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In a class of 30 students, 2 students did not borrow any books from the library, 12 students each borrowed 1 book, 10 students each borrowed 2 books, and the rest borrowed at least 3 books. If the average number of books per student was 2, what is the maximum number of books any single student could have borrowed?
Let's break down what we know in simple terms. We have a class of 30 students, and we need to figure out how many books each group borrowed.
From the problem, we can organize our students into groups:
First, let's figure out how many students are in that last group. We started with 30 students total, so: \(30 - 2 - 12 - 10 = 6\) students who borrowed at least 3 books each.
We also know that the average number of books per student across the entire class was 2 books per student.
Process Skill: TRANSLATE - Converting the word problem into clear numerical relationships
Since we have 30 students and the average is 2 books per student, we can find the total number of books borrowed by the entire class.
Think of it this way: if every single student borrowed exactly 2 books, we'd have \(30 \times 2 = 60\) books total. Since the average is 2, the actual total must also be 60 books, even though some students borrowed more and some borrowed less.
Total books borrowed by the class = 30 students × 2 books per student = 60 books
Now let's add up the books we already know about from the first three groups of students:
Books already accounted for = \(0 + 12 + 20 = 32\) books
We know the total must be 60 books, and we've already accounted for 32 books from the known groups.
This means the remaining 6 students (who each borrowed at least 3 books) must have borrowed: \(60 - 32 = 28\) books total among themselves.
So we need to distribute exactly 28 books among these 6 students, with each student getting at least 3 books.
Here's the key insight: to maximize the number of books that one student could have borrowed, we need to minimize what the other 5 students borrowed, while still meeting the constraint that each borrowed at least 3 books.
The minimum we can give to 5 of these students is exactly 3 books each:
\(5 \times 3 = 15\) books
This leaves us with: \(28 - 15 = 13\) books for the sixth student.
Let's verify this works: 5 students with 3 books each + 1 student with 13 books = \(15 + 13 = 28\) books total ✓
Process Skill: APPLY CONSTRAINTS - Ensuring we meet the "at least 3 books" requirement while maximizing one student's total
The maximum number of books any single student could have borrowed is 13 books.
Verification: Our total distribution would be:
The answer is D. 13
Students often misinterpret "the rest borrowed at least 3 books" as "the rest borrowed exactly 3 books." This leads them to calculate \(6 \times 3 = 18\) books for the remaining group, giving a total of 50 books instead of the required 60 books. They fail to recognize that "at least 3" means 3 or more, which is crucial for the maximization strategy.
When determining how many students borrowed at least 3 books, students may make arithmetic errors in subtraction: 30 - 2 - 12 - 10. Some students might incorrectly calculate this as 8 students instead of 6 students, which would completely change the distribution of the remaining 28 books.
Students may not realize that to maximize one student's books, they need to minimize what the other students in the "at least 3" group receive. Instead, they might try to distribute the 28 books equally among the 6 students, giving each about 4-5 books, missing the key insight about the optimization strategy.
When calculating the total books borrowed \(30 \times 2 = 60\), students might make simple multiplication errors. More commonly, when adding up books from known groups \(0 + 12 + 20 = 32\), students may incorrectly calculate this sum, leading to wrong values for the remaining books to be distributed.
After correctly identifying that 28 books need to be distributed among 6 students, students might incorrectly calculate the minimum for 5 students \(5 \times 3 = 15\) or make errors in the subtraction \(28 - 15 = 13\). These arithmetic mistakes lead to wrong final answers.
Students may arrive at their answer but forget to check that their distribution actually satisfies all constraints: each of the 6 students has at least 3 books, the total equals 60 books, and the average is 2 books per student.
Even after correctly calculating that 13 is possible, students might second-guess themselves and select a smaller option like 8 or 5, thinking these "seem more reasonable" or safer, without recognizing that the question specifically asks for the maximum possible value.
No likely faltering points