**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.