**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