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A train’s journey is disrupted due to an accident on its track after it has travelled 30 km. Its speed then comes down to 4/5th of its original and consequently it runs 45 min late. Had the accident taken place 18 km farther away, it would have been 36 min late. Find the original speed of the train.
- a)25 km/h
- b)36 km/h
- c)30 km/h
- d)20 km/h
Correct answer is option 'C'. Can you explain this answer?
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Problem: A train's journey is disrupted due to an accident on its track after it has travelled 30 km. Its speed then comes down to 4/5th of its original and consequently it runs 45 min late. Had the accident taken place 18 km farther away, it would have been 36 min late. Find the original speed of the train.
Solution:
Let the original speed of the train be x km/h and let the distance between the starting point and the accident spot be d km.
Case 1: Accident happens at d = 30 km
Let t be the original time taken by the train to cover the distance of 30 km at speed x km/h. Then,
t = 30/x
After the accident, the speed of the train reduces to (4/5)x km/h. Let the new time taken by the train to cover the remaining distance of (d - 30) km be t1. Then,
t1 = (d - 30)/(4/5)x = 5(d - 30)/(4x)
Due to the accident, the train runs 45 min late. Therefore,
t1 - t = 45/60 = 3/4 hours
Substituting the values of t and t1, we get
5(d - 30)/(4x) - 30/x = 3/4
Simplifying the above equation, we get
9x = 20(d - 30) ...(1)
Case 2: Accident happens at d = 48 km
Using similar calculations as above, we can write
t2 = 48/x
t2 - t = 36/60 = 3/5 hours
Substituting the values of t and t2, we get
4(d - 30)/(5x) - 30/x = 3/5
Simplifying the above equation, we get
3x = 10(d - 30) ...(2)
Solving equations (1) and (2), we get
d = 150 km and x = 30 km/h
Therefore, the original speed of the train is 30 km/h, which is option (c).
Solution:
Let the original speed of the train be x km/h and let the distance between the starting point and the accident spot be d km.
Case 1: Accident happens at d = 30 km
Let t be the original time taken by the train to cover the distance of 30 km at speed x km/h. Then,
t = 30/x
After the accident, the speed of the train reduces to (4/5)x km/h. Let the new time taken by the train to cover the remaining distance of (d - 30) km be t1. Then,
t1 = (d - 30)/(4/5)x = 5(d - 30)/(4x)
Due to the accident, the train runs 45 min late. Therefore,
t1 - t = 45/60 = 3/4 hours
Substituting the values of t and t1, we get
5(d - 30)/(4x) - 30/x = 3/4
Simplifying the above equation, we get
9x = 20(d - 30) ...(1)
Case 2: Accident happens at d = 48 km
Using similar calculations as above, we can write
t2 = 48/x
t2 - t = 36/60 = 3/5 hours
Substituting the values of t and t2, we get
4(d - 30)/(5x) - 30/x = 3/5
Simplifying the above equation, we get
3x = 10(d - 30) ...(2)
Solving equations (1) and (2), we get
d = 150 km and x = 30 km/h
Therefore, the original speed of the train is 30 km/h, which is option (c).
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A train’s journey is disrupted due to an accident on its track after it has travelled 30 km. Its speed then comes down to 4/5th of its original and consequently it runs 45 min late. Had the accident taken place 18 km farther away, it would have been 36 min late. Find the original speed of the train.a)25 km/hb)36 km/hc)30 km/hd)20 km/hCorrect answer is option 'C'. Can you explain this answer?
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Solutions for A train’s journey is disrupted due to an accident on its track after it has travelled 30 km. Its speed then comes down to 4/5th of its original and consequently it runs 45 min late. Had the accident taken place 18 km farther away, it would have been 36 min late. Find the original speed of the train.a)25 km/hb)36 km/hc)30 km/hd)20 km/hCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Quant.
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