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Question

A computer supports virtual memory and had 32-bit virtual addresses and 4KB pages. Suppose, a two-level page table was used, such that the first level page table was four times the size of each second level page table. Assume each page table entry occupies 4 bytes. Then, the sizes of the first and second level page tables are ........ and ........ respectively.
1. 4KB, 1KB
2. 16KB, 4KB
3. 8KB, 2KB
4. Not possible to determine

Comments

questions is ambiguous, if we solve it generally we get

Virtual memory size: 2^32B, Page size: 4KB, PTE: 4B

No. of pages in the first level: 2^32B/4KB = 2^20
First level page table size: (2^20)*4B = 2^22B = 4MB

No. of pages in the second level: 2^22B/4KB = 2^10
Second level page table size: (2^10B)*4B = 2^12B = 4KB

but how can we take into consideration that first level pt size is 4 times that of second level?

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sir, this is exactly what I did. but the given answer is option 3.

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what is the solution given and could you cross verify the question?