A rail replacement bus service and a direct express bus both cover the \(60\) mile route between two towns, as summarised in the table below. The express bu...
Question 1 Report
A rail replacement bus service and a direct express bus both cover the \(60\) mile route between two towns, as summarised in the table below. The express bus travels \(10\) mph faster than the replacement bus and completes the journey \(1\) hour sooner.
| Service | Distance (miles) | Speed (mph) | Time (hours) |
|---|---|---|---|
| Replacement bus | 60 | x | 60/x |
| Express bus | 60 | x + 10 | 60/(x + 10) |
- Show that \(x^2 + 10x - 600 = 0\). (3)
- Solve the equation to find \(x\), given that \(x \gt 0\). (2)
Answer Details
(a) The replacement bus takes \(1\) hour longer than the express bus, so \(\dfrac{60}{x}-\dfrac{60}{x+10}=1\). Multiplying every term by \(x(x+10)\) to clear both fractions: \(60(x+10)-60x=x(x+10)\). Expanding, \(60x+600-60x=x^{2}+10x\), so the \(60x\) terms cancel, leaving \(600=x^{2}+10x\). Rearranging gives \(x^{2}+10x-600=0\), as required. [3 marks]
(b) Using the quadratic formula with \(a=1\), \(b=10\), \(c=-600\): \(x=\dfrac{-10\pm\sqrt{100+2400}}{2}=\dfrac{-10\pm\sqrt{2500}}{2}=\dfrac{-10\pm50}{2}\). This gives \(x=20\) or \(x=-30\); since \(x>0\) (a speed cannot be negative), \(x=20\) mph. [2 marks]
Multiplying by \(x(x+10)\) clears both denominators in one step because it is the product of the two individual denominators; the resulting equation has no fractions left, which is what makes it solvable as an ordinary quadratic.
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