ITPEC FE Subject A October 2025 Question 16

Ā · B — the circuit outputs 1 only when A = 0 and B = 1.

Gate by gate:

• •Top: a NOT gate gives A, and a NAND of A and B gives P, which is 0 only when A · B = 1
• •Bottom: a NOR of A and B gives Q = A · B
• •Output: a NOR of P and Q gives Y = P · Q = (A · B) · (A + B) = A · B, because A · B = 1 already means B = 1

Truth table check:

Answer (b)

Why not others:

• •(a) — A · B is 1 at A = 1, B = 1, where the circuit gives 0
• •(c) — A · B is 1 at A = 1, B = 0, where the circuit gives 0
• •(d) — A · B is 1 at A = 0, B = 0, where both P and Q are 1, so the final NOR gives 0

Key rule: With two inputs, a four-row truth table taken through each gate settles any circuit; a bubble on a gate's output turns AND into NAND and OR into NOR.

Hardware · AI-written, may contain errors