A plane flies 405 miles with the wind and 315 miles against the wind in the same length of time. If the speed of the wind is \( 20 \mathrm{mph} \), find the speed of the plane in still air.
Answer
Cross-multiply and solve for x: \(405(x - 20) = 315(x + 20)\). Simplify and solve for x: \(405x - 8100 = 315x + 6300\) => \(90x = 14400\) => \(x = 160\).
Step 1 ：Let x be the speed of the plane in still air. Then, the speed with the wind is \(x + 20\) and against the wind is \(x - 20\).
Step 2 ：Since the time for both flights is the same, we can set up the equation: \(\frac{405}{x + 20} = \frac{315}{x - 20}\).
Step 3 ：Cross-multiply and solve for x: \(405(x - 20) = 315(x + 20)\). Simplify and solve for x: \(405x - 8100 = 315x + 6300\) => \(90x = 14400\) => \(x = 160\).
