Answer

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**Hint:**Here to solve this question we will use the concept of motion. First, we will compare the time differences between both the trains to reach their destination, and using this time and using the formula of average speed we will calculate the average speed of the first train.

Formula used:

The average speed formula

$v = \dfrac{s}{t}$

where $s$ is distance and $t$ is time.

**Complete step-by-step solution:**

We will first consider the given condition that the train left the station $\Delta {t_1} = 30\min $ earlier and on the next station the first train arrived $\Delta {t_2} = 20\min $ earlier than his train.

The passengers on the train’s speed ${v_p} = 60km/h$.

Distance between stations is $s = 20km$.

Hence time to reach the next station is given by

${v_P} = \dfrac{s}{t}$

$ \Rightarrow t = \dfrac{s}{{{v_P}}}$

Substituting the values of given speed ${v_p} = 60km/h$ and the distance $s = 20km$ gives

$t = \dfrac{{20}}{{60}} = \dfrac{1}{3}hr$

$ \Rightarrow t = 20\min $.

Hence we can conclude that when the passenger is at the first station, the first train reaches the station $2$. This shows that the first train must have left the station $1$ $30\min $ earlier and when passengers reach the next station time from the first station’s train arrival is $50\min $.

Therefore the time to reach the next station by the first train $ = \left( {50 - 20} \right)\min = 30\min $.

Hence the average speed of the train is given as

${v_{avg}} = \dfrac{s}{t}$

$ \Rightarrow {v_{avg}} = \dfrac{{20km}}{{0.5hr}}$

$\therefore {v_{avg}} = 40km/hr$.

Hence the average speed of the first train by using the calculation is given ${v_{avg}} = 40km/hr$.

**Note:**Here in solving such types of questions one must ensure that the units of the quantities should be in the proper format. In this question we first calculated the time in minutes then again for evaluating the average speed we again converted it into hours.

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