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JEE Main 2025 (4 Apr, Shift 1) · Logic Gates

Logic Gates question from JEE Main 2025 (4 Apr, Shift 1)

Q53JEE Main 2025 · 4 APR · SHIFT 1Difficult

Logic Gates

The Boolean expression Y=ABˉC+AˉCˉ\mathrm{Y} = A\bar{B}C + \bar{A}\bar{C} can be realised with which of the following gate configurations.

A. One 3-input AND gate, 3 NOT gates and one 2-input OR gate, one 2-input AND gate B. One 3 -input AND gate, 1 NOT gate, One 2 -input NOR gate and one 2 -input OR gate C. 3 -input OR gate, 3 NOT gates and one 2 -input AND gate

Choose the correct answer from the options given below:

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Answer: (c)

  1. A: NOTs give Bˉ\bar{B}, Aˉ\bar{A}, Cˉ\bar{C}; 3-input AND gives ABˉCA\bar{B}C; 2-input AND gives AˉCˉ\bar{A}\bar{C}; OR adds them. Works.
  2. B: AˉCˉ=A+C‾\bar{A}\bar{C} = \overline{A + C} from one 2-input NOR; one NOT gives Bˉ\bar{B} for the 3-input AND; OR adds them. Works.
  3. C: a 3-input OR with NOTs gives Aˉ+B+Cˉ=ABˉC‾\bar{A} + B + \bar{C} = \overline{A\bar{B}C}, the complement of the needed term, and no gate is left to invert it. Fails.
  4. So the answer is (c) A, B Only.
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