after **R join S **we have all 5000 different combination of **A E **in our new table say 1

now **1 join T **will give **80 **different combination present in **T, how – **as **A E **is key in **T **which will be compared with all possible combination of AE in best case our all 80 will match and will give 80 tuples , now name the result relation as 2

**now key point for max tuples -** for max tuples always think of duplicate values of common attribute in the relation in which our common attribute is not key .

here **G **is not key attribute in our both tables (2 and U ) so we can duplicates its value in any of the table for max number of tuple, now your aptitude will come in handy (think on your own now if possible)

**2 join U = **we have 80 tuples in 2 for maximum match i’m assuming all the G values are same (say 0) and also in U all G values are 0.when our 2’s G is compare with U’s G .

1st tuple of 2 matches with all of U because both have G as 0 and give u 10 tuples

for every 80 tuples there will be 10 tuples in join so resultant table will have **800 tuples.**