Problems on trains is an application of Time, speed and distance. The first section in this topic is an introduction where we look at the important points on time, speed and distance that will be helpful to solve problems on trains. The exercises contain solved question and answers on problems on trains. When solving exercises, we recommend you to try solving the problems orally and with out using equations.
Introduction to Time, Speed and Distance
As mentioned earlier, problems on trains is an application of problems based on Time, speed and distance. Hence lets look at the important aspects of Time, Speed and Distance to formulate an approach to solve problems on trains.
Points on Distance
- When a train is crossing a pole or a man, distance traveled by the train is equal to length of the train.
- When a train of length L is crossing a bridge or tunnel or platform of length B; the distance traveled by train is equal to L+B
- When a train of length L1 is crossing/ overtaking another train L2; then distance traveled is equal to L1+ L2
Points on Speed
- If two trains A and B are moving in same directions at speeds U & V respectively, then the speed of A with reference to B is U-V
- If two trains A and B are moving in opposite directions at speeds U & V respectively, then the speed of A with reference to B is U+V
Points on Time
- Time taken by a train of Length L to cross a stationary point such as pole is equal to the time taken by the train to cross the Length L
- Time taken by a train of Length L to cross a stationary object such as platform of length B is equal to the time taken by the train to cross the Length L+B
- Time taken by two trains of length L and B moving in same direction at speed U and V respectively is $\frac{L+B}{U-V}$
- Time taken by two trains of length L and B moving in opposite direction at speed U and V respectively is $\frac{L+B}{U+V}$
Problems
Problems on Trains - Exercise - Blog
15 questions
- Find the time taken by a train of length 180m to cross a pole at 36km/hr.
- 5 seconds
- 9 seconds
- 10 seconds
- 18 seconds
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Answer: 18 seconds
Time taken = length of the train / Speed of the train
= $\frac{180}{36 \text{x}\frac{5}{18}}$ = 18 seconds
- Find the time taken by a train of length 110m to cross a platform of length 40m at a speed of 60 km/hr.
- 5.4 seconds
- 9 seconds
- 12.5 seconds
- 15 seconds
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Answer: 9 seconds
Time taken = Length of the train + length of the platform / Speed of the train
= 110 + $\frac{40}{60 \text{x} \frac{5}{18}}$
= $\frac{150}{300}$ x 18
= 9 seconds
- Find the speed of the train of length 200m, if it takes 30 secs to cross a man moving in the same direction as the train at 9 km/hr.
- 122.4 km/hr
- 100 km/hr
- 120 km/hr
- 112.5 km/hr
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Answer: 122.4 km/hr
Relative speed = Speed of the train - speed of man (Since the man is moving in the same direction as train)
Let speed of train = x km/hr
Relative speed = ( x - 9 ) x $\frac{5}{18}$ m/s
Time taken to cross the man = Length of the train / Relative speed
30 = $\frac{200}{x-9}$ x $\frac{18}{5}$
5$x$ - 50 = 120.
$x$ = 34 m/s = 34 * $\frac{18}{5}$ = 122.4 km/hr
- A man is walking in the opposite direction of a train of length 240m at a speed of 6 km/hr. The train is travelling at speed 60 km/hr. If the distance between the man and train is 110 m, find the time taken by the train to cross the man.
- 18 seconds
- 12 seconds
- 24 seconds
- 30 seconds
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Answer: 18 seconds
Relative speed = Speed of train + Speed of man (Since man is cycling in the direction opposite to train)
= 60 km/hr + 10 km/hr = 70 x $\frac{5}{18}$ m/s
Total distance to be travelled = Length of the train + Distance between the man and train
= 240 + 110 = 350 m
Time taken to cross the man = $\frac{\text{Total distance to be travelled}}{\text{Relative Speed}}$
= $\frac{350}{70 \text{x} \frac{5}{18}}$
= 18 seconds.
- Find the length of the train, given that it takes 20 secs to cross a platform of length 150m at speed of 54 km/hr.
- 100 mts
- 150 mts
- 200 mts
- 250 mts
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Answer: 150 mts
In 20 seconds, the distance travelled by the train = length of the train + length of the platform
Let the length of the train be $x$ metres.
Length of the platform = 150m
Speed of the train = 54 x $\frac{5}{18}$ = 15 m/s
150 + $x$ = Speed of the train x Time taken to cross the platform
= 15 x 20
$x$ = 150 mts.
- If a train of certain length takes 12 secs to cross a pole and 30 secs to cross a platform of length 270 m at same speed. find the length of the train.
- 150 mts
- 180 mts
- 120 mts
- 210 mts
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Answer: 180 mts
Let the length of the train be $x$.
Since speed is constant
$\frac{\text{Length of the train}}{\text{Time taken to cross the pole}}$
= $\frac{\text{Length of the train + Length of the platform}}{\text{Time taken to cross the platform}}$
$\frac{x}{12} =\frac{x + 270}{30}$
15$x$ = 6$x$ + 1620
9$x$ = 1620
$x$ = 180m
Hence, length of the train = 180m
- Find the length of the platform, If a train travelling at a speed of 40 km/hr takes 27 secs to cross a platform and 10 secs to cross a man moving at 5 km/hr.
- 100 mts
- 125 mts
- 150 mts
- 175 mts
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Answer: 175 mts
Speed of the train = 40 km/hr
Speed of man = 5 km/hr
Relative speed = 40 + 5 = 45 km/hr = 45 x $\frac{5}{18}$ m/s
Length of the train = Relative speed x time taken to cross the man
= 45 x $\frac{5}{18}$ x 10
= 125 m
Let the length of the platform be $x$. Then,
length of the platform + length of the train = Speed of the train x Time taken to cross the platform
$x$ + 125 = 40 x $\frac{5}{18}$ x 27
$x$ + 125 = 300
$x$ = 175 m
Therefore, length of the platform = 175m
- A train overtakes one man walking at speed of 6 km/hr in the same direction as the train in 20 seconds and the crosses another man cycling in opposite direction in 10 seconds. What is the speed of the train?
- 79.2 km/hr
- 61.4 km/hr
- 86.5 km/hr
- 90.3 km/hr
Show the answer
Answer: 79.2 km/hr
Let the speed of the train be $s$.
Since the distance travelled by the train when overtaking both the people is same.
Relative Speed of train wrt to walking man x Time taken to cross the man walking = Relative Speed of train wrt to cycling man x Time taken to cross the man cycling
($s$-6) x 5/18 x 20 = ($s$+10) x 5/18 x 10
($s$-6) x 20 = ($s$+10) x 10
$s$ = 22 m/s
= 22 x 18/5 = 79.2 km/hr
- Two trains each of length 240m and 310m are traveling in opposite direction at speed of 40km/hr and 50 km/hr respectively. Find the time taken by the trains to cross each other.
- 22 seconds
- 12 seconds
- 25 seconds
- 16 seconds
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Answer: 22 seconds
Time taken by trains to cross each other = Sum of length of both the trains / Relative speed of the trains.
Sum of length of both the trains = 240 + 310 = 550m
Relative speed of the trains = 50 + 40 = 90 km/hr = 90 x $\frac{5}{18}$ = 25m/s
Time taken by trains to cross each other = $\frac{550}{25}$ = 22 seconds
- Two trains of length 240m and 300m take 54 secs and 36 secs to cross each other in same and opposite directions respectively. Find the speed of both the trains in km/hr.
- 45 km/hr and 9 km/hr
- 36 km/hr and 9 km/hr
- 45 km/hr and 12 km/hr
- 45 km/hr and 36 km/hr
Show the answer
Answer: 45 km/hr and 9 km/hr
Let speed of train of length 240 m be ‘u’ and speed of train length 300 m be ‘v’
Then,
u - v = $\frac{240 + 300}{54}$
= 10 m/s = 36 km/hr
u + v = $\frac{240 + 300}{36}$
= 15 m/s = 54 km/hr
Solving for u and v,
u = 45 km/hr
v = 9 km/hr
- Two trains traveling at a speed of 40 km/hr and 30 km/hr in opposite directions cross each other in 45 seconds. Find the sum of the length of both the trains.
- 765 mts
- 875 mts
- 820 mts
- 790 mts
Show the answer
Answer: 875 mts
Sum of length of both the trains = Relative speed of the faster train x time taken to cross the other train
= (40+30) x $\frac{5}{18}$ x 45
= 875 m
- A train T1 of length 200m traveling at a speed of 120 km/hr takes 63 secs to overtake train T2 traveling in the same direction at a speed 90 km/hr. If T2 takes 15 secs to cross a platform at the same speed. Find the length of the platform.
- 160 mts
- 150 mts
- 175 mts
- 180 mts
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Answer: 175 mts
Here, we first have to find the length of T2. Then using length of T2, we can find the length of the platform.
Let the length of T2 be $x$ and length of platform be $y$
$x$ + 200 = (120-90) x $\frac{5}{18}$ x 63
$x$ + 200 = 350
$x$ = 150 m
Therefore length of T2 is 150m
length of the platform + length of T2 = 90 x $\frac{5}{18}$ x 20
$y$ + 150 = 375
$y$ = 175 m
Therefore, Length of the platform = 175 m
- Two trains T1 and T2 of length 350 m and 450 m respectively traveling in opposite directions cross each other in 40 secs. If a person sitting in T2 notices that T1 crosses him 36 secs. Find the speed of both the trains.
- 35 km/hr & 35 km/hr
- 35 km/hr & 37 km/hr
- 30 km/hr & 35 km/hr
- 30 km/hr & 37 km/hr
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Answer: 35 km/hr & 37 km/hr
Since a person sitting in T2 notices that T1 crosses him in 36 secs,
Speed of T1 = $\frac{\text{Length of T1}}{\text{Time taken to cross the person in T2}}$
= $\frac{350}{36}$ x $\frac{18}{5}$ = 35 km/hr
Now, since the trains are travelling in opposite directions,
Speed of T1 + Speed of T2 = $\frac{350 + 450}{40}$ x $\frac{18}{5}$
Speed of T2 + 35 = 72
Speed of T2 = 37 km/hr
- Two trains of equal length take 20 secs and 30 secs to cross each other in opposite direction respectively. Find the ratio of their velocities.
- 2 : 1
- 2 : 3
- 1 : 1
- 1 : 2
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Answer: 1 : 1
Let the lengths of each train be l and their speeds be u and v m/s respectively. Then
Case 1: When the trains cross each other in opposite directions
$\frac{2l}{u+v}$ = 20 …… A
Case 2: When the trains cross each other in same directions
$\frac{2l}{u-v}$ = 30 …… B
Dividing A by B
$\frac{u - v}{u + v} = \frac{3}{2}$
2u - 2v = 3u + 3v
u = v.
Hence, the ratio of the velocity of the two trains in 1:1
- A train T1 of length 140m crosses pole in 10 secs, while another train T2 of length 180 m crosses the pole in 15 secs. Find the time taken by faster train to overtake slower train.
- 80 seconds
- 120 seconds
- 160 seconds
- 200 seconds
Show the answer
Answer: 160 seconds
Speed of train T1 = $\frac{140}{10}$ = 14 m/s
Speed of train T2 = $\frac{180}{15}$ = 12m/s
Time taken by faster train to overtake slower train = Distance travelled by faster train / Relative speed of faster train wrt slower train
= $\frac{140 + 180}{14 - 12}$
= 160 seconds
