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For R A  is key and B is key for S.B is foreign key of R refering B of S.If referential integrity is taken into consideration max and min no of tuples will be 100  and 0 (null values)if not considered max will be 100 and min will be 0(no value of B in R matches  B of S)
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In R(ABC) and relation S(BDE) we take care on solution we find that A is candiate key for R and B is foreign key for S then apply natural join we get max 100 records because all 200 values of S surly contain 100 records of R therefore maximum value will be 100 but min u can think no match as zero value.
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minmun 0 and maximum 100 tuple .may be no one tuple can be match then give 0 and if match all tuples then get 100.

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