Question

A pilot flew a jet from City A to City B, a distance of 2400 mi. On the return trip, the average speed was 20% faster than the outbound speed. The round-trip took 6 h 40 min. What was the speed from City A to City B?

Answer 1

Answer:

Explanation:

Let the velocity be Va for outbound journey and 1.2 Va for return journey .

Total time taken = 2400 / Va + 2400 / 1.2 Va

2400 / Va + 2400 / 1.2 Va = 20 / 3 h

1 / Va ( 2400 + 2000 ) = 20 / 3

4400 / Va = 20/3

Va = 4400 x 3 / 20 = 660 mi / h