**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