MadeEasy Subject Test: Operating System - Virtual Memory

Question

I am getting 7.

Answer 1

since each page table must be contained within the page, so the size of each page table = 1024 bytes

let there are 'n' rows in page table and size of each row is 2 bytes

so

2*n=1024

n=512

that means there are 512 rows in a page table.

So bits required to represent each row = log₂ 512 = 9

Now virtual address is 64 bytes out of which 10 bytes are used for page offset

So bits used to represent page table = 64-10 = 54 bits

So lvl of page tables = 54/9 = 6

Hence answer should be 6

Answer 2

4 | 10 | 10 | 10 | 10 | 10 | 10