ssc_speed, time and distance

1Speed, Time and Distance

KKUNDAN

Speed, Time and Distance1. Important Formulae

(i) Distance = Speed × Time

(ii) Time = DistanceSpeed

(iii) Speed = Distance

Time

(iv) x km/hr = 5

18x m/sec

(v) x m/sec = 185

x km/hr

2. Relative Speed(i) When objects are moving in opposite

directionsThe relative speed of one object with respectto the other will have magnitude greater thanindividual speed of each object. This is why,for example, a train A moving with speed 10km/hr will cross another train B moving inopposite directions with speed 25 km/hr, witha relative speed of (10 + 25 =) 35 km/hr whichis greater than the individual speed of eithertrain.

(ii) When objects are moving in same directionThe relative speed of one object with respect tothe other will have magnitude either less thanor greater than individual speed of each object.This is why, for example, a train A moving withthe speed of 20 km/hr will cross the anothertrain B moving in same direction with the speedof 15 km/hr, with a relative speed of (20 – 15 =)5 km/hr which is less than the individual speedof either train.

Take another example, a train A moving withthe speed of 60 km/hr will cross the anothertrain B moving in same direction with thespeed of 20 km/hr with a relative speedof (60 – 20 =) 40 km/hr which is less than thetrain A and greater than the train B.

3. Average Speed

Average Speed = Total Distance Covered

Total Time TakenFor example, a person divides his total route ofjourney into three equal parts and decides totravel the three parts with speed of 40, 30 and 15km/hr respectively. Find his average speed duringthe 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

Time taken to travel third part of the journey

= 15x

hour

Total time taken = 40 30 15x x x hours

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

Average Speed = Total Distance Travelled

Total Time Taken

=

40 30 15

xx x x

km/hr = 24 km/hr

1. For a journey, walking 76

of his usual speed, a

man becomes late by 25 minutes. What is hisusual time taken for the journey?

[SSC Section Officers’ (Audit) Exam–2005]2. A motorist covers a distance from A to B at a

speed of 20 km/hr and return journey from B to Aat a speed of 30 km/hr. If he takes 5 hours forthe whole journey, find the distance from A to B.

[SSC UDC (Mains) Exam–2001]3. Shivangi starts from her house for her school at

a certain fixed time. If she walks at the rate of

5 km/hr, she is late by 7 minutes. However, ifshe walks at the rate of 6 km/hr, she reachesthe school 5 minutes earlier than the scheduledtime. What is the distance of the school fromher house?

[SSC Section Officers’ (Audit) Exam–2005]4. If a scooterist drives at the rate of 24 km/hr, he

reaches his destination 5 minutes too late. If hedrives at the rate of 30 km/hr, he reaches hisdestination 4 minutes too soon. How far is hisdestination from the starting point?

[SSC Tax Assistant Excam–2004]

Exercise

2

KKUNDAN

5. Walking at a speed of 6 km/hr, a student reacheshis school 6 minutes early and walking at a speedof 4 km/hr, he reaches the school 4 minutes late.Find the distance of the school from the student’shouse.

[SSC UDC (Mains) Exam–2001]6. I will reach my destination 40 minutes late if I

walk at the rate of 3 km/hr. However, I will reach30 minutes before time if I walk at the rate of4 km/hr. Find the distance of my destination fromthe starting point.

[SSC UDC (Mains) Exam–2003;SSC Tax Assistant Excam–2008]

7. A student travels to his school at a speed of4 km/hr and reaches the school 15 minutes late.On travelling at a speed of 6 km/hr, he reachesthe school 5 minutes early. At what speed musthe travel to reach the school just in time?

[SSC Section Officers’ (Audit) Exam–2001]8. A person travels a certain distance on a bicycle

at a certain speed. Had he moved 3 km/hr faster,he would have taken 40 minutes less. Had hemoved 2 km/hr slower, he would have taken40 minutes more. Find the distance and originalspeed of the person.

[SSC UDC (Mains) Exam–2001]9. Two places A and B are 80 km apart from each

other on a highway. A car starts from A andanohter from B at the same time. If they move inthe same direction, they meet each other in8 hours. If they move in opposite directionstowards each other, they meet in 1 hour 20minutes. Determine the speeds of the cars.

[SSC Assistant Grade (Mains) Exam–2006]10. A train covers a distance between stations A and

B in 45 minutes. If the speed is reduced by5 km/hr, it will cover the same distance in48 minutes. What is the distance between thetwo stations A and B (in km)? Also, find the speedof the train.

11. A car covering half of a distance of 100 km developssome engine trouble and later travels at half ofits original speed. As a result, it arrives 2 hourslater than its normal time. What was the originalspeed of the car?

[SSC UDC (Mains) Exam–2004]

12. A train covers a distance of 31193 km in

414 hours with one stoppage of 10 minutes, two

of 5 minutes and one of 3 minutes on the way.Find the average speed of the train.

[SSC Section Officers’ (Audit) Exam–2001]13. Distance between two places X and Y is 90 km.

Two persons A and B start from X towards Y atthe same time. Speed of B is 4 km/hr less thanspeed of A. A reaches Y, returns at once and meets

B at a distance of 12 km from Y. What is thespeed of B?

[SSC UDC (Mains) Exam–2007]14. An army bomb squad man set a fuse for blasting

a rock to take place after one minute. He ranaway from the site at the speed of 13 m/s. Soundtravels at the speed of 325 m/s. Upto whatdistance could the army man run, before he heardthe sound of blast?

[SSC Graduate Level (Mains) Exam–2000]15. On a particular day a person starts walking from

a place X at 2 am and reaches place Y at 5 am.A second person starts walking from a place Y at4 am and reaches place X at 9 am on the sameday. At what time do they cross each other?

[SSC LDC (Mains) Exam–1999]16. A covers some distance in 50 days when he rests

9 hours a day. In how many days will he coverthe double distance by resting twice as before?

[SSC Assistant Grade (Mains) Exam–1995]17. A man travelled a total distance of 3990 km, part

of it by air, part by water and the rest by land.The time he spent in travelling by air, water andland was in ratio 1 : 16 : 2 respectively and theaverage speed of each mode of travel was in theratio 20 : 1 : 3 respectively. If his overall averagespeed was 42 km/hr, find the distance coveredby water.

[SSC Assistant Grade (Mains) Exam–1997]18. A goods train travelling from station A to station

B meets with an accident one hour after starting.After stopping there for 30 minutes, it proceeds

at 54

of its usual speed and arrives at B 2 hours

late. Had the train covered 80 km more beforethe accident, it would have been just one hourlate. Determine the original speed of the trainand the distance between A and B.

[SSC Assistant Grade (Mains) Exam–2001;SSC UDC (Mains) Exam–2004]

19. A train, an hour after starting, meets with anaccident which detains it for 30 minutes. After

this the train proceeds at 43

th of its former speed

and arrives 213 hours late. Had the accident

happened 90 km farther along the line, it wouldhave arrived only 3 hours late. Find the length ofthe journey.

[SSC Assistant Grade (Mains) Exam–1997;SSC Assistant Grade (Mains) Exam–2000]

20. A train after travelling 50 km meets with an

accident and then proceeds at 43

of its former

speed and arrives at its destination 25 minuteslate. Had the accident occurred 24 km behind, itwould have reached the destination only


KKUNDAN

35 minutes late. Find the speed of the train andthe distance travelled by the train.

21. Ravi can walk a certain distance in 40 days, whenhe rests 9 hours a day. How long will he take towalk twice the distance, twice as fast and resttwice as long each day?

[SSC UDC (Mains) Exam–199922. Two men set out at the same time to walk towards

each other from two points A and B, 72 km apart.The first man walks at the rate of 4 km/hr. The

second man walks 2 km in the first hour, 122 km

in the second hour, 3 km in the third hour andso on. Find the time after which the two men willmeet.

[SSC UDC (Mains) Exam–1999]23. Two trains start out towards each other from

points 650 km apart. If they start out at the sametime, they will meet in 10 hours, but if one ofthem starts out 4 hours and 20 minutes afterthe other, they will pass each other 8 hoursfollowing the departure of the latter. Determinethe average speed of each other.

[SSC UDC (Mains) Exam–1999]24. Distance between two stations X and Y is 220

km. Trains P and Q leave station X at 8 am and9.51 am respectively at the speed of 25 km/hrand 20 km/hr respectively for journey towards Y.A train R leaves station Y at 11.30 am at a speedof 30 km/hr for journey towards X. When andwhere will P be at equal distance from Q and R.

[SSC UDC (Mains) Exam–1999]25. Two places P and Q are 336 km apart. A train

leaves P for Q and at the same time another train

leaves Q for P. Both trains meet at the end of the6 hours. If one trains travels 8 km/hr faster thanthe other, find the speed of the other trains.

[SSC LDC (Mains) Exam–2007]26. On a 2-km road, a total number of 201 trees are

planted on each side of the road at equaldistances. How many such trees in all will beplanted on both sides of a 50-km road such thatthe distance between two consecutive trees isthe same as that of the consecutive trees on the2-km road?

[SSC UDC (Mains) Exam–2004]27. Two men A and B walk from P to Q a distance of

21 km, at 3 and 4 km an hour respectively.B reaches Q, returns immediately and meets Aat R. Find the distance from P to R.

[SSC UDC (Mains) Exam–2000;[SSC LDC (Mains) Exam–2000]

28. A train met with an accident 3 hours afterstarting, which detains it for one hour, after whichit proceed at 75% of its original speed. It arrivesat the destination 4 hours late. Had the accidenttaken place 150 km farther along the railway line,

the train would have arrived only 132 hours late.

Find the length of the trip and the original speedof the train.

[SSC UDC (Mains) Exam–1999]29. A man covers a certain distance on scooter. Had

he moved 3 km/hr faster, he would have taken40 minutes less. If he had moved 2 km/hr slower,he would have taken 40 minutes more. Find thedistance (in km) and original speed.

[SSC Assistant Grade (Mains) Exam–2007]

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

Now, according to the question,

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

1251

67D

x

25

6125D

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

5x = 60 × 5 = 300

x = 5300

= 60 km

4

KKUNDAN

3. Let the required distance be x km.

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

hour

=

60

5x

minutes = 12x minutes

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

hour

=

60

6x

minutes = 10x minutes

Since the difference between the two times taken is(7 + 5 =) 12 minutes 12x – 10x = 12 2x = 12

x = 212

= 6

Hence, the required distance is 6 km.Alternative Method:Let x km be the distance between her house andschool and t hours be the time required to reach theschool 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

121

6 tx

....(ii)

Subtracting equation (ii) from equation (i), we get

121

607

65ttxx

51

6012

6075

607

121

30

x

x = 530

= 6

Hence the required distance = 6 km4. Solve as Q.No. 3. Try yourself.

[Ans : 18 km]5. Solve as Q.No. 3. Try yourself.

[Ans : 2 km]6. Let the required distance be x km.

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

= 6070

hours = 67

hours

Times taken at 3 km/hr = 3x

hours

Times taken at 3 km/hr = 4x

hours

According to the question,

67

43

xx

67

1234

xx

67

12

x

x = 1267 = 14

Distance of the destination = 14 km7. Solve as Q.No.

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

12

124

1

xx

881212 xx 204 x 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.8. Let the original speed and distance be V 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

32

VD

3VD


KKUNDAN

32

3VD

VD

32

3VVDV3DDV

32

3VV3D

9

3V2VD ....(i)

When the person moves 2km/hr slower, then

6040

VD

2VD

32

VD

2VD

32

VD

2VD

32

2)V(V2DDVDV

32

2)V(V2D

32)V(VD

....(ii)

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

32)V(V

93)2V(V

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

we get D = 31012 = 40 km.

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

Let A and B be two places and C be the place ofmeeting.Let the speed of car starting from A be x km/hr andthe car starting form B be y km/hr.Relative speed = (x – y) km/hrAccording to the question, (x – y) × 8 = 80 x – y = 10 ...(i)Case II: When the cars are moving in the oppositedirections 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

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.10. 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

443

54

y

y = 1516204

= 80 km/hr

Therefore speed = 80 km/hr and distance

x = 8043 = 60 km

11. Half of the original speed means double the normaltime. It means that the car should have covered halfof the distance of 100 km, ie 50 km, in 2 hours.Hence, the original speed of the car

= 250

= 25 km/hr

12. 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 the abovedistance

= 6023

417

= 60

231517

= 60232

6023255

=

1558 hours

6

KKUNDAN

Average speed of the train =

15583

580

km/hr

= 35815580

= 50 km/hr

13.

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 A 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), we have

23

48724

x

x

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

14. 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 it catchesup with army man = 780 metresSpeed of the sound = 325 m/sec (given)Since the man and sound are travelling in the samedirection, 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 have travelledfurther. So, distance covered by man in 2.5 seconds= 2.5 × 13 = 32.5 mThe total distance travelled by man

= 780 + 32.5 = 812.5 metres.15. X P Y

l l l

Let the speed of the person who starts from X be xkm/hr and speed of the person who starts from Y bey km/hr.Time taken by the person who starts from X

= 5 am – 2 am = 3 hoursTime taken by the person who starts from Y

= 9 am – 4 am = 5 hoursAgain, let the distance between X and Y be D km.Now, according to the question,

x km/hr = 3D

and y km/hr = 5D

If the person starting from X reaches the meetingpoint after t hours, person starting from Y will reachthe meeting point after (t – 2) hours. Since the personstarting from X starts moving at 2 am while the personstarting from Y starts moving at 4 am. And thedifference of time = (4 am – 2 am)= 2 hours Distance (XP) travelled by the person starting from

X =

t

3D

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

D5

23

D

tt

15

23

tt

115

635

tt

8t – 6 = 15 8t = 15 + 6 = 21

t = 852

821

hours

Converting this in hours, minutes and seconds, weget 2 hours 37 minutes and 30 seconds.

[ 852 hours = 2 hours +

6085

minutes

= 2 hours + 21

372

75

minutes

= 2 hours + 37 minutes + 21

minutes

= 2 hours + 37 minutes +

6021

30 seconds = 2 hours 37 minutes 30 seconds]

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

Speed (in km/hr) = 1550x

..... (i)

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

= 6Y2

TimeDistance

x ..... (ii)


KKUNDAN

In both the cases, the speed remains the same

15502

6Y2

xx

Y × 6 = 50 × 15

Y = 61550

= 125 days

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

= 2183990 = 1520 km

18. Let the distance between station A and station B bed km.Again, let the initial speed of the goods train be xkm/hr.As the accident takes place after 1 hour distance covered in 1 hour by the goods train

= x kmRemaining distance = (d – x) kmTotal time taken, if no accident happened

=

xd

hours

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

=

5460

301 x

xd

=

xxd

4)(5

211 hours


24

)(5211

xd

xxd

21

45)(

xd

xxd

21

45)(

xd

xxd

21

4455

xdxd

21

45

x

xd

2d – 10x = 4x 2d = 14x d = 7x ....(i)Case II:If the goods train had covered 80 km more before theaccident, then the distance of site of 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

14

]80([521801

xd

xxd

x

xd

xxd

x

4)]80([5

2180

xd

xxd

21

440055320

xd

xxd

21

48055

xxdd

480554

21

2x = 5x – d + 80Putting the value of d from equation (i), we have 2x = 5x – 7x + 80 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.

19. Solve as Q.No. 18. Try yourself.[Ans: Speed = 60 km/hr

Length of the journey = 600 km]20. Let the distance be D km and speed be the x km/hr

From the question, we have

125D

6025D

34)50D(50

xxxx

xx

x 125D12

3200D4150

xx

x 125D12

350D4

x5D12200D16 4D – 5x = 200 ... (i) and

6035D

34)26D(2450

xxx

xx

x 127D12

127D

xx

xx 127D12

3104D426

xx

x 127D12

3104D478

8

KKUNDAN

xx

x 127D12

326D4

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

D = 4

440 = 110 km. (Ans)

21. 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 quantities : Speed, Distance, Workand Days. We have to calculate number of days.Hence, Days will be in the last column. Herefollowing relationships 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

2 × 1 × 6 : 1 × 2 × 15 :: 40 : x 2 × 1 × 6 × x = 1 × 2 × 15 × 40Product of extremes = Product of means

x = 612401521

= 100

Hence the required time = 100 days.22. 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 term and 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

47 2tt

+ 4t = 72

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

23. Let the trains A and B travel at speed of x and y km/hr respectively and meets 10 hours after departure.

From the figure it can be seen thatAC = (x × 10) kmBC = (y × 10) km AC + BC = xX × 10 + y × 10or, 650 = 10(x + x)or, x + y = 65In the second situation when the other train startsafter 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 A leaves P.Now if they meet at C1 thenPC1 = 8 × x = 8x kmBC1 = 8 × y = 8y kmAccording to the question,

8x + 8y = 313650 x

3

136508 xyx

313650658 x

1305206503

13

x

x = 133130

= 30 km/hr

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

= 35 km/hr24. 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 R1R

YX 33 km 25 t 30 t

20 t

87.5 km


KKUNDAN

Distance travelled by train P between 8:00 to 11:30

ie in 213 hours = 5.87

217525

27

km

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

1 hour 39 minutes = 20203320

60391

= 33 kmAssume that trains P and Q are at A and B respectivelyat 11 : 30 am. Also assume that t minutes after 11 :30 am, train P was equidistant from train Q andtrain R. At the equidistant position train P, Q and Rwere at P1, 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 kmP1R1 = 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

60 t = 78 or t = 606078

minutes

t = 78 minutesSo 78 minutes after 11 : 30 am ie at 12 : 48 pm trainP will be equidistant from train Q and R.

XP1 = 87.5 + 25 t = 87.5 + 25 × 6078

= 87.5 + 32.5

XP1 = 120 km At 120 km away from station X, trains would be atequal distances.

25.

Let R be the meeting point.Let the speed of train form P = x km/hr and thatform Q = (x + 8) km/hrBoth trains meet after 6 hours (x × 6) + (x + 8) × 6 = 336 6x + 6x + 48 = 336 12x = 336 – 48 = 288

x = 12288

= 44

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

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

=

1201

10002 = 10 m

Number of trees planted on both sides of a 50-kmroad

=

1101000502 = 10002

27. When B meets A at R, B has walked the distance PQ+ QR and A the distance PR. That is both of themhave together walked twice the distance from P toQ, ie 42 km.

Now the rates of A and B are 3 : 4 and they havewalked 42 km.Hence the distance PR travelled by A

= 37 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 = 42

4x

hours

Distance travelled by A = x km

Time taken by A = 3x

hours

42

3 4x x

4x = 126 – 3x 7x = 126

x = 126

7 = 18

required distance = 18 kmAlternative Method II:A’s speed = 3 kmB’s speed = 4 kmLet us consider that A and B meets after t hours.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 is twicethe distance between P and Q.So, 7t = 21 × 2

t = 2127

t = 6 hoursSo, the distance between P and R = Distance travelledby A = 3 × 6 = 18 km.

28. Let the length of the trip be d km and the originalspeed 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) kmTotal time taken by the train if no accident happens

=

xd

hours

Case I:Time taken by the train to cover the whole length ofthe trip

10

KKUNDAN

= ( 3 )3 1 75

100

d x

x

hours

= 4( 3 )4

3d x

x hours


4( 3 )4 43

d x dx x

4 12

3d x d

x x

4d – 12x = 3d d = 12x .... (i)Case II:If the train had covered 150 km more before theaccident then the distance of the accident = (3x +150) kmRemaining distance

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

3 150 (3 150)1 75

100

x d xx x


3 150 (3 150) 71 3 24

x d x dxx x

3 150 4 12 600 7 1

3 2x d x d

x x x

3 150 4 12 12 600

3x x x

x x

2512

xx

12d x

9 450 36 600 29

3 2x x

x

3 150 12 200 29

2x x

x

3 150 12 200 29

2x x

x

15 50 29

2xx

30x – 100 = 29x x = 100Hence, speed = 100 km/hr and the length of thetrip (d) = 12x = 12 × 100 = 1200 km

29. Suppose the distance is D km and the initial speedis x km/hr.

Then, we have 6040D

3D

xx and

6040D

2D

xx

32

3DD

xx

32

)3(3D

xx

.... (1)

and 32D

2D

xx

32

)2(2D

xx

.... (2)

From (1) and (2), we have

)2(D2

)3(D3

xxxx

)3(2)2(3 xx

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 the original speedis 12 km/hr.


KKUNDAN

ssc_speed, time and distance

Documents