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 the original speed be $x$ km/h. A 3-hour late start ended only 1 hour late, so 2 hours were saved:
$\dfrac{6300}{x} - \dfrac{6300}{x + 200} = 2$
$1260000 = 2x^2 + 400x \Rightarrow x^2 + 200x - 630000 = 0 \Rightarrow x = 700$
Speed with a further 100 km/h $= 700 + 300 = 1000$ km/h
Time $= \dfrac{6300}{1000} = 6.3$ h $= 6$ h 18 min
Leaving at 1:00 PM, arrival $= 7{:}18$ PM
$\dfrac{6300}{x} - \dfrac{6300}{x + 200} = 2$
$1260000 = 2x^2 + 400x \Rightarrow x^2 + 200x - 630000 = 0 \Rightarrow x = 700$
Speed with a further 100 km/h $= 700 + 300 = 1000$ km/h
Time $= \dfrac{6300}{1000} = 6.3$ h $= 6$ h 18 min
Leaving at 1:00 PM, arrival $= 7{:}18$ PM