+14 votes
821 views

What values of $A, B, C$ and $D$ satisfy the following simultaneous Boolean equations?

$\overline{A} + AB =0, AB=AC, AB+A\overline{C}+CD=\overline{C}D$

1. $A=1, B=0, C=0, D=1$

2. $A=1, B=1, C=0, D=0$

3. $A=1, B=0, C=1, D=1$

4. $A=1, B=0, C=0, D=0$

asked
edited | 821 views
0
AB+AC'+CD=C'D how we get this part can uh please explain !

## 4 Answers

+5 votes
Best answer
$A'+AB=0 \implies A'+B=0$

$\therefore A'=0$ and $B=0$

$A=1$

$AB=AC \implies B=C\implies C=0$

$AB+AC'+CD=C'D$

$\implies 0+1+0=D$

$\implies D=1$

Correct Answer: $A$
answered by Active (4k points)
edited ago
0

$AB=AC \rightarrow B=C$

Cancellation Law doesn't hold good for boolean algebra.
It holds only when -
$(AB=AC) \ \wedge (A=1) \rightarrow B=C$  \\doesn't hold when $A=0.$
and
$(A+B=A+C) \ \wedge (A=0) \rightarrow B=C$  \\doesn't hold when $A=1.$

+16 votes
Answer is A.

For verification, just put up the values and check for AND, OR operations and their outputs.
answered by Boss (19.9k points)
+1 vote

"Options make this question pretty straight forward what we need to do just put the options and see which one satisfy the given condition

What if the same question is given in " Numeric" value where we have to find the value of Boolean variable A, B, C, D or asking to find the value in decimal where A(MsB) & D(Lsb) which make bit "tricky" answered by Loyal (8.5k points)
0 votes
As, A' + AB = 0

A' can't be 1 because if A'=1 this implies 1+AB which is not 1 therefore, A'=0 => A=1

now AB has to be 0 to satisfy the equation, therefore B has to be 0 So , A=1, B=0

Now  AB=AC
as A=1, B=0 So C must be 0 then only AC will be 0. So, C=0

AB+AC'+CD=C'D

now AB=0 as B is 0 then AC'=1 (as A=1 , C'=1) , CD=0 as C is 0
So LHS is 1 now for RHS to be 1, C' and D both have to be 1 therefore D=1

So A=1, B=0, C=0, D=1
option a) is the ans.
answered by Junior (665 points)

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