ISRO Computer Science Previous Year Question Paper 2007 with Solutions

Therefore, Y = (1) $\cdot C' = C'$.

Key Points for Revision:
• Expand inside parenthesis: (A + B' + A'B).
• Note that A + A'$B = A + B (Distributive Law)$.
• Thus, (A + B') + B = A + (B' + B) = A + 1 = 1.
• Therefore, Y = (1) $\cdot C' = C'$.

The final AND gate combines both terms: ((A + B)' + C)(D' E').

Key Points for Revision:
• The NOR gate produces (A + B)'.
• The OR gate combines (A + B)' and C to yield ((A + B)' + C).
• Inverted inputs D and E entering an AND gate give D' E'.
• The final AND gate combines both terms: ((A + B)' + C)(D' E').

The logic diagram incorporates the required XOR and inverted-A AND gates corresponding to a Full Subtracter.

Key Points for Revision:
• A Full Subtracter generates Difference = A ⊕ B ⊕ Bin and Borrow = A'B + Bin(A ⊕ B)'.
• The logic diagram incorporates the required XOR and inverted-A AND gates corresponding to a Full Subtracter.

Subtracting 0011 (3 in binary) restores the correct excess-3 digit.

Key Points for Revision:
• Excess-3 addition adds two biases of 3 (total +6).
• If no carry occurs (sum $\le 9$), the result retains an extra bias of 3.
• Subtracting 0011 (3 in binary) restores the correct excess-3 digit.

K-map simplification yields $Q_{n+1}$ = S + R'$Q_{n}$ with the constraint SR = 0.

Key Points for Revision:
• From the SR truth table: Set condition enforces $Q_{n+1}=1$ when $S=1$, and hold state enforces $Q_{n+1}=Q_{n}$ when $S=0, R=0$.
• K-map simplification yields $Q_{n+1}$ = S + R'$Q_{n}$ with the constraint SR = 0.