Important Points and Formulas :1. Speed : The distance is covered in a unit time,

is called speed.

(i) Speed = Distance

Time (ii) Time =

Distance

Speed(iii) Distance = Speed x time2. Relative Speed : The distance changed in unit

time between two bodies, is called relative speed.(i) Two bodies which are going in the same direction,

then their relative speed is equal to the differenceof their speeds.

(ii) Two bodies are going in the opposite direction toeach other, then their relative speed is equal tothe sum of their speeds.

3. To Change the Unit of Speed :

(i) x kmph = x 5 x

18m/sec.

(ii) x meters/sec = x 18 x

5 km hr..

4. (i) Average speed =Total distance travelled

Total time taken

(ii) If a man covers a certain distance with xkm/h and he also covers the same distance withy km/h, then : the average speed for whole

journey = x

x2 y

+ ykm h

5. If the speeds of A and B are in the ratio of a : b,then the ratio of times in which they covers thesame distance is b : a

6. If a man covers same distance with differentspeeds in different times, then ; Required distance

=Product of speeds

x Difference of timesDifference of speeds

7. If a man goes to one place to another place at aspeed of x km/h and return back at a speed of ykm/h. If he takes h hours in whole journey, then :Distance between two places

=Product of speeds

x Total TimeSum of speeds

orx

xy

x h + y

,

Note : If two equal distances are coveredconsequently, then the required distance will be2 times of the above.

8. A man runs to a place after some time an anotherman chase him from the same place, then :

(i) Distance covered by 2nd man to catch 1st man

=Product of speeds

x Time IntervalDifference of speeds

(ii) Time taken by 2nd man to catch 1st man

=Speeds of 1 st man

x Time IntervalDifference of speeds

9. If two men run towards each other from twoplaces, then theTime taken to meet each other =

Distance between them

Sum of speeds

orDifference of covered distances by them

Difference of speeds

10. A man covered a distance at the speed of V1 inT1 time and the speeds V2 in T2 time. Then :V1 T1= V2 T2

11. Two men run toward each other and take thetime t1 and t2 to change their places by crossingeach other, then :

(speed of 1st) : (speed of 2nd) = 2t : 1t ,12. Important facts and formulas for boat :(i) If the boat is going in the same direction of the

current, then: the total speed of the boat= the speed of the boat in the still (standing)water + speed of the current.

(ii) If the boat is going in the opposite direction ofthe current, then :the total speed ot the board= the speed of the boat in still (standing)water - speed of the current.

(iii) If the speed of the boat in the direction of thecurrent is x km/h and in the opposite direction ofthe current is y km/h, then:(a) Speed of the boat in still water = ½ (x +y)km/h

TIME AND DISTANCE DAY - 16

(b) Speed of the current = ½ (x - y) km/h(iv) The speed of a boat in still water is u km/h and

the speed of the current is v km/h. If the boattakes t hours in up and down from a place then :

the distance of the place =u v

u

2 2-x t

2(v) A boat covered a distance in t1 time in the

direction of the current and takes t2 time forcoming back. If the speed of the current is v,

then: the speed of boat in still water = 2 1

2 1

t + tx

t - tv

13. Important facts and formulas for train :(i) The time taken by a x metre long train to cross a

standing man, a pole or a point is equal to thetime in which the train covers x metre longdistance by its speed.

(ii) The time taken by a x metre long train to pass ay metre long plateform, bridge, tunnel or anotherstanding train is equal to the time in which thetrain covers x + y metre long distance by itsspeed.

(iii) If two trains of lengths x km and y km are movingin the same direction at the speed of u kmphand v kmph, where u > v, then : the time takenby faster train to cross the slower train

=x

+ y

- u vhrs.

(iv) If two trains of lengths x km and y km are movingin the opposite directions at the speed of u kmphand v kmph, then : the time taken by the trains

to cross each other = x

+ y

+ u vhrs

(v) If a man and a train are going in the samedirection. Then:The time taken by the train to cross the man= Length of the train The difference betweenthe speeds of train and man.

(vi) If a train and a man are going the oppositedirections. Then:the time taken by the train to cross the man= length of the train The sum of the speedsthe train and man.

SOLVED EXAMPLESEx.1. A 80 m long train passed a 120 m long

platform in 20 seconds. Find the speed ofthe train per hour.

Sol. The total distance which the train has to pass= 80 + 120 = 200 mTime = 20 seconds, Speed = Distance time

= 200 20

1000 3600=

200 3600 x

1000 20= 36kmph. Ans.

Note: In mathematical calculation the speed iscalculated in kilometre per hour. Therefore themetres and seconds are made kilometres andhours respectively.

Ex.2. 160 m long train passed a 440 m longbridge at the speed of 90 km per hour.Find the time taken by the train in passingthe bridge.

Sol. Total distance = 440 + 160 = 600 m.

Speed = 90 km/h = 5

90 x18

= 25 m/sec.

Hence, time = distance speed= 600 25 = 24 sec. Ans.

Ex.3. A train with the speed of 63 km/h passesa pole in 8 sec. Find the length of the train.

Sol. The length of the train is equal to the distancecovered by the train to pass the pole.

The length of the train = Speed x Time

= 5

63 x x 818

= 140 mt. Ans.

Ex.4. If a train 110 m long passes a telephonepole in 3 sec., then find the time to crossa railway plateform 165 m long.

Sol. Now, 110m length is covered by train in = 3sec.

Hence, 275 m (110 + 165 effective distance) lengthis covered by train in

=3

x 275110

= 7.5 sec. Ans.

Ex.5. A person is standing on a railway bridgewhich is 50 metres long. He finds that atrain crosses the bridge in 4½ secondsbut himself in 2 seconds. Find the lengthof the train and its speed in km per hour.

Sol. For crossing the personThe train moves the distance equal to its lengthin = 2 seconds.For crossing the bridge.The train moves the distance equal to its length+ length of the bridge in = 4½ seconds.

So, the train moves the distance equal to the lengthof the bridge

i.e., 50 metres in = 4.5 - 2 = 2.5 seconds.

So, the train will move in 1 hour

=50 x 2 x 3600

1000 x 5 = 72km

Now, if the train moved in 2.5 seconds = 50 m.Then the train will move in 2 seconds

=50 x 2 x 2

5= 40 metree

Hence, the length of the train = 40m and the speedof the train = 72 km. Ans.

Ex.6. A thief covered a distance of 300m, thena Sepoy started chasing him. If the speedof the Sepoy is 100 m/minute and of thethief is 90m/minute, then in how muchtime and the distance the Sepoy will catchthe thief ?

Sol. Speed of the theif = 90 m/minute,Speed of the Sepoy = 100 m/minute.

Relative speed = 100 - 90 = 10 m/minute.Now, the required time

=Distance covered by the thief

Relative speed =

300

10

= 30 Minutesand, the required distance = (Speed of the Sepoy)

x 30 minutes.= 100 x 30 = 3000 m = 3 km.

Hence, the sepoy will catch the thief in 30 minutes ata distance of 3 km. Ans.

Ex.7. A train 125 m long overtakes a manwalking at the rate of 4 km/h parallel tothe line in the same direction and passedhim completely in 9 sec. Find the speedof the train.

Sol. Relative speed = 125

9m/sec.

=125 18

x 9 5

= 50 km/hr

Hence, the speed of train = 50 + 4 = 54 km/hr Ans.Ex.8. A boat goes 35 km in 5 hours along the

river and comes back at the starting placein 7 hours. Find the speedof the boat instill water.

Sol. Speed of the boat along the river =35

5= 7 km/h

Speed of the boat opposite of the river = 35

7

= 5 km/hNow, Speed of the boat in still water + speed of the

current =7Speed of the boat in still water - speed of thecurrent = 5

or 2 x speed of the boat in still water = 7 + 5 = Speed of the boat in still water

=12

2= 6 km/h Ans.

Ex.9. Rajesh covered a distance of 100 kms.from his village to Rewari at an averagespeed of 40 kms/h on motorcycle. Duringthe return journey he maintained anaverage speed of 60 kms/h. Find hisaverage speed per hour for the whole trip.

Sol. For quick solution :The average speed

=x

x2 y

+ y =

2 x 40 x 60

40 + 60 =

4800

100 = 48 km/h.

Ex.10. If a man walks at the rate of 5 kmph, hemisses a train by 7 minutes. However, ifhe walks at the rate of 6 kmph, he reachesthe station 5 minutes before the arrivalof the train. Find the distance covered byhim to reach the station.

Sol. Let the required distance be x km.Difference in the times taken at two speeds =

12 min = 1

5hr..

x x

5 6=

1

56x - 5x = 6 x = 6.

MULTIPLE CHOICE QUESTIONS1. A man covers a certain distance by car driving at

40 km/hr and he returns back to the starting pointriding on a scooter at 10km/hr. Find his averagespeed for the whole journey.(a) 8 km/hr (b) 16 km/hr(c) 12 km/hr (d) None of these

2. A motor car does a journey in 6 hrs, the first halfat 10 km/hr and the second half at 20 km/hr.Find the distance.(a) 90 km (b) 80 km(c) 60 km (d) None of these

3. A man covers a certain distance between hishouse and office on scooter. Having an averagespeed of 60 km/hr. he is late by 20 min. However,with a speed of 80 km/hr, he reaches his office

10 min earlier. Find the distance between hishouse and office.(a) 120 km (b) 90 km (c) 80 km (d) 60km

4. A start from Allahabad to Kanpur and walks atthe rate of 12 km an hour. B starts from Kanpur 2hours later and walks towards Allahabad at therate of 8 kilometres an hour, if they meet in 9hours after B started, find the distance fromAllahabad to Kanpur.(a) 204 km (b) 104 km(c) 140 km (d) 240 km

5. Walking 2/3 of his usual speed, a person is 15min late to his office. Find his usual time to coverthe distance.(a) 30 minutes (b) 25 minutes(c) 15 minutes (d) None of these

6. A man sets out to cycle from Delhi to Rohtak,and at the same time another man starts fromRohtak to cycle to Delhi; After passing eachother they complete their journeys in 16 and 25hours respectively. At what rate does/thesecond man cycle if the first cycle at 25 km perhour?(a) 21 km (b) 18 km(c) 12½km (d) 20 km

7. A carriage driving in a fog passed a man whowas walking at the rate of 6 km an hour in thesame direction. He could see the carriage for 8minutes and it was visible to him upto a distanceof 200 m. What was the speed of the carriage?(a) 9 km/hr (b) 7½ km/hr(c) 7 km/hr (d) 8½ km/hr

8. Without any stoppage a person travels a certaindistance at an average speed of 42 km/hr, andwith stoppages he covers the same distance atan average speed of 28 km/hr. How manyminutes per hours does he stop?(a) 15 minutes (b) 14 minutes(c) 28 minutes (d) 20 minutes

9. One aeroplane started I hour later than thescheduled time from a place 3000 km away fromits destination. To reach the destination at thescheduled time the pilot had to increase the speedby 500 km/hr. What was the speed of theaeroplane per hour during the journey?(a) 1500 km/hr (b) 1000 km/hr(c) 850 km/hr (d) None of these

10. One aeroplane staited 1½ hrs later than thescheduled time from a place 2400 km away fromits destination. To reach the destination at thescheduled, time the pilot had to increase the

speed by 800 km hr. What was the speed ofthe aeroplane per hour during the journey?(a) 1600 km/hr (b) 800 km/hr(c) 1200 km/hr (d) 1550 km/hr

11. Two bullets were fired at a place at an intervalof 38 minutes. A person approaching the firingpoint in his car hears the two sounds al an intervalof 36 minutes. The speed of sound is 330 in/sec. What is the speed of the car?(a) 66 km (b) 49 km(c) 99 km (d) 98 km

12. A policeman goes after a thief who has 100metres start. If the policeman runs a kilometerin 8 minutes, and the thief a kilometer in 10minutes, how far will the thief have gone beforehe is overtaken?(a) 350 metres (b) 400 metres(c) 450 metres (d) 460 metres

13. A thief steals a motor car at 1 PM and drives it at45 km an hour. The theft is discovered at 2 PMand the owner sets off in another car at 54 kman hour. When will he overtake the thief?(a) 5 PM (b) 3PM(c) 7 PM (d) Can't be determined

14. A train leaves Calcutta at 7.30 am and travels 40km an hour, another train leaves Calcutta at noonand travels 64 km an hour, when and where willthe second train overtake the first?(a) 480 km,7.30 pin (b) 480km, 2.30 pm(c) 840 km, 7.30 pm (d) 480 km, 6.30 pm

15. Two men A and B walk from P to Q, a distanceof 18 km, at 4 and 5 km an hour respectively. Breaches Q, returns immediately and meets A atR. Find the distance from P to R.(a) 15 km(b) 16 km(c) 12 km(d) Cant be determined

TRAINS

16. A train 110 m in length runs through a station atthe rate of 36 km per hour. How long will it taketo pass a given point?(a) 11 sec (b) 12 sec(c) 13 sec (d) 15 sec

17. A train crosses a platform in 60 seconds at aspeed of 45 km/hr. How much time will it taketo cross an electric pole if the length of theplatform is 100 metres?(a) 8 seconds (b) 1 minute(c) 52 seconds (d) None of these

18. Find the length of a bridge, which a train 130mlong, travelling at 45 km an hour, can cross in 30secs.(a) 240m (b) 235 m (c) 250 m (d) 215 m

19. Two trains 70 m and 80 m long respectively, runat the rates of 68 and 40 km an hour respectivelyon parallel rails in opposite directions. How longdo they take to pass each other?(a) 5 seconds (b) 10 seconds(c) 12 seconds (d) 16 seconds

20. A train 135 metres long is running with a speedof 49 km per hour. In what time will it pass aman who is walking at 5 km'hr in the directionopposite to that of ihe train?(a) 9 sec (b) 12 sec (c) 15 sec (d) 18 sec

21. Two trains 127 metres and 113 metres in lengthrespectively are running in opposite direction, oneal the rate of 46 knvhr and another at the rate 26km per hour. In what lime will they be clear ofeach other from the moment they meet?(a) 17 sec (b) 12 sec(c) 14 sec (d) None of these

22. Two trains of length 110 metres and 90 metresare running on parallel lines in the same directionwith a speed of 35 km per hour and 40 km perhour respectively. In what time will they passeach other.(a) 144 sec (b) 140 sec(c) 134 sec (d) 154 sec

23. A man sitting in the train which is travelling atthe rate of 50 km per hour observes that it takes9 seconds for a goods train travelling in theopposite direction to pass him. If the goods trainis 187.5 metres long, find its speed.(a) 25 km/hr (b) 40km/hr(c) 35 km/hr (d) 36 km/hr

24. Two trains of the same length but with differentspeeds-pass a static pole in 5 seconds and 6seconds respectively. In what time will they crosseach other when they are moving in the samedirection.(a) 1 hr (b) 50 sec (c) 1 min (d) 60 min

25. Two trains of the same length but with different

speeds pass a static pole in 12 seconds and 18seconds respectively. In what time will they crosseach other when they are moving in theoppositedirection.(a) 14 sec(b) 13.4 sec (c) 14.4 sec(d)15 sec

26. The rowing speed of man in still water is 20 km/hr. Going downstream, it moves at the rate of 25km/ hr. The rate of stream is(a) 45 km/hr (b) 2.5 km/hr(c) 12.5 km/hr (d) 5 km/hr

27. A boat's man goes 48 km downstream in 8 hoursand returns back in 12 hours, find the speed ofthe boat in still water and the rate of the stream.(a) 5km/hr, 1km/hr (b) 10 km/hr, 2km/hr(c) 6km/hr, 1.5km/hr (d) None of these

28. A motor boat can travel at 10 km hr in still water.It travelled 91 km downstream in a river and thenreturned, taking altogether 20 hours. Find therate of flow of river.(a) 6km/hr (b) 2 km/hr(c) 3 km/hr (d) 4 km/hr

29. Ajay can row a certain distance downstream in5 hours and return the same distance in 7 hours.If the stream flows at the rate of 2 km per hourfind the speed of Ajay in still water.(a) 12 km/hr (b) 10 km/hr(c) 18 km/hr (d) 16 km/hr

30. Rohit can row a certain distance downstream in8 hours and return the same distance in 12hours. If the stream flows at the rate of 5 kmper hour find the speed of Rohit in still water.(a) 20 km/hr (b) 30 km/hr(c) 15 km/hr (d) 25 km/hr

ANSWERS1. (b) 2. (b) 3. (a) 4. (a) 5. (a)6. (d) 7. (b) 8. (d) 9. (b) 10. (b)

11. (a) 12. (b) 13. (c) 14. (a) 15. (b)16. (a) 17. (c) 18. (d) 19. (a) 20. (a)21. (b) 22. (a) 23. (a) 24. (c) 25. (c)26. (d) 27. (a) 28. (c) 29. (a) 30. (d)

HINTS AND EXPLANATIONS1.b) If two equal distances are concerned at two

different speeds S1 & S2 Avg/Speeds = .1 2+1 2

2S 2S

S S

Avg. speed = 2 x 40 x 10

40 +10

= 2 x 400

50= 16 km/h.

2.b)Here also,

Avg.speed = 2 x 10 x 20

30=

40

3 km

Distance = T x S = 40

3x 6 = 80 km.

3.a) Let man office distance is x kmwhen he has avg. speed of 60 km/h he is lateby 20 min.when he has avg.speed of 80 km/h he is earlierby 10 min

i.e., time gap is 30 min = 30

60hrs

T1 = x

60, T2 =

x80

According to question,

x x

60 80=

30

60

x x4 3

240=

1

2

x = 240

=1202

km.

4.a)

A B12 km/hr 8 km/hr

(11) (9)Since the two person meet in 9 hrs. after thestart of BSo, ‘A’ would have run for 11 hrs.Distance covered by A = 12 x 11 = 132 kmDistance coverd by B = 8 x 9 = 72Total distance = 132 + 72 = 204 km.

5.a) Here

Speed S2

3s

or 3s 2sTime 2s 3sNow difference in timing is 1Part = 15 min Usual Time Taken is 2Part = 2 x15

= 30 mins

6.d) If two bodies A and B starts from P and Qtowards each other and after crossing each otherat R they take x hrs & y hrs respectively to reach

Q and P, then, Ratio of their speeds is A

B

=S yS x

Ratio speed of Delhi

Ratio speed from Rohtak= D

R

t 25 5=

4 t 16

5 part = 251 part = 54 part = 20 km/h.

7.b) Carriage covered extra 200m over man in 8m

Carriage over extra 60 min is = ×200

608

= 1500m or 1.5 kmCarriage could travel 1.5 km more than man in1hrSpeed of man = 6 km/hSpeed of Carriage = 6+1.5 = 7.5 km/hSpeed of man = 8 km/hSpeed of Carriage = 8+2 = 10 km/h

8.d) With out stoppage bus covers 42 km in 60 min

With stoppage bus covers 2.8 km in 60 minIn 2nd Case bus stop for the time in which itcould cover 14 kmNow, 42 km is cover in 60 min

14 km is covered in 60

1442

= 20 min.

Or

Difference in speed× 60

Actual speed

Difference in speed = 42-28 =14

14×60

42 = 20 min.

9.b) A B3000 kmSpeed S1 = S km/hSpeed S2 = S + 500 km/h

t1 =3000

S, t2 =

3000

S + 500According to Questiont1 - t2 = 1

3000

S-

3000

S + 500= 1

23000 (S+500) - 3000S

S +500S=1

3000S + 3000x500-3000S = S2 + 500S S2 + 500S 1500000 = 0 S2 + 1500S - 1000S - 1500000 = 0 S (S+ 15000) - 1000(S+1500)= 0 (S-1000) (S+1500) =0S = 1000, S = -1500Since Speedcan’t be -veS = 1000orAccording to question we will divie it into twocolumns Delay in time & Increase in speed thenwe write the figure in increasing order then we,multiply in cross taking 2 at a time and stop if weget the multiplication equal to the given distance

Delay in Time Increase in Speed

1 Hr

2 Hr

3 Hr

4 Hr

500

1000 - S1

1500 - S2

2000

3000t1t2

Using the shorter method to solve as discussedin the above question

10.b) Using the short cut method to solve as discusedin the above question

Delay in Time Increase in Speed

1½

3

800 - S1

1 00 - S26

t1t2

= 2400

Original speed = 800 km/h

11.a) According to the method discussed abovedistance covered by man in 36mins is covered bysound in 2mins.

Men SoundTime taken by 36 2Speed 2 36

36 part = 330 m/s

1 part =330

36m/s

1 part =330

36x

1833 km/h

52 part = 66 km/h

12.b) Policeman ThiefTime 8 10Speed 5 4Difference 1 part = 100m

1 parts = 1004 part = 400 m

13.c) Thief OwnerSpeed 45 54Time 54 45

6 51 part = 1 hour

6 parts = 6 hourTime of thief = 6 hrsSo thief is caught at 1 + 6 = 7pm

14.a) Let the two train meets ‘t’ hrs after 12 noon Distance travelled by 2nd train = 64 tAs the 1st train run for 4:30 mins more when thetwo

Distance travelled by 1st train =

940 t +

2

Distance covered by same in both the cases

64t =

940 t +

2

64t = 9

40t + 40 x2

24t = 180

t = 180

24=

30

40= 7½

time = 12 + hr = 7:30 PM

Distance covered = 30

64 x4

= 480 km.

15.b) Since, when A + B meets at R they were inopposite directionSo, we take R.Speed = 4 + 5 = 9Distance taken = 2 x 18 = 36

Time taken (t) = 36

9 = 4km/h

Now, distance from P to R is the distance coveredby A = 4 x 4 =16 km/h.

16.a) Distance = 110 m

Speed = 36 km/h = 36 x5

10=10 m/sec

T = ?

Time =Dist.

Speed

= 11 sec17.c) D = Train + Platform or T + 100

T = 60 sec

S = 45 k/h or 45 x5

18S = D x T

60 x 45 x 5

18= T + 100

T = 750 - 100 = 650mTrain Length = 650 m

S = 45 x 5

18 =

Time =650 x 18

45 x 5= 52 sec.

18.d) Distance = 130 + platform

Speed = 45 km/h or 45 x 5

18 =

25

2T = 30 secD = T x S

130 + P = 30 x 25

2P = 375 - 130P = 245

19.a) Distance = (70 + 80) = 150 mSince bodies are moving in opp. direction

R.Speed = (68 + 40) =108 km/h or 108 x 5

18ms

Time = 150

30 = 5 sec

20.a) Dist. = 135 mR.S = 49 + 5 = 54

Speed = 54 km/h or 54 x 5

18 = 15

Time = 1 3 5

1 5 = 9 sec.

21.b) Total Length = (127 + 113) = 240 metersR. Speed = (46 + 26) = 72 km/h

or 72 x 5

18= 20 m/s

Time = 240

20 = 12 sec.

22.a) Distance 110 + 90 = 200R. Speed = 40 - 35 = 5 km/h

Speed = 5 x 5

18

Time = 200

55 x

18

= 144 sec

23.a) Distance = 187.5Speed = (50 + x)km/h or 50 + x )

x5

18m

Time =Dist.

Speed

x

187518

10(50+ ) 5 = 9

1875

25= 50 + x

x = 75 - 50x = 25 km/h

24.c) Let length of the train be xmdist1 = x

time1 = 5 sec

speed1 =x

5dist2 = x

time2 = x

speed2 = x

6Since, the two bodies are moving in the samedirection dist. = x + x = 2x

speed = x x

-5 6

= x

30

time = 2x

30x

= 60 sec or 1 min.

25.c) Case IDistance = x

Let speed = x

12case II

Dist. = x

speed = x

18Since, two bodies are moving in opposite direction,

dist. = x + x

speed = x x 3x + 2x 5x

+ = =12 18 18 36

time = 2x 72

36 =5x 5

or 14.4 sec.

26.d) Speed in still water = 20 km/hGoing down stream = 25 km/hRate of stream = (25 - 20) = 5 km/h

27.a) Distance = 48 km/hspeed downstream = (x+y) km/htime = 8 h

D = 48

6 = 6 km/h

Distance2= 48time2 = 12 h

x - y = 48

12= 4km/h

x = 6 + 4

2, y =

6 - 4

2x = 5, y = 1

28.c) x = 10 km/hDown stream = 91

91 91+

10 + y 10 - y=20 --(i)

When ever there is quadiatic in question not

solution the zn put the otic in it to get right ans.put = y = 3will satisfy the above = ny = 3

29.a) dist. = 5x = ?y = 2U/S = (x - 2)D/S= (x + 2)Down stream speed = 5 km/hUp stream time = 7 km/hD.S. Dist. = (x + 2)5Dist. = 7(x - 2)Since Distance is same in both the cases5x + 10 = 7x -142x = 24x = 12 km/h

30.d) x = ?y = 5(x + 5) 8 = (x - 5) 128x + 40 = 12x -6012x - 8x = 60 + 404x = 100x = 25 km/h.