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TIME AND DISTANCE PROBLEMS

Aptitude

TIME AND DISTANCE -> IMPORTANT FACTS AND FORMULAE

1. Speed = [Distance/Time],
Time=[Distance/Speed],
Distance = (Speed*Time)
2. x km/hr = [x*5/18] m/sec.
3. If the ratio of the speeds of A and B is a:b, then the ratio of the times taken by them to cover the same distance is 1/a : 1/b or b:a.
4. x m/sec = [x*18/5] km/hr.
5. Suppose a man covers a certain distance at x km/hr and an equal distance at y km/hr. then, the average speed during the whole journey is [2xy/x+y] km/hr.

TIME AND DISTANCE -> EXAMPLES

1. How many minutes does John take to cover a distance of 400 m, if he runs at a speed of 20 km/hr?
  Sol. John’s speed = 20 km/hr
= [20*5/18] m/sec
= 50/9 m/sec.
2. While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minute, the distance covered by him was 5/7 of the remaining distance. What was his speed in metres per second?
  Sol. Let the speed be x km/hr.
Then, distance covered in 1 hr: 40 min. i.e., 1 2/3 hrs = 5x/3 km.
Remaining distance = [24-5x/3] km.
∴ 5x/3 = 5/7[24-5x/3] ⇔ 5x/3 = 5/7[72-5x/3] ⇔ 7x = 72-5x
⇔ 12x = 72 ⇔ x = 6
Hence, speed = 6 km/hr = [6*5/18] m/sec = 5/3m/sec = 1 2/3 m/sec
3. If a man walks at athe rate of 5 kmph, he misses a train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.
  Sol. Let the required distance be x km.
Difference in the times taken at two speeds = 12 min = 1/5 hr.
∴ x/5 - x/6 = 1/5 ⇔ 6x - 5x = 6 ⇔ x = 6.
Hence, the required distance is 6 km.

TIME AND DISTANCE -> EXERCISE

4. A and B walk around a circular track. They start at 9 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 9.30 a.m.?
 
  • A. 8
  • B. 7
  • C. 6
  • D. 5
Ans: B.
Sol.
Relative speed = (2 + 3) = 5 rounds per hour.
So, they cross each other 5 times in an hour and 2 times in half an hour.
Hence, they cross each other 7 times before 9.30 a.m.
 
5. Two guns were fired from the same place at an interval of 10 minutes and 30 seconds, but a person in the train approaching the place hears the second shot 10 minutes after the first. The speed of the trian (in km/hr), supposing that speed travels at 330 metres per second, is:
 
  • A. 50.4
  • B. 111.80
  • C. 19.88
  • D. 59.4
Ans: D.
Sol.
Let the speed of the train be x m/sec. Then,
Distance travelled by the train in 10 min. = Distance travelled by sound in 30 sec.
∴ Speed of the train = 16.5 m/sec = [16.5*18/5] km/hr = 59.4 km/hr
 
 
6. A thief is noticed by a policeman from a distance of 200m. The theif starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?
 
  • A. 100 m
  • B. 120 m
  • C. 110 m
  • D. 130 m
Ans: A.
Sol.
Relative speed of the thief and policeman
= (11-10) km / hr = 1 km / hr.
Distance coverd in 6 minutes = [1/60 * 6] km = 1/10 km = 100 m.
∴ Distance between the thief and policeman = (100-100) m = 100 m. (time taken to walk x km) + (Time taken to ride x km) = 23/4 hrs.