GATE Computer Science Engineering(CSE) 2027 Test: Synchronous & Asynchronous

f(x,y,z) = x' + y'x + xz
An implicant of a function is a product term that is included in the function. so x', y'x and xz ,all are implicants of given function.
A prime implicant of a function is an implicant that is not included in any other implicant of the function.
option a)   y'x is not a prime implicant as it is included in xz [ xy'z+ xyz]
option d) y is not a prime implicant as it include in both x' and xz.
a product term in which all the variables appear is called a minterm of the function

answer - A
using K map f = ac' + a'c

There are 4*8 = 32 DRAM chips to get 4MB from 1M x 1-bit chips. Now, all chips can be refreshed in parallel so do all cells in a row. So, the total time for refresh will be number of rows times the refresh time

A is 4 bit binary no A4A3A2A1
B is 4 bit binary no B4B3B2B1
M is result of multiplication M8M7M6M5M4M3M2M1     [check biggest no 1111 x 1111 =11100001]

So 4 bit of A 4 bit of B
input will consist of 8 bit need address 00000000 to 11111111 = 2⁸ address
output will be of 8 bits
so memory will be of
M = 256 , N = 8

When we multiply two 8 bit numbers result can go up to 16 bits. So, we need 16 bits for each of the multiplication result.
Number of results possible = 2⁸ ⨯ 2⁸ = 2¹⁶ = 64 K as we need to store all possible results of multiplying two 8 bit numbers. So, 64 K ⨯ 16 is the answer. 

multiplying 2 8 bit digits will give result in maximum 16 bits
total number of multiplications possible = 2⁸ x 2⁸
hence space required = 64K x 16 bits

A ROM cannot be written. So, to implement a 4 bit multiplier we must store all the possible combinations of input bits and  output bits giving a total of . So, (D) is the answer.

For R-S flip flop with NAND gates (inputs are active low) 11-no change 00-indeterminate..............so option A may make the system oscillate as "00" is the final input. In option D, after "00" flipflop output may oscillate but after "11", it will be stabilized.