GATE | GATE-CS-2015 (Set 1) | Question 65

Consider a system with byte-addressable memory, 32 bit logical addresses, 4 kilobyte page size and page table entries of 4 bytes each. The size of the page table in the system in megabytes is ___________
(A) 2
(B) 4
(C) 8
(D) 16

Answer: (B)
Explanation:

Number of entries in page table = 232 / 4Kbyte  
                                = 232 / 212 
                                        = 220

Size of page table = (No. page table entries)*(Size of an entry) 
                   = 220 * 4 bytes 
                   = 222 
                   = 4 Megabytes

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