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.
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
Let the speed of the faster train = x km/hr
∴ Speed of slower train = (x – 10) km/hr
Time taken by faster train = 200/x hours
Time taken by slower train = 200/(x – 10) hours
Since the faster train takes 1 hour less:
$$\frac{200}{x-10} - \frac{200}{x} = 1$$
$$200\left(\frac{x - (x-10)}{x(x-10)}\right) = 1$$
$$200 \times 10 = x(x - 10)$$
$$x^2 - 10x - 2000 = 0$$
Factorising:
$$x^2 - 50x + 40x - 2000 = 0$$
$$x(x - 50) + 40(x - 50) = 0$$
$$(x + 40)(x - 50) = 0$$
So, x = 50 or x = –40.
Since speed cannot be negative, x = 50.
∴ Speed of faster train = 50 km/hr
∴ Speed of slower train = 50 – 10 = 40 km/hr
Source: Chapter 4, Exercise 4.2
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Explanation
- Set up the variable for the faster train's speed; express the slower train's speed in terms of it.
- Form the equation using Time = Distance/Speed and the given 1-hour difference.
- Simplify to get a standard quadratic, then factorise (preferred in CBSE).
- Reject the negative root with a reason — examiners specifically look for this step.
- State both answers clearly at the end for full marks.