Solve the problem

Question

A twin-engined aircraft can fly 1216 miles from city A to city B in 4 hours with the wind and make the return trip in 8 hours against the wind. What is the speed of the wind?

Answer 1

My answer is based in two equations as follows:

4(Vw+Vp) = 1216 First trip with the wind

8(Vp-Vw) = 1216 Return trip against the wind

to eliminate one variable I multiply the first equation by 2 and then add it to the second equation.

8Vw + 8Vp = 2432

8Vp - 8Vw = 1216

adding them up, I got

16 Vp = 3648

Vp = 228 mph

Then plugging this speed into the first equation, I obtain:

4Vw = 1216 - 4*228= 304

Vw = 304/4 = 76 mph.

I think that the two answers are very similar and the results the same.

Answer 2

Speed on the direct flight is 1216/4=304 mph=V_a+V_w Here V_a is the speed of aircraft and V_w is the speed of wind.

 

Speed on the way back is 1216/8=152 mph=V_a-V_w

 

The speed of wind is half the difference between them, that is

V_w=(304-152)/2=76 mph.