Word Translation:Chapter 7: Q No 8

[matrikabp]

There are 26 students who have read a total of 56 books among them. The only books they have read, though, are Aye, Bee, Cod, and Dee. If 10 students have only read Aye,
and 8 students have read only Cod and Dee, what is the smallest number of books any of the remaining students could have read?

I was Convinced with the following explanation on the book:
2 books

Students|Books
----------------------------------
10 | 10
8 | 16
----------------------------------
18 | 26

Remaining (26-18)= 8 Students, (56-26)=30 books
Smallest No. of books read by any of the remaining students = 7x4 + 1x2
So, 2 books

But, one of the person in other forums posted following question and I got confused:

"I am not too convinced with the explanation; why can't 1 person read 23 of the 30 remaining books and let the rest 7 read 7 different books, 1 each. In that case; the smallest number of books that any of the remaining students reads is going to be 1. And that 1 is just not read by one person; but 7. There are no mandates that a student must read different books. Am I misinterpreting something?

_________________
~fluke "

Any Manhattan Staffs over there please clarify!

[mithunsam]

Question states that students read only 4 books. That means, there are only 4 books available - Aye, Bee, Cod, and Dee.

So, how can someone read 23 books? Maximum number of books that can be read by one person is 4 (Aye, Bee, Cod, and Dee)

[matrikabp]

Thank you very much!

[jnelson0612]

Thanks!