# TIME AND DISTANCE

Time and distance problems play a major role in quantitative aptitude test. It is bit difficult to score marks in competitive exams without knowing the shortcuts related to time and distance.

Here, we are going to have some time and distance problems. You can check your answer online and see step by step solution.

## Formula

(1)  Speed or Velocity = Distance / Time

(2)  Distance= Speed x Time

(3)  Time = Distance / Speed

(4)  a km/hr= (ax5/18) m/sec

(5)  b m/sec = (bx18/5) km/hr.

(6)  If the ratio of the speeds of A and B is a:b, then ratio of time taken by them to cover the same distance is b:a

(7)  If a certain distance is covered at p km/hr and an equal distance is covered at q km/hr, then the average speed of the whole journey is = (2pq)/(p+q) km/hr

Question 1 :

A person covers a certain distance at a certain speed. If he increases his speed by 33 1/3%, he takes 15 minutes less to cover the same distance. Find the time taken by him initially to cover the distance at the original speed.

Solution :

If the original speed is 100%, speed after increment is 133 1/3%.

Ratio of the speeds is

100%: 133 1/3%  -->  100%:(400/3)%

So, ratio of the speeds is 1 : 4/3

If the ratio of the speed is 1 : 4/3, ratio of time taken would be 1 : 3/4

When the speed is increased by 33 1/3%, 3/4 of the original time is enough to cover the same distance.

That is, when the speed is increased by 33 1/3%, 1/4 of the original time will be decreased.

The question says that when speed is increased by 33 1/3%, time is decreased by 15 minutes.

Therefore, 1/4 of the original time  =  15 minutes

Original time = 4X15  =  60 minutes

Hence, time taken by him initially = 60 mins or 1 hour

Question 2 :

A man traveled from the village to the post office at the rate of 25 k mph and walked back at the rate of 4 kmph. If the entire journey had taken 5 hours 48 minutes, find the required distance of the post office from the village.

Solution :

Average speed = 2pq/(p+q)

here p = 25,  q = 4

= (2X25X4)/(25+4)

Therefore, average speed  =  200/29 km/hr

And 5 hour 48 min = 54860hrs  =  29/5 hours

Distance = Speed X Time

Distance = (200/29)X(29/5) = 40 km

Distance covered in 5 hrs 48 min = 40 km

Hence, distance of the post office from the village

= 40/2 = 20km

Question 3 :

If a man walks at the rate of 5 km/hr, he misses a train by 7 minutes. However, if he walks at the rate of 6 km/hr, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.

Solution :

Let "x" be the distance which has to be found.

Difference between the times in walking at different speed

12 minutes = 12/60 hr = 1/5 hr

When the speed is 5kmph, time = x/5

When the speed is 6kmph, time = x/6

Difference in time taken = 1/5 hr

(x/5) - (x/6) = 1/5

By simplification, we get x = 6 km

Hence,the distance covered by him to reach the station is 6 km

Question 4 :

A person has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds of the total time; to cover the remaining distance in the remaining time, his speed(in kmph) must be

Solution :

From the given information, Already 3 km distance (one half of the distance 6km) has been covered in 30 minutes or 1/2 hr. (two third of the total time 45 minutes)

Remaining distance = 3 km

Time available = 15 minutes or 1/4 hr

Speed required = Distance / Time

= 3/(1/4) kmph = 3X4 kmph = 12 kmph

Hence, speed required to cover remaining distance is 12 kmph

Question 5 :

A is faster than B . A and B each walk 24 km. The sum of their speeds is 7 km / hr and the sum of their time taken is 14 hrs. Then A's speed and B's speed (in kmph) are

Solution :

Let "x" be the speed of A . Then speed of B = 7-x

Time taken by A = 24/x

Time taken by B = 24/(7-x)

Time(A)+Time(B) = 14 hours.

24/x + 24/(7-x) = 14

24(7-x)+24x = 14x(7-x)

14x- 98 x + 168 = 0  --->  x- 7 x + 12 = 0

(x-4)(x-3)=0--->x = 4 and x = 3

Hence, the speed of A and B are 4kmph and 3 kmph respectively.

After having gone through the stuff given above, we hope that the students would have understood "Time and distance"

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