NIELIT 2016 MAR Scientist B - Section C: 1

Question

Which of the following logic expression is incorrect?

• A. $1\oplus0=1$
• B. $1\oplus1\oplus0=1$
• C. $1\oplus1\oplus1=1$
• D. $1\oplus1=0$

Comments

ExOR operator is left associative and whenever there is odd nos of 1 ,the result is 1 ( true).

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Yes option b is the incorrect  as exor gate gives output on getting odd number of 1's else output will be 0,

A) contains odd number of 1's so output is 1

B)contains even number of 1's so output should be 0..but it is given 1.

C) contains odd number of 1's so output should be 1 yes it is 1.

D)even number of 1's so output should be 0..yes it is zero.

Answer 1

option is B
Always XOR of even number of 1's is zero
In option A)  There is odd number of 1 so the answer is 1
In option B) There are even number of 1's so the answer is 0
In option C) There are odd number of 1's so answer is 1
In option D) There are even number of 1's so answer is 0
It is clear that incorrect option is B

Another method is
As we know  A XOR B = ~A.B +A.~B
Use it to each option
option A)  1 XOR 0 = 0.0 + 1.1 = 1
option B) 1 XOR 1 XOR 0
As we know XOR is commutative as well as associative
(1 XOR 1) XOR 0
(0.1 + 1.0) XOR 0
0 XOR 0 = 1.0 + 0.1 =0
option C) 1 XOR 1 XOR 1 = (1 XOR 1) XOR 1
(0.1 + 1.0) XOR 1 = 0 XOR 1 = 1.1 + 0.0 = 1
option D) 1 XOR 1 = 0.1 + 1.0 = 0

Answer 2

In Exclusive-OR gate if number of 1 is odd then then the output is high(1)

so according tho this B is the answer.

Answer 3

B is incorrect as XOR operation will high when odd no.of inputs are high i.e., high (1) for only (0,1) or (1,0) so in option B the given expression is 1 Xor 1 Xor 0 = 1 Xor 1=0 Xor 0=0

So Option B is the answer.

Answer 4

option (B) is the answer.

short trick for this question👇👇

in xor gate problem we will apply (addition of 1’s/2) and remender will be the answer..

(A)  ½ remeinder will be 1.(correct)

(B) 2/2 remeinder will be 0.(incorrect)

(C)3/2 remeinder will be 1(correct)

(D) 2/2 remeinder will be 0.(correct)