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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

19. Walking 6/7th of his usual speed, a man is 12 minutes too late. The usual time taken by him to cover that distance is :
 
  • A. 1 hr 20 min
  • B. 1 hr 10 min
  • C. 1 hr 11 min
  • D. 1 hr 12 min
Ans: D.
Sol.
New speed = 6/7 of usual speed.
New time = 7/6 of usual time
∴ [7/6 of usual time] - (usual time) = 1/5 hr.
⇒ 1/6 of usual time = 1/5 hr ⇒ usual time = 6/5 hr = 1 hr 12 min.
 
20. A man can reach a certain place in 30 hours. If he reduces his speed by 1/15th, he goes 10 km less in that time. Find his speed.
 
  • A. 4 1/2km/hr
  • B. 5 km/hr
  • C. 5 1/2km/hr
  • D. 6 km/hr
Ans: B.
Sol.
Let the speed be x km/hr. Then,
30x-30*14/15x = 10
⇔ 2x = 10
⇔ x = 10/2
= 5 km/hr
 
 
21. A boy rides his bicycle 10 km at an average sped of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. His average speed for the entire trip is approximately.
 
  • A. 10.2 kmph
  • B. 10.4 kmph
  • C. 10.8 kmph
  • D. 11.8 kmph
Ans: C.
Sol.
Total distance travelled = (10+12) km/hr = 22 km/hr.
total time taken = [10/12 + 12/10] hrs = 61/30 hrs
∴ Average speed = [22*30/61] km/hr = 10.8 km/hr.