A man makes four trips of equal distances. His speed on first trip was 720 km/hr and in each subsequent trip his speed was half of the previous trip. What is the average speed of the man in these four trips ?
Answer & explanation
Correct answer: option 3
Let the distance of each trip be X
For first trip time can be given as
Time for first trip = \(\frac{X}{720}\)
For second trip, since speed is now 36 km/hr time can be given as
Time for second trip = \(\frac{X}{360}\)
For third trip, since speed is now 18 km/hr time can be given as
Time for second trip = \(\frac{X}{180}\)
For fourth trip, since speed is now 9 km/hr time can be given as
Time for second trip = \(\frac{X}{90}\)
Average speed = \(\frac{4X}{X/720 + X/360 + X/180 + X/90}\)
Average speed = \(\frac{4X × 720}{15x}\)
= \(\frac{2880}{15}\)
= 192 km/h
The average speed is 192 km/hr