A boatman rows to a place 48 km distant and back in 14 hours. He finds

Boats and streams use the concept of relative speed, where we have one moving object on a surface which is in motion.

In this case, Speeds are added when moving in the same direction, and subtracted when moving in the opposite direction.



In the question above, we are given that for every 4 kms traveled downstream, he moves 3 kms upstream i.e the ratio of speeds \(S_d : S_u = 4 : 3\).


Let \(S_d = 4x\) and \(S_u = 3x\)

\(Total \space Time = \frac{d_u}{S_u} + \frac{d_d}{S_d}\)

\(14 = \frac{48}{4x} + \frac{48}{3x} = \frac{12}{x} + \frac{16}{x} = \frac{28}{x}\)

Therefore x = 2.


Substituting for x, \(S_d = 8\) and \(S_u = 6\)


\(S_s = \frac{S_d \space - \space S_u}{2}= \frac{8 - 6}{2} = 1\)



Option A



It is good to remember to derive the equations when doing questions on Boats and Streams

There are 4 equations based on this.


If \(S_b\) is the Speed of Boat in still water and \(S_s\) is the speed of the stream, then


Speed Downstream, \(S_d = S_b \space + \space S_s\) .... (1)

Speed Upstream, \(S_d = S_b \space - \space S_s\) ..... (2)


Adding Equations (1) and (2), we get \(S_b = \frac{S_d \space + \space S_u}{2}\) ... (3)


Subtracting the Equations, we get \(S_s = \frac{S_d \space - \space S_u}{2}\) ... (4)


Arun Kumar

Let R = Speed of boat in still water

Let W = Speed of Stream

Given, that when time is Constant, he can travel 4 km in the same time he can travel 3 km

Rule: given constant time, the distance traveled is directly proportional to the speed traveled at


Ratio of——-speed with stream : speed against stream = 4 / 3

Now, the person is traveling the Same Distance of 48 km with the stream and 48 km back against the Stream in a round trip. Thus, Distance is Constant across these 2 parts of the Round Trip.

Rule: when Distance is Constant, the SPEED traveled at is INVERSELY Proportional to the TIME taken to travel over that Distance.

Ratios are also Inversely Proportional.


Speed With Stream : Speed Against Stream = 4 : 3

Time With Stream : Time Against Stream = 1/4 : 1/3 = 3 : 4

This means 3/7 of the Total travel Time of 14 hours is spent traveling WITH the stream—- (3/7) * 14 = 6 hours

And 4/7 of the Total Time of 14 hours is spent traveling AGAINST the stream —— (4/7) * 14 = 8 hours


Rule: Speed = (Distance traveled) / (travel Time taken)


Speed With Stream = R + W = 48 km / 6 hr = 8 km/hr

Speed Against Stream = R - W = 48 km / 8 hr = 6 km/hr


Rule: the following Speeds are in an Arithmetic Progression with the common difference = W = Speed of Stream


Speed WITH Stream = R + W = 8

Speed in Still Water = R = ?

Speed AGAINST Stream = R - W = 6

In any A.P., any 2 consecutive terms will have a Common Difference = d —— in this A.P. involving the Speeds, the Speed in Still Water = R = will equal:

-(W) from the Speed WITH the Stream —- and —- +(W) above the Speed AGAINST the Stream:

8 - W = R = W + 6

8 - W = W + 6

2 = 2*W

W = 1 km/hr = Speed of Stream

-Answer A-

Posted from my mobile device

is this correct?
d=96 kms
t=14 h
sboat = 96/14 = 7km/h

4/(speed boat+speed river) = 3(speed boat-speed river)
speed boat = 7 speed river
speed river = speed of the boat/7
speed river = 7/7= 1km/h

is this correct:

4/(b+s)= 3/(b-s)
=> b=7s

48/7 = 7s (speed of boat calculated using 96/14)
s = 48/49 ~1 km/hr
Bunuel Krunaal

hereiam2410 and miag

Your approach is incorrect since you consider 96/14 as the speed of boat, this nulls the effect of speed of stream. The boat travels 48kms upstream and downstream at different speeds, including the speed of stream.

Thus, \(\frac{48}{b+s}\) + \(\frac{48}{b-s}\) = 14 ................(i)

We got b = 7s, subs. in (i)

\(\frac{48}{8s}\) + \(\frac{48}{6s}\) = 14

Solving we get s = 1