ISRO2018-31

Question

A byte addressable computer has a memory capacity of $2$$^{m}$$KB$ ($k$ bytes) and can perform $2$$^{n}$ operations. An instruction involving $3$ operands and one operator needs maximum of:

1. $3m$ bits
2. $3m + n$ bits
3. $m + n$ bits
4. none of the above

Comments

How we can add byte + bits . Why byte is not converted into bits ?

Answer 1

Memory capacity = 2^(m+10) bytes

No. of bits to define 1 operator=(m+10) bits

No. of bits to define 1 operation = nbits 

Given an instruction have 3 operands and 1operator.

No. of bits needed by instructions = 3∗(m+10)+n

=3m+30+n

 so Option d) is the correct answer