The given truth table is for the combinationABY000010100111

The correct answer is Option (3)

The truth table has output 1 only when both inputs are 1, which is the truth table of an AND gate, \( Y = A\cdot B \). Each circuit is reduced with De Morgan’s theorem; a NAND or NOR gate with its two inputs joined acts as a NOT gate.

In Option (3) the first two NOR gates have joined inputs, so they give \( \overline{A} \) and \( \overline{B} \). The last NOR gate then gives

    \( \displaystyle Y = \overline{\overline{A}+\overline{B}} = \overline{\overline{A}}\cdot\overline{\overline{B}} = A\cdot B \)

which is the AND function.

Why the other options are incorrect

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  Option (1) — Incorrect. The first two NAND gates give \( \overline{A} \) and \( \overline{B} \), and the last NAND gives \( \overline{\overline{A}\cdot\overline{B}} = A+B \), an OR gate.

• 

  Option (2) — Incorrect. A NOR gate followed by a NOT (NOR with joined inputs) gives \( \overline{\overline{A+B}} = A+B \), an OR gate.

• 

  Option (4) — Incorrect. Two NOT gates feed a NAND gate, \( \overline{\overline{A}\cdot\overline{B}} = A+B \), an OR gate.

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
AND Gate – “An AND gate has two or more inputs and one output. The output Y of AND gate is 1 only when input A and input B are both 1.”