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At 9 am train X left from station A towards station B at uni
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13 Jul 2019, 08:13
13 Jul 2019, 08:13
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Question Stats:
62% (03:33) correct
37% (01:50) wrong based on 32 sessions
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At 9 am train X left from station A towards station B at uniform speed of 90km/hr. After 2 hours another train Y leaves from station B on a parallel track towards station A at uniform speed of 60km/hr. If train X reached station B at 2 pm, at what time will the two train cross each other?
A) 10 : 12 am
B) 10 : 48 am
C) 11 : 20 am
D) 12 : 48 am
E) 1 : 20 pm
Source : Green Book 1 Jamboree
A) 10 : 12 am
B) 10 : 48 am
C) 11 : 20 am
D) 12 : 48 am
E) 1 : 20 pm
Source : Green Book 1 Jamboree
Re: At 9 am train X left from station A towards station B at uni
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20 Nov 2019, 13:56
20 Nov 2019, 13:56
1
gagangm wrote:
Can anyone kindly let me know? Which way can we use to solve this sort of problems?
I am not quite sure if there is a method to solve this particular one. I solved it graphically.
It takes train A, 5 hours (9 am - 2 pm) for the journey from Station A to station B.
It takes train b 7,5 hours from station B to station A (based on the total distance of 450 km and the speed of train B).
Then, you can represent it graphically on 2 segment lines.
Divide the first segment into 5 parts and the segment below in 7.5. Each part of the segment represents one hour.
You can clearly see that both trains will cross each other between 12 pm and 1 pm.
The solution D.
At 9 am train X left from station A towards station B at uni
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15 Feb 2023, 11:27
15 Feb 2023, 11:27
1
The basic method for objects heading toward each other that leave at the same time is:
1. Calculate relative speed (sum the 2 objects' given speeds)
2. Apply t=d/s formula, using the above relative speed as s and the distance between the 2 objects at the starting point (i.e., the length of the track) as d
3. Manipulate the above per the question's demands.
When the objects depart at different times, you have to do 2 different things relative to the above:
1. For step 2, instead of simply using the distance of the track (i.e., distance b/w 2 stations), you would take this distance and subtract from it the distance that the earlier-departing object would have covered in the time elapsed between the 2 departures.
2. When calculating the total time, add to the time you get in the steps outlined above the time elapsed between the 2 departures.
Following the above in the case of this question,
1. Relative speed: 150 (90+60)
2. Time = Distance/Speed = 270/150 = 1.8 hours. We know the track distance is 450km because Train X left the station at 9am, reached the next station at 2pm (5 hrs later) and travelled at 90km/h: 5*90 = 450. Thus, the actual distance between them when Train Y left the station, 2 hours after Train x, is 270km (450-(90*2)).
3. 1.8 hrs + 2 hrs (time elapsed b/w the 2 departures) = 3.8 hrs. 3.8 hrs after 9am (earlier departure) is thus 12:48pm.
1. Calculate relative speed (sum the 2 objects' given speeds)
2. Apply t=d/s formula, using the above relative speed as s and the distance between the 2 objects at the starting point (i.e., the length of the track) as d
3. Manipulate the above per the question's demands.
When the objects depart at different times, you have to do 2 different things relative to the above:
1. For step 2, instead of simply using the distance of the track (i.e., distance b/w 2 stations), you would take this distance and subtract from it the distance that the earlier-departing object would have covered in the time elapsed between the 2 departures.
2. When calculating the total time, add to the time you get in the steps outlined above the time elapsed between the 2 departures.
Following the above in the case of this question,
1. Relative speed: 150 (90+60)
2. Time = Distance/Speed = 270/150 = 1.8 hours. We know the track distance is 450km because Train X left the station at 9am, reached the next station at 2pm (5 hrs later) and travelled at 90km/h: 5*90 = 450. Thus, the actual distance between them when Train Y left the station, 2 hours after Train x, is 270km (450-(90*2)).
3. 1.8 hrs + 2 hrs (time elapsed b/w the 2 departures) = 3.8 hrs. 3.8 hrs after 9am (earlier departure) is thus 12:48pm.
