GATE IT 2007 | Question: 38

Question

The following expression was to be realized using $2$-input AND and OR gates. However, during the fabrication all $2$-input AND gates were mistakenly substituted by $2$-input NAND gates. $(a.b).c + (a'.c).d + (b.c).d + a. d$

What is the function finally realized ?

• A. $1$
• B. $a' + b' + c' + d'$
• C. $a' + b + c' + d'$
• D. $a' + b' + c + d'$

Comments

Option (C) is correct!

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We can not minimize the expression and then replace with NAND gate. Always replace with appropriate gates then try to minimize. In this particular question, the answer will be the same but in general, first replace then minimize.

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@Nitesh Singh 2 thank bhaiyya,help ho gyi 

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Not in this even, answer will be different if we first minimize and then replace.

Answer 1

The final answer will come as:

$a'+c'+d'+a'c+ab+bc$

$= a'(c+1)+c'+d'+ab+bc$

$=a'+c'+d'+ab+bc$

$=(a'+a)(a'+b)+(c'+c)(c'+b)+d'$

$=a'+b+c'+b+d'$

$=a'+b+c'+d'$

Option is C.

Comments

Hira Thakur sir, there is a difference in both question, see the 3^rd term.

in GATE PYQ :  (b.c).d

in GO2017 Qus : (b¯.c).d

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Hey anyone ,Can I use kmaps anywhere like I am minimizing  a expression using properties  and got stuck  

Can I use kmaps now???

Or I have use original expression 

@Sachin Mittal 1 @arjun @Deepak Poonia sir

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Can’t we use Consensus Theorem here, for ab+bc+a’c??

Answer 2

replace AND with NAND
  ((a.b)'.c)' + ((a'.c)'.d)' + ((b.c)'.d)' + (a.d)'
=ab+c'+a'c+d'+bc+d'+a'+d'
=(ab+a')+c'+a'c+d'+bc
=b+c'+(a'+a'c)+d'+bc
=(b+bc)+c'+a'+d'
=a'+b+c'+d'

So ans is C

Comments

Best answer