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An 8085-microprocessor based system uses a 4K × 8 bit RAM whose starting address is AA00H. The address of the last byte in this RAM is:

An 8085-microprocessor based system uses a 4K × 8 bit RAM whose starting address is AA00H. The address of the last byte in this RAM is: Correct Answer B9FFH

Specification of RAM = 4K × 8 bit

Size of the RAM = 22 × 210 × 8 bit

= 212 × 8 bit

So 12 address lines will be engaged.

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Number of address locations = 0FFFH

Starting address = AA00H

So, final address = (AA00H + 0FFFH) = B9FFH

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