A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the journey, what was its first average speed ?
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
Let the first average speed of the train be $x$ km/h.
Setting up the equation:
Time = Distance ÷ Speed
- Time for first part: $\dfrac{54}{x}$ hours
- Time for second part: $\dfrac{63}{x+6}$ hours
Total time = 3 hours, so:
$$\frac{54}{x} + \frac{63}{x+6} = 3$$
Simplifying:
$$54(x+6) + 63x = 3x(x+6)$$
$$54x + 324 + 63x = 3x^2 + 18x$$
$$117x + 324 = 3x^2 + 18x$$
$$3x^2 - 99x - 324 = 0$$
$$x^2 - 33x - 108 = 0$$
Factorising:
$$x^2 - 36x + 3x - 108 = 0$$
$$x(x - 36) + 3(x - 36) = 0$$
$$(x + 3)(x - 36) = 0$$
So $x = -3$ or $x = 36$.
Since speed cannot be negative, $x = -3$ is rejected.
∴ The first average speed of the train is 36 km/h.
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Explanation
- The key step is forming the equation: sum of two time expressions equals total time (3 hours).
- After cross-multiplying and simplifying, you get a standard quadratic $x^2 - 33x - 108 = 0$. (Check your arithmetic here — examiners award marks for each correct step.)
- Always reject the negative root with a reason, as speed must be positive. CBSE awards 1 mark specifically for this reasoning.
- Source: Chapter 4, Quadratic Equations (application-type problems similar to Exercise 4.2, Q.2).