a,b,c,d - books.
10 ppl only A. total 10 books read.
only c&d = 8. total 16 books read.
remaining students = 8. remaining books = 30.
a person can read max 4. thus 8 ppl can read max 32.
if 7 ppl read 4 max, 28 books read. Thus left =2.
IMO A

26 students have read 4 books in such a way that they make 56 combinations
18 students have made 2 combinations
So 8 students should read a minimum (smallest value) of x book/books to make another 54 combinations

Minimum of x so that 8Cx <= 54

8C1 = 8
8C2 = 28
8C3 = 56
8C4 = 70

The answer is x=1 (Option C)

(If question asked is maximum (highest value), answer will be 2)

Hey pavand

it is correct that there are 30 other books to be covered by 8 students.. but why you suppose that there are 5 books to be covered while we have only 4 books (a,b,c,d).. Also i could not underestand how you conclude the last sentence

10 read only Aye (1 Book); 8 have read both Cod and Dee (2 Books). This means 18 of 26 students have read 26 books.
Students Remaining: 26-18=8
Books Remaining: 56-26=30

So we need to divy up 30 books, with four different books available to read, among 8 students and we want to find out who can read the least amount of books among those 8 students. Since we are looking for the smallest, we can pile on the reading among the other 7:
7*4=28
30-28=2
The least that one student can possibly read among the 8 remaining students is 2. Answer: A

10 students have read 10 A books
8 students have read 8C and 8D books (total-16 books)

then 8 students (26(total) -10-8) need to read 30 books (56(total)-10-16)

to minimize one point, we need to maximize other ones .so
30 total books minus -28 books (if each of 7 students have read 4 books (total 28) ) = 2 books

answ is A