Problem Solving

Final question.

Thanks.

0 books -> 2
1 book -> 12
2 books -> 10
3 or more -> 6 [ we will call this <3-or-more> ]

Since Arith. mean = 2

(2*0 + 1*12 + 2*10 + <3-or-more>) / 30 = 2

<3-or-more> = 60 - 12 - 20
<3-or-more> = 28

We know that 6 students borrowed >= 3 books. So, the max number
of books taken by a single student can be computed when 5 out of
6 students take only 3 books

5*3 + max = 28

max = 28 - 15 = 13

Is 13 the correct answer ?

Yes, 13 is the correct answer.

12+20+15+x = (2*30)
47 + x = 60
x = 13

To maximise the number of books borrowed by 1 student you have to minimise the books borrowed by the rest.

2 students did not borrow anything; Thus 30-2 = 28 borrowed atleast 1

12 borrowed 1 each

10 borrowed 2 each

Thus 28 - 10 - 12 = 6 borrowed atleast 3 each

Now assume that 5 out of these 6 borrowed the least possible (which is 3 in this case). Let us assume that the last student borrowed y books

Average books borrowed = 2

2 = (12*1 + 10*2 + 5*3 + 1*y)/30

60 = 12 + 20 + 15 + y

Therefore y = 13

IMO13