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Question: A steamer going downstream in a river, covers the distance between 2 towns in 15 hours. Coming back ...

A steamer going downstream in a river, covers the distance between 2 towns in 15 hours. Coming back upstream, it covers this distance in 20 hours. The speed of water is 3 kmhr\dfrac{km}{hr}. Find the distance between two towns.
(a) 360km
(b) 220km
(c) 180km
(d) 120km

Explanation

Solution

Hint: Downstream means stream and streamer are moving in the same direction and upstream means stream and streamer are moving in the opposite direction. So, suppose the speed of the streamer in still water as a variable and hence find the speed of streamer in downstream and upstream with the help of relation Speed = DistanceTime\text{Speed}\ \text{=}\ \dfrac{\text{Distance}}{\text{Time}}. Now form two equations in terms of distance and speed of streamer and hence solve it further.

Complete step-by-step answer:
Let the distance between both the towns is ‘d’km and the speed of streamers in still water is ‘u’ kmhr\dfrac{km}{hr}.
Now we know that streamer and flow of water (streamer) are in the opposite direction and downstream means that the direction of stream (water flow) and streamer are the same.

It means speed of streamer in downstream direction is the sum of speed of stream and streamer in still water and speed of streamer in upstream direction is difference of speed of streamer in still direction and speed of stream.

Where u is the speed of streamer in still water and v is the speed of stream.
Hence, the speed of stream in downstream direction is u+vu+v.
Speed of stream in upstream direction is uvu-v.

Now, coming to the question we have the speed of water as 3kmhr\dfrac{km}{hr}. And time taken by streamers downstream are 15 hours and 20 hours respectively.
So, we know the relation among speed, distance and time as
Speed = DistanceTime\text{Speed}\ \text{=}\ \dfrac{\text{Distance}}{\text{Time}} ………………………………………………………………………………………………(i)
So, for upstream, we have
Speed of streamer = u+v=\ u+v
Time taken to cover distance between towns = 15 hours=\ 15\ \text{hours}
Distance between towns = d km=\ d\ \text{km}
So, we get
u+v = d15u+v\ =\ \dfrac{d}{15}
Put v = 3 kmhrv\ =\ 3\ \dfrac{km}{hr} as the speed of water is 3 kmhr\dfrac{km}{hr}. from the problem. Hence, we get
u+3 = d15u+3\ =\ \dfrac{d}{15}
u = d153\Rightarrow u\ =\ \dfrac{d}{15}-3 ……………………………………………………………………………………….(ii)
And, for upstream, we have
Speed of streamer = uv=\ u-v
Time taken to cover the distance ‘d’ = 20 hours=\ 20\ \text{hours}
Distance between both towns = d km=\ d\ \text{km}
So, we get from the equation (i) as
uv = d20u-v\ =\ \dfrac{d}{20}
Now, put v = 3 kmhrv\ =\ 3\ \dfrac{km}{hr} to the above equation so, we get
u 3 = d20u-\ 3\ =\ \dfrac{d}{20}
u = d20+3\Rightarrow u\ =\ \dfrac{d}{20}+3 …………………………………………………………………………………(iii)

Now, we can equate the value of ‘u’ from the equation (ii) and (iii) as both are representing the same value. Hence, we get
d153 = d20+3\dfrac{d}{15}-3\ =\ \dfrac{d}{20}+3
d15d20 = 3+3\dfrac{d}{15}-\dfrac{d}{20}\ =\ 3+3
Take L.C.M. of denominators of fractions of L.H.S. to solve the expression; so, we get
4d3d60 = 6\dfrac{4d-3d}{60}\ =\ 6
d60 = 6\dfrac{d}{60}\ =\ 6
d = 360 kmd\ =\ 360\ \text{km}
Hence, the distance between the two towns is 360 km.
So, Answer (a) is correct.

Note: Terms ‘downstream’ and ‘upstream’ should be clear to solve these types of questions. Don’t get confused with the word stream, it is used for the speed of water flow.
One may take the velocities of streamers as vuv-u and v+uv+u as well, and if one finds the ‘u’ as negative value, don’t get confused. Negative sign represents the direction only.
Don’t get confused with the relation of Speed = DistanceTime\text{Speed}\ \text{=}\ \dfrac{\text{Distance}}{\text{Time}} . Students get confuse with these relation as well and may apply formula as Time = SpeedTime\text{Time}\ \text{=}\ \dfrac{\text{Speed}}{\text{Time}} or Time = Distance×Speed\text{Time}\ \text{=}\ \text{Distance}\times \text{Speed} which are wrong. So, be clear with the relation speed distance and time.