Find stopping time of train given the speed with and without stopping 
Without stopping for passengers, a train travels a certain distance with an average speed of 60 km/h, and when it stops, it covers the same distance with an average speed of 40 km/h. On average, how many minutes per hour does the train stop during the journey?

My approach: average speed = total distance/ total time;
Here it is given that the first train's speed is $60$ km/h.
$\therefore 60 = \frac{x}{t_1}$ where $x$ is the total distance and $t_1$ is total time.
Now average speed is 40 km/hr.
$\therefore 40 = \frac{x}{t_2}$ where $x$ is the total distance and $t_2$ is total time.
Now solving this I get $3t_1= 2t_2$ but I don't understand how to approach the solution from here.
 A: Travelling 40 km takes 40 min, hence 20 min of standstill per hour.
(In fact, the train travelled for an hour and waited 30 minutes.)

For an algebraic solution (though this seems overkill), let $t$ the stopping time per $60$ minutes, we have
$${40\text{ km/h}}\times{60\text{ min}}={60\text{ km/h}}\times{(60-t)\text{ min}}.$$
A: We have that
$$60=\frac d {t}\, (km/h), \quad 40=\frac d {t+t_0}\, (km/h)$$
then
$$60 t=40(t+t_0) \implies 20 t=40 t_0  \implies \frac{t}{t_0}=2$$
and then
$$\frac{t_0}{t+t_0}=\frac1{\frac t {t_0}+1}=\frac1 3$$
that is 20 minutes per hour.

By your way from here $$3t_1= 2t_2 \iff t_1=\frac23 t_2 =\frac23 (t_1+t_0) \iff t_1=2 t_0$$ which leads to the same result.
A: From $3t_1 = 2t_2$, we get $t_1 = \frac{2}{3}t_2$. This means that two-thirds of the time, the train is running, so during the other one-third of the time, the train is stopping.
This means the train stops for $\frac{1}{3} \times 60 = 20$ minutes every hour.
It might help to let the distance be a number, as the proportions will remain unchanged (minutes per hour). For example, if the distance is $120 \text{km}$, then the train runs for $2$ hours without stopping, and $3$ hours with stopping. Hence $1$ hour out of every $3$ hours is spent stopping, which is the same as $20$ minutes every $1$ hour.
