Question

A long distance train is scheduled to reach its destination in $ 19 $ hours. After a $ 10 $ hours of journey, due to disruption of rail traffic, the train has to be stationed for $ 1 $ hour. If the average speed of the train is $ 100 $ \[\dfrac{{km}}{{hr}}\], at what speed should it travel to cover the distance in the same amount of time?

Answer 1

Hint: The train in question has completed $ 10 $ hours of its journey at the required speed the required average speed as given in the question is $ 100 $ \[\dfrac{{km}}{{hr}}\], The total journey was to be completed in $ 19 $ hours. Now due to a $ 1 $ hours delay the journey has to be completed in the same amount of hours. So the rest of the journey has to be completed within the $ 8 $ hours. We will find the total distance and then find the speed required to travel that distance in $ 8 $ hours.

Complete step-by-step answer:
The total distance the train has to cover at a average speed of is $ 100 $ \[\dfrac{{km}}{{hr}}\] in $ 19 $ hours is,
 $ D = S \times T $
\[D = 100 \times 19\]
 $ D = 1900km $
The train has to cover $ 1900 $ km
When the train broke down it had covered
 $ D = 100 \times 10 $
 $ D = 1000 $ km
So the train has to cover the distance of
 $ = 1900 - 1000 $
 $ = 900km $
This distance has to be covered in the remaining $ 8 $ hours, so the required speed of the train should be
 $ Speed = \dfrac{{{\text{Distance}}}}{{Time}} $
 $ Speed = \dfrac{{900}}{8} $
 $ Speed = 112.5\dfrac{{km}}{{hr}} $
Hence the train has to travel at a speed of $ 112.5\dfrac{{km}}{{hr}} $ to reach the destination on time.
So, the correct answer is “$ 112.5\dfrac{{km}}{{hr}} $”.

Note: The three formulas of speed, time and distance in the context of this chapter must be learnt by heart, the formula of speed and distance as used above are given below,
 $ Speed = \dfrac{{{\text{Distance}}}}{{Time}} $
 $ {\text{Distance = Speed}} \times {\text{Time}} $
The formula of time is given by,
 $ Time = \dfrac{{{\text{Distance}}}}{{Speed}} $
These formulas will let you be able to solve the majority of the questions of this topic.