For the given logic gate circuit, which of the following is the correct truth table? (a) n m z 0 0 1 0 1 1 1 1 0 1 0 0 (b) n m z 0 0 1 0 1 0 1 1 1 1 0 0 (c) n m z 0 0 0 0 1 1 1 1 0 1 0 1 (d) n m z 0 0 1 0 1 0 1 1 0 1 0 0

Answer: (a). OR gate gives n + m; NAND gate: z = n · (n + m). Since n · (n + m) = n, z = n. So z = 1 when n = 0 and z = 0 when n = 1, whatever m is. Table (a) rows (0,0,1), (0,1,1), (1,1,0), (1,0,0) match, so the answer is (a).

For the given logic gate circuit, which of the following is the… | JEE Main 2026 (23 Jan, Shift 2)

Q20JEE Main 2026 · 23 JAN · SHIFT 2Medium

Logic Gates

For the given logic gate circuit, which of the following is the correct truth table?

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Answer: (a)

OR gate gives $n + m$ ; NAND gate: $z = \overline{n \cdot (n + m)}$ .
Since $n \cdot (n + m) = n$ , $z = \bar{n}$ .
So $z = 1$ when $n = 0$ and $z = 0$ when $n = 1$ , whatever m is.
Table (a) rows $(0,0,1), (0,1,1), (1,1,0), (1,0,0)$ match, so the answer is (a).

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