During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour...

1. Translate the problem requirements: Francine travels \(\mathrm{x}\%\) of total distance at \(40\) mph and \((100-\mathrm{x})\%\) at \(60\) mph. We need to find overall average speed using the fundamental relationship: average speed = total distance ÷ total time.
2. Set up distance portions using a concrete total: Choose total distance = \(100\) miles to avoid fractions. This gives us \(\mathrm{x}\) miles at \(40\) mph and \((100-\mathrm{x})\) miles at \(60\) mph.
3. Calculate time for each segment: Use time = distance ÷ speed to find time spent at each speed, then sum for total time.
4. Apply average speed formula and match to answer choices: Compute total distance ÷ total time, then algebraically manipulate to match one of the given expressions.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening in plain English: Francine takes a trip and drives part of it slowly (\(40\) mph) and part of it faster (\(60\) mph). Specifically, she drives \(\mathrm{x}\) percent of the total distance at \(40\) mph, which means the remaining part - that's \((100-\mathrm{x})\) percent - she drives at \(60\) mph.

We need to find her average speed for the entire trip. Remember, average speed isn't just adding the two speeds and dividing by \(2\). Average speed means: how far did she go in total, divided by how much time the whole trip took.

So our fundamental relationship is: Average Speed = Total Distance ÷ Total Time

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

2. Set up distance portions using a concrete total

Here's a smart trick to avoid messy fractions: let's say the total distance is \(100\) miles. This makes the percentages easy to work with.

If the total trip is \(100\) miles:

• Distance traveled at \(40\) mph = \(\mathrm{x}\) percent of \(100\) = \(\mathrm{x}\) miles
• Distance traveled at \(60\) mph = \((100-\mathrm{x})\) percent of \(100\) = \((100-\mathrm{x})\) miles

Notice how choosing \(100\) miles as our total makes the math clean - \(\mathrm{x}\) percent of \(100\) is just \(\mathrm{x}\)!

Process Skill: SIMPLIFY - Using a concrete value to avoid complex fractions

3. Calculate time for each segment

Now we need to figure out how long each part of the trip took. We use the basic relationship: Time = Distance ÷ Speed

For the first part of the trip:

• Distance = \(\mathrm{x}\) miles, Speed = \(40\) mph
• Time = \(\mathrm{x} ÷ 40 = \mathrm{x}/40\) hours

For the second part of the trip:

• Distance = \((100-\mathrm{x})\) miles, Speed = \(60\) mph
• Time = \((100-\mathrm{x}) ÷ 60 = (100-\mathrm{x})/60\) hours

Total time for the entire trip = \(\mathrm{x}/40 + (100-\mathrm{x})/60\) hours

4. Apply average speed formula and match to answer choices

Now we can calculate the average speed using our formula:

Average Speed = Total Distance ÷ Total Time

Average Speed = \(100 ÷ [\mathrm{x}/40 + (100-\mathrm{x})/60]\)

To simplify the denominator, let's find a common denominator for the fractions. The LCD of \(40\) and \(60\) is \(120\):

• \(\mathrm{x}/40 = 3\mathrm{x}/120\)
• \((100-\mathrm{x})/60 = 2(100-\mathrm{x})/120 = (200-2\mathrm{x})/120\)

So: \(\mathrm{x}/40 + (100-\mathrm{x})/60 = 3\mathrm{x}/120 + (200-2\mathrm{x})/120 = (3\mathrm{x} + 200 - 2\mathrm{x})/120 = (\mathrm{x} + 200)/120\)

Therefore:

Average Speed = \(100 ÷ [(\mathrm{x} + 200)/120] = 100 × 120/(\mathrm{x} + 200) = 12,000/(\mathrm{x} + 200)\)

Process Skill: MANIPULATE - Algebraic manipulation to match answer format

5. Final Answer

Our result is \(12,000/(\mathrm{x} + 200)\), which exactly matches answer choice E.

Let's verify this makes sense: if \(\mathrm{x} = 0\) (all driving at \(60\) mph), we get \(12,000/200 = 60\) mph ✓

If \(\mathrm{x} = 100\) (all driving at \(40\) mph), we get \(12,000/300 = 40\) mph ✓

Answer: E

Common Faltering Points

Errors while devising the approach

1. Misunderstanding average speed calculation: Students often think average speed is simply the arithmetic mean of the two speeds: \((40 + 60)/2 = 50\) mph. This is incorrect because average speed must account for the time spent at each speed, which depends on the distances traveled at each speed.

2. Confusion about percentage interpretation: Students may misinterpret "\(\mathrm{x}\) percent of the total distance" and think it means \(\mathrm{x}\) miles instead of \(\mathrm{x}\%\) of the total distance. This leads to incorrect setup of distance portions and subsequent errors throughout the solution.

3. Attempting to work with abstract total distance: Some students try to keep the total distance as a variable (like D) instead of using a concrete value like \(100\). This makes the algebraic manipulation much more complex and increases the likelihood of errors when simplifying fractions.

Errors while executing the approach

1. Arithmetic errors when finding common denominators: When combining fractions \(\mathrm{x}/40 + (100-\mathrm{x})/60\), students often make mistakes finding the LCD (\(120\)) or incorrectly convert the fractions. For example, they might write \(\mathrm{x}/40 = 2\mathrm{x}/120\) instead of \(3\mathrm{x}/120\).

2. Sign errors in algebraic manipulation: When expanding \((100-\mathrm{x})/60 = 2(100-\mathrm{x})/120\), students frequently make sign errors, writing \((200+2\mathrm{x})/120\) instead of \((200-2\mathrm{x})/120\), which leads to an incorrect final expression.

3. Incorrect division of complex fractions: Students often struggle with the step \(100 ÷ [(\mathrm{x} + 200)/120]\), either forgetting to flip and multiply or making errors when multiplying \(100 × 120\).

Errors while selecting the answer

1. Failing to match the exact form: Even when students arrive at the correct numerical expression, they might write it in a different form (like \(12000/(200+\mathrm{x})\) instead of \(12,000/(\mathrm{x}+200)\)) and incorrectly conclude their answer doesn't match any of the choices.

2. Not verifying with boundary cases: Students may select an answer without checking if it makes sense in extreme cases (like \(\mathrm{x}=0\) or \(\mathrm{x}=100\)), missing an opportunity to catch calculation errors that would make their chosen answer unreasonable.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a convenient value for x
Let's choose \(\mathrm{x} = 20\), meaning Francine travels \(20\%\) of the distance at \(40\) mph and \(80\%\) at \(60\) mph. This gives us clean percentages to work with.

Step 2: Set up concrete distances
Let total distance = \(100\) miles (chosen to make percentage calculations simple)
• Distance at \(40\) mph = \(20\%\) of \(100 = 20\) miles
• Distance at \(60\) mph = \(80\%\) of \(100 = 80\) miles

Step 3: Calculate time for each segment
• Time at \(40\) mph = \(20\) miles ÷ \(40\) mph = \(0.5\) hours
• Time at \(60\) mph = \(80\) miles ÷ \(60\) mph = \(4/3\) hours
• Total time = \(0.5 + 4/3 = 1/2 + 4/3 = 3/6 + 8/6 = 11/6\) hours

Step 4: Calculate average speed
Average speed = Total distance ÷ Total time = \(100 ÷ (11/6) = 100 × 6/11 = 600/11 ≈ 54.55\) mph

Step 5: Test answer choices with x = 20

1. \((180-20)/2 = 160/2 = 80\) ✗
2. \((20+60)/4 = 80/4 = 20\) ✗
3. \((300-20)/5 = 280/5 = 56\) ✗
4. \(600/(115-20) = 600/95 ≈ 6.32\) ✗
5. \(12,000/(20+200) = 12,000/220 = 600/11 ≈ 54.55\) ✓

Step 6: Verify with a second value
Let's try \(\mathrm{x} = 40\) for confirmation:
• \(40\) miles at \(40\) mph, \(60\) miles at \(60\) mph
• Time = \(40/40 + 60/60 = 1 + 1 = 2\) hours
• Average speed = \(100/2 = 50\) mph
• Choice E: \(12,000/(40+200) = 12,000/240 = 50\) ✓

The answer is E.