Question

ACE test series closed

Comments

please provide solution to this question

thank you very much

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@deepanshu sharma 3,

please $\underline{\color{red}{Provide}\text{ test-series name in }\color{red}{title}}$

Answer 1

For a 2 address instruction Instruction bit is divided as

4 bit-Opcode

6 bit- address1

6 bit- address2

Total instruction possible with 2 address =2^4=16

For one address instruction(given opcode expanding)

No of one address instruction= No of free 2 address opcode×2^6(Why? Only one address is used in one address so one of the 2, 6 bit address which was used for 2 address instruction can now be used to calculate new instructions in one address)

192(given)=No of free opcode×2^6

No of free opcode=192/64=3

Total opcode=16

No of two address instruction=16-3=13