Aaron will jog from home at x miles per hour and then walk back home by the same route at...

1. Translate the problem requirements: Aaron travels the same distance twice - first jogging away from home at speed \(\mathrm{x}\) mph, then walking back at speed \(\mathrm{y}\) mph. We need to find how far he can go so that his total travel time equals \(\mathrm{t}\) hours.
2. Define the distance variable: Let \(\mathrm{d}\) be the distance from home that Aaron jogs, which is also the distance he walks back.
3. Express individual travel times: Use the relationship time = distance/speed to write expressions for jogging time and walking time.
4. Set up the total time equation: Add the jogging time and walking time to equal the given total time \(\mathrm{t}\).
5. Solve for the distance: Algebraically manipulate the equation to isolate \(\mathrm{d}\) and match with the answer choices.

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what's happening in plain English: Aaron is taking a round trip from his home. He jogs away from home at one speed, then walks back home at a different speed. We need to figure out how far from home he can go so that his entire trip takes exactly \(\mathrm{t}\) hours.

Think of it like this: if you drive to a friend's house at 60 mph and walk back at 3 mph, how far can your friend live so that your total travel time is exactly 4 hours? It's the same type of problem.

The key insight is that Aaron travels the same distance twice - once jogging away, once walking back - but at different speeds, so the times will be different.

Process Skill: TRANSLATE - Converting the story into mathematical relationships we can work with

2. Define the distance variable

Let's call the distance from home that Aaron jogs "\(\mathrm{d}\)" miles.

Since Aaron takes the same route back home, he also walks \(\mathrm{d}\) miles on his return trip.

So Aaron's total journey involves:

- Jogging \(\mathrm{d}\) miles away from home at \(\mathrm{x}\) mph
- Walking \(\mathrm{d}\) miles back to home at \(\mathrm{y}\) mph

3. Express individual travel times

Now we use the fundamental relationship: Time = Distance ÷ Speed

For the jogging portion:

- Distance = \(\mathrm{d}\) miles
- Speed = \(\mathrm{x}\) miles per hour
- Time for jogging = \(\frac{\mathrm{d}}{\mathrm{x}}\) hours

For the walking portion:

- Distance = \(\mathrm{d}\) miles (same distance back)
- Speed = \(\mathrm{y}\) miles per hour
- Time for walking = \(\frac{\mathrm{d}}{\mathrm{y}}\) hours

4. Set up the total time equation

Aaron's total time is the sum of his jogging time and walking time, and this must equal \(\mathrm{t}\) hours:

Total time = Jogging time + Walking time
\(\mathrm{t} = \frac{\mathrm{d}}{\mathrm{x}} + \frac{\mathrm{d}}{\mathrm{y}}\)

This equation captures the constraint that Aaron's entire trip takes exactly \(\mathrm{t}\) hours.

Process Skill: MANIPULATE - Setting up the equation that represents our constraint

5. Solve for the distance

Now we solve for \(\mathrm{d}\) by getting it by itself on one side of the equation:

\(\mathrm{t} = \frac{\mathrm{d}}{\mathrm{x}} + \frac{\mathrm{d}}{\mathrm{y}}\)

To combine the fractions on the right side, we need a common denominator of \(\mathrm{xy}\):
\(\mathrm{t} = \frac{\mathrm{d} \times \mathrm{y}}{\mathrm{x} \times \mathrm{y}} + \frac{\mathrm{d} \times \mathrm{x}}{\mathrm{y} \times \mathrm{x}}\)
\(\mathrm{t} = \frac{\mathrm{dy} + \mathrm{dx}}{\mathrm{xy}}\)
\(\mathrm{t} = \frac{\mathrm{d}(\mathrm{y} + \mathrm{x})}{\mathrm{xy}}\)
\(\mathrm{t} = \frac{\mathrm{d}(\mathrm{x} + \mathrm{y})}{\mathrm{xy}}\)

Now multiply both sides by \(\mathrm{xy}\):
\(\mathrm{t} \times \mathrm{xy} = \mathrm{d}(\mathrm{x} + \mathrm{y})\)

Finally, divide both sides by \((\mathrm{x} + \mathrm{y})\):
\(\mathrm{d} = \frac{\mathrm{t} \times \mathrm{xy}}{\mathrm{x} + \mathrm{y}} = \frac{\mathrm{xyt}}{\mathrm{x} + \mathrm{y}}\)

6. Final Answer

The distance from home that Aaron can jog is \(\frac{\mathrm{xyt}}{\mathrm{x} + \mathrm{y}}\) miles.

Looking at our answer choices, this matches choice (C): \(\frac{\mathrm{xyt}}{\mathrm{x}+\mathrm{y}}\)

We can verify this makes sense: if the speeds are higher (larger \(\mathrm{x}\) or \(\mathrm{y}\)), Aaron can go farther in the same total time. If the total time \(\mathrm{t}\) is longer, he can also go farther. The denominator \((\mathrm{x}+\mathrm{y})\) represents how the two different speeds combine to affect the total journey time.

Answer: (C)

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the round-trip nature of the problem
Students often miss that Aaron travels the SAME distance twice - once jogging away from home and once walking back. They might incorrectly think Aaron jogs for some distance and walks for a different distance, or that the total distance traveled is what we're solving for, rather than the one-way distance from home.

2. Confusing which direction uses which speed
Students may incorrectly assign the speeds to the wrong portions of the trip. The problem states Aaron jogs FROM home at \(\mathrm{x}\) mph and walks BACK home at \(\mathrm{y}\) mph. Some students might reverse this, thinking he walks away from home and jogs back.

3. Setting up time equation with total distance instead of one-way distance
Students might define their variable as the total distance of the round trip (\(2\mathrm{d}\)) rather than the one-way distance (\(\mathrm{d}\)) from home, leading to incorrect equations and ultimately wrong answers.

Errors while executing the approach

1. Algebraic errors when combining fractions
When solving \(\mathrm{t} = \frac{\mathrm{d}}{\mathrm{x}} + \frac{\mathrm{d}}{\mathrm{y}}\), students often make mistakes finding the common denominator or combining the fractions. They might incorrectly write it as \(\frac{\mathrm{d}}{\mathrm{x}+\mathrm{y}}\) instead of the correct \(\frac{\mathrm{d}(\mathrm{x}+\mathrm{y})}{\mathrm{xy}}\), or make errors in the cross-multiplication steps.

2. Incorrect isolation of variable d
After setting up \(\mathrm{t} = \frac{\mathrm{d}(\mathrm{x}+\mathrm{y})}{\mathrm{xy}}\), students may make algebraic errors when solving for \(\mathrm{d}\). Common mistakes include forgetting to multiply both sides by \(\mathrm{xy}\), or incorrectly dividing by \((\mathrm{x}+\mathrm{y})\), leading to expressions like \(\mathrm{txy}\) or \(\mathrm{t}(\mathrm{x}+\mathrm{y})\) instead of the correct \(\frac{\mathrm{xyt}}{\mathrm{x}+\mathrm{y}}\).

Errors while selecting the answer

1. Selecting answer choice with incorrect variable arrangement
Even if students derive the correct expression \(\frac{\mathrm{xyt}}{\mathrm{x}+\mathrm{y}}\), they might hastily select a wrong answer choice due to not carefully checking the arrangement of variables. For instance, they might confuse their result with answer choice (B) which has a similar structure but different variables in the numerator and denominator.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose convenient smart numbers
Let's select values that will make calculations clean:
• \(\mathrm{x} = 6\) mph (jogging speed)
• \(\mathrm{y} = 3\) mph (walking speed)
• \(\mathrm{t} = 3\) hours (total time)

Step 2: Set up the problem with concrete numbers
Let \(\mathrm{d}\) = distance Aaron jogs from home (in miles)
• Time jogging = \(\frac{\mathrm{d}}{6}\) hours
• Time walking back = \(\frac{\mathrm{d}}{3}\) hours
• Total time = \(\frac{\mathrm{d}}{6} + \frac{\mathrm{d}}{3} = 3\) hours

Step 3: Solve for distance with concrete numbers
\(\frac{\mathrm{d}}{6} + \frac{\mathrm{d}}{3} = 3\)
\(\frac{\mathrm{d}}{6} + \frac{2\mathrm{d}}{6} = 3\)
\(\frac{3\mathrm{d}}{6} = 3\)
\(\frac{\mathrm{d}}{2} = 3\)
\(\mathrm{d} = 6\) miles

Step 4: Verify using answer choice (C)
Using formula \(\frac{\mathrm{xyt}}{\mathrm{x}+\mathrm{y}}\):
\(= \frac{(6)(3)(3)}{6+3}\)
\(= \frac{54}{9}\)
\(= 6\) miles ✓

Step 5: Test another set of smart numbers
Let's verify with \(\mathrm{x} = 4\), \(\mathrm{y} = 2\), \(\mathrm{t} = 6\):
• Time equation: \(\frac{\mathrm{d}}{4} + \frac{\mathrm{d}}{2} = 6\)
• Solving: \(\frac{\mathrm{d}}{4} + \frac{2\mathrm{d}}{4} = 6 \rightarrow \frac{3\mathrm{d}}{4} = 6 \rightarrow \mathrm{d} = 8\)
• Using formula: \(\frac{(4)(2)(6)}{4+2} = \frac{48}{6} = 8\) ✓

The smart numbers approach confirms that answer choice (C) \(\frac{\mathrm{xyt}}{\mathrm{x}+\mathrm{y}}\) is correct.