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During a trip on an expressway, Don drove a total of \(\mathrm{x}\) miles. His average speed on a certain \(\mathrm{5}\)-mile section of the expressway was \(\mathrm{30}\) miles per hour, and his average speed for the remainder of the trip was \(\mathrm{60}\) miles per hour. His travel time for the \(\mathrm{x}\)-mile trip was what percent greater than it would have been if he had traveled at a constant rate of \(\mathrm{60}\) miles per hour for the entire trip?
Let's break down what we're being asked to find. Don drove x miles total, but at two different speeds for different parts of the trip. We need to compare his actual travel time to what it would have been if he had driven the entire trip at 60 mph.
Think of it this way: if you're running late and have to drive slowly through a construction zone, you'll arrive later than if you could have driven at highway speed the whole way. We want to know how much longer Don's trip took because of that slow 5-mile section.
The question asks for the percent increase, which means we need:
Process Skill: TRANSLATE - Converting the complex percentage comparison into clear time calculations
Don's trip has two parts with different speeds, so let's find the time for each part separately.
For the slow section: Don drove 5 miles at 30 mph. Using the basic relationship that time equals distance divided by speed, this took him \(5 \div 30 = \frac{1}{6}\) hours.
For the fast section: Don drove the remaining \((\mathrm{x} - 5)\) miles at 60 mph. This took him \(\frac{(\mathrm{x} - 5)}{60}\) hours.
Don's total actual time = \(\frac{1}{6} + \frac{(\mathrm{x} - 5)}{60}\) hours
To add these fractions, we need a common denominator. Since \(60 = 6 \times 10\), we can rewrite \(\frac{1}{6}\) as \(\frac{10}{60}\):
Total actual time = \(\frac{10}{60} + \frac{(\mathrm{x} - 5)}{60} = \frac{(10 + \mathrm{x} - 5)}{60} = \frac{(\mathrm{x} + 5)}{60}\) hours
This is straightforward. If Don had driven the entire x miles at 60 mph, his time would have been:
Hypothetical time = \(\mathrm{x} \div 60 = \frac{\mathrm{x}}{60}\) hours
The extra time Don spent = Actual time - Hypothetical time
Extra time = \(\frac{(\mathrm{x} + 5)}{60} - \frac{\mathrm{x}}{60} = \frac{(\mathrm{x} + 5 - \mathrm{x})}{60} = \frac{5}{60} = \frac{1}{12}\) hours
Notice something important here: the extra time is always \(\frac{1}{12}\) hours, regardless of the total distance x. This makes sense because the only difference between the two scenarios is that 5-mile section where Don drove at 30 mph instead of 60 mph.
At 30 mph, those 5 miles took \(\frac{1}{6}\) hours. At 60 mph, they would have taken \(\frac{5}{60} = \frac{1}{12}\) hours. The difference is \(\frac{1}{6} - \frac{1}{12} = \frac{2}{12} - \frac{1}{12} = \frac{1}{12}\) hours.
To find the percent increase:
Percent increase = \((\text{Extra time} \div \text{Hypothetical time}) \times 100\%\)
Percent increase = \(\left(\frac{1}{12}\right) \div \left(\frac{\mathrm{x}}{60}\right) \times 100\%\)
= \(\left(\frac{1}{12}\right) \times \left(\frac{60}{\mathrm{x}}\right) \times 100\%\)
= \(\frac{60}{12\mathrm{x}} \times 100\%\)
= \(\frac{5}{\mathrm{x}} \times 100\%\)
= \(\frac{500}{\mathrm{x}}\%\)
Process Skill: SIMPLIFY - Recognizing that the extra time is constant regardless of x, which simplifies our calculation significantly
Don's travel time was \(\frac{500}{\mathrm{x}}\) percent greater than it would have been at constant 60 mph.
This matches answer choice E. \(\frac{500}{\mathrm{x}}\%\)
Verification: Let's check with a concrete example. If \(\mathrm{x} = 25\) miles, Don's actual time would be \(\frac{(25+5)}{60} = \frac{1}{2}\) hour, while constant 60 mph would take \(\frac{25}{60} = \frac{5}{12}\) hours. The percent increase is \(\left(\frac{1}{2} - \frac{5}{12}\right) \div \left(\frac{5}{12}\right) \times 100\% = \left(\frac{1}{12}\right) \div \left(\frac{5}{12}\right) \times 100\% = \frac{1}{5} \times 100\% = 20\%\). Using our formula: \(\frac{500}{25} = 20\%\). ✓
Faltering Point 1: Misinterpreting what "percent greater" means. Students often confuse this with finding the ratio or simple difference between times, rather than understanding they need to find: (actual time - hypothetical time) / hypothetical time × 100%. This leads them to set up incorrect formulas from the start.
Faltering Point 2: Incorrectly identifying the two scenarios to compare. Some students might compare Don's time on just the 5-mile section at 30 mph versus that same section at 60 mph, instead of comparing his total trip time versus the total trip time at constant 60 mph. This fundamental misunderstanding leads to completely wrong calculations.
Faltering Point 3: Failing to recognize that the trip consists of exactly two distinct parts with different speeds. Students might assume there are multiple sections with varying speeds or misread the problem structure, leading them to overcomplicate their approach unnecessarily.
Faltering Point 1: Making arithmetic errors when finding common denominators. When adding \(\frac{1}{6} + \frac{(\mathrm{x}-5)}{60}\), students often struggle with converting \(\frac{1}{6}\) to \(\frac{10}{60}\), or make mistakes in the fraction addition, leading to incorrect total actual time calculations.
Faltering Point 2: Incorrectly calculating the time for each section by mixing up distance, speed, and time relationships. For example, using \(30 \div 5\) instead of \(5 \div 30\) for the slow section, or getting confused about which distance corresponds to which speed.
Faltering Point 3: Making errors in the final percentage calculation, particularly when dividing fractions. Students often get confused when computing \(\left(\frac{1}{12}\right) \div \left(\frac{\mathrm{x}}{60}\right)\), either forgetting to flip the second fraction or making arithmetic mistakes that prevent them from reaching the simplified form \(\frac{500}{\mathrm{x}}\).
Faltering Point 1: Selecting answer choice D (\(\frac{60}{\mathrm{x}}\%\)) instead of E (\(\frac{500}{\mathrm{x}}\%\)) due to computational errors in the final simplification step. Students might incorrectly simplify \(\frac{60}{12\mathrm{x}} \times 100\%\) as \(\frac{60}{\mathrm{x}}\%\) instead of recognizing it equals \(\frac{500}{\mathrm{x}}\%\).
Faltering Point 2: Choosing a numerical answer choice like A (8.5%) or B (50%) by plugging in a specific value for x during their solution process and forgetting that the final answer must be expressed in terms of the variable x, not as a fixed percentage.
Step 1: Choose a convenient value for x
Since we need to work with a 5-mile section out of x total miles, let's choose \(\mathrm{x} = 65\) miles. This gives us clean numbers to work with:
• 5-mile section at 30 mph
• Remaining \((65 - 5) = 60\) miles at 60 mph
Step 2: Calculate Don's actual travel time
Time for 5-mile section: \(5 \text{ miles} \div 30 \text{ mph} = \frac{1}{6}\) hour
Time for remaining 60 miles: \(60 \text{ miles} \div 60 \text{ mph} = 1\) hour
Total actual time = \(\frac{1}{6} + 1 = \frac{7}{6}\) hours
Step 3: Calculate hypothetical time at constant 60 mph
Time for entire 65-mile trip at 60 mph: \(65 \text{ miles} \div 60 \text{ mph} = \frac{13}{12}\) hours
Step 4: Find the time difference
Extra time = \(\frac{7}{6} - \frac{13}{12}\)
Converting to common denominator: \(\frac{7}{6} = \frac{14}{12}\)
Extra time = \(\frac{14}{12} - \frac{13}{12} = \frac{1}{12}\) hour
Step 5: Calculate percent increase
Percent increase = \((\text{Extra time} \div \text{Hypothetical time}) \times 100\%\)
= \(\left(\frac{1}{12} \div \frac{13}{12}\right) \times 100\%\)
= \(\left(\frac{1}{12} \times \frac{12}{13}\right) \times 100\%\)
= \(\frac{1}{13} \times 100\% = \frac{100}{13}\%\)
Step 6: Verify with the answer formula
Using our smart number \(\mathrm{x} = 65\) in answer choice E:
\(\frac{500}{\mathrm{x}}\% = \frac{500}{65}\% = \frac{100}{13}\%\)
This matches our calculated result, confirming that answer choice E (\(\frac{500}{\mathrm{x}}\%\)) is correct.