MIA-20260531-C1337630

**Step 1 – Set up a timeline** | Event | Time (relative to 10 am) | |-------|--------------------------| | Train A leaves Chicago (southbound) | 0 h | | Train B leaves St. Louis (northbound) | 1 h (i.e., 11:00 am) | Let \[ t = \text{hours after 10:00 am when the two trains meet.} \] Thus * Train A has been moving for **\(t\) hours**. * Train B has been moving for **\(t-1\) hours**, because it starts one hour later (this expression is only valid once \(t\ge 1\); the meeting will indeed occur after 11 am). **Step 2 – Write the distance each train has covered** * Train A speed = 60 mph → distance covered = \(60t\) miles. * Train B speed = 80 mph → distance covered = \(80(t-1)\) miles. **Step 3 – Use the total separation** The cities are 300 miles apart (St. Louis is 300 mi south of Chicago). When the trains meet, the sum of the distances they have traveled must equal this separation: \[ 60t + 80(t-1) = 300. \] **Step 4 – Solve for \(t\)** \[ \begin{aligned} 60t + 80t - 80 &= 300\\[4pt] 140t &= 380\\[4pt] t &= \frac{380}{140}= \frac{19}{7}\ \text{hours}. \end{aligned} \] \[ \frac{19}{7}\text{ h}=2\frac{5}{7}\text{ h}=2\text{ h}+ \frac{5}{7}\times 60\text{ min} =2\text{ h}+42.\overline{857}\text{ min}\approx 2\text{ h}\,42\ \text{min}. \] **Step 5 – Convert to clock time** Starting from 10:00 am and adding 2 h 42 min gives \[ 10{:}00\ \text{am} + 2\text{ h}\,42\text{ min} \approx 12{:}42\ \text{pm}. \] (If we keep the seconds, the exact time is \(12{:}42{:}51\) pm, since \(0.857\text{ min}=51\text{ s}\).) **Answer** The two trains meet at **approximately 12:43 pm** (exactly 12 pm 42 min 51 s). *Verification:* * Train A distance: \(60 \times 2.7142857\approx 162.86\) mi. * Train B distance: \(80 \times (2.7142857-1) = 80 \times 1.7142857 \approx 137.14\) mi. * Sum ≈ 300 mi, confirming the calculation.