When a certain stretch of highway was rebuilt and straightened, the distance along the stretch was decreased by 20 percent...

1. Translate the problem requirements: We need to find the percent reduction in driving time when distance decreases by 20% and speed limit increases by 25%. This involves understanding how changes in distance and speed affect travel time.
2. Set up the relationship using concrete values: Use the fundamental relationship Time = Distance ÷ Speed with simple baseline values to make calculations straightforward.
3. Calculate the new time after both changes: Apply both the 20% distance reduction and 25% speed increase to determine the new travel time.
4. Determine the percent change in time: Compare the new time to the original time to find the percent reduction.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening in everyday terms. When they rebuilt the highway, two things changed:

• The distance became shorter by 20% (because they straightened out curves)
• The speed limit increased by 25% (safer straight road allows higher speeds)

We need to find how these changes affect the time it takes to drive this stretch. Think about it intuitively: if the distance gets shorter, you'll spend less time driving. If the speed limit goes up and you drive faster, you'll also spend less time driving. Both changes work together to reduce your driving time.

Process Skill: TRANSLATE - Converting the real-world scenario into mathematical relationships we can work with

2. Set up the relationship using concrete values

Let's use simple numbers to make this easy to follow. The key relationship we need is:

\(\mathrm{Time} = \mathrm{Distance} ÷ \mathrm{Speed}\)

Let's say originally:

• Distance = 100 miles (easy number to work with)
• Speed limit = 60 mph (another easy number)
• Original time = \(100 ÷ 60 = \frac{5}{3}\) hours

Now let's figure out what happens after the changes:

• New distance = 100 - (20% of 100) = 100 - 20 = 80 miles
• New speed = 60 + (25% of 60) = 60 + 15 = 75 mph

3. Calculate the new time after both changes

Now we can find the new driving time using our formula:

New time = New distance ÷ New speed

New time = 80 ÷ 75

To make this division easier, let's simplify:

\(80 ÷ 75 = \frac{80}{75} = \frac{16}{15}\) hours

So our times are:

• Original time = \(\frac{5}{3}\) hours
• New time = \(\frac{16}{15}\) hours

4. Determine the percent change in time

To find the percent reduction, we need to compare the new time to the original time.

First, let's find the difference in time:

Time saved = Original time - New time = \(\frac{5}{3} - \frac{16}{15}\)

To subtract these fractions, we need a common denominator. The LCD of 3 and 15 is 15:

\(\frac{5}{3} = \frac{25}{15}\)

So: Time saved = \(\frac{25}{15} - \frac{16}{15} = \frac{9}{15} = \frac{3}{5}\) hours

Percent reduction = (Time saved ÷ Original time) × 100%

Percent reduction = \((\frac{3}{5}) ÷ (\frac{5}{3}) \times 100\%\)

Percent reduction = \((\frac{3}{5}) \times (\frac{3}{5}) \times 100\%\)

Percent reduction = \(\frac{9}{25} \times 100\% = 36\%\)

Final Answer

The driving time along this stretch was reduced by 36%.

This matches answer choice B. The intuition makes sense: both the shorter distance and higher speed limit work together to significantly reduce driving time, and the 36% reduction reflects the combined effect of both improvements.

Common Faltering Points

Errors while devising the approach

• Misunderstanding the direction of change: Students may incorrectly think that since the speed limit increased by 25%, the time also increases by 25%. They fail to recognize the inverse relationship between speed and time - when speed increases, time decreases (for the same distance).
• Treating changes as additive instead of multiplicative: Students might think they can simply subtract 20% (distance reduction) and add 25% (speed increase) to get a net 5% change, rather than understanding that these percentage changes need to be applied multiplicatively to find the combined effect on time.
• Confusion about what quantity to find the percentage change for: Students may start calculating the percentage change in distance or speed instead of focusing on the time reduction, which is what the question actually asks for.

Errors while executing the approach

• Arithmetic errors with fractions: When calculating \((\frac{3}{5}) ÷ (\frac{5}{3})\), students commonly make the error of multiplying by the wrong reciprocal, calculating \((\frac{3}{5}) \times (\frac{5}{3})\) instead of \((\frac{3}{5}) \times (\frac{3}{5})\), leading to an incorrect result of 1 instead of \(\frac{9}{25}\).
• Finding common denominators incorrectly: When subtracting \(\frac{5}{3} - \frac{16}{15}\), students may struggle to find the correct common denominator of 15, or make errors when converting \(\frac{5}{3}\) to \(\frac{25}{15}\), leading to incorrect time difference calculations.
• Calculation errors in percentage conversions: Students may correctly get \(\frac{9}{25}\) but then make arithmetic mistakes when converting to percentage, such as calculating 9 ÷ 25 = 0.36 but forgetting to multiply by 100, or making decimal calculation errors.

Errors while selecting the answer

• Selecting the wrong magnitude: If students make calculation errors and arrive at 0.36 instead of 36%, they might look for 0.36% in the answer choices and get confused, or they might incorrectly select an answer choice that represents a decimal value rather than the percentage.
• Choosing the complement percentage: Students who correctly calculate that time is reduced to 64% of original time (since \(\frac{16}{15} ÷ \frac{5}{3} = 64\%\)) might mistakenly select 64% as their answer instead of recognizing that a 36% reduction means the time is now 64% of the original.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose convenient baseline values

To make calculations clean, let's use smart numbers that work well with the given percentages:

• Original distance = 100 miles (convenient for percentage calculations)
• Original speed limit = 80 mph (convenient because 25% increase gives us a clean number)

Step 2: Calculate original driving time

Using Time = Distance ÷ Speed:

Original time = \(100 \text{ miles} ÷ 80 \text{ mph} = 1.25 \text{ hours}\)

Step 3: Apply the changes and find new values

After reconstruction:

• New distance = 100 - (20% of 100) = 100 - 20 = 80 miles
• New speed limit = 80 + (25% of 80) = 80 + 20 = 100 mph

Step 4: Calculate new driving time

New time = \(80 \text{ miles} ÷ 100 \text{ mph} = 0.8 \text{ hours}\)

Step 5: Find the percent reduction in time

Time reduction = Original time - New time = 1.25 - 0.8 = 0.45 hours

Percent reduction = (Time reduction ÷ Original time) × 100%

Percent reduction = \((0.45 ÷ 1.25) \times 100\% = 0.36 \times 100\% = 36\%\)

Verification: The smart numbers approach gives us the same answer as the algebraic method, confirming that the driving time is reduced by 36%.