The boolean function $Y = A B + C D$ is to be realized using only 2-input NAND gates. The minimum number of gates required is

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Solution

The correct option is B 3
$Y = A B + C D = ¯ ¯¯¯¯¯¯¯¯¯ ¯ ¯ ¯¯¯¯¯¯¯¯¯ ¯ A . B + ¯ ¯¯¯¯¯¯¯¯¯ ¯ ¯ ¯¯¯¯¯¯¯¯¯ ¯ C . D$

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Similar questions

f = (¯ ¯¯ ¯ A + ¯ ¯¯ ¯ B) (C + D)

Z = A ¯ B C,

F = (¯ X + ¯ Y) (Z + W)

F (A, B, C, D) = B ¯ ¯¯¯ ¯ D + ¯ ¯¯ ¯ B C D + A B C + A B ¯ ¯¯ ¯ C D + ¯ ¯¯ ¯ B ¯ ¯¯¯ ¯ D

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