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Certain CPU uses expanding op-code . It has 16 bit instructions with 6-bit addresses . It has maximum 192 one-address Instructions.

Then number of 2-addresses instructions are supported by the system is __________________
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6 bit addresses ....2 address instructions are also there

so 12 bit for address and 4 bit for opcode

now let 2 address instructions be n so no.of unused instructions will be 2^4-n

now for expanding opcode we will have

(2^4-n)*2^6 on adress which will be equal to 192

(2^4-n)*2^6=192 solving it we get n as 13

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