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Two trains, X and Y, started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; Train Y, traveling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had Train X traveled when it met Train Y?
Let's start by understanding what's happening in this problem in everyday terms. Imagine two trains starting at the same time from opposite ends of a straight 100-mile track. They're heading directly toward each other on parallel tracks, so they will definitely meet somewhere along the way.
We need to find exactly how many miles Train X has traveled by the time these two trains meet each other. This is like asking: if two people start walking toward each other from opposite ends of a hallway, how far has the slower person walked when they meet?
The key insight is that both trains travel for the exact same amount of time until they meet - they started simultaneously and meet at the same moment.
Process Skill: TRANSLATE - Converting the word problem into a clear mathematical scenario
Now let's figure out how fast each train travels using simple reasoning.
Train X completes the entire 100-mile journey in 5 hours when traveling alone. This means Train X travels at a steady pace of:
- \(100 \text{ miles} ÷ 5 \text{ hours} = 20 \text{ miles per hour}\)
Train Y completes the same 100-mile journey in 3 hours when traveling alone. This means Train Y travels at:
- \(100 \text{ miles} ÷ 3 \text{ hours} = 33\frac{1}{3} \text{ miles per hour}\) (or 33.33... mph)
So Train Y is faster than Train X, which makes sense - it completes the same distance in less time.
Technical notation: \(\text{Speed} = \text{Distance} ÷ \text{Time}\), so \(\text{Speed}_X = 20 \text{ mph and Speed}_Y = 33\frac{1}{3} \text{ mph}\)
Here's the key insight: when the trains meet, the total distance they've covered together equals exactly 100 miles (the length of the route).
Think of it this way: if Train X travels some distance and Train Y travels some distance, and they meet, then those two distances must add up to the full 100 miles between their starting points.
Since they travel for the same amount of time (let's call it 't' hours), we can say:
- \(\text{Distance covered by Train X} = 20t \text{ miles}\)
- \(\text{Distance covered by Train Y} = 33\frac{1}{3}t \text{ miles}\)
- These distances must add up to 100 miles
So: \(20t + 33\frac{1}{3}t = 100\)
Combining: \(53\frac{1}{3}t = 100\)
Solving for t: \(t = 100 ÷ 53\frac{1}{3} = 100 ÷ \frac{160}{3} = 100 × \frac{3}{160} = \frac{300}{160} = \frac{15}{8} = 1.875 \text{ hours}\)
The trains meet after 1.875 hours (or 1 hour and 52.5 minutes).
Process Skill: INFER - Recognizing that combined distances equal total route length
Now we can find how far Train X has traveled when the trains meet.
Train X travels at 20 mph for 1.875 hours:
\(\text{Distance} = \text{Speed} × \text{Time} = 20 × 1.875 = 37.5 \text{ miles}\)
Let's verify this makes sense: if Train X traveled 37.5 miles, then Train Y should have traveled \(100 - 37.5 = 62.5 \text{ miles}\).
Check: Train Y at \(33\frac{1}{3}\) mph for 1.875 hours = \(33\frac{1}{3} × 1.875 = \frac{100}{3} × \frac{15}{8} = \frac{1500}{24} = 62.5 \text{ miles}\) ✓
Train X had traveled 37.5 miles when it met Train Y.
This matches answer choice (A) 37.5.
Students often struggle with the key insight that when two objects moving toward each other meet, the sum of distances they've each traveled equals the total distance between their starting points. Some students might think they need to find when the trains are closest to each other or use more complex geometric relationships, rather than recognizing this is simply a "combined distance equals total distance" problem.
Students may get confused about timing and think that since the trains have different speeds, they somehow travel for different amounts of time before meeting. The crucial insight that both trains start simultaneously and meet at the same moment (meaning they travel for identical time periods) is often missed.
Students might incorrectly think the 5-hour and 3-hour times given in the problem represent how long it takes each train to reach the meeting point, rather than understanding these are the times each train would take to complete the entire 100-mile journey if traveling alone.
The calculation involves working with \(33\frac{1}{3} \text{ mph}\) (which is \(\frac{100}{3}\)), and students often make mistakes when adding fractions like \(20t + \frac{100}{3}t = 100\). Converting between mixed numbers, improper fractions, and decimals creates multiple opportunities for computational errors.
When calculating Train Y's speed, students might incorrectly compute \(100 ÷ 3 = 30 \text{ mph}\) instead of the correct \(33\frac{1}{3} \text{ mph}\), or make similar basic division errors. These early mistakes compound throughout the rest of the solution.
The step \(t = 100 ÷ 53\frac{1}{3}\) requires converting \(53\frac{1}{3}\) to an improper fraction \(\frac{160}{3}\) and then dividing, which becomes \(100 × \frac{3}{160}\). Students frequently make errors in this multi-step division process, especially when dealing with the fraction manipulation.
After correctly calculating that trains meet when Train X has traveled 37.5 miles and Train Y has traveled 62.5 miles, students may accidentally select 62.5 (answer choice D) because Train Y is the faster train or because they worked with Train Y's distance last in their verification step.
Students who correctly calculate the meeting time as 1.875 hours might mistakenly look for this value among the answer choices, not realizing they need to take one additional step to multiply this time by Train X's speed to get the distance traveled.