Test by Bikram | Mock GATE | Test 1 | Question: 31

Question

The value of following Boolean expression:

$a\left ( a+b+c \right )$$\left ( \overline a+b+\overline c \right )$$\left ( a+b+\overline c \right )$$\left ( \overline a+\overline bc\right )$

• A. $a(a+b)$
• B. $\overline a$
• C. $( \overline a+b)$
• D. $0$

Comments

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

Draw K-Map of POS terms, all entries will turn out to be zero.

Answer 1

(a+ b+ c)(a + b+ ~c) = a+ b,  (~a + b+ ~c)(~a + ~bc) = ~a + ~bc(b + ~c) = ~a
Required expression becomes = a(a+b)~a = 0

Answer 2

$a(a+b+c)(\bar a+b+\bar c)(a+b+\bar c)(\bar a+\bar bc)$

$\implies(a+ab+ac)(\bar a.a+\bar ab+\bar a\bar c+ab+b+b\bar c+a\bar c+b\bar c+\bar c)(\bar a+\bar bc)$

$\implies a(1+b+c)(\bar ab+\bar ac+b(a+1+\bar c)+\bar c(a+b+1))(\bar a+\bar bc)  [\because (1+a=1)]$

$\implies a(\bar ab+\bar ac+b+\bar c)(\bar a+\bar bc)$

$\implies (ab+a\bar c)(\bar a+\bar bc)$

$\implies ab.\bar a+ab.\bar bc+a\bar c\bar a+a\bar c.\bar bc   [\because a.\bar a=0]$

$\implies 0$

Option (D) is correct.