Aptitude - Problems on Trains - Discussion

@Chai.

Answer : 2.15 pm. Is this correct?

One train will cover 20 km in one hour.

Now the distance remaining is 90 km.

So time take is equal to 90/20+25 = 2 hours means will meet 2 hours after 8 am i.e. 10 am.

A and B are two stations 390 km a part. A train start from A at 10 AM and travels towards B at 65 kmph. Another train start from B at 11 AM and travels towards A at 35 kmph. At what time do they meet? Solve this problem.

How 20x + 25(x-1) = 110?

A_________________110KM______________________B.

7.00 AM. 8.00 AM.

20 KMH. 25 KMh.

2nd train after 1 hour.

So 1st train gone to 20 km.

Because totally 110-20 = 90 km.

Now 20 kmh+25 kmh = 45 kmh.

90/45 = 2 hours.

Now we count 8.00 am+ 2 hours = 10.00 AM.

Hai @Poojitha.

x is the time taken, 20 is speed.

So Speed*Time = Distance.

For first train it is 20x. And as the second train started 1 hr later.

The time will be x-1. And distance = 25(x-1).

Since both the trains are yet to cover 110 km.

And travelling in opposite directions to meet each other and we have to calculate the meeting point of time. We have to add them. So,

It is 20x+25(x-1) = 110.

Hope it is clear now.

How is it 20x+25(x-1) = 110?

Second train leaves from B towards A. Many of us got confused that it starts from A towards B.

And we don't have to make it so complicated. We can generalize this question.
first train leaves at 7am.

By 8am it covers 20kms. Remaining distance is 90kms (110-20).
By 9am first train covers 20kms more and second train covers 25kms. Remaining distance is 45kms (90-(20+25)).

At 10am the first train covers another 20kms and second train covers 25kms more. (45-(20+25))=0 kms.

Distance between them is zero. Hence they meet at 10am.

@Easwar.

Distance is x and speed first train is = x/6 (because speed=distance/time).

And speed of train second train is = x/3.

First train start at 5pm. If it travel two hours up to 7pm, the distance covered by this train is ( (x/6) *2) is x/3 and the remaining distance is (x- (x/3) is 2x/3.

Now find relative speed (opp direction) is (x/6) + (x/3) =x/2.

Remaining distance is 2x/3 and relative speed is x/2 so,

( (2x/3) / (x/2) ) = 4/3hours.

Means after 7 pm both trains meet at 4/3 hours later. (4/3) *60=80 minutes so at 8:20 pm both trains meet.

Thank you.

So x = 3.

Then B start 7 a.m so count three numbers like 8,9,10 (after 7 we will get 8) :10 a.m.