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The J&K Express from Delhi to Srinagar was delayed by snowfall for 16 minutes and made upfor the delay on a section of 80 km travelling with a speed 10 km per hour higher than its normal speed. Find the original speed of the J&K Express (according to the schedule)
- a)60 km/h
- b)66.66 km/h
- c)50 km/h
- d)40 km/h
Correct answer is option 'C'. Can you explain this answer?
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Given information:
- The JK Express was delayed by snowfall for 16 minutes.
- It made up for the delay on a section of 80 km travelling with a speed 10 km per hour higher than its normal speed.
To find:
- The original speed of the JK Express (according to the schedule)
Solution:
Let the original speed of the JK Express be x km/h.
Due to the delay of 16 minutes, the train would have covered a distance of (16/60) * x = (4/15) * x km less than the scheduled distance during that time.
To make up for this delay, the train needs to cover the same distance in less time. So, the time taken to cover a distance of 80 km at the original speed x km/h is given by:
time = distance / speed
=> time = 80 / x
To cover the same distance of 80 km in less time, the train needs to travel with a higher speed. Let the new speed be (x + 10) km/h. Then, the time taken to cover the same distance of 80 km at this speed is given by:
time = distance / speed
=> time = 80 / (x + 10)
Since the train made up for the delay, the two times taken should be equal. So, we have:
80 / x = 80 / (x + 10)
=> x = 50 km/h
Therefore, the original speed of the JK Express (according to the schedule) is 50 km/h.
Answer: Option (c) 50 km/h
- The JK Express was delayed by snowfall for 16 minutes.
- It made up for the delay on a section of 80 km travelling with a speed 10 km per hour higher than its normal speed.
To find:
- The original speed of the JK Express (according to the schedule)
Solution:
Let the original speed of the JK Express be x km/h.
Due to the delay of 16 minutes, the train would have covered a distance of (16/60) * x = (4/15) * x km less than the scheduled distance during that time.
To make up for this delay, the train needs to cover the same distance in less time. So, the time taken to cover a distance of 80 km at the original speed x km/h is given by:
time = distance / speed
=> time = 80 / x
To cover the same distance of 80 km in less time, the train needs to travel with a higher speed. Let the new speed be (x + 10) km/h. Then, the time taken to cover the same distance of 80 km at this speed is given by:
time = distance / speed
=> time = 80 / (x + 10)
Since the train made up for the delay, the two times taken should be equal. So, we have:
80 / x = 80 / (x + 10)
=> x = 50 km/h
Therefore, the original speed of the JK Express (according to the schedule) is 50 km/h.
Answer: Option (c) 50 km/h
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