MIA-20260531-4DCADD09

The first train leaves Chicago at 10:00 AM traveling south at 60 mph. The second train leaves St. Louis at 11:00 AM traveling north at 80 mph. The distance between Chicago and St. Louis is 300 miles. Let \( t \) be the number of hours after 10:00 AM when the trains meet. - The first train travels for \( t \) hours at 60 mph, covering \( 60t \) miles. - The second train starts one hour later, so it travels for \( t - 1 \) hours at 80 mph, covering \( 80(t - 1) \) miles. The sum of the distances traveled equals the total distance between the cities: \[ 60t + 80(t - 1) = 300. \] Simplify and solve: \[ 60t + 80t - 80 = 300, \] \[ 140t = 380, \] \[ t = \frac{380}{140} = \frac{19}{7} = 2\frac{5}{7} \text{ hours}. \] Convert \( \frac{5}{7} \) hour to minutes: \[ \frac{5}{7} \times 60 = \frac{300}{7} \approx 42.857 \text{ minutes} = 42 \frac{6}{7} \text{ minutes}. \] Thus, the meeting time is 2 hours and \( 42\frac{6}{7} \) minutes after 10:00 AM, which is **12:42 and \( \frac{6}{7} \) minutes PM** (approximately 12:43 PM). We can verify by an alternative method: At 11:00 AM, the first train has traveled 60 miles, leaving \( 300 - 60 = 240 \) miles between them. Their combined speed is \( 60 + 80 = 140 \) mph, so the remaining time is \( \frac{240}{140} = \frac{12}{7} = 1\frac{5}{7} \) hours, which added to 11:00 AM gives the same result. \boxed{12:42 \frac{6}{7} \text{ PM}}