When an aircraft flies between two cities, 11 / 4H downwind and 3H upwind, and the known wind speed is 20km / h, what is the distance between the two cities?

When an aircraft flies between two cities, 11 / 4H downwind and 3H upwind, and the known wind speed is 20km / h, what is the distance between the two cities?

Time ratio of downwind to upwind: 11 / 4:3 = 11:12
So the speed ratio: 12:11
So headwind speed: 20 × 2 × 11 = 440 (km)
Distance: 440 × 3 = 1320 (km)
Let the speed of the plane be v
(v+20)*(11/4) =3*(v-20)
11/4(v+20)=3*(v-20)
11(v+20)=12*(v-20)
11v+220=12v-240
v=220+240
v=460
Distance: (460-20) × 3 = 440 × 3 = 1320 (km)

A plane flies between the two cities with a wind speed of 20 km / h, 2 h and 30 min downwind and 3 h upwind
When there is no wind, the speed of the plane and the distance between the two cities

A round trip takes 5 hours and 30 minutes, which has nothing to do with the wind speed, so the one-way trip is 2 hours and 45 minutes when there is no wind, which is 15 minutes different from that when there is wind
2q45min = 11 × 15min, obviously, the aircraft is 11 times faster than the wind, so the speed of the aircraft is 11 × 20 = 220km / h
The distance between the two places is 220 × 2h45min = 495km!
Solution 2:
Let the speed of the aircraft be x when there is no wind
(x+20)5/ 2＝(x－20)×3
The solution is x = 220, the result is the same!