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 ?
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jayhawk2001
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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
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
