MIA-20260531-B14C202A

**Assumption:** The first train is heading **south** from Chicago toward St. Louis (otherwise the trains would not meet on the Chicago–St. Louis route). --- ### Step 1: Positions at 11:00 AM - **Train 1** left Chicago at 10:00 AM traveling at 60 mph. By 11:00 AM it has traveled: $$60 \text{ mph} \times 1 \text{ hr} = 60 \text{ miles}$$ So at 11:00 AM, Train 1 is **60 miles south of Chicago**. - **Train 2** leaves St. Louis at 11:00 AM. St. Louis is **300 miles south of Chicago**. ### Step 2: Distance between the trains at 11:00 AM $$300 \text{ miles} - 60 \text{ miles} = \mathbf{240 \text{ miles}}$$ ### Step 3: Combined speed The trains are moving toward each other, so their speeds add: $$60 \text{ mph} + 80 \text{ mph} = \mathbf{140 \text{ mph}}$$ ### Step 4: Time to meet after 11:00 AM $$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{240}{140} = \frac{12}{7} \text{ hours}$$ Convert to hours and minutes: $$\frac{12}{7} \text{ hours} = 1\frac{5}{7} \text{ hours}$$ $$\frac{5}{7} \times 60 \text{ minutes} = \frac{300}{7} \text{ minutes} = 42\frac{6}{7} \text{ minutes}$$ So the trains meet **1 hour and $42\frac{6}{7}$ minutes after 11:00 AM**. ### Step 5: Meeting time $$11:00 \text{ AM} + 1 \text{ hour} = 12:00 \text{ PM (noon)}$$ $$12:00 \text{ PM} + 42\frac{6}{7} \text{ minutes} = \mathbf{12:42\frac{6}{7} \text{ PM}}$$ Or approximately **12:43 PM** (more precisely 12:42:51 PM). --- **Verification:** - Train 1 travels for $1 + \frac{12}{7} = \frac{19}{7}$ hours total. Distance: $60 \times \frac{19}{7} = \frac{1140}{7} \approx 162.86$ miles south of Chicago. - Train 2 travels for $\frac{12}{7}$ hours. Distance north from St. Louis: $80 \times \frac{12}{7} = \frac{960}{7} \approx