A certain truck traveling at 55 miles per hour gets 4.5 miles per gallon of diesel fuel consumed. Traveling at...

1. Translate the problem requirements: We need to find how many miles were traveled at 55 mph. The truck travels 500 total miles using 120 total gallons, with different fuel efficiencies at different speeds (4.5 mpg at 55 mph, 3.5 mpg at 60 mph).
2. Set up variables for the two portions of the trip: Define variables for miles traveled at each speed, ensuring they sum to the total 500 miles.
3. Connect fuel consumption to distance traveled: Use the fuel efficiency rates to express total fuel consumption in terms of the distances traveled at each speed.
4. Solve the system using the fuel constraint: Use the fact that total fuel used equals 120 gallons to find the specific distances traveled at each speed.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening in plain English first. We have a truck that takes a 500-mile trip and uses 120 gallons of fuel total. The truck drives at two different speeds during this trip: some miles at 55 mph and the remaining miles at 60 mph. At 55 mph, the truck is more fuel-efficient (gets 4.5 miles per gallon), but at 60 mph, it's less fuel-efficient (gets only 3.5 miles per gallon). We need to find how many of those 500 miles were traveled at 55 mph.

The key insight is that we know the total distance (500 miles) and total fuel used (120 gallons), so we can work backwards to figure out how the trip was split between the two speeds.

Process Skill: TRANSLATE - Converting the real-world scenario into mathematical relationships we can work with

2. Set up variables for the two portions of the trip

Since the trip is split into two parts, let's use simple language to define what we're looking for:

Let's say the truck traveled some number of miles at 55 mph. We'll call this distance \(\mathrm{d_1}\).
Since the total trip is 500 miles, the remaining distance must have been traveled at 60 mph. This remaining distance would be \(500 - \mathrm{d_1}\) miles.

So we have:

• Miles traveled at 55 mph = \(\mathrm{d_1}\)
• Miles traveled at 60 mph = \(500 - \mathrm{d_1}\)
• Total miles = \(\mathrm{d_1} + (500 - \mathrm{d_1}) = 500\) ✓

This makes sense because the two portions must add up to the complete 500-mile trip.

3. Connect fuel consumption to distance traveled

Now we need to figure out how much fuel was used for each portion of the trip. This is where the fuel efficiency rates come into play.

Think about it this way: if you travel a certain distance and you know your car's fuel efficiency, you can calculate how much fuel you used by dividing the distance by the efficiency.

For the portion at 55 mph:

• Distance = \(\mathrm{d_1}\) miles
• Efficiency = 4.5 miles per gallon
• Fuel used = \(\frac{\mathrm{d_1}}{4.5}\) gallons

For the portion at 60 mph:

• Distance = \(500 - \mathrm{d_1}\) miles
• Efficiency = 3.5 miles per gallon
• Fuel used = \(\frac{500 - \mathrm{d_1}}{3.5}\) gallons

The total fuel used for the entire trip is the sum of fuel used in both portions:
Total fuel = \(\frac{\mathrm{d_1}}{4.5} + \frac{500 - \mathrm{d_1}}{3.5} = 120\) gallons

4. Solve the system using the fuel constraint

Now we have our equation: \(\frac{\mathrm{d_1}}{4.5} + \frac{500 - \mathrm{d_1}}{3.5} = 120\)

To solve this, let's clear the fractions by finding a common approach. I'll multiply everything by a number that eliminates the decimals.

First, let's convert to fractions: \(4.5 = \frac{9}{2}\) and \(3.5 = \frac{7}{2}\)

So our equation becomes: \(\frac{\mathrm{d_1}}{\frac{9}{2}} + \frac{500 - \mathrm{d_1}}{\frac{7}{2}} = 120\)

This simplifies to: \(\frac{2\mathrm{d_1}}{9} + \frac{2(500 - \mathrm{d_1})}{7} = 120\)

To clear fractions, multiply everything by 63 (which is 9 × 7):
\(63 \times \frac{2\mathrm{d_1}}{9} + 63 \times \frac{2(500 - \mathrm{d_1})}{7} = 63 \times 120\)

\(14\mathrm{d_1} + 18(500 - \mathrm{d_1}) = 7560\)
\(14\mathrm{d_1} + 9000 - 18\mathrm{d_1} = 7560\)
\(-4\mathrm{d_1} + 9000 = 7560\)
\(-4\mathrm{d_1} = -1440\)
\(\mathrm{d_1} = 360\)

Let's verify:

• Miles at 55 mph: 360
• Miles at 60 mph: \(500 - 360 = 140\)
• Fuel at 55 mph: \(\frac{360}{4.5} = 80\) gallons
• Fuel at 60 mph: \(\frac{140}{3.5} = 40\) gallons
• Total fuel: \(80 + 40 = 120\) gallons ✓

Final Answer

The truck traveled 360 miles at 55 miles per hour.

Looking at our answer choices, this corresponds to choice E. 360.

Common Faltering Points

Errors while devising the approach

1. Confusing fuel efficiency with fuel consumption

Students often struggle with the concept that "miles per gallon" means how far you can travel on one gallon. They might incorrectly think 4.5 miles per gallon means the truck uses 4.5 gallons per mile, leading them to set up equations like: Fuel used = \(\mathrm{d_1} \times 4.5\) instead of Fuel used = \(\frac{\mathrm{d_1}}{4.5}\).

2. Misinterpreting what the question is asking for

The question asks for "how many miles did it travel at 55 miles per hour" but students might set up their variable to represent time spent at each speed or fuel used at each speed, rather than distance traveled at each speed. This leads to solving for the wrong quantity entirely.

3. Setting up incomplete constraint equations

Students might focus only on the distance constraint (total = 500 miles) and forget to use the fuel constraint (total = 120 gallons). Without both constraints, they cannot solve for the unique values of distance at each speed, leading to an incomplete system of equations.

Errors while executing the approach

1. Arithmetic errors when clearing fractions

When converting decimals like 4.5 and 3.5 to fractions (9/2 and 7/2) and then finding common denominators, students often make calculation mistakes. For example, they might incorrectly multiply by the wrong common multiple or make sign errors when simplifying: \(14\mathrm{d_1} + 9000 - 18\mathrm{d_1} = 7560\) could become \(14\mathrm{d_1} + 9000 + 18\mathrm{d_1} = 7560\).

2. Incorrectly handling the distance constraint

When students set up \(500 - \mathrm{d_1}\) for the distance traveled at 60 mph, they sometimes make substitution errors in the fuel equation, like writing \(\frac{500 + \mathrm{d_1}}{3.5}\) instead of \(\frac{500 - \mathrm{d_1}}{3.5}\), which completely changes the problem setup.

Errors while selecting the answer

1. Answering the wrong part of the question

After solving and finding \(\mathrm{d_1} = 360\), students might accidentally select the distance traveled at 60 mph (which is \(500 - 360 = 140\) miles) instead of the distance traveled at 55 mph. They solve correctly but select answer A (140) instead of E (360) because they lose track of which variable represents which speed.