come back, then the distance between two places, d is given by
Downstream: Let the time taken to go downstream be td
d = (b + s) * td
Upstream: Let the time taken to go upstream be tu
d = (b - s) * tu
td + tu = t
[d / (b + s)] + [d / (b - s)] = t
So, d = t * [(b2 - s2) / 2b]
OR
d = [t * (Speed to go downstream) * (Speed to go upstream)]/[2 * Speed of boat or man in still water]
d = (b + s) * td
Upstream: Let the time taken to go upstream be tu
d = (b - s) * tu
td + tu = t
[d / (b + s)] + [d / (b - s)] = t
So, d = t * [(b2 - s2) / 2b]
OR
d = [t * (Speed to go downstream) * (Speed to go upstream)]/[2 * Speed of boat or man in still water]
Scenario 4: A man can row in still water at bkm/h. In a stream flowing at s km/h, if it takes t hours more in upstream than to go downstream for the same distance, then the distance d is given by
Time taken to go upstream = t + Time taken to go downstream
(d / (b - s)) = t + (d / (b + s))
=> d [ 2s / (b2 - s2 ] = t
So, d = t * [(b2 - s2) / 2s]
OR
d = [t * (Speed to go downstream) * (Speed to go upstream)] / [2 * Speed of still water]
(d / (b - s)) = t + (d / (b + s))
=> d [ 2s / (b2 - s2 ] = t
So, d = t * [(b2 - s2) / 2s]
OR
d = [t * (Speed to go downstream) * (Speed to go upstream)] / [2 * Speed of still water]
