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Any more time saving approach? Trains are leaving station

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Any more time saving approach? Trains are leaving station [#permalink]

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New post 05 Sep 2007, 12:54
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Any more time saving approach?

Trains are leaving station A for station B every 15 minutes starting 8:00 am. The distance between the stations is 80 km. If a train leaves station B for station A at 11:05 am, how many oncoming trains will it meet. Trains travel in both directions at 40 km/h.

How would you solve this problem?
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New post 05 Sep 2007, 13:40
I get a total of 8 trains. The train is going 40 km/h and the distance is 80km so it will take the train 2 hours. There are trains leaving every 15 mins. Since the train leaves at 11:05 it should pass all oncoming trains leaving from 11:00 to 1:00.
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Re: PS_rate_distance_incoming trains [#permalink]

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New post 05 Sep 2007, 13:55
IrinaOK wrote:
Any more time saving approach?

Trains are leaving station A for station B every 15 minutes starting 8:00 am. The distance between the stations is 80 km. If a train leaves station B for station A at 11:05 am, how many oncoming trains will it meet. Trains travel in both directions at 40 km/h.

How would you solve this problem?


I get 16
Either train takes two hours to get to the destination.
Train B leaves at 11.05am. This means that it will meet all train A that did not reach station B before 11.05 am. It will meet anything after until it arrives at station A.
Since train A takes two hours to reach, if train A leave at 9.15am, then it will reach station B at 11.15am and will meet train B. If train A leaves at 9.00am, it will get there at 11.00am and won't meet train B. Therefore, we start counting from 9.15 am. and end at 1.00pm (since train B arrives at 1.05pm)
Every 15 mins including 9.15am to 1.00pm will get you 16 trains.
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New post 05 Sep 2007, 18:55
Putting it this way..the train B will get to destination A at 105 PM

Train B would encounter the trains which started from station A at 915 AM pr later//until it gets there..The trains that started from station A before 9:15..(9AM and before) would have already gotten to station B before Train B leaves the station. B for Station A..
So it would meet the 3 trains that started at 915-10AM..the ones before that would have already arrived
The three trains from 10:15-11 AM it would meet..
Same thing with the 3 trains from 11:15-12 PM
Last but not least, the 3 trains from 12:15 to 1: PM
It would get there before the 115 train leaves from station A..
I get a total of 12..
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New post 05 Sep 2007, 19:56
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Each train takes 2 hours to reach the destination.
2 hours before 11:05 is 9:05, so all trains leaving A for B before 9am can be ignored.

The train leaving B for A will reach A at 13:05. So the number of trains it meets are those leaving A for B between 9am and 1pm.

Time schedule for trains leaving A for B
9:15
9:30
9:45
10:00
10:15
10:30
10:45
11:00
11:15
11:30
11:45
12:00
12:15
12:30
12:45
13:00

Total = 16
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New post 05 Sep 2007, 19:59
I got the concept but miscounted..need more sleep:)
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New post 17 Oct 2007, 07:39
ywilfred wrote:
Each train takes 2 hours to reach the destination.
2 hours before 11:05 is 9:05, so all trains leaving A for B before 9am can be ignored.

The train leaving B for A will reach A at 13:05. So the number of trains it meets are those leaving A for B between 9am and 1pm.

Time schedule for trains leaving A for B
9:15
9:30
9:45
10:00
10:15
10:30
10:45
11:00
11:15
11:30
11:45
12:00
12:15
12:30
12:45
13:00

Total = 16


why do we choose the time frame for 2 hours BEFORE and not AFTER?
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Re: Any more time saving approach? Trains are leaving station [#permalink]

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New post 20 Apr 2016, 13:16
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Re: Any more time saving approach? Trains are leaving station   [#permalink] 20 Apr 2016, 13:16
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