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MHT CET 2025 (25 Apr, Shift 2) · Logic Gates

Logic Gates question from MHT CET 2025 (25 Apr, Shift 2)

Q38MHT CET 2025 · 25 APR · SHIFT 2Medium

Logic Gates

One of the following values of inputs A,B\mathrm{A}, \mathrm{B} and C respectively gives output (Y)(\mathrm{Y}) of the following combination of logic gates as ' 1 ' is

Figure

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Answer: (d)

  1. OR gate gives B+CB + C; AND with A gives Y=A⋅(B+C)Y = A \cdot (B + C).
  2. Y = 1 needs A=1A = 1 and at least one of B,CB, C equal to 1.
  3. (a), (b) have A=0A = 0; (c) has B+C=0B + C = 0; (d) 1⋅(0+1)=11 \cdot (0 + 1) = 1.
  4. So the answer is (d) 1,0,11, 0, 1.
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