Question & Answer
QUESTION

A train which travels at a uniform speed due to some mechanical fault after travelling for an hour goes at $\dfrac{3}{4}th$ of the original speed and reaches the destination 2 hours late. If the fault had occurred after travelling another 50 miles, the train would have reached 40 minutes earlier. What is the distance between the two stations?

ANSWER Verified Verified
Hint: According to this question, first let the uniform speed be x. Now calculate the decrease in speed due to the mechanical fault. By using the next condition, evaluate the original speed and decreased speed. Also evaluate the total time taken to complete the journey. At last, multiply time with speed to get the distance between two stations.

Complete step-by-step answer:

Let the uniform speed of the train be x and the decreased speed due to mechanical fault be $\dfrac{3}{4}x$ or 0.75x.
As we know that speed is the ratio of distance and time. Now, for traveling 50 miles the time taken would be: $time=\dfrac{dist.}{speed}$ where dist. is distance covered.
$\therefore \dfrac{50}{x}\text{ and }\dfrac{50}{0.75x}$.
Equating this time with 40 minutes we get,
$\begin{align}
  & \dfrac{50}{0.75x}-\dfrac{50}{x}=\dfrac{40}{60} \\
 & \dfrac{50}{3x}=\dfrac{2}{3} \\
 & x=25m{{s}^{-1}} \\
 & \dfrac{3}{4}x=18.75m{{s}^{-1}} \\
\end{align}$
Speed reduced due to fault = 25 – 18.75 = 6.25 m/s.
For the obtained speed due to delay, the distance travelled in 2 hours will be,
$\begin{align}
  & =18.75\times 2 \\
 & =37.5 \\
\end{align}$
Due to the reduction in speed, the actual time for would be:
$\dfrac{37.5}{6.25}=6$
So, due to 6.25 m/s reduction in speed, the train had to travel another 6 hours after the first hour of the journey.
Total time taken to complete the journey = 1 + 6 = 7 hours.
Therefore, the total distance would be $=7\times 25=175m$ with the speed of 25 m/s.

Note: The key step in solving this problem is the knowledge of the relationship between speed and distance. By using the statement provided in the question we can easily formulate expressions for speed of trains. In this way, we can calculate the distance between train stations.