In class of 30 students, 2 did not borrow any books from the liberary, 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 ( arithemtic mean) number of books borrowed per student was 2, what is the maximum number of books that any single student could have borrowed?
3
5
8
13
15
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How many books we could borrow
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Assuming your math is correct to this point, here's where you made your mistake.adam15 wrote:In class of 30 students, 2 did not borrow any books from the liberary, 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 ( arithemtic mean) number of books borrowed per student was 2, what is the maximum number of books that any single student could have borrowed?
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, my answer:
28 is the sum of books borrowed by the 6 students, the maximum book that each could have is
5*5+3; thus 5 of 6 will have 5, and one will have 3.
the number of books a single student could borrow will be:
5+3+2+1=11+2=13 (because the 2 students who did not borrow any book.
We want "the maximum number of books that any single student could have borrowed?"
Whenever you're asked to maximize one thing, you want to minimize everything else.
So, those 6 students must borrow 28 books and each one must borrow at least 3 books. To maximize a single student, we minimize the other 5 at 3 each, giving us:
28 - 5*3 = 28 - 15 = 13 books
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Hy Stuart,
you totally right. I was think about my solution, and it seems weird. and I tried backsolving here is my backsolution and by the way thank you again or you solution.
let's start with the first choice. between parethensis the number of students left
60-3(28)=57-12(15)=45-20 (5)=25=5*5 and max(3,5)=5 so wrong.
second choice
60-5=55-12=43-20=23=5*4+3 so we still have tree book left
60-8=52-12=40-20=20=5*4 no book left maybe
60-13=47-12=35-20=15=5*3 no book left and max(13,8)=13.
60-15=45-12=33-20=13=5*2+3 we still have tree book left
then best answer choice will be 13
please correct me if wrong
you totally right. I was think about my solution, and it seems weird. and I tried backsolving here is my backsolution and by the way thank you again or you solution.
let's start with the first choice. between parethensis the number of students left
60-3(28)=57-12(15)=45-20 (5)=25=5*5 and max(3,5)=5 so wrong.
second choice
60-5=55-12=43-20=23=5*4+3 so we still have tree book left
60-8=52-12=40-20=20=5*4 no book left maybe
60-13=47-12=35-20=15=5*3 no book left and max(13,8)=13.
60-15=45-12=33-20=13=5*2+3 we still have tree book left
then best answer choice will be 13
please correct me if wrong
12(1) + 10(2)=32 booksadam15 wrote:In class of 30 students, 2 did not borrow any books from the liberary, 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 ( arithemtic mean) number of books borrowed per student was 2, what is the maximum number of books that any single student could have borrowed?
3
5
8
13
15
12 +10 =22 students. 6 to divide remaining 28 books
Each of 6 must have 3 or 6x 3 =18 books. This leaves 10 extra books.
If one student takes all 10 in addition to his 3 he will have 13 books.
Choose D
