Aptitude - Problems on Trains - Discussion

Here,

The total distance between them = 110km.
(Making equal starting point)

By the time train B has started, the Train A has already covered 20km (since speed =20km/hr). Both at 8:00am now.
Now the distance between them is;
110-20=90km.

Relative speed= 20+25(since opposite direction) =45km/hr.
To cover 90 km time taken=90/45=2hrs.
2 hrs from 8:00am=10:00am.
(since we have made initial starting point at 8:00am for both the trains).

Let x be the meeting distance from A and B.

Then A travels (X/20) hrs and B travels (110-X/25) {time=distance/speed}.
But A stars 1 hour earlier than B.

So;

(X/20)-1 = (110-X)/25.
solving this;
we get X=60kms.

Now
A travels 60 kms at the speed of 20kmph to meet B in the midway of travel.

So now
Time is taken to travel to meet B 60 kms from starting point
A takes+ 60(distance) / 20(speed) = 3 hrs.

From question,

'A' starts to travel at 7 A.M and travels 3 hrs(from final answer) to meet B in the midway of travel.

So the time 'A' and 'B' at meet is [7+3]= 10 A.M.

Let they meet X hrs from the 8 a.m.
Then,
110 * 1000 = 20 * (5/18) * 3600 + 45(5/18) * X.
X=2.
So, time= 8+2 = 10 a.m.

The speed of one train is twice the other so let,
Speed of 1st train is Y.
Speed of 2nd train is 2Y.

Then distance between point A and B is 6Y.

Suppose they meet x hours after 5 a.m.

Distance covered by 1st train in X hours = YX km.

Distance covered by B in (X - 2) hours = 2Y(X - 2) km.

Therefore YX + 2Y(X - 2) = 6Y.

=> YX + 2YX - 4Y = 6Y.

=> 3YX = 10Y.

=> X =10/3 (in hours).
=> X =10/3*60 (because converting into minutes).
=> X =200 minutes.
200 minutes means 3 hour 20 minutes.

So, They meet at 8 : 20 p.m.

Here, how will get 110? Please explain.

Your values for the distance travelled by each train are mathematically correct based on their speeds and times.

But from 10:00 a.m. onward, your reasoning ignores that the trains would have already met at 10 a.m.

Because their combined distance = 110 km, which is the total distance between A and B.

So, continuing to add distances after the meeting point doesn't make logical sense in the context of this problem.

I dint get how x-1 term is used.

Thanks yuvraj.

A----------110---------B 7am

A---------90--------B 8am(Speed of the 1st train is 20 km/h)

A----45----B 9am(relative speed=20+25(45) km/h)

AB 10am

Yuvraj. Cool answer!