Against the wind a small plane flew 175 miles in 1 hour and 10 minutes.

The return trip took only 50 minutes.

What was the speed of the​ wind?

What was the speed of the plane in still​ air?

x = rate of the wind

v = speed of the small plane in still air

1 hour and 10 min = 1 10/60 hours = 7/6 hours

the speed of the plane against the wind is v - x = 175 / (7/6) miles per hour

the return trip flying in the direction of the wind with speed v + x took 50 min = 50/60 = 5/6 hours

thus we can write

v - x = 175 / (7/6) = 150

v + x = 175 / (5/6) = 210

by solving the system of equations

v = 180 mph

x = 30 mph

The speed of the plane in still air is 180 mph.

The speed of the wind is 30 mph.

Let s be the speed of the plane in still air.

Let w be the speed of the wind.

Against the wind, the net speed will be s-w.

With the wind, the net speed will be s+w.

Against the wind --> 70 minutes or 7/6 hours

With the wind --> 50 minutes or 5/6 hours

Equation against the wind:

175 = 7/6 * (s-w)

6/7 * 175 = s - w

150 = s - w

Equation with the wind:

175 = 5/6 * (s+w)

6/5 * 175 = s + w

210 = s + w

Add the last two equations to eliminate w:

360 = 2s

s = 180

210 = 180 + w

w = 30

Answer:

The plane travels at 180 mph (in still air) and the wind is blowing at 30 mph.

w = speed of wind

s = speed in still air

50 min = 5/6 hr and 1 hr 110 min = 7/6 hr. Distance = time times speed so

175 = 7/6 (s - w)

175 = 5/6 (s + w)

Multiply row 1 by 6/7 and row 2 by 6/5 to get rid of the fractions

150 = s - w

210 = s + w and now add the two together to eliminate w

360 = 2s so solve for s then plug in to get w