# of students(n) in the class = 30

# of students with 0 books = 2
# of students with 1 book = 12
# of students with 2 books = 10
# of students with 3+ books = 30-2-12-10 = 6

Average = (Total # books)/Total # students

2 = (2(0) + 12(1) + 10(2) + x)/30

x is the number of books that the 6 students have.

60 = 32 + x
28 = x

READ THIS CAREFULLY.."what is the maximum number of books that any single student could have borrowed?"

Assume that 5 of the 6 that borrowed at least three books only borrowed 3. That is 3 * 5 = 15.

Thus, the sixth of these kids has the rest of the 28 which is 28-15 = 13.

HTH

Use a table with two columns to solve this one. In the left column write down the number of students and in the right write down the number of books.



2 | 0
12 | 12
10 | 20

You now have 24 students with 32 books. Now, let 5 students have 3 books each for 15 books total. Then, set up an equation using the information you are given.

0 + 12 + 20 + 15 + x/30 = 2

47 + x = 60

x = 13

There is your answer.

Aaring23 wrote:Hey guy's I was wondering if someone could help me out with this problem. Here is the problem.

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 of the students each borrowed atleast 3 books. If the average (arithmetic mean) number of books borrowed per student was 2, what is the maximum number of books that any single student could have borrowed?

Answer is 13.

If someone demonstrate the steps involved to solve this problem I would appreciate it very much. Thank you.

where did you get this question from ?

AleksandrM wrote:Use a table with two columns to sole this one. In the left column write down the number of students and in the right write down the number of books.



2 | 0
12 | 12
10 | 20

You now have 24 students with 32 books. Now, let 5 students have 3 books each for 15 books total. Then, set up an equation using the information you are given.

0 + 12 + 20 + 15 + x/30 = 2

47 + x = 60

x = 13

There is your answer.

I like yoru approach and it makes a lot of sense. I follow all parts of your strategy EXCEPT the fact that you get 5 left over students. I gt 6 students. 10+12+2 and final 6 that had atleast 3 books.

I know that there are 6 students left over with at least 3 each. The AT LEAST part is what you should keep in mind.

If the remaining six have 3 each and that is the most any one student can have, then you can give 3 books to only five of the remaining students and provide the last student with x amount of books to find his maximum potential capacity.

If that last student can only have 3 books, then, when you solve for x, you will get 3, and that will be the answer to our question (the maximum number of books any one student can have is 3). However, in our case, when you solve for x, you see that the last student can have up to 13 books, maximum. This still holds for what we are asked to provide the last six students with. All of the six have AT LEAST three books each. Our last student has at least 3 and at most 13.

Hope this helps.