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Question 28

To get output '1' at R, for the given logic gate circuit the input values must be:

Solution

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$$P = \bar{X} + Y$$

$$Q = \overline{X \cdot \bar{Y}}$$

$$R = \overline{P + Q}$$

$$R = 1 \implies \overline{P + Q} = 1 \implies P + Q = 0$$

$$\implies P = 0 \quad \text{and} \quad Q = 0$$

$$\bar{X} + Y = 0 \implies \bar{X} = 0 \text{ and } Y = 0 \implies X = 1, \, Y = 0$$

Verifying, $$Q = \overline{1 \cdot \bar{0}} = \overline{1 \cdot 1} = \bar{1} = 0$$ 

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