JEE Main 2025 · 3 April · evening · Single correct · +4 / −1
The truth table corresponding to the circuit given below is:
- (a) A, B, C; 0, 0, 0; 0, 1, 0; 1, 0, 1; 1, 1, 1;
- (b) A, B, C; 0, 0, 1; 1, 0, 0; 0, 1, 0; 1, 1, 0;
- (c) A, B, C; 0, 0, 1; 0, 1, 0; 1, 0, 0; 1, 1, 0;
- (d) A, B, C; 0, 0, 0; 1, 0, 0; 0, 1, 0; 1, 1, 1;
Answer and solution
Answer (a) A, B, C; 0, 0, 0; 0, 1, 0; 1, 0, 1; 1, 1, 1;
Solution
Physics sir · 3 steps
Follow A along both its paths: C is A AND (A OR B), and that collapses to just A.
- Read the wiring. A and B go into the OR gate, so its output is A + B. A is also tapped off and sent straight to the AND gate. The AND gate's other input is the OR output.
C = A · (A + B) = A· A + A· B = A + A· B = A(1 + B) = A
- Why it reduces. A · A = A, and 1 + B = 1 in Boolean algebra. So the output just copies A, and B has no effect at all.
A B A + B C = A·(A+B) 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 - So. C is 0 whenever A is 0 and 1 whenever A is 1. That is the table in option (a).
The idea to keep. When one input feeds a gate directly and also through another gate, write the Boolean expression and simplify. A(A+B) = A and A + AB = A are the absorption rules JEE loves.
Answer from the NTA final answer key. Solution by PrepWiser.