Let us say box contain x number of apple, bag contain y number of apple and string bag contain z number of apple.

In first day shop keeper receives (9x+3y) apples.

When he repacks he use 11 string bags. At that time he had less than 97 apples left without packing in string bags.

So;

[Apples received > apples in string bags +97]

Because we have less than 97 apples left.

9x+3y > 11z+97------(1)

On second day he receives (6x+2y) apples.

When he repack he use 8 string bags. At that time he had lat least 56 apples left without packing in string bags.

So;

[Apples received >= apples in string bags +56]

Because we have at least 56 apples left.

6x+2y >= 8z+56

We can divide this by 2.

3x+2y >= 4z+28------(2)

(2)*3

9x+3y >= 12z+84------(3)

Now we have two expressions for (9x+3y). (1) is absolutely greater than (9x+3y) but (3) is greater than or equal to (9x+3y).

So we can be sure that (1)>=(3)

11z+97>=12z+84

13 >= z

So in each string bag we can have apples less than or equal to 13. It is given that a string bag contains more than 11 apples.

So z = 12 or z = 13

**So a string bag contains either 12 apples or 13 apples.**