A train running between two stations A and B arrives at its destination 10 minutes late when its speed is 50 km/hr and 50 minutes late when its speed is 30 km/hr. How far is station A from B?
A. 50 km
B. 20 km
C. 60 km
D. 15 km

Hint: We will first assume the distance between A and B as X and use the formula of, $\text{time =}\dfrac{\text{distance}}{\text{speed}}$. When the speed of the train is 50 km/hr, then the time taken would be $\dfrac{X}{50}$ hrs. And when the speed of the train is 30 km/hr, then the time taken would be $\dfrac{X}{30}$ hrs. So, we will subtract the time delayed form the obtained time and equate them, $\dfrac{X}{50}-\dfrac{10}{60}=\dfrac{X}{30}-\dfrac{50}{60}$ and then find the value of X and so the required answer.

Complete step-by-step answer:
It is given in the question that a train running between two stations A and B arrives at its destination 10 minutes late when its speed is 50 km/hr and 50 minutes late when its speed is 30 km/hr and we have been asked to find the distance between A and B. We will start by assuming the distance between the station A and B as X km. We know that the formula for speed is given as, $\text{speed =}\dfrac{\text{distance}}{\text{time}}$. So, we can say from this that, $\text{time =}\dfrac{\text{distance}}{\text{speed}}$. So, we will calculate the time in both the cases given. In the first case, when the speed of the train is 50 km/hr, then the time taken by the train to cover the X kms will be,
$\dfrac{Xkm}{50km/hr}\Rightarrow \dfrac{X}{50}hr$
Now, it is given that the train has a delay of 10 minutes, so we get the reporting time as,
$\begin{align}
  & \dfrac{X}{50}-\dfrac{10}{60} \\
 & \Rightarrow \dfrac{X}{50}-\dfrac{1}{6}\ldots \ldots \ldots \left( i \right) \\
\end{align}$
Now, in the second case, we have the speed of the train as 30 km/hr, so the time taken to cover the X km will be given as,
$\dfrac{Xkm}{30km/hr}\Rightarrow \dfrac{X}{30}hr$
Now, in this case, we have been given that the train has a delay of 50 minutes, so the reporting time will be,
$\begin{align}
  & \dfrac{X}{30}-\dfrac{50}{60} \\
 & \Rightarrow \dfrac{X}{30}-\dfrac{5}{6}\ldots \ldots \ldots \left( ii \right) \\
\end{align}$
Now, we know that the reporting time in both the cases must be the same. So, we will equate equation (i) and (ii). So, we get,
$\dfrac{X}{50}-\dfrac{1}{6}=\dfrac{X}{30}-\dfrac{5}{6}$
We will transpose $-\dfrac{1}{6}$ from the LHS to the RHS, so we get,
$\dfrac{X}{50}=\dfrac{X}{30}-\dfrac{5}{6}+\dfrac{1}{6}$
Now, we will transpose $\dfrac{X}{30}$ from the RHS to the LHS. So, we get,
$\begin{align}
  & \dfrac{X}{50}-\dfrac{X}{30}=-\dfrac{5}{6}+\dfrac{1}{6} \\
 & \Rightarrow \dfrac{X}{50}-\dfrac{X}{30}=\dfrac{-5+1}{6} \\
 & \Rightarrow \dfrac{30X-50X}{1500}=\dfrac{-4}{6} \\
 & \Rightarrow \dfrac{-20X}{1500}=\dfrac{-4}{6} \\
\end{align}$
On multiplying with - 1 on both the sides we get,
$\begin{align}
  & \dfrac{20X}{1500}=\dfrac{4}{6} \\
 & \Rightarrow \dfrac{20X}{1500}=\dfrac{2}{3} \\
 & \Rightarrow 60X=3000 \\
 & \Rightarrow X=\dfrac{3000}{60} \\
 & \Rightarrow X=50 \\
\end{align}$
Thus, the distance between station A and B will be 50 km.
Therefore, the correct answer is option A.

Note: Most of the students equate the reporting time without converting all the values in the same unit. They may form the equation as, $\dfrac{X}{50}-10=\dfrac{X}{30}-50$. This would be totally incorrect as $\dfrac{X}{50},\dfrac{X}{30}$ are given in hours and 10 and 50 are in minutes. Thus, it is very important to convert all the values into the same units.