Rate Time Distance

Question

Flying with the wind, a plane traveled 1080 miles in 6 hours. Flying against the wind, the plane covered only 2/3 of this distance in the same time. Find the rate of the plane in still air and the rate of the wind.

Answer 1

The actual rate of travel is the plane's rate plus the wind rate (positive if helping the plane, negative if hindering the plane). We'll need to use equation D = RT, but we'll need to use it twice since the plane makes two trips. Let's call the plane's rate p and the wind's rate w.

For the first trip, we have:

D = RT

1080 = (p + w)*6

For the second trip, we have:

D = RT

(2/3)*1080 = (p - w)*6

Now we have two equations and two variables (p and w), so we should be able to solve for each variable. We can expand the two equations to get:

1080 = 6p + 6w

720 = 6p - 6w

Now we can solve this system of equations using the elimination method. First, add the two equations together to eliminate the w variable. That will give you a single equation with one variable (p), which you can solve for p. Next, subtract the bottom equation from the top one to eliminate the p variable. That will give you a single equation with one variable (w), which you can solve for w. I think that you can do that, but let me know if you have any trouble with the algebra.