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Marla starts running around a circular track at the same time Nick starts walking around the same circular track. Marla completes \(\mathrm{32}\) laps around the track per hour and Nick completes \(\mathrm{12}\) laps around the track per hour. How many minutes after Marla and Nick begin moving will Marla have completed \(\mathrm{4}\) more laps around the track than Nick?
Let's start by understanding what's happening in plain English. Imagine Marla and Nick are both moving around the same circular track at the same time, but Marla is running much faster than Nick is walking.
Marla completes \(\mathrm{32}\) laps every hour, while Nick only completes \(\mathrm{12}\) laps in that same hour. We want to know: at what point in time will Marla have completed exactly 4 more complete laps than Nick?
Think of it like this: both start at the same starting line at the same time. As time passes, Marla pulls ahead because she's faster. We're looking for the specific moment when Marla has gone around the track exactly 4 more times than Nick has.
Process Skill: TRANSLATE - Converting the word problem into a clear mathematical relationship
Now let's figure out how much faster Marla gains on Nick. Since they're both moving forward, what matters is the difference in their speeds.
Marla's speed: \(\mathrm{32\,laps\,per\,hour}\)
Nick's speed: \(\mathrm{12\,laps\,per\,hour}\)
The key insight is that Marla gains on Nick at a rate equal to the difference in their speeds:
Relative rate = \(\mathrm{32 - 12 = 20\,laps\,per\,hour}\)
This means that every hour, Marla completes \(\mathrm{20}\) more laps than Nick does. In other words, the gap between them grows by \(\mathrm{20}\) laps every hour.
Now we can use simple reasoning: if the gap grows by \(\mathrm{20}\) laps every hour, how long does it take for the gap to reach exactly \(\mathrm{4}\) laps?
We can think of this as a proportion:
Using basic proportional thinking:
Time needed = \(\mathrm{(4\,laps) ÷ (20\,laps\,per\,hour) = \frac{4}{20} = \frac{1}{5}\,hour}\)
So it takes \(\mathrm{\frac{1}{5}}\) of an hour for Marla to get exactly 4 laps ahead.
The problem asks for the answer in minutes, not hours, so we need to convert:
\(\mathrm{\frac{1}{5}\,hour = \frac{1}{5} × 60\,minutes = 12\,minutes}\)
Let's verify this makes sense: In \(\mathrm{12}\) minutes (which is \(\mathrm{\frac{1}{5}}\) of an hour):
After \(\mathrm{12}\) minutes, Marla will have completed exactly 4 more laps than Nick. The answer is (C) 12.
Students often confuse the relative difference with absolute positions. They might think they need to find when Marla has completed exactly 4 laps total, rather than understanding that they need to find when the gap between Marla's lap count and Nick's lap count equals 4. This leads them to set up equations like "\(\mathrm{32t = 4}\)" instead of the correct "\(\mathrm{32t - 12t = 4}\)".
Some students struggle with the concept that both people are moving at the same time from the same starting point. They might attempt to solve this as if Nick is stationary while Marla moves, or think that one person needs to complete a full lap before the other starts moving. This misunderstanding leads to using only Marla's speed in their calculations.
Students may not recognize that this is a relative speed problem. Instead of finding the difference in speeds (\(\mathrm{20\,laps\,per\,hour}\)), they might try to work with the individual speeds separately, leading to more complex and incorrect approaches like setting up separate time equations for each person.
Even when students correctly identify that they need \(\mathrm{\frac{1}{5}}\) of an hour, they often make mistakes converting to minutes. Common errors include: \(\mathrm{\frac{1}{5} × 60 = 15\,minutes}\) (forgetting the fraction), or converting incorrectly by doing \(\mathrm{60 ÷ 5 = 15}\) instead of the correct \(\mathrm{\frac{1}{5} × 60 = 12\,minutes}\).
When verifying their answer, students sometimes get confused by the decimal lap counts (\(\mathrm{6.4}\) laps for Marla, \(\mathrm{2.4}\) laps for Nick). They may round these incorrectly or think that partial laps don't count, leading them to doubt their correct answer and recalculate unnecessarily.
No likely faltering points - Once students have correctly calculated \(\mathrm{12}\) minutes, the answer selection is straightforward as it directly matches option (C) without requiring additional interpretation or conversion.