A train traveled the first \(d\) miles of its journey at an average speed of 60 miles per hour, the next \(d\) miles of its journey at an average speed of \(y\) miles per hour, and the final \(d\) miles of its journey at an average speed of 160 miles per hour. If the train's average speed over the total distance was 96 miles per hour, what is the value of \(y\)?

Explanation: Total distance = \(d + d + d = 3d\). Total time = sum of time for each leg. Time = \(\frac{\text{Distance}}{\text{Speed}}\). Total Time = \(\frac{d}{60} + \frac{d}{y} + \frac{d}{160} = d(\frac{1}{60} + \frac{1}{y} + \frac{1}{160})\). Average Speed = \(\frac{\text{Total Distance}}{\text{Total Time}}\). \(96 = \frac{3d}{d(\frac{1}{60} + \frac{1}{y} + \frac{1}{160})}\). \(96 = \frac{3}{\frac{1}{60} + \frac{1}{y} + \frac{1}{160}}\). \(\frac{3}{96} = \frac{1}{60} + \frac{1}{y} + \frac{1}{160}\). \(\frac{1}{32} = \frac{1}{60} + \frac{1}{y} + \frac{1}{160}\). \(\frac{1}{y} = \frac{1}{32} - \frac{1}{60} - \frac{1}{160}\). LCM of 32, 60, and 160 is 480. \(\frac{1}{y} = \frac{15}{480} - \frac{8}{480} - \frac{3}{480} = \frac{15 - 8 - 3}{480} = \frac{4}{480} = \frac{1}{120}\). \(y = 120\). (Source solution uses \(96 = \frac{3d}{d(\frac{1}{60} + \frac{1}{y} + \frac{1}{160})}\) and skips to \(\frac{1}{60} + \frac{1}{y} + \frac{1}{160} = \frac{1}{32}\) and \(y = 120\)).