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A total of \(\mathrm{n}\) trucks and cars are parked in a lot. If the number of cars is \(\frac{1}{4}\) the number of trucks, and \(\frac{2}{3}\) of the trucks are pickups, how many pickups, in terms of \(\mathrm{n}\), are parked in the lot?
Let's break down what the problem is telling us in simple terms:
Think of it this way: if you walked through the parking lot and counted all vehicles, you'd get n. We want to know what fraction of that n represents just the pickup trucks.
Process Skill: TRANSLATE - Converting the word relationships into mathematical understanding
Since cars depend on the number of trucks, let's use trucks as our starting point. Let's call the number of trucks "T".
From the problem:
To add these together: \(T + \frac{1}{4}T = \frac{4}{4}T + \frac{1}{4}T = \frac{5}{4}T\)
So the total number of vehicles is \(\frac{5}{4}T\), which means our total \(n = \frac{5}{4}T\)
Now we need to flip this relationship around. We know that:
\(n = \frac{5}{4}T\)
To find T in terms of n, we multiply both sides by \(\frac{4}{5}\):
\(T = n \times \frac{4}{5} = \frac{4}{5}n\)
Let's verify this makes sense: if \(n = 15\) total vehicles, then \(T = \frac{4}{5} \times 15 = 12\) trucks, and cars = \(\frac{1}{4} \times 12 = 3\) cars. Total: \(12 + 3 = 15\) ✓
Now we can find the number of pickups. We know:
So: Number of pickups = \(\frac{2}{3} \times \frac{4}{5}n\)
Multiplying fractions: \(\frac{2}{3} \times \frac{4}{5} = \frac{2 \times 4}{3 \times 5} = \frac{8}{15}\)
Therefore: Number of pickups = \(\frac{8}{15}n\)
The number of pickups in terms of n is \(\frac{8}{15}n\).
Looking at our answer choices, this matches choice D: \(\frac{8}{15}n\)
To double-check: If \(n = 15\) vehicles, we'd have 12 trucks and 3 cars. Of the 12 trucks, \(\frac{2}{3}\) would be pickups: \(\frac{2}{3} \times 12 = 8\) pickups. And indeed, \(\frac{8}{15} \times 15 = 8\) pickups ✓
Students often confuse which variable depends on which. The problem states "the number of cars is \(\frac{1}{4}\) the number of trucks," but students might incorrectly interpret this as "the number of trucks is \(\frac{1}{4}\) the number of cars" or set up the equation backwards. This leads to expressing trucks in terms of cars instead of cars in terms of trucks, creating a fundamentally flawed approach.
Students frequently miss that pickups are a subset of trucks, not a separate category. They might treat the problem as having three distinct vehicle types (cars, trucks, pickups) instead of understanding that pickups are \(\frac{2}{3}\) of the trucks. This misunderstanding leads to incorrect total vehicle counts and wrong relationships.
Many students jump straight into calculations without first establishing what variable to use as their foundation. They might try to work directly with 'n' from the start instead of recognizing that expressing everything in terms of the number of trucks first makes the relationships clearer and the algebra more manageable.
When calculating the total vehicles as \(T + \frac{1}{4}T\), students often make mistakes converting to common denominators. They might incorrectly calculate this as \(\frac{5}{4}T\) by adding \(1 + \frac{1}{4} = \frac{5}{4}\), but fail to properly convert the "1" to \(\frac{4}{4}\) first, leading to errors like getting \(\frac{2}{4}T\) or \(1.25T\) instead of \(\frac{5}{4}T\).
When students have \(n = \frac{5}{4}T\) and need to find T in terms of n, they often make algebraic errors. Common mistakes include multiplying by \(\frac{5}{4}\) instead of \(\frac{4}{5}\), or incorrectly stating that \(T = \frac{5}{4}n\) instead of \(T = \frac{4}{5}n\). This error propagates through all subsequent calculations.
When calculating \(\frac{2}{3} \times \frac{4}{5}\), students frequently make multiplication errors. They might incorrectly multiply denominators and numerators in wrong combinations, getting results like \(\frac{2}{5} \times \frac{4}{3} = \frac{8}{15}\), or make arithmetic errors like \(\frac{2 \times 4}{3 \times 5} = \frac{8}{8} = 1\), leading to completely wrong final expressions.
Students who correctly calculate that trucks represent \(\frac{4}{5}n\) of the total vehicles might accidentally select an answer choice that matches this intermediate result rather than continuing to find pickups. They might look for \(\frac{4}{5}n\) among the choices, and when not finding it exactly, select the closest option instead of completing the final step.
After correctly calculating \(\frac{8}{15}n\), students might misread the answer choices and select a fraction that looks similar, such as \(\frac{8}{12}n = \frac{2}{3}n\) if it were available, or misidentify which letter corresponds to \(\frac{8}{15}\). This is especially common when students are rushing or not double-checking their fraction matching.
Instead of working with variables algebraically, we can assign a specific numerical value for the number of trucks that makes our fraction calculations clean.
Step 1: Choose a convenient value for trucks
Since we need to work with fractions \(\frac{1}{4}\) (for cars) and \(\frac{2}{3}\) (for pickups), let's choose trucks = 12, because 12 is divisible by both 4 and 3, making our calculations clean.
Step 2: Calculate the number of cars
Cars = \(\frac{1}{4} \times \text{trucks} = \frac{1}{4} \times 12 = 3\) cars
Step 3: Find the total number of vehicles (n)
\(n = \text{trucks} + \text{cars} = 12 + 3 = 15\) total vehicles
Step 4: Calculate the number of pickups
Pickups = \(\frac{2}{3} \times \text{trucks} = \frac{2}{3} \times 12 = 8\) pickups
Step 5: Express pickups as a fraction of total vehicles
Fraction of pickups = \(\frac{8}{15}\) of the total vehicles
Therefore, pickups = \(\frac{8}{15} \times n\)
This matches answer choice D: \(\frac{8}{15}n\)