Arithmetic Aptitude : Problems on Trains

Problems on Trains

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Hello everyone! In the previous blog, you saw that aptitude isn't just a simple term. it has many sections and multiple types of questions. First things first: if this is your first time visiting my blog, I strongly recommend you read the previous one to understand the modules and structure properly.

Now, let's dive straight into the very first concept in Arithmetic Aptitude - Problems on Trains. This is a really simple and easy topic to understand once you know the formulas. I'll show you a question first, then explain the concepts. that's my way of teaching!

Type 1: Train Crossing a Stationary Point Object

Let's start with a basic question to set the foundation.

Question 1: Divy is standing at a railway station. The Chennai Express crosses Divy in 20 seconds, and the length of the train is 500 meters. Can you find the speed of the train?

This is a beginner-level question — if you can't solve it right away, don't feel bad! I'll explain everything step by step. For this question, we need to know the basic formula:

Speed = Length / Time

So plugging in the values:

Speed = 500 m / 20 seconds = 25 m/s

Simple, right? But here's the thing — most aptitude questions play with unit conversion. For example, from Question 1, the Chennai Express speed is 25 m/s. Can we convert this into km/h? Yes! Just multiply by 18/5:

25 × (18/5) = 90 km/h

Now you're probably thinking — what if we need to convert km/h back to m/s? I've got you covered! Just reverse the formula:

90 × (5/18) = 25 m/s

This is just a Type 1 question. Examiners can twist it in multiple ways. For example, instead of speed, they might ask for the length of the train or the crossing time. They might also change the crossing element — like a stick or a car. If they don't mention the length of the element, we take it as zero.

The same formula applies — just rearrange it:

• If asked for length: 25 = Length / 20 → Length = 500 meters
• If asked for time: 25 = 500 / Time → 25T = 500 → T = 20 seconds

That's it! Now let's move to Type 2.

Type 2: Train Crossing a Platform or Bridge

In Type 1, we saw that when the crossing element (like a car or a person) has no significant length, we take it as zero. Type 2 is all about what happens when that element does have a length. Most commonly, they use a bridge or platform as the crossing point.

Question 2: A train crosses the Nine-Ark Bridge in 30 seconds. The train length is 500 meters, and the speed of the train is 180 km/h. Find the length of the bridge.

First, look at the units — meters, seconds, and km/h? We can't use them directly. Since the speed is in km/h, we need to convert it to m/s first:

180 × (5/18) = 50 m/s

Now here's the formula for this type:

Speed = [Length of Train + Length of Bridge] / Time

We have: Speed = 50 m/s, Length of train = 500 m, Time = 30 seconds

So plugging in:

50 = [500 + Bridge Length] / 30
50 × 30 = 500 + Bridge Length
1500 = 500 + Bridge Length
Bridge Length = 1000 meters

And just like Type 1, the question can swap what you need to find. Just apply what you have and do the math — the formula remains the same!

Practice Problems for Type 2:

1. A 200-meter train crosses a platform in 30 seconds at 54 km/h. Find the platform length.
2. A train crosses a 150-meter bridge in 25 seconds at 72 km/h. Find the train length.
3. A 300-meter train crosses a 200-meter platform at 60 km/h. Find the time taken.
4. A train crosses a 250-meter bridge in 40 seconds. If the train is 150 meters long, find its speed in km/h.

(Answers are at the end of the blog.)

Type 3: Two Trains Crossing in Opposite Directions

Now things get interesting! In this type, we have two trains moving toward each other (opposite directions). The key idea here is that their speeds add up because they're approaching each other.

Question 3: Two trains, 150 meters and 100 meters long, run toward each other at 54 km/h and 36 km/h. In what time do they cross each other completely?

Here's the formula for opposite directions:

Time = [Length of Train 1 + Length of Train 2] / [Speed of Train 1 + Speed of Train 2]

First, convert speeds to m/s:

54 × (5/18) = 15 m/s
36 × (5/18) = 10 m/s

Now apply the formula — note: we ADD both lengths and ADD both speeds (this is where many students make a mistake!):

Time = [150 + 100] / [15 + 10]
Time = 250 / 25
Time = 10 seconds

Got it? Opposite directions = speeds add up. Simple!

Type 4: Two Trains Crossing in the Same Direction

Alright, now let's flip the scenario. What if both trains are going the same way? This is about overtaking. The faster train catches up to the slower one, and the key idea here is that we use the difference of their speeds. From the slower train's perspective, the faster one creeps past at the speed difference. The distance to cover is still the sum of both lengths.

Question 4: Train A (200 meters, 72 km/h) overtakes Train B (150 meters, 54 km/h) running in the same direction. How long does the overtaking take?

Here's your formula for same direction:

Time = [Length of Faster Train + Length of Slower Train] / [Faster Speed − Slower Speed]

Let's solve step by step:

Step 1: Convert speeds to m/s:
72 × (5/18) = 20 m/s (Train A)
54 × (5/18) = 15 m/s (Train B)

Step 2: Find relative speed (same direction = subtract):
20 − 15 = 5 m/s

Step 3: Total distance to cover (sum of both lengths):
200 + 150 = 350 meters

Step 4: Apply the formula:
Time = 350 / 5 = 70 seconds

Let me show you one more twist — what if they ask for the length of the overtaken train instead?

Another Example: Train P (180 meters, 80 km/h) overtakes Train Q in the same direction in 54 seconds. Train Q runs at 44 km/h. Find the length of Train Q.

Step 1: Convert speeds:
80 × (5/18) = 200/9 ≈ 22.22 m/s
44 × (5/18) = 110/9 ≈ 12.22 m/s

Step 2: Relative speed:
200/9 − 110/9 = 90/9 = 10 m/s

Step 3: Total distance covered in overtaking:
Speed × Time = 10 × 54 = 540 meters

Step 4: Length of Train Q = Total distance − Length of Train P:
540 − 180 = 360 meters

Practice Problem 4:
Train X (160 meters) overtakes Train Y (140 meters) in 60 seconds. If Train Y runs at 36 km/h, find the speed of Train X in km/h.
(Answer at the end!)

Type 5: Train Passing a Man Sitting in Another Train

This is a clever variation that appears frequently in exams. When a train passes a passenger or object inside another moving train, the passenger has negligible length compared to the trains. So the moving train covers a distance equal to its own length only — not the sum! But we still use the relative speed between the two trains.

Question 5: A train 180 meters long runs at 63 km/h. It passes a man sitting in another train moving at 45 km/h. Find the time taken if the trains move (a) in opposite directions, (b) in the same direction.

Key insight: We use only the length of the train that is doing the passing — not both!

Step 1: Convert speeds to m/s:
63 × (5/18) = 17.5 m/s
45 × (5/18) = 12.5 m/s

(a) Opposite Directions

Relative speed = speeds add up:
17.5 + 12.5 = 30 m/s
Time = Length of passing train / Relative speed
Time = 180 / 30 = 6 seconds

(b) Same Direction

Relative speed = subtract:
17.5 − 12.5 = 5 m/s
Time = 180 / 5 = 36 seconds

Notice the dramatic difference — 6 seconds vs. 36 seconds! Direction really matters here.

Practice Problem 5:
A man in a train traveling at 60 km/h observes that a goods train, moving in the opposite direction, takes 9 seconds to pass him. If the goods train is 250 meters long, find its speed.
(Answer at the end!)

Type 6: Advanced — Finding Speed from Two Crossing Scenarios

Now we're getting into the advanced stuff! A common tricky problem gives you two different crossing scenarios for the same train and asks for its speed or length. You'll need to set up two equations and solve them simultaneously. Don't worry — it sounds harder than it is!

Question 6: A train crosses a pole in 10 seconds and a 200-meter platform in 20 seconds. Find the length and speed of the train.

Let's define our variables first:
Let L = length of train in meters
Let S = speed in m/s

Step 1: From crossing the pole (Type 1 scenario):
S = L / 10 → L = 10S (Equation 1)

Step 2: From crossing the platform (Type 2 scenario):
S = [L + 200] / 20 → 20S = L + 200 (Equation 2)

Step 3: Substitute Equation 1 into Equation 2:
20S = 10S + 200
10S = 200
S = 20 m/s

Step 4: Now find L:
L = 10 × 20 = 200 meters

Step 5: Convert speed to km/h:
20 × (18/5) = 72 km/h

So the train is 200 meters long and runs at 72 km/h. Two equations, two variables — solved!

Practice Problem 6:
A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/h, find the length of the platform.
(Answer at the end!)

Summary of All Formulas

Scenario	Distance Covered	Effective Speed	Time Formula
Train crosses pole/man	Length of train only	Train speed (V)	T = L / V
Train crosses platform	L_train + L_platform	Train speed (V)	T = (L₁ + L₂) / V
Two trains opposite	L₁ + L₂	V₁ + V₂	T = (L₁ + L₂) / (V₁ + V₂)
Two trains same direction	L₁ + L₂	V_fast − V_slow	T = (L₁ + L₂) / (V_f − V_s)
Train passes man in another train (opposite)	Length of passing train only	V₁ + V₂	T = L / (V₁ + V₂)
Train passes man in another train (same)	Length of passing train only	V_fast − V_slow	T = L / (V_f − V_s)

Solutions to Practice Problems

Type 2 Practice Solutions

a) Speed = 54 × 5/18 = 15 m/s. Total distance = 15 × 30 = 450 m. Platform = 450 − 200 = 250 m

b) Speed = 72 × 5/18 = 20 m/s. Total distance = 20 × 25 = 500 m. Train length = 500 − 150 = 350 m

c) Speed = 60 × 5/18 = 50/3 m/s. Total distance = 300 + 200 = 500 m. Time = 500 ÷ (50/3) = 30 seconds

d) Total distance = 250 + 150 = 400 m. Speed = 400/40 = 10 m/s. In km/h: 10 × 18/5 = 36 km/h

Practice Problem 4

Total distance = 160 + 140 = 300 m. Relative speed = 300/60 = 5 m/s. Train Y speed = 36 × 5/18 = 10 m/s. Train X speed = 10 + 5 = 15 m/s. In km/h: 15 × 18/5 = 54 km/h

Practice Problem 5

Let goods train speed = V km/h. Relative speed = (60 + V) km/h = (60 + V) × 5/18 m/s. Time = Length / Relative speed → 9 = 250 / [(60 + V) × 5/18] → 9 × (60 + V) × 5 = 250 × 18 → (60 + V) × 45 = 4500 → 60 + V = 100 → V = 40 km/h

Practice Problem 6

Speed = 54 × 5/18 = 15 m/s. Train length (from crossing man) = 15 × 20 = 300 m. Total distance crossing platform = 15 × 36 = 540 m. Platform length = 540 − 300 = 240 meters

Wrapping Up

And there you have it — Problems on Trains broken down into six clear types! Let me quickly recap what we've learned:

• Type 1 is the foundation — train crossing a stationary object with negligible length. Master the basic Speed = Length/Time formula and unit conversions (×18/5 for m/s→km/h, ×5/18 for km/h→m/s).
• Type 2 adds length to the crossing element (platform/bridge). Just add both lengths in the formula.
• Type 3 brings in two trains moving opposite — speeds add up, lengths add up.
• Type 4 is same direction overtaking — speeds subtract, lengths still add up.
• Type 5 is the tricky one — passing a man inside another train uses only the passing train's length, but still uses relative speed.
• Type 6 is advanced — set up two equations from two scenarios and solve simultaneously.

The most common mistakes students make are: forgetting unit conversions, mixing up when to add vs. subtract speeds, and using both train lengths in Type 5 problems. If you can avoid these three pitfalls, you'll ace every train problem that comes your way!

Practice each type until it becomes second nature. Start with the easy ones, build your confidence, then tackle the advanced problems. Remember — aptitude is all about patterns, and once you recognize the type, the formula practically writes itself!

Happy problem-solving, and I'll see you in the next blog where we tackle another exciting aptitude topic!