algebra help?

Question

An airplane took 2hour to fly 600km against a head wind. The return trip with the wind took 1 hour and 2/3. How fast is the plane going in still air.

Can sum1 give me the answer, equation and how to do it... thx

Answer 1

if in 2 hours(120 min) fly 600 km, then his speed is 300 km/hr, and if in 1 hour and 2/3 , that is, 100 minutes, fly the same distance, 300 x 120 / 100 = 360 km / hr

Answer 2

I'll assume that the wind was still blowing with the same strength and the same direction both ways the whole time.

When you have a plane moving with the presence of wind or a boat moving upstream or downstream, the current either moves against you and lowers your speed, or moves with you and increases the speed.

The speed leaving was 600/2 = 300 km/h. The time to return was 5/3 of an hour over the same 600km, so the speed returning was 600 / (5/3) = 600*3/5 = 360 km/h.

So if P is the plane speed in still air and W is the wind speed, we have
P - A = 300
P + A = 360

To solve this system, we can add the two equations togeter to get 2P = 660. So P = 330 km/hr.

Answer 3

330 km/hr in still air.

Going against the wind, the plane goes 300 km/hr (600 km in 2 hours = 300 km/hr). Coming back with the wind, the plane goes 360 km/hr (the same 600 km in 5/3 hours = 360 km/hr).

Say that X is the speed in still air and Y is the speed of the wind. X+Y = 360 (this is the coming back leg). X-Y = 300 (this is the going there leg).

If you add these two equations together, the Ys cancel and you get 2X=660, or X=330.

Answer 4

ok.. lets say spreed of plane in still air is X.. the speed of wind is Y.. so going against wind the actual speed of plane will be (X-Y).. X-Y as speed of plane is greater than that of wind.. okie.. let the distance is Z (600 km here) is constant.. the speed of the plane while coming back is X+Y.. We know the equation D = S*T where D = Distance covered
S = Speed, T is the time taken.
T for first trip is 2 hrs.. for return it is 5/3 hrs.. or 1 and 2/3 hrs.. so puttin the values in the equation:

600 = (X-Y)2 ------ 1
600 = (X+Y)5/3 ----- 2

ie 300= x-y
360=x+y
adding both of these we get 2x=660 giving x=330 and substituting it gives Y as 30.. So plane speed is 330 Km/hr and wind speed is 30Km/hr.

Answer 5

rate*time = distance
So rate against wind = 600/2 = 300 km/h
Rate with wind = 600/5/3 = 360 km /hour
Hence rate of plane in still air must be (300+360)/2 = 330 km/h