Staff Selection Commission Mathematics - TIME AND DISTANCE TYPE-II

Q.11

A train travelling at 48 km/hr crosses another train, having half its length and travelling in oppo- site direction at 42 km/hr, in 12 seconds. It also passes a railway platform in 45 seconds. The length of the railway platform is

1) 200 m

(2) 300 m

(3) 350 m

(4) 400 m

Ans .

(4) 400 m

Explanation :

Let the length of the train travelling at 48 kmph be x metres.
Let the length of the platform be y metres.
Relative speed of train = (48 + 42) kmph
\( \frac{90*5}{18}\)m/sec
= 25 m./sec. and 48 kmph
\( \frac{48*5}{18}\)=\( \frac{40}{3}\)m./sec.
According to the question,
\( \frac{x+\frac{x}{2}}{25}\)=12
\( \frac{3x}{2x*25}\)=12
3x = 2 × 12 × 25 = 600
x = 200 m.
Also,\( \frac{200+y}{\frac{40}{3}}\)=45
600 + 3y = 40 × 45
3y = 1800 – 600 = 1200
y=\( \frac{1200}{3}\)= 400 m.

Q.12

Two towns A and B are 500 km. apart. A train starts at 8 AM from A towards B at a speed of 70 km/ hr. At 10 AM, another train starts from B towards A at a speed of 110 km/hr. When will the two trains meet ?

(1) 1 PM

(2) 12 Noon

(3) 12.30 PM

(4) 1.30 PM.

Ans .

(2) 12 Noon

Explanation :

Let two trains meet after t hours when the train from town A leaves at 8 AM.
Distance covered in t hours at 70 kmph + Distance covered in (t – 2) hours at 110 kmph = 500km
70t + 110 (t – 2) = 500
70t + 110t – 220 = 500
180 t = 500 + 220 = 720
t=\( \frac{720}{180}\)=4 hours
Hence, the trains will meet at 12 noon.

Q.13

Two trains of length 70 m and 80 m are running at speed of 68 km/hr and 40 km/hr respectively on parallel tracks in opposite directions. In how many seconds will they pass each other ?

(1) 10 sec

2) 8 sec

(3) 5 sec

(4) 3 sec

Ans .

(3) 5 sec

Explanation :

Relative speed
= (68 + 40) kmph = 108 kmph
=\( \frac{108*5}{18}\) m/s or 30 m/s
Required time
\( \frac{Sum of the lengths of both trains}{Relative speed}\)
\( \frac{70+80}{30}\)second = 5 seconds

Q.14

Two trains of equal length take 10 seconds and 15 seconds re- spectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direc- tion ?

(1) 16

(2) 15

(3) 12

(4) 10

Ans .

(3) 12

Explanation :

When a train crosses a telegraph post, it covers its own length.
Speed of first train=\( \frac{120}{10}\)= 12 m/sec.
Speed of second train=\( \frac{120}{15}\)= 8 m/sec.
Relative speed = 12 + 8
= 20 m/sec.
Required time
\( \frac{Total length of trains}{Relative speed}\)
=\( \frac{2*120}{20}\)= 12 seconds

Q.15

Two trains of length 137 metre and 163 metre are running with speed of 42 km/hr and 48 km/hr respectively towards each other on papallel tracks. In how many seconds will they cross each other?

(1) 30 sec

(2) 24 sec

(3) 12 sec.

(4) 10 sec

Ans .

(3) 12 sec.

Explanation :

Relative speed = 42 + 48 = 90 kmph
\( \frac{90*5}{18}\)m/s = 25 m/s
Sum of the length of both trains = 137 + 163 = 300 metres
Required time =\( \frac{300}{25}\)= 12 seconds

Q.16

Two trains 150 m and 120 m long respectively moving from oppo- site directions cross each other in 10 secs. If the speed of the second train is 43.2 km/hr, then the speed of the first train is

(1) 54 km/hr

(2) 50 km/hr

(3) 52 km/hr

(4) 51 km/hr

Ans .

(1) 54 km/hr

Explanation :

Speed of second train = 43.2 kmph
=\( \frac{43.2*5}{18}\) m/sec
Let the speed of first train be x m per second, then \( \frac{150 + 120}{x + 12}\)=10
27 = x + 12
x = 15 m/s=15*\( \frac{18}{5}\)kmph = 54 kmph

Q.17

Two trains start from station A and B and travel towards each other at speed of 16 miles/ hour and 21 miles/ hour respectively. At the time of their meeting, the second train has travelled 60 miles more than the first. The dis- tance between A and B (in miles) is :

(1) 444

(2) 496

(3) 333

(4) 540.

Ans .

(1) 444

Explanation :

Let the trains meet after t hours
Then, 21t – 16t = 60
5t = 60 Þ t = 12 hours
Distance between A and B = (16 + 21) × 12
= 37 × 12 = 444 miles

Q.18

Two trains 108 m and 112 m in length are running towards each other on the parallel lines at a speed of 45 km/hr and 54 km/ hr respectively. To cross each other after they meet, it will take

(1) 12 sec

(2) 9 sec

(3) 8 sec

(4) 10 sec

Ans .

(3) 8 sec

Explanation :

Relative speed = 45 + 54
= 99 kmph
99*\( \frac{5}{18}\)m/sec
Required time =\( \frac{108 + 112}{\frac{55}{2}}\)
=\( \frac{220*2}{55}\)= 8 seconds

Q.19

A man standing on a platform finds that a train takes 3 seconds to pass him and another train of the same length moving in the opposite direction, takes 4 sec- onds. The time taken by the trains to pass each other will be

(1) 4.42 sec

(2) 3.45 sec

(3) 3.42 sec

(4) 3.30 sec

Ans .

(3) 3.42 sec

Explanation :

Let the length of each train be x metres
Then, Speed of first train = \( \frac{x}{3}\)m/sec.
Speed of second train =\( \frac{x}{4}\)m/sec.
They are moving in opposite di- rections Relaive speed = \( \frac{x}{3}\)+ \( \frac{x}{4}\)
=\( \frac{4x+3x}{12}\)=\( \frac{7x}{12}\)m/sec
Total length = x + x = 2 x m.
Time taken = \( \frac{2x}{\frac{7x}{12}}\)=\( \frac{24}{7}\)=3.42sec.

Q.20

Two trains, each of length 125 metre, are running in parallel tracks in opposite directions. One train is running at a speed 65 km/hour and they cross each other in 6 seconds. The speed of the other train is

(1) 75 km/hour

(2) 85 km/hour

(3) 95 km/hour

(4) 105 km/hour

Ans .

(2) 85 km/hour

Explanation :

To tal length of both trains = 250 metres
Let speed of second train = x kmph
Relative speed = (65 + x) kmph
=(65 + x )*\( \frac{5}{18}\)m/sec
Time=\( \frac{Sum of length of trains}{Relative speed}\)
6=\( \frac{250}{65 + x * \frac{5}{18}}\)
=6*\( \frac{5}{18}\)*(65 + x ) = 250
65+x=\( \frac{250*3}{5}\)
65 + x = 150
x = 150 – 65 = 85 kmph

Q.21

A train running at the speed of 84 km/hr passes a man walk- ing in opposite direction at the speed of 6 km/hr in 4 seconds. What is the length of train (in metre) ?

(1) 150

(2) 120

(3) 100.

(4) 90

Ans .

(3) 100.

Explanation :

Relative speed = (84 + 6)
= 90 kmph
=90*\( \frac{5}{18}\) m/sec.
= 25 m/sec.
Length of train = Relative speed × Time
= 25 × 4 = 100 metre

Q.22

Two trains X and Y start from Jodhpur to Jaipur and from Jaipur to Jodhpur respectively. After passing each other they take 4 hours 48 minutes and 3 hours 20 minutes to reach Jaipur and Jodhpur respectively. If X is moving at 45 km/hr, the speed of Y is

(1) 60 km/hr

(2) 58 km/hr

(3) 54 km/hr

(4) 64.8 km/hr

Ans .

(3) 54 km/hr

Explanation :

=\( \frac{Speed of X}{Speed of Y}\)
=\( \frac{Time taken by Y}{Time taken by X} ^\frac{1}{2} \)
=\( \frac{45}{y}\)=\( \frac{3 hours 20 min}{4 hours 48 min.} ^\frac{1}{2} \)
=\( \frac{45}{y}\)=\( \frac{200 minutes}{288 minutes.} ^\frac{1}{2} \)
=\( \frac{10}{12}\)
10y = 12 × 45
y=54 kmph.

Q.23

P and Q starting simultaneously from two different places proceed towards each other at a speed of 20 km/hour and 30 km/hour respectively. By the time they meet each other. Q has covered 36 km more than that of P. The distance (in km.) between the two places is

(1) 144

(2) 162

(3) 180

(4) 108

Ans .

(3) 180

Explanation :

Let P and Q meet after t hours.
Distance = speed × time
According to the question,
30t – 20t = 36
10t = 36
t=3.6 hours.
Distance between P and Q
= 30t + 20t
= 50t = (50 × 3.6) km.
= 180 km

Q.24

Two places P and Q are 162 km apart. A train leaves P for Q and simultaneously another train leaves Q for P. They meet at the end of 6 hours. If the former train travels 8 km/hour faster than the other, then speed of train from Q is

(1) 12.5 kmph

(2) 4.5 kmph

(3) 9.5 kmph

(4) 5.5 kmph

Ans .

(3) 9.5 kmph

Explanation :

Speed of train starting from Q = x kmph
Speed of train starting from P = (x + 8) kmph
According to the question, PR + RQ = PQ
(x + 8) × 6 + x × 6 = 162
[Distance = Speed × Time]
6x + 48 + 6x = 162
12x = 162 – 48 = 114
x=\( \frac{114}{12}\)
=9.5 kmph.

Q.25

Two trains start at the same time from A and B and proceed to- ward each other at the speed of 75 km/hr and 50 km/hr respec- tively. When both meet at a point in between, one train was found to have travelled 175 km more than the other. Find the distance between A and B.

(1) 875 km.

(2) 785 km.

(3) 758 km..

(4) 857 km

Ans .

(1) 875 km.

Explanation :

Let the trains meet after t hours.
Distance = Speed × Time
According to the question,
75t – 50t = 175
25t = 175
t=7hours.
Distance between A and B
= 75t + 50t = 125t
= 125 × 7 = 875 km.

Q.26

Two trains of lengths 150m and 180m respectively are running in opposite directions on parallel tracks. If their speeds be 50 km/ hr and 58 km/hr respectively, in what time will they cross each other?

(1) 22 seconds.

(2) 15 seconds.

(3) 30 seconds

(4) 11 seconds.

Ans .

(4) 11 seconds

Explanation :

Relative speed = (50 + 58) kmph
=108*\( \frac{5}{18}\) m/sec
= 30 m/sec
Required time
=\( \frac{Total length of trains}{Relative speed}\)
=\( \frac{150+180}{30}\) sec
=\( \frac{330}{30}\)=11 sec.

Q.27

Two trains start at the same time from Aligarh and Delhi and pro- ceed towards each other at the rate of 14 km and 21 km per hour respectively. When they meet, it is found that one train has travelled 70 km more than the other. The distance between two stations is

(1) 350 km

(2) 210 km

(3) 300 km.

(4) 140 km

Ans .

(1) 350 km.

Explanation :

Let the trains meet each other after t hours.
Distance = Speed × Time
According to the question,
21t – 14t = 70
7t = 70
t=10
Required distance
= 21t + 14t = 35t
= 35 × 10 = 350 km.

TYPE-5

Q.1

A train running at \( \frac{7}{11}\) of its own speed reached a place in 22 hours. How much time could be saved if the train would run at its own speed?

(1) 14 hours

(2) 7 hours

(3) 8 hours

(4) 16 hours

Ans .

(3) 8 hours

Explanation :

Since the train runs at \( \frac{7}{11}\)of its own speed, the time it takes is \( \frac{11}{7}\)of its usual speed.Let the usual time taken be t hours.
Then we can write,\( \frac{11}{7}\)t = 22
t=14 hours
Hence, time saved
= 22 – 14 = 8 hours

Q.2

A man with \( \frac{3}{5}\) of his usual speed reaches the destination \( \frac{5}{2}\) hours late. Find his usual time to reach the destination.

(1) 3.75 hours

(2) 7.5 hours

(3) 8 hours

(4) 16 hours

Ans .

(1) 3.75 hours

Explanation :

\( \frac{3}{5}\)of usual speed will take \( \frac{5}{3}\)of usual time. time & speed are inversely
proportional \( \frac{5}{3}\) of usual time
= usual time + \( \frac{5}{2}\)
=\( \frac{2}{3}\) of usual time = \( \frac{5}{2}\)
usual time
\( \frac{5}{2}\) * \( \frac{3}{2}\)=\( \frac{15}{4}\)=3.75 hours.

Q.3

A car travelling with \( \frac{5}{7}\) of its usual speed covers 42 km in 1 hour 40 min 48 sec. What is the usual speed of the car?

(1) 35 kmph

(2) 7 kmph

(3) 25 kmph

(4) 16 kmph

Ans .

(1) 35 kmph

Explanation :

1 hr 40 min 48 sec
1 hr 40 + \( \frac{48}{60}\)
1 hr 40 + \( \frac{4}{5}\)
1 hr \( \frac{204}{5}\)
1 + \( \frac{204}{300}\)hr=\( \frac{504}{300}\)hr
Speed = \( \frac{42}{\frac{504}{300}}\)= 25 kmph
Now \( \frac{5}{7}\)*usual speed = 25
Usual speed = \( \frac{25*7}{5}\)= 35 kmph

Q.4

Walking at three-fourth of his usual speed, a man covers a cer- tain distance in 2 hours more than the time he takes to cover the distance at his usual speed. The time taken by him to cover the distance with his usual speed is

(1) 4.5 hours

(2) 5.5 hours

(3) 6 hours

(4) 5 hours

Ans .

(3) 6 hours

Explanation :

\( \frac{4}{3}\)× usual time – usual time = 2
\( \frac{1}{3}\)usual time = 2
Usual time = 2 × 3 = 6 hours

Q.5

By walking at \( \frac{3}{4}\)of his usual speed, a man reaches his office 20 minutes later than his usual time. The usual time taken by him to reach his office is

(1) 75 minutes

(2) 60 minutes

(3) 40 minutes

(4) 30 minutes

Ans .

(2) 60 minutes

Explanation :

\( \frac{4}{3}\)of usual time
= Usual time + 20 minutes
\( \frac{1}{3}\)of usual time = 20 minutes
Usual time = 20 × 3
= 60 minutes

Q.6

Walking at \( \frac{3}{4}\) of his usual speed, a man is \( \frac{3}{2}\) hours late. His usual time to cover the same distance, (in hours) is

(1) 14 hours

(2) 4.5 hours

(3) 8 hours

(4) 16 hours

Ans .

(2) 4.5 hours

Explanation :

Time and speed are inversely proportional.
\( \frac{4}{3}\)of usual time –usual time
=\( \frac{3}{2}\)
\( \frac{1}{3}\) * usual time= \( \frac{3}{2}\)
Usual time=\( \frac{3*3}{2}\)=\( \frac{9}{2}\)=4.5 hours

Q.7

Walking at \( \frac{6}{7}\) of his usual speed a man is 25 minutes late. His usual time to cover this dis- tance is

(1) 2 hours 30 minutes

(2) 2 hours 15 minutes

(3) 2 hours 25 minutes

(4) 2 hours 10 minutes

Ans .

(1) 2 hours 30 minutes

Explanation :

Time and speed are inversely proportional.
\( \frac{7}{6}\)* Usual time – Usual time
= 25 minutes
Usual time \( \frac{7}{6}\)-1
= 25 minutes
Usual time × \( \frac{1}{6}\)
= 25 minutes
Usual time = 25 × 6
= 150 minutes
= 2 hours 30 minutes

Q.8

Walking \( \frac{6}{7}\) th of his usual speed, a man is 12 minutes late. The usual time taken by him to cover that distance is

(1) 1 hour

(2) 1 hour 12 minutes

(3) 1 hour 15 minutes

(4) 1 hour 20 minutes

Ans .

(2) 1 hour 12 minutes

Explanation :

Time and speed are inversely proportional.
Usual time * \( \frac{7}{6}\)– usual time
= 12 minutes
Usual time * \( \frac{1}{6}\)= 12 minutes
Usual time = 72 minutes
= 1 hour 12 minutes

Q.9

A car travels from P to Q at a constant speed. If its speed were increased by 10 km/h, it would have been taken one hour less- er to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/h. The dis- tance between the two cities is

(1) 540 km

(2) 420 km

(3) 600 km

(4) 620 km

Ans .

(2) 420 km

Explanation :

Fixed distance = x km and certain speed = y kmph (let).
Case I,
\( \frac{x}{y+10}\)=\( \frac{x}{y}\) - 1
=\( \frac{x}{y+10}\) + 1=\( \frac{x}{y}\) .....(1)
Case II,
\( \frac{x}{y+20}\) = \( \frac{x}{y}\) -1 -\( \frac{3}{4}\)
=\( \frac{x}{y}\)- \( \frac{4+3}{4}\)
\( \frac{x}{y+20}\)+\( \frac{7}{4}\)=\( \frac{x}{y}\).....(2)
From equations (i) and (ii),
\( \frac{x}{y+10}\)+1=\( \frac{x}{y+20}\)+\( \frac{7}{4}\)
\( \frac{x}{y+10}\)-\( \frac{x}{y+20}\)=\( \frac{7}{4}\)-1
x*( \frac{y + 20 - y - 10}{y + 10 )( y + 20 )} )
\( \frac{7-4}{4}\)=\( \frac{3}{4}\)
\( \frac{x *10}{( y + 10 )( y + 20 )}\)=\( \frac{3}{4}\)
3 (y + 10) (y + 20) = 40 x
\( \frac{3 ( y + 10 )( y + 20 )}{40}\)=x...(3)
From equation (i),
\( \frac{3 ( y + 10 )( y + 20 )}{40(y+10)}\) + 1
\( \frac{3 ( y + 10 )( y + 20 )}{40y}\)
3 (y +20) + 40
\( \frac{3 ( y + 10 )( y + 20 )}{y}\)
3y 2 + 60y + 40 y = 3(y 2 + 30y + 200)
3y 2 + 100y = 3y 2 + 90y + 600
10y = 600 Þ y = 60
Again from equation (i),
\( \frac{x}{y+10}\)+1=\( \frac{x}{y}\)
\( \frac{x}{60+10}\)+1=\( \frac{x}{60}\)
\( \frac{x}{70}\)+1=\( \frac{x}{60}\)
\( \frac{x+70}{70}\)+1=\( \frac{x}{60}\)
6x + 420 = 7x
7x – 6x = 420
x = 420 km.

Q.10

A car covers four successive 7 km distances at speeds of 10 km/hour, 20 km/hour, 30 km/ hour and 60 km/hour respec- tively. Its average speed over this distance is

(1) 30 km/hour

(2) 20 km/hour

(3) 60 km/hour

(4) 40 km/hour

Ans .

(2) 20 km/hour

Explanation :

Total distance
= 7 × 4 = 28 km.
Total time
\( \frac{7}{10}\)+\( \frac{7}{20}\)+\( \frac{7}{30}\)+\( \frac{7}{60}\) hours
\( \frac{42 + 21 + 14 + 7}{60}\)hours
=\( \frac{84}{60}\)hours=\( \frac{7}{5}\)hours
Average speed
=\( \frac{Total distance}{Total time}\)=\( \frac{28}{\frac{7}{5}}\)kmph
=20 kmph

Q.11

A car goes 20 metres in a sec-ond. Find its speed i n km/hr.

(1) 18

(2) 72

(3) 36

(4) 20

Ans .

(2) 72

Explanation :

1 m/sec = \( \frac{18}{5}\) kmph
20 m/sec =\( \frac{20*18}{5}\)
= 72 kmph

Q.12

The speed of a car is 54 km/hr. What is its speed in m/sec?

(1) 15 m/sec

(2) 19 - 44 m/sec

(3) 194.4 m/sec

(4) 150 m/sec

Ans .

(1) 15 m/sec

Explanation :

1 kmph =\( \frac{5}{18}\)m/sec
54 kmph =\( \frac{5}{18}\)*54
= 15 m/sec.

Q.13

A car covers a certain distance in 25 hours. If it reduces the speed by \( \frac{1}{5}\) th, the car covers 200 km. less in that time. The speed of car is

(1) 60 km./hr.

(2) 30 km./hr.

(3) 40 km./hr.

(4) 50 km./hr.

Ans .

(3) 40 km./hr.

Explanation :

Speed of car = x kmph.
Distance = Speed × Time
= 25x km.
Case II,
Speed of car =\( \frac{4x}{5}\)kmph
Distance covered =\( \frac{4x}{5}\)*25
= 20x km.
According to the question,
25x – 20x = 200
5x = 200
x=40 kmph.

Q.14

A car moving in the morning fog passes a man walking at 4 km/ h. in the same direction. The man can see the car for 3 minutes and visibility is upto a distance of 130 m. The speed of the car is :

(1) 14 km. per hour

(2) 7.6 km. per hour

(3) 8.6 km. per hour

(4) 6.6 km. per hour

Ans .

(4) 6.6 km. per hour

Explanation :

Speed of car = x kmph.
Relative speed = (x – 4) kmph
Time = 3 minutes =\( \frac{3}{60}\)hour=\( \frac{1}{20}\)hour
Distance = 130 metre
\( \frac{130}{1000}\)km=\( \frac{13}{100}\)km
Relative speed =\( \frac{Distance}{Time}\)
5x – 20 = 13
5x = 20 + 13 = 33
x=6.6 kmph.

TYPE-6

Q.1

A boy rides his bicycle 10km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. His average speed for the entire trip is approximately :

(1) 10.4 km/hr

(2) 10.8 km/hr

(3) 11.0 km/hr

(4) 12.2 km/hr

Ans .

(2) 10.8 km/hr

Explanation :

Total distance = 10 + 12
= 22 km
Total time = \( \frac{10}{12}\)+ \( \frac{12}{10}\)=\( \frac{244}{120}\)hours
Required average speed
\( \frac{Total Distance}{Total Time}\)=\( \frac{22}{\frac{244}{120}}\)=\( \frac{22}{244}\)*120
= 10.8 km/hr.

Q.2

A person travels 600 km by train at 80km/hr, 800 km by ship at 40 km/hr 500 km by aeroplane at 400 km/hr and 100 km by car at 50km/hr. What is the av- erage speed for the entire dis- tance ?

(1) 65.04 km/hr

(2) 61.8 km/hr

(3) 51.0 km/hr

(4) 72.2 km/hr

Ans .

(1) 65.04 km/hr

Explanation :

Total distance = 10 + 12
= 22 km
Total time
\( \frac{600}{80}\)+\( \frac{800}{40}\) +\( \frac{500}{400}\) +\( \frac{100}{50}\)
\( \frac{246}{8}\)hr
Average speed
\( \frac{600 + 800 + 500 + 100}{\frac{246}{8}}\)
\( \frac{2000*8}{246}\)
=65.04 km/hr.

Q.3

A train moves with a speed of 30 kmph for 12 minutes and for next 8 minutes at a speed of 45 kmph. Find the average speed of the train:

(1) 37.5 kmph

(2) 36 kmph

(3) 48 kmph

(4) 30 kmph

Ans .

(2) 36 kmph

Explanation :

Average speed
=\( \frac{Total distance}{Time taken}\)
=\( \frac{30* \frac{12}{60}+45*\frac{8}{60}}{\frac{12}{60}+\frac{8}{60}}\)
= 12 × 3 = 36 kmph

Q.4

A man covers half of his journey at 6km/hr and the remaining half at 3km/hr. His average speed is

(1) 9 km/hr

(2) 4.5 km/hr

(3) 4 km/hr

(4) 3 km/hr

Ans .

(3) 4 km/hr

Explanation :

If the same distance are covered at different speed of x kmph and y kmph, the average speed of the whole journey is given by =\( \frac{2xy}{x+y}\)kmph
Required average speed =\( \frac{36}{9}\)=4 kmph

Q.5

A man goes from A to B at a uni- form speed of 12 kmph and re- turns with a uniform speed of 4 kmph Hi s average speed (in kmph) for the whole journey is :

(1) 8

(2) 7.5

(3) 6

(4) 4.5

Ans .

(3) 6

Explanation :

If two equal distances are cov- ered at two unequal speed of x
kmph and y kmph, then average =\( \frac{2xy}{x+y}\)kmph
=\( \frac{96}{16}\)= 6 kmph

Q.6

A train covers a distance of 3584 km in 2 days 8 hours. If it cov- ers 1440 km on the first day and 1608 km on the second day, by how much does the average speed of the train for the remaining part of the journey differ from that for the entire journey ?

(1) 3 km/hour more

(2) 3 km/hour less

(3) 4 km/hour more

(4) 5 km/hour less

Ans .

(1) 3 km/hour more

Explanation :

Remaining distance
= (3584 – 1440 – 1608) km
= 536 km.
This distance is covered at the rate of \( \frac{536}{8}\)= 67 kmph.
Average speed of whole journey =\( \frac{3584}{56}\)=64 kmph
Required difference in speed = (67 – 64) kmph i.e. = 3 kmph more

Q.7

A man travels a distance of 24 km at 6 kmph. Another distance of 24 km at 8 kmph and a third distance of 24 km at 12 kmph. His average speed for the whole journey (in kmph) is

(1) 8

(2) 10

(3) 11

(4) 12

Ans .

(1) 8

Explanation :

Total distance
= 24 + 24 + 24 = 72 km.
Total time
=\( \frac{24}{6}\)+\( \frac{24}{8}\)+\( \frac{24}{12}\)
= (4 + 3 + 2) hours = 9 hours
\ Required average speed
=\( \frac{Total distance}{Total time}\)=8 kmph

Q.8

A constant distance from Chennai to Bangalore is covered by Ex- press train at 100 km/hr. If it returns to the same distance at 80 km/hr, then the average speed during the whole journey is

(1) 90.20 km/hr

(2) 88.78 km/hr

(3) 88.98 km/hr

(4) 88.89 km/hr

Ans .

(4) 88.89 km/hr

Explanation :

If same distance are covered at two different speed of x and y kmph, the average speed of journey =\( \frac{2xy}{x+y}\)
=\( \frac{2*100*80}{100+80}\)
= 88.89 kmph

Q.9

A person went from A to B at an average speed of x km/hr and returned from B to A at an aver- age speed of y km/hr. What was his average speed during the to- tal journey ?

(1) \( \frac{x+y}{2xy}\)

(2) \( \frac{2xy}{x+y}\)

(3) \( \frac{2}{x+y}\)

(4) \( \frac{1}{x}\)+\( \frac{1}{y}\)

Ans .

(2) \( \frac{2xy}{x+y}\)

Explanation :

Required average speed \( \frac{2xy}{x+y}\)
Since, can be given as corollary If the distance between A and B be z units, then
Average speed =\( \frac{Total speed}{Time taken}\)
\( \frac{z+z}{\frac{z}{x}+\frac{z}{y}}\)
=\( \frac{2xy}{x+y}\)

Q.10

A man goes from Mysore to Ban- galore at a uniform speed of 40 km/hr and co mes b ack to Mysore at a uniform speed of 60 km/hr. His average speed for the whole journey is

(1) 48 km/hr

(2) 50 km/hr

(3) 54 km/hr

(4) 55 km/hr

Ans .

(1) 48 km/hr

Explanation :

Average speed
\( \frac{2xy}{x+y}\)
\( \frac{2*40*60}{40+60}\)
= 48 kmph

Q.11

A man goes from a place A to B at a speed of 12 km/hr and re- turns from B to A at a speed of 18 km/hr. The average speed for the whole journey is

(1) 14*\( \frac{2}{5}\)km/hr

(2) 15 km/hr

(3) 15*\( \frac{1}{2}\) km/hr

(4) 16 km/hr

Ans .

(1) 14*\( \frac{2}{5}\)km/hr

Explanation :

Average speed
\( \frac{2xy}{x+y}\)
\( \frac{2*12*18}{12+18}\)
=14*\( \frac{2}{5}\)

Q.12

One third of a certain journey is covered at the rate of 25 km/ hour, one-fourth at the rate of 30 km/hour and the rest at 50 km/ hour. The average speed for the whole journey is

(1) 35 km/hour

(2) 33*\( \frac{1}{3}\) km/hr

(3) 30 km/hr

(4) 22 km/hr

Ans .

(2) 33*\( \frac{1}{3}\) km/hr

Explanation :

Let the total distance be x km
Total time =\( \frac{\frac{x}{3}}{25}\)+\( \frac{\frac{x}{4}}{30}\)+\( \frac{\frac{5x}{12}}{50}\)
=\( \frac{4x+5x}{300}\)
=\( \frac{3x}{100}\)
Average speed=\( \frac{Distance}{Time}\)
=\(\frac{x}{\frac{3x}{100}}\)
=33*\( \frac{1}{3}\) km/hr

Q.13

A man completes 30 km of a jour- ney at the speed of 6 km/hr and the remaining 40 km of the jour- ney in 5 hours. His average speed for the whole journey is

(1) 7 km/hr

(2) 7.5 km/hr

(3) 8 km/hr

(4) 12 km/hr

Ans .

(1) 7 km/hr

Explanation :

Time taken to cover 30km at 6 kmph=\( \frac{30}{6}\)= 5 hour
Time taken to cover 40 km = 5 hours
\ Average speed=\( \frac{Total Distance}{Time}\)
\( \frac{30+40}{10}\)
=7 km/hr

Q.14

A man covers the journey from a station A to station B at a uni- form speed of 36 km/hr and returns to A with a uniform speed of 45 km/hr. His average speed for the whole journey is :

(1) 40 km/hr

(2) 40.5 km/hr

(3) 41 km/hr

(3) 42 km/hr

Ans .

(1) 40 km/hr

Explanation :

Here same distances are covered at different speeds.
\ Average speed
\( \frac{2xy}{x+y}\)
=\( \frac{2*36*45}{36+45}\)
=40 kmph

Q.15

The speed of a train going from Nagpur to Allahabad is 100 kmph while its speed is 150 kmph when coming back from Allaha- bad to Nagpur. Then the average speed during the whole journey is :

(1) 120 kmph

(2) 125 kmph

(3) 140 kmph

(4) 135 kmph

Ans .

(1) 120 kmph

Explanation :

Here, the distances are equal.
\ Average speed=\( \frac{2*100*150}{100+150}\)
=120 kmph

Q.16

P travels for 6 hours at the rate of 5 km/ hour and for 3 hours at the rate of 6 km/ hour. The av- erage speed of the journey in km/ hour is

(1) 3*\( \frac{1}{5}\)

(2) 5*\( \frac{1}{3}\)

(3) 1*\( \frac{2}{9}\)

(4) 2*\( \frac{2}{5}\)

Ans .

(2) 5*\( \frac{1}{3}\)

Explanation :

Total distance = 5 × 6 + 3 × 6
= 30 + 18 = 48 km
Total time = 9 hours
\ Average speed
\( \frac{48}{9}\)
=5*\( \frac{1}{3}\)

Q.17

With an average speed of 40 km/ hr, a train reaches its destination in time. If it goes with an average speed of 35 km/hr, it is late by 15 minutes. The total journey is

(1) 30 km

(2) 40 km

(3) 70 km

(4) 80 km

Ans .

(3) 70 km

Explanation :

Let the length of journey be x km, then \( \frac{x}{35}\)-\( \frac{x}{40}\)=\( \frac{15}{60}\)=\( \frac{1}{4}\) x= 70 km

Q.18

A bus covers four successive 3 km stretches at speed of 10 km/ hr, 20 km/hr, 30 km/hr and 60 km/hr respectively. Its average speed over this distance is

(1) 30 km/hr

(2) 25 km/hr

(3) 20 km/hr

(4) 10 km/hr

Ans .

(3) 20 km/hr

Explanation :

Average speed =\( \frac{Distance}{Time}\) =\( \frac{12}{\frac{3}{10}+\frac{3}{20}+\frac{3}{30}+\frac{3}{60}}\) \( \frac{12*60}{3*12}\) =20 km/hr

Q.19

A train travelled at a speed of 35 km/hr for the first 10 minutes and at a speed of 20 km/hr for the next 5 minutes. The average speed of the train for the total 15 minutes is

(1) 30 km/hr

(2) 23 km/hr

(3) 31 km/hr

(4) 29 km/hr

Ans .

(1) 30 km/hr

Explanation :

Distance covered 35*\( \frac{10}{60}\)+20*\( \frac{5}{60}\)
=\( \frac{45}{6}\)km
Total time = 15 minutes=\( \frac{1}{4}\)hr
Required average speed =\( \frac{Distance}{Time}\)
=30 kmph

Q.20

On a journey across Kolkata, a taxi averages 50 km per hour for 50% of the distance, 40 km per hour for 40% of it and 20 km per hour for the remaining. The average speed (in km/hour) for the whole journey is :

(1) 42

(2) 40

(3) 35

(4) 45

Ans .

(2) 40

Explanation :

Total distance = 100 km.
Total time \( \frac{50}{50}\)+\( \frac{40}{40}\)+\( \frac{10}{20}\)
=\( \frac{5}{2}\)hr
Average speed =\( \frac{100*2}{5}\)
= 40

Q.21

A train goes from Ballygunge to Sealdah at an average speed of 20 km/hour and comes back at an average speed of 30 km/hour. The average speed of the train for the whole journey is

(1) 27 km/hr

(2) 26 km/hr

(3) 25 km/hr

(4) 24 km/hr

Ans .

(4) 24 km/hr

Explanation :

Required average speed \( \frac{2*30*20}{30+20}\)
= 24 km/hr

Q.22

A and B are 20 km apart. A can walk at an average speed of 4 km/ hour and B at 6 km/hr. If they start walking towards each other at 7 a.m., when they will meet ?

(1) 8.00 a.m.

(2) 8.30 a.m.

(3) 9.00 a.m.

(4) 10.00 a.m.

Ans .

(3) 9.00 a.m.

Explanation :

If A and B meet after t hours, then
4 t + 6 t = 20
10 t = 20
t = 2 hr
Hence, both will meet at 9 a.m.

Q.23

A train runs from Howrah to Ban- del at an average speed of 20 km/ hr and returns at an average speed of 30 km/hr. The average speed (in km/hr) of the train in the whole journey is

(1) 20

(2) 22.5

(3) 24

(4) 25

Ans .

(3) 24

Explanation :

Average speed =\( \frac{2*20*30}{20+30}\)
= 24

Q.24

A motorist travels to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/ hr. His average speed for the whole journey in km/hr is

(1) 37.5

(2) 37

(3) 35

(4) 40

Ans .

(1) 37.5

Explanation :

Average speed of whole journey \( \frac{2*50*30}{50+30}\)
= 37.5 kmph

Q.25

A man walks from his house at an average speed of 5 km per hour and reaches his office 6 minutes late. If he walks at an average speed of 6 km/h he reaches 2 minutes early. The dis- tance of the o ffice from his house is

(1) 6 km

(2) 4 km

(3) 5 km

(4) 12 km

Ans .

(2) 4 km

Explanation :

Required distance of office from house = x km. (let)
Time =\( \frac{Distance}{Speed}\)
According to the question,
\( \frac{x}{5}\)-\( \frac{x}{6}\)=\( \frac{2}{15}\)
\( \frac{x}{30}\)=\( \frac{2}{15}\)
x= 4 km

Q.26

A train runs at an average speed of 75 km/hr. If the distance to be covered is 1050 kms, how long will the train take to cover it ?

(1) 13 hrs

(2) 12 hrs

(3) 15 hrs

(4) 14 hrs

Ans .

(4) 14 hrs

Explanation :

Time =\( \frac{Distance}{Speed}\)=\( \frac{1050}{75}\)
= 14 hrs

Q.27

A train travels 500 m in first minute. In the next 4 minutes, it travels in each minute 125 m more than that in the previous minute. The average speed per hour of the train during those 5 minutes will be

(1) 30 km/hr

(2) 45 km/hr

(3) 50 km/hr

(4) 55 km/hr

Ans .

(2) 45 km/hr

Explanation :

Total distance covered by train in 5 minutes
= (500 + 625 + 750 + 875 + 1000)
metre = 3750 metre
= 3.75 km.
Time = 5 minutes =\( \frac{5}{60}\)=\( \frac{1}{12}\)hr
Speed of train=\( \frac{3.75}{\frac{1}{12}}\)
= (3.75 × 12) kmph
= 45 kmph

Q.28

A man covers a total distance of 100 km on bicycle. For the first 2 hours, the speed was 20 km/ hr and for the rest of the jour- ney, it came down to 10 km/hr. The average speed will be

(1) 12\( \frac{1}{2}\) km/hr

(2) 13 km/hr

(3) 20 km/hr

(4) 12 km/hr

Ans .

(1) 12\( \frac{1}{2}\) km/hr

Explanation :

Distance covered in first 2 hours = 2 × 20 = 40 km.
Remaining distance = 100 – 40 = 60 km.
Time taken in covering 60 km at
10 kmph \( \frac{60}{10}\)=6 hr
Required average speed=\( \frac{Distance}{Time}\)
\( \frac{100}{2+6}\)
=12\( \frac{1}{2}\) km/hr

Q.29

When Alisha goes by car at 50 kmph, she reaches her office 5 minutes late. But when she takes her motorbike, she reaches 3 minutes early. If her office is 25 kms away, what is the approxi- mate average speed at which she rides her motorbike ?

(1) 68 kmph

(2) 62 kmph

(3) 58 kmph

(4) 52 kmph

Ans .

(1) 68 kmph

Explanation :

Difference of time = 5 + 3 = 8 minutes
\( \frac{8}{60}\)=\( \frac{2}{15}\)hr
If the speed of motorbike be x kmph, then
\( \frac{25}{50}\)-\( \frac{25}{x}\)=\( \frac{2}{15}\)
11x = 25 × 30
x=68.18 kmph
x= 68 kmph

Q.30

A man goes to a place on bicycle at speed of 16 km/hr and comes back at lower speed. If the aver- age speed is 6.4 km/hr in total journey, then the return speed (in km/hr) is :

(1) 10

(2) 8

(3) 6

(4) 4

Ans .

(4) 4

Explanation :

Let the speed of cyclist while returning be x kmph.
\ Average speed \( \frac{2*16*x}{16+x}\)
6.4 × 16 + 6.4x = 32x
32x – 6.4x = 6.4 × 16
25.6x = 6.4 × 16
x= 4 kmph

Q.31

A car completed a journey of 400 km in 12\( \frac{1}{2}\) hrs. The first \( \frac{3}{4}\) of the journey was done at 30 km/hr. Calculate the speed for the rest of the journey.

(1) 45 km/hr

(2) 25 km/hr

(3) 40 km/hr

(4) 30 km/hr

Ans .

(3) 40 km/hr

Explanation :

Total distance covered = 400 km. Total time =\( \frac{25}{2 }\)hr \( \frac{3}{4}\)of total journey \( \frac{3}{4}\) * 400 = 300 km. Time=\( \frac{Distance}{Speed}\) \( \frac{300}{30}\)=10 Remaining time =\( \frac{25}{2}\)-10 =\( \frac{5}{2}\) Remaining distance = 100 km. \ Required speed of car \( \frac{100}{\frac{5}{2}}\) =40 km/hr

Q.32

Durga walks 5 km from her home to school in 60 minutes, then bi- cycles back to home along the same route at 15 km per hour. Her sister Smriti makes the same round trip, but does so at half of Durga’s average speed. How much time does Smriti spend on her round trip ?

(1) 120 minutes

(2) 40 minutes

(3) 160 minutes

(4) 80 minutes

Ans .

(3) 160 minutes

Explanation :

Durga’s average speed
\( \frac{2*5*15}{5+15}\)
=\( \frac{15}{2}\) kmph
Distance of School = 5 km.
Smriti’s speed =\( \frac{15}{4}\)
Required time =2*\( \frac{5}{\frac{15}{4}}\)
=\( \frac{8}{3}\)hr
\( \frac{8}{3}\)*60 =160 minutes

Q.33

Gautam travels 160 kms at 32 kmph and returns at 40 kmph. Then his average speed is

(1) 72 kmph

(2) 71.11 kmph

(3) 36 kmph

(4) 35.55 kmph

Ans .

(4) 35.55 kmph

Explanation :

Here, distances are equal.
\ Average speed
\( \frac{2*32*40}{32+40}\)
\( \frac{320}{9}\)
= 35.55 kmph

Q.34

A car travels from A to B at the rate of 40 km/h and returns from B to A at the rate of 60 km/ h. Its average speed during the whole journey is

(1) 48 km/h

(2) 50 km/h

(3) 45 km/h

(4) 60 km/h

Ans .

(1) 48 km/h

Explanation :

Here, distance is same.
Average speed=\( \frac{2xy}{x+y}\)
=\( \frac{2*40*60}{40+60}\)
=48 km/h

Q.35

A bus travels 150 km in 3 hours and then travels next 2 hours at 60 km/hr. Then the average speed of the bus will be

(1) 54 km/hr.

(3) 50 km/hr.

(3) 11.0 km/hr

(4) 60 km/hr.

Ans .

(1) 54 km/hr

Explanation :

Total distance covered by the bus = 150 km. + 2 × 60 km.
= (150 + 120) km.
= 270 km.
\ Average speed=\( \frac{Distance}{Time}\)
\( \frac{270}{5}\)
= 54 km/hr

Q.36

Gautam goes to office at a speed of 12 kmph and returns home at 10 kmph. His average speed is :

(1) 11 kmph

(2) 22 kmph

(3) 10.9 kmph

(4) 12.5 kmph

Ans .

(3) 10.9 kmph

Explanation :

Here distances are same
=\( \frac{2*12*10}{12+10}\)
=\( \frac{240}{22}\)
= 10.9 kmph

Q.37

A man travels 50 km at speed 25 km/h and next 40 km at 20 km/ h and there after travels 90 km at 15 km/h. His average speed is :

(1) 18 kmph.

(2) 25 kmph.

(3) 20 kmph.

(4) 15 kmph.

Ans .

(1) 18 kmph.

Explanation :

Total distance covered
= (50 + 40 + 90) km
= 180 km
Time = \( \frac{Distance}{Speed}\)
Total time taken
\( \frac{50}{25}\)+\( \frac{40}{20}\) +\( \frac{90}{15}\)hours
= (2 + 2 + 6) hours
= 10 hours
Average speed
=\( \frac{Total distance}{Total time taken}\)
=\( \frac{180}{10}\)
=18 kmph

Q.38

At an average o f 80 km/hr Shatabdi Express reaches Ran- chi from Kolkata in 7 hrs. The distance between Kolkata and Ranchi is

(1) 560 km.

(2) 506 km.

(3) 560 m.

(4) 650 m.

Ans .

(3) 560 m.

Explanation :

Distance = Speed × Time
= (80 × 7) km.
= 560 km.

Q.39

To cover a distance of 216 km in 3.2 hours, what should be the average speed of the car in me- tre/second?

(1) 67.5 metre/second

(2) 33.75 metre/second

(3) 37.5 metre/second

(4) 18.75 metre/second

Ans .

(4) 18.75 metre/second

Explanation :

Required speed of car=\( \frac{Distance}{Time}\)
\( \frac{216}{3.2}\)kmph
\( \frac{216}{3.2}\)*\( \frac{5}{18}\)m/sec
= 18.75 m./sec.

TYPE-7

Q.1

In covering a certain distance, the speed of A and B are in the ratio of 3 : 4. A takes 30 minutes more than B to reach the destination. The time taken by A to reach the destination is :

(1) 2 hours

(2) 2.5 hours

(3) 3 hours

(4) 4 hours

Ans .

(1) 2 hours

Explanation :

Let the distance of destination be D km
Let the speed of A = 3x km/hr
then speed of B = 4x km/hr
\ According to question,
\( \frac{D}{3x}\)-\( \frac{D}{4x}\)=30 min
=\( \frac{1}{2}\)hour
\( \frac{D}{12x}\)=\( \frac{1}{2}\)
\( \frac{D}{3x}\)=\( \frac{4}{2}\)= 2 hours
Hence, time taken by A to reach destination = 2hrs.

Q.2

The speed of A and B are in the ratio 3 : 4. A takes 20 minutes more than B to reach a destina- tion. In what time does A reach the destination ?

(1) 1.33 hour.

(1) 2.30 hour.

(1) 1.30 hour..

(1) 1 hour.

Ans .

(1) 1.33 hour..

Explanation :

Ratio of speed = 3 : 4
Ratio of time taken = 4 : 3
Let the time taken by A and B be 4x hours and 3 x hours respec- tively.
Then, 4x–3x =\( \frac{20}{60}\)
x=\( \frac{1}{3}\)
Time taken by A = 4x hours
4*\( \frac{1}{3}\)
=1.33 hour

Q.3

The ratio of length of two trains is 5 : 3 and the ratio of their speed is 6 : 5. The ratio of time taken by them to cross a pole is

(1) 5 : 6.

(2) 11 : 8

(3) 25 : 18

(4) 27 : 16

Ans .

(3) 25 : 18.

Explanation :

Required ratio
\( \frac{5}{6}\):\( \frac{3}{5}\)
\( \frac{5*30}{6}\):\( \frac{30*3}{5}\)
= 25 : 18

Q.5

Two trains started at the same time, one from A to B and the other from B to A. If they arrived at B and A respectively 4 hours and 9 hours after they passed each other, the ratio of the speed of the two trains was

(1) 2 : 1

(2) 3 : 2.

(3) 4 : 3

(4) 5 : 4

Ans .

(2) 3 : 2.

Explanation :

Required ratio of the speed of two trains
= \( \frac{√9}{√4}\)
3 : 2

Q.6

The speed of two trains are in the ratio 6 : 7. If the second train runs 364 km in 4 hours, then the speed of first train is

(1) 60 km/hr

(2) 72 km/hr

(3) 78 km/hr

(4) 84 km/hr.

Ans .

(3) 78 km/hr

Explanation :

Speed of second train
\( \frac{364}{4}\)
= 91 kmph
7x = 91
6x=\( \frac{91}{7x}\)*6x
=78 kmph

Q.7

A truck covers a distance of 550 metres in 1 minute whereas a bus covers a distance of 33 kms in 45 minutes. The ratio of their speed is :

(1) 4 : 3

(2) 3 : 5.

(3) 3 : 4

(4) 50 : 3

Ans .

(3) 3 : 4.

Explanation :

Speed of truck
= 550m/minute
Speed of bus =\( \frac{33000}{45}\)
Required ratio = 550 :\( \frac{2200}{3}\)
=3:4

Q.8

Three cars travelled distance in the ratio 1 : 2 : 3. If the ratio of the time of travel is 3 : 2 : 1, then the ratio of their speed is

(1) 3 : 9 : 1

(2) 1 : 3 : 9

(3) 1 : 2 : 4

(4) 4 : 3 : 2

Ans .

(2) 1 : 3 : 9

Explanation :

Required ratio =\( \frac{1}{3}\):\( \frac{2}{2}\):\( \frac{3}{1}\)
=\( \frac{1}{3}\):1:3
=\( \frac{1}{3}\)*3:1*3:3*3
= 1 : 3 : 9

Q.9

A and B run a 5 km race on a round course of 400 m. If their speed are in the ratio 5 : 4, the number of times, the winner passes the other, is

(1) 1

(2) 2

(3) 3

(4) 5

Ans .

(3) 3

Explanation :

The winner will pass the other, one time in covering 1600m. Hence, the winner will pass the other 3 times in completing 5km race

Q.10

A cyclist, after cycling a distance of 70 km on the second day, finds that the ratio of distance covered by him on the first two days is 4 : 5. If he travels a distance of 42 km. on the third day, then the ratio of distance travelled on the third day and the first day is :

(1) 4 : 3

(2) 3 : 2

(3) 3 : 4

(4) 2 : 3

Ans .

(3) 3 : 4

Explanation :

Distance covered on the first day
\( \frac{4}{5}\)*70= 56 km
Required ratio = 42 : 56
= 3 : 4

Q.11

A certain distance is covered by a cyclist at a certain speed. If a jogger covers half the distance in double the time, the ratio of the speed of the jogger to that of the cyclist is

(1) 1 : 4

(2) 4 : 1

(3) 1 : 2

(4) 2 : 1

Ans .

(1) 1 : 4

Explanation :

Let speed of cyclist = x kmph
& Time = t hours
Distance= \( \frac{xt}{2}\)while time = 2t
Required ratio =\( \frac{xt}{2*2t}\):x
= 1 : 4

Q.12

It takes 8 hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the car is

(1) 2 : 3

(2) 3 : 2

(3) 3 : 4

(4) 4 : 3

Ans .

(3) 3 : 4

Explanation :

Speed of train = x kmph
Speed of car = y kmph
Case 1:
\( \frac{120}{x}\)+\( \frac{600-120}{y}\)=8
\( \frac{15}{x}\)+\( \frac{60}{y}\)=1...(1)
Case 2
\( \frac{200}{x}\)+\( \frac{400}{y}\)= 8 hours 20 min
\( \frac{24}{x}\)+\( \frac{48}{y}\)=1...(2)
\( \frac{15}{x}\)+\( \frac{60}{y}\)=\( \frac{24}{x}\)+\( \frac{48}{y}\)
\( \frac{9}{x}\)=\( \frac{12}{y}\)
\( \frac{x}{y}\)=\( \frac{9}{12}\)
=3:4

Q.13

It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the car is :

(1) 3 : 5.

(2) 3 : 4

(3) 4 : 3

(4) 4 : 5

Ans .

(2) 3 : 4

Explanation :

Let the speed of train be x kmph. and the speed of car be y kmph
Time=\( \frac{Distance}{Speed}\)
\( \frac{120}{x}\)+\( \frac{480}{y}\)=8
\( \frac{15}{x}\)+\( \frac{60}{y}\)=1.....(1)
\( \frac{200}{x}\)+\( \frac{400}{y}\)=\( \frac{25}{3}\)
\( \frac{24}{x}\)+\( \frac{48}{y}\)=1....(2)
From equations (i) and (ii),
\( \frac{15}{x}\)+\( \frac{60}{y}\)=\( \frac{24}{x}\)+\( \frac{48}{y}\)
\( \frac{x}{y}\)=\( \frac{9}{12}\)
=3:4

Q.14

A truck covers a distance of 550 metre in one minute where as a bus covers a distance of 33 km in \( \frac{3}{4}\) hour. Then the ratio of their speeds is :

(1) 1 : 3 .

(2) 2 : 3

(3) 3 : 4

(4) 1 : 4

Ans .

(3) 3 : 4.

Explanation :

Speed of truck =\( \frac{550 metre}{60 second}\)
\( \frac{55}{6}\)m/sec
Speed of bus =\( \frac{33 * 1000 metre}{\frac{3}{4}*60*60sec}\)
=\( \frac{440}{36}\)
Required ratio =\( \frac{55}{6}\):\( \frac{440}{36}\)
= 55 × 6 : 440
= 3 : 4

Q.15

A car travels 80 km. in 2 hours and a train travels 180 km. in 3 hours. The ratio of the speed of the car to that of the train is :

(1) 2 : 3

(2) 3 : 2

(3) 3 : 4

(4) 4 : 3

Ans .

(1) 2 : 3

Explanation :

Speed =\( \frac{Distance}{Time}\)
Speed of car : Speed of train
=\( \frac{80}{2}\):\( \frac{180}{3}\)
= 40 : 60 = 2 : 3

Q.16

The speeds of three cars are in the ratio of 1 : 3 : 5. The ratio among the time taken by these cars to travel the same distance is

(1) 3 : 5 : 15

(2) 15 : 3 : 5

(3) 15 : 5 : 3

(4) 5 : 3 : 1

Ans .

(3) 15 : 5 : 3

Explanation :

Speed ∝ \( \frac{1}{Time}\)
Required ratio of time
1:\( \frac{1}{3}\):\( \frac{1}{5}\)
=15:\( \frac{1}{3}\)*15:\( \frac{1}{5}\)*15
= 15 : 5 : 3

TYPE-8

Q.1

A thief is noticed by a policeman from a distance of 200m. The thief starts running and the policeman chases him. The thief and the po- liceman run at the rate of 10 km./ hr and 11 km./hr respectively. What is the distance between them after 6 minutes ?

(1) 100 m

(2) 190 m

(3) 200 m

(4) 150 m.

Ans .

(1) 100 m.

Explanation :

Relative speed of police
= 11 – 10 = 1 kmph
=\( \frac{5}{18}\)
Distance decreased in 6 min- =\( \frac{5}{18}\) × 6×60 = 100 m
Distance remained between them = 200–100 = 100 m

Q.2

A moving train, 66 metres long, overtakes another train of 88 metres long, moving in the same direction in 0.168 minutes. If the second train is moving at 30 km/hr, at what speed is the first train moving ?

(1) 85 km/hr.

(2) 50 km/hr.

(3) 55 km/hr.

(4) 25 km/hr

Ans .

(1) 85 km/hr

Explanation :

Suppose the speed of first train be x kmph
Speed of second train
= 30 kmph
\( \frac{30*1000}{60}\)= 500 m per min.
According to question
\( \frac{Total distance}{Relative speed}\)
\( \frac{(66 + 88 )}{x-500}\)=0.168
0.168x – 84 = 154
0.168x = 238
x=\( \frac{238}{0.168}\)
\( \frac{238*1000}{168}\)*\( \frac{3}{50}\)
= 85 kmph

Q.3

A constable is 114 metres be- hind a thief. The constable runs 21 metres and the thief runs 15 metres in a minute. In what time will the constable catch the thief ?

(1) 19 minutes

(2) 18 minutes

(3) 17 minutes

(4) 16 minutes

Ans .

(1) 19 minutes.

Explanation :

The gap of 114 metre will be filled at relative speed. Required time
\( \frac{114}{21-15}\)
=19 minutes

Q.4

How much time does a train, 50 m long, moving at 68 km/ hour take to pass another train, 75 m long, moving at 50 km/ hour in the same direction ?

(1) 5 seconds .

(2) 10 seconds

(3) 20 seconds

(4) 25 seconds

Ans .

(4) 25 seconds.

Explanation :

Both trains are moving in the same direction.
\ Their relative speed
= (68 – 50) kmph = 18 kmph
=18*\( \frac{5}{8}\)= 5 m/sec
Total length = 50 + 75 = 125 m
\ Required time =\( \frac{Total length}{Relative speed}\)=\( \frac{125}{5}\)
=25 seconds.

Q.5

A constable follows a thief who is 200 m ahead of the constable. If the constable and the thief run at speed of 8 km/hour and 7 km/hour respectively, the con- stable would catch the thief in

(1) 10 minutes

(2) 12 minutes

(3) 15 minutes

(4) 20 minutes

Ans .

(2) 12 minutes

Explanation :

The constable and thief are running in the same direction
\ Their relative speed
= 8 – 7 = 1km.
1*\( \frac{5}{18}\)
Required time =\( \frac{200}{\frac{5}{18}}\)
=720 sec
=\( \frac{720}{60}\)
=12 minutes

Q.6

Two trains are running with speed 30 km/hr and 58 km/hr in the same direction. A man in the slower train passes the faster train in 18 seconds. The length (in metres) of the faster train is :

(1) 70

(2) 100.

(3) 128

(4) 140

Ans .

(4) 140

Explanation :

Relative speed
= (58 – 30) km/hr
28*\( \frac{5}{18}\)
\( \frac{70}{9}\)m/sec.
Length of train =\( \frac{70}{9}\)*18
= 140 metres

Q.7

Two trains travel in the same di- rection at the speed of 56 km/h and 29 km/h respectively. The faster train passes a man in the slower train in 10 seconds. The length of the faster train (in me- tres) is

(1) 100.

(2) 80

(3) 75

(4) 120

Ans .

(3) 75.

Explanation :

Relative speed
= 56 – 29 = 27 kmph
27*\( \frac{5}{18}\)
\( \frac{15}{2}\)
Distance covered in 10 sec- onds \( \frac{15}{2}\)*10
= 75 m

Q.8

A bus moving at a speed of 45 km/hr overtakes a truck 150 metres ahead going in the same direction in 30 seconds. The speed of the truck is

(1) 27 km/hr.

(2) 24 km/hr

(3) 25 km/hr

(4) 28 km/hr

Ans .

(1) 27 km/hr

Explanation :

Let the speed of the truck be x kmph
Relative speed of the bus
= 45 - x kmph
Time=\( \frac{Distance}{Relative speed}\)
\( \frac{30}{60*60}\)=\( \frac{\frac{150}{1000}}{45-x}\)
(45 – x ) = 18
x=27 kmph

Q.9

Two trains of equal length are run- ning on parallel lines in the same direction at 46 km/h and 36 km/h. The faster train passes, the slow- er train in 36 seconds. The length of each train is :

(1) 82 m

(2) 50 m

(3) 80 m

(4) 72 m

Ans .

(2) 50 m.

Explanation :

Let the length of each train be x metre.
Relative speed
= 46 – 36 = 10 kmph
=\( \frac{25}{9}\)
=\( \frac{2x}{\frac{25}{9}}\)=36
x = 50 metre

Q.10

Two trains start from a certain place on two parallel tracks in the same direction. The speed of the trains are 45 km/hr and 40 km/ hr respectively. The distance be- tween the two trains after 45 min- utes will be

(1) 2 km 500 m

(2) 2 km 750 m

(3) 3 km 750 m

(4) 3 km 250 m

Ans .

(3) 3 km 750 m

Explanation :

Relative speed
= 45– 40 = 5 kmph
Required distance
5*\( \frac{45}{60}\)
\( \frac{15}{4}\)km
= 3 km 750

Q.11

A boy started from his house by bicycle at 10 a.m. at a speed of 12 km per hour. His elder brother started after 1 hr 15 mins by scooter along the same path and caught him at 1.30 p.m. The speed of the scooter will be (in km/hr)

(1) 4.5

(2) 36.

(3) 18.6

(4) 15

Ans .

(3) 18.6

Explanation :

Let the speed of Scooter be x
Distance covered by cycling in 3\( \frac{1}{2}\)hours = Distance covered
by scooter in 2\( \frac{1}{4}\) hours
12*\( \frac{7}{2}\)=x*\( \frac{9}{4}\)
x=\( \frac{56}{3}\)
= 18.6 kmph

Q.12

A policeman goes after a thief who has 100 metres start, if the policeman runs a kilometre in 8 min, and the thief a km in 10 min, the dis- tance covered by thief before he is over-powered is

(1) 350 m

(2) 400 m

(3) 320 m

(4) 420 m.

Ans .

(2) 400 m

Explanation :

Relative speed
\( \frac{1000}{8}\)-\( \frac{1000}{10}\)=\( \frac{1000}{40}\)
Required time = 4 m/minute
Distance covered by the thief =\( \frac{1000}{10}\)*4
= 400 metres

Q.13

Two trains are running 40 km/hr and 20 km/hr respectively in the same direction. The fast train com- pletely passes a man sitting in the slow train in 5 seconds. The length of the fast train is

(1) 27.7 m

(2) 25 m.

(3) 20 m.

(4) 15 m.

Ans .

(1) 27.7 m

Explanation :

Relative speed = 40 – 20
= 20 km/hour
=\( \frac{20*5}{18}\)
Length of the faster train =\( \frac{250}{9}\)= 27.7 metres

Q.14

A train is moving at a speed of 80 km/h and covers a certain distance in 4.5 hours. The speed of the train to cover the same distance in 4 hours is

(1) 100 km/h.

(2) 70 km/h.

(3) 85 km/h

(4) 90 km/h

Ans .

(4) 90 km/h

Explanation :

Distance = Speed × Time
= 80 × 4.5 = 360 km
Required speed = \( \frac{360}{4}\)
= 90 kmph.

Q.15

Two trains 180 metres and 120 metres in length are running to- wards each other on parallel tracks, one at the rate 65 km/ hour and another at 55 km/hour. In how many seconds will they be clear of each other from the moment they meet ?

(1) 6.

(2) 9

(3) 12

(4) 15

Ans .

(2) 9

Explanation :

Required time =\( \frac{Sum of the lengths of trains}{Relative speed}\)
Relative speed = 65 + 55
= 120 kmph
\( \frac{120*5}{18}\)
Required time = \( \frac{180+120}{\frac{120*5}{18}}\)
= 9 seconds

Q.16

Two trains, of same length, are running on parallel tracks in the same direction with speed 60 km/hour and 90 km/hour respectively. T he l atter completely crosses the former in 30 seconds. The length of each train (in metres) is

(1) 125

(2) 150.

(3) 100

(4) 115

Ans .

(1) 125

Explanation :

When two trains cross each other, they cover distance equal to the sum of their length with relative speed.
Let length of each train = x metre
Relative speed = 90 – 60
= 30 kmph
\( \frac{30*5}{18}\)
=\( \frac{25}{3}\)m/sec
\( \frac{2x}{\frac{25}{3}}\)=30
2x = 250
x = 125 metres

Q.17

Two trains, 80 metres and 120 metres long, are running at the speed of 25 km/hr and 35 km/hr respectively in the same direction on parallel tracks. How many seconds will they take to pass each other ?

(1) 48

(2) 64

(3) 70

(4) 72.

Ans .

(4) 72.

Explanation :

Relative speed = 35 – 25
= 10 kmph
=\( \frac{10*5}{18}\)m/sec
Total length = 80 + 120
= 200 metres
Required time =\( \frac{Sum of the length of trains}{Relative speed}\)
=\( \frac{200*18}{10*5}\)
= 72 seconds

Q.18

A goods train starts running from a place at 1 P.M. at the rate of 18 km/hour. Another goods train starts from the same place at 3 P.M. in the same direction and overtakes the first train at 9 P.M. The speed of the second train in km/hr is

(1) 24

(2) 30

(3) 15

(4) 18.

Ans .

(1) 24

Explanation :

Distance covered by the first goods train in 8 hours = Distance covered by the second goods train in 6 hours.
18 × 8 = 6 * x
x=\( \frac{18*8}{6}\)
= 24 kmph

Q.19

Two trains 125 metres and 115 metres in length, are running to- wards each other on parallel lines, one at the rate of 33 km/ hr and the other at 39 km/hr. How much time (in seconds) will they take to pass each other from the moment they meet ?

(1) 8

(2) 10

(3) 12

(4) 15

Ans .

(3) 12.

Explanation :

Relative speed
= (33 + 39) kmph
= 72 kmph
\( \frac{72*5}{18}\)m/sec
= 20 m/sec.
\ Time taken in crossing
=\( \frac{Length of both trains}{Relative speed}\)
=\( \frac{240}{20}\)
=12 seconds

Q.20

A thief steals a car at 1.30 p.m. and drives it off at 40 km/hr. The theft is discovered at 2 p.m. and the owner sets off in anoth- er car at 50 km/hr. He will over- take the thief at

(1) 5 p.m.

(2) 4 p.m.

(3) 4.30 p.m.

(4) 6 p.m.

Ans .

(2) 4 p.m..

Explanation :

Distance covered by the thief in half an hour =\( \frac{1}{2}\)*40 =20 km
Relative speed of car owner
= 50 – 40 = 10 km
\ Required time
=\( \frac{Difference of distance}{Relative speed}\)
\( \frac{20}{10}\)
= 2 hours
i.e. at 4 p.m.

Q.21

Two trains of equal length are running on parallel lines in the same direction at the rate of 46 km/hr and 36 km/hr. The fast- er train passes the slower train in 36 seconds. The length of each train is

(1) 50 m

(2) 72 m

(3) 80 m

(4) 82 m

Ans .

(1) 50 m

Explanation :

Length of each train = x metre

Relative speed = 46 – 36
= 10 kmph
=\( \frac{10*5}{18}\)
=\( \frac{25}{9}\)m/sec
Time taken in crossing
\( \frac{Length of both trains}{Relative speed}\)
36=\( \frac{2x}{\frac{25}{9}}\)
x = 50 metre

Q.22

Two trains start from stations A and B and travel towards each other at speeds of 50 kmph and 60 kmph respectively. At the time of their meeting, the sec- ond train has travelled 120 km more than the first. The distance between A and B is

(1) 1200 km.

(2) 1440 km

(3) 1320 km

(4) 990 km

Ans .

(3) 1320 km

Explanation :

Let both trains meet after t hours.
\ Distance = speed × time
60t – 50t = 120
10t = 120 t = 12 hours
Required distance
= 60t + 50t
= 110t = 110 × 12
= 1320 km

Q.23

The distance between two plac- es A and B is 60 km. Two cars start at the same time from A and B, travelling at the speeds of 35 km/h and 25 km/h, respective- ly. If the cars run in the same direction, then they will meet af- ter ( in hours)

(1) 6.5

(2) 6.2

(3) 6.

(4) 6.52

Ans .

(3) 6.

Explanation :

Let both cars meet at C after t hours.
Distance covered by car A = AC = 35t km
Distance covered by car B
= BC = 25t km
AC – BC = AB = 60 km.
35t – 25t = 60
10t = 60
t = 6 hours

Q.24

A train ‘B’ speeding with 100 kmph crosses another train C, running in the same direction, in 2 minutes. If the length of the train B and C be 150 metre and 250 metre respectively, what is the speed of the train C (in kmph)?

(1) 75

(2) 88

(3) 95

(4) 110.

Ans .

(2) 88.

Explanation :

Let the speed of train C be x kmph.
Relative speed of B
= (100 – x ) kmph.
Time taken in crossing
\( \frac{Length of both trains}{Relative speed}\)
\( \frac{2}{60}\)=\( \frac{\frac{150+250}{1000}}{100-x}\)
100 – x = 12
x = 100 – 12 = 88 kmph.

Q.25

A passenger train running at the speed of 80 kms./hr leaves the railway station 6 hours after a goods train leaves and overtakes it in 4 hours. What is the speed of the goods train?

(1) 32 kmph

(2) 50 kmph

(3) 45 kmph

(4) 64 kmph

Ans .

(1) 32 kmph

Explanation :

Let the speed of goods train be x kmph.
Distance covered by goods train in 10 hour = distance cov- ered by passenger train in 4 hours
10x = 80 × 4
x = 32 kmph.

Q.26

Two trains start from a certain place on two parallel tracks in the same direction. The speed of the trains are 45 km/hr. and 40 km/ hr respectively. The distance be- tween the two trains after 45 min- utes will be

(1) 2.5 km..

(2) 2.75 km.

(3) 3.7 km.

(4) 3.75 km.

Ans .

(4) 3.75 km..

Explanation :

Relative speed = 45 – 40
= 5 kmph.
Gap between trains after 45 minutes = 5*\( \frac{45}{60}\)
= 3.75 km.

Q.27

A thief is stopped by a policeman from a distance of 400 metres. When the policeman starts the chase, the thief also starts run- ning. Assuming the speed of the thief as 5 km/h and that of po- liceman as 9 km/h, how far the thief would have run, before he is over taken by the policeman ?

(1) 400 metre

(2) 600 metre

(3) 500 metre

(4) 300 metre

Ans .

(3) 500 metre

Explanation :

Distance between thief and policeman = 400 metre
Relative speed of policeman with respect to thief
= (9 – 5) kmph
= 4 kmph
4*\( \frac{5}{18}\)
\( \frac{10}{9}\)m/sec
Time taken in overtaking the thief
\( \frac{400}{\frac{10}{9}}\)
= 360 second
Distance covered by thief
= Speed × Time
=5*\( \frac{5}{18}\)*360
= 500 metre

Q.28

Two trains of equal length are running on parallel lines in the same direction at 46 km/hour and 36 km/hour. The faster train passes the slower train in 36 sec- onds. The length of each train is

(1) 72 m.

(2) 80 m

(3) 82 m

(4) 50 m.

Ans .

(4) 50 m

Explanation :

Let the length of each train be x metre.
Relative speed = (46 – 36) kmph
= 10 kmph
10*\( \frac{5}{18}\)
=\( \frac{25}{9}\)m/sec
\( \frac{2x}{\frac{25}{9}}\)=36
x=50 metre

TYPE-9

Q.1

If a man walks 20 km at 5 km/ hr, he will be late by 40 minutes. If he walks at 8 km/hr, how early from the fixed time will he reach?

(1) 15 minutes

(2) 25 minutes.

(3) 40 minutes

(4) 50 minutes.

Ans .

(4) 50 minutes

Explanation :

Time taken to cover 20 km at the speed of 5km/hr = 4 hours.
\ Fixed time = 4 hours – 40 min- utes
= 3 hour 20 minutes
Time taken to cover 20 km at the speed of 8 km/hr =\( \frac{20}{8}\)=2 hours 30 minutes
Required time = 3 hours 20 minutes – 2 hours 30 minutes = 50 minutes

Q.2

If a man reduces his speed to 2/ 3, he takes 1 hour more in walk- ing a certain distance. The time (in hours) to cover the distance with his normal speed is :

(1) 2

(2) 1.

(3) 3

(4) 1.5

Ans .

(1) 2.

Explanation :

Since man walks at \( \frac{2}{3}\)of usual speed, time taken wil be \( \frac{3}{2}\) usual time.
=usual time + 1 hour.
\( \frac{3}{2}\)-1 of usual time = 1
usual time = 2 hours.

Q.3

A student rides on bicycle at 8 km/hour and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/ hour and reaches school 5 min- utes early. How far is the school from his house ?

(2) 8 km

(2) 2 km.

(3) 5 km

(3) 8 km

Ans .

(3) 5 km

Explanation :

Let x km. be the required dis- tance.
Difference in time
= 2.5 + 5 = 7.5 minutes
=\( \frac{7.5}{60}\)=\( \frac{1}{8}\)hr
\( \frac{x}{8}\)-\( \frac{x}{10}\)=\( \frac{1}{8}\)
x=\( \frac{40}{8}\)= 5 km.

Q.4

A man covered a certain distance at some speed. Had he moved 3 km per hour faster, he would have taken 40 minutes less. If he had moved 2 km per hour slower, he would have taken 40 minutes more. The distance (in km) is :

(1) 20

(2) 35

(3) 10

(4) 40

Ans .

(4) 40

Explanation :

Let the distance be x km and initial speed be y kmph. According to question,
\( \frac{x}{y}\)-\( \frac{x}{y+3}\)=\( \frac{40}{60}\).....(1)and
\( \frac{x}{y-2}\)-\( \frac{x}{y}\)=\( \frac{40}{60}\)......(2)
From equations (i) and (ii),
\( \frac{x}{y}\)-\( \frac{x}{y+3}\)=\( \frac{x}{y-2}\)-\( \frac{x}{y}\)
3 (y – 2) = 2 (y + 3)
Þ 3y – 6 = 2y + 6
Þ y = 12
From equation (i),\( \frac{x}{12}\)-\( \frac{x}{15}\)=\( \frac{40}{60}\)
x=40
Distance = 40 km.

Q.5

If a train runs at 40 km/hour, it reaches its destination late by 11 minutes. But if it runs at 50 km/ hour, it is late by 5 minutes only. The correct time (in minutes) for the train to complete the journey is

(1) 13

(2) 15

(3) 19

(4) 21

Ans .

(3) 19 .

Explanation :

If the distance be x km, then
\( \frac{x}{40}\)-\( \frac{x}{50}\)=\( \frac{6}{60}\)
x = 20 km.
Required time
\( \frac{20}{40}\)hr-11 mimnutes
= 19 minutes

Q.6

A student walks from his house at speed of 2\( \frac{1}{2}\) km per hour and reaches his school 6 min- utes late. The next day he in- creases his speed by 1 km per hour and reaches 6 minutes be- fore school time. How far is the school from his house ?

(1) 1.75 km

(2)2.75 km.

(3)1.25 km

(4) 1.15 km.

Ans .

(1) 1.75 km

Explanation :

Let the required distance be x km.
Difference of time
= 6 + 6 = 12 minutes = \( \frac{1}{5}\) hr
According to the question,
\( \frac{x}{\frac{5}{2}}\)-\( \frac{x}{\frac{7}{2}}\)=\( \frac{1}{5}\)
\( \frac{14x-10x}{35}\)=\( \frac{1}{5}\)
x=\( \frac{7}{4}\)=1.75 km

Q.7

A boy is late by 9 minutes if he walks to school at a speed of 4 km/hour. If he walks at the rate of 5 km/hour, he arrives 9 min- utes early. The distance to his school is

(1) 9 km.

(2) 5 km

(3) 4 km

(4) 6 km

Ans .

(4) 6 km

Explanation :

Let the required distance be x km.
According to the question,
\( \frac{x}{4}\)-\( \frac{x}{5}\)=\( \frac{18}{60}\)
x=\( \frac{3}{10}\)* 20= 6 km

Q.8

A car can cover a certain distance in 4\( \frac{1}{2}\) hours. If the speed is in creased by 5 km/hour, it would takes \( \frac{1}{2}\) hr less to cover the same distance. Find the slower speed of the car.

(1) 50 km/hour

(2) 40 km/hour.

(3) 45 km/hour

(4) 60 km/hour

Ans .

(2) 40 km/hour.

Explanation :

Let the initial speed of the car be x kmph and the distance be y km.
Then,y=\( \frac{9}{2}\)x
and, y = 4 (x + 5)
9x = 8x + 40
x = 40 kmph

Q.9

Shri X goes to his office by scooter at a speed of 30km/h and reaches 6 minutes earlier. If he goes at a speed of 24 km/h, he reaches 5 minutes late. The distance of his office is

(1) 20 km

(2) 21 km.

(3) 22 km

(4) 24 km.

Ans .

(3) 22 km

Explanation :

Let the distance of office be x km
\( \frac{x}{24}\)-\( \frac{x}{30}\)=\( \frac{11}{60}\)
\( \frac{x}{120}\)=\( \frac{11}{60}\)
x=22 km

Q.10

Walking at 5 km/hr a student reaches his school from his house 15 minutes early and walking at 3 km/hr he is late by 9 minutes. What is the distance between his school and his house ?

(1) 5 km.

(2) 8 km

(3) 3 km

(4) 2 km.

Ans .

(3) 3 km.

Explanation :

Let the required distance be x km.
\( \frac{x}{3}\)-\( \frac{x}{5}\)=\( \frac{24}{60}\)
\( \frac{2x}{3}\)=2
2x = 2 × 3
x = 3 km

Q.11

A student goes to school at the rate of 2\( \frac{1}{2}\) km/h and reaches 6 minutes late. If he travels at the speed of 3 km/h. he is 10 min- utes early. The distance (in km) between the school and his house is

(1) 5

(2) 4

(3) 3 .

(4) 1.

Ans .

(2) 4

Explanation :

Let the required distance be x
km \( \frac{x}{\frac{5}{2}}\)-\( \frac{x}{3}\)=\( \frac{16}{60}\)
\( \frac{6x-5x}{15}\)=\( \frac{4}{15}\)
x = 4 km.

Q.12

When a person cycled at 10 km per hour he arrived at his office 6 minutes late. He arrived 6 mi nutes early, when he increased his speed by 2 km per hour. The distance of his office from the starting place is

(1) 6 km

(2) 7 km.

(3) 12 km.

(4) 16 km

Ans .

(3) 12 km.

Explanation :

Let the distance be x km.
\( \frac{x}{10}\)-\( \frac{x}{12}\)=\( \frac{12}{60}\)
x=\( \frac{1}{5}\)*60
= 12 km.

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