Two planes leave Pittsburg and Philadelphia at the same time, each going to the other city. One plane flies 25 miles per hour faster than the other. Find the airspeed of each plane given that the cities are 275 miles apart and the planes pass each other after 40 minutes of flying time.

Answer:193.75 miles per hour and 218.75 miles per hour.Step-by-step explanation:Let x represent the speed of one plane.We have been given that one plane flies 25 miles per hour faster than the other, so the speed of 2nd plane would be [tex]x+25[/tex].We are also told that both cities are 275 miles apart.Since the planes are flying towards each other, so the speed to complete 275 miles would be the sum of speed of both planes:[tex]x+x+25=2x+25[/tex]We are also told that the planes pass each other after 40 minutes of flying time.[tex]40\text{ minutes}=\frac{40}{60}\text{ hour}=\frac{2}{3}\text{ hour}[/tex]We know [tex]\text{Speed}=\frac{\text{Distance}}{\text{Time}}[/tex].Upon substituting our given values, we will get:[tex]2x+25=\frac{275}{\frac{2}{3}}[/tex][tex]2x+25=\frac{275*3}{2}[/tex][tex]2x+25=\frac{825}{2}[/tex][tex]2x+25=412.5[/tex][tex]2x+25-25=412.5-25[/tex][tex]2x=387.5[/tex][tex]\farc{2x}{2}=\frac{387.5}{2}[/tex][tex]x=193.75[/tex]Therefore, the speed of one plane is 193.75 miles per hour.To find the speed of 2nd plane, we will substitute [tex]x=193.75[/tex] in [tex]x+25[/tex] as:[tex]x+25\Rightarrow 193.75+25=218.75[/tex]Therefore, the speed of 2nd plane is 218.75 miles per hour.