Aptitude Train Problems
3 Types Of Train Problems For Bank Exams
Type I: Time Taken by a Train to Cross a Platform or a Man or a Pole
In this type, you have to find the time taken by a train to cross a platform or man or pole. You have to learn an easy formula to solve this type of questions. See the below example.
Example Question 1: A train is moving at a speed of 120 km/hr. If the length of the train is 130 metres, how long will it take to cross a railway platform 170 metres long?
Solution:
From the question, you can write down the below values.
Length of the train = 130 m
Length of the platform = 170 m
Speed of the train = 120 km/h
You have to convert the speed in km/h to m/s. This is because the lengths of the train and platform are given in metres and not in kilometres.
To convert speed from km/h to m/s, you have to multiply it by 5/18.
Speed of the train in m/s = 120 x 5/18 = 100/3 m/s.
Speed of the train in m/s = 120 x 5/18 = 100/3 m/s.
Let Lt be the length of the train and Lp be the length of the platform.
Then, you can find the time taken by the train to cross the platform using the below formula.
Time to cross the platform = Lt + Lp / Speed of the train
You know Lt = 130 m, Lp = 170 m and Speed of the train = 100/3 m/s
Therefore, Answer = (130 + 170) / (100/3)
= 300 x 3 /100 = 9 sec
Type II: Time Taken for 2 Trains to Cross Each Other
In this type, lengths of 2 trains and their speeds will be given. You have to find the time taken for the first train to cross the second.You have to remember 2 simple formulas to solve this type.
Assume below values:
L1 = Length of the first train
L2 = Length of the second train
S1 = Speed of the first train
S2 = Speed of the second train
Case I: Trains travelling in opposite directions:
Time taken for 2 trains to cross each other = (L1 + L2) / (S1 + S2)
Case II: Trains travelling in same direction:
Time taken for 2 trains to cross each other =
(L1 + L2) / (S1 – S2) when S1 > S2
Or
(L1 + L2) / (S2 – S1) when S2 > S1
Here is your example question.
Example Question 2: Two trains of lengths 120 metres and 150 metres respectively are running towards each other on parallel lines. Speed of the first train is 40 km/h and that of the second is 46 km/h. In how much time the first train will cross the second?
Solution:
From the question, you can write down the below values
From the question, you can write down the below values
L1 = 120 m, L2 = 150 m, S1 = 40 km/h and S2 = 46 km/h
As you can see, the speeds are in km/h units but lengths are in metres. Therefore, you have to convert speeds to m/s units.
S1 = 40 x 5/18 m/s
S2 = 46 x 5/18 m/s
According to the formula (that we saw above),
Time taken for the trains to cross each other = 120 + 150 / (40 x 5/18 + 46 x 5/18)
= 120 + 150 / (215/9 )
= 170 x 9 / 215 = 7.12 sec
= 7 sec (approx)
2 Important Points to Note:
1. Instead of 2 trains, if you find a train and a man in question, you have to assume the length of man to be zero.
2. If pole is given in question and length of the pole is not given, you can assume its length to be zero as well.
Type III: Equations Based Train Problems
This type can be an extension to type 1 or type 2 or both. Based on data given, you have to form equations to solve this type. Below is an example.
Example Question 3: A train running at 60 km/h takes 10 seconds to pass a platform. Next it takes 5 seconds to pass a man walking at 6 km/h in the same direction. Find the length of the train and the length of the platform.
Solution:
From the question, you know that the train at 60 km/h takes 10 seconds to cross the platform. (Speed of train in m/s = 60 x 5/18 = 50/3 m/s)
From the question, you know that the train at 60 km/h takes 10 seconds to cross the platform. (Speed of train in m/s = 60 x 5/18 = 50/3 m/s)
Part 1:
In type 1, you saw the below formula.
Time taken for a train to cross a platform = Lt + Lp / Speed of the train
You have to substitute the values (from question) in above equation and simplify as shown below.
10 = Lt + Lp / (50/3)
500 = 3(Lt + Lp)
Or Lt + Lp = 500/3 … equation 1
Part 2:
From the second half of the question, you know that the train takes 5 seconds to cross a man walking 6 km/h in the same direction.
(Speed of man in m/s = 6 x 5/18 = 5/3 m/s)
You know the formula for time taken for 2 trains travelling in same direction to cross each other.
Time taken = (L1 + L2) / (S1 – S2)
If instead of 2 trains, 1 train and 1 man is given, you have to assume his length to be 0.
Therefore L2 = 0.
S1 = Speed of the train = 50/3 m/s,
S2 = man’s speed = 5/3 m/s
and time taken for train to cross the man = 5 seconds
You have to substitute the values in the formula, Time taken = (L1 + L2) / (S1 – S2).
You will get,
5 = L1 + 0 / (50/3 – 5/3)
5 = L1 / 15
L1 = 75 m
Part 3: (Finding Lp)
In part 2, you found L1 = 75m. L1 is nothing but the length of the train Lt.
If you substitute Lt = 75 m in equation 1, you will get,
75 + Lp = 500/3
Lp = 500/3 – 75 = 91.6m
Your answer is as follows.
Length of the train = 75 m and
Length of the platform = 91.6 m
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