A carrier pigeon flies distance between two cities in the absence of wind in t = 60 minutes

A carrier pigeon flies distance between two cities in the absence of wind in t = 60 minutes, and in a headwind in time t2 = 75 minutes. How long will it take t1 for a pigeon to cover this distance with a tailwind?

No wind:

S = Vg * t, where S is the distance between two cities, Vg is the pigeon’s own speed, t is the time the pigeon moves (t = 60 min).

Headwind:

S = (Vg – Vw) * t2, where Vw is the wind speed, t2 is the time the pigeon moves (t = 75 minutes).

Favourable wind:

S = (Vg + Vv) * t1.

1) Vg * t = (Vg – Vv) * t2.

Vg * 60 = (Vg – Vv) * 75.

60Vg = 75Vg – 75Vv.

15Vg = 75Vv.

Vg = 5Vv.

2) Vg * t = (Vg + Vv) * t1.

5Vv * t = (5Vv + Vv) * t1.

t1 = 5Vv * t / 6Vv = 5 * 60/6 = 50 min.