Express and Passenger Train Speed Problem | Class 10 Maths

Quick answer: Let the passenger train's speed be $x\text{ km/h}$, so the express train's speed is $(x + 11)\text{ km/h}$. The passenger train takes $1$ hour longer to cover $132\text{ km}$, so $\frac{132}{x} - \frac{132}{x + 11} = 1$, which gives $x^2 + 11x - 1452 = 0$, i.e. $(x + 44)(x - 33) = 0$. Speed cannot be negative, so $x = 33$: the passenger train travels at 33 km/h and the express train at 44 km/h.

When two vehicles cover the same distance at different speeds, the faster one takes less time. Using $\text{time} = \frac{\text{distance}}{\text{speed}}$ for each vehicle, and the given difference in time, gives an equation that becomes a quadratic.

In CBSE Class 10 Mathematics, Chapter 4 (Quadratic Equations), the express train and passenger train problem is a well-known NCERT question.

What You Will Learn

• How to write the time taken when the speed is unknown
• How to form the time-difference equation
• A full solution of the NCERT train problem
• How to factorise $x^2 + 11x - 1452 = 0$
• Why the negative root is rejected

1. Setting Up a Time-Difference Equation

If a distance $d$ is covered at speeds $x$ and $x + k$:

The slower vehicle takes more time, so:

$\frac{d}{x} - \frac{d}{x + k} = \text{difference in time}$

Key point: Always subtract the faster vehicle's time from the slower vehicle's time, so the difference is positive.

2. The NCERT Train Problem

Question

An express train takes $1$ hour less than a passenger train to travel $132\text{ km}$ between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is $11\text{ km/h}$ more than that of the passenger train, find the average speed of the two trains.

Solution

Step 1: Define the variables. Let the average speed of the passenger train be $x\text{ km/h}$. Then the express train's speed is $(x + 11)\text{ km/h}$.

Step 2: Write the times.

$\text{Passenger train: } \frac{132}{x}\text{ hours} \qquad \text{Express train: } \frac{132}{x + 11}\text{ hours}$

Step 3: Form the equation. The express train takes $1$ hour less:

$\frac{132}{x} - \frac{132}{x + 11} = 1$

Step 4: Simplify.

$\frac{132(x + 11) - 132x}{x(x + 11)} = 1 \implies \frac{1452}{x^2 + 11x} = 1$

$x^2 + 11x - 1452 = 0$

Step 5: Factorise. We need two numbers with product $-1452$ and sum $11$. Since $1452 = 44 \times 33$ and $44 - 33 = 11$, the numbers are $44$ and $-33$:

$x^2 + 44x - 33x - 1452 = 0$

$x(x + 44) - 33(x + 44) = 0$

$(x + 44)(x - 33) = 0$

Step 6: Choose the valid root. $x = -44$ or $x = 33$. Speed cannot be negative, so $x = -44$ is rejected and $x = 33$.

Answer: The passenger train's average speed is 33 km/h, and the express train's is 44 km/h.

Check: Passenger train: $\frac{132}{33} = 4$ hours. Express train: $\frac{132}{44} = 3$ hours. The difference is $1$ hour. ✓

3. Summary Table

4. Exam Tips and Common Mistakes

Exam Tip: Write a sentence rejecting the negative root: "speed cannot be negative, so $x = -44$ is rejected".

Exam Tip: Give both speeds in your final answer, with units.

Common Mistake: Subtracting the times the wrong way round. $\frac{132}{x + 11} - \frac{132}{x} = 1$ leads to $x^2 + 11x + 1452 = 0$, which has no real roots. The slower train's time, $\frac{132}{x}$, must come first.

Common Mistake: Forgetting to multiply $132 \times 11 = 1452$ correctly when simplifying.

Frequently Asked Questions

What are the speeds of the express and passenger trains in the NCERT problem?

The passenger train travels at 33 km/h and the express train at 44 km/h.

What quadratic equation do you get in the train problem?

x² + 11x − 1452 = 0, which factorises as (x + 44)(x − 33) = 0.

Why is x = −44 rejected?

Speed cannot be negative, so only x = 33 makes sense.

How do you set up a time-difference equation?

Write each time as distance ÷ speed, then subtract the faster vehicle's time from the slower vehicle's time and set this equal to the given difference.