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Guest
 
 

IN A CLASS OF 3O STUDENTS, 2

by Guest Wed May 14, 2008 2:46 pm

IN A CLASS OF 3O 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 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.

I WANT TO MAKE SURE I WORKED OUT CORRECTLY.
MP
 
 

I got 13 as the answer

by MP Wed May 14, 2008 9:04 pm

Students Books Borrowed Total books borrowed
====== =========== ==============
2 0 0
12 1 12
10 2 20
Rem: 6 >= 3 < will compute later in the answer >

So, we know that 22 students borrowed a total of 32 books.

Total students in class = 30

Since average books per student = 2,

(32 + n)/30 = 2; where n is the total number of books borrowed by remaining 6 students.

=> n = 28

Consider 5 students borrow the minimum number of books i.e. 3
=> 15 books => 28 - 15 = 13 books remaining which will be borrowed by the last student.

So, I get the ans as 13.
Guest
 
 

by Guest Thu May 15, 2008 7:39 pm

Could you explain how you solved the last step where you get 5...
Guest
 
 

I got 13 as the answer

by Guest Mon May 19, 2008 8:54 am

There are 6 people who took atleast 3 books. If we need to find the max no of books taken, we need to minimize the no of books taken by all of them except one and maximize that one guy's selection of books. So we arrive at 5 guys taking min books (>3) and 1 guy taking all the remaining books.
RonPurewal
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Joined: Tue Aug 14, 2007 8:23 am
 

by RonPurewal Fri May 23, 2008 5:27 am

Anonymous Wrote:Could you explain how you solved the last step where you get 5...


i'll lay out the key step more explicitly than is done in the other posts.

here's the deal:
if you want to maximize the number of books taken by one person, then you need to minimize the number of books taken by the other people.

incidentally, this is a really common theme in 'optimization' problems:
to maximize one quantity, minimize the others; to minimize one quantity, maximize the others.
this is supremely obvious in some circumstances - for instance, if a baseball team with a salary cap wants to pay superstar X as much as possible, it can only do so by paying all the other players as little as possible.

--

another common theme:
if you're given a statement about an average, then you should transform it into a statement about a sum.

--

if we follow both of the above points of advice, we arrive at the following solution:
first, realize that the 'average of 2 books' statement is really just a roundabout way of telling you that the 30 students took out a total of 60 books.
the mentioned quantities add up to 32 books, so you have to account for the other 28 books, among 6 students.
if you minimize the book count for five of the six students, that's 3 books per student = 15 books.
28 - 15 = 13 books for the lucky sixth student.