<PRE><B>Question 252978</B>
Speed of aircraft is X
On outward trip (X + 25.5) = 4.23/Distance
On inbound trip (X - 25.5) = 4.97/Distance 
Equating the distances
 Distance = (X + 25.5)/4.97 = (X - 25.5)/4.23
So 4.23(X + 25.5) = 4.97(X - 25.5)

4.23X + 107.865 = 4.97X - 126.735

234.6 = 0.74 X
X= 317.027 km/hr
































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