Question

A man flies a small airplane from Fargo to Bismarck, North Dakota --- a distance of 180 miles. Because he is flying into a head wind, the trip takes him 2 hours. On the way back, the wind is still blowing at the same speed, so the return trip takes only 1 hour 12 minutes. What is his speed in still air, and how fast is the wind blowing?

Answer 1

Change 1 hr 12 min: 1 + 12/60 = 1.2 hrs
:
Let s = his speed in still air
Let w = speed of the wind
then
(s-w) = effective speed against the wind
and
(s+w) = effective speed with the wind
:
Write a distance equation for each way; dist = time * speed

2(s - w) = 180
1.2(s+w) = 180
Simplify both equations, divide the 1st eq by 2, the 2nd eq by 1.2; results:
s - w = 90
s + w = 150
addition eliminates w, find s
2s = 240
s = 240/2
s = 120 mph is the speed in still air
Find w

s + w = 150
120 + w = 150
w = 150 - 120
w = 30 mph is the speed of the wind

Check in 1st original equation
2(120-30) = 180
2(90) = 180

Answer 2

r, t and d is rate, time and distance respectively.
\[\left\{r*t=d, t=\frac{d}{r}\right\} \]\[\text{Solve}\left[\left\{2=\frac{180}{r-w},1+\frac{12}{60}=\frac{180}{r+w}\right\}\right] \]\[\{r\to 120, w\to 30\} \]