On a 1-mile test course, a vehicle is travelling at a constant rate of r miles per hour. If it...

1. Translate the problem requirements: We need to find the original speed r when a 1-mile course takes 4 minutes less time at speed (r+20) compared to speed r
2. Set up the time relationship using distance/speed: Express the time taken at each speed and identify that the difference equals 4 minutes
3. Create and solve the equation: Set up the equation from the time difference and solve for r
4. Verify using answer choices: Check that our solution makes practical sense with the given options

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what's happening here in everyday terms. We have a vehicle traveling on a 1-mile test course at some speed r miles per hour. If that same vehicle could go 20 mph faster, it would complete the same 1-mile course in 4 minutes less time.

Think of it like this: imagine you're driving to a destination that's exactly 1 mile away. If you drive faster, you'll get there sooner. The question tells us that driving 20 mph faster saves exactly 4 minutes of travel time.

We need to find the original speed r.

Process Skill: TRANSLATE - Converting the problem's language into a clear mathematical relationship

2. Set up the time relationship using distance/speed

Now let's think about how time, speed, and distance work together. When you travel a certain distance, the time it takes equals the distance divided by your speed.

For our 1-mile course:

• At speed r mph: Time = \(\frac{1}{\mathrm{r}}\) hours
• At speed (r+20) mph: Time = \(\frac{1}{(\mathrm{r}+20)}\) hours

The key insight is that the difference between these two times equals 4 minutes. But we need to be careful about units - our speeds are in miles per hour, so our times will be in hours. Since 4 minutes = \(\frac{4}{60}\) hours = \(\frac{1}{15}\) hours, we can write:

Time at slower speed - Time at faster speed = \(\frac{1}{15}\) hours

This gives us: \(\frac{1}{\mathrm{r}} - \frac{1}{(\mathrm{r}+20)} = \frac{1}{15}\)

3. Create and solve the equation

Let's solve this step by step. We have: \(\frac{1}{\mathrm{r}} - \frac{1}{(\mathrm{r}+20)} = \frac{1}{15}\)

To subtract these fractions, we need a common denominator. The common denominator for \(\frac{1}{\mathrm{r}}\) and \(\frac{1}{(\mathrm{r}+20)}\) is r(r+20):

\(\frac{1}{\mathrm{r}} - \frac{1}{(\mathrm{r}+20)} = \frac{(\mathrm{r}+20)}{[\mathrm{r}(\mathrm{r}+20)]} - \frac{\mathrm{r}}{[\mathrm{r}(\mathrm{r}+20)]} = \frac{(\mathrm{r}+20-\mathrm{r})}{[\mathrm{r}(\mathrm{r}+20)]} = \frac{20}{[\mathrm{r}(\mathrm{r}+20)]}\)

So our equation becomes: \(\frac{20}{[\mathrm{r}(\mathrm{r}+20)]} = \frac{1}{15}\)

Cross-multiplying: \(20 \times 15 = \mathrm{r}(\mathrm{r}+20)\)

\(300 = \mathrm{r}^2 + 20\mathrm{r}\)

Rearranging: \(\mathrm{r}^2 + 20\mathrm{r} - 300 = 0\)

We can factor this quadratic. We need two numbers that multiply to -300 and add to 20. Those numbers are 30 and -10:

\((\mathrm{r} + 30)(\mathrm{r} - 10) = 0\)

This gives us \(\mathrm{r} = -30\) or \(\mathrm{r} = 10\)

Since speed cannot be negative, \(\mathrm{r} = 10\) mph.

Process Skill: MANIPULATE - Systematically working through the algebraic steps to isolate the variable

4. Verify using answer choices

Let's check our answer r = 10 mph makes sense:

• At 10 mph: Time = \(\frac{1}{10}\) hour = 6 minutes
• At 30 mph (10+20): Time = \(\frac{1}{30}\) hour = 2 minutes
• Time difference: \(6 - 2 = 4\) minutes ✓

Perfect! This matches exactly what the problem stated.

Looking at our answer choices, r = 10 corresponds to choice B.

Final Answer

The value of r is 10 miles per hour.

Answer: B. 10

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Unit Confusion
Students often miss that the 4-minute time difference needs to be converted to hours (\(\frac{1}{15}\) hours) to match the units of their speed calculation. They might incorrectly set up the equation as \(\frac{1}{\mathrm{r}} - \frac{1}{(\mathrm{r}+20)} = 4\), leading to a completely wrong solution path.

Faltering Point 2: Misunderstanding the Time Relationship
Some students incorrectly interpret "4 minutes less time" and set up the equation backwards, writing \(\frac{1}{(\mathrm{r}+20)} - \frac{1}{\mathrm{r}} = \frac{1}{15}\) instead of \(\frac{1}{\mathrm{r}} - \frac{1}{(\mathrm{r}+20)} = \frac{1}{15}\). This fundamental error stems from not clearly understanding that higher speed means lower time.

Faltering Point 3: Incorrect Distance-Speed-Time Application
Students might confuse the relationship and think that since distance is 1 mile, they should use time = speed/distance instead of time = distance/speed, leading them to write r and (r+20) instead of \(\frac{1}{\mathrm{r}}\) and \(\frac{1}{(\mathrm{r}+20)}\) for the time expressions.

Errors while executing the approach

Faltering Point 1: Common Denominator Errors
When subtracting fractions \(\frac{1}{\mathrm{r}} - \frac{1}{(\mathrm{r}+20)}\), students often make algebraic mistakes in finding the common denominator or in the numerator manipulation. They might incorrectly get (20-r) instead of 20 in the numerator, or use an incorrect common denominator.

Faltering Point 2: Cross-Multiplication and Arithmetic Errors
After setting up \(\frac{20}{[\mathrm{r}(\mathrm{r}+20)]} = \frac{1}{15}\), students may make errors during cross-multiplication (getting \(300 = \mathrm{r}^2 + 20\mathrm{r}\) correctly) or when rearranging to standard quadratic form. Simple arithmetic mistakes can lead to wrong coefficients in the quadratic equation.

Faltering Point 3: Quadratic Factoring Mistakes
When factoring \(\mathrm{r}^2 + 20\mathrm{r} - 300 = 0\), students might struggle to find the correct factor pair that multiplies to -300 and adds to 20. They could incorrectly factor as \((\mathrm{r} + 25)(\mathrm{r} - 12)\) or other incorrect combinations, leading to wrong values of r.

Errors while selecting the answer

Faltering Point 1: Accepting Negative Speed
After solving the quadratic and getting r = 10 or r = -30, some students might not recognize that speed cannot be negative in this context, potentially selecting an answer that doesn't make physical sense or getting confused about which solution to use.

Faltering Point 2: Skipping Verification
Students might arrive at r = 10 but fail to verify their answer by substituting back into the original problem conditions. This verification step is crucial to catch any errors made during the solution process and ensure the answer makes sense in the problem context.