- km/hr to m/s conversion: a km/hr = (a X 5/18) m/s
- m/s to km/hr conversion: a km/hr = (a X 18/5) km/hr
- Time taken by a train of length metres to pass a pole or standing man or a signal post is equal to the time taken by the train to cover l metres
- Time taken by a train of length metres to pass a stationery object of length b metres is the time taken by the train to cover (l + b) metres
- Suppose two trains or two objects bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relative speed is = (u – v) m/s
- Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s
- If two trains of length metres and b metres are moving in opposite directions at u m/s and v m/s, then: their relative speed is = (u – v) m/s
- Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s
- If two trains of length metres and b metres are moving in opposite directions at u m/s and v m/s, then: The time taken by the trains to cross each other = (a+b) / (u+v) sec
- The time taken by the faster train to cross the slower train = (a+b) / (u-v) sec
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Problems on trains – Question and Answers
Relative speed = (x + 50) km/hr
= [(x + 50) * 5 / 18] m/sec
=[(250 + 5x) / 18] m/sec
Distance covered = (108 + 112) = 220 m
Therefore, (250 + 5x) / 18 = 6
250 + 5x = 660 x = 82 km/hr
Time = 1 minute = 60 seconds
Let the length of the tunnel be x metres
Then, (800 + x) / 60 = 65 / 3
3(800 + x) = 3900 x = 500
Then, distance covered = 2xmetres
Relative speed = (46 – 36) km/hr
= (10 * 5 / 18) m/sec
= (25 / 9) m/sec
Therefore,( 2x / 36) = (25 / 9)
2x = 100 x = 50
= (25/2) m/sec
= (25 / 2 * 18 / 5) km/hr
= 45 km/hr
Let the speed of the train be x km/hr
Then, relative speed = (x – 5) km/hr
x – 5 = 45 => x = 50 km/hr
=(150 * 5 / 18) m/sec
= (125 / 3) m/sec
Distance covered = (1.10 + 0.9) km = 2 km = 2000 m
Required time = (2000 * (3/125) sec = 48 sec
Time = 26 sec
Let the length of the train be x metres
Then, (x + 250) / 26 = 20
x + 250 = 520 x = 270
Distance covered by A in x hours = 20xkm
Distance covered by B in (x – 1) hours = 25(x – 1) km
20x + 25(x – 1) = 110 45x = 135 x = 3.So, they meet at 10 a.m
Then, relative speed of the two trains = 2x m/sec
So, 2x = (120 + 120) / 12
2x = 20 x = 10
Speed of each train = 10 m/sec
=> (10 * 18/5)km/hr = 36 km/hr
Therefore, Speed = (45 * 5 /18) m/sec
= 25/2 m/sec
Total distance to be covered = (360 + 140) m = 500 m
Formula for finding Time = (Distance / Speed)
Required time = (500 x 2 / 25)sec = 40 sec
Time = 30 sec
Let the length of bidge be x metres
Then, (130 + x / 30) = 25 / 2
2(130 + x) = 750 x = 245 m
Time = 30 sec
Let the length of bidge be x metres
Then, (130 + x / 30) = 25 / 2
2(130 + x) = 750 x = 245 m
Therefore, Required time = (240 + 650) / 10sec = 89 sec
= (36 x (5 / 18)) m/sec
= 10 m/sec
Distance to be covered = (240 + 120) m = 360 m
Therefore, time taken = 360 /10 sec = 36 sec
= 60 km/hr
= (60 * 5 / 18) m/sec
= (50 / 3) m/sec
Therefore, time taken to pass the man = (500 * 3/50 ) sec
= 30 sec
Then, speed of the faster train = 2xm/sec
Relative speed = (x + 2x) m/sec = 3xm/sec
Hence (100 + 100) / 8 = 3x
24x = 200
x = 25 / 3
So, speed of the faster train = 50/3 m/sec
= (50 / 3) * (18 / 5) km/hr
= 60 km/hr
Let the length of the platform be x metres
Then, (x + 300) / 36 = 15
x + 300 = 540 x = 240 m
Then, length of the first train = 27x metres, and length of the second train = 17y metres
Therefore, (27x + 17y) / (x + y) = 23
27x + 17y = 23x + 23y 4x = 6y
x/y = 3/2
= 20 * 5/18 = 50/9 m/sec
Length of faster train = 50/9 * 5 = 250/9 = 27 7/9 m
Speed of the train relative to man = ( 63 – 3 ) km/hr = 60 km/hr
= (60 x 5/18) m/s = 50/3 m/s
Time taken to pass the man = distance/speed = (500 x 3/50) sec = 30 sec
Distance = 50m(Given)
Time = 50/120 = 5/12 hrs = 25 minutes to meet Now,
Relative length of train = 1/6 + 1/6 = 1/3m
Relative speed = 120m/h
Rime to cross both the trains = ( 1/3 )/120 = 1/360 Hours
Convert it into minutes = 1/6 minutes
Speed of passenger train = y
Sum of their length = s
So time = s/(x – y)when both are in same direction
Time = s/(x + y) when opposite direction
From question s/(x – y) = 2. s/(x + y) so solving this we get, x/y = 3/1
So x : y = 3 : 1
Distance covered by A in x hours = 20x km
Distance covered by B in (x – 1) hours = 25(x – 1) km
20x + 25(x – 1) = 110
45x = 135
x = 3
So, they meet at 10 a.m
t = 15/(60 * 60) = time taken in hours
length = t*rs = 75m
Since the trains are moving in the same direction,the time required to cross each other = (a + b)/(u – v) sec
Here, a = 300 m, b = 400 m and u = 90 km/hr, v = 80 km/hr
a + b = 300 + 400 = 700 m
And, u – v = (90 – 80)km/hr = 10 km/hr
Converting the unit of speed into m/Sec:
10 km/hr = 10 x 5/18 m/sec = 50/18 m/s
Now, the required time = 700 x 18/50 sec = 252 sec
Since the trains are moving in the opposite direction, the time required to cross each other = (a + b)/(u + v) sec
Here, a = 0.55 km = 550 meters, b = 0.45 km = 450 meters and u = 30 km/hr, v = 45 km/hr
a + b = 550 + 450 = 1000 m
And, u + v = (30 + 45)km/hr = 75 km/hr
Converting the unit of speed into m/Sec:
75 km/hr = 75 x 5/18 m/sec = 375/18 m/s
Now, the required time = 1000 x 18/375 sec = 48 sec
And, they are moving in the opposite direction, they take 25 seconds to cross each other
Let the required speed of the trains be X m/sec
Now, from the formula (a + b)/(u + v), we have a = b = 250 m and u = v = X m/sec
Then, a + b = 250 + 250 = 500 m and u + v = 2X m/sec,
25 = 500/2X sec
X = 10 m/sec = 10 x 18/5 km/hr = 36 km/hr
Hence, the speed of two trains is 36 km/hr
Time taken to collide = 200/1200 = 1/6 hours
The bee travels for 1/6 hours. Hence distance covered = 1500 * 1/6 = 250 miles
X + 120/15 = X + 180/18
6X + 720 = 5X + 900
X = 180m
S= 1200/120
S= 10M/Sec
Total length(D)= 1900m
T = D/S
T = 1900/10
T = 190Sec
Time = 30 sec
Let the length of bridge be x metres
Then, (130 + x)/30 = 25/2
2(130 + x) = 750
x = 245 m