A flight was scheduled to depart at 10:00 AM and travel 6300 km. Due to maintenance, departure was delayed until 1:00 PM. To make up time, the pilot increased the average speed by 200 km/h over the original plan, which limited the delay at the destination to just 1 hour. If instead the pilot had increased the speed by a further 100 km/h, at what time would the flight have arrived?
Answer and solution
Correct answer: C
Solution
Let original speed $= x$ km/h. Delayed 3 hours, arrived 1 hour late → saved 2 hours by flying faster:
$\dfrac{6300}{x} - \dfrac{6300}{x+200} = 2$
$6300 \times 200 = 2x(x+200) \Rightarrow x^2 + 200x - 630000 = 0$
$x = \dfrac{-200 + \sqrt{40000 + 2520000}}{2} = \dfrac{-200 + 1600}{2} = 700$ km/h.
At speed 700+200+100 = 1000 km/h: flight time $= \dfrac{6300}{1000} = 6.3\text{ h} = 6\text{ h }18\text{ min}$.
Departed 1:00 PM + 6h 18min = $7{:}18\text{ PM}$.
$\dfrac{6300}{x} - \dfrac{6300}{x+200} = 2$
$6300 \times 200 = 2x(x+200) \Rightarrow x^2 + 200x - 630000 = 0$
$x = \dfrac{-200 + \sqrt{40000 + 2520000}}{2} = \dfrac{-200 + 1600}{2} = 700$ km/h.
At speed 700+200+100 = 1000 km/h: flight time $= \dfrac{6300}{1000} = 6.3\text{ h} = 6\text{ h }18\text{ min}$.
Departed 1:00 PM + 6h 18min = $7{:}18\text{ PM}$.