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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 ?

  1. $1$
  2. $a' + b' + c' + d'$
  3. $a' + b + c' + d'$
  4. $a' + b' + c + d'$
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2 Answers

Best answer
34 votes
34 votes

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.

edited by
17 votes
17 votes
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
Answer:

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