A pilot flew a jet from City A to City B, a distance of 1800 mi. On the return trip, the average speed was 20% faster than the outbound speed

Question

A pilot flew a jet from City A to City B, a distance of 1800 mi. On the return trip, the average speed was 20% faster than the outbound speed. The round-trip took 9 h 10 min. What was the speed from City A to City B?

Answer 1

Time equals distance divided by speed.

Suppose the speed on the way out was X, then the speed on the way back was 1.2X.

Write the equation for the total time of the round-trip flight:

 

1800 mi / X + 1800 mi / 1.2 X = 9 h 10 min

 

1800 + 1500 = X * 9 h 10 min

 

X = 3,300 / 9 h 10 min = 360 mph

 

The means, the speed from city A to city B was 360 mph.