RRB Group D Maths Train Concept and Tricks

Train-based questions are one of the most scoring topics in RRB Group D Maths. These questions are easy once you understand the basic logic of speed, time, and distance along with a few important tricks. Shubham Sir’s approach focuses on simplifying concepts, using memory shortcuts, and solving questions quickly with accuracy. If mastered well, this topic can help you save time and boost your overall score in the exam.

Also Read:

• RRB Group D Maths Compound Interest
• RRB Group D Maths Ratio and Proportion

Train Concepts and Tricks

Train problems are an extension of Speed = Distance ÷ Time concepts. The only difference is how we interpret distance in such questions.

Core Idea

In train problems, distance is not always just a straight path—it often includes:

• Length of the train

• Length of the object (if applicable)

Fundamental Formula and its Application to Train Problems

The bedrock for all train problems is the formula: Distance = Speed × Time. The basic formula applies, but for trains, Distance is uniquely defined as the sum of:

1. Length of the Train

2. Length of the Object the train is crossing.

Length of Object in Train Problems

An object's length consideration depends on its size relative to the train.

Objects with Negligible Length:

For objects like a Person, Signal Post, Pole, or Tree, their lengths are not considered. As their length is negligible compared to the train's, the distance covered is only the train's length.

Objects with Significant Length:

For objects like another Train, a Tunnel, Platform, or Bridge, their lengths must be considered and added to the train's length for the total distance covered.

Solved Problem 1: Train Crossing a Platform and a Person

A bullet train crosses a platform in 55 seconds and a person in 30 seconds. Given its speed is 55 m/s, find the platform's length.

Solution:

1. Time to cross train's length: 30 seconds.

2. Time to cross platform's length only: 55 s - 30 s = 25 seconds. (Memory Tip: The difference in time for crossing the platform and a person directly gives the time taken to cover the platform's length.)

3. Length of Platform: 55 m/s × 25 s = 1375 meters.

Solved Problem 2: Train Crossing a Standing Man and a Platform

A train at 36 km/h crosses a standing man in 10 seconds. How much time to cross a 55-meter platform?

Solution:

1. Speed: 36 km/h = $36 \times \frac{5}{18}$ m/s = 10 m/s.

2. Train's Length: 10 m/s × 10 s = 100 meters.

3. Total Distance (Train + Platform): 100 m + 55 m = 155 meters.

4. Time to cross platform: 155 m / 10 m/s = 15.5 seconds.

Solved Problem 3: Train Crossing a Pole

A 300-meter train runs at 60 km/h. Find the time to cross a pole.

Solution:

1. Distance: 300 meters (train's length).

2. Speed: 60 km/h = $60 \times \frac{5}{18}$ m/s.

3. Time: $300 \text{ m} / (60 \times \frac{5}{18} \text{ m/s})$ = 18 seconds.

Solved Problem 4: Train Crossing a Platform and then an Electric Pole

A train crosses a 100-meter platform at 45 km/h in 60 seconds. Find the time to cross an electric pole.

Solution:

1. Speed: 45 km/h = $45 \times \frac{5}{18}$ m/s = 12.5 m/s.

2. Total Distance (Train + Platform): 12.5 m/s × 60 s = 750 meters.

3. Train's Length: 750 m - 100 m = 650 meters.

4. Time to Cross Pole: 650 m / 12.5 m/s = 52 seconds.

Solved Problem 5: Initial Distance of a Pole from Train's Front End

A 110-meter train at 36 km/h takes 53 seconds for its rear end to cross a pole. Find the pole's initial distance from the train's front end.

Solution:

1. Speed: 36 km/h = $36 \times \frac{5}{18}$ m/s = 10 m/s.

2. Total Distance Covered: 10 m/s × 53 s = 530 meters.

3. Initial Distance (x): 530 m = 110 m + x $\implies$ x = 420 meters.

Solved Problem 6: Train Overtaking Two Persons in the Same Direction

A train overtakes two persons (at 2 km/h and 4 km/h) in 9s and 10s respectively. Find its length and speed.

Solution:

Let train speed S km/h, length L meters.

1. Person 1 (2 km/h, 9s): L = $(S - 2) \times \frac{5}{18} \times 9$

2. Person 2 (4 km/h, 10s): L = $(S - 4) \times \frac{5}{18} \times 10$

3. Equating: $(S - 2) \times 9 = (S - 4) \times 10 \implies 9S - 18 = 10S - 40 \implies S = 22$ km/h.

• Train Speed = 22 km/h.

4. Length: L = $(22 - 2) \times \frac{5}{18} \times 9 = 20 \times \frac{5}{2}$ = 50 meters.

• Train Length = 50 meters.

Solved Problem 7: Train Crossing a Platform (Given Train Length)

A train at 54 km/h crosses a platform in 40 seconds. If the train is 150 meters long, what is the platform's length?

Solution:

1. Speed: 54 km/h = $54 \times \frac{5}{18}$ m/s = 15 m/s.

2. Total Distance (Train + Platform): 15 m/s × 40 s = 600 meters.

3. Platform's Length: 600 m - 150 m = 450 meters.

Solved Problem 8: Train Crossing a Tree and then a Tunnel

A 150-meter train crosses a tree in 12 seconds. How much time will it take to cross a 250-meter tunnel?

Solution:

1. Train's Speed: 150 m / 12 s = 12.5 m/s.

2. Total Distance (Train + Tunnel): 150 m + 250 m = 400 meters.

3. Time to Cross Tunnel: 400 m / 12.5 m/s = 32 seconds.

Solved Problem 9: Train Crossing a Man and a Platform (Find Train Length)

A train crosses a man in 8 seconds and a 264-meter platform in 20 seconds. Find the train's length.

Solution:

1. Time for platform's length only: 20 s - 8 s = 12 seconds.

2. Train's Speed: 264 m / 12 s = 22 m/s.

3. Train's Length: 22 m/s × 8 s = 176 meters.

Solved Problem 10: Two Trains Crossing Each Other (Same and Opposite Directions)

Two 225-meter trains. Overtake in 90s (same direction), cross in 10s (opposite). Find faster train's speed.

Solution:

Total Length = 225m + 225m = 450 meters.

1. Same Direction (S1 - S2): 450 = (S1 - S2) × 90 $\implies$ S1 - S2 = 5 m/s.

2. Opposite Direction (S1 + S2): 450 = (S1 + S2) × 10 $\implies$ S1 + S2 = 45 m/s.

3. Solving: Add equations $\implies$ 2S1 = 50 $\implies$ S1 = 25 m/s (or 90 km/h).

Solved Problem 11: Two Trains Crossing Each Other (Same and Opposite Directions) - Find Both Speeds

Two trains (100m, 95m) cross in 27s (same direction) and 9s (opposite). Find both speeds.

Solution:

Total Length = 100m + 95m = 195 meters.

1. Same Direction (S1 - S2): 195 = (S1 - S2) × 27 $\implies$ S1 - S2 = $65/9$ m/s.

2. Opposite Direction (S1 + S2): 195 = (S1 + S2) × 9 $\implies$ S1 + S2 = $65/3$ m/s.

3. Solving:

• Add equations: 2S1 = $65/9 + 65/3 = 260/9 \implies$ S1 = $130/9$ m/s = 52 km/h.

• Substitute S1: S2 = $65/3 - 130/9 = 65/9$ m/s = 26 km/h.

Solved Problem 12: Ratio of Speeds of Two Trains

Two trains running opposite cross a person in 30s and 18s. They cross each other in 25s. Find the ratio of their speeds.

Solution (Alligation Method):

For this specific problem type, alligation provides a quick solution:

• Times to cross a person: T1 = 30s, T2 = 18s

• Time to cross each other: 25s

  T1 (person)       T2 (person)
      30               18
        \             /
          \           /
          \         /
            25 (both)
          /         \
          /           \
        /             \
      |25-18|         |30-25|
        7               5

The ratio of speeds (S1:S2) is directly 7:5. (Memory Tip: For train problems involving individual crossing times for a person and combined crossing time, alligation quickly yields the speed ratio.)

Solved Problem 13: Counting Telegraph Poles

Telegraph poles are 50 meters apart. A train at 45 km/h. How many poles counted in 4 hours?

Solution:

1. Speed: 45 km/h = $45 \times \frac{5}{18}$ m/s = 12.5 m/s.

2. Time: 4 hours = $4 \times 3600$ s = 14400 seconds.

3. Total Distance Traveled: 12.5 m/s × 14400 s = 180000 meters.

4. Number of Intervals: 180000 m / 50 m = 3600 intervals.

5. Number of Poles: 3600 + 1 = 3601 poles. (Memory Tip: Always add 1 to the number of intervals for the total count of objects, as the first object doesn't require an interval.)

Solved Problem 14: Counting Telegraph Poles Per Minute

Poles 50 meters apart. Train at 48 km/h. How many pillars counted in one minute?

Solution:

1. Speed: 48 km/h = $48 \times \frac{5}{18}$ m/s.

2. Distance in 1 minute (60s): $(\frac{48 \times 5}{18}) \times 60 = 800$ meters.

3. Number of Intervals: 800 m / 50 m = 16 intervals.

4. Number of Poles: 16 + 1 = 17 poles. (Memory Tip: The 'first pole' means one more pole than the intervals covered.)

Solved Problem 15: Speed of Train from Counting Poles

A boy counts 31 telegraph poles in 60 seconds. Poles are 60 meters apart. Find the train's speed in km/h.

Solution:

1. Number of Intervals: 31 - 1 = 30 intervals.

2. Total Distance: 30 × 60 m = 1800 meters.

3. Speed in m/s: 1800 m / 60 s = 30 m/s.

4. Speed in km/h: $30 \text{ m/s} \times \frac{18}{5}$ km/h = 108 km/h.