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A combination of logic gates has the truth table as given below:
P Q Z
0 0 0
0 1 1
1 0 1
1 1 1
Among the following combinations, the one that has the truth table shown above is

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answer is C.

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Detailed Solution

The boolean expression of this arrangement is,

For NOR gate

$Y_{1} = \bar{P + Q}$

For NOT gate

$Y_{2} = \bar{Q}$

For NAND gate

$Z = \bar{\bar{Q} \cdot (\bar{P + Q})}$

Truth Table

P	Q	Z
0	0	0
0	1	1
1	0	1
1	1	1

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