Question

a train t1 left station s1 for station s2 at 11.00am at a speed of 72 km/hr. another train t2 left s2 for s1 at 10.00am at a speed of 108 km/hr. at what time would the trains be 24 km apart for the first time given that distance between s1 and s2 is 240 km?

Answer 1

Hi,

Answer:

36 minutes from 11:00 a.m. or at 11:36 a.m.

Step-by-step explanation:

Given data:

Speed of train t1 = 72 km/hr

Speed of train t2 = 108 km/hr

Distance between s1 and s2 = 240 km

To find: the time when the trains are 24 km apart for the first time

Let the distance traveled by train t1 be “x” km from s1 when the two trains are 24 km apart from each other.

Therefore,  

The time required by t1 = x / 72

And,

The time required by t2 = (240 – 24 – x) / 108  

We know that the train s2 started 1 hr early than train s2.

So,  

x / 72 = [(216 – x) / 108] - 1

or, x/72 = (108-x)/108

or, 180x = 7776 – 72x

or, x = 7776 / 180 = 43.2 km

∴ t1 = 43.2 / 72 = 0.6 hr = 36 minutes from 11:00 a.m.

Hence, at 11:36 a.m. the trains will be 24 km apart from each other for the first time.

Hope this helps!!!!!