Hi,
For the number of books a single student takes to be maximum, rest all should take minimum.
Mean of 30 elements is 2. So total = 30*2=60
1book - 12
2books - 10
3 books - 5
max. books(x) - 1
So. 1.(12)+2.(10)+3.(5)+x.1 = 60 => x=13
prob2
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Frankenstein
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cans
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total 30
0 book - 2 students. borrowed = 2*0 = 0
1 book - 12. borrowed = 12*1 = 12
2 books - 10 borrowed = 10*2=20
sum = 32
3 or 3+ books - 6 students ( 30-2-12-10)
Mean=2 books, thus total books borrowed = 30*2 = 60
remaining books = 60-32 = 28
thus 6 students borrowed 28 books overall
and each of them borrowed atleast 3.
To find max no of books any single student can borrow, 5 students borrow 3 only and 6th student borrows all which are left.
5*3=15
thus 28-15=13
IMO D
0 book - 2 students. borrowed = 2*0 = 0
1 book - 12. borrowed = 12*1 = 12
2 books - 10 borrowed = 10*2=20
sum = 32
3 or 3+ books - 6 students ( 30-2-12-10)
Mean=2 books, thus total books borrowed = 30*2 = 60
remaining books = 60-32 = 28
thus 6 students borrowed 28 books overall
and each of them borrowed atleast 3.
To find max no of books any single student can borrow, 5 students borrow 3 only and 6th student borrows all which are left.
5*3=15
thus 28-15=13
IMO D
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Average of books borrowed per student = 2
Hence, total number of books borrowed = 2*30 = 60
Number of students who borrowed at least 3 books each = 30 - (2 + 12 + 10) = 6
Total number of books borrowed by these 6 students = 60 - (2*0 + 12*1 + 10*2) = 60 - 32 = 28
As the total number of books borrowed is constant, to maximize the number of books borrowed by a single student, we have to minimize the number of books borrowed by other students. But we know that these 6 students have borrowed at least 3 books each. Hence, number of books borrowed by a single student will be maximum when other 5 students will borrow exactly 3 books each.
Hence, maximum number of books that a single student can borrow = 28 - 3*5 = 13
The correct answer is D.
Hence, total number of books borrowed = 2*30 = 60
Number of students who borrowed at least 3 books each = 30 - (2 + 12 + 10) = 6
Total number of books borrowed by these 6 students = 60 - (2*0 + 12*1 + 10*2) = 60 - 32 = 28
As the total number of books borrowed is constant, to maximize the number of books borrowed by a single student, we have to minimize the number of books borrowed by other students. But we know that these 6 students have borrowed at least 3 books each. Hence, number of books borrowed by a single student will be maximum when other 5 students will borrow exactly 3 books each.
Hence, maximum number of books that a single student can borrow = 28 - 3*5 = 13
The correct answer is D.
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breakkgmat
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Total students = 30.
2 did not buy any book..we left with 28 students.
12 students 1 book each = 12 books.
10 students 2 books each = 20 books.
Students left = 6 with atleast 3 books each. 6*3 = 18 book.
Mean book is 2.So total no of books.= 30 students * 2 = 60 books.
Now,there are (60-18+20+12) = 10 books left among 6 students who borrow atleast 3 books.
Max no of books will borrow by one student of this group if he/she borrows (3+10) = 13 books..
2 did not buy any book..we left with 28 students.
12 students 1 book each = 12 books.
10 students 2 books each = 20 books.
Students left = 6 with atleast 3 books each. 6*3 = 18 book.
Mean book is 2.So total no of books.= 30 students * 2 = 60 books.
Now,there are (60-18+20+12) = 10 books left among 6 students who borrow atleast 3 books.
Max no of books will borrow by one student of this group if he/she borrows (3+10) = 13 books..
