A plane flying against the wind flew 270 miles in 3 hours. Flying with a tailwind, the plane traveled 260 miles in 2 hours. Find the rate of the plane in calm air and the rate of the wind.

Start with the equation that time = distance / rate. We know the times of the two scenarios are equal, so we will set up two equations that equal each other. The wind speed when aiding the plan is added to the speed of the plane, and it is subtracted when going against the wind. For example, say the speed of the plane is x miles per hour (mph) and the wind speed is 5 mph (make sure you use the actual wind speed given in the problem). Then the total speed would be x + 5 mph. When going against the wind, the speed would be x – 5 mph. Since the Distance is given, we need to solve for the rate.

Set up the equation (assuming 5 mph is the wind speed, would need to change to match actual speed given in problem),

t = 270 miles / (x – 5) for the plane going against the wind t = 330 miles / (x +5) for the plane traveling with the wind

Since the times equal for these two distances, we can make the equations equal to each other and solve for x, which represents the speed of the plane without the wind.

270 / (x-5) = 330 / (x+5). Cross multiplying and solving for x gives you x = 50

So the speed of the plane with the wind would be 50 + 5= 55 miles per hour, and the speed of the plane against the wind would be 50 – 5 = 45 mph.

The time the plane traveled would then be solved by substituting x = 50 into either equation above and solving for t. This would give you 6 hours for t, or time traveled. This makes sense, as they both take the same time to travel the distances as the problem states.

Again make sure the value of 5 mph for wind I used is updated to match the wind speed given in the problem. Also I assumed the speed was measured in MPH, and this may need to change to match whatever speed unit the problem asks if different. Good luck!

Start with the equation that time = distance / rate. We know the times of the two scenarios are equal, so we will set up two equations that equal each other. The wind speed when aiding the plan is added to the speed of the plane, and it is subtracted when going against the wind. For example, say the speed of the plane is x miles per hour (mph) and the wind speed is 5 mph (make sure you use the actual wind speed given in the problem). Then the total speed would be x + 5 mph. When going against the wind, the speed would be x – 5 mph. Since the Distance is given, we need to solve for the rate.

Set up the equation (assuming 5 mph is the wind speed, would need to change to match actual speed given in problem),

t = 270 miles / (x – 5) for the plane going against the wind

t = 330 miles / (x +5) for the plane traveling with the wind

Since the times equal for these two distances, we can make the equations equal to each other and solve for x, which represents the speed of the plane without the wind.

270 / (x-5) = 330 / (x+5). Cross multiplying and solving for x gives you x = 50

So the speed of the plane with the wind would be 50 + 5= 55 miles per hour, and the speed of the plane against the wind would be 50 – 5 = 45 mph.

The time the plane traveled would then be solved by substituting x = 50 into either equation above and solving for t. This would give you 6 hours for t, or time traveled. This makes sense, as they both take the same time to travel the distances as the problem states.

Again make sure the value of 5 mph for wind I used is updated to match the wind speed given in the problem. Also I assumed the speed was measured in MPH, and this may need to change to match whatever speed unit the problem asks if different. Good luck!