Problems on Trains- Aptitude Questions and Answers

Time 12 min, Distance = 10 km
Speed of train = 10 / 12 kmpm = 10*60/12 = 50 kmph

New Speed of train after decreased by 10 kmph= 50-10 = 40 kmph
New Speed 40 kmph, Distance = 10 km
Time taken = 10/40 = 1/4 hr = 1/4 × 60 = 15 min

Speed of the train = 36 * (5/18) = 10 m/sec.
Length of the train, distance = (Time * Speed ) = (60 * 10 ) = 600 meters

Time taken by the train to cross the platform of length 250 meter
= Time taken by it to cover (600 + 250) m
= 850/10 = 85 sec

Relative speed when moving in same direction is S1 - S2
Speed of train relative to jogger = (48-12) = 36 kmph = 36*(5/18) = 10 m/sec

When a train of lengths L1 travelling at a speed S crosses a platform/bridge/tunnel of length L2 in time T,
then the equation is given as S = (L1+L2)/T

So,Time taken by train to pass the runner, T =Distance to cover(L1+L2)/Relative speed(S)
=> (200 + 160)/10= 360/10 = 36 Seconds

Time taken by train to pass the runner =36 Seconds

Speed= 54 km/hr =54 *(1000/3600)m/sec = 15 m/sec.
Distance = Time * Speed
Length of the train = (Speed * Time) = 15 * 30= 450 m.

Distance to cover(L1+L2)=>120+180 = 300 meter

Relative speed of the train= 300/10 =30 m/sec

Relative speed when moving in opposite direction is S1 + S2 = 30 m/sec
Relative speed when moving in same direction is S1 - S2= 10 m/sec

So,Time taken by faster train to pass other train,S1=1/2(30+10)
=>20 m/sec
=>20 * (3600/1000) km/h
=>72 km/h

When a train of lengths L travelling at a speed S crosses a pole/pillar/flag post in time T, then the equation is given as S =L/T
Hence, L=10*S m --------------------------(1)

When a train of lengths L1 travelling at a speed S crosses a platform/bridge/tunnel of length L2 in time T, then the equation is given as S = (L1+L2)/T
S = (L+200)/20
20*S=10*S + 200
10*S=200
S=20 m/s

So L= 10*S = 10*20 = 200 m

When two trains of lengths L1 and L2 respectively travelling at the speeds of S1 and D2 respectively cross each other in time T, then the equation is given as S1+S2 = (L1+L2)/T

S1+S2 = 36 km/hr + 54 km/hr = 90 km/hr = 90 km/hr * 5/18 = 25 m/sec
L1+L2 = 200m + 300m = 500 m

Then, S1+S2 = (L1+L2)/T
T = 500/25 sec = 20 sec

When a train of lengths L1 travelling at a speed S1 crosses a platform/bridge/tunnel of length L2 in time T, then the equation is given as S1 = (L1+L2)/T
L1+L2 = 330+330 =660 m
S1 = 264 km/hr = (264 * 5/18) m/sec = 220/3 m/sec

Then , S1 = (L1+L2)/T
220/3 = 660 /T
T = 660* 3 / 220 sec = 9 sec

Relative speed when moving in opposite direction, S1 + S2 = 50+80 = 130 kmph = 130*(5/18) m/sec
Total distance to cross to clear off each other, L1 + L2 = 120+160 = 280 meters

When a train of lengths L1 travelling at a speed S crosses a platform/bridge/tunnel of length L2 in time T, then the equation is given as S = (L1+L2)/T
So,Time taken cross each other after they meet, T = Distance to cover(L1+L2) / Relative speed(S1 + S2)
=> 280 / (130*(5/18)) = 7.75~ Seconds

Time taken by express train to cover 900 km = 900/90 = 10 hr
Time taken during stoppage= train stop 8 times in the whole journey after every 100 km= 8 *3 = 24 min.
So, total time taken by train including stoppage= 10 hr 24 min.

Speed of the train=60 m/s
Speed of the man=36 km/h = 36*1000/60*60 m/s= 10 m/s
relative speed=60-10 m/s(as they are moving in same direction)=50 m/s

length of the train=200 m

So,the time taken by a train = length of the train/relative speed
= 200/50
= 4 sec

Let the length of the bridge = X metre
Speed of train = 36 km/hr = 36* 5/18 = 10 m/sec

S=(L1 + L2)/T
10 = 200+X / 55
200+X = 550
X = 550-200 = 350 metre

Speed of the train = Length of (train+platform)/Time = 500+700 / 10 = 120 ft/ sec

Let the speed of passenger train= x km/hr
Speed of express train = (x + 15) km/hr
Then, As is given

6pm#-----540-----#-----540-----#9pm

(540 / x) - (540 / x + 15) = 3 [6 pm - 9 pm ]
540( 1/x - 1/x + 15) = 3
540( x + 15 - x / x*(x + 15) ) = 3
540*15 = 3x*(x + 15)
x2 + 15X - 2700 = 0
(x + 60)(x - 45) = 0
x = 45 [∵ x ≠ 60]

Time taken by passenger train to reach the meeting point = 540/x = 540/45 = 12 Hrs

Therefore, Both train will meet at (6 pm + 12 Hrs) = 6 a.m.

Relative Speed, in the same direction= Speed of train - Speed of man
=(30 - 3) km/h
=(27*1000/3600) m/s
=(27*5/18)
=15/2 m/s

Time taken to cross the man = Distance/ Relative Speed
= 225 / (15/2)
= 225*2 / 15
= 30 sec

Total distance between M and N= 569 km

First train leaves station at 8:00 am and second at 9:00 am. So first train i.e.,
from M to N has covered 53 km distance in 1 hour.

So, distance left between the stations = 569 − 53 = 516 km

trains are travelling in opposite directions. So, relative speed of trains = 53 + 76 = 129 km/h

Now, Time taken to cover 519 km = 519/129= 4 hours

Hence the time, at which both the trains will meet, is 4 hour after second train left i.e., 9:00 am + 4 h = 1:00 pm

Time taken to cross= 6 sec = 6/3600 sec = 1/600 hr
Let the speed of the second train be X km/hr
Relative speed = 40 + X km/hr (opposite directions)
Distance covered = 100 + 130 = 230 m = 0.23 km

Time taken to cross= Distance covered/Relative speed
1/600 = 0.23 / 40 + X
40 + X = 138
X = 98 km/hr