let we have 2(n) record of 5(M) byte each(since it is asking for first byte we can assume that it is byte oriented or stores byte wise)
record 1 |
0 |
1 |
2 |
3 |
4 |
record 2 |
5 |
6 |
7 |
8 |
9 |
to keep things simple: let we want 1st byte of record 2
which is 5 and we can get by (N-1)M =(2-1)*5=5
so for any N value :: (N-1)M gives the desired logical address.
so C is the ans