- TIME AND DISTANCE

# TYPE-I

Ans .

(2) 64 sec.

Explanation :

Time taken=$$\frac{Distance}{Time} = \frac{ \frac{4}{5}}{45} hour =\frac{4*60*60}{5*45} sec. = 64sec.$$

Ans .

(4) 720 km./hr.

Explanation :

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

Ans .

(4) 1250

Explanation :

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

Explanation :

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

Explanation :

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

Ans .

(4) 36

Explanation :

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

Ans .

(1) 6 hours

Explanation :

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

Ans .

(3) 20 km

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

Speed = 180 kmph
=$$\frac{180*5}{18}$$m/sec = 50 m/sec
..,1 km / hr =$$\frac{5}{18}$$m/s.

Ans .

(2) 21.6

Explanation :

Speed=$$\frac{150}{25}$$= 6 m/sec
= 6 *$$\frac{18}{5}$$=21.6 kmph

Ans .

(2) 54 km

Explanation :

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

Explanation :

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

Explanation :

Speed of bus = 72 kmph
=$$\frac{75*5}{18}$$metre/second
= 20 metre/second
Required distance= 20 × 5 = 100 metre

Ans .

(3) 6 km

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

1 m/sec = $$\frac{18}{5}$$kmph ..,$$\frac{10}{3}$$m/sec
$$\frac{18}{5}$$*$$\frac{10}{3}$$=12kmph

Ans .

(3) 10 km/hr

Explanation :

Remaining Time
=$$\frac{2}{5}$$*15=6 hours
Remaining Speed
=$$\frac{60}{6}$$=10kmph

Ans .

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

Explanation :

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.

Explanation :

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.

Explanation :

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

Ans .

(2) 24

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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) $$\frac{3}{2}$$

Explanation :

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.

Explanation :

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

Explanation :

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

Explanation :

Distance = Speed × Time
= 60 km.
Time taken at 40 kmph
=$$\frac{60}{40}$$=$$\frac{3}{2}$$hours

Ans .

(4) 4 km

Explanation :

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.

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

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.

Explanation :

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.

Explanation :

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.

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

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.

Explanation :

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

Ans .

(4) 12 km.

Explanation :

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.

# TYPE-2

Ans .

(3) 750

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

36 kmph ={36*$$\frac{5}{18}$$}m/sec
= 10 m/sec.
Required time=$$\frac{270+180}{10}$$= 45 seconds

Ans .

(3) 15

Explanation :

Speed of train
=$$\frac{Length of (train + platform)}{Time taken in crossing}$$
=$$\frac{50+100}{10}$$
=$$\frac{150}{10}$$=15m/sec

Ans .

(2) 54

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

# TYPE-3

Ans .

(3) 18 sec

Explanation :

Relative speed of man and train
= 20 – 10 = 10m/sec.
Required time =$$\frac{180}{10}$$
= 18 seconds

Ans .

(3) 12

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Ans .

(4) 500

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

# TYPE-4

Ans .

(3) 11 a.m.

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Ans .

(2) 12 Noon

Explanation :

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

Ans .

(3) 5 sec

Explanation :

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

Ans .

(3) 12

Explanation :

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

Ans .

(3) 12 sec.

Explanation :

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

Ans .

(1) 54 km/hr

Explanation :

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

Ans .

(1) 444

Explanation :

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

Ans .

(3) 8 sec

Explanation :

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

Ans .

(3) 3.42 sec

Explanation :

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

Ans .

(2) 85 km/hour

Explanation :

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

Ans .

(3) 100.

Explanation :

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

Ans .

(3) 54 km/hr

Explanation :

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

Ans .

(3) 180

Explanation :

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

Ans .

(3) 9.5 kmph

Explanation :

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

Ans .

(1) 875 km.

Explanation :

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

Ans .

(4) 11 seconds

Explanation :

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

Ans .

(1) 350 km.

Explanation :

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

# TYPE-5

Ans .

(3) 8 hours

Explanation :

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

Ans .

(1) 3.75 hours

Explanation :

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

Ans .

(1) 35 kmph

Explanation :

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

Ans .

(3) 6 hours

Explanation :

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

Ans .

(2) 60 minutes

Explanation :

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

Ans .

(2) 4.5 hours

Explanation :

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

Ans .

(1) 2 hours 30 minutes

Explanation :

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

Ans .

(2) 1 hour 12 minutes

Explanation :

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

Ans .

(2) 420 km

Explanation :

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

Ans .

(2) 20 km/hour

Explanation :

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

Ans .

(2) 72

Explanation :

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

Ans .

(1) 15 m/sec

Explanation :

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

Ans .

(3) 40 km./hr.

Explanation :

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

Ans .

(4) 6.6 km. per hour

Explanation :

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

# TYPE-6

Ans .

(2) 10.8 km/hr

Explanation :

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

Ans .

(1) 65.04 km/hr

Explanation :

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

Ans .

(2) 36 kmph

Explanation :

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

Ans .

(3) 4 km/hr

Explanation :

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

Ans .

(3) 6

Explanation :

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

Ans .

(1) 3 km/hour more

Explanation :

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

Ans .

(1) 8

Explanation :

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

Ans .

(4) 88.89 km/hr

Explanation :

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

Ans .

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

Explanation :

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

Ans .

(1) 48 km/hr

Explanation :

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

Ans .

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

Explanation :

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

Ans .

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

Explanation :

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

Ans .

(1) 7 km/hr

Explanation :

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

Ans .

(1) 40 km/hr

Explanation :

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

Ans .

(1) 120 kmph

Explanation :

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

Ans .

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

Explanation :

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

Ans .

(3) 70 km

Explanation :

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

Ans .

(3) 20 km/hr

Explanation :

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

Ans .

(1) 30 km/hr

Explanation :

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

Ans .

(2) 40

Explanation :

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

Ans .

(4) 24 km/hr

Explanation :

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

Ans .

(3) 9.00 a.m.

Explanation :

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

Ans .

(3) 24

Explanation :

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

Ans .

(1) 37.5

Explanation :

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

Ans .

(2) 4 km

Explanation :

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

Ans .

(4) 14 hrs

Explanation :

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

Ans .

(2) 45 km/hr

Explanation :

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

Ans .

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

Explanation :

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

Ans .

(1) 68 kmph

Explanation :

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

Ans .

(4) 4

Explanation :

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

Ans .

(3) 40 km/hr

Explanation :

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

Ans .

(3) 160 minutes

Explanation :

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

Ans .

(4) 35.55 kmph

Explanation :

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

Ans .

(1) 48 km/h

Explanation :

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

Ans .

(1) 54 km/hr

Explanation :

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

Ans .

(3) 10.9 kmph

Explanation :

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

Ans .

(1) 18 kmph.

Explanation :

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

Ans .

(3) 560 m.

Explanation :

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

Ans .

(4) 18.75 metre/second

Explanation :

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

# TYPE-7

Ans .

(1) 2 hours

Explanation :

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

Ans .

(1) 1.33 hour..

Explanation :

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

Ans .

(3) 25 : 18.

Explanation :

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

Ans .

(2) 3 : 2.

Explanation :

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

Ans .

(3) 78 km/hr

Explanation :

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

Ans .

(3) 3 : 4.

Explanation :

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

Ans .

(2) 1 : 3 : 9

Explanation :

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

Ans .

(3) 3

Explanation :

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

Ans .

(3) 3 : 4

Explanation :

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

Ans .

(1) 1 : 4

Explanation :

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

Ans .

(3) 3 : 4

Explanation :

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

Ans .

(2) 3 : 4

Explanation :

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

Ans .

(3) 3 : 4.

Explanation :

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

Ans .

(1) 2 : 3

Explanation :

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

Ans .

(3) 15 : 5 : 3

Explanation :

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

# TYPE-8

Ans .

(1) 100 m.

Explanation :

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

Ans .

(1) 85 km/hr

Explanation :

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

Ans .

(1) 19 minutes.

Explanation :

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

Ans .

(4) 25 seconds.

Explanation :

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

Ans .

(2) 12 minutes

Explanation :

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

Ans .

(4) 140

Explanation :

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

Ans .

(3) 75.

Explanation :

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

Ans .

(1) 27 km/hr

Explanation :

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

Ans .

(2) 50 m.

Explanation :

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

Ans .

(3) 3 km 750 m

Explanation :

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

Ans .

(3) 18.6

Explanation :

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

Ans .

(2) 400 m

Explanation :

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

Ans .

(1) 27.7 m

Explanation :

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

Ans .

(4) 90 km/h

Explanation :

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

Ans .

(2) 9

Explanation :

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

Ans .

(1) 125

Explanation :

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

Ans .

(4) 72.

Explanation :

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

Ans .

(1) 24

Explanation :

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

Ans .

(3) 12.

Explanation :

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

Ans .

(2) 4 p.m..

Explanation :

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

Ans .

(1) 50 m

Explanation :

Length of each train = x metre

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

Ans .

(3) 1320 km

Explanation :

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

Ans .

(3) 6.

Explanation :

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

Ans .

(2) 88.

Explanation :

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

Ans .

(1) 32 kmph

Explanation :

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

Ans .

(4) 3.75 km..

Explanation :

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

Ans .

(3) 500 metre

Explanation :

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

Ans .

(4) 50 m

Explanation :

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

# TYPE-9

Ans .

(4) 50 minutes

Explanation :

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

Ans .

(1) 2.

Explanation :

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

Ans .

(3) 5 km

Explanation :

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

Ans .

(4) 40

Explanation :

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

Ans .

(3) 19 .

Explanation :

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

Ans .

(1) 1.75 km

Explanation :

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

Ans .

(4) 6 km

Explanation :

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

Ans .

(2) 40 km/hour.

Explanation :

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

Ans .

(3) 22 km

Explanation :

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

Ans .

(3) 3 km.

Explanation :

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

Ans .

(2) 4

Explanation :

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

Ans .

(3) 12 km.

Explanation :

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

Ans .

(1) 60 km

Explanation :

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

Explanation :

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.

Explanation :

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.

Explanation :

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

Explanation :

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.
= 4 hours – 3 hours 40 minutes
= 20 minutes

Ans .

(3) 1 hour

Explanation :

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

# TYPE-10

Ans .

(4) 45 km/hr

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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.

Explanation :

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.

Explanation :

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.

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

Ans .

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

Explanation :

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.

Explanation :

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.

Explanation :

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.

Explanation :

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

Explanation :

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.

Explanation :

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.

Explanation :

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

# TYPE-11

Ans .

(2) 20 m.

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

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..

Explanation :

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

Explanation :

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..

Explanation :

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

Explanation :

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.

Explanation :

Relative speed
= 95 – 75 = 15 kmph
Required Time=$$\frac{Distance}{Relative speed}$$
$$\frac{5}{15}$$*60
=20 minutes

Ans .

(1) 15 minutes

Explanation :

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

Explanation :

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.

Explanation :

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.

Explanation :

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.

Explanation :

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

Explanation :

Relative speed
= 12 + 10 = 22 kmph Distance covered
= 55 – 11 = 44 km
\ Required time
$$\frac{44}{22}$$
= 2 hours

Ans .

(2) 200.

Explanation :

Required time = LCM of 40
and 50 seconds
= 200 seconds

Ans .

(1) 2 km.

Explanation :

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

# TYPE-12

Ans .

(2) 19 minutes

Explanation :

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.

Explanation :

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.

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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.

Explanation :

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.

Explanation :

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

Explanation :

Ans .

(2) 205 seconds

Explanation :

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.

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

Abhay’s speed = x kmph
Sameer’s speed = y kmph
$$\frac{30}{x}$$-$$\frac{30}{y}$$=12
$$\frac{30}{y}$$-$$\frac{30}{2x}$$=1
$$\frac{30}{y}$$-$$\frac{30}{2x}$$=3
$$\frac{30}{2x}$$=3
x = 5 kmph

Ans .

(3) 4 hours 45 minutes

Explanation :

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.

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

Ans .

(2) 800

Explanation :

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

Explanation :

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.

Explanation :

Ans .

(1) 1.20 am..

Explanation :

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.

Explanation :

Speed = 45 kmph
$$\frac{45*1000}{60*60}$$metre/second
$$\frac{45*5}{18}$$metre/second
= 12.5 metre/second

Ans .

(1) 610 m..

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

Consumption of petrol in cov- ering 540 km=$$\frac{540}{45}$$= 12 litres
Required expenses
= Rs. (12 × 20)
= Rs. 240

Ans .

(2) 6 cm

Explanation :

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.

Explanation :

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.

Explanation :

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

Explanation :

Distance = Speed × Time
= 330 × 10 = 3300 metre

Ans .

(2) 800 km.

Explanation :

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

Explanation :

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.

Explanation :

Ans .

(2) 5000.

Explanation :

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

# TEST YOURSELF

Ans .

(1) 10 m/sec.

Explanation :

36 km/hr.
36*$$\frac{5}{18}$$
= 10 m/sec.

Ans .

(2) 216 kmph

Explanation :

60 metres per sec.
60* $$\frac{18}{5}$$
= 216 km per hr.

Ans .

(3) 55 km.

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

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.

Explanation :

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.

Explanation :

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. t1=$$\frac{5}{3}$$ hr
Time taken in $$\frac{2}{3}$$ rd journey at 45 km per hr. t2=$$\frac{20}{9}$$ 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..

Explanation :

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.

Explanation :

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.

Explanation :

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
$$\frac{1500}{x+5x}$$=2
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.

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

$$\frac{5}{3}$$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

Explanation :

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.

Explanation :

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

Explanation :

Ans .

(1) 36 km

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

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.

Explanation :

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.

Explanation :

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

Explanation :

Ans .

(1) 210 leaps

Explanation :

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.

Explanation :

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.

Explanation :

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 = OA2+ 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

Explanation :

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.

Explanation :

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.

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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.

Explanation :

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

Explanation :

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+$$\frac{t2-t}{4}$$
Total distance covered by two men =4 t + 2 t +$$\frac{t2-t}{4}$$ =72
or 24t+t2-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

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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.

Explanation :

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..

Explanation :

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 .

Explanation :

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

Explanation :

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

Explanation :

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

Explanation :

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.

Explanation :

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.

Explanation :

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..

Explanation :

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.

Explanation :

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.

Explanation :

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.

Explanation :

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

Explanation :

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.

Explanation :

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.

Explanation :

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