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 o utbound speed. The round-trip took 6 h 40 min. What was the speed from City A to City B?
1 answer:
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
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