number of bytes available given address bus width and various information

Question

A processor has

• 24 bit address bus
• 16 bit data bus
• word contains 2 bytes
• byte addressable
  Peripherals and memory units will be connected and the entire memory space most likely will be used.

There are quite a few questions and I only ever use the fact that there is a 24 bit address bus.

What is the total number of addressable locations for the system? 2^24
1/4 of the address space is to be used for the peripherals, what is the total number of addresses for peripherals? 2^24/2^2
12/16 of the addresses are to be used for disk addressing, how many? (12/16)2^22
3/4 of the address space are to be used for memorey requirements of RAM and ROM, what is the total number of addresses avaliable? (3/4)2^24
This seems to easy

Answer 1

Except for one answer (which might be a typo), your math is correct.

1. What is the total number of addressable locations for the system? 2^24 = 16,777,216
2. 1/4 of the address space is to be used for the peripherals, what is the total number of addresses for peripherals? 2^24/2^2(?) = (1/4)2^24 = 4,194,304
3. 12/16 of the addresses are to be used for disk addressing, how many? (12/16)2^22 (12/16)2^24 = (3/4)2^24 = 12,582,912
4. 3/4 of the address space are to be used for memorey requirements of RAM and ROM, what is the total number of addresses avaliable? (3/4)2^24 = 12,582,912 (same as 3.)