A student pilot was required to fly to an airport and then return as part of his flight training. The average speed to the airport was 120 km/h, and the average speed returning was 150 km/h. If the total flight time was 3 hours, calculate the distance between the two airports.
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Correct Answer: Option B
Explanation:
Speed = \(\frac{Distance}{Time}\)
⇒ Time = \(\frac{Distance}{Time}\)
Let D = distance between the two airports
∴ Time taken to get to the airport = \(\frac{D}{120}\) and Time taken to return =\( \frac{D}{150}\)
Since total time of flight= 3hours,
⇒ \(\frac{D}{120} + \frac{D}{150}\) = 3
⇒ \(\frac{15D + 12D}{1800}\) = 3
⇒ \(\frac{27D}{1800}\) = 3
⇒ \(\frac{3D}{200} = \frac{3}{1}\)
⇒ 3D = 200 x 3
∴ D =\(\frac{ 200\times3}{3}\)= 200km
Speed = \(\frac{Distance}{Time}\)
⇒ Time = \(\frac{Distance}{Time}\)
Let D = distance between the two airports
∴ Time taken to get to the airport = \(\frac{D}{120}\) and Time taken to return =\( \frac{D}{150}\)
Since total time of flight= 3hours,
⇒ \(\frac{D}{120} + \frac{D}{150}\) = 3
⇒ \(\frac{15D + 12D}{1800}\) = 3
⇒ \(\frac{27D}{1800}\) = 3
⇒ \(\frac{3D}{200} = \frac{3}{1}\)
⇒ 3D = 200 x 3
∴ D =\(\frac{ 200\times3}{3}\)= 200km