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Asked in NEET 2012 Mains · Gate combinations and truth tables
Idea: collapse the circuit into a single Boolean expression, then test the four sets against it rather than against the drawing.
The first gate is an OR fed by A and B, so the signal leaving it is A+B.
That signal and C are the two inputs of the AND, so Y=(A+B)· C.
For Y=1 both factors must be 1: C must be high, and at least one of A and B must be high.
Only A=1, B=0, C=1 meets both conditions; the other three sets all have C=0.
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