Question

A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of the slower train is 10 km/hr less than that of the faster train, find the speeds of the two trains.

Hint:

In speed-distance problems, the equation is usually written as: \(\text{Time}_{\text{slow}} - \text{Time}_{\text{fast}} = \text{Difference}\).

Answer 1

Step 1: Understanding the Concept:
We use the relation:
Time = Distance / Speed

Both trains travel the same distance (200 km).
The slower train takes 1 hour more than the faster train.

Step 2: Let the Speeds be:
Let speed of faster train = x km/hr
Then speed of slower train = (x − 10) km/hr

Time taken by faster train = 200 / x
Time taken by slower train = 200 / (x − 10)

Given difference in time = 1 hour:
200/(x − 10) − 200/x = 1

Step 3: Solving the Equation:
Multiply both sides by x(x − 10):
200x − 200(x − 10) = x(x − 10)

Expand left side:
200x − 200x + 2000 = x² − 10x

So we get:
2000 = x² − 10x

Rearranging:
x² − 10x − 2000 = 0

Step 4: Factorising the Quadratic:
x² − 10x − 2000 = 0

(x − 50)(x + 40) = 0

x = 50 or x = −40

Since speed cannot be negative:
x = 50

Slower train speed = x − 10
= 50 − 10
= 40

Final Answer:
Speed of faster train = 50 km/hr
Speed of slower train = 40 km/hr