Ans .
(3) 12 km.
Let the distance be x km.
\( \frac{x}{10}\)-\( \frac{x}{12}\)=\( \frac{12}{60}\)
x=\( \frac{1}{5}\)*60
= 12 km.
Ans .
(1) 60 km
Let the distance between stations
be x km, then speed of train
=\( \frac{x}{\frac{45}{60}}\)=\( \frac{4x}{3}\)
\( \frac{3x}{4x-15}\)=\( \frac{4}{5}\)
16x – 60 = 15x
x = 60 km
Ans .
(1) 13.33 minutes
Speed of train =\( \frac{Distance}{Time}\)=\( \frac{10}{\frac{12}{60}}\)
= 50 kmph
New speed = 45 kmph
Required time =\( \frac{10}{45}\)
\( \frac{2}{9}\)*60 minutes
=\( \frac{40}{3}\)=13.33 minutes.
Ans .
(1) 4 km.
Let the distance of the office
be x km, then \( \frac{x}{5}\)-\( \frac{x}{6}\)=\( \frac{8}{60}\)
x = 2 × 2 = 4 km
Ans .
(2) 4 km.
Let the distance of school be
x km,
then \( \frac{x}{3}\)-\( \frac{x}{4}\)=\( \frac{20}{60}\)
\( \frac{x}{12}\)=\( \frac{1}{3}\)
x= 4 km
Ans .
(3) 20 minutes
Distance between stations X and
Y = Speed × Time
= 55 × 4 = 220 km.
New speed = 55 + 5 = 60 kmph
Required time =\( \frac{220}{60}\)=\( \frac{11}{3}\)
= 3 hours 40 minutes.
Required answer
= 4 hours – 3 hours 40 minutes
= 20 minutes
Ans .
(3) 1 hour
Distance of journey = x km
Difference of time = 12 – 3 = 9 minutes
\( \frac{9}{60}\) hr=\( \frac{3}{20}\)hr
\( \frac{x}{70}\)-\( \frac{x}{80}\)=\( \frac{3}{20}\)
\( \frac{x}{56}\)=\( \frac{3}{2}\)
x=84 km
Required correct time \( \frac{84}{70}\)hr-12 minutes
=72 – 12 = 60 minutes
= 1 hour
Ans .
(4) 45 km/hr
Let the length of train be x me-
tres
\ According to question
Speed of the train =\( \frac{x}{10}\)m / sec
Also, the speed of the train
\( \frac{x+50}{14}\)m / sec.
It passes the platform in 14
seconds]
Both the speeds should be equal,
i.e.,
\( \frac{x}{10}\)=\( \frac{x+50}{14}\)
or 14x = 10x + 500
or 14x – 10x = 500
or 4x = 500
\ x = 125 metres
Hence, Speed =\( \frac{125}{10}\)= 12 . 5 m / sec
\( \frac{12.5*18}{5}\)km / hr .
= 45 km/hr.
Ans .
(2) 176
Let length of train be x m
Speed of train \( \frac{x+264}{20}\)
Also, speed of train =\( \frac{x}{8}\)
\( \frac{x}{8}\)=\( \frac{x+264}{20}\)
5x = 2x + 528
5x – 2x = 528
x = 528 ÷ 3 = 176 m
Ans .
(4) 79.2 km/hr
Let the length of train be x me-
tres.
Then, speed of train when it passes a telegraph post = \( \frac{x}{8}\)m/sec
and speed of train, when it
passes the bridge =\( \frac{x+264}{20}\)
Clearly,
\( \frac{x}{8}\)=\( \frac{x+264}{20}\)
5x = 2x + 528
3x = 528
x=176 m
Speed of train
\( \frac{176}{8}\)= 22 m/sec
22*\( \frac{18}{5}\)kmph
= 79.2 kmph
Ans .
(1) 25.2 km/hour.
Let the length of train be x
metres.
When the train crosses the standing man, its speed = \( \frac{x}{9}\)
When the train crosses the plat-
form of length 84 m, its speed \( \frac{x+84}{21}\)
Obviously,\( \frac{x+84}{21}\)=\( \frac{x}{9}\)
21x – 9x = 9 × 84
12x = 9 × 84
x=63 m
Required speed =\( \frac{63}{9}\)=\( \frac{63}{9}\)*\( \frac{18}{5}\)= 25.2 kmph
Ans .
(4)45 km/hr.
Suppose length of train be x
According to question
\( \frac{x+50}{14}\) = \( \frac{x}{10}\)14x = 10x + 500
4x = 500
x=125 m
Therefore, speed \( \frac{125}{10}\)*\( \frac{18}{5}\)= 45 kmph
Ans .
(4) 21.6 kmph.
Let the length of the train be x
According to the question,
Speed of the train \( \frac{x+90}{30}\)=\( \frac{x}{15}\)
x + 90 = 2x
x = 90 m
Speed of train =\( \frac{90}{15}\)
6 m/s =6*\( \frac{18}{5}\)= 21.6 kmph
Ans .
(3) 20 seconds.
Let the length of the train be x
metre
Speed of train when it crosses man=\( \frac{x}{10}\)
Speed of train when it crosses platform =\( \frac{x+300}{25}\)
According to the question,
Speed of train=\( \frac{x}{10}\)=\( \frac{x+300}{25}\)
25x = 10x + 3000
15x = 3000
x=200 m
Length of train = 200 metre
Speed of train= \( \frac{x}{10}\)=\( \frac{200}{10}\)= 20 m/sce
Time taken in crossing a 200m long platform = \( \frac{200+200}{20}\)= 20 seconds
Ans .
(4) 330 m
Let the length of the train be x
metres.
Speed of train in crossing boy = \( \frac{x}{30}\)
Speed of train in crossing platform \( \frac{x+110}{40}\)
According to the question,
\( \frac{x}{30}\)=\( \frac{x+110}{40}\)
4x = 3x + 330
x = 330 metres
Ans .
(3) 150
Let the length of train be x metre
\( \frac{x}{15}\)=\( \frac{x+100}{25}\)
5x = 3x + 300
2x = 300
x=150 metres
Ans .
(2) 40, 30
Ans .
(2) 52 km/hr, 26 km/hr
Let the speed of trains be x
and y metre/sec respectively,
\( \frac{100+95}{x-y}\)=27
x-y=\( \frac{65}{9}\)..(1)
\( \frac{195}{x+y}\)=9
x+y=\( \frac{195}{9}\)...(2)
By equation (i) + (ii)
2x= \( \frac{65}{9}\)+\( \frac{195}{9}\)=\( \frac{260}{9}\)
x = \( \frac{130}{9}\)m/sec.
\( \frac{130}{9}\)*\( \frac{18}{5}\)kmph = 52 kmph
From equation (ii),
y = \( \frac{65}{9}\)m/sec
\( \frac{65}{9}\)*\( \frac{18}{5}\)
= 26 kmph
Ans .
(2) 130.
Let the length of train be x me-
tre, then
\ Speed of train
\( \frac{x}{7}\)=\( \frac{x+390}{28}\)
x=\( \frac{390}{3}\)
= 130 metres
Ans .
(3) 15.5 seconds.
Speed of train = 36 kmph
36 * \( \frac{5}{18}\)= 10 m/sec
Length of train = 10 × 10
= 100 metres
Required time= \( \frac{100+55}{10}\)
= 15.5 seconds
Ans .
(2) 300.
Speed of train = 60 kmph
60*\( \frac{5}{18}\)m/sec.
=\( \frac{50}{3}\)m/sec.
If the length of platform be
= x metre, then
Speed of train=\( \frac{Length of (train + platform)}{Time taken in crossing}\)
50 × 10 = 200 + x
x = 500 – 200 = 300 metre
Ans .
(4) 9:24 am
Let both trains meet after t
hours since 7 a.m.
Distance between stations A and
B = x Km.
\( \frac{x}{4}\)*t+\( \frac{x}{\frac{7}{2}}\)*(t-1)=x
Speed=\( \frac{Distance}{Time}\)
\( \frac{7t+8t-8}{28}\)=1
15 t – 8 = 28
15 t = 28 + 8 = 36
t=2 hours 24 minutes
Required time = 9 :24 a.m.
Ans .
(2) 12.1 seconds.
Speed of train = 72 kmph.
\( \frac{72*5}{18}\)m/sec
= 20 m./sec.
Required time
\( \frac{Length of train and bridge}{Speed of train}\)
\( \frac{242}{20}\)
= 12.1 seconds
Ans .
(2) 6 seconds.
Relative speed of train
= (60 + 6) kmph.
\( \frac{66*5}{18}\)m/sec
=\( \frac{55}{3}\)m/sec.
Length of train = 110 metre
Required time =\( \frac{110}{\frac{55}{3}}\)
= 6 seconds
Ans .
(2) 20 m.
Let the time taken to complete
the race by A,B, and C be x min-
utes
Speed of A =\( \frac{1000}{x}\)
B =\( \frac{1000-50}{x}\) =\( \frac{950}{x}\)
C =\( \frac{1000-69}{x}\)= =\( \frac{931}{x}\)
Now, time taken to complete the
race by
B=\( \frac{1000}{\frac{950}{x}}\)=\( \frac{1000*x}{950}\)
and distance travelled by C in
\( \frac{1000x}{950}\)min
\( \frac{1000x}{950}\)*\( \frac{931}{x}\)= 980 km.
B can allow C
= 1000 – 980 = 20 m
Ans .
(4) 12 minutes
Ratio of the speed of A, B and
C = 6 : 3 : 1
Ratio of the time taken
=\( \frac{1}{6}\):\( \frac{1}{3}\):1=1:2:6
Time taken by A
\( \frac{72}{6}\)= 12 minutes
Ans .
(1) 17.24 seconds.
Let A take x seconds in covering
1000m and b takes y seconds
According to the question,
x+20=\( \frac{900}{1000}y\)
x+20=\( \frac{9}{10}y\)...(1)
\( \frac{950}{1000}\)x + 25 = y....(2)
From equation (i),
\( \frac{10x}{9}\)+\( \frac{200}{9}\)=y
\( \frac{10x}{9}\)+\( \frac{200}{9}\)=\( \frac{950}{1000}\)x + 25
\( \frac{200x-171x}{180}\)=\( \frac{225-200}{9}\)
\( \frac{29x}{180}\)=\( \frac{25}{9}\)
x=\( \frac{500}{29}\)=17.24
Ans .
(3) 14.4 kmph
Time taken by Kamal
\( \frac{100}{18*\frac{5}{18}}\)
= 20 seconds
Time taken by Bimal
= 20 + 5 = 25 seconds
Bimal’s speed = \( \frac{100}{25}\)=4 m/sec
=\( \frac{4*18}{5}\)=14.4 kmph.
Ans .
(1) 95 m..
When A runs 1000m, B runs
900m.
\ When A runs 500m, B runs
450 m.
Again, when B runs 400m, C runs
360 m.
\ When B runs 450m, C runs
\( \frac{360}{400}\)*450 = 405 metres
Required distance = 500 – 405
= 95 metres
Ans .
(1) 11.9 metre
According to the question,
\ When A runs 800 metres, B
runs 760 metres
\ When A runs 200 metres, B runs= \( \frac{760}{800}\)*200= 190 metres
Again, when B runs 500 metres,
C runs 495 metres.
\ When B runs 190 metres, C runs =\( \frac{495}{500}\)*190= 188.1 metres
Hence, A will beat C by
200 – 188.1 = 11.9 metres in a
race of 200 metres.
Ans .
(3) 29 metres..
According to the question,
Q When B runs 200 m metres, A
runs 190 metres
\ When B runs 180 metres, A runs=\( \frac{190}{200}\)*180= 171 metres
When C runs 200m, B runs 180
metres.
Hence, C will give a start to A by
= 200 – 171 = 29 metres
Ans .
(1) 31.25 metre
According to the question,
When A covers 1000m, B covers
= 1000 – 40 = 960 m
and C covers =1000 – 70 = 930 m
When B covers 960m, C covers
930 m.
\ When B covers 1000m, C covers=\( \frac{930}{960}\)*1000
= 968.75 metre
Hence, B gives C a start of
= 1000 – 968.75 = 31.25 metre
Ans .
(2) 20 min.
Relative speed
= 95 – 75 = 15 kmph
Required Time=\( \frac{Distance}{Relative speed}\)
\( \frac{5}{15}\)*60
=20 minutes
Ans .
(1) 15 minutes
Time taken by C = t hours
Time taken by B =\( \frac{t}{3}\)hours
Time taken by A =\( \frac{t}{6}\)hours
Here,t=\( \frac{3}{2}\)hours
Required time taken by A
\( \frac{3}{\frac{2}{6}}\)=\( \frac{1}{4}\)
\( \frac{1}{4}\)*60= 15 minutes
Ans .
(2) 40 min
2 hours 45 minutes
2 + \( \frac{45}{60}\)hours=\( \frac{11}{4}\)hours
Distance = Speed × Time
4 * \( \frac{11}{4}\)= 11 km.
Time taken in covering 11 km
at 16.5 kmph
=\( \frac{11}{16.5}\)
=40 minutes.
Ans .
(2) 35 km.
Let the total distance be x km.
Time =\( \frac{Distance}{Speed}\)
According to the question,
\( \frac{10}{6}\) +\( \frac{20}{16}\) +\( \frac{x-30}{3}\)=4 \( \frac{35}{60}\)
=4\( \frac{7}{12}\)
\( \frac{5}{3}\) +\( \frac{5}{4}\) +\( \frac{x}{3}\)-10 =\( \frac{55}{12}\)
x=\( \frac{140}{12}\)*3=35 km
Ans .
(1) 1 hour.
Usual time = x minutes
New time =\( \frac{4x}{3}\)
Speed ∝ \( \frac{1}{Time}\)
According to the question,
\( \frac{4x}{3}\)– x = 20
x= 1hour
Ans .
(2) 4 km/hr.
Let, A’s speed = x kmph.
\ B’s speed = (7 – x) kmph
Time=\( \frac{Distance}{Speed}\)
According to the question,
\( \frac{24}{x}\)+\( \frac{24}{7-x}\)=14
\( \frac{24*7}{x(7-x)}\)=14
x (7 – x) = 12 = 4 × 3 or 3 × 4
Þ x (7 – x) = 4 (7 – 4) or 3 (7 – 3)
Þ x = 4 or 3
\ A’s speed = 4 kmph.
Ans .
(3) 2 hours
Relative speed
= 12 + 10 = 22 kmph
Distance covered
= 55 – 11 = 44 km
\ Required time
\( \frac{44}{22}\)
= 2 hours
Ans .
(2) 200.
Required time = LCM of 40
and 50 seconds
= 200 seconds
Ans .
(1) 2 km.
Distance between starting
point and multiplex = x metre
Time =\( \frac{Distance}{Speed}\)
According to the question,
\( \frac{x}{3}\) -\( \frac{x}{4}\)= \( \frac{5+5}{60}\)
\( \frac{x}{12}\) =\( \frac{1}{6}\)
x=2 km
Ans .
(2) 19 minutes
Two ways walking time
= 55 min...(i)
One way walking + One way
riding time = 37 min.....(ii)
By 2 × (ii) – (i),
2 ways riding time
= 2×37–55 = 19 minutes.
Ans .
(3) 50 km.
Let the distance be x km
Time taken by A =\( \frac{x}{40}\)hrs
Time taken by B =\( \frac{x}{50}\)hrs
\( \frac{x}{40}\)-\( \frac{x}{50}\)=\( \frac{15}{60}\)
\( \frac{5x-4x}{200}\)=\( \frac{15}{60}\)
x=50 km
Ans .
(1) 5 km/hr.
Let the speed of man be x
kmph
30 x – 30{x-\( \frac{x}{15}\)} = 10
30{x-x+\( \frac{x}{15}\)}=10
\( \frac{x}{15}\)=\( \frac{10}{30}\)
x= 5kmph
Ans .
(1) 46 minutes 12 seconds
Required time = LCM of 252,
308 and 198 seconds.
Now, 252 = 2 × 2 × 3 × 3 × 7
308 = 2 × 2 × 7 × 11
198 = 2 × 3 × 3 × 11
\ LCM = 2 × 2 × 3 × 3 × 7 × 11
= 36 × 77 seconds
\( \frac{36*77}{60}\)minutes
= 46 minutes 12 seconds
Ans .
(4) 6 hours
Suppose, time taken while
walking be x hours
And, time taken on riding be y
hours
\ According to question
x+y=4\( \frac{1}{2}\)hr
Then, 2y = 3 hours
y=1\( \frac{1}{2}\)hr
x=4\( \frac{1}{2}\) - 1\( \frac{1}{2}\)=3 hr
Time required to walk both ways
= 6 hours
Ans .
(4) 9 km
Let the required distance be
x km
\( \frac{x}{\frac{9}{2}}\)+\( \frac{x}{3}\)=5
x{\( \frac{2+3}{9}\)=5
x=9 km
Ans .
(4) 26.7 km.
Distance covered by A in 4
hours = 4 × 4 = 16 km
Relative speed of B with respect
to A = 10 – 4 = 6 km/hr
Time taken to catch A
\( \frac{16}{6}\)=\( \frac{8}{3}\)hr
Required distance
=\( \frac{8}{3}*10\)
= 26.67 km.= 26.7 km
Ans .
(2) 480 km.
Suppose distance be x km
\( \frac{x}{2*40}\)+\( \frac{x}{2*60}\)=10
\( \frac{3x+2x}{240}\)=10
x=480 km
Ans .
(1) 2 min 25 sec.
If A covers the distance of 1
km in x seconds, B covers the
distance of 1 km in (x + 25) sec-
onds. If A covers the distance of
1 km, then in the same time C
covers only 725 metres.
If B covers 1 km in (x + 25) sec-
onds, then C covers 1 km in (x +
55) seconds.
Thus in x seconds, C covers the
distance of 725 m.
\( \frac{x}{725}\)*1000=x+55
x = 145
A co vers the di stance of 1 km in 2 minutes 25 seconds.
Ans .
(4) 250 m
Ans .
(2) 205 seconds
A beats B by 30 seconds and
B beats C by 15 seconds.
Clearly, A beats C by 45 seconds.
Also, A beats C by 180 metres.
Hence, C covers 180 metres in
45 seconds
Speed of C =\( \frac{180}{45}\)= 4 m/sec
Time taken by C to cover 1000 m
=\( \frac{1000}{4}\)=250 sec
Time taken by A to cover 1000 m
= 250–45 = 205 sec
Ans .
(2) 27.
Difference of time
= 6 min. – 5 min. 52 sec.
= 8 seconds
Distance covered by man in 5 min.
52 seconds
= Distance covered by sound in
8 seconds
= 330 × 8 = 2640 m.
\ Speed of man
\( \frac{2640 m}{5 min. 52 sec.}\)=\( \frac{2640}{352}\)
\( \frac{2640}{352}\)*\( \frac{18}{5}\)kmph
= 27 kmph
Ans .
(1) 70 km
Let the required distance be x
km.
Difference of time
= 15 + 5 = 20 minutes
=\( \frac{1}{3}\)hr
According to the question,
\( \frac{x}{35}\)-\( \frac{x}{42}\)=\( \frac{1}{3}\)
\( \frac{x}{210}\)=\( \frac{1}{3}\)
x=70 km
Ans .
(3) 7 hours 30 minutes
1-\( \frac{5}{6}\) of time taken by B
=1 hour 15 minutes
\ Time taken by B
= 1 hour 15 minutes × 6
=7 hours 30 minutes
Ans .
(1) 5
Abhay’s speed = x kmph
Sameer’s speed = y kmph
\( \frac{30}{x}\)-\( \frac{30}{y}\)=12
\( \frac{30}{y}\)-\( \frac{30}{2x}\)=1
On adding,
\( \frac{30}{y}\)-\( \frac{30}{2x}\)=3
\( \frac{30}{2x}\)=3
x = 5 kmph
Ans .
(3) 4 hours 45 minutes
Time taken in walking both
ways = 7 hours 45 minutes ....(i)
Time taken in walking one way
and riding back = 6 hours 15
minutes
....(ii)
By equation (ii) × 2 – (i), we have
Time taken by the man to ride
both ways
= 12 hours 30 minutes – 7 hours
45 minutes
= 4 hours 45 minutes
Ans .
(1) 25 km/hr.
Let the total distance be 100
km.
Average speed
\( \frac{Total distance covered}{Time taken}\)
\( \frac{100}{ \frac{30}{20}+ \frac{60}{40}+ \frac{10}{10}}\)
\( \frac{100*2}{8}\)=25 kmph
Ans .
(2) 6 km/hr
Let the speed of A = x kmph and
that of B = y kmph
According to the question,
x × 6 + y × 6 = 60
x + y = 10
and \( \frac{2}{3}\)x* 5 + 2 y * 5 = 60
10x + 30y = 180
x + 3y = 18
...(ii)
From equations (i) × (3) – (ii)
3x + 3y – x – 3y = 30 – 18
12x = 12
x = 6 kmph.
Ans .
(2) 81 km
Let the trains meet after t hours,
then
24t – 18t = 27
6t = 27
t=\( \frac{9}{2}\)hours
QR = 18t = 18 * \( \frac{9}{2}\)= 81 km
Ans .
(3) 8 km/hr
Let the speed of Ravi be x kmph
then, Ajay’s speed = (x + 4) kmph
Distance covered by Ajay
= 60 + 12 = 72 km
Distance covered by Ravi
= 60 – 12 = 48 km.
According to the question
\( \frac{72}{x+4}\)=\( \frac{48}{8}\)
\( \frac{3}{x+4}\)=\( \frac{2}{x}\)
3x = 2x + 8
x = 8 kmph
Ans .
(2) 16 km
Let man walked for t hours.
then, t × 4 + (9 – t) × 9 = 61
4t + 81 – 9t = 61
81 – 5t = 61
5t = 20
t = 4
Distance travelled on foot
= 4 × 4 = 16 km.
Ans .
(1) 6
Let the required distance be x
km, then
\( \frac{x}{5}\)-\( \frac{x}{6}\)=\( \frac{1}{5}\)
\( \frac{x}{30}\)=\( \frac{1}{5}\)
x=6 km
Ans .
(4) 6 km.
Let the required distance be
x km.
\( \frac{x}{3}\)-\( \frac{x}{4}\)=\( \frac{30}{60}\)
\( \frac{x}{12}\)=\( \frac{1}{2}\)
x= 6km
Ans .
(2) 60 km/hr
Let the speed of train be x
kmph and that of car be y kmph,
then
\( \frac{60}{x}\)+\( \frac{240}{y}\)=4
and ,\( \frac{100}{x}\)+\( \frac{200}{y}\)=\( \frac{25}{6}\)
\( \frac{4}{x}\)+\( \frac{8}{y}\)=\( \frac{1}{6}\)
By equation (i ) – equation (ii) ×
30
\( \frac{60}{x}\)+\( \frac{240}{y}\)-\( \frac{120}{x}\)-\( \frac{240}{y}\)=4-5
-\( \frac{60}{x}\)=-1
x=60 kmph
Ans .
(2) 38 minutes.
Ratio of the speed of A and B
= A : B = 2 : 1 = 6 : 3
B : C = 3 : 1
A : B : C = 6 : 3 : 1
Ratio of their time taken
\( \frac{1}{6}\):\( \frac{1}{3}\):1=1:2:6
Time taken by B
\( \frac{2}{6}\)*114 minutes
= 38 minutes
Ans .
(3) 54 km/hr
Ans .
(2) 800
Total distance of trip
=\( \frac{1200*5}{2}\)= 3000 km
Part of journey covered by train
1-\( \frac{2}{5}\)-\( \frac{1}{3}\)=\( \frac{4}{15}\)
Distance covered by train
3000*\( \frac{4}{15}\)= 800 km
Ans .
(1) 1.6 minutes
A’s speed =\( \frac{1000}{5}\)
= 200 m/minute
B’s speed =\( \frac{1000}{8}\)
= 125 m/minute
C’s speed =\( \frac{1000}{10}\)
= 100 m/minute
Distance covered by C in 2 min-
utes = 200 metre
Distance covered by B in 1 minute
= 125 metre
Relative speed of A with respect
to C = 100 metre
Time=\( \frac{200}{100}\)= 2 minutes
Relative speed of A with respect
ot B = 75 metre
Time=\( \frac{125}{75}\)= \( \frac{5}{3}\) minutes
=1.6minutes
Ans .
(2) v 1 : v 2 = 4 : 5.
Ans .
(1) 1.20 am..
Time taken in covering 999km
\( \frac{999}{55.5}\)= 18 hours
Required time = 18 hours + 1
hour 20 minutes
= 19 hours 20 minutes
i.e. 1 : 20 am
Ans .
(1) 12.5 metre/seccond.
Speed = 45 kmph
\( \frac{45*1000}{60*60}\)metre/second
\( \frac{45*5}{18}\)metre/second
= 12.5 metre/second
Ans .
(1) 610 m..
Distance covered in 2nd
minute = 90 – 50 = 40 metre
Distance covered in 3rd minute
= 130 – 90 = 40 metre
\ Required distance
= 50 + 40 × 14
= 50 + 560 = 610 metre
Ans .
(3) 7: 56 AM
Here distance is constant.
Speed ∝ \( \frac{1}{Time}\)
Ratio of the speeds of A and B
\( \frac{\frac{7}{2}}{4}\)=7:8
A’s speed = 7x kmph (let)
B’s speed = 8x kmph
AB = 7x × 4 = 28x km.
Let both trains cross each other
after t hours from 7 a.m.
According to the question,
7x (t + 2) + 8x × t = 28x
7t + 14 + 8t = 28
15t = 28 – 14 = 14
t=\( \frac{14}{15}\)hours
\( \frac{14}{15}\)*60
=56 minutes.
Required time = 7 : 56 A.M.
Ans .
(4) 9 hours
Speed of plane = \( \frac{Distance}{Time}\)
=\( \frac{6000}{8}\)= 750 kmph
New speed = (750 + 250) kmph
= 1000 kmph
Required time =\( \frac{9000}{1000}\)
= 9 hours
Ans .
(1) 45
Let speed of train be x kmph.
Speed of car = y kmph.
Case I,
Time = \( \frac{Distance}{Speed}\)
\( \frac{240}{x}\) + \( \frac{210}{y}\)=8\( \frac{2}{3}\)
\( \frac{240}{x}\) + \( \frac{210}{y}\)=8\( \frac{26}{3}\)...(1)
Case II,
\( \frac{180}{x}\)+\( \frac{270}{y}\)=9....(2)
By equation (i) × 3 – (ii) × 4,
\( \frac{720}{x}\) +\( \frac{630}{y}\) -\( \frac{720}{x}\) -\( \frac{1080}{y}\)
= 26 – 36
=\( \frac{-450}{y}\)= –10
= –10
y = 45 kmph
Ans .
(3) 81 km/h.
Difference of time = 11 min-
utes 45 seconds – 11 minutes =
45 seconds
Distance covered by sound in 45
seconds = Distance covered by
train in 11 minutes
330 × 45 = 11 × 60 × Speed of train
Speed of train
\( \frac{330*45}{11*60}\)m/sec.
\( \frac{45}{2}\) \( \frac{18}{5}\)
=81 kmph
Ans .
(2) 45
Distance covered in 3 hours
36 minutes i.e. 3\( \frac{36}{60}\)hours
=5\( \frac{18}{5}\)= 18 km.
Time taken at 24 kmph.
\( \frac{18}{24}\)hours
\( \frac{18}{24}\)*minutes
=45 minutes
Ans .
(3) 2 hours
Let the origi nal speed of
aeroplane be x kmph.
According to the question,
\( \frac{1200}{x-300}\)-\( \frac{1200}{x}\)=2
1200{\( \frac{x-x+300}{x(x-300)}\)}=2
x (x – 300) =\( \frac{1200*300}{2}\)
x (x – 300) = 600 × 300
x (x – 300) = 600 (600 – 300)
x = 600 kmph.
Scheduled duration of flight =\( \frac{1200}{600}\)= 2 hours
Ans .
(4) 240.
Consumption of petrol in cov-
ering 540 km=\( \frac{540}{45}\)= 12 litres
Required expenses
= Rs. (12 × 20)
= Rs. 240
Ans .
(2) 6 cm
18 km o 1.5 cm
1 km= \( \frac{1.5}{18}\)cm
72=\( \frac{1.5*72}{18}\)cm = 6 cm
Ans .
(2) 16 km.
Length of journey on foot
= x km. (let).
\ Length of journey on cycle =
(61 – x ) km.
According to the question,
Time = \( \frac{Distance}{Speed}\)
\( \frac{x}{4}\)+\( \frac{61-x}{9}\)=9
\( \frac{9x+244-4x}{36}\)=9
5x + 244 = 36 × 9 = 324
5x = 324 – 244 = 80
x=16 km
Ans .
(1) 16 km.
Let the distance covered on
foot be x km.
Distance covered on cycle =
(61 – x) km.
Time=\( \frac{Distance}{Speed}\)
\( \frac{x}{4}\)+\( \frac{61-x}{9}\)=9
\( \frac{x}{4}\)-\( \frac{x}{9}\)=9-\( \frac{61}{9}\)
\( \frac{5x}{36}\)=\( \frac{20}{9}\)
x=16 km
Ans .
(4) 3300 metre
Distance = Speed × Time
= 330 × 10 = 3300 metre
Ans .
(2) 800 km.
Let total distance covered be
2x km.
Total time = 14 hours 40 min-
utes
=14\( \frac{40}{60}\)hours=\( \frac{44}{3}\)hours
Time = \( \frac{Distance}{Speed}\)
According to the question,
\( \frac{x}{60}\)\)+\( \frac{x}{50}\)=\( \frac{44}{3}\)
\( \frac{11x}{300}\)=\( \frac{44}{3}\)
x=400
Total distance
= 2x = 2 × 400 = 800 km
Ans .
(2) 80
Distance between both don-
keys = 400 metre.
Relative speed = (3 + 2) m./sec.
= 5 m./sec.
Required time
=\( \frac{Distance}{Relative Speed}\)
=\( \frac{400}{5}\)= 80 seconds