Aptitude - Problems on Trains - Discussion

As A train starts at 7 am at 20kmph and B starts at 8 am at 25kmph.

The distance between them is 110 km.

As A starts an hour early at 8 am, A has travelled 20 km.

Now at 8 am, the distance between them is 110–20 km.

As of now, we have a distance as 90 km.
The relative speed is 45kmph.

Time = Distance/Speed.
Time = 90/45 = 2 hrs.
i.e after 2 hours, so 8 a.m + 2 hours = 10 a.m.

It's very interesting and useful to learn. Thanks.

The correct option is B: 10 A.M.

We know, Time = Distance/speed.

Here, we have to consider the relative speed of two trains.

=> 20 km/hr + 25 km/hr = 45 km/hr (since they are moving in opposite directions)

Here, the distance between them cannot be taken as 110 km as train B didn't start at 7 A.M.

The distance train A would have covered from 7 A.M to 8 A.M will be =
20km/hr* 1hr= 20 km.

So, the distance between these two trains will become 110 km - 20 km = 90 km at 8 A.M.

So now, the time is taken for them to meet = 90 km/ 45 km/hr= 2 hrs.

They meet 2 hrs after 8 A.M i.e 10 A.M.

I didn't get anything. Please help me to get this.

As A train starts at 7 am at 20kmph and B starts at 8 am at 25kmph.

The distance between them is 110km.
As A starts an hour early at 8 am A has travelled 20 km.
Now at 8 am distance between them is (110-20)km.
As of now we have a distance as 90km.
The relative speed is 45kmph.
Time = Distance/Speed.
Time = 90/45 =2 hrs.

Tell me that these two trains meet at what distance. Please anyone explain to me.

@Chai.

I got 2.25 is it correct or not?

Here, how will get 110? Please explain.

Hey @Syed.

x-1 is used because the second train starts one hour after the first train.

When X=3 is the time in hours to meet. The starting time is 7 a.m so add 3 to it is 10 a.m when the trains meet.

Why x-1 is used in this sum, and how x=3 and how the answer is 10? Please explain.