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 ?

192 km/hr

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