Loading...
A certain truck traveling at \(55\) miles per hour gets \(4.5\) miles per gallon of diesel fuel consumed. Traveling at \(60\) miles per hour, the truck gets only \(3.5\) miles per gallon. On a \(500\)-mile trip, if the truck used a total of \(120\) gallons of diesel fuel and traveled part of the trip at \(55\) miles per hour and the rest at \(60\) miles per hour, how many miles did it travel at \(55\) miles per hour?
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
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:
This makes sense because the two portions must add up to the complete 500-mile trip.
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:
For the portion at 60 mph:
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
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:
The truck traveled 360 miles at 55 miles per hour.
Looking at our answer choices, this corresponds to choice E. 360.
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 = \(\frac{\mathrm{d_1}}{4.5}\).
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.
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.
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 \(14\mathrm{d_1} + 9000 + 18\mathrm{d_1} = 7560\).
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.
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.