Tom travels d miles by a car, whose consumption C gallons per hour is directly proportional to the square of...

1. Translate the problem requirements: Fuel consumption per hour is proportional to speed squared (\(\mathrm{C = k \times v^2}\)). Tom travels 80% of distance d at 40 mph and 20% at 20 mph, consuming 45 gallons total. Find fuel consumption if entire distance d is traveled at 30 mph.
2. Set up the proportionality relationship: Express fuel consumption rate in terms of speed using the constant of proportionality k, where consumption per hour = \(\mathrm{k \times (speed)^2}\)
3. Calculate fuel consumption for the mixed-speed journey: Find time spent at each speed, then calculate fuel used at each segment using the consumption rates, totaling to 45 gallons
4. Solve for the distance and proportionality constant: Use the total fuel consumption equation to find the relationship between k and d
5. Apply the relationship to constant speed scenario: Calculate fuel consumption for traveling entire distance d at 30 mph using the established relationship

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're given in everyday terms:

Imagine Tom's car is like a sports car - the faster he drives, the more fuel it guzzles per hour. Specifically, if he doubles his speed, his car uses four times as much fuel per hour (since consumption is proportional to speed squared).

For his actual trip:

• He travels 80% of distance d at 40 mph
• He travels the remaining 20% of distance d at 20 mph
• Total fuel consumed = 45 gallons

We need to find: How much fuel if he travels the entire distance d at a steady 30 mph?

Technically, we can express the fuel consumption rate as: \(\mathrm{C = k \times v^2}\) (where k is our proportionality constant and v is speed)

Process Skill: TRANSLATE - Converting the proportional relationship into mathematical understanding

2. Set up the proportionality relationship

Since fuel consumption per hour is proportional to the square of speed, let's think about what this means:

• At 20 mph: consumption rate = \(\mathrm{k \times (20)^2 = 400k}\) gallons per hour
• At 40 mph: consumption rate = \(\mathrm{k \times (40)^2 = 1600k}\) gallons per hour
• At 30 mph: consumption rate = \(\mathrm{k \times (30)^2 = 900k}\) gallons per hour

Notice that at 40 mph, the car uses exactly 4 times as much fuel per hour compared to 20 mph (\(\mathrm{1600k ÷ 400k = 4}\)). This makes intuitive sense with the "square" relationship.

3. Calculate fuel consumption for the mixed-speed journey

Now let's figure out how much time Tom spends at each speed and how much fuel he uses:

Time calculations:

• Time for first 80% of journey: \(\mathrm{(0.8d\text{ miles}) ÷ (40\text{ mph}) = 0.02d\text{ hours}}\)
• Time for last 20% of journey: \(\mathrm{(0.2d\text{ miles}) ÷ (20\text{ mph}) = 0.01d\text{ hours}}\)

Fuel consumption calculations:

• Fuel for first segment: \(\mathrm{(1600k\text{ gallons/hour}) \times (0.02d\text{ hours}) = 32kd\text{ gallons}}\)
• Fuel for second segment: \(\mathrm{(400k\text{ gallons/hour}) \times (0.01d\text{ hours}) = 4kd\text{ gallons}}\)
• Total fuel = \(\mathrm{32kd + 4kd = 36kd\text{ gallons}}\)

Since we know the total fuel consumed was 45 gallons:

\(\mathrm{36kd = 45}\)

4. Solve for the distance and proportionality constant

From our equation \(\mathrm{36kd = 45}\), we can find the relationship we need:

• \(\mathrm{kd = 45 ÷ 36 = 5/4 = 1.25}\)

This relationship (\(\mathrm{kd = 1.25}\)) is the key insight - we don't need to find k and d separately, just their product!

Process Skill: SIMPLIFY - Recognizing we only need the product kd, not individual values

5. Apply the relationship to constant speed scenario

Now for the main question: fuel consumption at constant 30 mph for the entire distance d.

Time and fuel calculations:

• Time for entire journey at 30 mph: \(\mathrm{d ÷ 30 = d/30\text{ hours}}\)
• Fuel consumption rate at 30 mph: \(\mathrm{900k\text{ gallons per hour}}\)
• Total fuel needed: \(\mathrm{(900k\text{ gallons/hour}) \times (d/30\text{ hours}) = 30kd\text{ gallons}}\)

Substituting our known relationship \(\mathrm{kd = 1.25}\):

\(\mathrm{\text{Total fuel} = 30 \times 1.25 = 37.5\text{ gallons}}\)

4. Final Answer

Tom will consume 37.5 gallons of fuel if he travels the entire distance at 30 mph.

This corresponds to answer choice B.

Quick verification: This makes intuitive sense because 30 mph is between his two original speeds (20 and 40 mph), and 37.5 gallons is less than his original 45 gallons, which makes sense since he's avoiding the very fuel-hungry high-speed portion of his original journey.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the proportionality relationship
Students often confuse "consumption C gallons per hour is directly proportional to the square of speed" and instead set up consumption as proportional to just speed (\(\mathrm{C = kv}\)) rather than speed squared (\(\mathrm{C = kv^2}\)). This fundamental misunderstanding leads to completely incorrect fuel consumption rates at different speeds.

2. Setting up consumption per mile instead of per hour
The problem states consumption is in "gallons per hour," but students frequently assume it means "gallons per mile." This confusion leads them to skip time calculations entirely and work directly with distance ratios, missing the critical insight that higher speeds consume more fuel per hour but less time overall.

3. Misunderstanding the distance breakdown
Students sometimes misinterpret "80% of distance at 40 mph and remaining at 20 mph" as meaning 80% of time rather than 80% of distance. This leads to incorrect time and fuel calculations since time spent at each speed depends on both distance and speed.

Errors while executing the approach

1. Arithmetic errors in time calculations
When calculating time = distance/speed, students frequently make errors like: \(\mathrm{0.8d ÷ 40 = 0.8d/40 = 0.02d\text{ hours}}\) (correct) vs incorrectly getting \(\mathrm{0.8d \times 40}\) or other variations. These seemingly small arithmetic mistakes compound through the entire solution.

2. Incorrectly combining fuel consumption terms
After calculating fuel for each segment (\(\mathrm{32kd + 4kd}\)), students often make algebraic errors when combining like terms, getting results like \(\mathrm{32kd + 4kd = 32k + 4k = 36k}\) instead of the correct \(\mathrm{36kd}\). This error prevents them from properly solving for the kd relationship.

3. Mixing up the proportionality constant relationships
When working with different speeds, students sometimes incorrectly assume the same proportionality constant applies to different speed calculations, failing to properly square each speed value (using \(\mathrm{k \times 30}\) instead of \(\mathrm{k \times 30^2 = 900k}\)).

Errors while selecting the answer

1. Forgetting to substitute the kd relationship
Students correctly derive that fuel needed = \(\mathrm{30kd\text{ gallons}}\) but then forget to substitute \(\mathrm{kd = 1.25}\), leaving their answer as "30kd" or trying to solve for k and d individually instead of recognizing they only need their product.

2. Confusing which scenario the final answer represents
After calculating both scenarios (mixed speeds = 45 gallons, constant speed = 37.5 gallons), students sometimes select 45 gallons thinking the question asks for the original journey's fuel consumption rather than the constant 30 mph journey.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose convenient values for distance
Since we're dealing with 80% and 20% splits, let's set d = 100 miles for clean calculations.
• 80% of distance = \(\mathrm{0.8 \times 100 = 80\text{ miles}}\) at 40 mph
• 20% of distance = \(\mathrm{0.2 \times 100 = 20\text{ miles}}\) at 20 mph

Step 2: Calculate time spent at each speed
• Time at 40 mph = \(\mathrm{80\text{ miles} ÷ 40\text{ mph} = 2\text{ hours}}\)
• Time at 20 mph = \(\mathrm{20\text{ miles} ÷ 20\text{ mph} = 1\text{ hour}}\)
• Total time = 3 hours

Step 3: Set up fuel consumption using C = k×v²
• Fuel rate at 40 mph = \(\mathrm{k \times (40)^2 = 1600k\text{ gallons/hour}}\)
• Fuel rate at 20 mph = \(\mathrm{k \times (20)^2 = 400k\text{ gallons/hour}}\)

Step 4: Calculate total fuel consumption
• Fuel at 40 mph = \(\mathrm{1600k \times 2 = 3200k\text{ gallons}}\)
• Fuel at 20 mph = \(\mathrm{400k \times 1 = 400k\text{ gallons}}\)
• Total fuel = \(\mathrm{3200k + 400k = 3600k = 45\text{ gallons}}\)

Step 5: Solve for k
\(\mathrm{3600k = 45}\)
\(\mathrm{k = 45/3600 = 1/80}\)

Step 6: Calculate fuel for 100 miles at 30 mph
• Time at 30 mph = \(\mathrm{100 ÷ 30 = 10/3\text{ hours}}\)
• Fuel rate at 30 mph = \(\mathrm{k \times (30)^2 = (1/80) \times 900 = 11.25\text{ gallons/hour}}\)
• Total fuel = \(\mathrm{11.25 \times (10/3) = 37.5\text{ gallons}}\)

Why this approach works:
The choice of d = 100 is logical because it simplifies percentage calculations (80% and 20% become whole numbers), making the arithmetic clean while preserving all the proportional relationships in the original problem.