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JEE Main 2019 (12 Apr, Shift 1) · Logic Gates

Logic Gates question from JEE Main 2019 (12 Apr, Shift 1)

Q145JEE Main 2019 · 12 APR · SHIFT 1Medium

Logic Gates

The truth table for the circuit given in the fig. is:

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Answer: (c)

OR gives $A + B$ ; NAND gets $A$ and $A + B$ .
$Y = \overline{A(A + B)} = \overline{A}$ .
Rows: $00 \to 1$ , $01 \to 1$ , $10 \to 0$ , $11 \to 0$ .
So the answer is (c).

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