Aptitude - Problems on Trains - Discussion

Thank you @Manish Dabhi.

A train travels in one houe=20 km.

Now at 8 o'clock distance between tow trains = 90 km.
It is relavent to these two trains speed =20+25km/h = 45km/h.
Now time = distance/speed.
= 90/45 = 2h.

Hence trains meet after 2hours that is 8+2 = 10 a.m.

For equal starting time of two trains (ie 8am, 8am).

Train A cover a distance in one hour = 20km.
So total distance covered by two trains =110-20.
= 90km.
The relative speed of trains =20+25.
= 45kmph.

Two trains are meeting at a time of X hr.
The formula is;
90/X = 45,
X=2,
8am+2 = 10am.

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.

2 station A and B are 110km apart.
1 train start from A at 7 am towards B at 20 kmph 110-20 = 90.

Next one start from B at 8 a.m towards A at 25kmph.
20+25 =45.

Now 90/45 = 2hrs.
.`. 8+2 = 10 a.m.

How we are obtained the 10 ?

Please explain it.

To solve this we can take a very simple approach.
As we have total distance between them 110 km and A is starting at 7 a.m.
So after 1 hr that is at 8 a.m he will cover 20 km so;
110-20 = 90km.

Now for both the trains we have 90 km and starting at 8 a.m so we can find there crossing time, as they will cross each other
T = d / s(a train ) + s (b train).
T = 90/ (20 + 25),
T = 90/ 45 = 2 hrs,
i.e after 2 hrs so 8 a.m + 2hrs = 10 a.m.