# Aptitude - Problems on Trains - Discussion

Discussion Forum : Problems on Trains - General Questions (Q.No. 30)

30.

Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?

Answer: Option

Explanation:

Suppose they meet *x* hours after 7 a.m.

Distance covered by A in *x* hours = 20*x* km.

Distance covered by B in (*x* - 1) hours = 25(*x* - 1) km.

20*x* + 25(*x* - 1) = 110

45*x* = 135

*x* = 3.

So, they meet at 10 a.m.

Discussion:

90 comments Page 1 of 9.
Ravi said:
8 years ago

Here, the first train is started at 7:00 am at 20 km/h and the second train started at 8 am at the speed of 25 km/ h.

Train first started at 7 am @ 20 km/h.

7 am to 8 am covers =20 km

8 am to 9 am covers = 40 km

9 am to 10 am covers =60 km

10 am to 11am covers = 80km

11 am to 12 it covers 100 km

Train second started at 8am @ 25 km/h.

8 am to 9 am train covers = 25 km

9 am to 10 am it covers = 50 km

10 am to 11 am it covers = 75 km

11 am to 12 it covers = 100km

Now, both it covers 100 km at 12 o'clock. then how these 2 train meets each other at 10 am please help me, sir.

Train first started at 7 am @ 20 km/h.

7 am to 8 am covers =20 km

8 am to 9 am covers = 40 km

9 am to 10 am covers =60 km

10 am to 11am covers = 80km

11 am to 12 it covers 100 km

Train second started at 8am @ 25 km/h.

8 am to 9 am train covers = 25 km

9 am to 10 am it covers = 50 km

10 am to 11 am it covers = 75 km

11 am to 12 it covers = 100km

Now, both it covers 100 km at 12 o'clock. then how these 2 train meets each other at 10 am please help me, sir.

Nikhil_vats said:
7 years ago

Platform A ____110km____ Platform B.

Difference between train A and train B is =1hr.

Relative speed of both the trains are = (20kmph+25kmph)= 45kmph.

Now think as if you are the boss (what if you can make both the trains depart at 8 AM how much easy it would have become).

So, the train A is moving at 20km in 1 hour at 7 AM it means it already has completed 20kms by 8 AM from the given distance of 110 km = so (110-20=90 km) left for train A.

And being a boss you now calculate from 8AM for both the trains.

Assume,

If they both depart at 8 AM

Train A now have to go 90 km more

Train B have to go 110 km

So we will calculate

D=S*T ( distance still left = relative speed *time

90=45*time

90/45=2.

2 hours from the time which we shifted like a boss 8 AM so add 2 hours more after 8.

Its 10 AM.

Difference between train A and train B is =1hr.

Relative speed of both the trains are = (20kmph+25kmph)= 45kmph.

Now think as if you are the boss (what if you can make both the trains depart at 8 AM how much easy it would have become).

So, the train A is moving at 20km in 1 hour at 7 AM it means it already has completed 20kms by 8 AM from the given distance of 110 km = so (110-20=90 km) left for train A.

And being a boss you now calculate from 8AM for both the trains.

Assume,

If they both depart at 8 AM

Train A now have to go 90 km more

Train B have to go 110 km

So we will calculate

D=S*T ( distance still left = relative speed *time

90=45*time

90/45=2.

2 hours from the time which we shifted like a boss 8 AM so add 2 hours more after 8.

Its 10 AM.

Rahul said:
8 years ago

Easy way:

d = 110 km

train A = starts @ 7.00 with speed = 20 km

train B = starts @ 8.00 with speed = 25 km *(5kms difference A & B)

So, A+B=45 kms && d = 110

---> Time = d/s = 110/45 (after cancellation = 22/9)

---> Now add 5 (difference kms between 2 trains speed) to 22 = 27.

---> Now divide 27/9 = 3

---> Add 3 am to 7 am = 10 am.

**Here the difference in time is only 1 hr so added only 5 km's extra.

If the time difference was 2 hrs then we 10 and so on. If the answer you get doesn't match with the options then choose the closest options. The simple reason is time has 60 min where as when you are actually calculating it is of 100. Hope you'll are understanding.

HOPE THIS WORKS I WAS ACTUALLY TRYING TO MAKE SOME EASY SHIT SO THAT THIS TYPE OF SUMS BECOMES EASY FOR THE US :-)

d = 110 km

train A = starts @ 7.00 with speed = 20 km

train B = starts @ 8.00 with speed = 25 km *(5kms difference A & B)

So, A+B=45 kms && d = 110

---> Time = d/s = 110/45 (after cancellation = 22/9)

---> Now add 5 (difference kms between 2 trains speed) to 22 = 27.

---> Now divide 27/9 = 3

---> Add 3 am to 7 am = 10 am.

**Here the difference in time is only 1 hr so added only 5 km's extra.

If the time difference was 2 hrs then we 10 and so on. If the answer you get doesn't match with the options then choose the closest options. The simple reason is time has 60 min where as when you are actually calculating it is of 100. Hope you'll are understanding.

HOPE THIS WORKS I WAS ACTUALLY TRYING TO MAKE SOME EASY SHIT SO THAT THIS TYPE OF SUMS BECOMES EASY FOR THE US :-)

Randhir said:
1 decade ago

@Gayatri, nithi, nayana

Speed=distance/time

1st train leaves a point A at 5 p.m. and reaches another point B a 11 p.m.that means time taken by train 6 hours.

Time = 6 hour

Speed of the 1st train is Y

Speed = Y km/h

Distance= speed*time (because speed=distance/time)

Distance= Y * 6 = 6Y

if you are still not getting my point then comes to train 2nd

2nd train leaves point B at 7 p.m. and reaches point A at 10 p.m. that means time taken by train 3 hours.

Time = 3 hour

Speed of the 1st train is 2Y

Speed = 2Y km/h

Distance= speed*time (because speed=distance/time)

Distance= 2Y * 3 = 6Y

Speed=distance/time

1st train leaves a point A at 5 p.m. and reaches another point B a 11 p.m.that means time taken by train 6 hours.

Time = 6 hour

Speed of the 1st train is Y

Speed = Y km/h

Distance= speed*time (because speed=distance/time)

Distance= Y * 6 = 6Y

if you are still not getting my point then comes to train 2nd

2nd train leaves point B at 7 p.m. and reaches point A at 10 p.m. that means time taken by train 3 hours.

Time = 3 hour

Speed of the 1st train is 2Y

Speed = 2Y km/h

Distance= speed*time (because speed=distance/time)

Distance= 2Y * 3 = 6Y

Gayatri said:
1 decade ago

@Easwar

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.

Hii Randhir How come distance between point A and B is 6Y? Please explain.

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.

Hii Randhir How come distance between point A and B is 6Y? Please explain.

Nandakumar said:
5 years ago

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.

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.

Sumanth kumar said:
1 year ago

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.

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.

(79)

Rambabu said:
1 decade ago

@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.

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.

Randhir said:
1 decade ago

hi Easwar,

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.

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.

Dezosa said:
5 years ago

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.

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.

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