**Ans . **

(2) 64 sec.

Time taken=\( \frac{Distance}{Time}

= \frac{ \frac{4}{5}}{45} hour

=\frac{4*60*60}{5*45} sec.

= 64sec.\)

**Ans . **

(4) 720 km./hr.

Let the required speed is x km/
hr Then, \(240 * 5=\frac{5}{3}*x .., x = 720 km/hr\)

**Ans . **

(4) 1250

Speed of the man =\( 5km/hr =5 * \frac{1000}{60} m / min =\frac{250}{3} m / min.\) Time taken to cross the bridge
= 15 minutes
Length of the bridge
= speed × time = \( \frac{250}{3}*15m= 1250m.\)

**Ans . **

(2) 12

Speed = \( \frac{Distance}{Time} = \frac{250}{75}= \frac{10}{3}\) m/sec= \( \frac{10}{3}\) * \( \frac{18}{5} \)km/hr ..,1 m/s = \( \frac{18}{5}\) km / hr = 2 × 6 km/hr. = 12 km/hr.

**Ans . **

(4) 30

Speed =\( \frac{Distance}{Time}\)= \( \frac{200}{24}\)m/s. =\( \frac{200}{24}\) * \( \frac{18}{5}\) = 30 km/h

**Ans . **

(4) 36

Speed of car = 10 m/sec.
Required speed in kmph =\( \frac{10*18}{5}\)= 36 km/hr.

**Ans . **

(1) 6 hours

= Speed × Time
= 40 × 9 = 360 km.
The required time at 60 kmph= \( \frac{360}{60}\)=6hours.

**Ans . **

(3) 20 km

Total time = 5 hours 48 minutes=5 + \( \frac{48}{60}\)={5+ \( \frac{4}{5}\) }hours.=\( \frac{29}{5}hours\)..,\( \frac{x}{25}\) + \( \frac{x}{4}\)=\( \frac{29}{5}\)

\( \frac{4x+25x}{100}\)=\( \frac{29}{5}\)

5 × 29x = 29 × 100

x=\( \frac{29*100}{5*29}\)= 20 km.

**Ans . **

(1) 6 km

Let the required distance be x
km.
Then,\( \frac{x}{3}\) + \( \frac{x}{2}\) =5

\( \frac{2x+3x}{6}\)=5

5x = 6 × 5

=6km.

**Ans . **

(1) 2 hours

The boy covers 20 km in 2.5
hours.
Speed=\( \frac{20}{2.5}\)= 8 km/hr..

New speed = 16 km/hr

..,Time=\( \frac{32}{16}\)=2hours.

**Ans . **

(4) 50

Speed = 180 kmph

=\( \frac{180*5}{18}\)m/sec = 50 m/sec

..,1 km / hr =\( \frac{5}{18}\)m/s.

**Ans . **

(2) 21.6

Speed=\( \frac{150}{25}\)= 6 m/sec

= 6 *\( \frac{18}{5}\)=21.6 kmph

**Ans . **

(2) 54 km

Let the distance between A and
B be x km, then \( \frac{x}{9}\)-\( \frac{x}{10}\)=\( \frac{36}{60}\)=\( \frac{3}{5}\)

\( \frac{x}{90}\)=\( \frac{3}{5}\)

x =\( \frac{3}{5}\)*90 = 54 km.

**Ans . **

(4) 6.30 a.m

Difference of time

= 4.30 p.m – 11.a.m.
=5*\( \frac{1}{2}\)hours=\( \frac{11}{2}\)hours

\( \frac{5}{6}\)-\( \frac{3}{8}\)=\( \frac{20-9}{24}\)=\( \frac{11}{24}\)part

\( \frac{11}{24}\)part of the journey is covered in \( \frac{11}{2}\)hours

\( \frac{3}{8}\)part of the journey is covered in

\( \frac{11}{2}\) * \( \frac{24}{11}\)* \( \frac{3}{8}\)=\( \frac{9}{2}\)hours

4*\( \frac{1}{2}\)hours.

Clearly the person started at 6.30a.m.

**Ans . **

(1) 100 m

Speed of bus = 72 kmph

=\( \frac{75*5}{18}\)metre/second

= 20 metre/second

Required distance= 20 × 5 = 100 metre

**Ans . **

(3) 6 km

If the required distance be x km, then
\( \frac{x}{3}\)-\( \frac{x}{4}\)=\( \frac{1}{2}\)

\( \frac{4x-3x}{12}\)=\( \frac{1}{2}\)

\( \frac{x}{12}\)=\( \frac{1}{2}\)=x = 6 km

**Ans . **

(1) 420 km

Time=10*\( \frac{1}{2}\)hours

\( \frac{21}{2}\)hours

Speed = 40 kmph

Distance = Speed × Time

40*\( \frac{21}{2}\)= 420 km

**Ans . **

(2) 54.55 minutes

Distance covered on foot
4*3\( \frac{3}{4}\)km=15km.

Time taken on cycle

\(\frac{Distance}{Speed}\)=\( \frac{15}{16.5}\)hour

\(\frac{15*60}{16.5}\)minutes

= 54.55 minutes

**Ans . **

(2) 13 minutes 20 sec

Speed of Train=\( \frac{Distance}{Time}\)

\( \frac{10}{\frac{12}{60}}\)kmph

\( \frac{10*60}{12}\)=50kmph

New speed = 45 kmph

Required time =\( \frac{10}{45}\)hour

=\( \frac{2}{9}*60 minutes\)

=\( \frac{40}{3}\)minutes

= 13 minutes 20 seconds

**Ans . **

(2)\( \frac{b}{5a}\)hours

Man's speed=\( \frac{Distance}{Time}\)

\( \frac{a}{b}\)kmph

\( \frac{1000a}{b}\)m/hour

Time taken in walking 200metre
=\( \frac{200}{\frac{1000a}{b}}\)=\( \frac{b}{5a}\)hours

**Ans . **

(4) 12

1 m/sec = \( \frac{18}{5}\)kmph
..,\( \frac{10}{3}\)m/sec

\( \frac{18}{5}\)*\( \frac{10}{3}\)=12kmph

**Ans . **

(3) 10 km/hr

Remaining Time

=\( \frac{2}{5}\)*15=6 hours

Remaining Speed

=\( \frac{60}{6}\)=10kmph

**Ans . **

(2) 2\( \frac{5}{8}\)hours.

Speed of train = 60 kmph

Time = 210 minutes

\( \frac{210}{60}\)hours or \( \frac{7}{2}\)hours

Distance covered

=60*\( \frac{7}{2}\)=210km

Time taken at 80 kmph

=\( \frac{210}{80}\)=\( \frac{21}{8}\)hours

=2\( \frac{5}{8}\)hours

**Ans . **

(2) 6 hrs. 12 min.

90 km = 12 × 7km + 6 km. To
cover 7 km total time taken =\( \frac{7}{18}\)hours+6 min.=\( \frac{88}{3}\)min.So,(12 × 7 km) would be covered in 12* \( \frac{88}{3}\) min. and remaining 6 km. is \( \frac{6}{18}\)hrs or 20 min.

Total time=\( \frac{1056}{3}\)+20

=\( \frac{1116}{3*60}\)hours=6*\( \frac{1}{5}\)hours

= 6 hours 12 minutes.

**Ans . **

(1) 8.5 m/sec.

30.6 kmph

= (30.6*\( \frac{5}{18}\) )m/sec.
= 8.5 m/sec

**Ans . **

(2) 24

Let the total journey be of x
km, then \( \frac{2x}{15}\)+\( \frac{9x}{20}\)+10=x

x-\( \frac{2x}{15}\)-\( \frac{9x}{20}\)=10

\( \frac{60x-8x-27x}{60}\)=10

\( \frac{25x}{60}\)=10

x=\( \frac{60*10}{25}\)=24km.

**Ans . **

(2) 5

If the required distance be = x
km, then\( \frac{x}{4}\)-\( \frac{x}{5}\)=\( \frac{10+5}{60}\)

\( \frac{5x-4x}{20}\)=\( \frac{1}{4}\)

\( \frac{x}{20}\)=\( \frac{1}{4}\)

x=\( \frac{1}{4}\)*20=5km

**Ans . **

(1) 4.0 hrs.

Relative speed

{\( \frac{5}{2}\)+2}kmph=\( \frac{9}{2}\)kmph

Time=\( \frac{Distance}{Relative speed}\)=\( \frac{18}{ \frac{9}{2}}\)

\( \frac{18*2}{9}\)=4hours

**Ans . **

(1) 32 km

Journey on foot = x km

Journey on cycle = (80 – x) km

\( \frac{x}{8}\)+\( \frac{80-x}{16}\)=7

\( \frac{2x+80-x}{16}\)=7

x + 80 = 16 × 7 = 112

x = 112 - 80 = 32 km.

**Ans . **

(4) 30 km/hr

Distance covered by car in 2 hours=\( \frac{300*40}{100}\)=120 km

Remaining distance

= 300 – 120 = 180 km

Remaining time = 4 – 2

= 2 hours

Required speed=\( \frac{180}{2}\)=90kmph

Original speed of car =\( \frac{120}{2}\)=60kmph

Required increase in speed

= 90 – 60 = 30 kmph

**Ans . **

(2) 50 minutes

Time taken in covering 5 Km =\( \frac{5}{10}\)=\( \frac{1}{2}\)hour
= 30 minutes
That person will take rest for four times.Required time= (30 + 4 × 5) minutes= 50 minutes

**Ans . **

(2) \( \frac{40}{3}\)minutes

Time = 12 minutes

= \( \frac{12}{60}\)hours=\( \frac{1}{5}\)hours

=Speed of train=\( \frac{10}{\frac{1}{5}}\)=50kmph

New speed = 50 – 5 = 45 kmph

Required time=\( \frac{Distance}{Speed}\)

=\( \frac{10}{45}\)=\( \frac{2}{9}\)hours

\( \frac{2}{9}\)*60minutes

=\( \frac{40}{3}\)minutes.

**Ans . **

(2) \( \frac{3}{2}\)

Distance covered by motor cyclist
P in 30 minutes =30*\( \frac{1}{2}\)=15km
Relative speed
= 40 – 30 = 10 kmph
\ Required speed = Time taken
to cover is km at 10 kmph
=\( \frac{15}{10}\)=\( \frac{3}{2}\)hours.

**Ans . **

(2) 64 sec.

Speed of B = x kmph (let)
Speed of A = 2x kmph
Speed of C =\( \frac{x}{3}\)kmph
\( \frac{Speed of A }{Speed of C}\)=\( \frac{2x}{\frac{x}{3}}\)=6
Required time =\( \frac{1}{6}\)of\( \frac{3}{2}\)
\( \frac{1}{4}\)hour = 15 minutes

**Ans . **

(4) 4 hrs

Distance covered by truck in \( \frac{3}{2}\)hours

= Speed × Time

=90*\( \frac{3}{2}\)=135km.

Remaining distance

= 310 – 135 = 175 km

Time taken at 70 kmph

\( \frac{175}{70}\)=2.5hours

Total time = 1.5 + 2.5

= 4 hours

**Ans . **

(3) \( \frac{3}{2}\)hours

Distance = Speed × Time

= 60 km.

Time taken at 40 kmph

=\( \frac{60}{40}\)=\( \frac{3}{2}\)hours

**Ans . **

(4) 4 km

Distance of school = x km

Difference of time

= 16 minutes =\( \frac{16}{60}\)hours

\( \frac{x}{\frac{5}{2}}\)-\( \frac{x}{3}\)=\( \frac{16}{60}\)

=\( \frac{2x}{5}\)-\( \frac{x}{3}\)=\( \frac{4}{15}\)

=\( \frac{6x-5x}{15}\)=\( \frac{4}{15}\)

=\( \frac{x}{15}\)=\( \frac{4}{15}\)

x=\( \frac{4}{15}\)*15=4km.

**Ans . **

(1) 44\( \frac{4}{9}\)km/hr.

Average speed of journey

={\ \frac{2xy}{x+y}\)kmph

=\( \frac{2*40*50}{40+50}\)=\( \frac{2*40*50}{90}\)

\( \frac{400}{9}\)=44\( \frac{4}{9}\)kmph.

**Ans . **

(1) 5 hours

60 kmph =\( \frac{60*5}{18}\)m/sec

\( \frac{50}{3}\)m/sec.

Speed=\( \frac{1}{Time}\)

S 1 × T 1 = S 2 × T 2

\( \frac{50}{3}\)*\( \frac{9}{2}\)= 15 × T 2

75 = 15 × T 2

t2= \( \frac{75}{15}\)=5hours.

**Ans . **

(2) 3 kmph

Speed of Romita = x kmph

(let)
Distance = Speed × Time

According to the question,

4 × 6 + x × 6 = 42

Þ 6x = 42 – 24 = 18

Þ x = 18 ÷ 6 = 3 kmph

**Ans . **

(1) 16 km

Distance travelled by farmer

on foot = x km (let)

\ Distance covered by cycling

= (61–x ) km.

Time=\( \frac{Distance}{Speed}\)

According to the question,

\( \frac{x}{4}\)+\( \frac{61-x}{9}\)=9

\( \frac{9x + 61 * ́ 4 – 4x}{9*4}\)=9

5x + 244 = 9 × 9 × 4 = 324

5x = 324 – 244 = 80

x=\( \frac{80}{5}\)= 16 km.

**Ans . **

(4) 5 hrs.

Distance = Speed × TIme

[40*6\( \frac{1}{4}\)]km

[\( \frac{40*25}{4}\)]km = 250 km

New speed = 50 kmph

Required time

\( \frac{Distance}{Speed}\)=\( \frac{250}{50}\)=5 hours

**Ans . **

(4) \( \frac{7}{4}\)km

Distance between school and

house = x km (let)

Time=\( \frac{Distance}{Speed}\)

According to the question,

\( \frac{x}{\frac{5}{2}}\)-\( \frac{x}{\frac{7}{2}}\)=\( \frac{6+6}{60}\)=\( \frac{1}{5}\)

(Difference of time = 6 + 6 =12
minutes

\( \frac{2x}{5}\)-\( \frac{2x}{7}\)=\( \frac{1}{5}\)

\( \frac{14x-10x}{35}\)=\( \frac{2x}{5}\)

\( \frac{4x}{35}\)=\( \frac{1}{5}\)

4x=\( \frac{35}{5}\)=7

x=\( \frac{7}{4}\)km.

**Ans . **

(3) 80 km

Let the total distance be 4x km.

Time=\( \frac{Distance}{Speed}\)

According to the question,

\( \frac{x}{10}\)+\( \frac{3x}{12}\)=7

\( \frac{x}{10}\)+\( \frac{x}{4}\)=7

\( \frac{2x+5x}{20}\)=7

7x = 7 × 20

x=\( \frac{7*20}{7}\)= 20 km.

PQ = 4x = 4 × 20 = 80 km.

**Ans . **

(1) 4 km

Let the distance of school be
x km.

Difference of time = 6 + 10

= 16 minutes=\( \frac{16}{60}\)hour

\( \frac{4}{15}\)hour

Time=\( \frac{Distance}{Speed}\)

\( \frac{x}{\frac{5}{2}}\)-\( \frac{x}{3}\)=\( \frac{4}{15}\)

\( \frac{6x-5x}{15}\)=\( \frac{4}{15}\)

x = 4 km.

**Ans . **

(4) 270 km.

Let the distance covered be 2x
km.

Time=\( \frac{Distance}{Speed}\)

According to the question,

\( \frac{x}{60}\)+\( \frac{x}{45}\)=5\( \frac{15}{60}\)=5\( \frac{1}{4}\)

5\( \frac{3x+4x}{180}\)=5\( \frac{21}{4}\)

7x=\( \frac{21}{4}\)*180

x=\( \frac{21*180}{4*7}\)=135km

Length of total journey

= 2 × 135 = 270 km.

**Ans . **

(3) 18

Distance covered by car

= 42 × 10 = 420 km.

New time = 7 hours

Required speed\( \frac{420}{7}\)

= 60 kmph.

Required increase

= (60 – 42) kmph

= 18 kmph

**Ans . **

(1) 9.6 kmph.

Distance of the office= x km.Difference of time = 2 hours

\( \frac{x}{8}\)-\( \frac{x}{12}\)=2

\( \frac{3x-2x}{24}\)=2

\( \frac{x}{24}\)=2

x = 48 km.

Time taken at the speed of 8kmph=\( \frac{48}{8}\)= 6 hours

Required time to reach the office at 10 a.m. i.e., in 5 hours

=\( \frac{48}{5}\)kmph

= 9.6 kmph

**Ans . **

(1) 10 metre.

Speed of bus = 36 kmph.

36*\( \frac{5}{18}\)m/sec.

= 10 m/sec.

\ Distance covered in 1 second

= 10 metre

**Ans . **

(3) 990 km.

Time taken by bus moving at

60 kmph = t hours

Distance = Speed × Time

60*t=45*(t+\( \frac{11}{2}\)

60t – 45t =\( \frac{45*11}{2}\)

15t=\( \frac{45*11}{2}\)

t=\( \frac{45*11}{15*2}\)=\( \frac{33}{2}\)hours

Required distance

\( \frac{60*33}{2}\)= 990 km

**Ans . **

(4) 580 metre

Speed of train = 116 kmph

116*\( \frac{5}{18}\)m./sec.

\( \frac{580}{18}\)m./sec.

Required distance

= Speed × Time

\( \frac{580}{18}\)*18metre

= 580 metre

**Ans . **

(4) 24 km.

Part of journey covered by bus

and rickshaw,\( \frac{3}{4}\)+\( \frac{1}{6}\)=\( \frac{9+2}{12}\)=\( \frac{11}{12}\)

Distance covered on foot

=1-\( \frac{11}{12}\)=\( \frac{1}{12}\)part

Total journey

= 12 × 2 = 24 km.

**Ans . **

(3) 75 km/h

Distance covered by train in

15 hours = Speed × Time

= (60 × 15) km. = 900 km.

Required speed to cover 900 km.

in 12 hours =\( \frac{900}{12}\)= 75 kmph

**Ans . **

(2) 3.3 km.

Distance = Speed × Time
= 330 × 10 = 3300 metre
\( \frac{3300}{1000}\) km. = 3.3 km.

**Ans . **

(4) 12 km.

Let the required distance be
x km.

Time = 2 hours 20 minutes

=2\( \frac{1}{3}\)hours

According to the question,

\( \frac{x}{12}\)+\( \frac{x}{9}\)=\( \frac{7}{3}\)

\( \frac{7x}{36}\)=\( \frac{7}{3}\)

x=\( \frac{7}{3}\)*\( \frac{36}{7}\)= 12 km.

**Ans . **

(3) 750

Let the length of train be x metre
Speed = 90 km/hr

\( \frac{90*5}{18}\)metre / sec.
= 25 metre/sec.

Distance covered in 60 sec.
= 25 × 60 = 1500 metres

Now, according to question,
2x = 1500

x = 750 metre

**Ans . **

(3) 200 m

When a train crosses a bridge it
covers the distance equal to
length of Bridge & its own length
Let the length of the train be = x
Speed of the train
\( \frac{x + 800}{100}\)m/s

Since train passes the 800 m
bridge in 100 seconds.
Again, train passes the 400 m
bridge in 60 seconds.

\( \frac{400 + x}{\frac{x + 800}{100}}\)= 60

\( \frac{(400+x)*100}{x+800}\)=60

40000 + 100x

= 60x + 48000

100x – 60x = 48000 – 40000

40x = 8000

x= \( \frac{8000}{40}\)= 200 m

**Ans . **

(3) 20 seconds

In crossing the bridge, the
train travels its own length plus
the length of the bridge.
Total distance (length)

= 300 + 200 = 500 m.

Speed = 25m/sec.

The required time

= 500 ÷ 25 = 20 seconds

**Ans . **

(2) 500

When a train crosses a tun-
nel, it covers a distance equal to
the sum of its own length and
tunnel.

Let the length of tunnel be x
Speed = 78 kmph

\( \frac{78*1000}{60*60}\)m/sec. =\( \frac{65}{3}\)m/sec.

Speed=\( \frac{Distance}{Time}\)

\( \frac{65}{3}\)=\( \frac{800+x}{60}\)

(800 + x ) × 3 = 65× 60

800 + x = 65 × 20 m

x = 1300 – 800 = 500

Length of tunnel = 500 metres.

**Ans . **

(2) 7.5 seconds

When a train crosses a rail-
way platform, it travels a distance
equal to sum of length of platform
and its own length.
Speed = 132 kmph
=132*\( \frac{5}{18}\)=\( \frac{110}{3}\)m/sec.
Required time
\( \frac{110+165}{\frac{110}{3}}\)sec.
=\( \frac{275*3}{110}\)= 7 . 5 seconds

**Ans . **

(3) 90

Let the length of the train be x
metres.

When a train corsses a platform
it covers a distance equal to the
sum of lengths of train and plat-
form. Also, the speed of train is
same.

\( \frac{x+162}{18}\)=\( \frac{x+120}{15}\)

6x + 720 = 5x + 810

6x – 5x = 810 – 720

x = 90

\ The length of the train = 90m

**Ans . **

(3) 24 secs

When a train croses a bridge,
distance covered = length of
(bridge + train).

Speed of train

=\( \frac{150+500}{30}\)

\( \frac{650}{30}\)=\( \frac{65}{3}\)m/sec.

Time taken to cross the 370m long platform

=\( \frac{370 + 150}{\frac{65}{3}}\)

=\( \frac{520*3}{65}\)= 24 seconds

**Ans . **

(4)14 seconds

Speed of train = 90 kmph

90*\( \frac{5}{18}\)= 25 m/sec

Distance covered
= 230 + 120 = 350 m

Time taken=\( \frac{350}{25}\)

= 14 seconds

**Ans . **

(4) 300

Let the length of train be x
According to the question,

\( \frac{x+600}{30}\)=30

x + 600 = 900

x = 900 – 600 = 300 m

**Ans . **

(4) 39.6 km/hour

Let the length of the train be x
According to the question,

\( \frac{x+122}{17}\)=\( \frac{x+210}{25}\)

25x + 3050 = 17x + 3570

25x – 17x = 3570 – 3050

8x = 520

x=\( \frac{520}{8}\)= 65 metres

Speed of the train
=\( \frac{65+122}{17}\)

\( \frac{187}{17}\)metre/second

= 11 metre/second

\( \frac{11*18}{5}\)kmph

= 39.6 kmph

**Ans . **

(3) 50.4 km/hr

Let the Length of the train be x

Then, \( \frac{x+162}{6}\)=\( \frac{x+120}{5}\)

6x + 720 = 5x + 810

x = 810 – 720 = 90

Speed of the train
\( \frac{90+162}{18}\)m/sec

\( \frac{252}{18}\)*\( \frac{18}{5}\)kmph

= 50.4 kmph

**Ans . **

(1) 72 km/hr

Let the length of the train be x
Speed of train

\( \frac{x+300}{21}\)=\( \frac{x+240}{18}\)

\( \frac{x+300}{7}\)=\( \frac{x + 240}{6}\)

7x + 1680 = 6x + 1800

x = 120

Speed of train
\( \frac{x+300}{21}\)=\( \frac{420}{21}\)= 20 m/sec

\( \frac{20*18}{5}\)kmph = 72 kmph

**Ans . **

(2) 16.8 sec

Speed of Train,

=\( \frac{Sum of length of both trains}{Time taken}\)

\( \frac{60*5}{18}\)=\( \frac{110+170}{t}\)=\( \frac{280}{t}\)

t=\( \frac{280*18}{60*5}\)= 16.8 seconds.

**Ans . **

(4) 120 ft/second

Speed of Train,

\( \frac{Length of ( train + platform )}{Time taken to cross}\)

\( \frac{500+700}{10}\)feet/second

= 120 feet/second

**Ans . **

(3) 350 m.

Speed of train = 36kmph

=36*\( \frac{5}{18}\)= 10 m/sec.

If the length of bridge be x me-
tre, then

10=\( \frac{200+x}{55}\)

200 + x = 550

x = 550 – 200 = 350 metre.

**Ans . **

(2) 45 sec

36 kmph ={36*\( \frac{5}{18}\)}m/sec

= 10 m/sec.

Required time=\( \frac{270+180}{10}\)= 45 seconds

**Ans . **

(3) 15

Speed of train

=\( \frac{Length of (train + platform)}{Time taken in crossing}\)

=\( \frac{50+100}{10}\)

=\( \frac{150}{10}\)=15m/sec

**Ans . **

(2) 54

Speed of train

\( \frac{Length of platform and train}{Time taken in crossing}\)

=\( \frac{100+50}{10}\)metre/second

= 15 metre/second

15*\( \frac{18}{5}\)kmph

= 54 kmph

**Ans . **

(1) 48 sec

Speed of train = 36 kmph

{\( \frac{36*5}{18}\)m/sec.

=10 m/sec.

Required time

=\( \frac{Length of train and bridge}{Speed of train}\)

=\( \frac{120+360}{10}\)=\( \frac{480}{10}\)

= 48 seconds

**Ans . **

(4) 1250 metres

Time = 5 minutes

=\( \frac{1}{12}\)hour

Length of bridge = Speed × Time

=15*\( \frac{1}{12}\)=(\ \frac{5}{4}\)km

=\( \frac{5}{4}\)*1000metre

= 1250 metre

**Ans . **

(1) 50 seconds

Speed of train = 72 kmph

=\( \frac{72*5}{18}\)m/sec.

= 20 m/sec.

Required time

\( \frac{Length of train and bridge}{Speed of train}\)

=\( \frac{200+800}{20}\)

=\( \frac{1000}{20}\)=50 sec.

**Ans . **

(2) 125 m

Length of train = x metre (let)

Speed of train

=\( \frac{Length of train and bridge}{Time taken in crossing}\)

\( \frac{x+500}{100}\)=\( \frac{x+250}{60}\)

\( \frac{x+500}{5}\)=\( \frac{x+250}{3}\)

5x + 1250 = 3x + 1500

5x – 3x = 1500 – 1250

2x = 250

x=125metre

**Ans . **

(3) 750

Let the length of train be x metre
Speed = 90 km/hr

\( \frac{90*5}{18}\)metre / sec.
= 25 metre/sec.

Distance covered in 60 sec.
= 25 × 60 = 1500 metres

Now, according to question,
2x = 1500

x = 750 metre

**Ans . **

(4) 45

Let the length of train be x
metre.

When a train crosses a platform,
distance covered by it = length of
train and platform.
Speed of Train,

=\( \frac{x+50}{14}\)=\( \frac{x}{10}\)

=\( \frac{x+50}{7}\)=\( \frac{x}{5}\)

7x = 5x + 250

7x – 5x = 250

2x = 250 Þ x =\( \frac{250}{2}\)

= 125 metre

Speed of Train=\( \frac{x}{10}\)

\( \frac{125}{10}\)m./sec.

\( \frac{125}{10}\)*\( \frac{18}{5}\)kmph

=45 kmph

**Ans . **

(3) 750

Let, length of train = length

of platform = x metre

Speed of train = 90 kmph

\( \frac{90*5}{18}\)m/sec.

= 25 m/sec.

Speed of train

=\( \frac{Length of train and platform}{Time taken in crossing}\)

25=\( \frac{2x}{60}\)=2x = 25 × 60

x=\( \frac{25*60}{2}\)= 750 metre

**Ans . **

(2) 74.16

Speed of train,

\( \frac{Length of train and platform}{Time taken in crossing}\)

=\( \frac{221+500}{35}\)m/sec

=\( \frac{721}{35}\)m/sec

=\( \frac{721*18}{35*5}\)kmph

= 74.16 kmph

**Ans . **

(3) 25

Speed of train
= 54 kmph

\( \frac{54*5}{18}\)m/sec

= 15 m/sec.

Required time

\( \frac{Length of train and bridge}{Speed of train}\)

\( \frac{200+175}{15}\)sec

\( \frac{375}{15}\)sec

= 25 seconds

**Ans . **

(3) 18 sec

Relative speed of man and train

= 20 – 10 = 10m/sec.

Required time =\( \frac{180}{10}\)

= 18 seconds

**Ans . **

(3) 12

In this situation, the train covers
it length.

Required time

\( \frac{100}{30*1000}\)hr

\( \frac{100 *60 * 60}{30*1000}\)= 12 seconds

**Ans . **

(3) 30 sec

Relative speed of train
= 63–3 = 60 kmph

60*\( \frac{5}{18}\)m/sec

Time=\( \frac{Length of train}{Relative speed}\)

=\( \frac{500*18}{60*5}\)=30sec.

**Ans . **

(2) 15 km/hr

Speed=\( \frac{Distance}{time}\)

\( \frac{125}{30}\)= 4 . 16 m / s

= 4 . 16 m / s = 4.16*\( \frac{18}{5}\)

= 15 km/hr

**Ans . **

(1) 12 m/sec.

In crossing a man standing on
platform, train crosses its own
length.

Speed of train
\( \frac{120}{10}\)= 12 m/s

**Ans . **

(4) 13.5 seconds

Speed of train (in m/s)

20*\( \frac{5}{18}\)=\( \frac{50}{9}\)m/sec

Required time=\( \frac{75}{50}\)*9

= 13.5 seconds

**Ans . **

(1) 2.5 seconds

Speed of the train

= 144 kmph =144*\( \frac{5}{18}\)

= 40 m/s

When a train crosses a pole, it
covers a distance equal to its own
length.

The required time =\( \frac{100}{40}\)=\( \frac{5}{2}\)= 2 . 5 seconds.

**Ans . **

(3) 48

Speed of train
=\( \frac{120}{9}\)*\( \frac{18}{5}\)= 48 kmph

**Ans . **

(4) 500

Speed of train = 60 kmph
=60*\( \frac{5}{18}\)=\( \frac{50}{3}\)m/sec

Length of train
= Speed × Time

=\( \frac{50}{3}\)*30= 500 m

**Ans . **

(3) 25 km/hour

Let the speed of train be x
kmph and its length be y km.
When the train crosses a man, it
covers its own length
According to he question,

\( \frac{y}{(x-3)*\frac{5}{18}}\)=10

18 y = 10 × 5(x –3)

18y = 50x –150

and \( \frac{y}{(x-3)*\frac{5}{18}}\)=11

18y = 55(x–5)

18y = 55x –275

From equations (i) and (ii),

55x –275 = 50x–150

55x –50x = 275 – 150

5x = 125

x=25

Speed of the train = 25 kmph

**Ans . **

(3) 20 sec

Relative speed of train
= (36 – 9) kmph = 27 kmph

=\( \frac{27*5}{18}\)m/sec

=\( \frac{15}{2}\)m/sec

Required time
\( \frac{Length of the train}{Relative speed}\)

=\( \frac{150*2}{15}\)= 20 seconds

**Ans . **

(3) 5 km/hr

Distance covered in 10 min-
utes at 20kmph = distance cov-
ered in 8 minutes at (20 + x )
kmph

=20*\( \frac{10}{60}\)=\( \frac{8}{60}\)20 + x

200 = 160 + 8x

8x = 40

x=5 kmph

**Ans . **

(3) 35.72 kmph

If the speed of the train be x
kmph, then relative speed
= (x – 3) kmph.

=\( \frac{300}{x-3*\frac{5}{18}}\)= 33

5400 = 33 × 5 (x – 3)

360 = 11 (x – 3)

11x – 33 = 360

x=\( \frac{393}{11}\)kmph.
=35.72kmph

**Ans . **

(3) 83.4 kmph

If the speed of train be x kmph
then,
Its relative speed = (x + 3) kmph
Time=\( \frac{Length of the train}{Relative speed}\)
\( \frac{10}{3600}\)=\( \frac{\frac{240}{1000}}{x+3}\)=\( \frac{240}{1000 ( x + 3 )}\)
x + 3 = 86.4
x = 83.4 kmph

**Ans . **

(2) 250 m

Speed of train = 36 kmph

\( \frac{36*5}{18}\)m/sec = 10 m/sec.

Length of train

= Speed × time

= 10 × 25 = 250 metre

**Ans . **

(4) 250

Speed of train = 90 kmph
=\( \frac{90*5}{18}\)metre/second

= 25 metre/second

If the length of the train be x
then,

Speed of train

\( \frac{Length of train}{Time taken in crossing the signal}\)

25=\( \frac{x}{10}\)

x = 250 metre

**Ans . **

(1) 45 km/hr

Let speed of train be x kmph
Relative speed = (x + 5) kmph
Length of train=\( \frac{100}{1000}\)km
=\( \frac{1}{10}\)km
\( \frac{\frac{1}{10}}{x+5}\)=\( \frac{36}{5 *60 *60}\)
\( \frac{1}{10(x+5)}\)=\( \frac{1}{500}\)
x + 5 = 50
x = 45 kmph

**Ans . **

(1) 72 km/hr

Speed of train
=\( \frac{Length of train}{Time taken in crossing the pole}\)=\( \frac{120}{6}\)= 20 m/sec
=20*\( \frac{18}{5}\)
= 72 kmph

**Ans . **

(1) 20 seconds

Speed of train = 54 kmph
=\( \frac{54*5}{18}\)m/sec = 15 m/sec
Required time
=\( \frac{Length of trains}{Speed of train}\)
=\( \frac{300}{15}\)= 20 seconds

**Ans . **

(4) 7.2 secs

Speed of train = 90 kmph

=90*\( \frac{5}{18}\)m/sec.

= 25 m/sec.

When a train crosses a post, it
covers a distance equal to its own
length

Required time=\( \frac{Distance}{Speed}\)=\( \frac{180}{25}\)= 7.2 seconds

**Ans . **

(1) 6 km

Let the required distance be
x km.

Difference of time = 7 + 5 = 12

minutes=\( \frac{1}{5}\)hour

Time=\( \frac{Distance}{Speed}\)

According to the question,

\( \frac{x}{5}\)-\( \frac{x}{6}\)=\( \frac{1}{5}\)

\( \frac{6x-5x}{30}\)=\( \frac{1}{5}\)

\( \frac{x}{30}\)=\( \frac{1}{5}\)

x=\( \frac{30}{5}\)= 6 km.

**Ans . **

(2) 200m

If the length of train be x metre,
then speed of train

\( \frac{x}{20}\)=\( \frac{x+250}{45}\)

\( \frac{x}{4}\)=\( \frac{x+250}{9}\)

9x = 4x + 1000

9x – 4x = 1000

5x = 1000

x=\( \frac{1000}{5}\)

= 200 metre

**Ans . **

(4) 18

Speed of train

\( \frac{Length of train}{Time taken in crossing}\)

\( \frac{250}{50}\)= 5 m/sec.

5*\( \frac{18}{5}\)kmph

= 18 kmph

**Ans . **

(1) 90 km/hr

Speed of train A
\( \frac{150}{30}\)= 5 m/sec.

Speed of train B = x m/sec.

Relative speed = (5+x) m/sec.
\ Length of both trains = Rela-
tive speed × Time
300 = (5 + x) × 10

5 + x =\( \frac{300}{10}\)= 30

x = 30 – 5 = 25 m/sec.

\( \frac{25*18}{5}\)kmph.

= 90 kmph.

**Ans . **

(1) 8 seconds

Distance covered in crossing
a pole = Length of train

Speed of train = 72 kmph

\( \frac{72*5}{18}\)m./sec.

= 20 m./sec.

Required Time,\( \frac{160}{20}\)

= 8 seconds

**Ans . **

(3) 7.2 seconds

Speed of train = 50 kmph

\( \frac{50*5}{18}\)m./sec.

\( \frac{125}{9}\)m./sec.

Required Time:\( \frac{\frac{100}{125}}{9}\)seconds

\( \frac{100*9}{125}\)seconds

=7.2seconds

**Ans . **

(4) 45

Distance covered by train in
crossing a telegraphic post =
length of train..

Speed of Train=\( \frac{Distance}{Time}\)

=\( \frac{150}{12}\)m/sec

=\( \frac{150}{12}\)*\( \frac{18}{5}\)kmph

= 45 kmph

**Ans . **

(4) 6 seconds

Speed of train = 36 kmph

\( \frac{36*5}{18}\)m./sec

= 10 m./sec.

Required time

\( \frac{Length of train}{Speed of train}\)

=\( \frac{60}{10}\)= 6 seconds

**Ans . **

(3) 54 km/hr

When a train crosses a pole it
travels a distance equal to its
length.

Speed of train

\( \frac{240}{16}\)= 15 m./sec.

15 *\( \frac{18}{5}\)kmph

= 54 kmph.

**Ans . **

(2) 4.5 seconds

Distance covered by train
= Length of train

Speed of train = 60 kmph

\( \frac{60*5}{18}\)m./sec.

\( \frac{50}{3}\)m./sec.

Required Time=\( \frac{Distance}{Speed}\)

\( \frac{75}{\frac{50}{3}}\)sec

\( \frac{75*3}{50}\)seconds

= 4.5 seconds

**Ans . **

(2) 3 seconds

Speed of train = 120 kmph. \( \frac{120*5}{18}\)m./sec. \( \frac{100}{3}\)m./sec. Required time =\( \frac{Length of train}{Speed of train}\) \( \frac{100}{ \frac{100}{3}}\)seconds \( \frac{100}{100}\)*3 seconds = 3 seconds

**Ans . **

(3) 11 a.m.

Distance travelled by first train in one hour = 60 × 1 = 60km

Therefore, distance between two train at 9 a.m. = 330 – 60 = 270 km

Now, Relative speed of two trains = 60 + 75 = 135 km/hr

Time of meeting of two trains =\( \frac{270}{135}\)= 2 hrs.

Therefore, both the trains will meet at 9 + 2 = 11 A.M.

**Ans . **

(2) 80 minutes

Men are walking in opposite di-
rections. Hence, they will cover
the length of bridge at their rela-
tive speed.

Required time

\( \frac{1200}{5+10}\) = 80 minutes

**Ans . **

(2) 7.5 seconds

If two trains be moving in oppo-
site directions at rate u and v
kmph respectively, then their
relative speed

= (u + v) kmph.

Further, if their length be x and

y km. then time taken to cross

each other =\( \frac{x+y}{u+v}\) hours.

Here,

Total length = 160 + 140

= 300m.

Relative speed = (77 + 67) kmph

== 144 kmph = 144*\( \frac{5}{18}\)m/s

Time=\( \frac{300}{40}\)=\( \frac{15}{2}\)=7.5seconds

**Ans . **

(3) 36

Let the speed of each train be x
kmph.

Their relative speed

= x + x = 2x kmph.

Time taken

\( \frac{Total length of trains}{Relative Speed}\)

\( \frac{12}{60*60}\)=\( \frac{240*\frac{1}{1000}}{2x}\)

\( \frac{1}{300}\)=\( \frac{120}{1000x}\)

x=\( \frac{300*120}{1000}\)=36

The required speed = 36 kmph.

**Ans . **

(2) 10.8 sec.

Total length of trains

= 140 + 160 = 300 m.

Relative speed = 60 + 40

= 100 kmph

=100*\( \frac{5}{18}\)m/sec

Time taken to cross each other

=\( \frac{300}{\frac{250}{9}}\)=\( \frac{300*9}{250}\)= 10 . 8 sec.

**Ans . **

(3) 1320 km

Let train A start from station
A and B from station B.

Let the trains A and B meet after
t hours.

Distance covered by train A in
t hours = 50t

Distance covered by train B in t
hours = 60t km

According to the question,

60t – 50t = 120

t=\( \frac{120}{10}\)= 12 hours.

Distance AB = 50 × 12 + 60 ×

12 = 600 + 720 = 1320 km

**Ans . **

(2) 120 km/hr

Let the speed of second train
be x m/s.

80 km/h=\( \frac{80*5}{18}\)m/s

According to the question

\( \frac{1000}{x+\frac{80*5}{18}}\)=18

1000 = 18x + 400

x=\( \frac{600}{18}\)m/s

\( \frac{600}{18}\)*\( \frac{18}{5}\)km/h = 120 km/h

**Ans . **

(2) 6 seconds

Length of both trains
= 105 + 90 = 195 m.

Relative speed = (45 + 72)
= 117 kmph

117*\( \frac{5}{18}\)m/sec

Time taken=\( \frac{195}{\frac{65}{2}}\)=\( \frac{195*2}{65}\)

= 6 seconds

**Ans . **

(1) 14.4 seconds

Let the length of each train be x
metre.

Speed of first train =\( \frac{x}{18}\)m/sec

Speed of second train =\( \frac{x}{12}\)m/sec

When both trains cross each oth-
er, time taken

=\( \frac{2x}{\frac{x}{18}+\frac{x}{12}}\)

\( \frac{2x}{\frac{2x+3x}{36}}\)=\( \frac{2x*36}{5x}\)

\( \frac{72}{5}\)

= 14.4 seconds

**Ans . **

(4) 54 km./hr

Let the speed of the second train
be x m/s

Speed of first train
=\( \frac{150}{15}\)= 10 m/sec

Relative speed of trains
= (x + 10) m/s

Total distance covered
= 150 + 150 = 300 metre

Time taken = \( \frac{300}{x+10}\)

\( \frac{300}{x+10}\)=12

12x + 120 = 300

12x = 300 – 120 = 180

x=\( \frac{180}{12}\)= 15 m/s

\( \frac{15*18}{5}\)or 54 kmph

**Ans . **

(4) 400 m

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.

**Ans . **

(2) 12 Noon

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.

**Ans . **

(3) 5 sec

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

**Ans . **

(3) 12

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

**Ans . **

(3) 12 sec.

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

**Ans . **

(1) 54 km/hr

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

**Ans . **

(1) 444

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

**Ans . **

(3) 8 sec

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

**Ans . **

(3) 3.42 sec

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.

**Ans . **

(2) 85 km/hour

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

**Ans . **

(3) 100.

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

**Ans . **

(3) 54 km/hr

=\( \frac{Speed of X}{Speed of Y}\)

=\( \frac{Time taken by Y}{Time taken by X} ^\frac{1}{2} \)

=\( \frac{45}{y}\)=\( \frac{3 hours 20 min}{4 hours 48 min.} ^\frac{1}{2} \)

=\( \frac{45}{y}\)=\( \frac{200 minutes}{288 minutes.} ^\frac{1}{2} \)

=\( \frac{10}{12}\)

10y = 12 × 45

y=54 kmph.

**Ans . **

(3) 180

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

**Ans . **

(3) 9.5 kmph

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.

**Ans . **

(1) 875 km.

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.

**Ans . **

(4) 11 seconds

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.

**Ans . **

(1) 350 km.

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.

**Ans . **

(3) 8 hours

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

**Ans . **

(1) 3.75 hours

\( \frac{3}{5}\)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.

**Ans . **

(1) 35 kmph

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

**Ans . **

(3) 6 hours

\( \frac{4}{3}\)× usual time – usual time = 2

\( \frac{1}{3}\)usual time = 2

Usual time = 2 × 3 = 6 hours

**Ans . **

(2) 60 minutes

\( \frac{4}{3}\)of usual time

= Usual time + 20 minutes

\( \frac{1}{3}\)of usual time = 20 minutes

Usual time = 20 × 3

= 60 minutes

**Ans . **

(2) 4.5 hours

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

**Ans . **

(1) 2 hours 30 minutes

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

**Ans . **

(2) 1 hour 12 minutes

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

**Ans . **

(2) 420 km

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.

**Ans . **

(2) 20 km/hour

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

**Ans . **

(2) 72

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

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

= 72 kmph

**Ans . **

(1) 15 m/sec

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

54 kmph =\( \frac{5}{18}\)*54

= 15 m/sec.

**Ans . **

(3) 40 km./hr.

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.

**Ans . **

(4) 6.6 km. per hour

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.

**Ans . **

(2) 10.8 km/hr

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.

**Ans . **

(1) 65.04 km/hr

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.

**Ans . **

(2) 36 kmph

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

**Ans . **

(3) 4 km/hr

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

**Ans . **

(3) 6

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

**Ans . **

(1) 3 km/hour more

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

**Ans . **

(1) 8

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

**Ans . **

(4) 88.89 km/hr

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

**Ans . **

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

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 =\( \frac{Total speed}{Time taken}\)

\( \frac{z+z}{\frac{z}{x}+\frac{z}{y}}\)

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

**Ans . **

(1) 48 km/hr

Average speed

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

\( \frac{2*40*60}{40+60}\)

= 48 kmph

**Ans . **

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

Average speed

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

\( \frac{2*12*18}{12+18}\)

=14*\( \frac{2}{5}\)

**Ans . **

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

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

**Ans . **

(1) 7 km/hr

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

**Ans . **

(1) 40 km/hr

Here same distances are covered
at different speeds.

\ Average speed

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

=\( \frac{2*36*45}{36+45}\)

=40 kmph

**Ans . **

(1) 120 kmph

Here, the distances are equal.

\ Average speed=\( \frac{2*100*150}{100+150}\)

=120 kmph

**Ans . **

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

Total distance
= 5 × 6 + 3 × 6

= 30 + 18 = 48 km

Total time = 9 hours

\ Average speed

\( \frac{48}{9}\)

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

**Ans . **

(3) 70 km

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

**Ans . **

(3) 20 km/hr

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

**Ans . **

(1) 30 km/hr

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

**Ans . **

(2) 40

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

**Ans . **

(4) 24 km/hr

Required average speed \( \frac{2*30*20}{30+20}\)

= 24 km/hr

**Ans . **

(3) 9.00 a.m.

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.

**Ans . **

(3) 24

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

= 24

**Ans . **

(1) 37.5

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

= 37.5 kmph

**Ans . **

(2) 4 km

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

**Ans . **

(4) 14 hrs

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

= 14 hrs

**Ans . **

(2) 45 km/hr

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

**Ans . **

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

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

**Ans . **

(1) 68 kmph

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

**Ans . **

(4) 4

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

**Ans . **

(3) 40 km/hr

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

**Ans . **

(3) 160 minutes

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

**Ans . **

(4) 35.55 kmph

Here, distances are equal.

\ Average speed

\( \frac{2*32*40}{32+40}\)

\( \frac{320}{9}\)

= 35.55 kmph

**Ans . **

(1) 48 km/h

Here, distance is same.

Average speed=\( \frac{2xy}{x+y}\)

=\( \frac{2*40*60}{40+60}\)

=48 km/h

**Ans . **

(1) 54 km/hr

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

**Ans . **

(3) 10.9 kmph

Here distances are same

=\( \frac{2*12*10}{12+10}\)

=\( \frac{240}{22}\)

= 10.9 kmph

**Ans . **

(1) 18 kmph.

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

**Ans . **

(3) 560 m.

Distance = Speed × Time

= (80 × 7) km.

= 560 km.

**Ans . **

(4) 18.75 metre/second

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.

**Ans . **

(1) 2 hours

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.

**Ans . **

(1) 1.33 hour..

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

**Ans . **

(3) 25 : 18.

Required ratio

\( \frac{5}{6}\):\( \frac{3}{5}\)

\( \frac{5*30}{6}\):\( \frac{30*3}{5}\)

= 25 : 18

**Ans . **

(2) 3 : 2.

Required ratio of the speed of two trains

=
\( \frac{√9}{√4}\)

3 : 2

**Ans . **

(3) 78 km/hr

Speed of second train

\( \frac{364}{4}\)

= 91 kmph

7x = 91

6x=\( \frac{91}{7x}\)*6x

=78 kmph

**Ans . **

(3) 3 : 4.

Speed of truck

= 550m/minute

Speed of bus =\( \frac{33000}{45}\)

Required ratio = 550 :\( \frac{2200}{3}\)

=3:4

**Ans . **

(2) 1 : 3 : 9

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

**Ans . **

(3) 3

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

**Ans . **

(3) 3 : 4

Distance covered on the first day

\( \frac{4}{5}\)*70= 56 km

Required ratio = 42 : 56

= 3 : 4

**Ans . **

(1) 1 : 4

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

**Ans . **

(3) 3 : 4

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

**Ans . **

(2) 3 : 4

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

**Ans . **

(3) 3 : 4.

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

**Ans . **

(1) 2 : 3

Speed =\( \frac{Distance}{Time}\)

Speed of car : Speed of train

=\( \frac{80}{2}\):\( \frac{180}{3}\)

= 40 : 60 = 2 : 3

**Ans . **

(3) 15 : 5 : 3

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

**Ans . **

(1) 100 m.

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

**Ans . **

(1) 85 km/hr

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

**Ans . **

(1) 19 minutes.

The gap of 114 metre will be filled
at relative speed. Required time

\( \frac{114}{21-15}\)

=19 minutes

**Ans . **

(4) 25 seconds.

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.

**Ans . **

(2) 12 minutes

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

**Ans . **

(4) 140

Relative speed

= (58 – 30) km/hr

28*\( \frac{5}{18}\)

\( \frac{70}{9}\)m/sec.

Length of train =\( \frac{70}{9}\)*18

= 140 metres

**Ans . **

(3) 75.

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

**Ans . **

(1) 27 km/hr

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

**Ans . **

(2) 50 m.

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

**Ans . **

(3) 3 km 750 m

Relative speed

= 45– 40 = 5 kmph

Required distance

5*\( \frac{45}{60}\)

\( \frac{15}{4}\)km

= 3 km 750

**Ans . **

(3) 18.6

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

**Ans . **

(2) 400 m

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

**Ans . **

(1) 27.7 m

Relative speed = 40 – 20

= 20 km/hour

=\( \frac{20*5}{18}\)

Length of the faster train =\( \frac{250}{9}\)= 27.7 metres

**Ans . **

(4) 90 km/h

Distance = Speed × Time

= 80 × 4.5 = 360 km

Required speed = \( \frac{360}{4}\)

= 90 kmph.

**Ans . **

(2) 9

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

**Ans . **

(1) 125

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

**Ans . **

(4) 72.

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

**Ans . **

(1) 24

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

**Ans . **

(3) 12.

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

**Ans . **

(2) 4 p.m..

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.

**Ans . **

(1) 50 m

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

**Ans . **

(3) 1320 km

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

**Ans . **

(3) 6.

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

**Ans . **

(2) 88.

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.

**Ans . **

(1) 32 kmph

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.

**Ans . **

(4) 3.75 km..

Relative speed = 45 – 40

= 5 kmph.

Gap between trains after 45 minutes = 5*\( \frac{45}{60}\)

= 3.75 km.

**Ans . **

(3) 500 metre

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

**Ans . **

(4) 50 m

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

**Ans . **

(4) 50 minutes

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

**Ans . **

(1) 2.

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.

**Ans . **

(3) 5 km

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.

**Ans . **

(4) 40

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.

**Ans . **

(3) 19 .

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

**Ans . **

(1) 1.75 km

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

**Ans . **

(4) 6 km

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

**Ans . **

(2) 40 km/hour.

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

**Ans . **

(3) 22 km

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

**Ans . **

(3) 3 km.

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

**Ans . **

(2) 4

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.

**Ans . **

(3) 12 km.

Let the distance be x km.

\( \frac{x}{10}\)-\( \frac{x}{12}\)=\( \frac{12}{60}\)

x=\( \frac{1}{5}\)*60

= 12 km.

**Ans . **

(1) 60 km

Let the distance between stations
be x km, then speed of train

=\( \frac{x}{\frac{45}{60}}\)=\( \frac{4x}{3}\)

\( \frac{3x}{4x-15}\)=\( \frac{4}{5}\)

16x – 60 = 15x

x = 60 km

**Ans . **

(1) 13.33 minutes

Speed of train =\( \frac{Distance}{Time}\)=\( \frac{10}{\frac{12}{60}}\)

= 50 kmph

New speed = 45 kmph

Required time =\( \frac{10}{45}\)

\( \frac{2}{9}\)*60 minutes

=\( \frac{40}{3}\)=13.33 minutes.

**Ans . **

(1) 4 km.

Let the distance of the office

be x km, then \( \frac{x}{5}\)-\( \frac{x}{6}\)=\( \frac{8}{60}\)

x = 2 × 2 = 4 km

**Ans . **

(2) 4 km.

Let the distance of school be
x km,

then \( \frac{x}{3}\)-\( \frac{x}{4}\)=\( \frac{20}{60}\)

\( \frac{x}{12}\)=\( \frac{1}{3}\)

x= 4 km

**Ans . **

(3) 20 minutes

Distance between stations X and

Y = Speed × Time

= 55 × 4 = 220 km.

New speed = 55 + 5 = 60 kmph

Required time =\( \frac{220}{60}\)=\( \frac{11}{3}\)

= 3 hours 40 minutes.

Required answer

= 4 hours – 3 hours 40 minutes

= 20 minutes

**Ans . **

(3) 1 hour

Distance of journey = x km

Difference of time = 12 – 3 = 9 minutes

\( \frac{9}{60}\) hr=\( \frac{3}{20}\)hr

\( \frac{x}{70}\)-\( \frac{x}{80}\)=\( \frac{3}{20}\)

\( \frac{x}{56}\)=\( \frac{3}{2}\)

x=84 km

Required correct time \( \frac{84}{70}\)hr-12 minutes

=72 – 12 = 60 minutes

= 1 hour

**Ans . **

(4) 45 km/hr

Let the length of train be x me-
tres

\ According to question

Speed of the train =\( \frac{x}{10}\)m / sec

Also, the speed of the train
\( \frac{x+50}{14}\)m / sec.
It passes the platform in 14
seconds]

Both the speeds should be equal,
i.e.,

\( \frac{x}{10}\)=\( \frac{x+50}{14}\)

or 14x = 10x + 500

or 14x – 10x = 500

or 4x = 500

\ x = 125 metres

Hence, Speed =\( \frac{125}{10}\)= 12 . 5 m / sec

\( \frac{12.5*18}{5}\)km / hr .

= 45 km/hr.

**Ans . **

(2) 176

Let length of train be x m

Speed of train \( \frac{x+264}{20}\)

Also, speed of train =\( \frac{x}{8}\)

\( \frac{x}{8}\)=\( \frac{x+264}{20}\)

5x = 2x + 528

5x – 2x = 528

x = 528 ÷ 3 = 176 m

**Ans . **

(4) 79.2 km/hr

Let the length of train be x me-
tres.

Then, speed of train when it passes a telegraph post = \( \frac{x}{8}\)m/sec
and speed of train, when it
passes the bridge =\( \frac{x+264}{20}\)

Clearly,

\( \frac{x}{8}\)=\( \frac{x+264}{20}\)

5x = 2x + 528

3x = 528

x=176 m

Speed of train

\( \frac{176}{8}\)= 22 m/sec

22*\( \frac{18}{5}\)kmph

= 79.2 kmph

**Ans . **

(1) 25.2 km/hour.

Let the length of train be x
metres.

When the train crosses the standing man, its speed = \( \frac{x}{9}\)

When the train crosses the plat-
form of length 84 m, its speed \( \frac{x+84}{21}\)

Obviously,\( \frac{x+84}{21}\)=\( \frac{x}{9}\)

21x – 9x = 9 × 84

12x = 9 × 84

x=63 m

Required speed =\( \frac{63}{9}\)=\( \frac{63}{9}\)*\( \frac{18}{5}\)= 25.2 kmph

**Ans . **

(4)45 km/hr.

Suppose length of train be x

According to question

\( \frac{x+50}{14}\) = \( \frac{x}{10}\)14x = 10x + 500

4x = 500

x=125 m

Therefore, speed \( \frac{125}{10}\)*\( \frac{18}{5}\)= 45 kmph

**Ans . **

(4) 21.6 kmph.

Let the length of the train be x

According to the question,

Speed of the train \( \frac{x+90}{30}\)=\( \frac{x}{15}\)

x + 90 = 2x

x = 90 m

Speed of train =\( \frac{90}{15}\)

6 m/s =6*\( \frac{18}{5}\)= 21.6 kmph

**Ans . **

(3) 20 seconds.

Let the length of the train be x
metre

Speed of train when it crosses man=\( \frac{x}{10}\)

Speed of train when it crosses platform =\( \frac{x+300}{25}\)

According to the question,

Speed of train=\( \frac{x}{10}\)=\( \frac{x+300}{25}\)

25x = 10x + 3000

15x = 3000

x=200 m

Length of train = 200 metre

Speed of train= \( \frac{x}{10}\)=\( \frac{200}{10}\)= 20 m/sce

Time taken in crossing a 200m long platform = \( \frac{200+200}{20}\)= 20 seconds

**Ans . **

(4) 330 m

Let the length of the train be x
metres.

Speed of train in crossing boy = \( \frac{x}{30}\)

Speed of train in crossing platform \( \frac{x+110}{40}\)

According to the question,

\( \frac{x}{30}\)=\( \frac{x+110}{40}\)

4x = 3x + 330

x = 330 metres

**Ans . **

(3) 150

Let the length of train be x metre

\( \frac{x}{15}\)=\( \frac{x+100}{25}\)

5x = 3x + 300

2x = 300

x=150 metres

**Ans . **

(2) 40, 30

**Ans . **

(2) 52 km/hr, 26 km/hr

Let the speed of trains be x
and y metre/sec respectively,

\( \frac{100+95}{x-y}\)=27

x-y=\( \frac{65}{9}\)..(1)

\( \frac{195}{x+y}\)=9

x+y=\( \frac{195}{9}\)...(2)

By equation (i) + (ii)

2x= \( \frac{65}{9}\)+\( \frac{195}{9}\)=\( \frac{260}{9}\)

x = \( \frac{130}{9}\)m/sec.

\( \frac{130}{9}\)*\( \frac{18}{5}\)kmph = 52 kmph

From equation (ii),

y = \( \frac{65}{9}\)m/sec

\( \frac{65}{9}\)*\( \frac{18}{5}\)

= 26 kmph

**Ans . **

(2) 130.

Let the length of train be x me-
tre, then

\ Speed of train

\( \frac{x}{7}\)=\( \frac{x+390}{28}\)

x=\( \frac{390}{3}\)

= 130 metres

**Ans . **

(3) 15.5 seconds.

Speed of train = 36 kmph

36 * \( \frac{5}{18}\)= 10 m/sec

Length of train = 10 × 10

= 100 metres

Required time= \( \frac{100+55}{10}\)

= 15.5 seconds

**Ans . **

(2) 300.

Speed of train = 60 kmph

60*\( \frac{5}{18}\)m/sec.
=\( \frac{50}{3}\)m/sec.

If the length of platform be

= x metre, then

Speed of train=\( \frac{Length of (train + platform)}{Time taken in crossing}\)

50 × 10 = 200 + x

x = 500 – 200 = 300 metre

**Ans . **

(4) 9:24 am

Let both trains meet after t
hours since 7 a.m.

Distance between stations A and
B = x Km.

\( \frac{x}{4}\)*t+\( \frac{x}{\frac{7}{2}}\)*(t-1)=x

Speed=\( \frac{Distance}{Time}\)

\( \frac{7t+8t-8}{28}\)=1

15 t – 8 = 28

15 t = 28 + 8 = 36

t=2 hours 24 minutes

Required time = 9 :24 a.m.

**Ans . **

(2) 12.1 seconds.

Speed of train = 72 kmph.

\( \frac{72*5}{18}\)m/sec

= 20 m./sec.

Required time

\( \frac{Length of train and bridge}{Speed of train}\)

\( \frac{242}{20}\)

= 12.1 seconds

**Ans . **

(2) 6 seconds.

Relative speed of train

= (60 + 6) kmph.

\( \frac{66*5}{18}\)m/sec

=\( \frac{55}{3}\)m/sec.

Length of train = 110 metre

Required time =\( \frac{110}{\frac{55}{3}}\)

= 6 seconds

**Ans . **

(2) 20 m.

Let the time taken to complete
the race by A,B, and C be x min-
utes

Speed of A =\( \frac{1000}{x}\)

B =\( \frac{1000-50}{x}\) =\( \frac{950}{x}\)

C =\( \frac{1000-69}{x}\)= =\( \frac{931}{x}\)

Now, time taken to complete the
race by

B=\( \frac{1000}{\frac{950}{x}}\)=\( \frac{1000*x}{950}\)

and distance travelled by C in

\( \frac{1000x}{950}\)min

\( \frac{1000x}{950}\)*\( \frac{931}{x}\)= 980 km.

B can allow C

= 1000 – 980 = 20 m

**Ans . **

(4) 12 minutes

Ratio of the speed of A, B and

C = 6 : 3 : 1

Ratio of the time taken

=\( \frac{1}{6}\):\( \frac{1}{3}\):1=1:2:6

Time taken by A

\( \frac{72}{6}\)= 12 minutes

**Ans . **

(1) 17.24 seconds.

Let A take x seconds in covering

1000m and b takes y seconds
According to the question,

x+20=\( \frac{900}{1000}y\)

x+20=\( \frac{9}{10}y\)...(1)

\( \frac{950}{1000}\)x + 25 = y....(2)

From equation (i),

\( \frac{10x}{9}\)+\( \frac{200}{9}\)=y

\( \frac{10x}{9}\)+\( \frac{200}{9}\)=\( \frac{950}{1000}\)x + 25

\( \frac{200x-171x}{180}\)=\( \frac{225-200}{9}\)

\( \frac{29x}{180}\)=\( \frac{25}{9}\)

x=\( \frac{500}{29}\)=17.24

**Ans . **

(3) 14.4 kmph

Time taken by Kamal

\( \frac{100}{18*\frac{5}{18}}\)

= 20 seconds

Time taken by Bimal

= 20 + 5 = 25 seconds

Bimal’s speed = \( \frac{100}{25}\)=4 m/sec

=\( \frac{4*18}{5}\)=14.4 kmph.

**Ans . **

(1) 95 m..

When A runs 1000m, B runs
900m.

\ When A runs 500m, B runs
450 m.

Again, when B runs 400m, C runs
360 m.

\ When B runs 450m, C runs

\( \frac{360}{400}\)*450 = 405 metres

Required distance = 500 – 405

= 95 metres

**Ans . **

(1) 11.9 metre

According to the question,

\ When A runs 800 metres, B
runs 760 metres

\ When A runs 200 metres, B runs= \( \frac{760}{800}\)*200= 190 metres

Again, when B runs 500 metres,
C runs 495 metres.

\ When B runs 190 metres, C runs =\( \frac{495}{500}\)*190= 188.1 metres

Hence, A will beat C by
200 – 188.1 = 11.9 metres in a
race of 200 metres.

**Ans . **

(3) 29 metres..

According to the question,

Q When B runs 200 m metres, A
runs 190 metres

\ When B runs 180 metres, A runs=\( \frac{190}{200}\)*180= 171 metres
When C runs 200m, B runs 180
metres.

Hence, C will give a start to A by
= 200 – 171 = 29 metres

**Ans . **

(1) 31.25 metre

According to the question,

When A covers 1000m, B covers
= 1000 – 40 = 960 m

and C covers =1000 – 70 = 930 m

When B covers 960m, C covers
930 m.

\ When B covers 1000m, C covers=\( \frac{930}{960}\)*1000
= 968.75 metre

Hence, B gives C a start of
= 1000 – 968.75 = 31.25 metre

**Ans . **

(2) 20 min.

Relative speed

= 95 – 75 = 15 kmph

Required Time=\( \frac{Distance}{Relative speed}\)

\( \frac{5}{15}\)*60

=20 minutes

**Ans . **

(1) 15 minutes

Time taken by C = t hours

Time taken by B =\( \frac{t}{3}\)hours

Time taken by A =\( \frac{t}{6}\)hours

Here,t=\( \frac{3}{2}\)hours

Required time taken by A

\( \frac{3}{\frac{2}{6}}\)=\( \frac{1}{4}\)

\( \frac{1}{4}\)*60= 15 minutes

**Ans . **

(2) 40 min

2 hours 45 minutes

2 + \( \frac{45}{60}\)hours=\( \frac{11}{4}\)hours

Distance = Speed × Time

4 * \( \frac{11}{4}\)= 11 km.

Time taken in covering 11 km
at 16.5 kmph

=\( \frac{11}{16.5}\)

=40 minutes.

**Ans . **

(2) 35 km.

Let the total distance be x km.

Time =\( \frac{Distance}{Speed}\)

According to the question,

\( \frac{10}{6}\) +\( \frac{20}{16}\) +\( \frac{x-30}{3}\)=4 \( \frac{35}{60}\)

=4\( \frac{7}{12}\)

\( \frac{5}{3}\) +\( \frac{5}{4}\) +\( \frac{x}{3}\)-10 =\( \frac{55}{12}\)

x=\( \frac{140}{12}\)*3=35 km

**Ans . **

(1) 1 hour.

Usual time = x minutes

New time =\( \frac{4x}{3}\)

Speed ∝ \( \frac{1}{Time}\)

According to the question,

\( \frac{4x}{3}\)– x = 20

x= 1hour

**Ans . **

(2) 4 km/hr.

Let, A’s speed = x kmph.

\ B’s speed = (7 – x) kmph

Time=\( \frac{Distance}{Speed}\)

According to the question,

\( \frac{24}{x}\)+\( \frac{24}{7-x}\)=14

\( \frac{24*7}{x(7-x)}\)=14

x (7 – x) = 12 = 4 × 3 or 3 × 4

Þ x (7 – x) = 4 (7 – 4) or 3 (7 – 3)

Þ x = 4 or 3

\ A’s speed = 4 kmph.

**Ans . **

(3) 2 hours

Relative speed

= 12 + 10 = 22 kmph
Distance covered

= 55 – 11 = 44 km

\ Required time

\( \frac{44}{22}\)

= 2 hours

**Ans . **

(2) 200.

Required time = LCM of 40

and 50 seconds

= 200 seconds

**Ans . **

(1) 2 km.

Distance between starting

point and multiplex = x metre

Time =\( \frac{Distance}{Speed}\)

According to the question,

\( \frac{x}{3}\) -\( \frac{x}{4}\)= \( \frac{5+5}{60}\)

\( \frac{x}{12}\) =\( \frac{1}{6}\)

x=2 km

**Ans . **

(2) 19 minutes

Two ways walking time
= 55 min...(i)

One way walking + One way
riding time = 37 min.....(ii)

By 2 × (ii) – (i),

2 ways riding time
= 2×37–55 = 19 minutes.

**Ans . **

(3) 50 km.

Let the distance be x km

Time taken by A =\( \frac{x}{40}\)hrs

Time taken by B =\( \frac{x}{50}\)hrs

\( \frac{x}{40}\)-\( \frac{x}{50}\)=\( \frac{15}{60}\)

\( \frac{5x-4x}{200}\)=\( \frac{15}{60}\)

x=50 km

**Ans . **

(1) 5 km/hr.

Let the speed of man be x
kmph

30 x – 30{x-\( \frac{x}{15}\)} = 10

30{x-x+\( \frac{x}{15}\)}=10

\( \frac{x}{15}\)=\( \frac{10}{30}\)

x= 5kmph

**Ans . **

(1) 46 minutes 12 seconds

Required time = LCM of 252,

308 and 198 seconds.

Now, 252 = 2 × 2 × 3 × 3 × 7

308 = 2 × 2 × 7 × 11

198 = 2 × 3 × 3 × 11

\ LCM = 2 × 2 × 3 × 3 × 7 × 11

= 36 × 77 seconds

\( \frac{36*77}{60}\)minutes

= 46 minutes 12 seconds

**Ans . **

(4) 6 hours

Suppose, time taken while
walking be x hours
And, time taken on riding be y
hours

\ According to question

x+y=4\( \frac{1}{2}\)hr

Then, 2y = 3 hours

y=1\( \frac{1}{2}\)hr

x=4\( \frac{1}{2}\) - 1\( \frac{1}{2}\)=3 hr

Time required to walk both ways
= 6 hours

**Ans . **

(4) 9 km

Let the required distance be
x km

\( \frac{x}{\frac{9}{2}}\)+\( \frac{x}{3}\)=5

x{\( \frac{2+3}{9}\)=5

x=9 km

**Ans . **

(4) 26.7 km.

Distance covered by A in 4

hours = 4 × 4 = 16 km

Relative speed of B with respect

to A = 10 – 4 = 6 km/hr

Time taken to catch A

\( \frac{16}{6}\)=\( \frac{8}{3}\)hr

Required distance

=\( \frac{8}{3}*10\)

= 26.67 km.= 26.7 km

**Ans . **

(2) 480 km.

Suppose distance be x km

\( \frac{x}{2*40}\)+\( \frac{x}{2*60}\)=10

\( \frac{3x+2x}{240}\)=10

x=480 km

**Ans . **

(1) 2 min 25 sec.

If A covers the distance of 1
km in x seconds, B covers the
distance of 1 km in (x + 25) sec-
onds. If A covers the distance of
1 km, then in the same time C
covers only 725 metres.

If B covers 1 km in (x + 25) sec-
onds, then C covers 1 km in (x +
55) seconds.

Thus in x seconds, C covers the
distance of 725 m.

\( \frac{x}{725}\)*1000=x+55

x = 145

A co vers the di stance of 1 km in 2 minutes 25 seconds.

**Ans . **

(4) 250 m

**Ans . **

(2) 205 seconds

A beats B by 30 seconds and
B beats C by 15 seconds.
Clearly, A beats C by 45 seconds.
Also, A beats C by 180 metres.
Hence, C covers 180 metres in
45 seconds

Speed of C =\( \frac{180}{45}\)= 4 m/sec

Time taken by C to cover 1000 m
=\( \frac{1000}{4}\)=250 sec

Time taken by A to cover 1000 m
= 250–45 = 205 sec

**Ans . **

(2) 27.

Difference of time

= 6 min. – 5 min. 52 sec.

= 8 seconds

Distance covered by man in 5 min.
52 seconds

= Distance covered by sound in
8 seconds

= 330 × 8 = 2640 m.

\ Speed of man

\( \frac{2640 m}{5 min. 52 sec.}\)=\( \frac{2640}{352}\)

\( \frac{2640}{352}\)*\( \frac{18}{5}\)kmph

= 27 kmph

**Ans . **

(1) 70 km

Let the required distance be x
km.

Difference of time

= 15 + 5 = 20 minutes

=\( \frac{1}{3}\)hr

According to the question,

\( \frac{x}{35}\)-\( \frac{x}{42}\)=\( \frac{1}{3}\)

\( \frac{x}{210}\)=\( \frac{1}{3}\)

x=70 km

**Ans . **

(3) 7 hours 30 minutes

1-\( \frac{5}{6}\) of time taken by B

=1 hour 15 minutes

\ Time taken by B

= 1 hour 15 minutes × 6

=7 hours 30 minutes

**Ans . **

(1) 5

Abhay’s speed = x kmph

Sameer’s speed = y kmph

\( \frac{30}{x}\)-\( \frac{30}{y}\)=12

\( \frac{30}{y}\)-\( \frac{30}{2x}\)=1

On adding,

\( \frac{30}{y}\)-\( \frac{30}{2x}\)=3

\( \frac{30}{2x}\)=3

x = 5 kmph

**Ans . **

(3) 4 hours 45 minutes

Time taken in walking both
ways = 7 hours 45 minutes ....(i)

Time taken in walking one way
and riding back = 6 hours 15
minutes
....(ii)

By equation (ii) × 2 – (i), we have

Time taken by the man to ride
both ways

= 12 hours 30 minutes – 7 hours

45 minutes

= 4 hours 45 minutes

**Ans . **

(1) 25 km/hr.

Let the total distance be 100
km.

Average speed

\( \frac{Total distance covered}{Time taken}\)

\( \frac{100}{ \frac{30}{20}+ \frac{60}{40}+ \frac{10}{10}}\)

\( \frac{100*2}{8}\)=25 kmph

**Ans . **

(2) 6 km/hr

Let the speed of A = x kmph and
that of B = y kmph

According to the question,

x × 6 + y × 6 = 60

x + y = 10

and \( \frac{2}{3}\)x* 5 + 2 y * 5 = 60

10x + 30y = 180

x + 3y = 18
...(ii)

From equations (i) × (3) – (ii)

3x + 3y – x – 3y = 30 – 18

12x = 12

x = 6 kmph.

**Ans . **

(2) 81 km

Let the trains meet after t hours,
then

24t – 18t = 27

6t = 27

t=\( \frac{9}{2}\)hours

QR = 18t = 18 * \( \frac{9}{2}\)= 81 km

**Ans . **

(3) 8 km/hr

Let the speed of Ravi be x kmph

then, Ajay’s speed = (x + 4) kmph

Distance covered by Ajay

= 60 + 12 = 72 km

Distance covered by Ravi

= 60 – 12 = 48 km.

According to the question

\( \frac{72}{x+4}\)=\( \frac{48}{8}\)

\( \frac{3}{x+4}\)=\( \frac{2}{x}\)

3x = 2x + 8

x = 8 kmph

**Ans . **

(2) 16 km

Let man walked for t hours.

then, t × 4 + (9 – t) × 9 = 61

4t + 81 – 9t = 61

81 – 5t = 61

5t = 20

t = 4

Distance travelled on foot

= 4 × 4 = 16 km.

**Ans . **

(1) 6

Let the required distance be x
km, then

\( \frac{x}{5}\)-\( \frac{x}{6}\)=\( \frac{1}{5}\)

\( \frac{x}{30}\)=\( \frac{1}{5}\)
x=6 km

**Ans . **

(4) 6 km.

Let the required distance be
x km.

\( \frac{x}{3}\)-\( \frac{x}{4}\)=\( \frac{30}{60}\)

\( \frac{x}{12}\)=\( \frac{1}{2}\)

x= 6km

**Ans . **

(2) 60 km/hr

Let the speed of train be x
kmph and that of car be y kmph,
then

\( \frac{60}{x}\)+\( \frac{240}{y}\)=4

and ,\( \frac{100}{x}\)+\( \frac{200}{y}\)=\( \frac{25}{6}\)

\( \frac{4}{x}\)+\( \frac{8}{y}\)=\( \frac{1}{6}\)

By equation (i ) – equation (ii) ×
30

\( \frac{60}{x}\)+\( \frac{240}{y}\)-\( \frac{120}{x}\)-\( \frac{240}{y}\)=4-5

-\( \frac{60}{x}\)=-1

x=60 kmph

**Ans . **

(2) 38 minutes.

Ratio of the speed of A and B

= A : B = 2 : 1 = 6 : 3

B : C = 3 : 1

A : B : C = 6 : 3 : 1

Ratio of their time taken

\( \frac{1}{6}\):\( \frac{1}{3}\):1=1:2:6

Time taken by B

\( \frac{2}{6}\)*114 minutes

= 38 minutes

**Ans . **

(3) 54 km/hr

**Ans . **

(2) 800

Total distance of trip

=\( \frac{1200*5}{2}\)= 3000 km

Part of journey covered by train

1-\( \frac{2}{5}\)-\( \frac{1}{3}\)=\( \frac{4}{15}\)

Distance covered by train

3000*\( \frac{4}{15}\)= 800 km

**Ans . **

(1) 1.6 minutes

A’s speed =\( \frac{1000}{5}\)
= 200 m/minute

B’s speed =\( \frac{1000}{8}\)
= 125 m/minute

C’s speed =\( \frac{1000}{10}\)
= 100 m/minute

Distance covered by C in 2 min-
utes = 200 metre

Distance covered by B in 1 minute
= 125 metre

Relative speed of A with respect
to C = 100 metre

Time=\( \frac{200}{100}\)= 2 minutes

Relative speed of A with respect

ot B = 75 metre

Time=\( \frac{125}{75}\)= \( \frac{5}{3}\) minutes
=1.6minutes

**Ans . **

(2) v 1 : v 2 = 4 : 5.

**Ans . **

(1) 1.20 am..

Time taken in covering 999km

\( \frac{999}{55.5}\)= 18 hours

Required time = 18 hours + 1

hour 20 minutes

= 19 hours 20 minutes

i.e. 1 : 20 am

**Ans . **

(1) 12.5 metre/seccond.

Speed = 45 kmph

\( \frac{45*1000}{60*60}\)metre/second

\( \frac{45*5}{18}\)metre/second

= 12.5 metre/second

**Ans . **

(1) 610 m..

Distance covered in 2nd

minute = 90 – 50 = 40 metre

Distance covered in 3rd minute

= 130 – 90 = 40 metre

\ Required distance

= 50 + 40 × 14

= 50 + 560 = 610 metre

**Ans . **

(3) 7: 56 AM

Here distance is constant.

Speed ∝ \( \frac{1}{Time}\)

Ratio of the speeds of A and B

\( \frac{\frac{7}{2}}{4}\)=7:8

A’s speed = 7x kmph (let)

B’s speed = 8x kmph

AB = 7x × 4 = 28x km.

Let both trains cross each other
after t hours from 7 a.m.
According to the question,

7x (t + 2) + 8x × t = 28x

7t + 14 + 8t = 28

15t = 28 – 14 = 14

t=\( \frac{14}{15}\)hours

\( \frac{14}{15}\)*60

=56 minutes.

Required time = 7 : 56 A.M.

**Ans . **

(4) 9 hours

Speed of plane = \( \frac{Distance}{Time}\)

=\( \frac{6000}{8}\)= 750 kmph

New speed = (750 + 250) kmph

= 1000 kmph

Required time =\( \frac{9000}{1000}\)

= 9 hours

**Ans . **

(1) 45

Let speed of train be x kmph.

Speed of car = y kmph.

Case I,

Time = \( \frac{Distance}{Speed}\)

\( \frac{240}{x}\) + \( \frac{210}{y}\)=8\( \frac{2}{3}\)

\( \frac{240}{x}\) + \( \frac{210}{y}\)=8\( \frac{26}{3}\)...(1)

Case II,

\( \frac{180}{x}\)+\( \frac{270}{y}\)=9....(2)

By equation (i) × 3 – (ii) × 4,

\( \frac{720}{x}\) +\( \frac{630}{y}\) -\( \frac{720}{x}\) -\( \frac{1080}{y}\)

= 26 – 36

=\( \frac{-450}{y}\)= –10

= –10

y = 45 kmph

**Ans . **

(3) 81 km/h.

Difference of time = 11 min-
utes 45 seconds – 11 minutes =
45 seconds

Distance covered by sound in 45
seconds = Distance covered by
train in 11 minutes

330 × 45 = 11 × 60 × Speed of train

Speed of train

\( \frac{330*45}{11*60}\)m/sec.

\( \frac{45}{2}\) \( \frac{18}{5}\)

=81 kmph

**Ans . **

(2) 45

Distance covered in 3 hours

36 minutes i.e. 3\( \frac{36}{60}\)hours

=5\( \frac{18}{5}\)= 18 km.

Time taken at 24 kmph.

\( \frac{18}{24}\)hours

\( \frac{18}{24}\)*minutes

=45 minutes

**Ans . **

(3) 2 hours

Let the origi nal speed of
aeroplane be x kmph.

According to the question,

\( \frac{1200}{x-300}\)-\( \frac{1200}{x}\)=2

1200{\( \frac{x-x+300}{x(x-300)}\)}=2

x (x – 300) =\( \frac{1200*300}{2}\)

x (x – 300) = 600 × 300

x (x – 300) = 600 (600 – 300)

x = 600 kmph.

Scheduled duration of flight =\( \frac{1200}{600}\)= 2 hours

**Ans . **

(4) 240.

Consumption of petrol in cov-
ering 540 km=\( \frac{540}{45}\)= 12 litres

Required expenses

= Rs. (12 × 20)

= Rs. 240

**Ans . **

(2) 6 cm

18 km o 1.5 cm

1 km= \( \frac{1.5}{18}\)cm

72=\( \frac{1.5*72}{18}\)cm = 6 cm

**Ans . **

(2) 16 km.

Length of journey on foot
= x km. (let).

\ Length of journey on cycle =
(61 – x ) km.

According to the question,

Time = \( \frac{Distance}{Speed}\)

\( \frac{x}{4}\)+\( \frac{61-x}{9}\)=9

\( \frac{9x+244-4x}{36}\)=9

5x + 244 = 36 × 9 = 324

5x = 324 – 244 = 80

x=16 km

**Ans . **

(1) 16 km.

Let the distance covered on
foot be x km.

Distance covered on cycle =
(61 – x) km.

Time=\( \frac{Distance}{Speed}\)

\( \frac{x}{4}\)+\( \frac{61-x}{9}\)=9

\( \frac{x}{4}\)-\( \frac{x}{9}\)=9-\( \frac{61}{9}\)

\( \frac{5x}{36}\)=\( \frac{20}{9}\)

x=16 km

**Ans . **

(4) 3300 metre

Distance = Speed × Time

= 330 × 10 = 3300 metre

**Ans . **

(2) 800 km.

Let total distance covered be
2x km.

Total time = 14 hours 40 min-
utes

=14\( \frac{40}{60}\)hours=\( \frac{44}{3}\)hours

Time = \( \frac{Distance}{Speed}\)

According to the question,

\( \frac{x}{60}\)\)+\( \frac{x}{50}\)=\( \frac{44}{3}\)

\( \frac{11x}{300}\)=\( \frac{44}{3}\)

x=400

Total distance
= 2x = 2 × 400 = 800 km

**Ans . **

(2) 80

Distance between both don-
keys = 400 metre.

Relative speed = (3 + 2) m./sec.

= 5 m./sec.

Required time

=\( \frac{Distance}{Relative Speed}\)

=\( \frac{400}{5}\)= 80 seconds

**Ans . **

(2) 18 km./hr.

**Ans . **

(2) 5000.

Speed of person = 3 kmph

=\( \frac{3000}{60}\)m./min

= 50 m./min.

Length of the diagonal of square
field

= 50 × 2 = 100 metre

Required area = \( \frac{1}{2}\)*( 100 ) ^{2}

= 5000 sq. metre

**Ans . **

(1) 10 m/sec.

36 km/hr.

36*\( \frac{5}{18}\)

= 10 m/sec.

**Ans . **

(2) 216 kmph

60 metres per sec.

60* \( \frac{18}{5}\)

= 216 km per hr.

**Ans . **

(3) 55 km.

Distance = 20 kms

Time = 2 hours

Speed = \( \frac{Distance}{Time}\)

\( \frac{20}{2}\)=10 km per hr..

Now, we have, Speed = 10 km
per hr

Time=\( \frac{11}{2}\) hour

Distance = Speed × Time

10*\( \frac{11}{2}\)=55 km

**Ans . **

(4) 5 km

Man’s speed =\( \frac{1}{3}\)of the speed of car=\( \frac{1}{3}\)* 60= 20 km per hr..

Time taken to reach office= 15 minutes =\( \frac{15}{60}\)=\( \frac{1}{4}\) hour

Distance between his house and office
= Speed × Time

20*\( \frac{1}{4}\)= 5 km.

**Ans . **

(2) 37.5 hour

Speed = 6 km/hr

Time taken = 5 hours

\ Distance covered

= 6 × 5 = 30 kms

\ Time required to cover 30 kms

at the speed of 8 km/hr.

\( \frac{Distance}{Speed}\)=\( \frac{30}{8}\)=3\( \frac{3}{4}\)=37.5 hour

**Ans . **

(2) 5.16 kmph

Case I.

Distance = 10 kms

Speed = 4 km/hr.

Time taken (t_{ 1 }) =\( \frac{10}{4}\)=\( \frac{5}{2}\)hr.

Case II.

Distance = 21 kms

Speed = 6 km/hr

.
Time taken (t_{ 2 }) =\( \frac{21}{6}\)=\( \frac{7}{2}\)hr.

Total time taken =\( \frac{5}{2}\)+\( \frac{7}{2}\)= 6 hrs.

Total distance covered

= 10 + 21 = 31 kms

\ Average Speed

\( \frac{Total distance}{Total time}\)=\( \frac{31}{6}\)=5\( \frac{1}{6}\)=5.16 km per hr..

**Ans . **

(2) 13.4 kmph.

Let the speed between P and
Q be x km.

Then time taken to cover x km.

P to Q =\( \frac{x}{20}\)

Time taken to cover x km from

Q to P at 10 km per hr. P to Q=\( \frac{x}{10}\)

Total distance covered

= x + x = 2x km.

Time taken to cover 2x km

\( \frac{x}{20}\) + \( \frac{x}{10}\)=\( \frac{3x}{20}\)

Average Speed

\( \frac{2x}{\frac{3x}{20}}\)=\( \frac{2x*20}{3x}\)=\( \frac{40}{3}\)=13.4 km per hr..

**Ans . **

(2) 8.1 kmph

Here, the man covers equal
distance at different speeds. Us-
ing the formula, the Average
Speed is given by

\( \frac{3}{\frac{1}{5}+\frac{1}{10}+\frac{1}{15}}\)=\( \frac{90}{11}\)= 8.1 km per hour..

**Ans . **

(1) 640 kmph.

As distance is covered along
four sides (equal) of a square at
different speeds, the average
speed of the aeroplane

\( \frac{4}{\frac{1}{400}+\frac{1}{600}+\frac{1}{800}+\frac{1}{1200}}\)
\( \frac{48000}{75}\)= 640 km per hr..

**Ans . **

(2) 38.57 kmph.

Length of journey = 150 kms

\( \frac{1}{3}\)rd of journey =\( \frac{150}{3}\)= 50 kms

Remaining \( \frac{2}{3}\) journey
= 150 – 50 = 100 kms

Time taken in \( \frac{1}{3}\)rd journey at 30 km per hr.
t_{1}=\( \frac{5}{3}\) hr

Time taken in \( \frac{2}{3}\) rd journey at 45 km per hr.
t_{2}=\( \frac{20}{9}\) hr

Total time taken in whole jour-
ney = t_{ 1} + t_{ 2}

\( \frac{5}{3}\)+\( \frac{20}{9}\)=\( \frac{35}{9}\)hr

Average Speed
\( \frac{150}{\frac{35}{9}}\)=\( \frac{270}{7}\)=38.57 km per hr..

**Ans . **

(3) 150 km..

Let time taken to reach of
fice at 50 kmph be x hrs

Then time taken to reach office at 60 kmph =x+ \( \frac{30}{60}\)hrs

As, distance covered is same, x* 50 =60 {x+\( \frac{30}{60}\)}

50x = 60x + 30

x = 3 hrs

Hence, distance = 3 × 50

= 150 km

**Ans . **

(4) 5 km.

Let time taken to reach school
at 4 kmph be x hrs.

Then time taken to reach school at 5 kmph =x+ \( \frac{15}{60}\)hr

Since, distance is equal.

4x= 5{x+\( \frac{15}{60}\)}

x=\( \frac{5}{4}\)hr.

Hence, distance between school
& house =4*\( \frac{5}{4}\)km = 5 km

**Ans . **

(1) 25 kmph.

Let the original speed of the
car = x km per hr.

When it is increased by 5 km
per hr, the speed = x + 5 km per
hr.

As per the given information in
the question,

\( \frac{300}{x}\)-\( \frac{300}{x+5}\)=2

\( \frac{1500}{x

x 2 + 5x = 750

x 2 + 5x – 750 = 0

x 2 + 30x – 25x – 750 = 0

x (x + 30) – 25 (x + 30) = 0

(x + 30) (x – 25) = 0

x = – 30 or 25

The negative value of speed is
inadmissible.

Hence, the required speed = 25
km per hr

**Ans . **

(2) 12 kmph.

Time = 10 hours,

Speed = 48 km per hr.

Distance = Speed × Time

= 48 × 10 = 480 km

Now, this distance of 480 kms
is to be covered in 8 hours.
Hence, the required Speed

\( \frac{Distance}{New time}\)=\( \frac{480}{8}\)

= 60 km per hr.Increase in speed = 60 – 48 = 12 km per hr.

**Ans . **

(3) 4 km

Let the distance be x kms.

Time taken at 4 km per hr. t _{1}=\( \frac{x}{4}\)hr

Time taken at 3 km per hr. t _{2}=\( \frac{x}{3}\)hr

Difference in timings
= 10 + 10 = 20 minutes

or \( \frac{20}{60}\)=\( \frac{1}{3}\)hour

\( \frac{x}{3}\)-\( \frac{x}{4}\)=\( \frac{1}{3}\)

\( \frac{x}{12}\)=\( \frac{1}{3}\)

x = 4 km.

Hence the required distance
= 4 kms.

**Ans . **

(4) 8 kmph

Let the speed of Rickshaw be
'x ' .

Then, time taken to cover 16 km
on foot and 24 km on Rikshaw =\( \frac{16}{4}\)+\( \frac{24}{x}\)hr

and time taken to travel 24 km

on foot & 16 km on Rikshaw=\( \frac{16}{x}\)+\( \frac{24}{4}\)hr

According to question,

\( \frac{16}{4}\)+\( \frac{24}{x}\)+1=\( \frac{16}{x}\)+\( \frac{24}{4}\)

\( \frac{24-16}{x}\)=1

x = 8 km/hr

**Ans . **

(1) 30 minutes

Since I walk at \( \frac{3}{4}\)of my usual speed the time taken is \( \frac{4}{3}\)
of my usual time.

\( \frac{4}{3}\)of usual time

= Usual time + Time I reach late

\( \frac{1}{3}\)of usual time

= 10 minutes

Usual time
= 10 × 3 = 30 minutes.

**Ans . **

(2) 50 minutes

\( \frac{5}{3}\)of usual speed means\( \frac{3}{5}\)
of usual time as he reaches
earlier.

\( \frac{3}{5}\)usual time + 20 minutes=Usual time

20 minutes= 1-\( \frac{3}{5}\)usual time

=\( \frac{2}{5}\) usual time

Usual time
\( \frac{20*5}{2}\)=50 minutes

**Ans . **

(2) 7.5 hours

New speed is\( \frac{3}{4}\)of the usual speed

New time taken =\( \frac{4}{3}\)of the usual time

\( \frac{4}{3}\)of the usual time – Usual time =\( \frac{5}{2}\)

\( \frac{1}{3}\)of the usual time =\( \frac{5}{2}\)

Usual time =\( \frac{5}{2}\)*3

=\( \frac{15}{2}\)hours or 7.5 hrs

**Ans . **

(3) 35 km.

When B meets A at R,

by
then B has walked a distance (XY
+ YR) and A,the distance XR.

That is both of them have togeth-
er walked twice the distance
from X to Y, i.e., 42 kms.

**Ans . **

(4) 18 km

**Ans . **

(1) 36 km

Let the total distance travelled
be x kms.

Case I :
Speed for the first one-third distance \( \frac{x}{3}\)kms =10 km per hr.

Time taken =\( \frac{x}{30}\)hours

Similarly, time taken for the next
one-third distance=\( \frac{x}{27}\)hour

and time taken for the last one third distance
=\( \frac{x}{24}\)hour

Total time taken to cover x kms
\( \frac{x}{30}\)+\( \frac{x}{27}\)+\( \frac{x}{24}\)hour
Case II :

Time taken for one-half distance
at the speed of 10 km per hr.
\( \frac{x}{20}\)hr

and time taken for remaining \( \frac{1}{2}\) of distance \( \frac{x}{16}\)hrs. at 8 km per hr.

Total time taken
\( \frac{x}{20}\)+\( \frac{x}{16}\)hr

Time taken in (Case II – Case I)

1 minute=\( \frac{1}{60}\)hr

According to the question

\( \frac{x}{20}\)+\( \frac{x}{16}\)-\( \frac{x}{30}\)+\( \frac{x}{27}\)+\( \frac{x}{24}\)hour=\( \frac{1}{60}\)

\( \frac{x}{2160}\)=\( \frac{1}{60}\)

x=36 km

Hence the required distance
= 36 km.

**Ans . **

(3) 8 hours

hours at 4 km per hr. and y
hours at 5 km per hr. and cov-
ers a distance of 35 kms.

Distance = 4x + 5y = 35 ...(i)

Now, he walks at 5 km per hr.

for x hours and at 4 km per hr.

for y hours and covers a distance
(35 + 2) = 37 kms

Distance = 5x + 4y = 37...(ii)

By 5 × (i) – 4 × (ii) we have

20x + 25y = 175

20x + 16y = 148

By solving these equations, y=3

Putting the value of (y) in equa-
tion (i), we have

4x + 5 × 3 = 35

4x = 35 – 15 = 20

x = 5

Total time taken

= x + y = 5 + 3 = 8 hours.

**Ans . **

(4) 100 km.

Obviously,\( \frac{4}{5}\)of total time in train = 2 hour

Total time in train=\( \frac{5}{4}\)*2=\( \frac{5}{2}\)hr

Total time to cover 400 km is 4
hours

\ Time spent in travelling by air= 4-\( \frac{5}{2}\)=\( \frac{3}{2}\)hr

If 400 kms is travelled by air,
then time taken = 2 hours

\ In 2 hours, distance covered
by air = 400 kms

In \( \frac{3}{2}\)hr distance covered \( \frac{400}{2}\)*\( \frac{3}{2}\)
= 300 kms

Distance covered by the train
= 400 – 300 = 100 kms.

**Ans . **

(1) 40 kmph

Let the original speed be x
km/hr
then, increased speed

= (x + 10) km/hr

According to question,

\( \frac{100}{x}\)-\( \frac{100}{x+10}\)=\( \frac{30}{60}\)

100[\( \frac{1}{x}\)-\( \frac{1}{x+10}\)]=\( \frac{1}{2}\)

10 × 200 = x (x + 10)

x 2 + 10x – 2000 = 0

x 2 + 50x – 40x – 2000 = 0

x (x + 50) – 40 (x + 50) = 0

x = – 50, 40

Speed can’t be negative.

Hence, Original speed = 40 kmph

**Ans . **

(2) 100 days.

Working hours per day= 24
– 9 = 15 hrs.

Total working hours for 40 days
= 15 × 40 = 600 hrs.

On doubling the distance, the
time required becomes twice but
on walking twice as fast, the time
required gets halved. Therefore,
the two together cancel each
other with respect to time re-
quired. Increasing rest to twice
reduces walking hours per day
to
24 – (2 × 9) = 6 hrs.

\ Total number of days required
to cover twice the distance, at
twice speed with twice the rest.
\( \frac{600}{6}\)=100 days

**Ans . **

(3)16.58 minutes.

In 1 minute the monkey
climbs 12 metres but then he
takes 1 minute to slip down 5
metres. So, at the end of 2 min-
utes the net ascending of the
monkey is 12 – 5 = 7 metres.

So, to cover 63 metres the above
process is repeated \( \frac{63}{7}\)=9
times. Obviously, in 9 such hap-
penings the monkey will slip 8
times, because on 9th time, it
will climb to the top.

Thus, in climbing 8 times and
slipping 8 times, he covers 8 ×
7 = 56 metres.

Time taken to cover 56 metres
\( \frac{56*2}{7}\)= 16 minutes
= 16 minutes

\( \frac{7}{12}\)minutes

Total time taken = 16 + \( \frac{7}{12}\)
=16.58 minutes

**Ans . **

(4) 1100 metres

**Ans . **

(1) 210 leaps

Grey hound and hare make
3 leaps and 4 leaps respective-
ly.

This happens at the same time.
The hare goes 1.75 metres in 1
leap.

Distance covered by hare in 4 leaps = 4 × 1.75 = 7 metres
The grey hound goes 2.75 metres
in one leap

Distance covered by it in 3 leaps
= 3 × 2.75 = 8.25 metres

Distance gained by grey hound in 3 leaps=(825-7)
= 1.25 metres

Distance covered by hare in 50
leaps = 50 × 1.75 metres
= 87.5 metres

Now, 1.25 metres is gained by
grey hound in 3 leaps
87.5 metres is gained in
\( \frac{3}{1.25}\)*87.5

= 210 leaps.

**Ans . **

(2) 1 hour.

Let the original speed be x
kmph
then,

new speed = (x – 200) kmph

According to question,

Time taken with new speed –
time taken with original speed =30 min. i.e \( \frac{1}{2}\)

\( \frac{600}{x-200}\)-\( \frac{600}{x}\)=\( \frac{1}{2}\)

\( \frac{x-x+200}{x(x-200)}\)=\( \frac{1}{1200}\)

24000 = x (x – 200)

x 2 – 200x – 24000 = 0

x 2 – 600x + 400x – 24000
= 0

x (x – 600) + 400 (x – 600) = 0

(x – 600) (x + 400) = 0

x = 600, – 400

Speed cannot be negative
Hence, original speed = 600
kmph and duration of flight
=1 hour

**Ans . **

(2) 15 kmph.

Let the speed of the second
train be x km per hr. Then the
speed of the first train is x + 5
km per hr.

Let O be the position of the rail-
way station from which the two
trains leave. Distance travelled
by the first train in 2 hours =
OA = 2 (x + 5) km.

Distance travelled by the 2nd
train in 2 hours= OB = 2x km.

By Pythagoras theorem, AB ^{2} =
OA^{2}+ OB^{ 2}

50 ^{2} = [2 (x + 5)]^{ 2} + [2x] ^{2}

2500 = 4 (x + 5) ^{2 }+ 4x ^{2}

2500 = 4 (x ^{2} + 10x + 25) +
4x ^{2}

8x^{ 2} + 40x – 2400 = 0

x ^{ 2} + 5x – 300 = 0

x ^{ 2} + 20x – 15x – 300 = 0

x (x + 20) – 15 (x + 20) = 0

(x – 15) (x + 20) = 0

x = 15, – 20

But x cannot be negative

x = 15

The speed of the second train is 15 km per hr. and the speed
of the first train is 20 km per hr.

**Ans . **

(4) 9 kmph

The distance covered by man
in 4 minutes \( \frac{6*1000*4}{60}\)= 400 metres

The distance covered by carriage
in 4 minutes
= 200 + 400 = 600 metres

Speed of carriage Speed of carriage=\( \frac{600}{4}\)*\( \frac{60}{1000}\)
= 9 km per hr.

**Ans . **

(1) 34 kmph.

If the car were not moving,
the person would have heard the
two sounds at an interval of 12
minutes.

Therefore, the distance
travelled by car in 11 minutes
40 seconds is equal to the dis-
tance that could have been cov-
ered by sound in 12 min – 11
min. 40 seconds = 20 seconds.
Distance covered by sound in 20
seconds

= 330 × 20 = 6600 m

In 11 min 40 seconds
or 700 seconds the car travels
6600 m.

In 1 second the car will travel

\( \frac{6600}{700}\)=\( \frac{66}{7}\)metre

Speed of the car = \( \frac{66}{7}\)metre per second

\( \frac{66}{7}\)*\( \frac{18}{5}\)

=34 kmph

**Ans . **

(2) 40 km.

When A and B cross each other
at M for the first time, they have
together covered the whole dis-
tance PQ = 180 km.

When they meet again at N, they
have together covered total dis-
tance equal to 3 times of PQ = 3
× 180 = 540 km.

PM=\( \frac{5}{5+4}\)*180= 100 km

QP + PN =\( \frac{4}{5+4}\)*540

= 240 km

or PN = 240 – QP = 240 – 180
= 60 km.

Then, MN = PM – PN
= 100 – 60 = 40 km.

**Ans . **

(2) 6.6 kmph

Distance covered by man in
3 minutes

[\( \frac{4*1000}{60}\)]\( \frac{m}{minutes}\)*3 minutes
=200 metres

Total distance covered by the car
in 3 min.

= (200 + 130) m = 330 metres

Speed of the car \( \frac{330}{3}\)
= 110 m per minutes

\( \frac{\frac{110}{1000}}{\frac{1}{60}}\)=6.6 kmph

**Ans . **

(3) 25 kmph

Suppose that Ram and Mohan
meet at A. Let Ram’s speed be x
km per hr.

and Mohan’s speed
be y km per hr. Then AP=\( \frac{25}{4}\)x

km and AB = 4y km.

Now, time taken by Ram in going from B to A =\( \frac{4y}{x}\)

and the time taken by Mohan in going from P to A =\( \frac{25x}{4y}\)

Obviously time taken is equal

\( \frac{4y}{x}\)=\( \frac{25x}{4y}\)

16y 2 = 25x 2

\( \frac{y}{x}\)=\( \frac{5}{4}\)

y=\( \frac{5}{4}\)x

Here, x = 20 km per hr.

y = Mohan’s speed

\( \frac{5}{4}\)*20=25 km per hr..

**Ans . **

(4) 120 km ; 30 kmph

Let the original speed be x
and distance be y

Case I.

Time taken by train to travel 30 km= \( \frac{30}{x}\)

Time taken by train after acci-
dent=\( \frac{y-30}{\frac{4}{5}x}\)

Total time taken =\( \frac{30}{x}\)+\( \frac{y-30}{\frac{4}{5}x}\)

Case II :

Time taken by train to travel 48 km =\( \frac{48}{x}\)

Time taken by train after accident =\( \frac{y-48}{\frac{4}{5}x}\)

Total time taken =\( \frac{48}{x}\)+\( \frac{y-48}{\frac{4}{5}x}\)

According to question,

\( \frac{30}{x}\)+\( \frac{y-30}{\frac{4}{5}x}\) - \( \frac{48}{x}\)+\( \frac{y-48}{\frac{4}{5}x}\)=\( \frac{9}{60}\)

\( \frac{90-72}{4x}\)=\( \frac{9}{60}\)

x=30

Hence, original speed = 30 kmph

Also \( \frac{30}{x}\)+\( \frac{y-30}{\frac{4}{5}x}\) =\( \frac{y}{x}\)=\( \frac{45}{60}\)

3x – y = –30

3(30) – y = –30

y = 120 km

i.e. Distance = 120 km

**Ans . **

(2) 1200 km ; 100 kmph

Let A be the starting point, B
the terminus. C and D are points
where accidents take place.

0.75=\( \frac{3}{4}\)

By travelling at \( \frac{3}{4}\) of its original speed,

the train would take \( \frac{4}{3}\)of its usual time i.e.,\( \frac{1}{3}\)more of the usual time.

\( \frac{1}{3}\)of the usual time taken to travel the distance CB.
= 4 – 1 = 3 hrs

and \( \frac{1}{3}\)of the usual time taken to travel the distance

DB=3\( \frac{1}{2}\)-1=2\( \frac{1}{2}\)

Subtracting equation (ii) from (i)

we can write,\( \frac{1}{3}\)of the usual time taken to travel the distance

CD=3-2\( \frac{1}{2}\)=\( \frac{1}{2}\)hr

Usual time taken to travel CD=\( \frac{\frac{1}{2}}{\frac{1}{2}}\)=\( \frac{3}{2}\)

Usual sp eed of the train =\( \frac{150}{\frac{3}{2}}\)= 100 km per hr.

Usual time taken to travel CB =\( \frac{3}{\frac{1}{3}}\)= 9 hrs.

Total time = 3 + 9 = 12 hrs.

Length of the trip = 12 ×
100 = 1200 km.

**Ans . **

(3) 80 kmph; 460 km

Let P be the starting point, Q the
terminus, M and N the places
where accidents occur.

At \( \frac{3}{4}\)th of the original speed,

the train will take \( \frac{4}{3}\)of its usual time to cover the same distance

i.e.,\( \frac{1}{3}\)rd more than the usual time.

\( \frac{1}{3}\)rd of the usual time to travel a distance of 60 kms between MN
= 15 min.

Usual time to travel 60 kms

= 15 × 3 = 45 min. =\( \frac{3}{4}\)hr

Usual speed of the train per hour = 60 *\( \frac{4}{3}\)= 80 km per hr..

Usual time taken to travel MQ=90 × 3

= 270 minor \( \frac{9}{2}\)

The distance MQ 80*\( \frac{9}{2}\)= 360 km.

Therefore, the total distance PQ

= PM + MQ

= 100 + 360 = 460 kms.

**Ans . **

(4) 10 : 00 a.m.

Let they meet x hrs after 7
am.

Di stance covered by A i n x
hours = 20x km

Distance covered by B in (x –1) hr.

= 25 (x – 1) km

20x + 25 (x – 1) = 110

20x + 25x – 25 = 110

45x = 110 + 25 = 135

x = 3

Trains meet at 10 a.m.

**Ans . **

(1) 467 th line.

Writing ratio = 200 : 150 = 4
: 3
In a given time first boy will be
writing the line number \( \frac{4}{7}\)*817
\( \frac{3268}{7}\)th line

=466\( \frac{6}{7}\)

Hence, both of them shall meet
on 467th line

**Ans . **

(2) 9 hours

Let the two men meet after t
hours.

Distance covered by the first man
starting from A = 4 t km.

Distance covered by the second
man starting from B

= 2 + 2 . 5 + 3 + ..... +[2+\( \frac{t-1}{2}\)]

This is an arithmetic series of t terms with \( \frac{1}{2}\)as common difference

By applying formula S=\( \frac{n}{2}\)[2 a + (n – 1) d]

Where, n = no. of terms

a = first term

d = common difference

We have its sum =\( \frac{t}{2}\)[2*2+(t-1)*\( \frac{1}{2}\)]

2t+\( \frac{t^{2}-t}{4}\)

Total distance covered by two men =4 t + 2 t +\( \frac{t^{2}-t}{4}\) =72

or 24t+t^{2}-t=288

or t 2 – 9t + 32t – 288 = 0

or t (t–9) + 32 (t – 9) = 0

or (t – 9) (t + 32) = 0

Either t – 9 = 0

t = 9,-32

Time cannot be negative. Hence,
the two men will meet after 9 hrs.

**Ans . **

(3) 40 m ; 20 m/sec

Let the length of the train be
x metres

Then, the time taken by the train to cover (x + 50) metres is 4\( \frac{1}{2}\)sec

Speed of the train \( \frac{x+50}{\frac{9}{2}}\)m/s

Again, the time taken by the train
to cover x metres in 2 seconds.

Speed of the train = \( \frac{x}{2}\)metre
per second
..(ii)

From equations (i) and (ii), we
have

\( \frac{}{}\)=\( \frac{x}{2}\)

4x + 200 = 9x

5x = 200

x = 40

Length of the train

= 40 metre

Speed of the train=\( \frac{40}{2}\)=20 m/sec

**Ans . **

(4) 17.5 kmph

Bo th trains meet after 6
hours.

The relative speed of two trains =\( \frac{162}{6}\)== 27 km per hr..

The speed of the slower train
starting from B =\( \frac{19}{2}\) km per hr..

The speed of the faster train =w\( \frac{35}{2}\)= 17.5 km per hr..

**Ans . **

(1) 25 metres

Let the length of train be x
metres and the length of platform
be y metres.

Speed of the train
25*\( \frac{5}{18}\)=\( \frac{125}{18}\)

Time taken by train to pass the
platform

x+y *\( \frac{18}{125}\)
or, x + y = 125
...(i)

Speed of train relative to man
= (25 + 5) km per hr.

30*\( \frac{5}{18}\)=\( \frac{25}{3}\)

Time taken by the train to pass
the man

x*\( \frac{3}{25}\)

x=100 metres

length of the platform = 25
metres

**Ans . **

(2) 108 kmph ; 72 kmph

Let the speed of the train be
x metre per sec. and y metre per
sec. respectively.

Sum of the length of the trains =
200 + 175 = 375 metres

Case : I

When the trains are moving in
opposite directions
Relative speed = (x + y) m per sec.

In this case the time taken by
the trains to cross each other
=\( \frac{375}{x+y}\)

\( \frac{375}{x+y}\)=\( \frac{15}{2}\)

x + y = 50

Case : II

When the trains are moving in
the same direction.

Relative speed = (x – y) m per sec.

In this case, the time taken by
the trains to cross each other
\( \frac{375}{x-y}\)=\( \frac{75}{2}\)

x – y = 10

Now, x + y = 50

x – y = 10

x=30

Putting this value in equation (i),
we have

y = 50 – 30 = 20

\ Speed of trains = 30 m per sec.

=30*\( \frac{18}{5}\) = 108 km per hr.

and 20 m per sec. = 20 * \( \frac{18}{5}\)

= 72 km per hr.

**Ans . **

(3) 4.5 km.

Trains are running in oppo-
site direction Relative speed of the two trains
= 90 + 60 = 150 km per hr.

Distance travelled in 4\( \frac{1}{2}\)seconds onds with speed of 150 km per hr

=150*\( \frac{5}{18}\)=150*\( \frac{5}{18}\)*( \frac{9}{2}\)=\( \frac{375}{2}\)

Let the length of the first train
be x metres.

Then the length of the second train be \( \frac{x}{2}\)

\( \frac{3x}{2}\)=\( \frac{375}{2}\)

3x = 375

x = 125 metres

Hence, the length of the first
train = 125 metres

Speed of the first train = 60 km
per hr.

60*\( \frac{5}{18}\)=\( \frac{50}{3}\)

Time taken by the first train to
cross the tunnel = 4 minutes
and 37\( \frac{1}{2}\)sec

240+\( \frac{75}{2}\)sec=\( \frac{555}{2}\)sec

Speed of first train =\( \frac{50}{3}\)

Distance covered by it in \( \frac{555}{2}\) sec

=\( \frac{50}{3}\)*\( \frac{555}{2}\)

= 4625 metres

Hence, length of tunnel
= 4625 – 125 = 4500 metres

= 4.5 km

**Ans . **

(4) 50 metres

Let the length of the train be
x km and its speed y km per hr.

Case I : When it passes the man
walking at 2 km per hr. in the
same direction

Relative speed of train
= (y – 2) km per hr.

\( \frac{x}{y-2}\)=9 sec

=\( \frac{1}{400}\)hr

Case II : When the train crosses
the man walking at 4 km per hr.
in the same direction.

Relative speed of train= (y – 4)
km per hr.

\( \frac{x}{y-4}\)=10 sec

\( \frac{x}{y-4}\)=\( \frac{1}{360}\)hr

On dividing equation (i) by (ii),
we have

\( \frac{y-4}{y-2}\)=\( \frac{\frac{1}{400}}{\frac{1}{360}}\) =\( \frac{360}{400}\) =\( \frac{9}{10}\)

10y – 40 = 9y – 18

10y – 9y = 40 – 18

y = 22 km per hr.

\ From equaton (i), we have

\( \frac{x}{22-2}\)=\( \frac{1}{400}\)

x=50 metres

**Ans . **

(1) 50.4 kmph

Let the length of the train be
x metres

Then, in 18 sec. the train trav-
els (x + 162) metres
...(i)

and in 15 sec. the train travels
(x + 120) metres

In (18 – 15) = 3 sec. the train travels (x + 162)
– (x + 120) = 42m.

In 1 sec the train travels =14 metres

In 18 sec. the train travels

= 14 × 18 = 252 metres ...(iii)

From equations (i) and (iii)

\ x + 162 = 252

Þ x = 252 – 162 = 90

\ Length of the train = 90
metres

Also, from equation (ii) we see
that in 1hr. the train travels
= 14 × 60 × 60 metres

\( \frac{14*60*60}{1000}\)=50.4

The speed of the train
= 50.4 km per hr.

**Ans . **

(2) 20 m/sec.

Let the length of trains be x
m and (x + 50)m and the speed
of other train be y m per sec.

The speed of the first train
= 90 km per hr.

90*\( \frac{5}{18}\)= 25 m per sec.

Case I : Opposite direction,
Their relative speed
= (y + 25)m per sec.

Distance covered = x + x + 50

= 2x + 50 metres

Time taken= \( \frac{2 x + 50}{y + 25}\)=10

2x + 50 = 10y + 250 ...(i)

Case II. Direction is Same
Their relative speed
= (25 – y) m per sec.

Distance covered = x + x + 50

= 2x + 50m

Time taken= \( \frac{2 x + 50}{25-y}\)=90

2x + 50 = 90 (25 – y)

From equations (i) and (ii)

10y + 250 = 2250 – 90y

10y + 90y = 2250 – 250

y=20

Putting y = 20 in equation (i), we
have
2x + 50= 10 × 20 + 250 = 450

x=200

x + 50 = 200 + 50

= 250 metres.

Hence,
The length of the 1st train = 200
metres.

The length of the 2nd train
= 250 metres.

The speed of the 2nd train
= 20 m per sec

**Ans . **

(1)12.59 m/sec.

Let the length of the train be
x m and its speed y m/sec.
Distance covered in crossing the
platform
= 170 + x metres

and time taken = 21 seconds
Speed y= \( \frac{170+x}{21}\)

Distance covered to cross the
man = x metres
and time
taken = \( \frac{15}{2}\)sec

Speed y=\( \frac{2x}{15}\)

From equations (i) and (ii),

\( \frac{170+x}{21}\)=\( \frac{2x}{15}\)

2550 + 15x = 42x

Þ 42x – 15x = 2550

Þ 27x = 2550

x=94.44m/sec

and y=\( \frac{340}{27}\)

y=12.59m/sec

**Ans . **

(2) 4 hours 21.6 sec..

The goods train leaves Delhi
at 6 am and mail train at 12
noon, hence after 6 hours

The distance covered by the
goods train in 6 hours at 32 km
per hr. = 32 * 6 = 192 kms

The relative velocity of mail train
with respect to goods train = 80
– 32 = 48 km per hr.

To completely cross the goods
train, the mail train will have to
cover a distance

= 192 km + 158m + 130m

= 192km + 0.158 km + 0.130 km

= 192.288 km more

Since, the mail train goes 48 kms
more in 1 hour.

\ The mail train goes 192.288
kms more in
\( \frac{192288}{1000}\)*\( \frac{1}{48}\)=\( \frac{2003}{500}\)
= 4 hours 21.6 sec.

**Ans . **

(3) 9 kmph .

Let the speed of the motor-
boat in still water be Z km per
hr.

Downstream speed= (Z + 3) km
per hr.
Upstream speed
= (Z – 3) km per hr.

Total journey time

= 30 minutes =\( \frac{1}{2}\)hr

We can write,

\( \frac{2}{z-3}\)+\( \frac{2}{z+3}\)=\( \frac{1}{2}\)

Z ^{2}– 9 = 8Z

Z ^{2} – 8Z – 9 = 0

Z^{ 2} + Z – 9Z – 9 = 0

Z(Z + 1) – 9 (Z + 1) = 0

(Z + 1) (Z – 9) = 0

Z = – 1 or 9.

Since speed can’t be negative
Therefore, the speed of the mo-
tor-boat in still water = 9 km per
hr.

**Ans . **

(4) 2 kmph

Let the upstream speed be x
km per hr. and downstream speed
be y km per hr.

Then, we can write,

\( \frac{32}{x}\)+\( \frac{60}{y}\)=9and

\( \frac{40}{x}\)+\( \frac{84}{y}\)=12

Let \( \frac{1}{x}\)=m and \( \frac{1}{y}\)=n

The above two equations can now
be written as

32 m + 60 n = 9
...(i)

and, 40 m + 84 n = 12
...(ii)

7 × (i) – 5 × (ii) gives 24 m = 3

or x=8
4 × (ii) – 5 × (i) gives 36 n = 3

y=12km per hr..

Rate of current
\( \frac{y-x}{2}\)

= 2 km. per hr..

**Ans . **

(1) 2 kmph

Let the speed of boat and riv-
er be x km per hr. and y km per
hr. respectively. Then,

The speed of boatman down-
stream = (x + y) km per hr.

and the speed of boatman up-
stream = (x – y) km per hr.

Time taken by boatman in going
21 km downstream=\( \frac{21}{x+y}\)hours

Time taken by boatman in going 21 km upstream =\( \frac{21}{x-y}\)hours

According to the question,

\( \frac{21}{x+y}\)+\( \frac{21}{x-y}\)=10

Now, time taken for 7 kms downstream =\( \frac{7}{x+y}\)

and time taken for 3 kms upstream =\( \frac{3}{x-y}\)

\( \frac{7}{x+y}\)-\( \frac{3}{x-y}\)=0

Therefore,x + y = 7 and x – y = 3
On adding (iii) and (iv), we have

2x = 10

Þ x = 5

\ y = 7 – x = 7 – 5 = 2

\ Speed of river = 2 km per hr.

**Ans . **

(2) 12 kmph

Let the speed of the cyclist
be x km per hr.

Speed of the motorist= (x + 15)
km per hr.

Time taken by the motorist to cover half of the distance=\( \frac{9}{x+15}\)hr

After covering 9 kms, the speed
of motorist gets reduced by 20%

New speed = x + 15*\( \frac{80}{100}\)=\( \frac{4(x+15)}{5}\)

Time taken by the motorist to
cover the remaining half distance=\( \frac{45}{4(x+15)}\)

Total time taken by the motorist=\( \frac{9}{x+15}\)+\( \frac{1}{2}\)+\( \frac{45}{4*x+15}\)

Total time taken by the cyclist=\( \frac{18}{x}\)

Motorist reaches 15 minutes, i.e.,\( \frac{1}{4}\)

\( \frac{18}{x}\)-\( \frac{9}{x+15}\)-\( \frac{1}{2}\)-\( \frac{45}{4*x+15}\)=\( \frac{1}{4}\)

72x + 1080 – 36x – 2x 2 –

30x – 45x = x 2 + 15x

Þ 3x 2 + 54x – 1080 = 0

Þ x 2 + 18x – 360 = 0

Þ x 2 + 30x – 12x – 360 = 0

Þ x (x + 30) – 12 (x + 30) = 0

Þ (x + 30) (x – 12) = 0

Þ x = – 30, 12

The speed cannot be negative.

\ The speed of the cyclist = 12 km
per hr.

**Ans . **

(3) 1520 km.

Total distance travelled
= 3990 km

Distance = Time × Speed

Ratio of time spent = 1 : 16 : 2

Ratio of speed = 20 : 1 : 3

Ratio of time × speed

= 20 × 1 : 16 × 1 : 2 × 3

= 20 : 16 : 6

Sum of the ratios

= 20 + 16 + 6 = 42

Distance covered by sea=\( \frac{3990}{42}\)*16=1520 kms

**Ans . **

(4) 28 km.

Relative speed of insect
= 30 + 42 = 72 km per hr.

Distance between railway engine
and insect = 20 km.

Engine and insect will meet for the first time after =\( \frac{20}{72}\)hr.

Distance covered in this period \( \frac{20}{72}\)* 42 =\( \frac{35}{3}\) km returning to A.

The distance covered by engine in this period=\( \frac{20}{72}\)* 30 =\( \frac{25}{3}\)

Remaining distance between

A and engine 20- \( \frac{25}{3}\)+\( \frac{25}{3}\)=\( \frac{10}{3}\)

Again, engine and insect will meet after =\( \frac{5}{108}\)hr

The distance covered by the in-
sect in this period \( \frac{5}{108}\)*42 =\( \frac{35}{18}\)

and again the insect will cover \( \frac{35}{18}\) km in returning.

Total distance covered by the insect =\( \frac{70}{3}\)+\( \frac{70}{18}\) +....

[\( \frac{35}{3}\)+\frac{35}{3}\)=\frac{70}{3}\)and \frac{35}{18}\)+\frac{35}{18}\)=\frac{70}{18}\)....]

=\( \frac{70}{3}\)[1+\( \frac{1}{6}\)........]

It is a Geometric Progression to infinity with common ratio \( \frac{1}{6}\)

=\( \frac{70}{3}\) *\( \frac{1}{\frac{5}{6}}\)=28 km

**Ans . **

(1) 12:48 pm..

Let P be at equal distance from
Q and R after t hours.

(87.5 – 33) + 5t

t= 1 hr 18 minutes

11.30 am + 1 hr. 18 min.
= 12.48 pm

At 12.48 pm, P would have cov-
ered a distance

= (12.48 pm – 8 am) × 25

= 120 km

Therefore, P will be at equal dis-
tance from Q and R at 12.48 pm

**Ans . **

(2) 40 km.

Let the original speed of the
person be x km/hr. and the dis-
tance be y km.

Case 1:

\( \frac{y}{x}\)-\( \frac{y}{x+3}\)=40 minutes

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

or, 2 x (x + 3) = 9y...(i)

Case II :

\( \frac{2y}{x(x-2)}\)=\( \frac{2}{3}\)
or, x (x – 2) = 3y... (ii)

On dividing equation (i) by (ii) we
have,

\( \frac{2(x+3)}{x-2}\)=\( \frac{2}{3}\)
or, x = 12 km/hr.

Original speed of the person = 12 km/hr.

Putting the value of x in equa-
tion (ii)

12 (12 – 2) = 3y

or, 3y = 12 × 10

y=40

The required distance =40 km

**Ans . **

(3) 80 km.

Let the speed of steamer in
still water = x kmph
\ Rate downstream
= (x + 2) kmph

Rate upstream = (x – 2) kmph
Obviously, distance covered
downstream and upstream are
equal

Þ 4 (x + 2) = 5 (x – 2)

4x + 8 = 5x – 10

Þ 5x – 4x = 10 + 8 Þ x = 18

\ Rate downstream

= 18 + 2 = 20 kmph

Therefore, the required distance

= Speed downstream × Time

= 20 × 4 = 80 km.

**Ans . **

(2) 145 metre.

According to the question,
when A covers the distance of
200 metres, B covers only 200–
20 = 180 metres

Again, in 100 metre race, B beats
C by 5 metres.

Hence, if B runs 100 metres, C
runs 100–5 = 95 metres
Q If B runs 100 m, C runs
= 95 m

If B runs 180 m, C runs \( \frac{95*180}{100}\)= 171 m

A : B : C = 200 : 180 : 171

Hence, A will beat C by = 200–171 = 29 m in 200 m race.

i.e., 29 × 5 = 145 m in 1 km
race.

**Ans . **

(3) 35 kmph

Case I : When the cars are
moving in the same direction.

Let A and B be two places and C
be the place of meeting.

Let the speed of car starting from
A be x kmph, and that of car
starting from B be y kmph.

Relative speed = (x – y) kmph

According to the question.

(x – y) × 8 = 80

x – y = 10

Case II : When the cars are mov-
ing in the opposite directions and
they meet at point C.

Relative speed = (x + y) kmph

Time taken = 1 hour 20 minutes

1+\( \frac{1}{3}\)=\( \frac{4}{3}\) hr

x+y *\( \frac{4}{3}\)=80

x + y = 60

Adding equations (i) and (ii),

2x = 70

x = 35 Þ From equation (ii),

x + y = 60

35 + y = 60

Þ y = 60 – 35 = 25

\ Speed of the faster car
= 35 kmph

**Ans . **

(4) 8.3 m/sec.

Let B take x seconds to run
1000 m.

\ Time taken by C

= (x + 15) seconds

\( \frac{x}{x+15}\)=\( \frac{9}{10}\)

10x = 9x + 135

Þ x = 135 seconds

Now in a one kilometre race, A
beats B by 15 seconds.

It means A covers 1000 m in 135 – 15 = 120 seconds

Speed of A =\( \frac{1000}{120}\)=8.3 m/sec.

**Ans . **

(1) 3510 metre.

Trains are running in oppo-
site directions.

Relative speed = 72 + 90
= 162
kmph

=162*\( \frac{5}{18}\)=45

Let the length of the first train
be = x metre.

\ Length of the second train \( \frac{3}{4}\)x
distance travelled in 3 \( \frac{1}{2}\)onds at 45 m/sec
=\( \frac{315}{2}\)

This distance is equal to sum of
the lengths of trains.

x+\( \frac{3x}{4}\)=\( \frac{315}{2}\)

x=90

Hence, the length of the first
train = 90 metre.

Speed of first train = 72 kmph

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

Time taken by the first train to
cross the tunnel

= 3 minutes = 180 seconds

\ Distance covered by it in 180
seconds

= 180 × 20 = 3600 metre

\ Length of (first train + tunnel)
= 3600 metre

\ Length of tunnel
= 3600 – 90 = 3510 metre