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

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In a class of 30 students, 2 students did not borrow any [#permalink]

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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?

A. 3
B. 5
C. 8
D. 13
E. 15
[Reveal] Spoiler: OA

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Re: max books [#permalink]

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New post 30 Sep 2010, 11:22
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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?

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.
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Re: max books [#permalink]

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New post 30 Sep 2010, 11:28
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(D), 13

We know that 2 students borrowed 0 books, 12 borrowed 1 book, 10 borrowed 2 books, and 6 borrowed 3 or more books. The average of books per student is 2.

Let a, b, c, d, e, and f be the number of books borrowed by each of the 6 students who borrowed 3 or more.

So we have:

\(\frac{12 + 20 + a + b + c + d + e + f}{30} = 2\)

\(a + b + c + d + e + f = 28\)

Now, to maximize one of these, we need to minimize the other 5. So set them all equal to 3:

\(3 + 3 + 3 + 3 + 3 + f = 28\)

\(f = 28 - 15 = 13\)

So the maximum is 13.
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Re: max books [#permalink]

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New post 30 Sep 2010, 11:55
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)
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Re: In a class of 30 students, 2 students did not borrow any [#permalink]

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New post 17 Feb 2014, 08:32
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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Re: In a class of 30 students, 2 students did not borrow any [#permalink]

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New post 26 Jan 2015, 12:06
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
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Re: In a class of 30 students, 2 students did not borrow any [#permalink]

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New post 02 Feb 2015, 22:51
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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
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Re: In a class of 30 students, 2 students did not borrow any [#permalink]

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Re: In a class of 30 students, 2 students did not borrow any   [#permalink] 11 Feb 2016, 23:12
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