Introduction to Problems on Trains for RRB Exams
Problems on Trains are among the most frequently tested concepts under the Quantitative Aptitude section of Indian Railway exams, including RRB NTPC (CBT-1 & CBT-2), RRB Group D, RRB ALP, and RRB Technician (Grade I & III). Although it is fundamentally rooted in the core concept of Speed, Distance, and Time, train problems introduce critical variables such as the physical length of the train, stationary obstacles with negligible or \textended lengths (like poles vs. platforms), and the relative speed of moving objects.
Because Indian Railways operates on rail networks, questions evaluating your understanding of train speeds, crossing times, and directional mechanics are considered foundational. Mastering this topic gives you guaranteed scoring opportunities in competitive railway tests.
Topic Weightage and Importance
In almost every shift of RRB exams, train-related arithmetic problems make an appearance. Here is a breakdown of the typical distribution and weightage across key railway recruitment tests:
| Exam Name | Section | Expected Questions | Difficulty Level |
|---|---|---|---|
| RRB NTPC (CBT-1) | Mathematics | 1 to 2 Questions | Easy to Moderate |
| RRB NTPC (CBT-2) | Mathematics | 2 to 3 Questions | Moderate to High |
| RRB Group D | Mathematics | 1 to 2 Questions | Easy to Moderate |
| RRB Technician Grade I & III | Basic Math / Physics & Math | 1 to 2 Questions | Moderate |
With just a handful of straightforward formulas and standard relative speed rules, you can solve these questions in less than 40 seconds per question.
Key Concepts and Formulas
The standard equation governing all motion problems is:
$$\text{Distance} = \text{Speed} \times \text{Time}$$
However, when dealing with trains, the definition of "Distance" and "Speed" adapts depending on the physical scenario.
1. Conversion of Units (Crucial First Step)
In railway exams, speed is typically given in $\text{km/h}$ while distance and train lengths are given in meters ($\text{m}$), and time in seconds ($\text{s}$).
- To convert from $\text{km/h}$ to $\text{m/s}$: Multiply by $\frac{5}{18}$
$$\text{Speed in m/s} = \text{Speed in km/h} \times \frac{5}{18}$$ - To convert from $\text{m/s}$ to $\text{km/h}$: Multiply by $\frac{18}{5}$
$$\text{Speed in km/h} = \text{Speed in m/s} \times \frac{18}{5}$$
2. Crossing Stationary Objects of Negligible Length
When a train passes a point object (e.g., an electric pole, a signal post, a standing person, or a milestone), the length of the object is taken as zero ($0$).
- $$\text{Total Distance} = L_T \text{ (Length of the Train)}$$
- $$\text{Time to cross} = \frac{L_T}{S_T}$$ (where $S_T$ is the speed of the train in $\text{m/s}$)
3. Crossing Stationary Objects of Extended Length
When a train passes a platform, bridge, tunnel, or a stationary train, the train must cover its own length plus the length of that structure to completely clear it.
- $$\text{Total Distance} = L_T + L_P \text{ (Length of Train + Length of Platform/Bridge/Tunnel)}$$
- $$\text{Time to cross} = \frac{L_T + L_P}{S_T}$$
4. Relative Speed: Two Moving Trains / Objects
When two objects (e.g., two trains, or a train and a running man) are moving with speeds $u$ and $v$:
| Case | Direction | Relative Speed | Total Distance to Cross Each Other |
|---|---|---|---|
| Opposite Directions | Moving towards or away from each other ($\rightarrow \leftarrow$) | $$S_{\text{rel}} = u + v$$ | $$D = L_1 + L_2$$ |
| Same Direction | Moving in the same direction ($\rightarrow \rightarrow$) | $$S_{\text{rel}} = u - v \text{ (where } u > v)$$ | $$D = L_1 + L_2$$ |
5. Advanced Shortcut: Meeting Point Problem
If two trains start at the same time from points $A$ and $B$ towards each other, and after meeting they take $T_1$ and $T_2$ hours respectively to reach their destinations $B$ and $A$, their speeds are related by:
$$\frac{S_1}{S_2} = \sqrt{\frac{T_2}{T_1}}$$
Solved Examples (Step-by-Step)
Example 1: Crossing a Pole
Question: A train 240 meters long is running at a speed of $72\text{ km/h}$. How much time will it take to cross a telegraph pole?
Step-by-Step Solution:
- Step 1: Convert the speed into $\text{m/s}$.
$$\text{Speed} = 72 \times \frac{5}{18} = 4 \times 5 = 20\text{ m/s}$$ - Step 2: Identify the distance. Since a pole has negligible length, Distance = Length of Train = $240\text{ m}$.
- Step 3: Calculate time.
$$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{240}{20} = 12\text{ seconds}$$
Answer: 12 seconds
Example 2: Crossing a Bridge
Question: A train running at $54\text{ km/h}$ takes 30 seconds to cross a bridge of length 250 meters. Find the length of the train.
Step-by-Step Solution:
- Step 1: Convert speed to $\text{m/s}$.
$$\text{Speed} = 54 \times \frac{5}{18} = 15\text{ m/s}$$ - Step 2: Use the formula for crossing a bridge:
$$\text{Distance} = L_T + L_B = L_T + 250$$ - Step 3: Calculate total distance covered in 30 seconds:
$$\text{Total Distance} = \text{Speed} \times \text{Time} = 15 \times 30 = 450\text{ m}$$ - Step 4: Solve for $L_T$:
$$L_T + 250 = 450 \implies L_T = 450 - 250 = 200\text{ m}$$
Answer: 200 meters
Example 3: Two Trains Crossing Each Other (Opposite Directions)
Question: Two trains of lengths $160\text{ m}$ and $140\text{ m}$ are moving in opposite directions on parallel tracks at speeds of $40\text{ km/h}$ and $32\text{ km/h}$ respectively. Find the time taken by them to completely cross each other.
Step-by-Step Solution:
- Step 1: Calculate total distance to be covered:
$$\text{Total Distance} = L_1 + L_2 = 160 + 140 = 300\text{ m}$$ - Step 2: Calculate relative speed in opposite directions:
$$S_{\text{rel}} = 40 + 32 = 72\text{ km/h}$$ - Step 3: Convert relative speed to $\text{m/s}$:
$$72 \times \frac{5}{18} = 20\text{ m/s}$$ - Step 4: Calculate crossing time:
$$\text{Time} = \frac{\text{Total Distance}}{\text{Relative Speed}} = \frac{300}{20} = 15\text{ seconds}$$
Answer: 15 seconds
Example 4: Meeting Point Shortcut
Question: Train A starts from Delhi to Mumbai, and Train B starts from Mumbai to Delhi at the same time. After meeting each other on the way, Train A takes 4 hours to reach Mumbai, and Train B takes 9 hours to reach Delhi. If Train A's speed is $60\text{ km/h}$, find the speed of Train B.
Step-by-Step Solution:
- Step 1: Apply the formula $\frac{S_A}{S_B} = \sqrt{\frac{T_B}{T_A}}$.
- Step 2: Substitute the given values: $T_A = 4\text{ hrs}$, $T_B = 9\text{ hrs}$, $S_A = 60\text{ km/h}$.
$$\frac{60}{S_B} = \sqrt{\frac{9}{4}} = \frac{3}{2}$$ - Step 3: Solve for $S_B$:
$$S_B = \frac{60 \times 2}{3} = 40\text{ km/h}$$
Answer: $40\text{ km/h}$
Common Mistakes to Avoid
- Ignoring Unit Inconsistencies: Mixing $\text{km/h}$ with meters or seconds without converting via $\frac{5}{18}$ or $\frac{18}{5}$ is the #1 reason candidates lose marks.
- Forgetting Train's Own Length: When a train crosses an \textended object like a platform or another train, candidates often forget to add both lengths together ($L_1 + L_2$).
- Confusing Relative Speed Operations: Remember: Same direction requires subtraction ($u - v$), while Opposite direction requires addition ($u + v$).
- Man Inside a Moving Train: When a train passes a man sitting inside another moving train, the distance covered is ONLY the length of the passing train, NOT the sum of both trains!
Practice Questions with Solutions
Practice Questions
Q1: A train 180 meters long takes 9 seconds to pass a standing person. What is the speed of the train in km/h?
Q2: A train passes a 300 m long platform in 38 seconds and a signal pole in 18 seconds. Find the length and speed of the train.
Q3: A train 150 m long crosses a man walking at $6\text{ km/h}$ in the same direction in 15 seconds. Find the speed of the train in km/h.
Q4: Two trains running at $50\text{ km/h}$ and $30\text{ km/h}$ in the same direction cross each other in 36 seconds. If the faster train is 120 m long, find the length of the slower train.
Q5: Two stations X and Y are 390 km apart. A train starts from X at 8:00 AM at $65\text{ km/h}$ towards Y. Another train starts from Y at 9:00 AM at $70\text{ km/h}$ towards X. At what time will they meet?
Detailed Solutions
Solution to Q1:
$$\text{Speed in m/s} = \frac{180}{9} = 20\text{ m/s}$$
$$\text{Speed in km/h} = 20 \times \frac{18}{5} = 72\text{ km/h}$$
Solution to Q2:
Time taken to cross the pole = $\frac{L}{S} = 18\text{ s} \implies L = 18S$
Time taken to cross the platform = $\frac{L + 300}{S} = 38\text{ s}$
Substitute $L = 18S$:
$$18S + 300 = 38S \implies 20S = 300 \implies S = 15\text{ m/s}$$
$$\text{Speed} = 15 \times \frac{18}{5} = 54\text{ km/h}$$
$$\text{Length of train } L = 18 \times 15 = 270\text{ meters}$$
Solution to Q3:
Let the speed of the train be $S\text{ km/h}$. Since both move in the same direction, relative speed is $(S - 6)\text{ km/h}$.
$$\text{Relative speed in m/s} = \frac{150\text{ m}}{15\text{ s}} = 10\text{ m/s}$$
$$\text{Convert to km/h} = 10 \times \frac{18}{5} = 36\text{ km/h}$$
$$S - 6 = 36 \implies S = 42\text{ km/h}$$
Solution to Q4:
Relative speed in same direction = $50 - 30 = 20\text{ km/h} = 20 \times \frac{5}{18} = \frac{50}{9}\text{ m/s}$.
Total distance = $L_1 + L_2 = 120 + L_2$
$$\text{Distance} = \text{Relative Speed} \times \text{Time} \implies 120 + L_2 = \frac{50}{9} \times 36 = 200\text{ m}$$
$$L_2 = 200 - 120 = 80\text{ meters}$$
Solution to Q5:
In the first hour (from 8:00 AM to 9:00 AM), Train 1 travels alone at $65\text{ km/h}$.
$$\text{Distance covered in 1 hr} = 65\text{ km}$$
$$\text{Remaining distance at 9:00 AM} = 390 - 65 = 325\text{ km}$$
From 9:00 AM onwards, both trains move towards each other (opposite direction):
$$\text{Relative speed} = 65 + 70 = 135\text{ km/h}$$
$$\text{Time to meet after 9:00 AM} = \frac{325}{135} = \frac{65}{27}\text{ hours} \approx 2\text{ hours } 24\text{ minutes}$$
$$\text{Meeting time} = 9:00\text{ AM} + 2\text{ hours } 24\text{ minutes} = 11:24\text{ AM}$$
Frequently Asked Questions (FAQs)
1. How do I know when to add or subtract speeds in train problems?
If two moving entities travel in opposite directions (towards each other or away from each other), they close or open the gap faster, so you add their speeds ($u + v$). If they travel in the same direction, the gap closes slowly, so you subtract the slower speed from the faster speed ($u - v$).
2. Why is the length of a person or a pole taken as zero?
Compared to the physical length of an entire train (which is often 100–500 meters), the width of a standing man, pole, or tree is negligible (under a meter). Hence, it is counted as zero distance in quantitative aptitude problems.
3. What should I do if a question involves a man sitting inside a moving train?
Treat the man as a moving point object with zero length. His speed is equal to the speed of the train he is sitting in. The distance to cross him is solely the length of the other train that is passing him.
Conclusion and Final Tips
Problems on Trains are high-scoring, predictable, and simple once you master unit conversions ($5/18$ and $18/5$) and relative speed rules. To secure maximum marks in your upcoming RRB NTPC, Group D, or Technician exams, practice 20–30 varied problems, keep unit conversions at your fingertips, and always draw a quick mental timeline for two-train departure problems. Consistent practice will guarantee speed and accuracy on exam day!
