Welcome to aptitude tricks

## 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 * [(b

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 * [(b

^{2}– s^{2}) / 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 / (b

So, d = t * [(b

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 / (b

^{2}– s^{2}] = tSo, d = t * [(b

^{2}– s^{2}) / 2s]OR

d = [t * (Speed to go downstream) * (Speed to go upstream)] / [2 * Speed of still water]