Let the speed of the plane in still air be x and speed of the wind be y.
For the first trip the situation can be formulated as:
x + y = 1010/2
or x+y = 505 ----------------(1)
And for return journey:
x - y = 1010/ (5/2)
or x - y = 404 ----------------------(2)
On adding eq(1) and (2) we get:
2x = 909
or x =454.5
and putting the value of x in eq(1) we get:
454.5 + y = 505
or y = 50.5
Therefore speed of the plane in still air is 454.5 kmph and speed of the wind is 50.5 kmph