A VERY FAST WAS TO CALCULATE++++++++++++++++++
for first access there is 50*50/64 = 39.0625= with ceil 40 misses
now cache has 32 lines already given..
so after getting 32 misses it will overlao with first accessed 8 lines or blocks
so ultimately for 2nd access we can understand that 8 blocks or lines will be replaced.
now which 8?
0 to 7. and 4 to 11 are only possible answers..
if given starting address was 0000H or 0x0000 (both same)
answer would have been 0 to 7
so by trial we can know answer is 4 to 11 (A)
another way to be sure of it..
staring address 1100H or 0x1100
i.e. (bitwise) 0001000100000000
if you have noticed main memory is 16 bits..word offset or block 0ffset 6 , no. of lines 5 so tah is 5
00010-00100-000000 so block index is 00100 which is 4
so staring block or line of cache where first block of main memory is mapped is 4
so, answer is 4 to 11(A)