Boolean Algebra Practice Questions

A group of 8 cells in a 4-variable K-map eliminates 3 variables (since 2^3=8), leaving one variable. So the product term has 1 literal.

m5 = 0101, m13 = 1101. They differ only in the most significant bit (A). So A is the differing variable.

Implicant $\overline{B}D$ means B=0, D=1. A and C are free → 2×2 = 4 cells.

Only minterm 7 (111) is missing. So F = complement of m7 = $\overline{A}+\overline{B}+\overline{C}$? Actually that's three literals. But K-map grouping: all cells except 111 form two pairs: ones except 111 cover $\overline{B}+\overline{C}$. Let's test: if B=0 or C=0, then output 1. Indeed, the only case where F=0 is when B=1 and C=1 and A=1 (m7). So $F = \overline{B} + \overline{C}$. Option C.

With don't cares, the groups can expand. m0,m1,m4,m5? But m5 not given. Using don't cares 2(0010),6(0110) etc. The optimal grouping yields $\overline{B}\overline{D} + \overline{A}\overline{C}$.

A self-dual function satisfies $f(A,B,C) = \overline{f(\overline{A},\overline{B},\overline{C})}$. For n variables, the number of self-dual functions is $2^{2^{n-1}}$. For n=3, it is $2^{4}=16$. Wait, the known result: number of self-dual functions of n variables is $2^{2^{n-1}}$. For n=3, that's $2^{4}=16$. But some references say for odd n, the count is $2^{2^{n-1}}$. I'll correct: The correct answer is 16. However, in GATE 2015, the answer for 3 variables was 16. I'll put 16 and ensure it's NAT.

$\overline{A} \oplus \overline{B} = A \oplus B$ (since complement of inputs gives same XOR). Then $F = (A \oplus B) \cdot (A \oplus B) = A \oplus B$, which is not always 1. Wait: $\overline{A} \oplus \overline{B} = A \oplus B$. So expression becomes $(A \oplus B) \cdot (A \oplus B) = A \oplus B$. That is not 0. But is that simplification correct? Let's test: A=0,B=0: LHS = (0⊕0)·(1⊕1)=0·0=0, RHS A⊕B=0. A=0,B=1: (1)·(0⊕? Actually 0⊕1=1, 1⊕0=1, 1·1=1, A⊕B=1. So it equals A⊕B. So answer should be C. I need to correct: $\overline{A} \oplus \overline{B} = A \oplus B$ indeed. So product = $(A \oplus B)^2 = A \oplus B$. Thus the simplified expression is $A \oplus B$. So answer is C.

All minterms have C=0. So $F = \overline{C}$.

Number of cells = 2^n = 2^4 = 16.

Σm(0, 2, 4, 6) corresponds to terms where C is always 0. Simplifying this gives f = C'. Options B and D also simplify to C' (e.g., A'C' + AC' = C'(A'+A) = C').

NAND with inverted inputs: $\overline{\overline{A} \cdot \overline{B}} = \overline{\overline{A}} + \overline{\overline{B}} = A + B$, not XOR. That's OR. Check: Inverting inputs of NAND gives OR. For XOR, the correct equivalence is using four NANDs. None of the above is directly XOR. This question might be tricky. Actually XOR can be realized as $\overline{A\overline{B}} \cdot \overline{\overline{A}B}$? Not that. The correct answer might be none, but given options, perhaps they intend NOR with inverted? Let's skip this one. I'll replace with a better one.

The two product terms are prime implicants, and both are essential. So number of prime implicants in the minimal SOP = 2.

A group of 4 cells in a 4-variable K-map covers 2^2=4 cells, eliminating 2 variables, leaving 2 literals.

Using Boolean identity: A + A'B = A + B.