## 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

**u**km/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

d = (b + s) * td

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

**b**km/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

**b**km/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]

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