GATE CSE 1996 | Question: 2.22

Question

Consider the circuit in figure. $f$ implements

• A. $\overline{A} \overline{B}C + \overline{A}B \overline{C} + ABC$

• B. $A + B + C$

• C. $A \oplus B \oplus C$

• D. $AB + BC + CA$

Comments

Here we can see that there are three length string of 1s and 0s. So if we have a single bit with XOR and XNOR then we can analyze with even number of 0s and odd number of 1s. But do you want to say that aggregatedly if the odd number of 1s then we can say the XOR is there?

like ,  1 xor 0 xor 1 xor 0 , in this we can Easily identify even numbers of ones

but in : 110 XOR 101 xor 011 xor 111 , in this do we can identify? do we have to see total 1's ??

 

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Detailed Video Solution: https://youtu.be/wEhjAAs8iqk?t=2768&feature=shared 

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Option C is correct

repeated question

Answer 1

• $0 - C$ will be selected for $A = 0, B=0$.
• $1 - \bar C$ will be selected for $A = 0, B = 1.$
• $2 -\bar C$ will be selected for $A = 1, B = 0.$
• $3 -C$ will be selected for $A = 1, B = 1.$

So, $f = \bar A \bar B C + \bar A B \bar C +A \bar B \bar C + ABC$

$\qquad = \bar A (\bar BC + B \bar C) + A (\bar B \bar C + BC)$

$\qquad = \bar A (B \oplus C) + A (B \odot C)$

$\qquad = \bar A (B \oplus C) + A (\overline{B \oplus C})$

$\qquad =  A \oplus B \oplus C$

Correct Answer: $C$

Answer 2

See the last part...