Find stopping time of train given the speed with and without stopping

Question

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.

Answer 1

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.

Answer 2

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.)

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For an algebraic solution (though this seems overkill), let $t$ the stopping time per $60$ minutes, we have

$${40\text{ km/h}}\times{60\text{ min}}={60\text{ km/h}}\times{(60-t)\text{ min}}.$$

Answer 3

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.

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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.