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In an auditorium, 360 chairs are to be set up in a rectangular arrangement with \(\mathrm{x}\) rows of exactly \(\mathrm{y}\) chairs each. If the only other restriction is that \(10 < \mathrm{x} < 25\), how many different rectangular arrangements are possible?
Let's start by understanding what we're looking for in everyday terms. We have an auditorium with exactly 360 chairs that need to be arranged in a rectangle. Think of it like arranging desks in a classroom - we need rows and columns that form a perfect rectangle.
The key pieces of information are:
We need to find how many different ways we can choose the number of rows (x) such that when we multiply by the number of chairs per row (y), we get exactly 360 chairs.
Process Skill: TRANSLATE - Converting the rectangular arrangement language into the mathematical constraint \(\mathrm{x \times y = 360}\)
Since \(\mathrm{x \times y = 360}\), this means x must be a factor of 360. Let's find what 360 breaks down into by finding its prime factorization first.
\(\mathrm{360 = 36 \times 10 = (6^2) \times (2 \times 5) = (2 \times 3)^2 \times (2 \times 5) = 2^3 \times 3^2 \times 5}\)
Now we can find all the factors of 360. The factors are all the numbers that divide 360 evenly:
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360
Each of these numbers could potentially be our number of rows (x), as long as it satisfies our constraint.
Now we need to check which factors of 360 satisfy our constraint that \(\mathrm{10 < x < 25}\).
Looking at our list of factors: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360
We need factors that are greater than 10 but less than 25:
Let's verify each: \(\mathrm{12 \times 30 = 360}\) ✓, \(\mathrm{15 \times 24 = 360}\) ✓, \(\mathrm{18 \times 20 = 360}\) ✓, \(\mathrm{20 \times 18 = 360}\) ✓, \(\mathrm{24 \times 15 = 360}\) ✓
Process Skill: APPLY CONSTRAINTS - Filtering the complete list of factors using the given range restriction
From our analysis, we found exactly 5 values for x (number of rows) that satisfy all our requirements:
Therefore, there are exactly 5 different rectangular arrangements possible.
The answer is B. Five
We found exactly 5 valid values for the number of rows (12, 15, 18, 20, and 24) that satisfy both the requirement that \(\mathrm{x \times y = 360}\) and the constraint that \(\mathrm{10 < x < 25}\).
Faltering Point 1: Misinterpreting the constraint boundaries
Students often misread "\(\mathrm{10 < x < 25}\)" as "\(\mathrm{10 \leq x \leq 25}\)", incorrectly including x = 10 and x = 25 in their solutions. This would lead them to consider additional factors like x = 10 (\(\mathrm{10 \times 36 = 360}\)) and potentially others, resulting in an incorrect count.
Faltering Point 2: Confusing which variable represents rows vs. chairs per row
Some students may set up the equation correctly as \(\mathrm{x \times y = 360}\) but then apply the constraint "\(\mathrm{10 < x < 25}\)" to the wrong variable (y instead of x). This fundamental misunderstanding would lead them to look for factors of 360 that could represent chairs per row rather than number of rows.
Faltering Point 3: Forgetting that arrangements are determined by factor pairs, not individual factors
Students might think they need to count both (x,y) and (y,x) as separate arrangements, essentially double-counting. For example, they might count both "12 rows of 30 chairs" and "30 rows of 12 chairs" as different solutions, when the constraint only allows one of these possibilities.
Faltering Point 1: Incomplete factor identification
When finding all factors of 360, students may miss some factors, especially the larger ones or those that result from less obvious combinations. Missing even one factor within the constraint range (like 18 or 24) would lead to an undercount of possible arrangements.
Faltering Point 2: Arithmetic errors in verification
Students may correctly identify potential values of x but make multiplication errors when verifying that \(\mathrm{x \times y = 360}\). For example, they might incorrectly calculate \(\mathrm{18 \times 20}\) or \(\mathrm{24 \times 15}\), leading them to reject valid solutions or accept invalid ones.
Faltering Point 1: Including boundary values in the final count
Even after correctly solving most of the problem, students may accidentally include x = 10 or x = 25 in their final count, forgetting that these boundary values are excluded by the strict inequality "\(\mathrm{10 < x < 25}\)". This would lead them to choose "Six" instead of "Five".