Contents

# TIME AND DISTANCE

## IMPORTANT FACTS AND FORMULAE

Distance                    Distance

1. Speed = Time ,  Time=      Speed      , Distance  =  (Speed *  Time)

1. x km / hr = x *  5

18

1. x m/sec = (x * 18/5) km /hr

1. 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: 1                                                                                                                                                                              a   b

or b:a.

1. 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    km/ hr.

x+y

## SOLVED EXAMPLES

Ex. 1. How many minutes does Aditya take to cover a distance of 400 m, if he runs at a speed of 20 km/hr?

Sol. Aditya’s speed = 20 km/hr  = {20 * 5} m/sec  =   50 m/sec

• 9

\Time taken to cover 400 m= { 400 * 9 } sec =72 sec = 1 12  min 1 1 min.

50                             60           5

Ex. 2. A cyclist covers a distnce of 750 m in 2 min 30 sec. What is the speed in km/hr of the cyclist?

Sol. Speed = { 750 } m/sec  =5 m/sec  = { 5  *  18 } km/hr =18km/hr

• 5

Ex. 3. A dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to 4 leaps of the hare. Compare their speeds.

Sol. Let the distance covered in 1 leap of the dog be x and that covered in 1 leap of the hare by y.

Then , 3x = 4y => x = 4 y  =>  4x = 16  y.

• 3

\ Ratio of speeds of dog and hare = Ratio of distances covered by them  in the same time

= 4x : 5y = 16 y : 5y  =16  : 5  = 16:15

• 3

Ex. 4.While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minutes, the distance covered by him was 5 of the remaining distance. What was his speed in metres per second?

7

Sol. Let the speed be x km/hr.

Then, distance covered in 1 hr. 40 min. i.e., 1  2  hrs  = 5x  km

• 3

Remaining distance = { 24 – 5x } km.

3

• 5x 5 {  24 –  5x  } ó  5x  =  5 {  72-5x  }  ó  7x  =72 –5x

3      7              3             3       7        3

ó 12x = 72  ó  x=6

Hence speed = 6 km/hr ={ 6 * 5 } m/sec  =  5  m/sec = 1 2

18                  3                  3

Ex. 5.Peter can cover a certain distance in 1 hr. 24 min. by covering two-third of the distance at 4 kmph and the rest at 5 kmph. Find the total distance.

Sol.   Let the total distance be x km . Then,

2 x        1 x

3      +   3     =   ó  x  +  x  = 7    ó  7x  = 42  ó  x = 6

4         5          5         6     15    5

Ex. 6.A man traveled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village.

Sol.    Average speed   = { 2xy  } km/hr  ={  2*25*4  } km/hr  = 200  km/hr

x+y                        25+4                     29

Distance traveled in 5 hours 48 minutes i.e., 5 4  hrs.  =  { 200  *  29 } km  = 40 km

5                29          5

Distance of the post-office from the village ={  40  }  = 20 km

2

Ex. 7.An aeroplane files along the four sides of a square at the speeds of 200,400,600 and 800km/hr.Find the average speed of the plane around the field.

Sol. :

Let each side of the square be x km and let the average speed of the plane around the field by y km per hour then ,

x/200+x/400+x/600+x/800=4x/yó25x/2500ó4x/yóy=(2400*4/25)=384

hence average speed =384 km/hr

Ex. 8.Walking at 5 of its usual speed, a train is 10 minutes too late. Find its usual time to cover the journey.

7

Sol. :New speed =5/6 of the usual speed

New time taken=6/5 of the usual time

So,( 6/5 of the usual time )-( usual time)=10 minutes.

=>1/5 of the usual time=10 minutes.

• usual time=10 minutes

Ex. 9.If a man walks at the 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 time taken at two speeds=1 min =1/2 hr

Hence x/5-x/6=1/5<=>6x-5x=6

óx=6

Hence, the required distance is 6 km

Ex. 10. A and B are two stations 390 km apart. A train starts from A at 10 a.m. and travels towards B at 65 kmph. Another train starts from B at 11 a.m. and travels towards A at 35 kmph. At what time do they meet?

Sol. Suppose they meet x hours after 10 a.m. Then,

(Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs]=390.

65x + 35(x-1) = 390  => 100x = 425  => x = 17/4

So, they meet 4 hrs.15 min. after 10 a.m i.e., at 2.15 p.m.

Ex. 11. A goods train leaves a station at a certain time and at a fixed speed. After ^hours, an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph. This train catches up the goods train in 4 hours. Find the speed of the goods train.

Sol.  Let the speed of the goods train be x kmph.

Distance covered by goods train in 10 hours= Distance covered by express train in 4 hours

10x = 4 x 90 or x =36.

So, speed of goods train = 36kmph.

Ex. 12. A thief is spotted by a policeman from a distance of 100 metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 8km/hr and that of the policeman 10 km/hr, how far the thief will have run before he is overtaken?

Sol. Relative speed of the policeman = (10-8) km/hr =2 km/hr.

Time taken by police man to cover 100m       100   x  1  hr = 1  hr.

1000     2         20

In 1  hrs, the thief covers a distance of 8  x  1  km = 2  km  = 400 m

20                                                              20          5

Ex.13. I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways?

Sol. Let the distance be x km. Then,

( Time taken to walk x km) + (time taken to ride x km) =37 min.

( Time taken to walk 2x km ) + ( time taken to ride 2x km )= 74 min.

But, the time taken to walk 2x km = 55 min.

Time taken to ride 2x km = (74-55)min =19 min.