probability.

max books that can be borrowed by 30 students is 30 * 2 =60

Now 2 students dont borrow any book,
remaining student = 28
remaining books = 60

12 students each borrow 1 book,
remaining student = 16
remaining books = 48

10 students each borrow 2 book,
remaining student = 6
remaining books = 28

now 28 books to be borrowed among 6 students.
These students can't borrow 1 or 2 books.... This is the key...

So lets Assume 15 books by a student
remaining student = 5
remaining books = 13

not possible, as someone have to borrow 2 books then...

So next highest 13 books borrowed by a student
again...
remaining student = 5
remaining books = 15

this will do, as avg book borrowed among 5 students can be 3 so 15 books

ans D

i think its like this
max books that can be borrowed by 30 students is 30 * 2 =60

Now 2 students dont borrow any book,
remaining student = 28
remaining books = 60

12 students each borrow 1 book,
remaining student = 16
remaining books = 48

10 students each borrow 2 book,
remaining student = 6
remaining books = 28

it is given that the remaining students borrow atleast 3 buks each
So 6 * 3 =18 are distributed among them
Now remaining buks =10

So the max that can be be taken = 3(already considered) + 10(remain)
So answer is 13

I framed an eqn as follows
[2*0 + 12*1 + 10*2 + 6*3 + X] / 30 = 2

so x=10

maximun number of books a student can borrow = 3 + 10 = 13