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Question

Answer 1

no of main memory blocks = 400
no of cache blocks =20
no of sets = no of cache blocks / 2
                = 20/2

              =10

mod function realised during mapping of block no 232 is 

K MOD S= i
where K is any block no

S is any SET no .so 232 MOD 10=2

therefore value of Xis 2