In a class of 30 students, 2 students did not borrow any

GIVEN: The average number of books per student was 2
There are 30 students in total.
We can write: (total number of books borrowed)/30 = 2
Multiply both sides by 30 to get: total number of books borrowed = 60

2 students borrowed 0 books. (2)(0) = 0 books borrowed
12 students borrowed 1 book. (12)(1) = 12 books borrowed
10 students borrowed 2 books. (10)(2) = 20 books borrowed
6 students borrowed 3 or more books. We'll leave this one for now

0 + 12 + 20 = 32

60 - 32 = 28
So there are 28 borrowed books that are unaccounted for

This means the six students who borrowed three or more books must have borrowed a total of 28 books.
Our goal is to MAXIMIZE the number of books a single student borrowed.

To do this, we will MINIMIZE the number of books five of those students borrowed, which will leave all of the remaining books for the sixth student.
Since the six students borrowed three or more books, 3 is the minimum number of books one can borrow.
So, if five of the students borrowed 3 books each, this will account for 15 books.

28 - 15 = 13
So the remaining (the 6th) student must have borrowed 13 books

Answer: D

Cheers,
Brent

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?

A. 3
B. 5
C. 8
D. 13
E. 15

"The average number of books per student was 2" means that total of 2*30=60 books were borrowed;
2+12+10=24 students borrowed total of 2*0+12*1+10*2=32 books;
So 60-32=28 books are left to distribute among 30-24=6 students, these 6 are "the rest who borrowed at least 3 books";

To maximize the number of books one student from above 6 could have borrowed we should minimize the number of books other 5 students from 6 could have borrowed. Minimum these 5 students could have borrowed is 3 books per student, so total number of books they could have borrowed is 5*3=15 books. So the 6th student could have borrowed is 28-15=13 books.

Answer: D.

Hope it's clear.

Total book borrowed = 30*2 = 60
2 borrowed 0 books, 12 borrowed 1 book, 10 borrowed 2 books; So that accounts for 32 books, leaving 28 books for the 6 students
And we know that each of these had atleast 3 books

To maximize the highest number of books, give all but one student 3 books (the minimum they must have) and all the rest to the last student
So 5 students get a total of 15 books; leaving 13 for the last student.

Answer = (D)

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 at least 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 ?

A. 3
B. 5
C. 8
D. 13
E. 15

thanks Bunuel, explanation very clear.

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Option D.
A total of 2(no book)+12(1 book)+10(2 books)=24 students borrowed 32 books.
Total books borrowed=30*2=60
Books left=28
Rest of the students borrowed at least 3 books each=>6*3=18
Now to maximize books borrowed by any one individual we'll suppose,28-18=10 have been borrowed by one person only.
Max total becomes=3+10=13

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D - 13.....Distribute 3 books to remaining 6....total will come 50....but our total should be 60......so 60-50 = 10....add original 3....10+3 = 13

Average for 30 students = 2, so total books = 60

Un-allocated books =

60 - (1*12 + 2*10+3*6) = 60 - 50 = 10

Maximum books per individual possible = 10+3 = 13

Answer = D

X/30 (total students) = 2 --> x = 60 (# of books checked out)

We have 12+2(10)+6(at least 3), leaving us with 10 to account for, thus one of the students who borrowed at least 3, could add on an additional 10 books, walking out with a total of 13.

here is the solution of this problem,

Hope it is clear

Amazing Official Question.
Here is what i did in this Question =>


12*1=> 12
10*2=> 20
So 32 books are borrowed by 22 students
Sum(30)=2*30=60

Hence 6 students borrow 60-32 = 28 books
Now the least books borrows by them => 3
So for the maximum number => 3,3,3,3,3,3,x => x= 28-15 = 13


Hence D

Since the average number of books per student was 2, the total number of books borrowed was 2 x 30 = 60 books.

The number of books borrowed by the students who borrowed two or fewer books is 2 x 0 + 12 x 1 + 10 x 2 = 32.

These 32 books were borrowed by 2 + 12 + 10 = 24 students.

So we have 60 - 32 = 28 books left for the remaining 6 students. Since we want to maximize the number of books borrowed by a single student, we must minimize the number of books borrowed by the other 5 students. We see that if 5 of these students borrowed 3 books each, then we have 28 - 15 = 13 books for the last student.

Answer: D

How to determine this question is not a overlapping set besides the fact that the data is given in such a way to drop hint?

2 Students -----> 0

12 Students ------> 1 -- (12 * 1 = 12 books)

10 Students ------>2 --- (10 * 2 = 20 books)
_______________________________________
24 Students ------> 32 books

Remaining 6 Students --------> X books


Avg. No. of Books per Student = (32 + X) / 30 = 2

32 + X = 60

X = 28 --------> 28 books that the last 6 Students must have taken out.

What is the MOST any 1 Student can take?

5 of the 6 students must take AT LEAST 3 Books ------> 5 * 3 = 15

28 - 15 = 13 books remaining for the 6th Student to take out


13 = Max

-D-

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