After driving to a riverfront parking lot, Bob plans to run south along the river, turn around, and return to...

1. Translate the problem requirements: Bob has already run \(3.25\) miles south and has exactly \(50\) minutes remaining. He wants to run farther south (let's call this additional distance '\(\mathrm{x}\)' miles) and then return to the parking lot within his \(50\)-minute time limit. Since he runs at \(8\) minutes per mile, we need to find the maximum value of \(\mathrm{x}\).
2. Set up the time constraint equation: Bob's remaining journey consists of running \(\mathrm{x}\) more miles south (taking \(8\mathrm{x}\) minutes) plus running back the total distance of \((3.25 + \mathrm{x})\) miles north (taking \(8(3.25 + \mathrm{x})\) minutes). The total time must equal \(50\) minutes.
3. Solve for the maximum additional distance: Create the equation \(8\mathrm{x} + 8(3.25 + \mathrm{x}) = 50\) and solve for \(\mathrm{x}\) to find how much farther south Bob can run.
4. Verify the solution makes sense: Check that the calculated distance allows Bob to complete his return journey within the time constraint.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what Bob is doing step by step. Bob has already run \(3.25\) miles south from the parking lot. Now he wants to run even farther south, but he only has \(50\) minutes left for his entire remaining journey. The key insight is that Bob needs to get back to where he started - the parking lot - within those \(50\) minutes.

So Bob's remaining journey has two parts:

• Run some additional distance south (let's call this '\(\mathrm{x}\)' miles)
• Turn around and run all the way back north to the parking lot

If he runs \(\mathrm{x}\) more miles south, his total distance from the parking lot will be \((3.25 + \mathrm{x})\) miles. This means he'll need to run exactly \((3.25 + \mathrm{x})\) miles north to get back to the parking lot.

Process Skill: TRANSLATE - Converting the word problem into clear mathematical relationships

2. Set up the time constraint equation

Now let's think about the time Bob will spend. He runs at a constant rate of \(8\) minutes per mile, which means:

• Running \(\mathrm{x}\) miles south will take him: \(8\mathrm{x}\) minutes
• Running \((3.25 + \mathrm{x})\) miles north back to the parking lot will take him: \(8(3.25 + \mathrm{x})\) minutes

Since Bob has exactly \(50\) minutes total for both parts of his remaining journey, we can say:
Time going south + Time coming back north = \(50\) minutes

In plain English: "The time to run \(\mathrm{x}\) miles south plus the time to run \((3.25 + \mathrm{x})\) miles back north must equal exactly \(50\) minutes."

This gives us the equation: \(8\mathrm{x} + 8(3.25 + \mathrm{x}) = 50\)

3. Solve for the maximum additional distance

Let's solve our equation step by step:
\(8\mathrm{x} + 8(3.25 + \mathrm{x}) = 50\)

First, let's distribute the \(8\) on the left side:
\(8\mathrm{x} + 8(3.25) + 8\mathrm{x} = 50\)
\(8\mathrm{x} + 26 + 8\mathrm{x} = 50\)

Combining like terms:
\(16\mathrm{x} + 26 = 50\)

Subtracting \(26\) from both sides:
\(16\mathrm{x} = 50 - 26\)
\(16\mathrm{x} = 24\)

Dividing both sides by \(16\):
\(\mathrm{x} = \frac{24}{16} = 1.5\)

So Bob can run \(1.5\) additional miles south.

4. Verify the solution makes sense

Let's check our answer by working through Bob's complete remaining journey:

If Bob runs \(1.5\) more miles south:

• Time to run \(1.5\) miles south: \(1.5 \times 8 = 12\) minutes
• Total distance from parking lot: \(3.25 + 1.5 = 4.75\) miles
• Time to run \(4.75\) miles back north: \(4.75 \times 8 = 38\) minutes
• Total time for remaining journey: \(12 + 38 = 50\) minutes ✓

This matches exactly with Bob's \(50\)-minute time limit, confirming our answer is correct.

Final Answer

Bob can run \(1.5\) miles farther south and still return to the parking lot in \(50\) minutes. The answer is (A) \(1.5\).

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what distance Bob needs to cover on the return trip
Students often think Bob only needs to run back the additional distance '\(\mathrm{x}\)' that he plans to go further south, forgetting that he must return ALL THE WAY back to the parking lot. The return distance is not just '\(\mathrm{x}\)' miles, but the total distance from his farthest point back to the parking lot, which is \((3.25 + \mathrm{x})\) miles.

2. Incorrectly setting up the time constraint
Some students may think the \(50\) minutes applies to the entire run including the initial \(3.25\) miles already completed, rather than understanding that the \(50\) minutes is ONLY for the remaining portion of Bob's journey (running \(\mathrm{x}\) miles further south + returning to the parking lot).

3. Confusion about the rate conversion
Students might misinterpret "\(8\) minutes per mile" and incorrectly think Bob runs \(8\) miles per minute, leading them to set up equations with the rate inverted. The correct interpretation is that each mile takes \(8\) minutes to complete.

Errors while executing the approach

1. Algebraic distribution and combination errors
When expanding \(8(3.25 + \mathrm{x})\), students may forget to distribute the \(8\) to both terms, writing \(8(3.25) + \mathrm{x}\) instead of \(8(3.25) + 8\mathrm{x}\). Or when combining like terms \(8\mathrm{x} + 8\mathrm{x}\), they might incorrectly get \(8\mathrm{x}\) instead of \(16\mathrm{x}\).

2. Arithmetic calculation mistakes
Common errors include incorrectly calculating \(8 \times 3.25 = 26\) (might get \(24\) or \(28\)), or when dividing \(\frac{24}{16}\), not simplifying to \(1.5\) correctly (might get \(1.6\) or leave it as an improper fraction).

Errors while selecting the answer

1. Answering with the wrong quantity
Students might calculate correctly but then select the total distance from the parking lot \((3.25 + 1.5 = 4.75)\) instead of the additional distance \(\mathrm{x} = 1.5\) that the question specifically asks for. This would lead them to incorrectly choose answer choice (E) \(4.75\).