The time taken by a person to travel an upward distance of 150 km was $2\dfrac{1}{2}$ hours more than the time taken in the downward return journey. If he returned at a speed of 10 km/h more than the speed while going up, find the speeds in each direction.
Generated by claude-sonnet-4-6 · 2026-06-15 10:29 · grounding rag
Model Answer
Let the speed while going up = $x$ km/h
∴ Speed while coming down = $(x + 10)$ km/h
Time taken going up = $\dfrac{150}{x}$ hours; Time taken coming down = $\dfrac{150}{x+10}$ hours
Given condition:
$$\frac{150}{x} - \frac{150}{x+10} = \frac{5}{2}$$
$$150(x+10) - 150x = \frac{5}{2} \cdot x(x+10)$$
$$1500 = \frac{5}{2}x(x+10)$$
$$600 = x^2 + 10x$$
$$x^2 + 10x - 600 = 0$$
Factorising:
$$x^2 + 30x - 20x - 600 = 0$$
$$(x + 30)(x - 20) = 0$$
So $x = 20$ or $x = -30$.
Since speed cannot be negative, $x = 20$.
∴ Speed going up = 20 km/h; Speed coming down = 30 km/h.
Source: Chapter 4, Quadratic Equations
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Explanation
- Let the upward speed be $x$; downward speed is $x+10$ (10 km/h more).
- Time difference gives the equation; cross-multiply carefully to form a quadratic.
- $2\frac{1}{2}$ hours = $\frac{5}{2}$ hours — don't forget this conversion.
- Reject the negative root ($x = -30$) since speed must be positive.
- State both answers clearly in the final line — examiners award a mark for this conclusion.