A plane flies directly from city A to city B, which are separated by 2300 mi. From A to B, the plane flies into a 65 mi/h headwind. On the return trip from B to A, the wind velocity is unchanged. The trip from B to A takes $65 \mathrm {~min}$ less than the trip from A to B. What is the airspeed (assumed constant) of the plane relative to the air (that is, the airspeed)?
A) 530 mi/h
B) 400 mi/h
C) 480 mi/h
D) 610 mi/h

Question

Quizplus · Accepted Answer

The airspeed of the plane is the same in both directions, so the difference in travel time must be due to the headwind. The headwind reduces the plane's speed by 65 mi/h, so the airspeed must be 65 mi/h greater than the speed of the plane relative to the ground. The speed of the plane relative to the ground is 2300 mi / (65 min + 65 min) = 353.85 mi/h, so the airspeed is 353.85 mi/h + 65 mi/h = 418.85 mi/h, which is closest to 400 mi/h.

A Plane Flies Directly from City a to City B 65 min65 \mathrm {~min}65 min

A Plane Flies Directly from City a to City B $65 \mathrm {~min}$