Speed, Time and Distance

Chapter-26

Speed, Time and Distance

SpeedIf the location of an object changes with time thenit is said to be in motion. A bus running on a road,an ant crawling on a floor, a monkey climbing upa greased pole etc all are the examples of objectsin motion because the locations of these objectskeep on changing with respect to theirsurroundings.

Let an object move from a point A to the point Bthrough any path, then the actual length of thepath followed by the object is called the distancetravelled by the object. The rate at which anymoving body covers a particular distance is calledits speed.

Speed = taken Timetravelled Distance

If the distance is constant, Speed Time1

Time = SpeedDistance

If the time is constant, Distance SpeedDistance = Time Speed.If the speed is constant, Distance TimeWe can say that for constant distance travelled,

speed is inversely proportional to the time taken.This can be explained by a simple example. Tocover a distance of 100 kms, if a person goes atthe speed of 25 kmph, he will require 4 hours tocomplete the journey and travelling at a speed of50 kmph, 2 hours will be required.

(a) Uniform SpeedIf the object covers equal distance in equal timeintervals, howsoever small the interval may be thenits speed is called the uniform speed.

(b) Variable SpeedIf the object travels different distances in equalintervals of time, then its speed is called a variablespeed. In this case the speed changes frominstance to instance.

The Relative Speed

(i) Objects are moving in opposite directionsThe relative speed of one object with respect tothe other, will have magnitude greater thanindividual speed of each object. This is why, forexample, a train A moving with speed 10 km/hrwill cross another train B moving in oppositedirections with speed 25 km/hr, with a relativespeed of (10 + 25 =) 35 km/hr which is greaterthan the individual speed of either train.

(ii) Objects are moving in same directionThe relative speed of one object with respect to theother, will have magnitude either less than or greaterthan individual speed of each object. This is why,for example, a train A moving with the speed of 20km/hr will cross the another train B moving insame direction with the speed of 15 km/hr, with arelative speed of (20 15 =) 5 km/hr which is lessthan the individual speed of either train.

Take another example, a train A moving withthe speed of 60 km/hr will cross the another trainB moving in same direction with the speed of20 km/hr with a relative speed of (60 - 20 =)40 km/hr which is less than the train A andgreater than the train B.

Average Speed

Average Speed = Taken TimeTotalCovered Distance Total

For example, A person divides his total routeof journey into three equal parts and decides totravel the three parts with speed of 40, 30 and15 km/hr respectively. Find his average speedduring the whole journey.

Let the three equal parts of journey be x km.Time taken to travel first part of the journey

=

40x

hour

Time taken to travel second part of the journey

=

30x

hour

450 Concept of Arithmetic

KKUNDAN

An Important ResultIf two persons (or trains) A and B start at the sametime in opposite directions from two points, andarrive at the point in a and b hours respectivelyafter having met, then

As rate : Bs rate = ab : .Proof: Suppose A starts from P and B starts from

Q and they meet at R.

Let As rate be x km per hour, and Bs ratebe y km per hour.QR = ax km, PR = by km.Now time taken by A in travelling the

distance by km is xby

hours.

And time taken by B in travelling the

distance ax km is yax

hours.

Since both start at the same time and meet,those two times must be equal. Hence

xby

yax

ab

yx

2

2

ab

yx

For example, A man sets out to cycle fromBhiwani to Rohtak, and at the same time anotherman starts from Rohtak to cycle to Bhiwani. Afterpassing each other they complete their journeys

in 313 and 5

44 hours respectively. At what rate

does the second man cycles if the first cycles 8km per hour?

We proceed as follows:

56

313

544

rate sman' Secondrate sman' First

or 56

rate sman' Second8

second mans rate = 865 = 3

26 km per hour.

Time taken to travel third part of the journey

=

15x

hour

Total time taken =

153040xxx

hours

Total distance travelled = x + x + x = 3x km

Average Speed = Taken TimeTotal TravelledDistance Total

=

153040xxx

x km/hr

= 24 km/hr

Units of MeasurementTime is measured in seconds (s), minutes (min) orhours (hr)

Distance is usually measured in metres (m),kilometres (km), miles, yards or feet.

Speed is usually measured in metres per second(mps), kilometres per hour (kmph orkm/hr) or miles per hour (mph).

Conversion of units1 hour = 60 minutes = (60 60) seconds.1 kilometre = 1000 metres8 kilometres = 5 miles1 yard = 3 feet

1 kilometre per hour = hour 1kilometre 1

= 185

seconds 60)(60metres 1000

metre per second

18 kilometres per hour = 5 metres per second[To convert kilometres per hour into metresper second we multiply the given speed in

kilometres per hour with 185

. And to convert

metres per second into kilometres per hourwe multiply the given speed in metres per

second with 518

.]

1 kilometre per hour = 85

mile per hour.

1 mile per hour = 1522

feet per second.

451Speed, Time and Distance

KKUNDAN

Ex. 1: A man walks 22.5 km in 5 hours. Howmuch he will walk in 4 hours?

Soln: We have, distance covered = 22.5 km,time taken = 5 hours.

Speed = TimeDistance

or, Speed =

5

5.22 km/hr = 4.5 km/hr

Now, Distance = Speed Timeor, Distance covered in 4 hours

= (4.5 4) km = 18 km.Hence, the man will walk 18 km in 4hours.

Ex. 2: A car travels at the speed of 72 km/hr.How many metres will it travel in onesecond?

Soln: We have, speed of the car = 72 km/hr

Time = 1 sec = 601

minute

minute601sec1

1minuteseconds 60

=

60601

hour

hour6060

1minute601

hour601

minute1hour1minutes60

Now, Distance = Speed Time Distance covered in one second

= Speed Time

=

3600

172 km

=

10003600

172 m

[ 1 km = 1000 m]

=

1036172 m = 20 m.

Thus, the car will travel 20 metres in 1sec.

Ex. 3: The speed of a goods train is 4 m/sec.What is its speed in km/hr?

Soln: In order to find the speed of the goodstrain in km/hr, we have to find thedistance travelled (in km) by it in one hour.We have,Speed = 4 m/sec and Time = 1 hour.

Now, 1 hour = 60 minutes = (60 60) seconds = 3600 seconds

Distance covered by the goods train inone hour = (Speed Time)

= (4 3600) metres

=

100036004 km

= 572

km = 14.4 km

Thus, the distance covered by the goodstrain in 1 hour = 14.4 km.Hence, the speed of the train is 14.4km/hr.

Ex. 4: A man travels some distance at a speedof 12 km/hr and returns at 9 km/hr. Ifthe total time taken by him is 2 hours20 minutes, find the distance.

Soln: Let the distance be x km. Then,

time taken at 12 km/hr = 12x

hour

=

6012x

minutes = 5x minutes.

Time taken at 9 km/hr = 9x

hour

=

609x

minutes = 320x

minutes.

Since the total time taken is 2 hours 20minutes ie 140 minutes.Now, according to the question,

1403205 xx

or, 14032015

xx

or, 140335

x

or, 35x = 3 140

or, x = 354103

x = 12Hence, the distance is 12 km.

Ex. 5: Walking at 4 km/hr, a person reacheshis office 5 minutes late. If he walksat 5 km/hr, he will be 4 minutes tooearly. Find the distance of his officefrom his residence.

Solved Examples


KKUNDAN

Soln: Let the required distance be x km.


hours

=

604x

minutes = 15x minutes


hours

=

605x


Since the difference between the twotimes taken is (5 + 4) = 9 minutes. 15x - 12x = 9or, 3x = 9

or, x = 39

= 3.

Hence, the required distance is 3 km.Alternative Method I:Let the distance to the office be 1 km.Then, time taken to cover 1 km at the rateof 4 km/hr

= 41

SpeedDistance

hour

=

6041

= 15 minutes

Time taken to cover 1 km at the rate of 5km/hr

= 51

SpeedDistance

hour

=

6051

= 12 minutes

Difference in time taken= (15 12) = 3 minutes

But the actual difference in time= (5 + 4) = 9 minutes

Thus, when the difference in time takenis 3 minutes, the distance to office = 1 km If the difference in time is 9 minutes,

the distance to office =

931

= 3 km

Hence the distance to office = 3 km.Alternative Method II:Let x km be the distance of office of theperson and t hours be the time requiredto reach the office by the person.When the person walks at 4 km/hr, then

605

4 tx

or, 121

4 tx ....(i)

When the person walks at 5 km/hr, then

604

5 tx

or, 151

5 tx ....(ii)

Now, subtracting equation (ii ) fromequation (i), we have,

54xx

=

151

121 tt

= 151

121

= 6045

= 609

or, 609

2045

xx

or, 609

20

x

x = 60209

= 3

Hence, the required distance is 3 km.Ex. 6: A gun is fired at a distance of 3.32 km

away from Rohit. He hears the sound10 seconds later. Find the speed of thesound.

Soln: Since Rohit is at a distance of 3.32 kmfrom the gun and he hears the sound 10seconds later. This means that in 10seconds sound covers a distance of 3.32km. Distance covered = 3.32 km

= (3.32 1000) metres= 3320 metres

and, time taken = 10 seconds.

Now, Speed = TimeDistance

Speed = 103320

m/sec = 332 m/sec.

Ex. 7: A man travels a distance of 18 km fromhis house to an exhibition by tanga at15 km/hr and returns back on cycle at10 km/hr. Find his average speed forthe whole journey.

Soln: Time taken by a man to reach exhibitionfrom his house

= SpeedDistance

= 1518

hours = 56

hours

Time taken by a man to reach his housefrom the exhibition

= SpeedDistance

= 1018

hours = 59

hours


KKUNDAN

Total time of journey

=

59

56

hours = 515

hours = 3 hours.

Total distance covered= (18 + 18) km = 36 km

Average speed for the whole journey

= TakenTime TotalCovered Distance Total

= 336

km/hr = 12 km/hr

Ex. 8: Two men A and B start from a place P

walking at 3 km and 213 km an hour

respectively. How many km will they

be apart at the end of 212 hours,

(i ) if they walk in oppositedirections?

(i i) if they walk in the same direction?What time will they take to be 16km apart,

(i ii ) if they walk in the oppositedirections?

(iv) if they walk in the same direction?Soln: (i) When they walk in opposite directions,

Their relative velocity will be

216

2133

km

ie, They will be 216 km apart in one hour.

in 212 hours they will be

4116

212

216

km apart.

(ii) When they walk in same direction,Their relative velocity will be

213

213

km

ie, They will be 21

km apart in one hour.

in 212 hours they will be 4

11212

21

km apart. (iii) From (i), discussed above,

They will be 216

2133

km apart in

one hour.

They are 16 km apart in

1362

1332

216

16 hours.

(iv) From (ii), discussed above,

They are 213

213

km apart in 1 hour.

They are 16 km apart in

21

16 = 32 hours.

Ex. 9: A train travelling 25 km an hour leavesDelhi at 9 am and another traintravelling 35 km an hour starts at 2pm in the same direction. How manykm from Delhi will they be together?

Soln: Let the required distance be x km.A train leaves Delhi at 9 am and anothertrain leaves Delhi at 2 pm in the samedirection.ie difference of time = 5 hours.Distance travelled by the first train in 5hours = (25 5) = 125 kmNow, according to the question,Time taken by the train to cover (x 125)km is equal to the time taken by the secondtrain to cover the distance of x km.

25125

35

xx

or, 25x = 35x 35 125

x = 21437

2875

1012535

The distance from Delhi after which

they will be together = 21437 km.

Alternative Method :The first train has a start of 25 5 kmand the second train gains (35 25) or 10km per hour. the second train will gain 25 5 km in

10525

or 2112 hours.

the required distance from Delhi

= 352112 = 2

1437 km.

Ex. 10: Walking 43

of his usual speed, a person

is 211 hours late to his office. Find his

usual time to cover the distance.Soln: It is easy to see that if the speed of a train

or man be changed in the ratio of a : b,


KKUNDAN

then the time required to travel a certaindistance will be changed in the ratio ofb : a.

Since the man walks at 43

of his usual

rate, the time that he takes is 34

of his

usual time.

34

of usual time = usual time + 211 hours.

31

of usual time = 211 hours.

usual time = 3211 hours = 2

14 hours.

Alternative Method :Let the usual speed of the person be x km/hr and the distance of his office = D km.His usual time to cover the distance

=

xD

hours

Now, according to the question,

Speed = 43

of his usual speed

=

x43

km/hr

Time taken to cover the distance D km

= xx 3

4D

43D

hours

Again,

211D

34D

xx hours = 2

3 hours

or, 231

34D

x

or, 23

31D

x

or, 29D

x

Usual time = 29

= 214 hours.

Ex. 11: I have to be at a certain place at acertain time and find that I shall be40 minutes too late, if I walk at 3 kman hour and 30 minutes too soon, if Iwalk at 4 km an hour. How far have Ito walk?

Soln: Suppose I have to walk 1 km.

To walk 1 km, I require 31

hour in the

first case, and 41

hour in the second case.

Therefore, I save

41

31

hour in the first

case, and 41

hour in the second case.

Thefore, I save

41

31

hour or 5 minutes

in the second case. But , by the question,I save (40 + 30) or 70 minutes.Hence the required distance

= (70 5 =) 14 km.Alternative Method :Let the required distance be D km.Time taken to cover D km at 3 km/hr

=

3D

hour

Time taken to cover D km at 4 km/hr

=

4D

hour

Total difference in time= 40 minutes late + 30 minutes early

= 70 minutes = 67

6070

hours


67

4D

3D

or, 67

41

31D

or, 67

12D

D = 6127

= 14 km

Hence the required distance = 14 km.Ex. 12: Two men A and B walk from P to Q a

distance of 21 km, at 3 and 4 km anhour respectively. B reaches Q, returnsimmediately and meets A at R. Find thedistance from P to R.

Soln: When B meets A at R, B has walked thedistance PQ + QR and A the distance PR.That is both of them have together walkedtwice the distance from P to Q, ie 42 km.


KKUNDAN

Now the rates of A and B are 3 : 4 andthey have walked 42 km.Hence the distance PR travelled by A

= 73

of 42 km. = 18 km.

Alternative Method I:

Let the required distance be x km.Now, according to the question,A and B both walk for the same distance Distance travelled by B

= (21 + 21 x) = (42 x) km

Time taken by B =

442 x

hours

Distance travelled by A = x km

Time taken by A =

3x

hours

442

3xx

or, 4x = 126 3xor, 7x = 126

or, x = 7126

= 18

required distance = 18 kmAlternative Method II:As speed = 3 kmBs speed = 4 kmLet us consider that A and B meets after thours.Distance covered by A in t hours = 3t kmDistance covered by B in t hours = 4t kmTotal distance covered by A and B

= 3t + 4t = 7t kmBut the total distance covered by A and B istwice the distance between P and Q.So, 7t = 21 2

t = 7212

t = 6 hoursSo, the distance between P and R = Distancetravelled by A = 3 6 = 18 km.

Ex. 13: A and B start at the same time from Pand Q (55 km apart) to Q and P at 3

and 212 km an hour respectively, meet

at R, reach Q and P, return immediatelyand meet again at S. Find the distancefrom R to S.

Soln: When A and B meet at R for the first time,they have together covered the whole

distance PQ (= 55 km) and when they meetat S for the second time, they have togethercovered three times the distance PQ or165 km.

Now PR =

2123

3

of PQ = 5511

23

km

= 30 km.QP + PS is the distance covered by B whenhe meets A for the second time.

QP + PS =

2123

212

of 165 km = 75 km.

PS = 75 - QP = (75 - 55) km = 20 km. SR = PR - PS = (30 - 20) km = 10 km.

Ex. 14: Points A and B are 90 km apart fromeach other on a highway. A car startsfrom A and another from B at the sametime. If they go in the same directionthey meet in 9 hours and if they go in

opposite directions they meet in 79

hours. Find their speeds.Soln: Let X and Y be two cars starting from

points A and B respectively. Let the speedof car X be x km/hr and that of car Y be ykm/hr.Case I: When two cars move in the samedirection:

Suppose two cars meet at point Q. Then,distance travelled by car X = AQ,distance travelled by car Y = BQ.It is given that two cars meet in 9 hours. Distance travelled by car X in 9 hours

= 9x km or AQ = 9x kmDistance travelled by car y in 9 hours

= 9y km BQ = 9y kmClearly, AQ - BQ = AB 9x - 9y = 90

[ AB = 90 km]or, x - y = 10 .... (i)Case II: When two cars move in oppositedirections:Suppose two cars meet at point P. Then,distance travelled by car X = APdistance travelled by car Y = BP

In this case, two cars meet in 79

hours.


KKUNDAN

Distance travelled by car X in

79

hours = x79

km or AP = x79

km

Distance travelled by car Y in 79

hours

= y79

km or BP = y79

km

Clearly, AP + BP = AB

9079

79

yx

or, 90)(79

yx

or, x + y = 70 ... (ii)Solving (i) and (ii), we get

x = 40 and y = 30.Hence, speed of car X is 40 km/hr andspeed of car Y is 30 km/hr.

Ex. 15: Ved travels 600 km to his home partlyby train and partly by car. He takes 8hours if he travels 120 km by train andthe rest by car. He takes 20 minuteslonger if he travels 200 km by trainand the rest by car. Find the speed ofthe train and the car.

Soln: Let the speed of the train be x km/hr andthe speed of the car be y km/hr.Case I: When he travels 120 km by train andthe rest by car:If Ved travels 120 km by train, thendistance covered by car is (600 - 120) km= 480 km.Now, time taken to cover 120 km by train

= x120

hours

speedDistance

Time

Time taken to cover 480 km by car

= y480

hours

It is given that the total time of the journeyis 8 hours.

8480120

yx

or, 86015

8

yx

or, 16015

yx

or, 016015

yx .... (i)

Case II: When he travels 200 km by trainand the rest by car:If Ved travels 200 km by train, thendistance travelled by car is (600 - 200) km= 400 km.Now, time taken to cover 200 km by train

= x200

hours

Time taken to cover 400 km by train

= y400

hours

In this case the total time of journey is 8hours 20 minutes

yx400200

= 8 hours 20 minutes

or, 31

8400200

yx

[ 8 hours 20 minutes

= 60208 hours = 3

18 hours]

or, 325400200

yx

or, 325168

25

yx

or, 31168

yx

or, 14824

yx

or, 014824

yx .... (ii)

Putting ux

1 and vy

1

in (i) and (ii),

we get15u + 60v - 1 = 0 .... (iii)24u + 48v - 1 = 0 .... (iv)

On solving equations (iii) and (iv), we have

u = 601

and v = 801

Now, u = x1

x1

601

x = 60,

and v = y1

y1

801

y = 80.

Hence, speed of train = 60 km/hr andspeed of car = 80 km/hr.


KKUNDAN

Ex. 16: X takes 3 hours more than Y to walk30 km. But, if X doubles his pace, he is

ahead of Y by 211 hours. Find their

speed of walking.Soln: Let the speeds of X and Y be x km/hr and

y km/hr respectively.Then, time taken by X to cover 30 km

= x30

hours

And, time taken by Y to cover 30 km

= y30

hours

By the given conditions, we have

33030

yx

or, 11010

yx .... (i)

If X doubles his pace, then speed of X= 2x km/hr

Time taken by X to cover 30 km

= x230

hours

Time taken by Y to cover 30 km

= y30

hours

According to the given conditions, wehave

21

123030

xy

or, 23

23030

xy

or, 21510

xy

or, 12010

yx .... (ii)

Putting ux

1 and vy

1

, equations (i)

and (ii) become: 10u 10v = 1 .... (iii) 10u + 20v = 1 ... (iv)

Adding (iii) and (iv), we get:

10v 2 = 0 51

v .

Putting 51

v in (iii), we get:

10u 3 = 0 u = 103

Now, u = 103

1031

x

x = 310

and v = 51

511

y y = 5.

Hence, Xs speed = 310

km/hr and

Ys speed = 5 km/hr.Ex. 17: After covering a distance of 30 km with

a uniform speed there is some defectin a train engine and therefore, its

speed is reduced to 54

of its original

speed. Consequently, the train reachesits destination late by 45 minutes.Had it happened after covering 18kilometres more, the train would havereached 9 minutes earlier. Find thespeed of the train and the distance ofjourney.

Soln: Let the original speed of the train be xkm/hr and the distance of the journey be

y km. Then, time taken = xy

hours.

Case I: When defect in the engine occursafter covering 30 km.Speed for first 30 km = x km/hrand, speed for the remaining (y - 30) km

= x54

km/hr

Time taken to cover 30 km = x30

hours

Time taken to cover (y - 30) km

=

54

30x

y hours = x4

5(y - 30) hours

According to the given condition, we have

6045)30(

4530

xyy

xx

or, 43

4150530

xy

xy

x

or, xxy

xy

434

41505120

or, 5y 30 = 4y + 3xor, 3x y + 30 = 0


KKUNDAN

Case II: When defect in the engine occursafter covering 48 km.Speed for first 48 km = x km/hrSpeed for the remaining (y 48) km

= 54x

km/hr

Time taken to cover 48 km = x48

hours

Time taken to cover (y - 48) km

=

54

48x

yhour =

xy4

)48(5 hour

According to the given condition, the trainnow reaches 9 minutes earlier ie it is 36minutes late.

6036

4)48(548

xy

xy

x

or, 53

4240548

xy

xy

x

or, xxy

xy

535

42405192

or, 535

4485 xyy

or, 25y - 240 = 20y + 12xor, 12x - 5y + 240 = 0Thus, we have the following system ofsimultaneous equations:3x - y + 30 = 0 ... (i)12x - 5y + 240 = 0 ... (ii)On solving equations (i) and (ii), we have,

x = 30 and y = 120Hence, the original speed of the train is30 km/hr and the length of the journeyis 120 km.

Ex. 18: A train met with an accident 3 hoursafter starting, which detains it for onehour, after which it proceed at 75% ofits original speed. It arrives at thedestination 4 hours late. Had theaccident taken place 150 km fartheralong the railway line, the train would

have arrived only 213 hours late. Find

the length of the trip and the originalspeed of the train.

Soln: Let the length of the trip be d km and theoriginal speed of the train be x km/hr.As the accident takes place after 3 hours. distance covered in 3 hours by the train= (3 x) = 3x kmRemaining distance = (d - 3x) km

Total time taken by the train if no accident

happens =

xd

hours

Case I:Time taken by the train to cover the wholelength of the trip

=

10075

)3(13x

xd hours

=

x

xd3

)3(44 hours


43

)3(44 xd

xxd

or, xd

xxd

3124

or, 4d 12x = 3dor, d = 12x .... (i)Case II:If the train had covered 150 km morebefore the accident then the distance ofthe accident = (3x + 150) kmRemaining distance

= (d (3x + 15)) kmTime taken to cover the whole length ofthe trip

10075

)1503(1

1503

x

xdx

x


27

43

)1503(1

1503

xd

xxd

xx

or, 127

36001241503

xd

xxd

xx

or, xxx

xx

3600121241503

2512

xx

xd 12

or, 229

3600364509

xxx

or, 229200121503

xxx

or, 229200121503

xxx


KKUNDAN

or, 2295015

xx

or, 30x - 100 = 29xor, x = 100Hence, speed = 100 km/hr and the lengthof the trip (d) = 12x = 12 100 = 1200 km

Ex. 19: A train covered a certain distance at auniform speed. If the train would havebeen 6 km/hr faster, it would havetaken 4 hours less than the scheduledtime. And, if the train were slower by 6km/hr, it would have taken 6 hoursmore than the scheduled time. Find thelength of the journey.

Soln: Let the actual speed of the train bex km/hr and the actual time taken be yhours. Then,Distance = (xy) km ... (i)

[ Distance = speed time]If the speed is increased by 6 km/hr, thentime of journey is reduced by 4 hours iewhen speed is (x + 6) km/hr, t ime ofjourney is (y - 4) hours. Distance = (x + 6) (y - 4) kmor, xy = (x + 6) (y 4) [Using (i)]or, 4x + 6y 24 = 0or, 2x + 3y 12 = 0 ... (ii)When the speed is reduced by 6 km/hr,then the time of journey is increased by 6hours ie when speed is (x - 6) km/hr,time of journey is (y + 6) hours Distance = (x 6) (y + 6)or, xy = (x 6) (y + 6) [Using (i)]or, 6x 6y 36 = 0or, x y 6 = 0 .... (iii)On solving equations (ii) and (iii), we get

x = 30 and y = 24.Putting the values of x and y in (i), weobtainDistance = (30 24) km = 720 km.Hence, the length of the journey is 720km.Alternative Method:Let the original speed and distance be Vkm/hr and D km respectively.Time taken to complete the whole journey

= VD

hours

When the train moves 6 km/hr faster,then

4VD

6VD

or, 4VD

6VD

or, 46)V(V6)VD(V

or, 46)V(V6D

D = 66)4V(V

....(i)

When the person moves 6 km/hr slower,then

6VD

6VD

or, 6VD

6VD

or, 66)V(V6)VD(V

or, D = V(V 6) ....(ii)Combining equations (i) and (ii), we get

6)V(V6

6)4V(V

or, 4V + 24 = 6V 36or, 2V = 60

V = 260

= 30

Hence original speed = 30 km/hrPutting the value of V in equation (ii), wehaveD = 30 (30 6) = (30 24) = 720 km

Ex. 20: A hare sees a dog 100 metres awayfrom her and scuds off in the oppositedirection at a speed of 12 km an hour.A minute later the dog perceives herand gives chase at a speed of 16 kmper hour. How soon will the dogovertake the hare, and at whatdistance from the spot whence the haretook flight?

Soln: Suppose the hare at H sees the dog at D.

DH = 100 metres.Let K be the position of the hare wherethe dog sees her. HK = the distance gone by the hare in 1minute.

= 160100012

m = 200 m.

DK = 100 m + 200 m = 300 mThe hare thus has a start of 300 m.Now the dog gains (16 - 12) or 4 km in anhour.

the dog will gain 300 m in 1000430060

or

214 minutes.


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Again, the distance gone by the hare in

214 minutes.

= 214

60100012

m = 900 m.

distance of the place where the hare iscaught from the spot H whence the haretook flight

= (200 + 900) m = 1100 m.Ex. 21: A hare, pursued by a grey-hound, is 50

of her own leaps ahead of him. Whilethe hare takes 4 leaps the grey-houndtakes 3 leaps. In one leap the hare goes

431 metres and the grey-hound 4

32

metres. In how many leaps will the grey-hound overtake the hare?

Soln: 50 leaps of the hare = 43150 m = 2

175 m

the grey-hound should gain 2175

m over

the hare.Now the grey-hound takes 3 leaps whilstthe hare takes 4 leaps. the grey-hound takes 1 leap whilst the

hare takes 34

leaps.

the grey-hound goes 432 m whilst the

hare goes 431

34 m

the grey-hound gains

431

34

432 or

125

m in one leap.

the grey-hound gains 2175

m in

5

122

175 210 leaps.

Ex. 22: A hare starts to run at 12 km per hourwhen a dog is 100 metres off. Afterhalf a minute the dog sees hare andpursues at 16 km per hour. How soonwill he catch hare?

Soln: Suppose the hare H sees the dog at D.

DH = 100 metres.Let K be the position when the dog seeshare.

HK = distance gone by hare in 21

min

= 21

60100012

= 100 metres

DK = 100 metres + 100 metres = 200 metresThe hare thus has a strat of 200 metres.Now the dog gains (16 12) or 4 km in anhour Distance gained by dog in 1 min

= 3200

6010004

metres

Now 3200

metres is covered in time

= 1 min200 metres is covered in time

= 20020031

= 3 min.

Hence, dog will catch hare in 3 minutes.Ex. 23: Two guns were fired from the same

place at an interval of 13 minutes buta person in a train approaching theplace hears the second report 12minutes 30 seconds after the first. Findthe speed of the train, supposing thatsound travels 330 metres per second.

Soln: It is easy to see that the distance travelledby the train in 12 min 30 seconds couldbe travelled by sound in (13 min - 12 min30 seconds) or 30 seconds. the train travels 330 30 metres in

2112 min.

the speed of the train per hour

= 10002560230330

= 251188

or 251347 km per hour

Ex. 24: A carriage driving in a fog passed aman who was walking at the rate of 3km an hour in the same direction. Hecould see the carriage for 4 minutesand it was visible to him upto adistance of 100 m. What was the speedof the carriage?

Soln: The distance travelled by man in 4minutes

= 46010003

metres = 200 metres.


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distance travelled by carriage in 4minutes

= (200 + 100) or 300 metres. speed of carriage

= 100060

4300

km per hour

= 214 km per hour.

Ex. 25: Two trains start at the same time, onefrom A to B and the other from B to A.If they arrive at B and A respectively 5hours and 20 hours after they passedeach other. Show that one travels twiceas fast as the other.

Soln: Let the two trains be P and Q.

Let the train P starts from A and Q startsfrom B and they meet at C.Let Ps speed be x km per hour and Qsspeed be y km per hour. BC = 5x km; AC = 20y kmNow time taken by P to travel a distance

20y = xy20

hours

And time taken by Q to travel a distance

5x = yx5

hours

Since both start at the same time and meet,those two times must be equal.

yx

xy 520

or, 5x2 = 20y2

or, 22

yx

= 4

or, 12

yx

Ex. 26: A can walk a certain distance in 50days when he rests 9 hours each day,how long it will take for him to walktwice as far if he walks twice as fastand rests twice as long each day?

Soln: Suppose initially he covers x km in 50days

Distance covered in 1 day = 50x

km

Since he rests for 9 hours, Distance covered in (24 9 =) 15 hours

= 50x

km

Distance covered in 1 hour

= 1550x

km

In the second case:Distance to be covered = 2x km

Speed = 152515502

xx km/hr

Number of hours he walks per day= (24 2 9 =) 6 hours

Distance covered in 1 day

= 15256

x km

Now 15256

x km is covered in 1 day

2x km shall be covered in

= xx

615252

= 125 days.

Ex. 27: A person walks from A to B at the rateof 3 kmph and after transacting somebusiness which occupies him an hour,returns to A by tramway at the rate of5 kmph. He then finds that he has beenabsent for 2 hours 20 minutes. Findthe distnace from A to B.

Soln: We have average speed

=

53532

3.75 kmph

The time for which he travels

= 3111

312

hours

Distance =

31175.3

21

= 2.5 km

Note: In calculating distance we have

multiplied

31175.3 by 2

1. Because we

will get twice the distance otherwise.Ex. 28: A person has to reach a place 40 km

away. He walks at the rate of 4 km/hrfor the first 16 km and then travels in ascooter for the rest of the journey.However, if he had travelled by scooterfor the 16 km and covered the remainingdistance on foot at 4 km/hr, he wouldhave taken an hour longer to completethe journey. Find the speed of thescooter.

Soln: Total distance covered = 40 kmDistance covered on foot = 16 km Distance covered in a scooter

= 40 km 16 km = 24 kmSuppose speed of the scooter

= x km per hour


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Now, according to the question,speed of the man = 4 km per hourThen time taken in the first case

=

xx

244244

16 hours

Also time taken in the second case

=

61642416

xx hours

Since the time taken in the second caseis one hour longer

1244616

xx

or, xx244616 = 1

or, xx2416

= 1 6 + 4

or, 18

x

or, x8

=1

or, x = 8Hence, speed of the scooter

= 8 km per hour.Ex. 29: A boy goes to school at a speed of 3

km/hr and returns to the village at aspeed of 2 km/hr. If he takes 5 hoursin all, what is the distance between thevillage and the school?

Soln: Let the required distance be x km.Then time taken during the first journey

= 3x

hour.

And time taken during the second journey

= 2x

hour.

523

xx

or, 5632

xx

or, 305 x x = 6 required distance = 6 km

Ex. 30: A motor car does a journey in 10 hours,the first half at 21 km/hr and thesecond half at 24 km/hr. Find thedistance.

Soln: Let the distance be x km.

Then 2x

km is travelled at a speed of 21

km/hr and 2x

km at a speed of 24 km/hr.

Then, time taken to travel the whole

journey =

242212

xxhours


242212

xx = 10

or, 224212124

xx = 10

x = 22424212421102

km.

Ex. 31: A monkey tries to ascend a greased pole14 metres high. He ascends 2 metresin first minute and slips down 1 metrein the alternate minute. If he continuesto ascend in this fashion, how longdoes he take to reach the top?

Soln: In every 2 minutes he is able to ascend(2 1 =) 1 metre. This way he ascendsupto 12 metres because when he reachesat the top, he does not slip down. Thus,upto 12 metres he takes 12 2 = 24minutes and for the last 2 metres he takes1 minute. Therefore, he takes (24 + 1 =)25 minutes to reach the top. That is, in26th minute he reaches the top.

Ex. 32: Two runners cover the same distanceat the rate of 15 km and 16 km perhour respectively. Find the distancetravelled when one takes 16 minuteslonger than the other.

Soln: Let the distance be x km.

Time taken by the first runner = 15x

hours

Time taken by the second runner

= 16x

hours


6016

1615

xx

or, 6016

1615)1516(

x

x = 16156016

= 64 km.

Ex. 33: Without any stoppage a person travelsa certain distance at an average speedof 80 km/hr and with stoppages hecovers the same distance at an averagespeed of 60 km/hr. How many minutesper hour does he stop?


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Soln: Let the total distance be x km.Time taken at the speed of 80 km/hr

= 80x

hours.

Time taken at the speed of 60 km/hr

= 60x

hours.

he rested for

8060xx

hours

= 240806020 xx

hours

his rest per hour

= xxxx 60

24060240

= 41

hours = 15 minutes.

Ex. 34: A man rode out a certain distance bytrain at the rate of 25 km an hour andwalked back at the rate of 4 km perhour. The whole journey took 5 hoursand 48 minutes. What distance did heride?

Soln: Let the distance be x km.Then time spent in journey by train

= 25x

hours.

And time spent in journey by walking

= 4x

hours.

Therefore, 5425

xx hours 48 minutes.

or, 529

60485

10029

x

205100

x km

Ex. 35: A person travels for 3 hours at thespeed of 40 km/hr and for 4.5 hoursat the speed of 60 km/hr. At the end of

it, he finds that he has covered 53

of

the total distance. At what averagespeed should he travel to cover theremaining distance in 4 hours?

Soln: Total distance covered in (3 + 4.5) hours= 3 40 + 4.5 60 = 390 km.


53

of the distance = 390 km

Remaining distance = 52

531

of the

total distance

52

of the total distance= 52

35390

= 260 km. average speed for the remaining

distance =

4

260 65 km/hr.

Ex. 36: When a man travels equal distance atspeeds V1 and V2 km/hr, his averagespeed is 4 km/hr. But when he travelsat these speeds for equal time hisaverage speed is 4.5 km/hr. Find thedifference of the two speeds and alsofind the values of V1 and V2.

Soln: Suppose the equal distance = D kmThen time taken with V1 and V2 speeds

are 1V

D hours and

2VD

hours respectively.

average speed = time Totaldistance Total

4VVV2V

VD

VD

2D

21

21

21

km/hr

In second case,

average speed = 4.52VV 21 km/hr

or, V1 + V2 = 9 and V1 V2 = 18

Now, 21221221 V4VVVVV = 81 72 = 9

3VV 21 km/hr .....(i) and V1 + V2 = 9 km/hr .....(ii)

On solving equations (i) and (ii), we haveV1 = 6 km/hr and V2 = 3 km/hr.

Ex. 37: A man takes 8 hours to walk to acertain place and ride back. However,he could have gained 2 hours, if he hadcovered both ways by riding. How longwould he have taken to walk bothways?

Soln: Walking time + Riding time= 8 hours .... (1)

2 Riding time = (8 2 =) 6 hours .... (2)2 (1) (2) gives the result2 walking time = (2 8 6 =) 10 hours. both ways walking will take 10 hours.

Ex. 38: A person travelled 120 km by steamer,450 km by train and 60 km by horse.It took 13 hours 30 minutes. If the rateof the train is 3 times that of the horse


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and 1.5 times that of the steamer, findthe rate of horse, train and steamerper hour.

Soln: Suppose the speed of horse = x km/hr.Then speed of the train = 3x km/hrand speed of the steamer = 2x km/hrNow, according to the question,

5.13603450

2120

xxx hours

(Since 13 hours 30 minutes = 13.5 hours)

or, 5.136360900360

x

205.1361620

x

Hence, speed of horse = 20 km/hr Speed of train = 3x = 3 20

= 60 km/hrSpeed of steamer = 2x = 2 20

= 40 km/hrEx. 39: A man covers a certain distance on

scooter. Had he moved 3 km/hr faster,he would have taken 40 minutes less.If he had moved 2 km/hr slower, hewould have taken 40 minutes more.Find the distance (in km) and originalspeed.

Soln: Suppose the distance is D km and theinitial speed is x km/hr.

Then, we have 6040D

3D

xx and

6040D

2D

xx

or, 32

3DD

xx

or, 32

)3(3D

xx .... (1)

and 32D

2D

xx

or, 32

)2(2D

xx .... (2)

From (1) and (2), we have

)2(D2

)3(D3

xxxx

or, )3(2)2(3 xx

or, 6263 xx

12x km/hrNow, if we put this value in (1), we get

D = 31512

32 = 40 km.

Hence, the distance is 40 km and theoriginal speed is 12 km/hr.

Ex. 40: A person covers a distance in 40minutes if he runs at a speed of 45 kmper hour on an average. Find the speedat which he must run to reduce the timeof journey to 30 minutes.

Soln: Let the distance be D km.

45D

= 40 minutes = 32

6040

D = 4532 = 30 km

Let the required speed be x km per hour.Now, according to the question,

21

x = 30

hour21minutes30

x = 30 2 = 60 km/hr.Ex. 41: The distance between two stations,

Delhi and Amritsar, is 450 km. A trainstarts at 4 pm from Delhi and movestowards Amritsar at an average speedof 60 km/hr. Another train starts fromAmritsar at 3.20 pm and movestowards Delhi at an average speed of80 km/hr. How far from Delhi will thetwo trains meet and at what time?

Soln: Suppose the trains meet at a distance of xkm from Delhi. Let the trains from Delhiand Amritsar be A and B respectively.Then,[Time taken by B to cover (450 x) km] [Time taken by A to cover x km]

= 6040

..... (see note)

or, 6040

6080450

xx

3(450 x) 4x = 160or, 7x = 1190or, x = 170.Thus, the trains meet at a distance of 170km from Delhi.Time taken by A to cover 170 km

=

60

170 hours = 2 hours 50 min.

So, the trains meet at 6.50 pmNote: RHS = 4 : 00 pm 3.20 pm

= 40 minutes = 6040

hour

LHS comes from the fact that the trainfrom Amritsar took 40 minutes more totravel up to the meeting point because ithad started its journey at 3.20 pm whereas


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the train from Delhi had started its journeyat 4 pm and the meeting time is the samefor both the trains.

Ex. 42: A man leaves a point P at 6 am andreaches the point Q at 10 am. Anotherman leaves the point Q at 8 am andreaches the point P at 12 noon. At whattime do they meet?

Soln: Let the distance PQ be A km.And they meet x hours after the first manstarts.Average speed of first man

= 4A

610A

km/hr.

Average speed of second man

= 4A

812A

km/hr.

Distance travelled by first man = 4Ax

km

They meet x hours after the first manstarts. The second man, as he starts 2hours late, meets after (x - 2) hours fromhis start. Therefore, the distance travelledby the second man

= 4)2(A x

km

Now, 4)2(A

4A

xx

km = A

or, 2x 2 = 4 x = 3 hours. They meet at 6 am + 3 hours = 9 am

Ex. 43: A train travels a distance of 300 kmat constant speed. If the speed of thetrain is increased by 5 km an hour,the journey would have taken 2 hoursless. Find the original speed of thetrain.

Soln: Let x km/hr be the constant speed of thetrain.Then, time taken to cover 300 km

= x300

hours.

Time taken to cover 300 km when thespeed is increased by 5 km/hr

= 5300x hours.

It is given that the time to cover 300 kmis reduced by 2 hours.

25300300

xx

or, 2)5(300)5(300

xxxx

or, 25

30015003002

xxxx

or, 2x2 + 10x = 1500or, x2 + 5x - 750 = 0or, x2 + 30x - 25x - 750 = 0or, x(x + 30) - 25 (x + 30) = 0or, (x - 25) (x + 30) = 0or, x = 25 or x = -30.Negative value of x should be rejected.Hence, the original speed of the train is25 km/hr.

Ex. 44: A fast train takes 3 hours less than aslow train for a journey of 600 km. Ifthe speed of the slow train is 10 km/hrless than that of the fast train, find thespeeds of the two trains.

Soln: Let the speed of the slow train bex km/hr. Then, speed of the fast train is(x + 10) km/hr.Time taken by the slow train to cover 600

km = x600

hours.

Time taken by the fast train to cover 600

km = 10600x hours.


310

600600

xx

or, 3)10(600)10(600

xxxx

or, 310

60002 xx

or, 6000)10(3 2 xx

or, 02000102 xx

or, 0200040502 xxx

or, 0504050 xxxor, (x + 50) (x - 40) = 0or, x = 50 or x = 40or, x = 40

[ x cannot be negative]or, Hence, the speeds of two trains are 40km/hr and 50 km/hr.

Ex. 45: A plane left 30 minutes later than theschedule time and in order to reach itsdestination 1500 km away in time ithas to increase its speed by 250 km/hrfrom its usual speed. Find its usualspeed.

Soln: Let the usual speed of the plane bex km/hr. Then,


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Time taken to cover 1500 km with the

usual speed = x1500

hours

Time taken to cover 1500 km with the

speed of (x + 250) km/hr = 250150x

21

25015001500

xx

or, 21

25015001500

xx

or, 21

)250(150025015001500

xxxx

or, 21

2502501500

2

xxor, 750000 = x2 + 250xor, x2 + 250x 750000 = 0or, x2 + 1000x 750x 750000 = 0or, x(x + 1000) - 750 (x + 1000) = 0or, (x + 1000) (x - 750) = 0or, x = -1000 or x = 750or, x = 750

[ speed cannot be negative]Hence, the usual speed of the plane is750 km/hr.

Ex. 46: In a flight of 600 km, an aircraft wasslowed down due to bad weather. Itsaverage speed for the trip was reducedby 200 km/hr and the time of flightincreased by 30 minutes. Find theduration of flight.

Soln: Let the original speed of the aircraft be xkm/hr.Then, new speed = (x 200) km/hrDuration of flight at original speed

=

x

600hour

Duration of flight at reduced speed

=

200600

x hour

21600

200600

xx

or, 21

)200()200(600600

xxxx

or, 21

2001200002 xx

or, x2 200x 240000 = 0or, x2 - 600x + 400x - 240000 = 0or, x(x - 600) + 400 (x - 600) = 0or, (x - 600) (x + 400) = 0or, x - 600 = 0 or x + 400 = 0

or, x = 600or, x = -400or, x = 600

[ x cannot be negative]So, the original speed of the aircraft was600 km/hr.

Hence, duration of flight =

x

600hour

=

600600

hour = 1 hour.

Ex. 47: Two trains leave a railway station atthe same time. The first train travelsdue west and the second train duenorth. The first train travels 5 km/hrfaster than the second train. If aftertwo hours, they are 50 km apart, findthe average speed of each train.

Soln: Let the speed of the second train bex km/hr. Then, the speed of the first trainis (x + 5) km/hr.Let O be the position of the railway stationfrom which the two trains leave.Distance travelled by the first train in 2hours

= OA = speed time = 2(x + 5) kmDistance travelled by the second train in2 hours

= OB = speed time = 2x kmBy Pythagoras Theorem,

AB2 = OA2 + OB2

or, 502 = [2(x + 5)]2 + {2x}2or, 2500 = 4(x + 5)2 + 4x2or, 8x2 + 40x 2400 = 0or, x2 + 5x 300 = 0or, x2 + 20x 15x 300 = 0or, x(x + 20) 15 (x + 20) = 0or, (x + 20) (x 15) = 0or, x = 20 or x = 15

[ x cannot be negative]Hence, the speed of the second train is15 km/hr and the speed of the first trainis 20 km/hr.

Ex. 48: A walks half a kmph faster than B andthree quarters of a kmph faster thanC. To walk a certain distance C takes


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three quarters of an hour more than Band two hours more than A. Find thedistance covered and the time takenby B. Also find his speed.

Soln: Let the speed of B be y kmph

Speed of A =

21y kmph and

speed of B =

21y kmph and

speed of C = 41

43

21

yy kmph

Let the distance travelled be d km andtime for A is T.

Now, for A, T = 12

2

21

y

d

y

d ... (1)

For C, T + 2 = 14

4

41

y

d

y

d

or, 14

2842

144

T

yyd

yd

... (2)

and for B,

yd

45

T or, 45

T yd

or, yd4

544T

... (3)

From (1) and (2),

14284

122

yyd

yd

or, d (4y - 1) = (2y + 1) (2d - 4y + 1)or, 4dy - d = 4dy - 8y2 + 2y + 2d - 4y + 1or, 8y2 + 2y - 3d - 1 = 0

or, d = 3

128 2 yy .... (4)

and from (2) and (3)

yyd

yyd

454

14284

or, ydydyyydy 54201683216 22

or, 04312 2 dyy

or, 4

312 2 yyd .... (5)

From (4) and (5),

4312

3128 22 yyyy

or, yyyy 9364832 22

or, 04174 2 yy

or, 41,4y

Now from (5), d = 45 and from (3),

T = 10 hours. (y = 41

is not possible).

Ex. 49: A man started from his house to hisworkplace 8 km away at the rate of 4kmph so as to reach just in time. After5 minutes he realised that he had leftsome important documents at home,so he turned back, and now walkingat an increased speed, still succeededin reaching his workplace in time. Whatwas his increased speed?

Soln:

Let A be the house and B be the workplace.The workplace is 8 km away. Had hewalked all the way at 4 kmph he wouldhave taken 2 hours to reach hisworkplace, which would be just in time.Therefore, total time = 2 hoursReferring to the above diagram, the totaltime is made up(i) walking from A to C for 5 minutes

= 605

hours = 121

hours

now, 4 km in 1 hour

31

km in 121

hours

Distance AC = 31

km

(ii) If x is the increased speed, time forwalking from C to A

= xAC

hours = x31

hours

(iii) Walking from A to B time taken

= xx8AB

hours


xx8

31

1212

x = 2384 kmph


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Practice Exercise

1. For a journey, walking 76

of his usual speed,

a man becomes late by 25 minutes. What ishis usual time taken for the journey?

2 . A motorist covers a distance from A to B at aspeed of 20 km/hr and return journey from Bto A at a speed of 30 km/hr. If he takes 5hours for the whole journey, find the distancefrom A to B.

3. Shivangi starts from her house for her schoolat a certain fixed time. If she walks at therate of 5 km/hr, she is late by 7 minutes.However, if she walks at the rate of 6 km/hr,she reaches the school 5 minutes earlier than

the scheduled time. What is the distance ofthe school from her house?

4. I will reach my destination 40 minutes late ifI walk at the rate of 3 km/hr. However, I willreach 30 minutes before time if I walk at therate of 4 km/hr. Find the distance of mydestination from the starting point.

5. A student travels to his school at a speed of 4km/hr and reaches the school 15 minuteslate. On travelling at a speed of 6 km/hr, hereaches the school 5 minutes early. At whatspeed must he travel to reach the school justin time?

6. A person travels a certain distance on a bicycleat a certain speed. Had he moved 3 km/hr

Ex. 50: Dinky is picked up by his father by carfrom school everyday and they reachhome at 5.00 pm. One day, since schoolgot over an hour earlier than usual, hestarted walking towards home at 3kmph. He met his father on the wayand they reached home 15 minutesearlier than their usual time. What isthe speed of the car?

Soln:

Let the speed of car be ds and the speedof Dinky be sw sw = 3 kmphReferring to the above diagram, if Arepresents the home and B represents theschool, the father starts at his usual timebut meets his son on the way at C. So, ingoing from A to C and back to A he savesthe time he would have used commutingfrom C to B and back to C. Since theyreach back 15 minutes earlier than usual,ie the time saved = 15 minutes. Thereforetime taken to drive from C to B and backis 15 minutes. Therefore time taken to

drive the distance BC = td = 215

td = 7.5 minutesDinky starts 1 hour earlier than usual.Had he moved at driving speed, ie speedof the car, he would have reached 1 hourearlier. But he reaches only 15 minutesearlier. Therefore he loses 45 minutes, ie

by travelling partly by foot and partly bycar, he takes 45 minutes longer than ifhe would have travelled the wholedistance by car.Therefore time taken to walk the distanceBC= tw = (7.5 + 45 =) 52.5 minutes. tw = 52.5 minutesBut for a constant distance BC,

d

w

w

dtt

ss

5.75.52

3d

s

sd = 21 kmphEx. 51: Two places P and Q are 162 km apart.

A train leaves P for Q and at the sametime another train leaves Q for P. Boththe trains meet 6 hours after they startmoving. If the train travelling from Pto Q travels 8 km/hr faster than theother train, find the speed of the twotrains.

Soln: Suppose the speeds of the two trains arex km/hr and y km/hr respectively.Now,Total distance travelled by both the trainsin 6 hours = (6x + 6y) kmNow, according to the question,

(6x + 6y) = 162

or, 272162

yx ....(i)

and x y = 8 ....(ii)Solving equations (i) and (ii), we have

x = 17.5 km/hr andy = 9.5 km/hr.

Hence, speed of the two trains are 17.5km/hr and 9.5 km/hr.


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faster, he would have taken 40 minutes less.Had he moved 2 km/hr slower, he would havetaken 40 minutes more. Find the distanceand original speed of the person.

7. Two places A and B are 80 km apart fromeach other on a highway. A car starts from Aand another from B at the same time. If theymove in the same direction, they meet eachother in 8 hours. If they move in oppositedirections towards each other, they meet in1 hour 20 minutes. Determine the speeds ofthe cars.

8. A train covers a distance between stations Aand B in 45 minutes. If the speed is reducedby 5 km/hr, it will cover the same distancein 48 minutes. What is the distance betweenthe two stations A and B (in km)? Also, findthe speed of the train.

9. A car covering half of a distance of 100 kmdevelops some engine trouble and later travelsat half of its original speed. As a result, itarrives 2 hours later than its normal time.What was the original speed of the car?

10. A train covers a distance of 31193 km in 4

14

hours with one stoppage of 10 minutes, twoof 5 minutes and one of 3 minutes on theway. Find the average speed of the train.

11. Distance between two places X and Y is 60km. Two persons A and B start from X towardsY at the same time. Speed of B is 4 km/hrless than the speed of A. A reaches Y, returnsat once and meets B at a distance of 12 kmfrom Y. What is the speed of B?

12. An army bomb squad man set a fuse forblasting a rock to take place after one minute.He ran away from the site at the speed of 13m/s. Sound travels at the speed of 325 m/s.Upto what distance could the army man run,before he heard the sound of blast?

13. On a particular day a person starts walkingfrom a place X at 2 am and reaches place Y at5 am. A second person starts walking from aplace Y at 4 am and reaches place X at 9 amon the same day. At what time do they crosseach other?

14. A covers some distance in 50 days when herests 9 hours a day. In how many days willhe cover the double distance by resting twiceas before?

15. A man travelled a total distance of 3990 km,part of it by air, part by water and the rest byland. The time he spent in travelling by air,water and land was in ratio 1 : 16 : 2respectively and the average speed of eachmode of travel was in the ratio 20 : 1 : 3respectively. If his overall average speed was42 km/hr, find the distance covered by water.

16. A goods train travelling from station A to

station B meets with an accident one hourafter starting. After stopping there for 30

minutes, it proceeds at 54

of its usual speed

and arrives at B 2 hours late. Had the traincovered 80 km more before the accident, itwould have been just one hour late. Determinethe original speed of the train and the distancebetween A and B.

17. A train after travelling 50 km meets with an

accident and then proceeds at 43

of its former

speed and arrives at its destination 25minutes late. Had the accident occurred 24km behind, it would have reached thedestination only 35 minutes late. Find thespeed of the train and the distance travelledby the train.

18. Ravi can walk a certain distance in 40 days,when he rests 9 hours a day. How long willhe take to walk twice the distance, twice asfast and rest twice as long each day?

19. Two men set out at the same time to walktowards each other from two points A and B,72 km apart. The first man walks at the rateof 4 km/hr. The second man walks 2 km in

the first hour, 212 km in the second hour, 3

km in the third hour and so on. Find thetime after which the two men will meet.

20. Two trains start out towards each other frompoints 650 km apart. If they start out at thesame time, they will meet in 10 hours, but ifone of them starts out 4 hours and 20 minutesafter the other, they will pass each other 8hours following the departure of the latter.Determine the average speed of each other.

21. Distance between two stations X and Y is220 km. Trains P and Q leave station X at 8am and 9.51 am respectively at the speed of25 km/hr and 20 km/hr respectively forjourney towards Y. A train R leaves station Yat 11.30 am at a speed of 30 km/hr for journeytowards X. When and where will P be at equaldistance from Q and R.

22. Two places P and Q are 336 km apart. A trainleaves P for Q and at the same time anothertrain leaves Q for P. Both trains meet at theend of the 6 hours. If one train travels 8 km/hrfaster than the other, find the speeds of theother trains.

23. On a 2-km road, a total number of 201 treesare planted on each side of the road at equaldistances. How many such trees in all willbe planted on both sides of a 50-km road suchthat the distance between two consecutivetrees is the same as that of the consecutivetrees on the 2-km road?


KKUNDAN

Answers and explanations1. Let the usual speed of the person be x km/hr

and the distance of his journey be D km.

His usual time to cover the distance =

xD

hour


Speed = 76

of his usual speed =

x76

km/hr

Time taken to cover the distance D km

= xx 6

7D

76D

hour

Again,

6025D

67D

xx

or, 1251

67D

x

or, 256

125D

x hours

Usual time = 212 hours.

2 . Let the distance from A and B is x km. Time taken to cover the distance from A to

B at 20 km/hr = 20x

hours.

And time taken to cover the distance from B

to A at 30 km/hr = 30x

hours.

Total time taken = 5 hours. (Given)

53020

xx

560

23

xx

or, 5x = 60 5 = 300

or, x = 5300

= 60 km

3. Let the required distance be x km.

Time taken to walk at 5 km/hr = 5x

hours

=

605x


Time taken to walk at 6 km/hr = 6x

hours

=

606x


Since the difference between the two timestaken is (7 + 5 =) 12 minutes 12x 10x = 12or, 2x = 12

x = 212

= 6

Hence, the required distance is 6 km.Alternative Method:Let x km be the distance between her houseand school and t hours be the time requiredto reach the school from her house.When Shivangi walks at 5 km/hr, then

607

5 tx ....(i)

When Shivangi walks at 6 km/hr, then

605

6 tx

or 121

6 tx ....(ii)

Subtracting equation (ii) from equation (i), weget

121

607

65ttxx

or, 51

6012

6075

607

121

30

x

x = 530

= 6

Hence the required distance = 6 km4. Let the required distance be x km.

Difference of time taken at different speeds= (40 + 30) minutes = 70 minutes

= 6070

hours = 67

hours


hours


hours

According to the question,

67

43

xx

or, 67

1234

xx

or, 67

12

x

or, x = 1267 = 14

Distance of the destination = 14 km


KKUNDAN

5. Let the distance of the school be x km.

Time taken in first case = 4x

hours

But this time is 15 minutes late or 41

6015

hours late Actual time for reaching the school in time

should be

41

4x

hours

Time taken in second case = 6x

hours

But this time is 5 minutes early or 121

605

hours early. Actual time for reaching the school in time

should be

121

6x

hours

From the above, we have

121

641

4xx

or, 12

124

1

xx

or, 881212 xxor, 204 xor, x = 5 km The distance of the school be 5 km and

actual time to reach school in time = 4

)1( x

= 1 hour The required speed is 5 km/hr.

6. Let the original speed and distance beV km/hr and D km respectively.Time taken to complete the whole journey

= VD

hours.

When the person moves 3 km/hr faster, then

6040

VD

3VD

or, 32

VD

3VD

or, 32

3VD

VD

or, 32

3VVDV3DDV

or, 32

3VV3D

or, 9

3V2VD ....(i)

When the person moves 2km/hr slower, then

6040

VD

2VD

or, 32

VD

2VD

or, 32

VD

2VD

or, 32

2)V(V2DDVDV

or, 32

2)V(V2D

or, 32)V(VD ....(ii)

Combining equations (i) and (ii), we get

32)V(V

93)2V(V

or, 2(V + 3) = 3(V 2)or, 2V + 6 = 3V 6or, 3V 2V = 6 + 6or, V = 12 km/hrPutting the value of V in equation (ii),

we get D = 31012 = 40 km.

7. Case I: When the cars are moving in the samedirection.

Let A and B be two places and C be the placeof meeting.Let the speed of car starting from A be x km/hrand the car starting form B be y km/hr.Relative speed = (x y) km/hrAccording to the question, (x y) 8 = 80or, x y = 10 ...(i)Case II: When the cars are moving in theopposite directions and they meet at point C.

Relative speed = (x + y) km/hrTime taken = 1 hour 20 minutes

= 34

311

hours

Again, according to the question,

8034)( yx


KKUNDAN

or, x + y = 60 ...(ii)Solving equations (i) and (ii), we have

x = 35 and y = 25 Speeds of the cars

= 35 km/hr and 25 km/hr.8. Suppose the distance is x km and the speed

of the train is y km/hr.Thus we have two relationships:

(1) yxyx

43

43

6045

(2) )5(54

54

6048

5

yx

yx

From (1) and (2), we have

)5(54

43

yy

or, 443

54

y

or, y = 1516204

= 80 km/hr

Therefore speed = 80 km/hr and distance

x = 8043 = 60 km

9. Half of the original speed means double thenormal time. It means that the car shouldhave covered half of the distance of 100 km,ie 50 km, in 2 hours.Hence, the original speed of the car

=

2

50 = 25 km/hr

10. Distance covered by train

= 31193 km = 3

580 km

Time taken by the train to cover this distance

= 414 hours = 4

17 hours

Total stoppage during the journey= 10 1 + 5 2 + 3 1

= 23 minutes = 6023

hours

Actual time taken by the train to cover theabove distance

= 6023

417

= 60

231517

= 60232

6023255

=

1558 hours

Average speed of the train =

15583

580

km/hr

= 35815580

= 50 km/hr

11.

Let A and B meet after t hours.Let the speed of B be x km/hr. Speed of A = (x + 4) km/hrDistance covered by A in t hours = 60 + 12

= 72 kmDistance covered by B in t hours = 60 12

= 48 kmNow, according to the question,xt = 48 ....(i)(x + 4)t = 72 ....(ii)On dividing equation (ii) by equation (i), wehave

23

48724

xx

or, 2x + 8 = 3xor, x = 8 Speed of A = 8 km/hr

12. Time after which the bomb is set to explode= 1 minute = 60 seconds

Speed of the man = 13 m/secDistance covered by man in 60 sec

= 13 60 = 780 metresSo, distance to be travelled by sound before itcatches up with army man = 780 metresSpeed of the sound = 325 m/sec (given)Since the man and sound are travelling inthe same direction, the relative speed of sound

= (325 13 =) 312 m/secTime taken by sound to travel 780 metres

= 312780

= 2.5 sec

Now, during this time man would havetravelled further. So, distance covered by manin 2.5 seconds = 2.5 13 = 32.5 mThe total distance travelled by man

= 780 + 32.5 = 812.5 metres.13. X P Y

l l lLet the speed of the person who starts from Xbe x km/hr and speed of the person who startsfrom Y be y km/hr.Time taken by the person who starts from X

= 5 am 2 am = 3 hours


KKUNDAN

Time taken by the person who starts from Y= 9 am 4 am = 5 hours

Again, let the distance between X and Y be Dkm.Now, according to the question,

x km/hr = 3D

and y km/hr = 5D

If the person starting from X reaches themeeting point after t hours, person startingfrom Y will reach the meeting point after(t 2) hours. Since the person starting fromX starts moving at 2 am while the personstarting from Y starts moving at 4 am. Andthe difference of time = (4 am 2 am)= 2 hours Distance (XP) travelled by the person

starting from X =

t3D

km

and the distance (YP) travelled by the person

starting from Y =

)2(5D t km

Total distance travelled by both before meeting= Distance travelled by person from X + Distance travelled by person from Y

= D)2(5D

3D

tt

or, D52

3D

tt

or, 152

3

tt

or, 115635

tt

or, 8t 6 = 15or, 8t = 15 + 6 = 21

or, t = 852

821

hours

Converting this in hours, minutes andseconds, we get 2 hours 37 minutes and 30seconds.

[ 852 hours = 2 hours +

6085

minutes

= 2 hours + 2137

275

minutes

= 2 hours + 37 minutes + 21

minutes

= 2 hours + 37 minutes +

6021

30 seconds = 2 hours 37 minutes and 30 seconds]

14. Let the distance for A be x kmNumber of hours A walks daily = (24 9 =) 15hoursNumber of days = 50 days

Speed (in km/hr) = 1550x

..... (1)

In second situationLet the number of days be YDistance = 2xNumber of hours for which A walks daily = 6hours Speed in second case (in km/hr)

= 6Y2

TimeDistance

x ..... (2)

In both the cases, the speed remains the same

15502

6Y2

xx

or, Y 6 = 50 15

or, Y = 61550

= 125 days

15. Total distance travelled = 3990 kmRatio of time spent in travelling by air, waterand land = 1 : 16 : 2Ratio of respective speeds = 20 : 1 : 3From the given fact, the ratio of respectivedistances will be 20 : 16 : 6 = 10 : 8 : 3Sum of the ratios = 10 + 8 + 3 = 21Distance travelled by steamer will be

= 2183990 = 1520 km

16. Let the distance between station A and stationB be d km.Again, let the initial speed of the goods trainbe x km/hr.As the accident takes place after 1 hour distance covered in 1 hour by the goodstrain = x kmRemaining distance = (d x) kmTotal time taken, if no accident happened

=

xd

hours

Case I:Time taken by the goods train to cover thedistance

=

5460

301 x

xd

=

x

xd4

)(5211 hours


24)(5

211

xd

xxd


KKUNDAN

or, 21

45)(

xd

xxd

or, 21

45)(

xd

xxd

or, 21

4455

xdxd

or, 21

45

xxd

or, 2d 10x = 4xor, 2d = 14xor, d = 7x ....(i)Case II:If the goods train had covered 80 km morebefore the accident, then the distance of siteof the accident = (x + 80) kmRemaining distance = [d (x + 80)] kmTime taken to cover the whole of the distance

=

54

)80(6030)80(

xxd

xx hours

According to the question,

1

54

)80(603080

xd

xxd

xx

or, 14]80([5

21801

xd

xxd

x

or, xd

xxd

x

4)]80([5

2180

or, xd

xxd

21

440055320

or, xd

xxd

21

48055

or, xxdd

480554

21

or, 2x = 5x d + 80Putting the value of d from equation (i), wehave 2x = 5x 7x + 80or, 4x = 80

x = 480

= 20

Hence original speed of the train= 20 km/hr.

Distance between the stations A and B= d = 7x (From i)= (7 20) = 140 km.

17. Let the distance be D km and speed be the xkm/hrFrom the question, we have

125D

6025D

34)50D(50

xxxx

or, xx

x 125D12

3200D4150

or, xx

x 125D12

350D4

or, x5D12200D16 4D 5x = 200 ... (i) and

6035D

34)26D(2450

xxx

xx

x 127D12

127D

or, xx

xx 127D12

3104D426

or, xx

x 127D12

3104D478

or, xx

x 127D12

326D4

or, 1047D4 x .... (ii)Now, subtracting equation (ii) from equation(i), we have 2x = 96 x = 48 km/hrPut the value of x in equation (i) and find thedistance (D)or, 4D 5 48 = 200or, 4D = 200 + 240 = 440

D = 4

440 = 110 km.

18. Time for work per day in first condition= (24 9 =) 15 hours

Time for work per day in second condition= (24 9 2 =) 6 hours

Here we have four quantit ies : Speed,Distance, Work and Days. We have tocalculate number of days. Hence, Days willbe in the last column. Here followingrelationships exist:

More speed, less days (Inverse)More distance, more days (Direct)Less hours of work, more days (Inverse)

Hence,

x:40::15:62:11:2

or, 2 1 6 : 1 2 15 :: 40 : x(Compounding the ratios)


KKUNDAN

or, 2 1 6 x = 1 2 15 40(Product of extreme terms =

Product of mean terms)

x = 612401521

= 100

Hence the required time = 100 days.19. Let A starts from point X, B starts from point

Y and they meet after t hours.

PYX

A B

XP = 4 t = 4t kmYP = 2 + 2.5 + 3 + .... t termsThis is an AP.

Sum of an AP = dnan 122

where n = number of terms, a = first termand d = common difference

YP =

21

24

221)1(22

2tttt

= 4

744

744

2222 ttttttt

But it is given that XY = 72 or XP + PY = 72

4

7 2tt + 4t = 72

or, 7t + t2 + 16t = 288or, t2 + 23 t - 288 = 0or, t2 + 32 t 9t - 288 = 0or, t (t + 32) 9 (t + 32) = 0or, (t + 32) )(t 9) = 0 t + 32 = 0or, t = 32 (Not possible) t 9 = 0or, t = 9They meet after 9 hours.

20. Let the trains A and B travel at speed of x andy km/hr respectively and meets 10 hours afterdeparture.

From the figure it can be seen thatAC = (x 10) kmBC = (y 10) km AC + BC = x 10 + y 10or, 650 = 10(x + y)or, x + y = 65In the second situation when the other trainstarts after 4 hours and 20 minutes

4 hours and 20 minutes

= 313

314

60204 hours

Distance covered by train A in 313

hours

= AP = x 313

= 313x

Both the trains meet 8 hours after train Aleaves P. Now if they meet at C1 thenPC1 = 8 x = 8x kmBC1 = 8 y = 8y kmAccording to the question,

8x + 8y = 313650 x

or, 3136508 xyx

or, 313650658 x

or, 130520650313

x

or, x = 133130

= 30 km/hr

Speed of train A = 30 km/hrSpeed of train B = (65 30) km/hr

= 35 km/hr21. As given, speed of the train P = 25 km/hr

Speed of the train Q = 20 km/hrSpeed of the train R = 30 km/hr

PQ

B A Q1 P1 R1 R

YX 33 km 25 t 30 t

20 t

87.5 km

Distance travelled by train P between 8:00 to11:30

ie in 213 hours = 5.872

1752527

km

Distance travelled by train Q between 9 : 51to 11 : 30 ie. in

1 hour 39 minutes = 20203320

60391

= 33 kmAssume that trains P and Q are at A and Brespectively at 11 : 30 am. Also assume that tminutes after 11 : 30 am, train P wasequidistant from train Q and train R. At theequidistant position train P, Q and R were atP1, Q1 and R1. XP1 = XA + AP1 = (87.5 + 25 t) kmXQ1 = XB + BQ1 = (33 + 20 t) kmP1 Q1 = XP1 XQ1 = (87.5 + 25 t) - (33 + 20 t)

= (54.5 + 5 t) kmDistance RR1 = 30 t km


KKUNDAN

P1R1 = Total distance - XP1 - RR1 = 220 - (87.5 + 25 t) - 30 t = (132.5 - 55 t) km

P1Q1 = P1R1 5t + 54.5 = 132.5 - 55 t

or, 60 t = 78 or t = 606078

minutes

or, t = 78 minutesSo 78 minutes after 11 : 30 am ie at 12 : 48pm train P will be equidistant from train Qand R.

XP1 = 87.5 + 25 t = 87.5 + 25 6078

= 87.5 + 32.5XP1 = 120 km At 120 km away from station X, trains wouldbe at equal distances.

22.

Let R be the meeting point.

Let the speed of train from P = x km/hr andthat from Q = (x + 8) km/hrBoth trains meet after 6 hours (x 6) + (x + 8) 6 = 336or, 6x + 6x + 48 = 336or, 12x = 336 48 = 288

or, x = 12288

= 44

Speed of one train = 24 km/hrSpeed of the other train = (24 + 8 =)32 km/hr

23. Distance between 2 trees on a 2-km road

=

120110002

= 10 m

Number of trees planted on both sides of a50-km road

=

1101000502 = 10002

Documents

Speed, Time and Distance