a. If they allow 0.4 hour for takeoff and landing, then the approximate flight time from Dulles to D.C. is about 5.5−0.4=5.1 hours and the approximate flight time from D.C. to Dulles is 4.5−0.4=4.1 hours.

Let r be the average airplane speed and ww be the average wind speed.

The flight from Dulles to D.C. is scheduled for a longer time because the wind will slow down the plane. The rate of the plane will then be r−w for this trip. Since distance = (rate)(time), if the distance is 2,300 miles, the rate is r−w, and the time is 5.1 hours, then 2,300=5.1(r−w). Dividing both sides by 5.1 gives r−w=451, rounded to the nearest whole number.

For the flight from D.C. to Dulles, the wind will make the plane travel faster so the plane's rate will be r+wr+w. Since the distance is 2,300 miles, the rate is r+w, and the time is 4.1 hours, then 2,300=4.1(r+w). Dividing both sides by 4.1 gives r+w=561, rounded to the nearest whole number.

Add the two equations to eliminate ww and solve for r:

r+w = 561

r−w = 451

2r = 1012

r = 506

Substituting r=506 into r+w=561 gives 506+w=561. Subtracting 506 on both sides gives w=55. The average wind speed to the nearest 5 mph is then 55 mph.

b. From part (a), r=506. Rounding to the nearest 5 mph then gives an average airplane speed of 505 mph.

Let r be the average airplane speed and ww be the average wind speed.

The flight from Dulles to D.C. is scheduled for a longer time because the wind will slow down the plane. The rate of the plane will then be r−w for this trip. Since distance = (rate)(time), if the distance is 2,300 miles, the rate is r−w, and the time is 5.1 hours, then 2,300=5.1(r−w). Dividing both sides by 5.1 gives r−w=451, rounded to the nearest whole number.

For the flight from D.C. to Dulles, the wind will make the plane travel faster so the plane's rate will be r+wr+w. Since the distance is 2,300 miles, the rate is r+w, and the time is 4.1 hours, then 2,300=4.1(r+w). Dividing both sides by 4.1 gives r+w=561, rounded to the nearest whole number.

Add the two equations to eliminate ww and solve for r:

r+w = 561

r−w = 451

2r = 1012

r = 506

Substituting r=506 into r+w=561 gives 506+w=561. Subtracting 506 on both sides gives w=55. The average wind speed to the nearest 5 mph is then 55 mph.

b. From part (a), r=506. Rounding to the nearest 5 mph then gives an average airplane speed of 505 mph.