ACE TEST SERIES

Question

HOW??

Comments

Number of maximum possible encodings with 20 bits = 2²⁰  

Number of possible encodings with 1 address instructions = 1024 *2⁷  

Number of possible encodings remaining for 2 address instructions= 2²⁰-1024*2⁷ 

                                                                                                                         =917504

Number of possible 2 address instructions = N

N* 2¹⁴  = 917504

N =56 will be answer

 

Answer 1

See total required bits in opcode for 2 address feild is 6 so we can have 2^6 instruction in which some or allof them have been in 2 address instruction which we don't know .Take it as x

now total no. of 1 address insttruction will be (2^6-x)*2^7

as given 1 address instruction is 1024 we equate the above one with this value

(2^6-n)*2^7=1024

solving this will give you 56 which is the correct answer