Question

A train started its journey at 9:00. Its average speed between
9:00 and 13:00 was 80 km/h. It travelled 1/5 of its journey by
13:00. If the train was required to arrive at the destination at
21:00, what average speed must the train travel at for the rest
of the journey?

Answer 1

Answer:

Explanation:

In the four hours before 13:00 the train traveled 4(80) = 320 km.

It needs to travel another 4(320) = 1280 km in the next 21:00 - 13:00 = 8 hrs.

1280 / 8 = 160 km/hr

Answer 2

We can define average speed as:

s = (distance traveled)/(time it takes)

Using this, we will find that the average speed at which the train must travel the rest of the journey is 160km/h.

Here the given information is:

The train starts its journey at 9:00

The average speed between 9:00 and 13:00 was 80km/h, and it travels 1/5 of the journey by 13:00.

Then, if we define D as the total distance of the whole journey, in that lapse of 4 hours the train traveled 1/5 of D, or just D/5.

Then we can write the average speed as:

$80km/h = \frac{(D/5)}{4h}$

$80km/h*4h = D/5$

$320km = D/5$

$5*320km = D = 1,600km$

This means that the total distance of the journey is 1,600km

And at this time (13:00) the train already traveled 320km, so the distance left is:

$d = 1,600km - 320km = 1,280km$

And we know that the train needs to finish the journey at 21:00

The time between 13:00 and 21:00 is 8 hours.

So the train needs to travel a distance of 1,280km in 8 hours, then the average speed that the train needs is:

$s = (1,280km)/8h = 160 km/h$

s = 160km/h

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