Answer will be 14
there are 3 sets
(1) For every item A, I must buy one of item C
so (A,C) in one set and total costs of (A,C) is 110+70=180
(2) Similarly (D,2B) in another set , total costs is 40+180=220
(3) Similarly (E,2D,B) in another set , total costs is 45+80+90=215
Now, to maximize my point , I have to maximize my perchase , so small amount left in my hand
Now, if we take set (3) two times we get a perchase of Rs.430
Again if we take set (1) 3 times total cost will be Rs. 540
Total spend 430 + 540 =970 Rs.
That is most close range for total Rs.1000
So, total no of items 4*2 + 2*3 =14