A pilot flew a jet from City A to City B a distance of 2400

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 5 h 20 min. What was the speed from City A to City B?

Solution

let the speed from city A to city B = S miles/min

Distance between cities = 2400 mi

Time taken T₁ = Distance /speed = 2400/S

Return speed is 20% faster ==> Return speed = 1.2S

Time taken T₂ = Distance / speed = 2400/1.2S= 2000/S

Total time taken for the trip = 5h 20 min ==> 320 min

==> T₁ + T₂ = 320 min

==> 2400/S + 2000/S = 320

==> 4400/S = 320

==> S = 4400/320 = 13.75 miles/min

Hence speed from city A to city B = 13.75 miles/ min