PROBE-CAP-01

Here's how to solve this problem step-by-step: **1. Analyze the movement of the first train (from Chicago):** * **Speed:** 60 mph * **Departure Time:** 10:00 AM **2. Analyze the movement of the second train (from St. Louis):** * **Speed:** 80 mph * **Departure Time:** 11:00 AM * **Starting Position:** 300 miles south of Chicago **3. Account for the head start of the first train:** The train from Chicago has a one-hour head start before the train from St. Louis even begins moving. * **Distance covered by Chicago train in one hour:** 60 mph * 1 hour = 60 miles **4. Determine the remaining distance between the trains when both are moving:** At 11:00 AM (when the second train starts), the train from Chicago has traveled 60 miles south. This means the effective distance separating them at 11:00 AM is less than the initial 300 miles. * **Remaining distance:** 300 miles (initial distance) - 60 miles (Chicago train's head start) = 240 miles **5. Calculate their relative speed:** Since the trains are moving towards each other, their speeds add up to determine how quickly they close the distance between them. * **Relative speed:** 60 mph (Chicago train) + 80 mph (St. Louis train) = 140 mph **6. Calculate the time it takes for them to meet after 11:00 AM:** Now we need to find how long it takes for them to cover the remaining 240 miles at their relative speed. * **Time = Distance / Speed** * **Time to meet (after 11:00 AM):** 240 miles / 140 mph = 1.714 hours (approximately) **7. Convert the decimal hours into minutes:** * **Minutes:** 0.714 hours * 60 minutes/hour = 42.84 minutes (approximately, let's round to 43 minutes for practical purposes) **8. Determine the exact meeting time:** They start closing the remaining distance at 11:00 AM. Add the calculated time to this. * **Meeting time:** 11:00 AM + 1 hour + 43 minutes = 12:43 PM **Therefore, the trains will meet at approximately 12:43 PM.**

**Assumption:** Since the second train is heading north from St. Louis (toward Chicago), the first train must be heading south from Chicago (toward St. Louis) in order for them to meet. --- ### Step 1: Calculate the first train’s head start The first train travels alone from 10:00 AM to 11:00 AM (1 hour). \[ \text{Distance} = 60 \text{ mph} \times 1 \text{ hr} = 60 \text{ miles} \] At **11:00 AM**, Train 1 is **60 miles south of Chicago**, while Train 2 is just leaving St. Louis, **300 miles south of Chicago**. ### Step 2: Find the remaining distance between them at 11:00 AM \[ 300 \text{ miles} - 60 \text{ miles} = 240 \text{ miles} \] ### Step 3: Determine their combined closing speed The trains are now moving toward each other: - Train 1 southbound: **60 mph** - Train 2 northbound: **80 mph** \[ \text{Relative speed} = 60 + 80 = 140 \text{ mph} \] ### Step 4: Calculate the time needed to meet after 11:00 AM \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{240}{140} = \frac{12}{7} \text{ hours} \] Convert \(\frac{12}{7}\) hours to hours and minutes: \[ \frac{12}{7} = 1\frac{5}{7} \text{ hours} = 1 \text{ hour} + \frac{5}{7}\times 60 \text{ minutes} \] \[ \frac{5}{7}\times 60 = \frac{300}{7} = 42\frac{6}{7} \text{ minutes} \] So the trains meet **\(1\) hour and \(42\frac{6}{7}\) minutes after 11:00 AM**. ### Step 5: Add the time to 11:00 AM \[ 11:00 \text{ AM} + 1 \text{ hr } 42\frac{6}{7} \text{ min} = \mathbf{12:42\frac{6}{7} \text{ PM}} \] --- ### Algebraic Check Let \(t\) = hours after **11:00 AM** that they meet. - Train 1’s distance south of Chicago: \(60 + 60t\) - Train 2’s distance south of Chicago: \(300 - 80t\) Set positions equal: \[ 60 + 60t = 300 - 80t \] \[ 140t = 240 \] \[ t = \frac{12}{7} \text{ hours} \quad \checkmark \] --- **Answer:** The trains meet at **12:42 and \(\frac{6}{7}\) minutes PM** (approximately **12:42:51 PM**).

Here is the step-by-step solution. ### 1. Establish the initial conditions * **Train A** (from Chicago): Leaves at 10:00 AM traveling south at **60 mph**. * **Train B** (from St. Louis): Leaves at 11:00 AM traveling north at **80 mph**. * **Distance between cities:** 300 miles. ### 2. Calculate the situation at 11:00 AM In the first hour (from 10:00 AM to 11:00 AM), only Train A is moving. * Distance Train A covers by 11:00 AM: $$60 \text{ mph} \times 1 \text{ hour} = 60 \text{ miles}$$ At 11:00 AM: * Train A is 60 miles south of Chicago. * Train B is just leaving St. Louis, 300 miles south of Chicago. Therefore, the **remaining distance between the two trains** at 11:00 AM is: $$300 \text{ miles} - 60 \text{ miles} = \mathbf{240 \text{ miles}}$$ ### 3. Calculate the time to meet after 11:00 AM The trains are now moving toward each other, so their speeds add up. * **Combined speed:** $60 \text{ mph} + 80 \text{ mph} = \mathbf{140 \text{ mph}}$ They need to close a 240-mile gap at 140 mph. * **Time to meet:** $$t = \frac{\text{Distance}}{\text{Speed}} = \frac{240}{140} = \frac{12}{7} \text{ hours}$$ Convert $\frac{12}{7}$ hours to hours and minutes: $$\frac{12}{7} = 1\frac{5}{7} \text{ hours} = 1 \text{ hour} + \frac{5}{7} \text{ of an hour}$$ $$\frac{5}{7} \times 60 \text{ minutes} = \frac{300}{7} \text{ minutes} = 42\frac{6}{7} \text{ minutes}$$ So they meet **$1$ hour and $42\frac{6}{7}$ minutes after 11:00 AM**. ### 4. Determine the meeting time $$11:00 \text{ AM} + 1 \text{ hour } 42\frac{6}{7} \text{ minutes} = \mathbf{12:42\frac{6}{7} \text{ PM}}$$ This is approximately **12:42:51 PM**. ### 5. Verification * **Train A** travels for $2\frac{5}{7}$ hours at 60 mph: $$60 \times \frac{19}{7} = \frac{1140}{7} \approx 162.86 \text{ miles south of Chicago}$$ * **Train B** travels for $1\frac{5}{7}$ hours at 80 mph: $$80 \times \frac{12}{7} = \frac{960}{7} \approx 137.14 \text{ miles north from St. Louis}$$ * Position of Train B south of Chicago: $$300 - 137.14 = 162.86 \text{ miles}$$ The positions match. **Answer:** The trains meet at **approximately 12:43 PM** (exactly $12:42\frac{6}{7}$ PM).

**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.

Here is the step-by-step reasoning to solve this problem: **Step 1: Determine the head start distance of the first train.** The first train leaves Chicago at 10:00 AM, while the second train doesn't leave St. Louis until 11:00 AM. This means the first train has a 1-hour head start. * Distance = Speed × Time * Distance traveled by Train 1 in 1 hour = 60 mph × 1 hour = **60 miles** **Step 2: Calculate the remaining distance between the trains at 11:00 AM.** The total distance between Chicago and St. Louis is 300 miles. Since Train 1 has already traveled 60 miles south by the time Train 2 starts, we need to subtract that from the total distance. * Remaining distance = 300 miles - 60 miles = **240 miles** **Step 3: Calculate the combined (closing) speed of the two trains.** At 11:00 AM, both trains are now moving toward each other. When two objects move toward each other, their speeds add together to determine how fast they are closing the gap. * Closing speed = Speed of Train 1 + Speed of Train 2 * Closing speed = 60 mph + 80 mph = **140 mph** **Step 4: Calculate the time it takes to close the remaining distance.** Now we divide the remaining distance by the closing speed to find out how long they travel before meeting. * Time = Distance ÷ Speed * Time = 240 miles ÷ 140 mph = **12/7 hours** (or approximately 1.714 hours) **Step 5: Convert the time into hours and minutes.** To convert 12/7 hours into a more readable format: * 12/7 hours = 1 whole hour and 5/7 of an hour. * To convert 5/7 of an hour into minutes, multiply by 60: * (5/7) × 60 = 300/7 minutes ≈ **42.86 minutes** (or 42 minutes and 51 seconds) **Step 6: Add this travel time to the 11:00 AM starting point.** * Start time: 11:00 AM * Add 1 hour: 12:00 PM * Add 42.86 minutes: **12:42:51 PM** **Answer:** The two trains will meet at approximately **12:43 PM** (exactly at 12:42:51 PM).

Here's a step-by-step breakdown of the solution: **1. Understand the setup** - Distance between Chicago and St. Louis: `300 miles` - Train 1 (Chicago): Speed = `60 mph`, Departs at `10:00 AM` - Train 2 (St. Louis): Speed = `80 mph`, Departs at `11:00 AM` - Since Train 2 heads north and Train 1 is coming from Chicago (which is north of St. Louis), they are traveling **toward each other** along the same route. **2. Account for the head start** Train 1 leaves 1 hour before Train 2. - Distance covered by Train 1 from 10:00 AM to 11:00 AM: `60 mph × 1 hour = 60 miles` - Remaining distance between the two trains at 11:00 AM: `300 miles - 60 miles = 240 miles` **3. Calculate the closing speed** Because they are moving toward each other, their relative (closing) speed is the sum of their individual speeds: - `60 mph + 80 mph = 140 mph` **4. Find the time until they meet (after 11:00 AM)** - Time = Distance ÷ Speed - `Time = 240 miles ÷ 140 mph = 24/14 = 12/7 hours` **5. Convert `12/7 hours` to hours, minutes, and seconds** - `12/7 = 1 + 5/7 hours` - `5/7 hour × 60 minutes/hour = 300/7 ≈ 42.857 minutes` - `0.857 minutes × 60 seconds/minute ≈ 51.4 seconds` - So, `12/7 hours = 1 hour, 42 minutes, and 51.4 seconds` **6. Add this time to 11:00 AM** - `11:00:00 AM + 1 hour = 12:00:00 PM (noon)` - `12:00:00 PM + 42 minutes 51.4 seconds = 12:42:51 PM` ✅ **Final Answer:** The trains meet at **12:42:51 PM** (approximately **12:43 PM**). *(Note: If exact fractional form is preferred, the meeting time is exactly 12:42 and 6/7 minutes PM.)*

**Solution Overview** We have two cities, Chicago and St Louis, that are **300 mi** apart. - Train A leaves Chicago at **10 am**, heading **south** (toward St Louis) at **60 mph**. - Train B leaves St Louis at **11 am**, heading **north** (toward Chicago) at **80 mph**. We must find the clock time when the two trains meet. --- ## 1. Set up a time reference Let \[ t = \text{hours after 10:00 am} \] Thus - Train A starts at \(t=0\). - Train B starts at \(t=1\) hour (11 am). --- ## 2. Write the distance each train has traveled (in the north‑positive direction) If we call the northward direction **positive**: | Train | Start position | Velocity | Distance travelled (as a function of \(t\)) | |-------|----------------|----------|---------------------------------------------| | A (Chicago) | \(0\) mi (Chicago) | \(-60\) mph (south) | \(-60t\) | | B (St Louis) | \(-300\) mi (south of Chicago) | \(+80\) mph (north) | \(-300 + 80(t-1)\) for \(t\ge 1\) | *Why the “\(t-1\)” factor for Train B?* Train B does not start moving until 11 am, i.e. one hour after the reference time. --- ## 3. Equate positions – the meeting condition At the meeting moment the two trains are at the same coordinate: \[ -60t = -300 + 80(t-1), \qquad t\ge 1 \] --- ## 4. Solve for \(t\) \[ \begin{aligned} -60t &= -300 + 80t - 80 \\ -60t - 80t &= -300 - 80 \\ -140t &= -380 \\ t &= \frac{380}{140} = \frac{19}{7}\;\text{hours} \\ t &\approx 2.7142857\ \text{hours after 10:00 am} \end{aligned} \] --- ## 5. Convert the decimal hour to clock time \[ \frac{19}{7}\text{ h}=2\text{ h}+\frac{5}{7}\text{ h} \] \[ \frac{5}{7}\text{ h}= \frac{5}{7}\times 60\ \text{min}= \frac{300}{7}\ \text{min} = 42\ \text{min}+ \frac{6}{7}\ \text{min} \] \[ \frac{6}{7}\ \text{min}= \frac{6}{7}\times 60\ \text{s}= \frac{360}{7}\ \text{s} \approx 51.4\ \text{s} \] So the meeting occurs at \[ 10{:}00\ \text{am} + 2\ \text{h}\;42\ \text{min}\;51\ \text{s} \approx 12{:}42{:}51\ \text{pm}. \] Rounded to the nearest minute, **they meet at about 12:43 pm**. --- ## 6. Quick sanity‑check (relative‑speed method) 1. At 11 am (one hour after Train A starts) Train A has already covered \[ 60\ \text{mph}\times1\ \text{h}=60\ \text{mi}. \] Distance still between them: \(300-60=240\) mi. 2. After 11 am both trains are moving toward each other, so their **relative speed** is \[ 60\ \text{mph}+80\ \text{mph}=140\ \text{mph}. \] 3. Time needed to close the remaining 240 mi is \[ \frac{240\ \text{mi}}{140\ \text{mph}} = \frac{12}{7}\ \text{h} \approx 1.7142857\ \text{h}=1\ \text{h}\;42\ \text{min}\;51\ \text{s}. \] 4. Adding that to 11 am gives the same result: 12:42 pm (≈12:43 pm). Both approaches agree. --- ## 7. Final Answer **The trains meet at approximately 12 : 42 : 51 pm (about 12:43 pm).** --- *All reasoning steps have been shown, including the algebraic set‑up, the arithmetic, and a quick relative‑speed verification.*

To determine what time the two trains meet, we need to calculate the distance each train covers and track their positions over time. Here is the step-by-step reasoning: **Step 1: Understand the starting conditions** * The distance between Chicago and St. Louis is 300 miles. * Train 1 (Chicago) leaves at 10:00 AM traveling South at 60 mph. * Train 2 (St. Louis) leaves at 11:00 AM traveling North at 80 mph. **Step 2: Calculate Train 1's head start** Train 1 leaves exactly one hour before Train 2. During this hour, Train 2 is sitting still in St. Louis. * Distance covered by Train 1 between 10:00 AM and 11:00 AM = Speed × Time = 60 mph × 1 hour = **60 miles**. **Step 3: Calculate the remaining distance at 11:00 AM** At 11:00 AM, Train 2 begins to move. We need to find out how far apart they are at this exact moment. * Remaining distance = Total distance - Train 1's head start distance * Remaining distance = 300 miles - 60 miles = **240 miles**. **Step 4: Calculate their combined (relative) speed** From 11:00 AM onward, both trains are moving toward each other. When two objects move toward each other, their speeds add up to determine how fast the gap between them is closing. * Combined speed = Speed of Train 1 + Speed of Train 2 * Combined speed = 60 mph + 80 mph = **140 mph**. **Step 5: Calculate the time it takes to close the gap** Now we divide the remaining distance by their combined speed to find out how long it takes them to meet after 11:00 AM. * Time = Distance ÷ Speed * Time = 240 miles ÷ 140 mph = **12/7 hours** (or roughly 1.714 hours). **Step 6: Convert the time into hours and minutes** To make this readable, we convert the fraction of an hour into minutes: * 12/7 hours = 1 full hour and 5/7 of an hour. * (5/7) × 60 minutes = 300/7 minutes = **42 and 6/7 minutes** (approximately 42 minutes and 51 seconds). **Step 7: Add the travel time to the 11:00 AM start time** * Start time: 11:00 AM * Add 1 hour: 12:00 PM (Noon) * Add 42 minutes and 51 seconds: 12:42 PM and 51 seconds. **Conclusion:** The two trains will meet at exactly **12:42 and 6/7 PM** (or approximately **12:42:51 PM**).

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}}

**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

To solve this, we first assume the first train is heading **south** from Chicago toward St. Louis. (If it were heading any other direction, it would not meet the second train, which is heading north from St. Louis.) Here is the step-by-step reasoning: ### 1. Account for the head start The first train leaves at 10:00 AM and travels alone for **1 hour** before the second train departs at 11:00 AM. *Distance covered by the first train in that hour:* $$60 \text{ mph} \times 1 \text{ hour} = 60 \text{ miles}$$ So, by 11:00 AM, the gap between the two trains has shrunk from 300 miles to: $$300 - 60 = 240 \text{ miles}$$ ### 2. Determine their combined closing speed After 11:00 AM, the trains are moving toward each other. Their speeds add up: $$60 \text{ mph} + 80 \text{ mph} = 140 \text{ mph}$$ ### 3. Calculate how long it takes to close the remaining gap With 240 miles left and a combined speed of 140 mph, the time required to meet is: $$\text{Time} = \frac{\text{Remaining Distance}}{\text{Combined Speed}} = \frac{240}{140} = \frac{12}{7} \text{ hours}$$ ### 4. Convert the fractional hours to minutes $$\frac{12}{7} \text{ hours} = 1\frac{5}{7} \text{ hours}$$ Convert the fractional part to minutes: $$\frac{5}{7} \times 60 = \frac{300}{7} \approx 42.857 \text{ minutes}$$ So they meet **1 hour and about 42.86 minutes** after 11:00 AM. ### 5. Find the exact meeting time $$11:00 \text{ AM} + 1 \text{ hour} = 12:00 \text{ PM (noon)}$$ $$12:00 \text{ PM} + 42\frac{6}{7} \text{ minutes} = 12:42\frac{6}{7} \text{ PM}$$ The fraction of a minute is: $$\frac{6}{7} \times 60 \approx 51.4 \text{ seconds}$$ --- ## Answer The trains meet at exactly **12:42 and $\frac{6}{7}$ minutes PM**, or approximately **12:42:51 PM**. *(Verification: The first train travels for $1 + \frac{12}{7} = \frac{19}{7}$ hours, covering $60 \times \frac{19}{7} = \frac{1140}{7}$ miles. The second train travels for $\frac{12}{7}$ hours, covering $80 \times \frac{12}{7} = \frac{960}{7}$ miles. Together: $\frac{1140+960}{7} = \frac{2100}{7} = 300$ miles. ✓)*

Let’s carefully break down the problem. **Step 1: Define positions and speeds** - Train A leaves Chicago at 10:00 AM, traveling **south** at 60 mph. - Train B leaves St. Louis at 11:00 AM, traveling **north** at 80 mph. - Distance between Chicago and St. Louis: **300 miles**. --- **Step 2: Account for the head start of Train A** From 10:00 AM to 11:00 AM, Train A travels alone for 1 hour: \[ 60 \text{ mph} \times 1 \text{ hour} = 60 \text{ miles south.} \] So at 11:00 AM, Train A is **60 miles south of Chicago**. The remaining distance between the two trains at 11:00 AM is: \[ 300 - 60 = 240 \text{ miles.} \] --- **Step 3: Relative speed after 11:00 AM** They are moving toward each other, so their combined speed is: \[ 60 + 80 = 140 \text{ mph.} \] Time needed to close the 240-mile gap: \[ \frac{240}{140} = \frac{24}{14} = \frac{12}{7} \text{ hours.} \] That’s: \[ 1 \text{ hour and } \frac{5}{7} \text{ of an hour} \approx 1 \text{ hour } 42 \text{ minutes } 51 \text{ seconds.} \] --- **Step 4: Find meeting time** They start closing the gap at 11:00 AM. Time added: 1 hour 42 minutes and approximately 51 seconds. Meeting time: \[ 11:00 + 1:42:51 \approx 12:42:51 \text{ PM}. \] So they meet approximately at **12:43 PM**. --- \[ \boxed{12:43 \text{ PM}} \]

**Step‑by‑step reasoning** 1. **Define the situation** * Train A (Chicago → St. Louis) leaves Chicago at 10:00 AM, speed = **60 mph**. * Train B (St. Louis → Chicago) leaves St. Louis at 11:00 AM, speed = **80 mph**. * Distance between the cities = **300 mi** (south‑north line). We’ll work in *hours* after **10:00 AM**. 2. **How far is Train A when Train B starts?** Train A travels for 1 hour before Train B departs: \[ d_A(1\text{ h}) = 60\ \text{mph} \times 1\ \text{h}=60\ \text{mi} \] So at 11:00 AM the gap between the two trains is \[ 300\ \text{mi} - 60\ \text{mi}=240\ \text{mi}. \] 3. **Relative speed after 11:00 AM** Both trains are now moving toward each other, so their *relative* speed is the sum of their individual speeds: \[ v_{\text{rel}} = 60\ \text{mph} + 80\ \text{mph}=140\ \text{mph}. \] 4. **Time needed to close the 240‑mile gap** \[ t_{\text{close}} = \frac{\text{gap}}{v_{\text{rel}}} = \frac{240\ \text{mi}}{140\ \text{mph}} = \frac{12}{7}\ \text{h} = 1.7142857\ \text{h}. \] Convert the fractional hour to minutes and seconds: \[ 0.7142857\ \text{h}\times60\ \frac{\text{min}}{\text{h}} = 42.8571\ \text{min} \] \[ 0.8571\ \text{min}\times60\ \frac{\text{s}}{\text{min}} \approx 51.43\ \text{s}. \] So the closing phase lasts **1 hour 42 minutes 51 seconds**. 5. **Add this to the starting time of the second train** * Second train departs at **11:00 AM**. * Add the closing time: \[ 11{:}00\ \text{AM} + 1\ \text{h}\;42\ \text{min}\;51\ \text{s} = 12{:}42\ \text{min}\;51\ \text{s}\ \text{PM}. \] (If you round to the nearest minute, this is **12:43 PM**.) --- ### Verification with an algebraic approach Let * \(t\) = time after **10:00 AM** (in hours) when the trains meet. - Position of Train A (southward from Chicago): \[ d_A(t) = 60t. \] - Train B starts one hour later, so for \(t\ge 1\) its position (northward from St. Louis) is \[ d_B(t) = 300 - 80(t-1). \] Set the positions equal when they meet: \[ 60t = 300 - 80(t-1) \;\Longrightarrow\; 60t = 300 - 80t + 80 \;\Longrightarrow\; 140t = 380 \;\Longrightarrow\; t = \frac{380}{140}=2.7142857\text{ h}. \] **\(t=2.7142857\) h after 10:00 AM** equals \(2\) h \(+0.7142857\) h = 2 h 42 min 51 s → **12:42:51 PM**. Both methods give the same answer. --- ## Answer The two trains meet **approximately at 12:42 PM (more precisely 12 h 42 min 51 s after midnight, i.e., 12:42:51 PM).**