An airplane flying into a travels the flying distance betwee

An airplane flying into a travels the flying distance between two cities in 3 hours and. On the return flight, the airplane travels this distance in 2 hours. Find the of the and the speed of the

Solution

The print is faint. The figures in red are not clear. Hence the answer is on best effort basis.

An airplane, against headwind, travels a distance of 1640 miles between 2 cities in 2 hours, 36 minutes (i.e. 156 minutes)and , on the return journey, with tailwind, it travels 1640 miles in 2 hours. Let the speeds of the airplane and the wind be x and y mph respectively. Then, against the headwind, and with the tailwind, the speeds of the airplane are x-y mph and x+y mph respectively. Now, since distance = time *speed, we have (x-y)* 156/60 = (x+y)*2 or, 156(x-y) = 120(x+y) or, 156x- 120x = 120y+156y or, 36x = 276y so that x = (276/36)y= 23y/3. Then, the airplane’s speed with the tailwind is x +y = 23y/3 +y = 26y/3. Further, since, with the tailwind, the airplane travels 1640 miles in 2 hours, its speed, with tailwind, is 1640/2 = 820 mph. Hence 26y/3 = 820 so that y = 820*3/26 =2460/26= 94.62 mph ( on rounding off to 2 decimal places). Then x = 23y/3 = (23/3)*2460/26 = 725.38 mph (on rounding off to 2 decimal places). Thus, the speeds of the airplane and the wind are 725.38 mph and 94.62 mph respectively.

Note: If the figure for the time taken by the airplane, against headwind, in travelling a distance of 1640 miles between 2 cities is not 2 hours, 36 minutes, the calculations will change, but the method will be same.