chapter 18

Time and Distance

Rule 1 (i) Distance = Speed * Time

Distance (ii)Time =

(iii) Speed=

(iv)x km/hr = I * x jg I m/sec

Speed

Distance Time

Ex. 5: Express 10 m/sec in km/hr. Soln: Applying the above formula (v), we have,

10m/sec= [ 1 0 x y ] km/hr = 36km/hr,

Exercise 1. A train runs at the rate of 45 km an hour. What is its

speed in metres per second?

a) 12 m/sec 2

b)25 m/sec

(v) x m/sec = x x - km/hr

Illustrative Examples Ex. 1: Find the distance covered by a man walking for 10

minutes at a speed of 6 km/hr. Soln: Applying the above formula (i), we have

10 10 Distance = 6 km/hr x - hrf-.- 10 minutes = ~ hrl

60 60 = l km

Ex. 2: Find the time taken to cover a distance of 125 km by a train moving at a speed of 50 km/hr.

Soln: Applying the above formula (ii). we have,

125 Time = hours = 2.5 hours

Ex. 3: A train covers a distance of 1250 km in 25 hours. Find the speed of the train.

Soln: Applying the above formula (iii),

Speed = - ^ = 50 km/hr 25

Ex. 4: Express a speed of 18 km/hr in metres per second. Soln: Applying the above formula (iv), we have

18 km/hr ~ 1 I m/sec = 5 m/sec.

c) 10 m/sec 2

d) None of these

2. A motor car takes 50 seconds to travel 500 metres. What is its speed in km per hour? a) 32 km/hr b) 36 km/hr c) 34 km/hr d) 38 km/hr

3. How many km per hour does a man walk who passes through a street 600 m long in 5 minutes? a) 3.6 km/hr b) 7.2 km/hr c) 8 km/hr d) None of these

4. Compare the rates of two trains, one travelling at 45 km an hour and the other at 10 m a second. a)4:5 b)3:5 c)5:4 d)5:3

5. The distance of the sun from the earth is one-hundred forty-three million four hundred thousand km and light travels from the former to the latter in seven minutes and fifty-eight seconds. Find the velocity of light per sec-ond. a) 300000 km/sec b) 30000 km/sec c) 3 x 106 km/sec d) Can't be determined

. A2 6. The wheel of an engine 4 metres in circumference

makes seven revolutions in 4 seconds. Find the speed of the train in km per hour. a) 18 km/hr b) 24 km/hr c) 36 km/hr d) 27 km/hr

7. Sound travels 330 metres a second. How many kilometres is a thunder-cloud distant when the sound follows the flash after 10 seconds?

4 2 0 P R A C T I C E B O O K ON Q U I C K E R MATHS

a) 3.3 km b)3km c) 3.2 km d) 3.5 km 8. What is the length of the bridge which a man riding 15

km an hour can cross in 5 minutes?

a) 1 km b) 2 km c) 2 km d) 1 km

9. A man crosses a street 600 m long in 5 minutes. His speed in kilometres per hour is

IClerical Grade Exam, 19911

a) 7.2 b) 5_

36 c)3.6 / X d) 10

10. I f a man covers 10 km in 3 hours, the distance cov-

ered by him in 5 hours is [Asst. Grade Exam, 1987j

a) 18km b)15km c)16km d)17km 11. A train travels 92.4 km/hr. How many metres will it travel

inlOminutes? [BankPO Exam, 1991] a) 15400 b) 1540 c)154 d) 15.40

12. xA train covers a distance in 50 minutes i f it runs at a \ speed of 48 km per hour on an average. The speed at

which the train must run to reduce the time of journey to 40 minutes, will be

|Central Excise and I Tax Exam, 1988] a) 50 km/hr b) 55 km/hr c) 60 km/hr d) 70 km/hr

Answers 1. a; Hint: Use the given rule (iv)

ie required answer = 45x5 25

18 12

1 m per sec

500 2. b; Hint: Speed of the motor car in m/sec = -TJT =10 m/sec

10x18 10 m/sec = a = JO km/hr (Use rule (v))

time = 7 x 60 + 58 = 478 seconds

143400000 .-. speed =

478 = 300000 km/sec

6. d; Hint: Distance travelled by the train in 4 seconds

30 = x7 = 30 metres

15 m/sec 2 (using the given rule (

.-. speed of the train = =

15 15 18 m/sec = ~ x " =27 km/hr 2 2 5

(Using the given rule (v))

330x10 7. a; Hint: Distance = Speed * Time = = 3.3 km

. . . 5 5 . 1 8. d; Hint: Required answer = > 5 x = = 1 - km

^ 60 4 4

(Using the given rule (i))

18 5 x 6 0 X 5 600 18

9. a; Hint: Required answer = - ~ x = 7.2 km/hr

10. d; Hint: Distance covered in 5 hours = ,5x3

17km

x5 km

I i . a: Hint: speed of nam t * | g j m/sec = - m/sec

( 77 Distance covered in (10 * 60) sec = ^ ~ x 10x60

metres = 15400 metres

600 _ 3. b; Hint: Man's speed in m/sec = -- = 2 m/sec

5x60 18x2 _

2 m/sec = : = 7.2 km/hr

4. c; Hint: Speed of the first train = 45 km/hr

18 Speed of the second train = 10 m/sec = j 0 * = 36 km/

hr

.-. ratio of the speeds of the trains = 45 : 36 = 5 : 4

Distance 5. a; Hint: Speed =

Time Here, distance between the sun and the earth

= 143400000 km and

12. c; Hint: Distance = (speed x time) =

Distance (40x60

48 x 1 = 40 km 60 J

Speed = Time 40 km/hr = 60 km/hr

Rule 2 Theorem: If a certain distance is covered at x km/hr and the same distance is covered at y km/lir then the average

Ixy

speed during the whole journey is x + km/hr.

Illustrative Example Ex.: A man covers a certain distance by car driving at 70

km/hr and he returns back to the starting point riding on a scooter at 55 km/hr. Find his average speed for

Time and Distance -1\

the whole journey. Soln: Applying the above theorem, we have,

2x70x55 Average speed = 7 0 + 5 5 km/hr = 61.5 km/hr.

2.

Exercise 1. A man covers a certain distance by car driving at 40 km/

hr and he returns back to the starting point riding on a scooter at 10 km/hr. Find his average speed for the whole journey. a) 8 km/hr b) 16 km/hr c) 12 km/hr d) None of these A man covers a certain distance by car driving at 30 km/ hr and he returns back to the starting point riding on a scooter at 20 km/hr. Find his average speed for the whole journey. a) 24 km/hr b) 27 km/hr c) 15 km/hr d) 12 km/hr A man covers a certain distance by car driving at 15 km/ hr and he returns back to the starting point riding on a scooter at 35 km/hr. Find his average speed for the whole journey. a) 20 km/hr b) 18 km/hr c) 21 km/hr d) 24 km/hr A man covers a certain distance by car driving at 40 km/ hr and he returns back to the starting point riding on a scooter at 20 km/hr. Find his average speed for the whole journey.

a) 1 6 - km/hr b)26km/hr

d) 2 6 - km/hr

3

c) 36 km/hr

A train goes from a station A to another station B at a speed of 64 km/hr but returns to A at a slower speed. I f its average speed for the trip is 56 km/hr, the return speed of the train is nearly. | Hotel Management, 19911 a) 48 km/hr b) 50 km/hr c) 52 km/hr d) 47.4 km/hr On a tour a man travels at the rate of 64 km an hour for the first 160 km, then travels the next 160 km at the rate of 80 km an hour. The average speed in km per hour for the first 320 km of the tour, is a)35.55 b)71.11

Answers 2. a 3.c

[SBIPO Exam, 1988| c)36 d)72

4.d b; Hint: Let the return speed of the train be y km/hr

Then, applying the given rule, we have, 2x64x_y _ ^

"(64+v) ~~ or,2 x64 xy = 56(64+y) or, y = 49.8 = 50 km/hr (nearly)

( 2x64x80^ b; Hint: .-. Average speed = I ^ + go I km/hr

= 71.11 km/hr

Rule 3 Theorem: A person is walking at a speed ofx km/hr. After every kilometre, if he takes restfor t hours, then the time h e

will take to cover a distance ofy km is hr. Or.

in other words, required time

_ Distance to be covered Speed

+ Number of rest x

Time for each rest

Illustrative Example Ex.: A man is walking at a speed of 12 km per hour. After

every km he takes rest for 12 minutes. How much time will he take to cover a distance of 36 km.

Soln: Detail Method: Rest time = Number of rest x Time for each rest

12 -= 3 5 x - = 7hr

[ v To cover 36 km, the man has to take rest 35 times, as he takes rest every hour]

36 , Total time to cover 36km = + 7 = 10 hours.

12 Quicker Method: Applying the above theorem, Here, x= 12 km/hr y=36km

12 t = 12 minutes :

60 5 hr.

2.

36 , \Total time = + v->6 -1) =3 + 7=10 hours.

Exercise 1. A man is walking at a speed of 6 km per hour. After every

km he takes rest for 6 minutes. How much time will he take to cover a distance of 18 km. a) 5 hrs b) 5 hrs 42 min c) 4 hrs 42 min d) None of these A man is walking at a speed of 18 km per hour. After every km he takes rest for 18 minutes. How much time will he take to cover a distance of 54 km. a) 18 hrs b) 18 hrs 54 min c) 16 hrs 54 min d) None of these A man is walking at a speed of 9 km per hour. After every km he takes rest for 9 minutes. How much time will he take to cover a distance of 27 km. a) 6 hrs b) 6 hrs 45 min c) 6 hrs 54 min d) None of these A man rides at the rate of 18 km an hour, but stops 6 minutes to change hours at the end of every 7th kilometre How long will he take to go a distance of 90 kilometres0

3.

4.


a) 6 hrs b) 6 hrs 12 min c) 6 hrs 15 min d) None of these

5. A man rides at the rate of 20 kilometres an hour. But stops 10 minutes to change horses at the end of every 25th kilometre. How long will he take to go a distance of 175 kilometres? a) 9 hrs 15 min o) 9 hrs 45 min c) 10 hrs d) 8 hrs 45 min

Answers l .c 2.b 3.c

Hint: Here instead of every kilometre, every 7th kilometre is given. Hence, we should first calculate

4.b;

no. of rest ie 90

12

Now applying the given rule, we have

the required answer: + 12x 18 60

= 5 hrs+ 1 hr 12 min = 6 hrs 12 min

5.b; Hint: No. of rest : 175 ~25

- 1 = 6

175 , U 1 .-. required answer = -r + 6 x - y hrs 45 min

20 60

Rule 4 Theorem: A person covers a certain distance between two

points. Having an average speed of x km/lir, he is late by xx

hours. However, with a speed of y km/hr he reaches his

destination ys hours earlier. The distance between the two

points is given by

Required distance

*r(*i + y\) ( ? - * ) J

km. Or,

Product of two speeds Difference of two speeds

x Difference between

arrival times

(See note)

Illustrative Example Ex.: A man covers a certain distance between his house

and office on scooter. Having an average speed of 30 km/hr, he is late by 10 min. However, with a speed of 40 km/hr, he reaches his office 5 min earlier. Find the distance between his house and office.

Soln: Detail Method Let the distance be A: km.

Time taken to cover* km at 30 km/hr = hrs.

Time taken to cover x km at 40 km/hr = 40 hrs.

Difference between the time taken = 15 min = hr. 4

or, 4x - 3x = 30 or, x = 30 x x _ 1

' 3 0 ~ 4 0 ~ 4 Hence, the required distance is 30 km. Quicker Method: Applying the above theorem, we hove

30x40 10 + 5 the required distance = ~ ~ x - 30 km.

40-30 60 Note: 10 minutes late and 5 minutes earlier make a differ-

ence of 10 + 5 = 15 minutes. As the other units are in km/hr, the difference in time should also be changed into hours.

15 1 .-. 15 minutes = ~TZ ~r hr.

60 4 Exercise 1. A man covers a certain distance between his house and

office on scooter. Having an average speed of 15 km/hr, he is late by 5 min. However, with a speed of 20 km/hr, he

. 1 reaches his office 2 min earlier. Find the distance be-

2 tween his house and office.

a)15km b)15^-km c)7km 1

d ) 7 - km 2

2. A man covers a certain distance between his house and office on scooter. Having an average speed 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 his house and office. a) 120km b)90km c)80km d)60km

3. A man covers a certain distance between his house and office on scooter. Having an average speed of 45 km/hr, he is late by 15 min. However, with a speed of 60 km/hr,

he reaches his office 7 min earlier. Find the distance 2

between his house and office.

a) 45^- km b) 6 7 k m c)45km d) 37^ km

4. Ram travels at the rate of 3 km/hr and he reaches 15 minutes late. I f he travels at the rate of 4 km/hr, he reaches 15 minutes earlier. The distance Ram has to travel is

|CDS Exam, 1989| a) l k m b)6km c)7km d)12km

Answers l . d 2. a 3.b 4.b

Time and Distance 423

Rule 5 Theorem: A person walking at a speed ofx km/hr reaches

his destination x] hrs late. Next time he increases his speed

by y km/hr, but still he is late by y, hrs. The distance of his

destination from his house is given by

km.

(x, -y^x + y)-

Illustrative Example Ex.: A boy walking at a speed of 10 km/hr reaches his

school 15 minutes late. Next time he increases his speed 2 km/hr, but still he is late by 5 minutes. Find the distance of his school from his house.

Soln: Applying the above theorem, we have the required distance

l ^ | ( 1 0 + 2 ) x H = 10km. 60 Jv ' 2

Exercise 1. A boy walking at a speed of 15 km/hr reaches his school

20 minutes late. Next time he increases his speed 5 km/hr, but still he is late by 10 minutes. Find the distance of his school from his house. a) 15km b)10km c)18km d)20km

2. A boy walking at a speed of 45 km/hr reaches his school 10 minutes late. Next time he increases his speed 15 km/ hr, but still -he is late by 5 minutes. Find the distance of his school from his house. a) 10km b)12km c)25km d)15km

3. A boy walking at a speed of 30 km/hr reaches his school 40 minutes late. Next time he increases his speed 10 km/ hr, but still he is late by 20 minutes. Find the distance of his school from his house. a) 30 km b)25km c) 40 km d) None of these

4. A man walking with a speed of 5 km/hr reaches his target 5 minutes late. I f he walks at a speed of 6 km/hr, he still reaches 3 minutes late. Find the distance of his target from his house.

a) 3 km b)2km c) ! km d ) l k m

Answers l .b 2.d 3.c 4. d;Hint: Herey = 6km/hr-5km/hr = 1 km/hr

Now applying the given rule we have

the required distance =

Rule 6 Theorem: A person goes to a destination at a speed of x km/hr and returns to his place at a speed of y km/hr. If he takes T hours in all, the distance between his place and

( destination is

xy xT yx + y

Required distance

km. In other words,

Total time taken x Product of the two speeds Addition of the rwo speeds

Illustrative Example Ex: A boy goes to school at a speed of 3 km/hr and re-

turns to the village at a speed of 2 km/hr. I f he takes 5 hrs in all, what is the distance between the village and the school?

Soln: Detail Method: Let the required distance be x km.

Then time taken during the first journey = hr.

and time taken during the second journey = hr.

x x . 2x + 2>x .-. + = 5 => = 5 => 5x = 30

3 2 6 .-. x = 6 :. required distance = 6 km Quicker Method: Applying the above rule, we have

3x2 the required distance = 5 x = 6 km.

3 + 2

Exercise

1 1. A man walks to a town at the rate of 5 kilometres an

hour and rides back at the rate of 10 kilometres an hour, how far has he walked, the whole time occupied having been 6 hours 12 minutes? a)31km b)29km c)22km d)17km

2. A man walks to a town at the rate of 4 ki lometres an hour and rides back at the rate of 5 kilometres an hour, how far has he walked, the whole time occupied having been 9 hours? a)20km b)25km c)18km d)15km

3. A man walks to a town at the rate of 5 kilometres an hour and rides back at the rate of 7 kilometres an hour, how far has he walked, the whole time occupied having been 6 hours? a)17km b)35km c)17.5km d)35.2km

4. A man walks to a town at the rate of 8 kilometres an hour and rides back at the rate of 6 kilometres an hour, how far


has he walked, the whole time occupied having been 7 hours? a) 14km b)18km c)22km d)24km

5. A boy goes to school with the speed of 3 km an hour and returns with a speed of 2 km/hr. I f he takes 5 hours in all, the distance in km between the village and the school is |NDA 1990| a) 6 b)7 c)8 d)9

6. A car completes a certain journey in 8 hours. It covers half the distance at 40 km/hr and the rest at 60 km/hr. The length of the journey in kilometres is

IClerical Grade, 19911 a) 192 b)384 c)400 d)420

Answers

11 , 12 31 km/hr, y = 10 km/hr, T = + = hours 2 60 5

l.c;Hint: x


the required distance: xlO a, 2 w j l 11

22 km

2. a 3.c

1

4.d

+ 10 J

5.a

40x60 6. b; Hint: ofthejourney

2 - ^ 4 0 + 60 x8 = 192 km

.-. The total length of the journey = 192 x 2 = 384 km

Rule 7 Theorem: If a person does a journey in Thours and thefirst

half at S, km/It r andthe second halfat S2 km/hr, then the

2 x Time x S, x S2 distance S,+S,

Where, S, = Speed during first half and S2 = Speed during second half of journey

Illustrative Example Ex.: A motor car does a journey in 10 hrs, the first half at

21 km/hr and the second half at 24 km/hr. Find the distance.

Soln: Detail Method: Let the distance be x km.

x x Then km is travelled at a speed of 21 km/hr and

km at a speed of 24 km/hr. Then, time taken to travel the whole journey

= 10 hrs

So..v =

2x21 2x24

2x10x21x24

Quicker Method: Applying the above theorem, we have

2x10x21x24 a Distance = = 224 km.

21 + 24 Note: Remember that half of the journey means half of the

distance and not the time. Exercise 1. A motor car does a journey in 12 hrs,

kjn/hr and the second half at 30 km/hr. a) 190 km b) 280 km c) 240 km

2. A motor car does a journey in 6 hrs, km/hr and the second half at 20 km/hr. a) 90 km b)80km c)60km these

3. A motor car does a journey in 27 hrs, km/hr and the second half at 13 km/hr. a)264km b)351km c)251km

4. A motor car does a journey in 9 hrs, km/hr and the second half at 15 km/hr. a) 120 km b) 100 km c) 124 km

Answers l .c 2.b 3.d 4. a

the first half at 15 Find the distance.

d)210km the first half at 10 Find the distance.

d) None of

the first half at 14 Find the distance.

d) 364 km the first half at 12 Find the distance.

d)96km

Rule 8 Theorem: The distance between two stations, A and R is D km. A train starts from A and moves towards B at an aver-age speed of x km/lir. If an another train starts from B, t hours earlier than the train at A, and moves towards A at an average speed of y km/hr, then the distance from A,

where the two trains will meet, is {D-ty x + y) km.

21 + 24 = 224 km.

Illustrative Example Ex.: The distance between two stations, Delhi and

Amritsar, is 450 km. A train starts at 4 pm from Delhi and moves towards Amritsar at an average speed of 60 km/hr. Another train starts from Amritsar at 3.20 pm and moves towards Delhi at an average speed of 80 km/hr. How for from Delhi will the two trains meet and at what time?

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

40 - [Time taken by A to cover x km] = (see note)

60 450-x x 40

80 60 " 6 0


.-. 3(450-x)-4x = 160 7x = 1190 => x = 170. Thus, the trains meet at a distance of 170 km from Delhi.

Time taken by A to cover 170 km : 170 60 hrs

So, the trains meet at 6.50 p.m. = 2 hrs 50 min.

40 Note: RHS = 4:00 p.m. - 3.20 p.m. = 40 minutes = hr

LHS comes from the fact that the train from Amritsar took 40 minutes more to travel up to the meeting point because it had started its journey at 3.20 p.m. whereas the train from Delhi had started its journey at 4 p.m. and the meeting time is the same for both the trains. Quicker Method: Applying the above theorem, we have the distance from Delhi at which two trains meet

= 1 4 5 0 - ^ x 8 0 M^-60 A 60+ 80

190 170 km.

170 Time taken by A to cover 170 km = hrs

= 2 hrs 50 min.

So, the trains meet at 6.50 p.m.

Exercise 1. At what distance from Delhi will a train, which leaves

Delhi for Amritsar at 2.45 pm and goes at the rate of 50 km an hour, meet a train which leaves Amritsar for Delhi at 1.35 pm and goes at the rate of 60 km per hour, the distance between the two towns being 510 km? a) 150km b) 170km c)200km d)210km

2. The distance between two stations A and B is 900 km. Atrain starts from A and moves towards B at an average speed of 30 km/hr. Another train starts from B, 20 min-utes earlier than the train at A, and moves towards A at an average speed of 40 km/hr. How far from A will the two trains meet? a) 380 km b) 320 km c) 240 km d) None of these

3. The distance between two stations A and B is 450 km. Atrain.starts from A and moves towards B at an average speed of 15 km/hr. Another train starts from B, 20 min-utes earlier than the train at A, and moves towards A at an average speed of 20 km/hr. How far from A will the two trains meet? a) 180 km b) 320 km c) 190 km d) 260 km

4. Two stations A and B are 110 km apart on a straight line. A train starts from A and travels towards B at 40 km/hr.

Another train, starting from B, 2 hrs earlier, travels to-wards A at 50 km/hr. Find the distance from station A at which two trains meet?

a) 5 km

Answers l .c 2.a

b) 5 km c) 4 km d) 4 - km

3.c 4.d

Rule 9 Theorem: The distance between two stations A and B, is D km. A train starts from A and moves towards B at an aver-age speed of x km/hr. If an another train starts from B, t hours later than the train at A, and moves towards A at an average speed ofy km/hr, then the distance from A, where

the two trains will meet, is {D + ty x + y) km.

Illustrative Example Ex.: The distance between two stations, Delhi and

Amritsar, is 450 km. A train starts at 3.10 pm from Delhi and moves towards Amritsar at an average speed of 20 km/hr. Another train starts from Amritsar at 4 pm and moves towards Delhi at an average speed of 60 km/hr. How far from Delhi will the two trains meet and at what times. Detail Method: Suppose the trains meet at a distance of x km from Delhi. Let the trains from Delhi and Amritsar be A and B respectively. Then,

[Time taken by A to cover x km] - [Time taken by B to

50

Soln:

60 .. (see note) cover (450 - x) km] =

x 450 - x _ 50 20 60 ~ 60

3x-450 + x = 50

or, 4x = 500 x = = 125 km. 4

Thus, the trains meet at a distance of 125 km from Delhi.

125

Time taken by A to cover 125 km = -ZT - 6 hrs 15

minutes. So the trains meet at 9.25 p.m.

50 Note: RHS = 4 p.m.-3.10 p.m. = 50 minutes or, 77 hrs.

60 LHS comes from the fact that the train from Delhi took 50 minutes more to travel up to the meeting point

1


because it had started its journey at 3.10 p.m. whereas the train from Amritsar had started its journey at 4 p.m. and the meeting time is the same for both the trains. Quicker Method: Applying the above theorem, we have distance from Delhi at which two trains meet

450-t ~^x< Y 20 A 20+ 60

500 = 1 2 5 k m

Time taken by A to cover 125 km

125 20

= 6 hrs 15 min.

So the trains meet at 9.25 pm.

Exercise 1. A train which travels at the uniform rate of 10 metres per

second leaves Patna for Kanpur at 7 am. At what dis-tance from Patna will it meet a train which leaves Kanpur for Patna at 7.20 am and travels one-third faster than it does, the distance from Patna to Kanpur being 68 km? a)28km b)42km c)36km d)40km

2. A train going 50 km an hour leaves Calcutta for Allahabad (900 km) at 9 pm. Another train going 70 km an hour leaves Allahabad for Calcutta at the same time, when and where will they pass each other? a) 375 km from Calcutta, 4.30 am b) 525 km from Calcutta, 4.30 pm c) 525 km from Allahabad, 4.20 am d) None of these

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

4. The distance between Delhi and Patna is 1000 km. A train leaves Delhi for Patna at 5 pm at 150 km/hr. Another train leaves Patna for Delhi at 6.30 pm at 100 km/hr. How far from Delhi will the two trains meet? a)690km b)310km c)590km d)410km

5. The distance between two stations A and B is 220 km. A train leaves A towards B at an average speed of 80 km per hr. After half an hour, another train leaves B towards A at an average speed of 100 km/hr. Find the distance from A of the point where the two trains meet. a) 180km b) 120km c) 160km d)80km

Answers 1

l .c; Hint: t = 7.20am-7am = 20minutes = A 3

v' \ A 18 x = 10m/sec=\10xy =36 km/hr

hr

y = 36 + 36* j =48km/hr

Now, applying the given rule we have the required answer

68 + 1x48 I f - * -3 J148+36J

3 6 km

2. a; Hint: Here / = 0, and applying the given rule, we get

50 5 + 70

x 900 = 375 km from Calcutta

375 _ 1 1 time = - - - - - I hrs = 9 pm + 7 hrs = 4.30 am

Note: This problem can also be solved by Rule 9. Try to solve by that rule also.

3. a; Hint: The distance from B where the two meet = 9 x 8 = 72 km

.-. the distance from A where they meet = D - 72 km Let the D be the distance between Allahabad and Kanpur.

(D + 2*8) 12 20 = D - 7 2

or,3D + 48 = 5D-360 or, 2D = 408 .-. D = 204km

4. a 5.b

Rule 10 a

Theorem: If the new speed of a person is of the usual

speed, then the change in time taken to cover the same

distance is 1 1 x usual time or, usual time is given by

Change in time

a hrs.

Note: A person improves his timing or becomes late de-pends on the -ve or +ve sign respectively of the above expression.

Illustrative Examples

_ . 3 Ex.1: Walking of his usual speed, a person is 10 min late

to his office. Find his usual time to cover the dis-tance.

Soln: Detail Method: Let the usual time be .v min.

"lme and Distance 427

3 4x Time taken at of the usual speed = min

4 3 [Note: I f the speed of a body is changed in the ratio a : b, then the ratio of the time taken changes in the ratio b: a.]

-X X = 10 : = 10=>x = 30 mm.


Usual time =

utes.

Change in time 10 10 = 30 min-

1 -3

Here sign is (+ve) and hence in the question late time has been given.

4 " -E\.2: Running of his usual speed, a person improves

his timing by 10 minutes. Find his usual time to cover the distance.

Soln: Direct Formula: Applying the above theorem,

Usual time : Change in time 10

= -40 minutes 2 _ , _ I 4 4

Here (-ve) sign shows that the person improves his timing. Hence in the question improved time has been given.

.-. correct answer = 40 minutes.

Exercise

Walking of his usual speed, a person is 15 min late to

his office. Find his usual time to cover the distance, a) 30 minutes b) 25 minutes c) 15 minutes d) None of these

Walking of his usual speed, a person is 6 min late to

his office. Find his usual time to cover the distance. a) 12 minutes b) 18 minutes c) 24 minutes d) Data inadequate

1 Walking - of his usual speed, a person is 12 min late to

his office. Find his usual time to cover the distance. a) 36 minutes b) 18 minutes c) 6 minutes d) Can't be determined

Walking - of his usual speed, a person is 18 min late to

his office. Find his usual time to cover the distance, a) 27 minutes b) 25 minutes c) 24 minutes d) 20 minutes

3

By walking at of his usual speed, a man reaches of-

fice 20 minutes later than usual. His usual time is [Railways, 19911

a) 30 minutes b) 60 minutes c) 75 minutes

Answers l .a 2.c 3.c

. 1 d) 1 hours

2

4. a 5.b

Rule 11 Theorem: If a train travelling x km an hour leaves a place and t hours later another train travelling y km an hour, where y > x, in the same direction, then they will be to-

gether after travelling t(xy)

v-km from the starting place.

Illustrative Example Ex.: A train travelling 25 km an hour leaves Delhi at 9 a.m.

and another train travelling 35 km an hour starts at 2 p.m. in the same direction. How many km from Delhi will they be together?

Soln: Detail Method I: Let the required distance be x km From the question,

x x

- = 2p.m.-9a.m. = 5 hours.

x(35-25) or, 35x25

35x25x5 x = = 437^-km. 10

Detail Method II: The first train has a start of 25 x 5 km and the second train gains (35 - 25) or 10 km per hour.

25x5 10

the seconcrtrain will gain 25 * 5 km in

or 12 hours. 2

.. the required distance from Delhi = 12 * 35 km

= 4 3 7 - k m . 2

428 P R A C T I C E B O O K ON Q U I C K E R MATHS


the required distance 25x35x(2 p.m.-9 a.m.)

35-25

25x35x5 1 ' = = 4 3 7 - km

10 2

Exercise 1. A train leaves Calcutta at 7.30 am and travels 40 km an

hour, another train leaves Calcutta at noon and travels 64 km an hour, when and where wil l the second train overtake the first? a) 480 km, 7.30 pm b) 480 km, 2.30 pm c) 840 km, 7.30 pm d) 480 km, 6.30 pm

A man starts at 7 am and travels at the rate of 4 km an 4

hour. At 8.15 am a coach starts from the same place and

, 1

follows the man travelling at the rate of 6 km an hour,

at what o'clock will the coach overtake the man? a) 12.50 pm

c) 12.53 pm

b) 11.38- am

d) None of these

3. A starts from Delhi to Alwar (24 km) at 6 am walking 4 kilometres an hour. B starts from Delhi an hour later and reaches Alwar one hour before A, where did they meet? a) 12 km from Delhi b) 10 km from Alwar c) 10 km from Delhi d) Data inadequate

Answers

1. a; Hint: t = 7.30 am - 12 noon = 4 = hrs 2 2


64x40 9 . . . x - = 480 km

60-40 2

480 15 _ 1 Required time = = = 7 hrs

64 2 2

.-. required answer = 12 noon + 7 hrs = 7.30 pm.

2. b; Hint: Here t = 7 am - 8.15 am = 1 hrs 4

.-. Applying the given rule we have the

required distance =

19 13 x _4 ?_ 19 _ 13 4 12

19x13x5 4 x 2 x 7

km

19x13x5 4 _ 65 r ' t i m e = 7 ^ 7 ~ X ! 9 = .4 h r S

required answer = 7 am + 4 hrs 14

9x60 4 = 11 am + min =11.38 am.

14 7 3. a; Hint: Here,* = 4 km/hr

24 Time taken by A to cover the distance of 24 km = = 6

hrs Time taken by B to cover the same distance = 6 - 2 = 4 hrs (according to the question)

24 r ' v = = 6 km/hr and / = 1 hr

4


6x4 the required answer; 6 - 4 x l =12 km from Delhi.

Rule 12 Theorem: If two persons A and B start from a place walk-ing at x km/hr and y km/hr respectively, at the end of t hours, when they are moving in same direction andx


: Two men A and B start from a place P walking at 6 km and 8 km an hour respectively. How many km will they be apart at the end of 2 hours, i f they walk in same direc-tion?

a) 2 km b)8km c)6km d)4km

Answers d:Hint: 100 steps = 5 x 100 = 500 dm = 50 m

50 _ 5 Speed of the first man = 50 m per minute = ~rz - ~ m/sec

60 6 5 5 18

.-. m/sec = 7 x t - =3 km/hr 6 6 5

Now, applying the given rule, we have 75

(4 -3) / = 1000

-x60

.-. t = =4 min 30 sec 2

2,b 3.d

Rule 13 Theorem: If two persons A and B start from a place walk--ig at x km/lir and y km/hr respectively, at the end of t hours, when they are moving in opposite directions, they ill he (x +y)t km apart.

Illustrative Example Ex.: Two men A and B start from a place P walking at 3 km

and 3.5 km an hour respectively. How many km will they be apart at the end of 3 hrs, i f they walk in the opposite directions?

Soln: Applying the above theorem, we have the required distance = (3.5 + 3) x 3 = 19.5 km.

Exercise Two men A and B start from a place P walking at 2 km and 5 km an hour respectively. How many km will they be apart at the end of 2 hrs, i f they walk in the opposite directions? a)7km b)8km c)12km d)14km

1 Two men A and B start from a place P walking at 3.5 km and 4.5 km an hour respectively. How many km will they be apart at the end of 3 hrs, i f they walk in the opposite directions? a)21km b)24km c) 18 km d) Data inadequate

Two men A and B start from a place P walking at 5.5 km and 6.5 km an hour respectively. How many km will they be apart at the end of 5 hrs, i f they walk in the opposite directions? a)60km b)50km c)45km d)30km

Answers I d 2.b

Rule 14 Theorem: Two men A and B walk from PtoQ,a distance of 'D' km, at the speed of 'a' km/hr and 'b' km/hr respectively. IfB reaches Q, returns immediately and meets A at R, then the distance travelled by A (or distance from P to R) is

2D ~~Z I km and the distance travelled by B (or PQ +

QR) is 2D km.

Illustrative Example Ex.: 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 A at R. Find the dis-tance from P to R.

Soln: Detail Method: When B meets A at R, B has walked the distance PQ + QR and A the distance PR. That is, both of them have together walked twice the distance from P to Q, i.e. 42 km. Now the rates of A and B are 3 : 4 and they have walked 42 km.

Hence the distance PR travelled by A

3 = -j of42km=18km

Quicker Method: When the ratio of speeds of A and B is a : b, then in this case: Distance travelled by A " 2 * Distance of two points

a + b

and distance travelled by B = 2 x Distance of two

P i n t s [Jtb, Thus, distance travelled by A (PR)

= 2x21 3 + 4 = 18km.

3.a

Exercise 1. Two men A and B walk from P to Q, a distance of 18 km,

at 4 and 5 km an hour respectively. B reaches Q, returns immediately and meets A at R. Find the distance from P toR. a) 15km b)16km c)12km d) Can't be determined

2. Two men A and B walk from P to Q. a distance of 22 km. at 5 and 6 km an hour respectively. B reaches Q, returns immediately and meets A at R. Find the distance from P toR. a) 16km b)18km c)20km d)15km


3. Two men A and B walk from P to Q, a distance of 30 km, at 7 and 8 km an hour respectively. B reaches Q, returns immediately and meets A at R. Find the total distance travelled by B. a)35km b)36km c)32km d)33km

4. Two men A and B walk from P to Q, a distance of 24 km, at 11 and 13 km an hour respectively. B reaches Q, re-turns immediately and meets A at R. Find the total dis-tance travelled by B.

a)26km b)28km c)30km d)32km

Answers l .b 2.c 3.c 4.a

Rule 15 t Theorem: If two persons A and B start at the same time in

opposite directions from two points and after passing each

other they complete the journeys in 'a' and 'b' hrs respec-

tively then A's speed: B's speed = 4b \4a

Illustrative Example Ex.: A man sets out to cycle from Delhi to Rohtak, and at

the same time another man starts from Rohtak to cycle to Delhi. After passing each other they complete their

journeys in i- and 4 hours respectively. At what

rate does the second man cycle i f the first cycle at 8 km per hour?

Soln: I f two persons (or train) A and B start at the same time in opposite direction from two points, and arrive at the point a and b hrs respectively after having met, then j

A's rate : B's rate = : 4ct (from the theorem)

Thus in the above case

1st man's rate 2nd man's rate

5 _ ,2 .-.2nd man's rate = 7 x 8 _ 0 ~ " km/hr.

6 i Exercise 1. A man sets out to cycle from Delhi to Rohtak, and at the

same time another man starts from Rohtak to cycle to Delhi. After passing each other they complete their jour-neys in 4 and 9 hours respectively. At wjiat rate does the second man cycle i f the first cycle at 9 km per hour? a) 4 km b) 6 km c) 8 km d) Data inadequate

2. A man sets out to cycle from Delhi to Rohtak, and at the same time another man starts from Rohtak to cycle to

Delhi. After passing each other they complete their jour-neys in 4 and 16 hours respectively. At what rate does the second man cycle i f the first cycle at 18 km per hour? a) 8 km b)9km c)12km d)6km

3. A man sets out to cycle from Delhi to Rohtak, and at the same time another man starts from Rohtak to cycle to Delhi. After passing each other they complete their jour-neys in 16 and 25 hours respectively. At what rate does the second man cycle i f the first cycle at 25 km per hour?

a)21km b)18km c) 1 2 - k m d)20km 2

4. A man sets out to cycle from Delhi to Rohtak, and at the same time another man starts from Rohtak to cycle to Delhi. After passing each other they complete their jour-neys in 9 and 16 hours respectively. At what rate does the second man cycle i f the first cycle at 16 km per hour? a) 12km b)15km c)8km d)6km

Answers l .b 2.b 3.d 4. a

Rule 16 V Report of guns Theorem: If two guns were firedfrom the same place at an

interval of f, minutes but a person in a train approaching

the place hears the second report t2 minutes after the first, then the speed of the train, supposing that sound travels at

330 metres per second, is 1188 km/hr.

Illustrative Example Ex.: Two guns were fired from the same place at an inter-

val of 13 minutes but a person in a train approaching the place hears the second report 12 minutes 30 sec-onds after the first. Find the speed of the train, sup-posing that sound travels at 330 metres per second.

Soln: Detail Method: It is easy to see that the distance trav-elled by the train in 12 min. 30 seconds could be trav-elled by sound in (13 min - 12 min 30 seconds) = 30 seconds

.-. the train travels 330 * 30 metres in 12 min.

the speed of the train per hour 330x30x2x60

* 25x1000

1188 13 = 1 f 0 T 4 7 25 k m -



speed of the train per hour =1188

f 25^ 13-

2_ 25 2

1188 ,.,13 or 4 7 km. 25 25

Exercise 1. Two bullets were fired at a place at an interval of 12

minutes. A person approaching the firing point in his car hears the two sounds at an interval of 11 minutes 40 seconds. The speed of sound is 330 m/sec. What is the speed of the car?

594 a)

1188 35

km/hr b) 35 km/hr

1881 c) km/hr d) Data inadequate

2. Two bullets were fired at a place at an interval of 34 minutes. A person approaching the firing point in his car hears the two sounds at an interval of 33 minutes. The speed of sound is 330 m/sec. What is the speed of the car? a)72km b)36km c)45km d)51km

3. Two bullets were fired at a place at an interval of 28 minutes 30 seconds. A person approaching the firing point in his car hears the two sounds at an interval of 27 minutes. The speed of sound is 330 m/sec. What is the speed of the car? a)44km b)66km c)64km d)54km

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

Answers l.a 2.b 3.b 4. a

Rule lfr Theorem: If two runners A and B cover the same distance it the rate of x km/lir andy km/lir respectively, then the distance travelled, when A t~kes t hours longer than the B,

xy xt

Distance =

km. or

Multiplication of speeds Difference of Speeds

x Difference in time

to cover the distance (See Note)

Illustrative Example Ex.:

Soln:

Two runners cover the same distance at the rate of 15 km and 16 km per hour respectively. Find the distance travelled when one takes 16 minutes longer than the other. Detail Method: Let the distance bex km.

Time taken by the first runner = hrs

Time taken by the second runner = hrs

x x _ 16 N o w ' T T T 6 = ^

or, JC(16- 15) _ 16

15x16

16

60

Note:

:.x = x l 5 x l 6 = 64 km 60

Quicker Method: Applying the above theorem,

15x16 16 . , Distance = * = 6 4 km.

16-15 60 The above theorem may be written as, " I f two run-ners A and B cover the same distance at the rate of x km/hr and y km/hr respectively, then the distance trav-elled, when B takes t hours less than the A, is

km."

Ex.:

xy y-x

See the following example, Two cars run to a place at the speeds of 45 km/hr and 60 km/hr respectively. I f the second car takes 5 hrs less than the first for the journey, find the length of the journey.

Soln: Distance: 45x60 60-45

x5 = 900 km.

Exercise 1. Two men start together to walk a certain distance, one at

3 3 km an hour, and the other at 3 km an hour. The

4 former arrives half an hour before the latter. Find the distance.

1 a) 7 - km b)15km c) 15 : d)7km

Two bicyclists do the same journey by travelling respec-tively at the rates of 9 and 10 km an hour. Find the length of the journey when one takes 32 minutes longer than the other.


a) 3 8 km b)36km c)48km d)24km 3. A man walks from A to B and back in a certain time at the

rate of 3 km per hour. But if he had walked from A to B

at the rate of 3 km an hour and back from B to A at the rate of 4 km an hour, he would have taken 5 minutes longer. Find the distance between A and B. a) 14km b)12km c)6km d)7km

4. A car starts from P for Q travelling 20 kilometres an hour.

1 hours later another car starts from P and travelling

1 at the rate of 30 kilometres an hour reaches Q 2 hrs

before the first car. Find the distance from P to Q. a) 90 km b) 150 km c) 240 km d) 270 km

Answers I . a 2.c

3. d; Man's speed in the first case = 3^- km/hr

Man's average speed in the second case

2 x 3 x 4 24 3 + 4

Now, applying the given rule,

we have 2(AB) =

7 24 X 2 7

24 X 6 0

km/hr (See Rule 2)

14 km

Illustrative Example Ex.: A carriage driving in a fog passed a man who was

walking at the rate of 3 km an hour in the same direc-tion. He could see the carriage for 4 minutes and it was visible to him upto a distance of 100 m. What was the speed of the carriage?

Soln: Detail Method: The distance travelled by the man in 4

minutes 3x1000

60 x4 =200 metres.

distance travelled by the carriage in 4 minutes = (200+100) = 300 metres.

speed of the carriage 300 60

1000 km per hour

= 4 km per hour 2


x = 3 km/hr 4 1

t = 4 minutes = - 77 hrs 60 15

d= 100 metres = 100

km 1000 10

Now, applying the above rule, we have

the required answer = 3+ ^0- = 3 + ~ = 4 km/hr

15

A .-. AB = = 7

2

^ 1 , 1 4. c; Hint: Herer= 2+ 1 = 4 hours 2 2

Now applying the given rule we have

30x20 the required distance = -30-20

x4 = 240 km

Rule 18t Carriage driving in a fog Theorem: A carriage driving in a fog passed a person who was walking at the rate ofx km/hr in the same direction. If he could see the carriage for 1 hours and it was visible to him upto a distance of'd' km, then the distance travelled

by the carriage in t hours is (xt + d) metres and speed of

the carriage is \ + ~ | km/hr.

Exercise 1. A carriage driving in a fog passed a man who was walk-

ing at the rate of 6 km an hour in the same direction. He could see the carriage for 8 minutes and it was visible to him upto a distance of200 m. What was the speed of the carriage?

a)9km/hr b) 7 - km/hrc)7km/hr d) 8~ km/hr

2. A carriage driving in a fog passed a man who was walk-ing at the rate of 5 km an hour in the same direction. He could see the carriage for 6 minutes and it was visible to him upto a distance of 120 m. What was the speed of the carriage?

a) 6 j km/hr 1

b) 5 - km/hr

c) 6 km/hr d) None of these

3. A carriage driving in a fog passed a man who was walk-

rime and Distance

ing at the rate of 6 km an hour in the same direction. He could see the carriage for 12 minutes and it was visible to him upto a distance of 150 m. What was the speed of the carriage?

^ 3 3 a) 7 - km/hr b) 6 - km/hr c) 5 - km/hr d) 8 km/hr

4. A carriage driving in a fog passed a man who was walk-ing at the rate of 8 km an hour in the same direction. He could see the carriage for 15 minutes and it was visible to him upto a distance of 500 m. What was the speed of the carriage?

a) 10 km/hr

c) 12 km/hr

Answers l .b 2. a

b) 8 - km/hr 2

d) None of these

3.b 4. a

Rule 19 Theorem: A person takes x hours to walk to a certain place and ride back. However, he could have gained t hours, if he had covered both ways by riding, then the time taken by him to walk both ways is (x +1) hours.

or, Both ways walking = One way walking and one way riding rime + gain in time

Illustrative Example Ex.: A man takes 8 hours to walk to a certain place and

ride back. However, he could have gained 2 hrs, i f he had covered both ways by riding. How long would he have taken to walk both ways?

Soln: Detail Method: Walking time + Riding time = 8 hrs (1) 2 Riding time = 8 - 2 = 6 hrs (2) 2 x ( l ) _ ( 2 ) gives the result 2 x walking time = 2 * 8 - 6 = lOhrs. .-. both ways walking will take 10 hrs. Quicker Approach: Two ways riding saves a time of 2 hrs. It simply means that one way riding takes 2 hrs less than one way walking. It further means that one way walking takes 2 hrs more than one way riding. Thus, both way walking wil l take 8 + 2 = 10 hrs. Quicker Method: Applying the above theorem, we have Both ways walking = One way walking and one way riding time + Gain in time = 8 + 2 = 10 hrs.

Exercise L A man takes 6 hrs 30 min in walking to a certain place

and riding back. He would have gained 2 hrs 10 min by riding both ways. How long would he take to walk both ways?

2.

4.

433

a) 8 hrs b) 8 hrs 20 min c) ^ hrs 20 min d) 8 hrs 40 min A man takes 5 hrs 42 min in walking to a certain place and riding back. He would have gained 1 hr 18 min b> riding both ways. How long would he take to walk both ways? a) 6 hours b) 6 hours 50 min c) 7 hours d) Data inadequate A man takes 7 hours in walking to a certain place and riding back. He would have gained 3 hours by riding both ways. How long would he take to walk both ways? a} 10 hours b) 12 hours c) 8 hours d) 9 hours A man takes 8 hrs 32 min in walking to a certain place and riding back. He would have gained 2 hrs 14 min by riding both ways. How long would he take to walk both ways? a) 10 hrs c) 10 hrs 46 min

b) 10 hrs 36 min d) 10 hrs 40 min

Answers l . d 2.c 3. a 4.c

Rule 20 Theorem: A man takes x hours to walk to a certain place and ride back. However, if he walks both ways he needs t hours more, then the time taken by him to ride both ways is (x-t) hours.

Illustrative Example Ex.: A man takes 12 hrs to walk to a certain place and ride

back. However, i f he walks both ways he needs 3 hours more. How long would he have taken to ride both ways?

Soln: Applying the above theorem, we have required time = 12 - 3 = 9 hrs.

Exercise

1. I walk a certain distance and ride back by taking 6

hours altogether. 1 could walk both ways in 7 hours.

How long would it take me to ride both ways?

1 3 3 1 a) 4 - hrs b) 5 - hrs c) 4 - hrs d) 4 - hrs

' 4 4 4 2

2. A man takes 5 hrs to walk to a certain place and ride

back. However, i f he walks both ways he needs 1 hours 4

more. How long would he have taken to ride both ways?

, 1 , 1 1 . 1 1 . 1 a) 3 hrs b) 3 hrs c) 4 hrs d) 4 hrs


3. A man takes 8 hrs to walk to a certain place and ride back. However, i f he walks both ways he needs 1 hour more. How long would he have taken to ride both ways? a) 9 hrs b)7hrs c)6hrs d) None of these

Answers

^ 3 ^ 1 3 , 1. c; Hint: Here t = 1 6 = hrs

4 4 2 Applying the given rule, we have

1 3 3 the required answer = 6 = 4 hrs

4 2 4 2, b 3.b

Rule 21 Theorem: A person A leaves a point P and reaches Qinx hours. If another person B leaves the point Q, t hours later than A and reaches the point P in y hours, then the time in

which A meets to Bis{y-{f+ y

hrs.

Illustrative Example Ex.: A man leaves a point P and reaches the point Q in 4

hours. Another man leaves the point Q, 2 hours later and reaches the point P in 4 hours. Find the time in which first man meets to the second man.

Soln: Detail Method: Let the distance PQ = A km. And they meet* hrs after the first man starts.

Average speed of first man : km/hr

Average speed of second man = km/hr

Distance travelled by first man Ax

km

They meet x hrs after the first man starts. The second man, as he starts 2 hrs late, meets after ( x - 2 ) hrs from his start. Therefore, the distance travelled by the sec-

4.x-2) ond man = km

4

Ax A(X-2) Now, + ' km = A

4 4 or, 2 x - 2 = 4 .-. x = 3 hrs Quicker Method: Applying the above theorem, we have x = 4 hrs, y = 4 hrs and t = 2 hrs

.-. the required time - (4 + 2\ | =3 hrs. \ + 4 j

Note: The above example may be written as "A man leaves a point P at 6 a.m. and reaches the point Q at 10 a.m. Another man leaves the point Q at 8 a.m. and reaches the point P at 12 noon. At what time do they meet?"

Soln: Applying the above theorem, we have x = 10 am - 6 am = 4 hrs y = 12 noon - 8 am = 4 hrs First man starts at 6 am and second man starts at 8 am Therefore, second man starts (8 am - 6 am = 2 hrs) late than the first. .-. t = 2hrs.

.-. the required time = I 1

Hence, they meet at 6 am + 3 am = 9 am.

Exercise 1. A man leaves a point P and reaches the point Q in 3

hours. Another man leaves the point Q, 1 hour later and reaches the point P in 3 hours. Find the time in which first man meets to the second man. a) 2 hrs b) 3 hrs

1 c) 1- hrs

2.

d) Data inadequate

A man leaves a point P and reaches the point Q in 5 hours. Another man leaves the point Q, 2 hours later and reaches the point P in 7 hours. Find the time in which first man meets to the second man.

a) 4 hrs , 1

b) 4 - hrs c) j hrs d) ? | hrs

3. A man leaves a point P and reaches the point Q in 6 hours. Another man leaves the point Q, 4 hours later and reaches the point P in 6 hours. Find the time in which first man meets to the second man. a) 6 hrs b)5hrs c)4hrs d)3hrs

4. A man leaves a point P at 8 am and reaches the point Q at 12 noon. Another man leaves the point Q at 9 am and reaches the point P at 1 pm. At what time do they meet1? a) 10.30 am b) 11.30 am c)10am d)They will never meet

Answers L a 2.c 3.b 4. a

Rule 22 Theorem: A person A leaves a point P and reaches Q in x hours. If another person B leaves the point Q, t hours ear-lier than A and reaches the point P in y hours, then the time

in which A meets to B is {y-t x + y

hrs.

Illustrative Example Ex.: A man leaves a point P and reaches the point Q in 4


hours. Another man leaves the point Q. 2 hours ear-lier and reaches the point P in 4 hours. Find the time in which the first man meets to the second man.

Soln: Detail Method: Let the distance PQ = A km And they meet x hrs after the first man starts.

A Average speed of first man = km/hr

A Average speed of second man = km/hr

Ax Distance travelled by first man = km

4 They meet x hrs after the first man starts. The second man, as he starts 2 hrs late, meets after (x + 2) hrs from his start. Therefore, the distance travelled by the second man

= ^ k m 4

Ax A(x + 2) Now, + - km = A

p a s i ' 4 4 2x + 2 = 4 .-. x = l h r Quicker Method: Applying the above theorem, we have x = 4 hrs, y = 4 hrs and t = 2 hrs

.-. the required time = ( 4 - 2 4 + 4 = 1 hr

Note: The above example may be written as "A man leaves a point P at 8 am and reaches the point Q at 12 noon. Another man leaves the point Q at 6 am and reaches the point P at 10 am. At what time do they meet?"

Soln: Applying the above theorem, we have x = 12 noon - 8 a.m. = 4 hrs y = 10 a.m. - 6 a.m. = 4 hrs First man starts at 8 a.m. and second man starts at 6 a.m. Therefore, second man starts (8 a.m. - 6 a.m. = 2 hrs) earlier than the first. t = 2 hrs

.-. the required time = (4 - 2 4 + 4

= 1 hr

Hence, they meet at 8 a.m. + 1 hour = 9 a.m.

Exercise 1. A man leaves a point P and reaches the point Q in 3

hours. Another man leaves the point Q. 1 hour earlier and reaches the point P in 3 hours. Find the time in which the first man meets to the second man. a) 1 hr b) 2 hrs c) 30 min d) Can't be determined

2. A man leaves a point P and reaches the point Q in 6

hours. Another man leaves the point Q. 2 hours earlier and reaches the point P in 6 hours. Find the time in which the first man meets to the second man.

a) 1 hr b) 3 hrs c) 2 hrs d) 1 - hrs 2

A man leaves a point P and reaches the point Q in 5 hours. Another man leaves the point Q. 3 hours earlier and reaches the point P in 7 hours. Find the time in which the first man meets to the second man.

a ) l h r

c) 2 - hrs

Answers L a 2. c

b) 1 - hrs

d) None of these

3.b

Rule 23 Theorem: Speed and time taken are inversely proportional.

Therefore, SxTr = S2T2 = S 37 3....

Where, Sv S2, S 3 ... are the speeds

and Tx, T2, 73 .... are the time taken to travel the same dis-tance.

Note: Also see Rule 32.

Illustrative Example Ex.: A person covers a distance in 40 minutes if he runs at

a speed of 45 km per hour on an average. Find the speed at which he must run to reduce the time of journey to 30 minutes.

Soln: Required speed = 45 x 40 = S 2 x 30

45x40 r n , . S, = = 60 km/hr.

2 30 Exercise 1. A person covers a distance in 8 minutes i f he runs at a

speed of 9 km per hour on an average. Find the speed at which he must run to reduce the time of journey to 6 minutes. a) 12 km/hr b) 10 km/hr c) 9 km/hr d) None of these

2. A person covers a distance in 24 minutes i f he runs at a speed of 27 km per hour on an average. Find the speed at which he must run to reduce the time of journey to 18 minutes. a) 27 km/hr b) 36 km/hr c) 45 km/hr d) 48 km/hr

3. A person covers a distance in 12 minutes i f he runs at a

1

speed of 13 km per hour on an average. Find the speed

at which he must run to reduce the time of journey to 9


minutes. a) 16 km/hr b) 21 km/hr c) 24 km/hr d) 18 km/hr

4. A person covers a distance in 16 minutes i f he runs at a speed of 18 km per hour on an average. Find the speed at which he must run to reduce the time of journey to 12 minutes. a) 16 km/hr b) 18 km/hr c) 24 km/hr d) None of these

Answers l .a 2.b 3.d 4.c

Rule 24 Theorem: Without any stoppage if a person travels a cer-tain distance at an average speed of x km/hr, and with stoppages he covers the same distance at an average speed

ofy km/hr, then he stops

Time of rest per hour =

minutes.

f S 60 x-y

{ y J minutes per hour.

Or

Difference of speed Speed without stoppage

x60

Illustrative Example Ex.: Without any stoppage a person travels a certain dis-

tance at an average speed of 80 km/hr, and with stop-pages he covers the same distance at an average speed of 60 km/hr. How many minutes per hour does he stop?

Soln: Detail Method: Let the total distance be x km.

Time taken at the speed of 80 km/hr = hrs. 80

Time taken at the speed of 60 km/hr = hrs. 60

Exercise 1. Without any stoppage a person travels a certain dis-

tance at an average speed of 40 km/hr, and with stop-pages he covers the same distance at an average speed of 30 km/hr. How many minutes per hour does he stop?

_ 1 a) 7 min b) 7 min c) 15 min d)Noneofthese

2. Without any stoppage a person travels a certain dis-tance at an average speed of 42 km/hr, and with stop-pages he covers the same distance at an average speed of 28 km/hr. How many minutes per hour does he stop? a) 15 minutes b) 14 minutesc) 28 minutes d) 20 minutes

3. Without any stoppage a person travels a certain dis-tance at an average speed of 27 km/hr, and with stop-pages he covers the same distance at an average speed of 15 km/hr. How many minutes per hour does he stop? a) 26 min 40 sec b) 30 min c) 26 min d) 16 min 40 sec

4. Without any stoppage a person travels a certain dis-tance at an average speed of 15 km/hr, and with stop-pages he covers the same distance at an average speed of 12 km/hr. How many minutes per hour does he stop? a)15min b)12min c)16min d)18min

Answers l .c 2.d 3.a 4.b

Rule 25 Theorem: A person has to cover a distance of x km in t hours. If he covers nth part of the journey in mth of the total

fx \-n) time, then his speed should be I ~ x \m J km/hr to cover

the remaining distance in the remaining time.

Illustrative Example Ex: A person has to cover a distance of 80 km in 10 hrs. I f

he rested for x x 6 0 8 0

hrs : 20x _ x

60x80 ~~240 hrs

x x x 60 1 his rest per hour = - . _ x ^ - t a

= 15 minutes. Quicker Method: Applying the above theorem, Time of rest per hour

Difference of speed Speed without stoppage

x60 minutes

80-60 80

x 60 = - x 60 hr = 15 minutes. 4

he covers half of the journey in th time, what should

be his speed to cover the remaining distance in the time left?

Soln: Detail Method:

.elf di: lance 80 1-^-j =40 km

Time left = 10 1--5)

-4 hrs

.-. required speed: 40 T

10 km/hr


Time and Distance - 3 "

the required speed: 80

10 km/hr. 10

And time spent in journey by walking = hrs. 4

x x Therefore, "^" + 7" = 5 hrs 48 minutes.

Exercise 1. A person has to cover a distance of 15 km in 3 hrs. I f he

1 2 covers of the journey in rd time, what should be

his speed to cover the remaining distance in the time left? a)12km/hr b)10km/hr c)5km/hr d)15km/hr

2. A person has to cover a distance of 60 km in 15 hrs. I f he

1 2 covers of the journey in th time, what should be

his speed to cover the remaining distance in the time left? a)5km/hr b)9km/hr c)6km/hr d)12km/hr

3. A person has to cover a distance of 40 km in 5 hrs. I f he

1 3 covers ~ ofthe journey in th time, what should be his

speed to cover the remaining distance in the time left? a) 18 km/hr b) 15 km/hr c) 16 km/hr d) 14 km/hr

4. Laxman has to cover a distance of 6 km in 45 min. I f he

2 covers one half of the distance in rd time, what should

be his speed in km/hr to cover the remaining distance in the remaining time? [Bank PO Exam, 1991 ] a) 12 b) 16 c)3 d)8

Answers l .b 2. a 3.c 4. a

Rule 26 Theorem: A man rode out a certain distance by train at the rate of x km/hr and walked back at the rate of y km/hr. If the whole journey took 't'hours, then the distance he rode

xy x + y tkm.

Illustrative Example Ex.: A man rode out a certain distance by train at the rate

of 25 km an hour and walked back at the rate of 4 km per hour. The whole journey took 5 hours and 48 min-utes. What distance did he ride?

Soln: Detail Method: Let the distance be x km.

29x , 48 29 100 o n or = 5 = . \ = = 20 km

' 100 60 5 5 Quicker Method: Applying the above theorem, the required distance

,48 25x4 = 5 x

60 25 + 4

29 25x4 x 5 29

20 km

Exercise 1. A man rode out a certain distance by train at the rate of

30 km an hour and walked back at the rate of 5 km per hour. The whole journey took 7 hrs. What distance did he ride? a) 30 km b)25km c)28km d)35km

2. A man rode out a certain distance by train at the rate of 13 km an hour and walked back at the rate of 12 km per hour. The whole journey took 5 hours. What distance did he ride? a) 32 km b)26km c) 31.2 km d) 26.5 km

3. A man rode out a certain distance by train at the rate of 15 km an hour and walked back at the rate of 12 km per hour. The whole journey took 9 hours. What distance did he ride? a) 48 km b)60km c)36km d) None of these

Answers La 2.c 3.b

Rule 27 Theorem: A man travels D km in x hours, partly by air and partly by train. If he had travelled all the way by air, he

would have saved of the time he was in train and would b

have arrived at his destination y hours early, then the dis-

tance he travelled by air is D ajb

x-v km.

Then time spent in journey by train 25

hrs.

Illustrative Example Ex.: A man travels 360 km in 4 hrs, partly by air and partly

by train. I f he had travelled all the way by air, he would 4

have saved of the time he was in train and would have arrived at his destination 2 hours early. Find the distance he travelled by air and train.


Soln: Detail Method: oftotal time in train = 2 hours.

2x5 5 .-. Total time in train = = hrs.

4 2

5 3 Total time spent in air = 4 = hrs.

K 2 2 By the given hypothesis, i f 360 km is covered by air, then time taken is (4 - 2 =) 2 hrs.

.-. when hrs is spent in air, distance covered

360 3 = x - = 270 km.

2 2 .-. Distance covered by train = 360 - 270 = 90 km. Quicker Method: Applying the above theorem,

Distance covered by air =360 4 -

5x2

4 - 2

= 3 6 0 x - = 270 km. 4

.-. distance covered by train = 360 - 270 = 90 km.

Exercise 1. A man travels 480 km in 6 hrs, partly by air and partly by

train. If he had travelled all the way by air, he would have

3 saved of the time he was in train and would have

4 arrived at his destination 3 hours early. Find the dis-tance he travelled by train. a) 320 km b) 160 km c) 260 km d) 220 km

2. A man travels 120 km in 5 hrs, partly by air and partly by train. I f he had travelled all the way by air, he would have

2 saved of the time he was in train and would have

arrived at his destination 2 hours early. Find the dis-tance he travelled by air. a)80km b)40km c)85km d)90km

3. A man travels 400 km in 8 hrs, partly by air and partly by train. I f he had travelled all the way by air, he would have

4 saved ~ of the time he was in train and would have

arrived at his destination 4 hours early. Find the ratio between distances travelled by the air and the train. a)T:3 b) 3 : 1 c ) 2 : l d) 1:2

Answers l .b 2. a

3. b; Hint: Distance travelled by the air = 400

4 N 4/5

8 -4

= 300 km .-. distance travelled by the train = 400 - 300 = 100 km .-. required ratio = 3 : 1 .

Rule 28 Theorem: One aeroplane starts t hours later than the sched-uled time from a place D km away from its destination. To reach the destination at the schedued time the pilot has to increase the speed by 'p' km/hr. Then the plane takes

2 4Dt t + +t

hours in the normal case. And the normal

speed of the aeroplane is 2D.

2 4Dt r + + /

km/hr.

Note: Normal case indicates the original case in which speed of the aeroplane has not been changed.

Illustrative Example Ex: One aeroplane started 3 0 minutes later than the sched-

uled time from a place 1500 km away from its destina-tion. To reach the destination at the scheduled time the pilot had to increase the speed by 250 km/hr. What was the speed of the aeroplane per hour during the journey?

Soln: Detail Method: Let it take x hrs in second case.

Then speed 1500 1500

x + -1

+ 250

1500|x + - j - 1 5 0 0 x or, ^ , 2 \ =250

x \ + -

or,750 = 250x x + t X _ _ or, x1 + 3 = 0 2

or, 2x1+x-6 = 0

Time and Distance

or, 2x2 + 4 x - 3 x - 6 = 0

or, x ( 2 x - 3 ) + 2 ( 2 x - 3 ) = 0

or, (x + 2X2x-3) = 0

2.?

Therefore, the plane takes hrs in second case, ie

3 1 + = 2 hrs in normal case. Thus, normal speed = 2 2 1500

= 750 km/hr.

Quicker Method: Applying the above theorem, Normal speed of the aeroplane

2x1500 2x1500

1_ 4x1500x1 J_ 4 + 2x250 + 2

= 750 km/hr.

Exercise 1. One aeroplane started 1 hour later than the scheduled

time from a place 3000 km away from its destination. To reach the destination at the scheduled time the pilot had to increase the speed by 500 km/hr. What was the speed of the aeroplane per hour during the journey? a) 1500 km/hr b) 1000 km/hr c) 850 km/hr d) None of these

2. One aeroplane started 1 hrs later than the scheduled

time from a place 2400 km away from its destination. To reach the destination at the scheduled time the pilot had to increase the speed by 800 km/hr. What was the speed of the aeroplane per hour during the journey? a) 1600 km/hr b) 800 km/hr c) 1200 km/hr d) 1550 km/hr

3. One aeroplane started 30 minutes later than the sched-uled time from a place 1800 km away from its destination. To reach the destination at the scheduled time the pilot had to increase the speed by 300 km/hr. What was the speed of the aeroplane per hour during the journey? a) 600 km/hr b) 800 km/hr c) 1200 km/hr d) 900 km/hr

Answers l .b 2.b 3.d

Rule 29 Theorem: A train leaves the station t hours before the sched-uled time. The driver decreases its speed by p km/lir. At the next station D km away, the train reached on time. Then

the train takes

2 4Dt V + 1

hours in the normal case

(original case) and the normal speed (original speed) of

( \

2D the train is

2 4Dt r + 1

km/hr.

Illustrative Example Ex.:

Soln:

A train leaves the station 1 hour before the sched-uled time. The driver decreases its speed by 4 km/hr. At the next station 120 km away, the train reached on time. Find the original speed of the train. Detail Method: Let it takes x hours in second case.

Then speed = 120 120

x-\

120 120 or, = 4 x-l x

or, 120 = 4x2 -4x

or, x 2 - x - 3 0 = 0

or, 120(x-x + l )

or, 4x^ - 4 x - 1 2 0 = 0

or, xl - 6 x + 5x-30 = 0

or, x ( x - 6 ) + 5 ( x - 6 ) = 0

or,x = -5,6 .-. x = 6 Therefore, the train takes 6 hours in second case, ie ( 6 - 1 = 5 ) hours in original case.

120

.-. Original speed = ~ r ~ - 24 km/hr

Quicker Method: Applying the above theorem, 2x120 Original speed of the train :

, 2 , 4 * 1 2 0 x 1 ]

2x120 11-1

= 24 km/hr.

Exercise 1. A train leaves the station 1 hour before the scheduled

time. The driver decreases its speed by 50 km/hr. At the next station 300 km away, the train reached on time. Find the original speed of the train. a)100km/hr b)150km/hr


c)200 km/hr d) None of these

2. A train leaves the station 1 hours before the sched-2

uled time. The driver decreases its speed by 80 km/hr. At the next station 240 km away, the train reached on time. Find the original speed of the train, a) 180 km/hr b) 160 km/hr c) 200 km/hr d) 120 km/hr

3. A train leaves the station 30 minutes before the sched-uled time. The driver decreases its speed by 30 km/hr. At the next station 180 km away, the train reached on time. Find the original speed of the train. a)140km/hr b)125km/hr c)120km/hr d)100km/hr

Answers l .b 2.b 3.c

Rule 30 Theorem: When a person travels equal distance at speed

and V2 km/hr, his average speed is x km/lir. But when

he travels at these speeds for equal times his average speed

isy km/hr, then the values of K, and V2 are [y + yjy(y - x))

km/hr and (y - yjy(y - x)) km/hr respectively. And the dif-

ference of the two speeds is given by 2^y(y-x) km/hr.

Illustrative Example Ex.: When a man travels equal distance at speeds V and

V 2 km/hr, his average speed is 4 km/hr. But when he travels at these speeds for equal times his average speed is 4.5 km/hr. Find the difference of the two speeds and also find the values of V, and V,.

Soln: Detail Method: Suppose the equal distance = D km Then time taken with V. and V 2 speeds are

D D Y hrs and y hrs respectively.

.-. average speed Total distance Total time

2D

+

2V,V I v 2 v, + v 2

= 4 km/hr

In second case,

average speed = v,+v 2

= 4.5 km/hr

Thatis; V ^ V ^ Q a n d V , V , = 18

Now, (v, - V 2 ) 2 = (V, + V, ) 2 - 4V, V 2

= 81-72 = 9

V 2 =3 km/hr and V V = 6 km/hr and V,

V = 9 km/hr i 2

3 km/hr Quicker Method: Applying the above theorem.

V, - V 2 = 27(4.5X4.5-4) = 3 km/hr

v , = v + V v ( y - * )

= 4.5+ =6 km/hr and 2

v 2 - y - Jy(y~x)

3 = 4 .5- - =3km/hr.

2 Exercise 1. When a man travels equal distance at speeds V and V,

km/hr, his average speed is 8 km/hr. But when he travels at these speeds for equal times his average speed is 9 km/hr. Find the difference of the two speeds. a) 6 km/hr b) 5 km/hr c) 4 km/hr d) 8 km/hr

2. When a man travels equal distance at speeds V and V 2 km/hr, his average speed is 6 km/hr. But when he travels at these speeds for equal times his average speed is 8 km/hr. Find the difference of the two speeds. a) 7 km/hr b) 6 km/hr c) 8 km/hr d) 10 km/hr

3. When a man travels equal distance at speeds V! and V, km/hr, his average speed is 5 km/hr. But when he travels at these speeds for equal times his average speed is 9 km/hr. Find the difference of the two speeds. a) 10 km/hr b) 12 km/hr c) 14 km/hr d) 8 km/hr

Answers l .a 2.c 3.b

Rule 31 Theorem: A person travels for Tx hours at the speed of

km/hr and for T2 hours at the speed of R2 km/hr. At the end of it, if he finds that he has covered f of the total distance, then his average speed, to cover the remaining

distance in T hours, should be

hr.

{W+R2T2}y-\ km/

Illustrative Example Ex.: A person travels for 3 hrs at the speed of 40 km/hr

and for 4.5 hrs at the speed of 60 km/hr. At the end of 3

it, he finds that he has covered - of the total dis-

Time and Distance

km/

tance. At what average speed should he travel to cover the remaining distance in 4 hrs?

Soln: Detail Method: Total distance covered in (3 +4.5) hrs. = 3 x40 + 4.5 x60 = 390 km.

3 Now, since of the distance = 390 km

2 5 2 ' - of the distance = 390 * - * - = 260 km.

.-. average speed for the remaining distance =

260 4

Quicker Method: Applying the above theorem, we

: 65 km/hr.

have the average speed for the remaining distance

\ 1

(40x3+60x4.5 | - - 1

390x2 4x2

= 65 km/hr.

Exercise 1. A person travels for 4 hrs at the speed of 30 km/hr and

for 6 hrs at the speed of 40 km/hr. At the end of it, he

3 finds that he has covered of the total distance. At

4 what average speed should he travel to cover the re-maining distance in 3 hrs? a) 40 km/hr b) 35 km/hr c) 45 km/hr d) 60 km/hr

2. A person travels for 2 hrs at the speed of 15 km/hr and for 3 hrs at the speed of 20 km/hr. At the end of it, he


what average speed should he travel to cover the re-maining distance in 5 hrs? a) 8 km/hr b) 9 km/hr c)6km d) None of these

3. A person travels for 3 hrs at the speed of 12 km/hr and for 4 hrs at the speed of 15 km/hr. At the end of it, he


what average speed should he travel to cover the re-maining distance in 6 hrs? a)6km/hr b)8km/hr c)4km/hr d)5km/hr

Answers l .a 2.b 3.c

Rule 32 Theorem: If a person A walking at the rate of 5, kmhr.

takes t\ to cover a distance and another person B,

walking at the rate of S2 km/hr, takes t2 hours to cover the

same distance, then iS*,/, =S2t2 =Distanceor, S]t] = S2t2 = constant. Thus we see that both speed and time are inversely propor-tional to each other. That is, if the speed increases to 4

1 times, the time will decrease to times.

Application of the above theorem, Case I: I f A takes 8 hours to cover a distance and B is four

times faster than A, then what time will B take to cover the same distance?

We have, S\tx = S2t2

=> S,r, = 4S,r2

Case II: I f A takes 8 hours to cover a distance and he is 4 times faster than B, then what time will B take to cover the same distance? We have,

Sxtx=S2t2 =>45 2 x8 = S 2 xt2

:. t2 = 32 hrs. Case III: I f B is 20% faster than A, then what time will he take

to travel the distance which A travels in 20 minutes?

We have, 5,f, = S2t2

120 _ -5, xt2 5, x20

100 20x100

120 5 0 I f i 2 = l o y min

Case IV: I f B takes 30% less time than A, to cover the same distance. What should be the speed of B i f A walks at a rate of 7 km/hr?

Again, 7 x r ;

.:S2 = 7x100

70

100-30 100

= 10 km/hr.

Exercise 1. I f A takes 6 hours to cover a distance and B is 2 times

faster than A, then what time will B take to cover the


same distance?

a)2 hrs b) 3 hrs c) hrs 2

d) 4 hrs

2. I f A takes 4 hours to cover a distance and B is four times faster than A, then what time will B take to cover the same distance?

_ 1 a) 2 hrs b) 2 hrs c) 1 hr d) l | hrs

3. I f A takes 3 hours to cover a distance and he is 3 times faster than B, then what time will B take to cover the same distance? a) 9 hrs b) 6 hrs c) 8 hrs d) None of these

4. I f A takes 4 hours to cover a distance and he is 2 times faster than B, then what time will B take to cover the same distance? a) 6 hrs b) 8 hrs c) 9 hrs d) None of these

5. I f B is 25% faster than A, then what time will he take to travel the distance which A travels in 25 minutes? a) 25 minutes b) 20 minutes c) 30 minutes d) None of these

6. I f B is 30% faster than A, then what time wil l he take to travel the distance which A travels in 26 minutes? a) 15 minutes b) 20 minutes c) 25 minutes d) None of these

7. I f B takes 10% less time than A, to cover the same dis-tance. What should be the speed of B i f A walks at a rate of 9 km/hr? a)10km/hr b)15km/hr c)20km/hr d)5km/hr

8. I f B takes 40% less time than A, to cover the same dis-tance. What should be the speed of B i f A walks at a rate of 15 km/hr? a) 15 km/hr b) 20 km/hr c) 25 km/hr d) None of these

Answers l .b 2.c 3.a 4.b 5.b 6.b 7.a 8.c

Rule 33 Theorem: A person travels x} km by steamer, x2 km by

train and x 3 km by horse. It took T hours. If the rate of train is 'n' times that of the horse and'm' times that of the

steamer, then the rate of the train is TOC] + X2 + X 3

km/ J

hr, rate of the horse is 1 ( mx\2+ x 3

n \

rate of the steamer is m

mxx + x2 + roc3

km/ltr and the

km/hr.

Illustrative Example Ex.: A person travelled 120 km by steamer, 450 km by train

and 60 km by horse. It took 13 hours 30 minutes. I f the rate of the train is 3 times that of the horse and 1.5 times that of the steamer, find the rate of the train per hour.

Soln: Detail Method: Suppose the speed of horse = x km/ hr. Then speed of the train = 3x km/hr and speed of the steamer = 2x km/hr

120 450 60 Now, + + = 13.5 hours

2x 3x x

(Since 13 hrs 30 minutes =13.5 hrs)

360 + 900 + 360 or,

6JC

1620

= 13.5

= 20 6x13.5

.-. Speed of train = 3x = 3 x 20 = 60 km/hr Quicker Method: Applying the above theorem, we have

Speed of the train = 1.5x120 + 450 + 3x60

13.5 60 km/hr.

1 Speed of the horse = x 60 3

20 km/hr.

1 2 Speed of the steamer = - x 60 = - x 60 = 40 km/hr.

3/2 3

Exercise 1. A person travelled 60 km by steamer, 225 km by train and

30 km by horse. It took 15 hours. I f the rate of the train is 3 times that of the horse and 2 times that of the steamer. Find the speed of the train per hour.

29 a) km/hr

29 b) k m / h r

2.

3.

c) 29 km/hr d) None of these A person travelled 50 km by steamer, 60 km by train and 60 km by horse. It took 15 hours. I f the rate of the train is 4 times that of the horse and 3 times that of the steamer. Find the rate of the steamer. a)10km/hr b)30km/hr c)15km/hr d)18km/hr A person travelled 25 km by steamer, 40 km by train and 30 km by horse. It took 7 hours. I f the rate of the train is 4 times that of the horse and 2 times that of the steamer. Find the rate of the horse.

a)15km/hr b) 7 - km/hr c)30km/hr d)16km/hr

Answers l .c 2. a 3.b

Time and Distance

Rule 34 Theorem: A person covers a certain distance on scooter.

Had he moved ,x; km/hr faster, he would have taken f,

hours less. If he had moved x2 km/hr slower, he would have

taken t2 hours more, then the original speed (S) is

xxx2{tx +t2)~\ , t km/hr and the distance is given by

L '2*1 -hX2 J * J

\ + xx) km.

Illustrative Example Ex.: A man covers a certain distance on scooter. Had he

moved 3 km/hr faster, he would have taken 40 min-utes less. I f he had moved 2 km/hr slower, he would have taken 40 minutes more. Find the distance (in km) and original speed.

Soln: Detail Method: Suppose the distance is D km and the initial speed is x km/hr.

Then, we have D

x + 2 D 40 D

and x 60 x- 2

D D or,

and

x + 3

D D 2 x-2 x 3

3D _ 2 r' xjx^3)~3

2D _ 2 o r ' 4 c ^ 2 ) ~ 3

D 40 x 60

2 0)

(2)

From (1) and (2) we have 3D 2D

x(x + 3) x[x-l)

or, 3 (x-2)=2(x + 3) or, 3 x - 6 = 2x + 6

.-. x = 12 km/hr

Now, i f we put this value in (1) we get

2 12x15 D = T X -3 3

= 40 km.

Quicker Method I: Applying the above theorem, we have the required speed

v \3 3) S :

2 i 2 \ x3 x2 3 3

-12 km/hr and

the required distance: 1 2 x - ( l 2 + 3)

= 40 km.

Quicker Method II: In the above question when time

reduced in arrival (40 minutes) is equal to the time increased in arrival (40 minutes) then Speed

2 x (increase in speed x Decrease in speed) Difference in increase and decrease in speeds

2x(3x2) , = / x -12 km/hr

(3 -2 )

(l2 + 3 ) x ( l 2 - 2 ) Now, Distance = . \^ v \ Diff. between (l2 + 3 ) - ( l 2 - 2 ) '

arrival time

15x10 40 + 40 -x = 40 km. 5 60

Note: 1.40 minutes late and 40 minutes earlier make a differ-

80 ence of40 + 40 = 80 minutes = ~ hrs.

60 2. This question is the special case of the above theo-rem.

Here, = t2 ) , put this into the formula and get the

above method.

Exercise 1. A man covers a certain distance on scooter. Had he

moved 4 km/hr faster, he would have taken 30 minutes less. I f he had moved 3 km/hr slower, he would have taken 30 minutes more. Find the original speed. a)24 km/hr b)20 km/hr c) 28 km/hr d)18km/hr

2. A man covers a certain distance on scooter. Had he moved 6 km/hr faster, he would have taken 30 minutes less. I f he had moved 4 km/hr slower, he would have taken 1 hr 30 minutes more. Find the original speed.

a) 7 km/hr

c) 6 y km/hr

b) 6 y km/hr

d) None of these

3. A man covers a certain distance on scooter. Had he moved 8 km/hr faster, he would have taken 30 minutes less. I f he had moved 6 km/hr slower, he would have taken 15 minutes more. Find the distance (in km) anu original speed. a) 36 km/hr b) 24 km/hr c) 18 km/hr d) 30 km/hr

Answers l .a 2.c 3.a

Rule 35 Theorem: A thief is spotted by a policeman from a distance ofd km. When the policeman starts the chase, the thief also


starts running. Assuming the speed of the thiefx kilometres an hour, and that of the policeman y kilometres an hour,

x then the thief will run before ne is overtaken = d

km Or,

The distance covered by the thief before he gets caught

Lead of distance Relative speed

- x Speed of thief

Illustrative Example Ex: A thief is spotted by a policeman from a distance of

200 metres. When the policeman starts the chase, the thief also starts running. Assuming the speed of the thief 10 kilometres an hour, and that of the policeman 12 kilometres an hour, how far will have the thief run before he is overtaken?

Soln: Detail Method: Relative speed = 12- 10 = 2 km/hr

the thief will be caught after = 0.2

hr. 2 10

.-. distance covered by the thief before he gets caught

= 1 0 x - ^ = 1 km


0.2 km the required distance - x 1 U km/hr = 1 km.

2 km/hr

Exercise 1. A policeman goes after a thief who has 100 metres'start.

If the policeman runs a kilometre in 8 minutes, and the thief a kilometre in 10 minutes, how far will the thief have gone before he is overtaken? a) 350 metres b) 400 metres c) 450 metres d) 460 metres

2. A thief is spotted by a policeman from a distance of 400 metres. When the policeman starts the chase, the thief also starts running. Assuming the speed of the thief 8 kilometres an hour, and that of the policeman 10 kilometres an hour, how far will have the thief run before he is overtaken?

a) 2 km b) 1 km . 2

c) l km d) V 5 ' 5 3. A thief is spotted by a policeman from a distance of 150

metres. When the policeman starts the chase, the thief also starts running. Assuming the speed of the thief 12 kilometres an hour, and that of the policeman 18 kilometres an hour, how far will have the thief run before he is overtaken?

a) 150 metres b) 200 metres c) 300 metres d) 1 km

4. A thief is spotted by a policeman from a distance of 300 metres. When the policeman starts the chase, the thief also starts running. Assuming the speed of the thief 12 kilometres an hour, and that of the policeman 15 kilometres an hour, how far will have the thief run before he is overtaken?

a) l k m b) 1.2km c) 1.5km d)2km

Answers

1. b; Hint: d= 100 metres = km 100 _ 1 1000 10

x = 1 kilometre in 10 minutes = 6 km/hr

y = 1 kilometre in 8 minutes = 7 km/hr 3 2 Now applying the given rule, we have

f \

100 4 15

I 2 - 6

2.d 3.c

1000 10 km = 400 metres

4.b

Rule 36 To find the average speed Theorem: If a moving body travels xux2, x3,... xn, metres, moving with different speeds Sr S^ S J f S n metres per second in time Tt, T2, T } , F seconds respectively, then it is necessary to calculate the average speed of the body throughout the journey. If the average speed is denoted by K, then,

Total distance travelled

Case I =

Case 11 =

Total time taken

Xj + X2 " X^ + . . . *r XN

Case 111 =

T]+T2+T3+... + Tn SiTl+S2T2+SzTi+... + ST

Tx+T2+Ti+... + Tn

X, +X-, +X-, + ... + x

*1 . X 2

s, s2 + + ...+

S\

Illustrative Examples Ex. 1: A train travels 225 km in 3.5 hours and 370 km in |

hours. Find the average speed of train. Soln: Applying the above theorem (case-I)

225 + 370 Average speed = _, _ + ^ - 70 km/hr

MATHS Time and Distance 445

Ex.2:

Soln:

A car during its journey travels 30 minutes at a speed of 40 km/hr, another 45 minutes at a speed of 60 km/hr and 2 hours at a speed of 70 km/hour. Find its average speed of the car. Applying the above theorem (case-II) Average speed

^ x 4 0 | + 60

~ j x 6 0 | + (2x70)

- = 63 km/hr.

Ex.3:

Soln:

30 45 , + + 2 60 60

A man walks 3 km at a speed of 3 km/hr, runs 4 km at a speed of 4 km/hr and goes by bus another 16 km. Speed of the bus is 16 km/hr. I f the speed of the bus is considered as the speed of the man, find the average speed of the man. Applying the above theorem (case-Ill)

Average speed : 3 + 4 + 16 23 _2

~ T 6 = T = 7 3 k m / h r 3 4 - + + 3 4 16

Exercise 1 A train travels 150 km in 5 hours and 250 km in 3 hours.

Find the average speed of train. a) 50 km/hr b) 60 km/hr c) 100 km/hr d) 80 km/hr

2 A train travels 220 km in 6 hours and 340 km in 2 hours. Find the average speed of train. a) 80 km/hr b) 70 km/hr c) 50 km/hr d) 35 km/hr

I A car during its journey travels 2 hrs at a speed of 25 km/ hr, another 4 hrs at a speed of 30 km/hr and 4 hours at a speed of 35 km/hour. Find its average speed of the car. a) 62 km/hr b) 31 km/hr c) 29 km/hr d) 58 km/hr

I A car during its journey travels 40 minutes at a speed of 30 km/hr, another 50 minutes at a speed of 60 km/hr and 1 hour at a speed of 30 km/hr. Find its average speed of the car. a) 40 km/hr b) 35 km/hr c) 45 km/hr d) None of these

& A man walks 6 km at a speed of 1 km/hr, runs 8 km at a

speed of 2 km/hr and goes by bus another 32km. Speed of the bus is 8 km/hr. I f the speed of the bus is consid-ered as the speed of the man, find the average speed of the man.

a) 15 - km/hr

c) 3 km/hr 6

Answers La l.b 3.b

b) 7 y km/hr

d) None of these

4. a 5.c

Rule 37 Theorem: A train does a journey without stopping in T hours. If it had travelled x km an hour faster, it would have done the journey in t hours, then the origninal speed is

( ' V ) j _ J x * J km/hr and the length of the journey is given by

Tt T-t km.

Illustrative Example Ex: A train does a journey without stopping in 8 hours. I f

it had travelled 5 km an hour faster, it would have done the journey in 6 hours 40 minutes. What is its original speed?

Soln: Applying the above theorem, we have, the required original speed

r 2 20 6

= ^ - x 5 = 4 - = 25 km/hr. 4 3

8 -6 -

Exercise 1. A car finishes a journey in 10 hours at the speed of 80

km/hr. I f the same distance is to be covered in eight hours how much more speed does the car have to gain?

|BSRB Delhi PO, 2000] a)8km/hr b)10km/hr c)20km/hr d)16km/hr

2. A train does a journey without stopping in 8 hours. I f it had travelled 6 km an hour faster, it would have done the journey in 6 hours. What is its original speed? a)20km/hr b)15km/hr c)21 km/hr d)18km/hr

3. A train does a journey without stopping in 5 hours. I f it had travelled 3 km an hour faster, it would h

chapter 18

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