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A computer uses RAM Chips of 512X8 and ROM Chips of 2048x8. The computer needs 2K Bytes of RAM, 4K Bytes of ROM and 4 interface units each with 4 registers. An I/O mapped I/O configuration is used. How many RAM Chips and ROM Chips are required?
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size of RAM chip = 512 x 8

Memory size required = 2x 1024 bytes = (2 x 1024  x 8) bits

total number of RAM required =  $\frac{2 X 1024 X 8}{512 X 8} = 4 chips$

similarly ,

size of ROM chip = 2048 x 8

Memory size required = 4 x 1024 x 8 bits

total number of ROM required = (4 x 1024 x 8) /( 2048 x 8) = 2 chips

4 RAM and 2 ROM are required
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