Physics · Semiconductor and Electronic Devices

JEE Main 2025 — 4 April, Morning Shift — Question 47

The Boolean expression Y=ABˉC+AˉCˉY=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 given below.

  1. Option A:

    A, C only

  2. Option B:

    A, B, C only

  3. Option C:

    A, B only

    Correct
  4. Option D:

    B, C only

Answer: C

Step-by-step solution

Y=ABˉC+AˉCˉY=A \bar{B} C+\bar{A} \bar{C}

A. ABˉC+AˉCˉ=ABˉC+A+BA \bar{B} C+\bar{A} \bar{C}=A \bar{B} C+A+B

B. ABˉC+A+BA \bar{B} C+A+B

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Physics
Chapter
Semiconductor and Electronic Devices
Topic
Boolean Algebra, Truth Tables and Logic Gates
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