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.

xm/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?

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 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.