NIELIT 2018-66

Question

____ number of gates are required to implement the Boolean function $(AB+C)$ with using only $2$-input NOR gates.

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Comments

is it 4?

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should be 3. $AB + C = (A+C)(B+C) = \overline{\overline{(A+C)}+\overline{(B+C)}}$

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@Shubhgupta

The answer to (A+B).(B+C) = AB + AC + BC + C.  

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It's (A+C)(B+C) = AB+AC+CB+C = AB +(A+B+1)C and we know 1+ anything =1, so = AB+C moreover its a distributive law.

Answer 1

3 NOR GATES can implement the given function. but the official key says 2.

https://gateoverflow.in/1298/gate2009-6

Answer 2

NOR is Complement of OR
AB+C
= (A+C)(B+C) ← Distribution of + over
= ((A+C)’+(B+C)’)’
1st NOR- (A+C)’. Let X = (A+C)’
2nd NOR- (B+C)’. Let Y = (B+C)’
3rd NOR- (X+Y)’